Nothing can be adorable than booking Thailand Honeymoon Package for enjoying your honeymoon period. Go for this customizable Thailand tour itinerary and experience Phuket Fantasea Show, beaches, beaches, water sports adventures at Phi Phi Island and dinner cruise in Bangkok. Book for this Thailand trip and explore the land of smiles. The basic offering of this tour is comfortable stays, quick transfers, and sightseeing tours.
Upon your arrival at the Bangkok Airport, board the back to back flight to Phuket from the domestic terminal. You will be welcomed by our representative at Phuket Airport and then transferred to the pre-booked hotel.
Complete check-in formalities and relax for a while. Later, in the evening you will be taken to Phuket Fantasea – a 3000 seated theatre where you will enjoy a cultural show. When you are done with the entertaining show, you can enjoy delectable dinner at theme buffet restaurant in Phuket FantaSea. The entertaining day ends here and you will be taken back to the hotel for a comfortable overnight stay in Phuket.
Wake up to sumptuous breakfast and you will be taken to Phi Phi Island tour as per your Thailand Tour Itinerary. You will be transferred to the pier and a speedboat will introduce you to the picturesque beauty of Phi Phi Island. There are numerous water activities that can be enjoyed by both of you.
Snorkelling, fish feeding, and swimming are the main activities to enjoy at Phi Phi Island. Later, you both can enjoy sumptuous lunch by the beach which definitely adds charm to the romantic day of your honeymoon. Post lunch, you can spend some time relaxing with your partner on the stunning beaches of Phi Phi Island. Later, in the evening you will be taken back to the hotel in Phuket for an overnight stay.
Relish on a sumptuous breakfast at the hotel and you will be transferred to the Phuket Airport to catch your flight to Bangkok. Our agent's representative will pick you up there and he will transfer you to the hotel. Check into hotel and come over the jet lag for a while.
A beautiful morning welcomes you with a healthy breakfast at the hotel and later you will be taken to Bangkok city tour as per this Thailand honeymoon tour itinerary. Visit the popular Bangkok Wat Traimit and Wat Pho which is also known as the temple of reclining Buddha.
Make your stop at a local restaurant for lunch and post-lunch; you will visit Grand Palace and Chapel Royal of the Emerald Buddha. Not only these, this Bangkok city tour will also give you chance to visit other popular tourist attractions of Bangkok such as City Pillar Shrine, National Museum, Golden Mountain, Thai Boxing Stadium and Ananta Samakhom Palace with your love.
Day 4 of Thailand trip ends with a comfortable overnight stay.
Gorge on a lip-smacking breakfast at the hotel and then check out from the hotel and later, you will be transferred to Bangkok Airport as this is your last day in Bangkok. Catch your flight back home with lots of sweet memories.


Built on an ALUXX SL aluminium frame, Fathom 29 features updated geometry that produces a stable ride quality for challenging XC trails. The frame is optimized for a 100mm suspension fork giving you efficiency on the climbs and total control on the descents.
ALUXX SL aluminium frameset is hand-built in-house by the world leader in aluminum engineering.


text:
Career Research » Career Development » Part-Time Employment
Part-Time Employment
Part-time employment is a formalized work arrangement where an employee works fewer hours than what an employer judges to be customary for a full-time employee. For legal and comparative purposes, the U.S. Department of Labor defines part-time employees as those working 1-34 hours during a typical work week and full-time employees as those working 35 or more hours during a typical work week. In 2004, the Bureau of Labor Statistics reported that part-time employment status was held by about 17 percent of the U.S. labor force. This percentage is likely to increase and is particularly strong in the service and retail employment sectors. In addition, universities report that increasing numbers of their students now hold part-time positions while attending school.
Part-time work is also evident in many other countries, although the number of hours used to define part-time status is variable. European countries demonstrate a part-time employment rate of around 16%, comparable to the U.S. rate, but it has increased more rapidly as some nations have encouraged part-time work as a means of alleviating unemployment.
Historically speaking, the majority of part-time workers in industrialized countries have been female (e.g., 65 percent in the United States, 90 percent in Germany and France, 80 percent in the United Kingdom and Japan). Domestic responsibilities (e.g., housework, child care, elder care) are cited as explanations for these gender differences. Younger people, aged 16 to 19, also account for about 13 percent of U.S. part-timers, many of whom are students at the high school or university level. Few differences in part-time versus full-time employment have been observed around racial lines. In the future, retirees are anticipated to increasingly work part time. A return to employment is predicted for several reasons. Among these reasons are better health (and longer life expectancy) of older persons, the expected financial needs of many retirees, and possible revision of governmental policies to no longer tie hours of work through earnings tests to government retirement programs (e.g., Social Security benefits).
Changes In Employment Relationships
Part-time work is one of many new "nonstandard" forms of employment relationships. The conditions under which employees and employers agree to partner are no longer characterized by time and location or by implications regarding job security or advancement. Accordingly, new models of human resource staffing and individual employee career development are evolving, and these models entail part-time employment.
Motives for part-time employment from an employer's point of view include insufficient justification for hiring full-time employees because of seasonal variation in demand (peak workload time frames such as harvest or holidays), needs for employee staffing (coverage) that are shorter than customary employment periods (i.e., less than eight hours), and insufficient demand for employee skills to justify full-time employment. Employers usually further benefit from such arrangements by paying lower wages and by not having to provide employee benefits available to full-time employees. In addition, employers may benefit by securing a trial employment of a potential full-time employee. In short, employers accrue higher levels of staffing flexibility and lower human resource costs when employing part-time employees.
Motives for part-time employment from an employee's point of view include a desire to work fewer hours than a full-time employee (e.g., job sharing) for personal reasons (e.g., to care for children or elders, to attend school or training, health or medical limitations) or a perceived need to supplement one's personal or family income (i.e., moonlighting). Over 7 million Americans, for example, hold two jobs, at least one of which is almost necessarily a part-time job. Others are motivated to accept part-time work until full-time work with a current part-time employer (or another full-time opportunity with another employer) becomes available. Unfortunately, this latter category, termed involuntary part-time employment, characterizes about a fourth of all part-timers.
In addition to the voluntary versus involuntary sub-categories of part-time work, other distinctions can also be made. For example, part-time work may entail on-call, seasonal, or year-round employment, may be permanent or temporary, may be a primary or a secondary job, and may or may not be arranged through a third party (i.e., contract employment). Moreover, there are difficulties in defining part-time work, because the comparative standard regarding what constitutes full-time work is also a changing phenomenon. The historical "40 hours over five days" is a moving benchmark in several ways. Many Americans routinely work more than 40 hours a week. Work schedules are increasingly based on business availability to customers around the world 24 hours a day, 7 days a week. Accordingly, many new work scheduling patterns have evolved, making the definition of full-time work a more and more arbitrary reference point (e.g., many nurses now work three 12-hour shifts per week, and this is considered full time).
A case example of this phenomenon can be seen in the increasing number of part-time professionals. These jobs are typically permanent, year-round positions that an organization has elected to fill with an employee seeking a job with reduced hours. Professions that have demonstrated a willingness to engage part-timers with some frequency include law, medicine, accounting, and computer programming. However, defining a part-time professional becomes problematic, because full-time professional work usually exceeds 40 hours per week, leaving the appropriate hours of reduction for part-time status open to debate. In this case, full- and part-time distinctions become social contractions more than distinctions based on time worked.
Similarities and Differences Between Full-Time and Part-Time Workers
Both academic scholars and practitioners have long speculated that part-time and full-time employees are "different." One reason for this speculation is based on the idea that because part-timers spend less time at work, they are regarded and/or perceive themselves to be only partially included in the workplace. Another reason is that part-time work status alters these employees' frames of reference such that they identify more with other part-timers than full-time employees. More recently, it has been suggested that part-time employees formulate a different psychological contract than full-time employees do with their employers. None of these explanations, however, has yet to achieve widespread consensus.
Empirical studies seeking to determine whether part-time employees are materially different from full-timers with respect to work-related attitudes and behaviors have yielded inconsistent results. However, this body of research was recently examined using meta-analytic procedures (i.e., techniques that assess the combined findings of multiple studies) and revealed the following: First, there is little difference between part-timers and full-timers with respect to their level of job satisfaction (including satisfaction with specific facets of work), organizational commitment, and intentions to quit. Second, full-timers do manifest higher levels of job involvement than part-timers. Third, and perhaps most important, these conclusions did not appear to be affected by the type of job (professional vs. nonprofessional) or employee gender, and only a very small difference was detected based on voluntary versus nonvoluntary part-time status. Employees working part time on a voluntary basis reported higher job satisfaction than part-timers working part time on an involuntary basis.
As this latter finding suggests, what may be more critical to understanding whether part-time workers and full-time workers are different is whether an employee's part- or full-time work status reflects employee decision making or whether work status is organizationally imposed. The term decision can even be misleading if the employee freely decides to work part time but would prefer not to work at all. And few studies have posed questions about preferences for part-time work among those choosing to work full time. In other words, the motives for accepting part-time (and full-time) work may be more complex than the simple voluntary/involuntary dichotomy suggests.
The quality of the part-time job may also have a bearing on part-time employee attitudes and behaviors as well as the employee's career stage. When the least intrinsically rewarding or challenging work assignments are relegated to part-time employees, it is unrealistic to expect such job holders to demonstrate equally favorable attitudes and behaviors as their full-time counterparts. Moreover, individuals seeking post-retirement employment may approach a part-time job with a very different mind-set than a non-retired individual (e.g., opportunities for advancement may not be a relevant component of this individual's job satisfaction).
Based on scientific research, managers have a limited basis upon which to argue that part-time employees are less desirable than full-time employees. Only the job involvement of part-timers has consistently been shown to be lower than that of their full-time counterparts. However, additional research on this topic (e.g., motivation, organizational citizenship behavior, perceptions of organizational support) is needed to further substantiate this conclusion, as part-time and full-time work status remains confounded with other characteristics of work previously noted (e.g., temporary work status).
Career Development Issues
Part-time work is likely to increase in the coming years as a result of two converging trends: (1) the ever-increasing rates of women's labor force participation and (2) a shift in employee career patterning away from traditional, organizationally based careers emphasizing vertical progression to protean career models. As already acknowledged, women are more likely than men to be engaged in part-time work in order to accommodate domestic responsibilities. The increased labor force participation rate of women desiring employment has thus fueled the increase in the number of part-time workers. In addition, the growing ranks of women in managerial and professional jobs have independently contributed to the increase in part-time employees by their encouragement of employers to adopt more family-friendly work-life policies that include part-time work options, along with on-site day care, telecommuting, extended family leave, and so on.
Variability in career patterns continues to increase, and these changes have made the choice to work part-time far more socially acceptable for both men and women. As employers have moved away from lifelong employment models and, consequently, responsibility for their employees' career development, employees themselves have concomitantly taken on the responsibility for their career management. The term protean career, which asserts that individuals can and should self-direct their careers, is used to capture this change in orientation. Women and men who assume responsibility for their own career management have recognized that there is no longer a single "best" career model and that part-time work can be part of a protean career. "Cutting back," "stepping off the career track," or "following a different career trajectory" by working part-time allows for retraining, changes in career interests, opportunities to pursue personal goals such as travel, volunteer work, taking care of relatives, and so on. Indeed, career experts assert that periodic departure and reentry into the workforce may become much more commonplace.
In the near term, however, moving to part-time work is not without disadvantages. Requests to move from full- to part-time status are frequently denied, and the request itself can derail career prospects. Firms are most likely to approve requests from high-performing and difficult-to-replace employees. Women are more likely to be approved than men, especially for child care reasons as opposed to other personal reasons (e.g., to write a novel). Part-time workers are often provided less challenging assignments and experience fewer opportunities to network, which can initiate a spiral of perceptions that such employees are less valuable to the company.
Despite these disadvantages, nonstandard work options like part-time work are likely to continue to grow. Evidence is already beginning to accumulate to suggest that definitions of career success are rapidly changing away from definitions that are rooted in measures linked to organizational success, to definitions emphasizing the successful integration of personal and work-related goals. Human resource professionals, especially in the service, retail, and leisure and hospitality sectors, who recognize this shift may gain competitive advantage for their firms by demonstrating more employment scheduling flexibility to attract and retain the best employees. Indeed, securing a cadre of loyal, hard-working, part-time employees may be just the solution to employment situations characterized by low-skill jobs, limited opportunities for advancement, and where, accordingly, employee turnover has traditionally been high and costly. Such a scenario would be win-win for employees and employers alike.
Family-responsive workplace practices
Job sharing
Organizational staffing
Protean career
Barling, J. and Gallagher, D. G. 1996. "Part-time Employment." Pp. 243-277 in International Review of Industrial and Organizational Psychology, vol. 11, edited by C. I. Cooper and I. T. Robertson. Chichester, England: Wiley.
Gakovic, A. and Tetrick, L. E. 2003. "Perceived Organizational Support and Work Status: A Comparison of the Employment Relationships of Part-time and Full-time Employees Attending University Classes." Journal of Organizational Behavior 24:649-666.
Kalleberg, A. L. 2000. "Nonstandard Employment Relations: Part-time, Temporary and Contract Work." Annual Review of Sociology 26:341-365.
Klein, K., Berman, L. M. and Dickson, M. W. 2000. "May I Work Part-time? An Exploration of Predicted Employer Responses to Employee Requests for Part-time Work." Journal of Vocational Behavior 57:85-101.
Lawrence, T. B. and Corwin, V. 2003. "Being There: The Acceptance and Marginalization of Part-time Professional Employees." Journal of Organizational Behavior 24:923-943.
MacDermid, S. M., Lee, M. D., Buck, M. and Williams, M. L. 2001. "Alternative Work Arrangement among Professionals and Managers: Rethinking Career Development and Success." Journal of Management Development 20:305-317.
Morrow, P. C., McElroy, J. C. and Elliot, S. 1994. "The Effect of Preference for Work Status, Schedule, and Shift on Work-related Attitudes." Journal of Vocational Behavior 45:202-222.
Thorsteinson, T. J. 2003. "Job Attitudes or Part-time vs. Full-time Workers: A Meta-analytic Review." Journal of Occupational and Organizational Psychology 76:151-177.
U.S. Department of Labor, Bureau of Labor Statistics. 2004. Labor Force Statistics from the Current Population Survey. Washington, DC: Author.
← Pay Compression Outsourcing and Offshoring →

redpajama_set_name: RedPajamaCommonCrawl


text:
Matthew Ragan
Media Maker and Interactive System Designer
AME 394 | Compositional and Computational Principles for Media Arts
THP 494 & 598 | Generative Media for Live Performance
Python in TouchDesigner
TouchDesigner | Common Operations Cheat Sheet
TouchDesigner | Keyboard Shortcuts
Taxonomy of Media Installations
Vesturport's Woyzeck | A Case Study
Case Study: Vesturport's Woyzeck
The challenge of re-imagining a classic work often lies in finding the right translation of ideas, concepts, and imagery for a modern context. Classic pieces of theatre carry many pieces of baggage to the production process: their history, the stories of their past incarnations, the lives of famous actors and actresses who performed in starring roles, the interpretation of their designers, and all the flotsam and jetsam that might be found with any single production of the piece in question. A classic work, therefore, is not just the text of the author but a historical thread that traces the line of the work from its origin to its current manifestation. The question that must be addressed in the remounting of a classic work is, why: why this classic work, why now, why does this play matter more than any other?
In 2008 Iceland's Vesturport theatre company presented their re-imagining of Büchner's Woyzeck, a work about class, status, and madness. Written between 1836 – 1837, Büchner's play tells the story of Woyzeck, a lowly soldier stationed in a German town. He lives with Marie, with whom he has had a child. For extra pay Woyzeck performs odd jobs for the captain and is involved in medical experiments for the Doctor. Over the course of the play's serialized vignettes Woyzeck's grasp on the world begins to break apart as the result of his confrontation with an ugly world of betrayal and abuse. The end of the play is a jealous, psychologically crippled, and cuckolded Woyzeck who ruthlessly lures Marie to the pond in the woods where he kills her. There is some debate about the actual ending to Büchner's play. While the version that is most frequently produced has a Woyzeck who is unpunished, there is some speculation that one version of the play ended with the lead character facing a trial for his crime. As a historical note, Büchner's work is loosely based upon the true story of Johann Christian Woyzeck, a wigmaker, who murdered the window with whom he lived. Tragically, Büchner's died in 1837 from typhus and never saw Woyzeck performed. It wasn't, in fact, performed until 1913. In this respect, Woyzeck has always been a play that is performed outside of its original time in history. It has always been a window backwards to a different time, while simultaneously being a means for the theatre to examine the time in which it is being produced.
It therefore comes as no surprise that in 2008 a play offering a commentary on the complex social conditions of class and status opens in a country standing at the edge of a financial crisis that would come to shape the next three years of its economic standing in the world. A play about the use and misuse of power in a world where a desperate Woyzeck tries to explain to a bourgeoisie captain that the poor are "flesh and blood… wretched in this world and the next…" (Büchner) rings as a warning about what that corner of the world was soon to face.
The Response to Vesturport's Aesthetic
From the moment of its formation, Vesturport has been a company that often appropriates material and looks to add an additional element of spectacle – early in their formation as a troupe they mounted productions of Romeo and Juliet and Titus Andronicus. This additional element of spectacle is specifically characterized by a gymnastic and aerial (contemporary circus) aesthetic. The company's connection to a circus aesthetic is often credited to Gisli Örn Gardarsson's, the company's primary director, background as a gymnast (Vesturport). The use of circus as a mechanism for story telling is both compelling and engaging. Peta Tait captures this best as she talks about what circus represents:
Circus performance presents artistic and physical displays of skillful action by highly rehearsed bodies that also perform cultural ideas: of identity, spectacle, danger, transgression. Circus is performative, making and remaking itself as it happens. Its languages are imaginative, entertaining and inventive, like other art forms, but circus is dominated by bodies in action [that] can especially manipulate cultural beliefs about nature, physicality and freedom. (Tait 6)
The very nature of circus as a performance technique, therefore, brings a kind of translation to Vesturport's work that is unlike the work of other theatre companies. They are also unique in their use of language, as their productions frequently feature translations that fit the dominant language of a given touring venue. More than a company that features the use of circus as a gimmick, Vesturport uses the body's relationship to space as a translation of ideas into movement, just as their use of language itself is a constant flow of translation.
Vesturport's production of Woyzeck invites the audience to play with them as "Gardarsson's gleefully physical staging of Büchner's masterpiece … is played out on an industrial set of gleaming pipes, green astroturf, and water-filled plexiglass tanks" (Vesturport). Melissa Wong, in writing for Theatre Journal sees a stage that "resembled a swimming pool and playground" that fills the stage with a "playful illusion." The playful atmosphere of the production, however, is always in flux as a series of nightmarish moments of abuse are juxtaposed against scenes of slapstick comedy and aerial feats. Wong later sees a Woyzeck who "possessed a vulnerability that contrasted with the deliberately grotesque portrayals of the other characters." Wong's ultimate assessment of the contrasting moments of humor and spectacle is that they "served to emphasize the pathos of the play, especially at the end when the fun and frolicking faded away to reveal the broken man that Woyzeck had become." Not all American critics, however, shared her enthusiasm for Vesturport's production. Charles Isherwood in writing for the New York Times sees the use of circus as a distraction, writing that, "the circus is never in serious danger of being spoiled by that party-pooping Woyzeck…it's hard to fathom what attracted these artists to Büchner's deeply pessimistic play, since they so blithely disregard both its letter and its spirit." Jason Best shares a similar frustration with the production, writing "by relegating Büchner's words to second place, the production ends up more impressive as spectacle than effective as drama." Ethan Stanislawski was frustrated by a lack of depth in Gardarsson's production saying "this Woyzeck is as comical, manic, and intentionally reckless as it is intellectually shallow."
Circus as an Embodied Language
Facing such sharp criticism, why does this Icelandic company use circus as a method for interrogating text? Certainly one might consider the mystique of exploring new dimensions of theatricality, or notions of engaging the whole body in performance. While these are certainly appealing suggestions, there is more to the idea of circus as a physical manifestation of idea. Facing such sharp criticism, why does this Icelandic company use circus as a method for interrogating text? Certainly one might consider the mystique of exploring new dimensions of theatricality, or notions of engaging the whole body in performance. While these are certainly appealing suggestions, there is more to the idea of circus as a physical manifestation of idea. Tait writes "… aerial acts are created by trained, muscular, bodies. These deliver a unique aesthetic that blends athleticism and artistic expression. As circus bodies, they are indicative of highly developed cultural behavior. The ways in which spectators watch performers' bodies – broadly, socially, physical and erotically – come to the fore with the wordless performance of an aerial act." Spivak reminds us that:
Logic allows us to jump from word to word by means of clearly indicated connections. Rhetoric must work in the silence between and around words in order to see what works and how much. The jagged relationship between rhetoric and logic, condition and effect of knowing, is a relationship by which a world is made for the agent, so that the agent can act in an ethical way, a political way, a day-to-day way; so that the agent can be alive in a human way, in the world. (Spivak 181)
Woyzeck's challenge is fundamentally about understanding how to live in this world. A world that is unjust, frequently characterized by subjugation, and exploitative. Gardarsson uses circus to depict a world that is both ugly and beautiful. He uses circus to call our attention to these problems as embodied manifestation. The critics miss what's happening in the production, and this is especially evident when looking at what Tait has to say the role of new circus as a medium:
New circus assumes its audience is familiar with the format of traditional live circus, and then takes its artistic inspiration from a cultural idea of circus as identity transgression and grotesque abjection, most apparent in literature [and] in cinema. Early [new circus in the 1990's] shows reflected a trend in new circus practice to include queer sexual identities and expand social ideas of freakish bodies. Artistic representation frequently exaggerates features of traditional circus…. (Tait 123)
What Isherwood misses is that the use of garish spectacle that makes light of an ugly world is, in fact, at the very heart of what Gardarsson is trying to express. The working-poor Woyzeck who questions, and thinks, and is criticized for thinking is ruining the Captain and the Doctor's circus-filled party. Woyzeck's tragedy lies in his fight to survive, to be human, in the inhuman world that surrounds him – what could be more "deeply pessimistic" (as Isherwood calls it) than a vision of the world where fighting to be human drives a man to destroy the only anchor to the world (Marie) that he ever had?
Melissa Wong best sums up the production in seeing the tragedy in a Woyzeck "who seemed in some ways to be the most humane character in the production…the one who failed to survive." Her assessment of Gardarsson's use of levity is that it points "to the complicity of individuals [the audience] who, as part of society, had watched Woyzeck's life as entertainment without fully empathizing with the depth of his existential crisis" (Wong). She also rightly points out that the use of humor in the play "enabled us to access questions that in the bleakness of their full manifestation might have been too much to bear" (Wong). Tait also reminds us that the true transformative nature of circus as a medium is not what is happening with the performer, but how the experience of viewing the performer is manifest in the viewer.
Aerial motion and emotion produce sensory encounters; a spectator fleshes culturally identifiable motion, emotionally. The action of musical power creates buoyant and light motion, which corresponds with reversible body phenomenologies in the exaltation of transcendence with and of sensory experience. The aerial body mimics the sensory motion of and within lived bodies in performance of delight, joy, exhilaration, and elation. Aerial bodies in action seem ecstatic in their fleshed liveness. (Tait 152)
Here circus functions as a mechanism for translation and confrontation in a play whose thematic elements are difficult to grapple with. Vesturport's method and execution look to find the spaces between words, and while not perfect, strive to push the audience into a fleshed and lived experience of Büchner's play rather than a purely intellectual theatrical exercise.
Büchner, Georg. Woyzeck. Trans. Eric Bentley. New York: Samuel French, 1991.
Best, Jason."Woyzeck | Review." 14 October 2005. The Stage. The Stage Meida Company Limited. 3 October 2013 <www.thestage.co.uk/reviews/review.php/10047/woyzeck>.
Isherwood, Charles. Outfitting Woyzeck With a Pair of Rose-Colored Glasses. 17 October 2008. 2 October 2013 <theater.nytimes.com/2008/10/17/theater/reviews/17woyz.html>.
Pareles, Jon. "Shaking Up 'Woyzeck' With early Rock and Flying Trapeze." 13 October 2008. The New York Times. <www.nytimes.com/2008/10/14/arts/music/14cave.html?_r=2&scp=1&sq=woyzeck&st=cse&oref=slogin&>.
Richarsdon, Stan. Woyzec nytheatre.com review. 15 October 2008. The new York Theatre Experience. 2 October 2013 <www.nytheatre.com/Review/stan-richardson-2008-10-15-woyzeck>.
Spivak, Gayatri Chakravorty. Outside in the Teaching Machine. New York: Routledge, 1993.
Stanislawski, Ethan. Theatre Review (NYC): Woyzeck by George Buchner at UNDER St. Marks and BAM. 21 October 2008. 4 October 2013 <blogcritics.org/theater-review-nyc-woyzeck-by-georg/>.
Tait, Peta. Circus Bodies: Cultural Idenity in Aerial Performance. New York: Routledge, 2005.
Thielman, Sam. Review: "Woyzeck". 16 October 2008. 5 October 2013 <http://variety.com/2008/legit/reviews/woyzeck-3-1200471537/>.
Vesturport. Woyzeck by Georg Buchner | A Vesturport and Reykavik City Theatre production. 15 Janruary 2000. 7 October 2013 <http://vesturport.com/theater/woyzeck-georg-buchner/>.
Wong, Melisa Wansin. "Woyzeck (review)." Theatre Journal 61.4 (2009): 638-640.
Woyzeck. Dir. Gisli Örn Gardarsson. Vesturport. Vesturport and Reykjavik City Threatre. Vesturport, 2009.
This entry was posted in Class, Observation, Theatre History, Theory and tagged ASU, case study, circus, Grad School, graduate school, Theatre History, vesturport on October 16, 2013 by raganmd.
← Sending and Receiving OSC Values with TouchDesigner TouchDesigner | Buttons for Interfaces →


Practical and reliable information about giving medicine to your child - written by healthcare professionals, checked by parents and carers.
Our vision is that any parent, wherever they are, have information on their child's medicines that they need and can trust.
If you are a parent or carer of a child who needs to take medicines – whether once every few years or every day – we know you need good quality and reliable information.
We have also produced a new FAQs (Freqently Asked Questions) document which covers many of the questions asked by our users about Medicines for Children.
You can download our FAQs page by clicking here.
Our medicines leaflets cover many of the medicines that are prescribed or recommended to children by children's doctors (paediatricians), pharmacists or nurses.
Medicines come in different forms. Your child's medicine may be available as a tablet, capsule or liquid (to give using an oral syringe or medicine spoon). You may need to inject a medicine, use a suppository or enema (medicines in the bottom), or give medicine in the eye or ear.
Our general information leaflets give you tips about medicines, antibiotics and unlicensed medicines.
Our information is produced by experts in children's medicines – with real experience of the concerns and issues parents face when giving a child medicine at home.
Paediatricians (children's doctors), pharmacists and other healthcare professionals, including nurses, write and review our leaflets. They make sure that the advice in our leaflets is up-to-date and accurate. With their experience in the clinic and understanding of children's medicines, they advise on the medicines and topics we need to cover.
Parents and carers are the other experts in children's medicines - they have been at the heart of this programme since the beginning. Through surveys and focus groups, parents and carers across the country told us what information they need to know about giving medicines to children, and how they wanted it. A panel now reviews every leaflet before it is published.
Thank you to all who volunteer their time to develop our information! Medicines for Children couldn't exist without the knowledge and enthusiasm of our experts.
The Information Standard is a certification scheme set up by the Department of Health, to help you identify trusted sources of health and social care information. To gain the Standard, we had to show that the way we produce our information meets the criteria of the Standard. There are five steps.
*Royal College of Paediatrics and Child Health shall hold responsibility for the accuracy of the information they public and neither the Scheme Operator nor the Scheme Owner shall have any responsibility whatsoever for cost, losses or direct or indirect damages or cost arising from inaccuracy of information or omissions in information published on the website on behalf of Royal College of Paediatrics and Child Health.
Medicines for Children is a partnership of three organisations that work in child health - Royal College of Paediatrics and Child Health (RCPCH), Neonatal and Paediatric Pharmacists Group (NPPG), and national child health charity WellChild.
The Joint RCPCH and NPPG Standing Committee on Medicines leads and provides clinical expertise on this project.
"My little girl has six different medications in the morning and seven in the evening. As a parent trying to learn the names, dosages and side effects of all these drugs it is vital to have a simple yet thorough guide to medications, written in terms that non-medical professionals (i.e. most parents) can understand."


text:
Browse:Home Frostburg Frostburg youth hoops sign-ups Nov. 16-18
Frostburg youth hoops sign-ups Nov. 16-18
Published by AlleganyPlayground.com on November 6, 2021 | 0 Comment
The Frostburg Youth Basketball League has scheduled three nights for registration later this month.
From 6 p.m. to 7:30 p.m. Tuesday, Nov. 16, as well as Nov. 17 and Nov. 18, players and parents can register at the Frostburg Community Center, located at 27 S. Water St., Frostburg.
The league is open to boys and girls in grades 1 through 5. Registration is $40 per child. The season runs through February 2022.
Instruction and competition is broken into age groups — first grade, a combined second and third grade, as well as a combined fourth and fifth grade. Parents should note that league practices and games do not follow the school schedule for inclement weather closings.
Practice nights by age group, as specified below, will not change. However, start/finish times could be adjusted based on registration numbers and volunteers.
For more information, contact Adam Patterson at 240-727-2961.
1st grade details
Play is on Tuesdays and Thursdays at the Frostburg Armory from 5:30 p.m. to 6:30 p.m. Skills training begins Tuesday, Dec. 7. Coaches will be Pat Jones and Ryan Harper.
2nd and 3rd grade details
Players will play at the Frostburg Armory on Tuesdays and Thursdays from 6:30 p.m. to 7:30 p.m. kills training begins Tuesday, Dec. 7. Teams will be set in late January and games will begin in February.
Coaches will be Kacey Stafford, Dan DeWitt, Justin Patterson and Josh Bittner.
4th and 5th grade details
Skills training will be Monday and Wednesday nights at the Frostburg Armory. Fourth-graders will take the floor from 5:30 p.m. to 6:30 p.m.while fifth-graders will play from 6:30 p.m.to 7:30 p.m. twice a week. Games will begin in February.
Coaches are John Shaw, Dan DeWitt, Shawn McCusker, Ryan Harper, Justin Patterson and Adam Patterson.
Filed in: Frostburg, Latest Headlines, Sports


There are moments in life that are frustrating, confusing, scary, and difficult to endure. There are other times where we feel negative emotions like anger, loneliness, jealous, anxiety, and fear.
In these times we want to avoid these painful feelings. We may push them down, block them out, drown them out through overworking and other forms of busyness. We distract ourselves with activity, food, drinking, or worse.
All of these strategies add to our suffering in the long run. Avoidance keeps us from living fully and when we are finally forced to confront these difficult emotions, they may feel more intense and be more difficult to sort out.
As an example, I was speaking with a woman this week whose mother has Alzheimer disease and she shared with me the story of her mother's illness. Over the past year her mom has slowly been forgetting her life, her memory, her history, and her children. She has also been forgetting her bodily needs. She doesn't remember to eat, or dress, or brush her hair. She doesn't recognize her body's sensations when she has to go to the bathroom. She needs constant care. And her daughters are heart-broken as they watch this very slow decline.
This woman began to share with me a conversation that happened between her and her sister.
Sister: I'm so afraid of mom dying.
Woman: Don't think about it.
Sister: What will we do without her?
Woman: Stop thinking about it!
Sister: I can't imagine life and our family without her.
Woman: Just focus on now. She's alive now. We'll figure out the rest when it happens.
What do you think of this conversation? Do you see the sister's desire to grieve her mother's current situation and the clear ending? Maybe you've been in a situation yourself where someone has told you to "don't think about it", "stop being emotional".
As this story was related to me, I just silently listened. This woman didn't ask for my opinion or counsel and so I kept it to myself.
Each woman in this story is having a slightly different inner experience with regards to their mother and her illness. While the woman I spoke with may be pushing away her painful feelings of grief, she is also trying to make the most of each moment she does have left with her mother. The other sister, though, is deep in grief and may be feeling unacknowledged and chastised for her suffering.
Instead of 'turning away' pain, we can learn to gently 'turn towards' what we're experiencing. We can bring a caring open attention toward the parts of ourselves that are wounded and hurting and make wise choices about how to respond to ourselves and to others.
We can often feel immediate relief by naming the emotion. For example "I am sad", or "I feel afraid". "I feel mortal".
This process of naming helps disentangle the emotions from the situation. In the story above, there are many emotions rising in these two women. Almost at the level of flooding. These first two steps can help calm the nervous system and help you regain equilibrium.
Now that you've named the emotion, allow yourself to feel it. Cry it out, scream it out, whatever it's requiring. You'll notice an ebb and flow. The emotion may start out slow, then peak like a wave, then begin to subside.
After it has subsided, you can check to see if any other emotions need to be named and felt or move to the next step.
We often short-change emotions in our culture when we don't recognize them as data. Emotions tell us that something needs to be felt in order that we can know the wise choice that's waiting for us.
Our emotions may be telling us to rest, to take care of ourselves while we're caring for others. They may be trying to connect us with our inner knowing of what can alleviate a situation, improve it, or support those (including ourselves) who are involved. They may also be surfacing an old wound that needs healing.
As women, we often stop short of giving ourselves what we need. We may feel rushed and pulled in many directions. Attending to ourselves may feel like a luxury.
Consider this. When going through the safety review on an airplane, flight attendants remind you to put on your own oxygen mask first, and then for those of your children. Why? Because if you can't breathe, then you can't actually help anyone else.
Now that you have turned inwards, named the emotion, felt it, mined the wisdom, and taken care of what you need – you are more prepared to resume your life and face whatever comes next.
We cannot control what life presents to us, however, we can use our tools to respond choicefully and wisely.


text:
\section{Introduction}
\label{sec:intro}

In his last letter to G.~H.~Hardy, Ramanujan described new functions that
he called mock theta functions and listed mock theta functions of order
$3$, $5$ and $7$. Watson studied the behaviour of the third order
functions under the modular group, but was unable to find similar transformation
properties for the fifth and seventh order functions. The first substantial
progress towards finding such transformation properties was made by
Andrews \cite{An86},
who found double sum representations for the fifth and seventh order
functions. These double sum representations were reminiscent of certain
identities for modular forms found by Hecke and Rogers. Andrews results
for the fifth and seventh order mock theta functions were crucial to
Zwegers \cite{Zw-thesis},
who later showed how to complete these functions to harmonic Maass forms.
For more details on this aspect see Zagier's survey \cite{Zag06}.

Throughout this paper we use following standard notation:
$$
(a;q)_\infty = (a)_\infty = \prod_{n=1}^\infty (1-aq^n),
$$
\begin{align*}
(a;q)_n &= (a)_n = (a;q)_\infty /  (a q^n;q)_\infty\\
&\left(=(1-a)(1 -aq) \cdots (1- a q^{n-1})\quad
\mbox{for $n$ a nonnegative integer}\right).
\end{align*}

Andrews \cite{An86}
found Hecke-Rogers indefinite binary theta series  identities for all the fifth order
mock theta functions except for the following two:
\begin{align*}
\chi_0(q) &= \sum_{n=0}^\infty \frac{q^n}{(q^{n+1};q)_n}
           = \sum_{n=0}^\infty \frac{q^n \,(q)_n}{(q)_{2n}}\\
&=1+q+{q}^{2}+2\,{q}^{3}+{q}^{4}+3\,{q}^{5}+2\,{q}^{6}+3\,{q}^{7}+\cdots,
\end{align*}
and
\begin{align*}
\chi_1(q) &= \sum_{n=0}^\infty \frac{q^n}{(q^{n+1};q)_{n+1}}
           = \sum_{n=0}^\infty \frac{q^n \,(q)_n}{(q)_{2n+1}}\\
&=1+2\,q+2\,{q}^{2}+3\,{q}^{3}+3\,{q}^{4}+4\,{q}^{5}+4\,{q}^{6}+
6\,{q}^{7}+\cdots.
\end{align*}
Zwegers \cite{Zw09} found triple
sum identities for $\chi_0(q)$ and $\chi_1(q)$.  Zagier \cite{Zag06}
stated indefinite binary theta series identities for these two functions but gave
few details.  We find new
Hecke-Rogers indefinite binary theta series identities for these two functions.
In Section \sect{zagier} we compare our results with Zagier's.

\begin{thm}
\label{thm:chi01}
\begin{align}
&(q)_\infty (\chi_0(q) - 2) \nonumber\\
&=\sum_{j=0}^\infty
\sum_{\substack{-j \le 3m \le j}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+1)/2 }(1-q^{2j+1})
\label{eq:chi01a}\\
&+
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+11)/2 -1}(1-q^{2j+1}),
\nonumber
\end{align}
and
\begin{align}
&(q)_\infty \chi_1(q) \nonumber\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+7)/2 -1}(1+q^{j})
\label{eq:chi01b}\\
&+
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+13)/2 -2}(1+q^{j}),
\nonumber
\end{align}
where
$$
\textnormal{sgn}(m) =\begin{cases} 1 & \mbox{if $m\ge0$},\\
                       -1 & \mbox{if $m<0$}. \end{cases}
$$
\end{thm}

\textit{Idea of Proof}.
We need the following conjugate Bailey pair (with $a=q$):
\begin{align*}
\delta_n &= \frac{q^n (q)_n (q)_\infty}{(1-q)},\\
\gamma_n &= \sum_{j=n+1}^\infty (-1)^{j+n+1} q^{j(3j-1)/2 - 3n(n+1)/2 -1}(1+q^j).
\end{align*}
The proof of this only uses Heine's transformation \cite[Eq.(III.I)]{Ga-Ra-book}
and an exercise from
Andrews's book \cite[Ex.10,p.29]{An-book}.
The rest of the proof of Theorem \refthm{chi01} uses this conjugate Bailey pair,
the Bailey transform
and Slater's Bailey pairs A(4) and A(2) (with $a=q$) \cite[p.463]{Sl51}.
The necessary background on conjugate Bailey pairs, Bailey pairs and
the Bailey transform is given in Section \sect{bailey}.
In Section \sect{proofthm1} the proof of Theorem \refthm{chi01} is completed.

Using the same conjugate Bailey pair and Slater's A(7*), A(8) and A(6) (with
$a=q$) lead to new
Hecke-Rogers indefinite binary theta series identities
for Ramanujan's three seventh order mock theta functions.
A(7*) is actually a variant
of A(7) adjusted to work with $a=q$ instead of $a=1$. The three identities
given below in Theorem \refthm{mock7} appear to be new.
The following are Ramanujan's three seventh order mock theta functions:
\begin{align*}
\mathcal{F}_0(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q^{n+1};q)_n}\\
&=1+q+{q}^{3}+{q}^{4}+{q}^{5}+2\,{q}^{7}+{q}^{8}+2\,{q}^{9}+\cdots,\\
\mathcal{F}_1(q) &= \sum_{n=1}^\infty \frac{q^{n^2}}{(q^{n};q)_n}\\
&=q+{q}^{2}+{q}^{3}+2\,{q}^{4}+{q}^{5}+2\,{q}^{6}+2\,{q}^{7}+2\,{q}^{8}+\cdots,\\
\mathcal{F}_2(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q^{n+1};q)_{n+1}}\\
&=1+q+2\,{q}^{2}+{q}^{3}+2\,{q}^{4}+2\,{q}^{5}+3\,{q}^{6}+2\,{q}^{7}+\cdots.
\end{align*}

We have the following theorem
\begin{thm}
\label{thm:mock7}
\begin{align}
&(q)_\infty \mathcal{F}_0(q) \nonumber\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+13)/2 -1}\nonumber\\
&{\hphantom{XXXXXXXXXXXXXXXX}}(1+q^{j})(1-q^{6m+1}),
\label{eq:F0id}\\
&(q)_\infty \mathcal{F}_1(q) \nonumber\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+5)/2}(1+q^{j})
\label{eq:F1id}\\
& +
\sum_{j=2}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+19)/2 -2}(1+q^{j})\nonumber\\
&(q)_\infty \mathcal{F}_2(q) \nonumber\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+11)/2 -1}(1+q^{j})
\label{eq:F2id}\\
& +
\sum_{j=2}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+17)/2 -2}(1+q^{j}).
\nonumber
\end{align}
\end{thm}

We prove this theorem in Section \sect{proofthm2}.
In his last letter to Hardy, all that Ramanujan said about the seventh order
functions was that there were not related to each other.
Surprisingly we show that the coefficients of the three seventh order
functions are indeed related, although this is probably not the kind of
relationship that Ramanujan had in mind. For example we find for $n\ge0$ that
\begin{align}
f_0(25 n + 8) &= f_2(n),
\label{eq:f025}\\
f_1(25 n + 1) &= f_0(n),
\label{eq:f125}\\
f_2(25 n - 3) &= -f_1(n),
\label{eq:f225}
\end{align}
where
we define $f_j(n)$ by
$$
\sum_{n=0}^\infty f_j(n) q^n = (q)_\infty \,\mathcal{F}_j(q),
$$
for $j=0$, $1$, $2$. This and more general results including analogous
results for the fifth order functions are proved in Section \sect{zagier}.

\section{The Bailey Transform and Conjugate Bailey Pairs}
\label{sec:bailey}

\begin{thm}[The Bailey Transform]
\label{thm:baileytrans}
Subject to suitable convergence conditions, if
\begin{equation}
\beta_n = \sum_{r=0}^n \alpha_r u_{n-r} v_{n+r},\quad\mbox{and}
\qquad
\gamma_n = \sum_{r=n}^\infty \delta_r u_{r-n} v_{r+n},
\label{eq:bgdef}
\end{equation}
then
\begin{equation}
\sum_{n=0}^\infty \alpha_n \gamma_n = \sum_{n=0}^\infty \beta_n \delta_n.
\label{eq:bsum}
\end{equation}
\end{thm}

When applying his transform, Bailey \cite{Ba48}
chose $u_n = 1/(q)_n$ and $v_n=1/(aq;q)_n$. This motivates the following
definitions:
\begin{defn}
A pair of sequences $(\alpha_n,\beta_n)$ is a \textbf{Bailey pair}
relative to $(a,q)$ if
\begin{equation}
\beta_n = \sum_{r=0}^n \frac{\alpha_r}{(q)_{n-r} (aq)_{n+r}},
\end{equation}
for $n\ge0$.
\end{defn}
\begin{defn}
A pair of sequences $(\gamma_n,\delta_n)$ is a \textbf{conjugate Bailey pair}
relative to $(a,q)$ if
\begin{equation}
\gamma_n = \sum_{r=n}^\infty \frac{\delta_r}{(q)_{r-n} (aq)_{r+n}},
\end{equation}
for $n\ge0$.
\end{defn}

The basic idea is to find a suitable conjugate Bailey pair and apply
the Bailey Transform using known Bailey pairs.

\begin{thm}
\label{thm:conjbpair}
The sequences
\begin{align}
\delta_n &= \frac{q^n (q)_n (q)_\infty}{(1-q)},
\label{eq:dn}\\
\gamma_n &= \sum_{j=n+1}^\infty (-1)^{j+n+1} q^{j(3j-1)/2 - 3n(n+1)/2 -1}(1+q^j),
\label{eq:gn}
\end{align}
form a conjugate Bailey pair relative to $(q,q)$; i.e.\ $a=q$.
\end{thm}
\begin{rem}
We note that this result can be deduced from a special case of a result of Lovejoy
\cite[Thm1.1(4),p.53]{Lo12}.
We give a simple proof that uses only Heine's transformation and
a combinatorial result of Andrews \cite[Ex.10,p.29]{An-book}.
\end{rem}
\begin{proof}
We let
\begin{equation*}
\delta_n = (q)_n (q)_\infty \frac{q^n}{(1-q)},
\end{equation*}
and
\begin{equation*}
\gamma_n = \sum_{r=n}^\infty \frac{\delta_r}{(q)_{r-n} (q^2;q)_{r+n}}
= (q)_\infty \sum_{r=n}^\infty \frac{(q)_r q^r}{(q)_{r-n} (q;q)_{r+n+1}}.
\end{equation*}
We must show that $\gamma_n$ is given by \eqn{gn}.
\begin{align*}
&\sum_{r=n}^\infty \frac{(q)_r q^r}{(q)_{r-n} (q;q)_{r+n+1}}
=\sum_{r=0}^\infty \frac{(q)_{r+n} q^{r+n}}{(q)_{r} (q)_{r+2n+1}}\\
&= q^n\frac{(q)_n}{(q)_{2n+1}}
\sum_{r=0}^\infty \frac{(q^{n+1};q)_{r} q^{r}}{(q)_{r} (q^{2n+2};q)_{r}}
= q^n\frac{(q)_n}{(q)_{2n+1}}
\qbas{0}{q^{n+1}}{q}{q}{q^{2n+2}}\\
&= q^n\frac{(q)_n}{(q)_{2n+1}} \frac{(q^{n+1};q)_\infty}{(q^{2n+2};q)_\infty (q)_\infty} \, \qbas{q^{n+1}}{q}{q}{q^{n+1}}{0}\\
&=q^n \frac{1}{(q)_\infty} \sum_{j=0}^\infty (q^{n+1};q)_j q^{(n+1)j},
\end{align*}
by Heine's transformation \cite[Eq.(III.I)]{Ga-Ra-book}, so that
\begin{equation}
\gamma_n = q^n \sum_{j=0}^\infty (q^{n+1};q)_j q^{(n+1)j}.
\label{eq:gnb}
\end{equation}
From Andrews \cite[Ex.10,p.29]{An-book}
we have
\begin{equation}
\sum_{j=0}^\infty (xq)_j x^{j+1} q^{j+1}
=\sum_{m=1}^\infty (-1)^{m-1} q^{m(3m-1)/2} x^{3m-2}(1 + x q^m).
\label{eq:andex10}
\end{equation}
Using \eqn{gnb} and \eqn{andex10} with $x=q^n$ we have
\begin{align*}
\gamma_n &= q^n \sum_{j=0}^\infty (q^{n+1};q)_j q^{(n+1)j} \\
&=\sum_{m=1}^\infty (-1)^{m-1} q^{m(3m-1)/2 + n(3m-2)-1}(1+q^{m+n})\\
&=\sum_{m=n+1}^\infty (-1)^{m+n+1} q^{m(3m-1)/2 - 3n(n+1)/2 -1}(1+q^{m}),
\end{align*}
as required.
We note that Subbarao \cite{Su71} gave
a combinatorial proof of \eqn{andex10} by using a variant of Franklin's
involution \cite[pp.10--11]{An-book}.
\end{proof}

\section{Proof of Theorem \refthm{chi01}}
\label{sec:proofthm1}

To prove Theorem \refthm{chi01} we will apply the Bailey Transform,
with $u_n= 1/(q)_n$, $v_n=1/(q^2;q)_n$, using the conjugate Bailey pair
in Theorem \refthm{conjbpair}, and Slater's Bailey pairs $A(4)$ and $A(2)$.
By \cite[p.463]{Sl51}, the following gives Slater's $A(4)$ Bailey pair
relative to $(q,q)$:
\begin{equation}
\beta_n = \frac{q^n}{\aqprod{q^2}{q}{2n}}
        ,\qquad
        \alpha_n =
        \left\{\begin{array}{ll}
                 q^{6m^2-4m}& \mbox{ if } n = 3m-1,
                \\
                q^{6m^2+ 4m}& \mbox{ if } n = 3m,
                \\
                -q^{6m^2 + 8m + 2}-q^{6m^2+4m}& \mbox{ if } n = 3m+1.
        \end{array}\right.
\label{eq:A4}
\end{equation}
By \cite[Eq.(A${}_0$),p.278]{Wa36b} we have
\begin{align*}
\chi_0(q) &= \sum_{n=0}^\infty \frac{q^n}{(q^{n+1};q)_n}
          =1 +  \sum_{n=0}^\infty \frac{q^{2n+1}}{\aqprod{q^{n+1}}{q}{n+1}}\\
&=1 + q \sum_{n=0}^\infty
      \frac{q^n}{\aqprod{q^2}{q}{2n}}\cdot \frac{q^n(q)_n}{(1-q)}
=1 + \frac{q}{(q)_\infty} \sum_{n=0}^\infty \beta_n \delta_n,
\end{align*}
where $\delta_n$ is given in \eqn{dn}.
Thus by the Bailey Transform and \eqn{A4} we have
\begin{align}
&(q)_\infty \Parans{\chi_0(q) - 1}
= q\sum_{n=0}^\infty \beta_n \delta_n
= q\sum_{n=0}^\infty \alpha_n \gamma_n
\label{eq:chi0id1}\\
&=\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j} q^{j(3j-1)/2 - m(15m-1)/2}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j+1} q^{j(3j-1)/2 - m(15m+1)/2}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+2}^\infty (-1)^{m+j+1}
      \left\{ q^{j(3j-1)/2 - m(15m+11)/2-1} \right.\nonumber\\
&\left.\hphantom{XXXXXXXXXXXXXXX} + q^{j(3j-1)/2 - m(15m+19)/2-3}\right\} (1+q^j)\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+1)/2 }(1+q^{j})\nonumber\\
& \qquad +
\sum_{j=2}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+11)/2 -1}(1+q^{j}),
\nonumber
\end{align}
by noting that
$$
-(-m-1)(15(-m-1)+19)/2-3 = -m(15m+11)/2-1.
$$
Now from Euler's Pentagonal Number Theorem \cite[p.11]{An-book} we have
\begin{equation}
(q)_\infty = \sum_{n=-\infty}^\infty (-1)^n q^{n(3n-1)/2}
=\sum_{m=-\infty}^\infty q^{6m^2+m} - \sum_{m=-\infty}^\infty q^{6m^2+5m + 1}.
\label{eq:EPT}
\end{equation}
By \eqn{chi0id1} and \eqn{EPT} we have
\begin{align*}
&(q)_\infty \Parans{\chi_0(q) - 2} =
(q)_\infty \Parans{\chi_0(q) - 1} - (q)_\infty\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+1)/2 }(1+q^{j})\\
& \qquad +
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+11)/2 -1}(1+q^{j}),\\
&\qquad -\sum_{m=-\infty}^\infty q^{6m^2+m} + \sum_{m=-\infty}^\infty q^{6m^2+5m + 1}\\
&=\sum_{j=1}^\infty
\sum_{\substack{-j+1 \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+1)/2 }\\
& \qquad +
 \sum_{j=0}^\infty
\sum_{\substack{-j \le 3m \le j}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+1)/2 }\\
& \qquad+
\sum_{j=2}^\infty
\sum_{\substack{-j \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+11)/2 -1} \\
& \qquad +
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+11)/2 -1}.
\end{align*}
On the right side of the last equation above replace $j$ by $j+1$ in
the first and third double sums to obtain
\begin{align*}
&(q)_\infty (\chi_0(q) - 2) \\
&=\sum_{j=0}^\infty
\sum_{\substack{-j \le 3m \le j}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+1)/2 }(1-q^{2j+1})
\\
&+
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j+1)/2 - m(15m+11)/2 -1}(1-q^{2j+1}),
\end{align*}
which is \eqn{chi01a}.

To prove \eqn{chi01b} we need Slater's \cite[p.463]{Sl51} A(2) Bailey pair relative to $(q,q)$:
\begin{equation}
\beta_n = \frac{1}{\aqprod{q^2}{q}{2n}}
        ,\qquad
        \alpha_n =
        \left\{\begin{array}{ll}
                 q^{6m^2-m}& \mbox{ if } n = 3m-1,
                \\
                q^{6m^2+ m}& \mbox{ if } n = 3m,
                \\
                -q^{6m^2 + 5m + 1}-q^{6m^2+7m+2}& \mbox{ if } n = 3m+1.
        \end{array}\right.
\label{eq:A2}
\end{equation}
We have
\begin{align*}
\chi_1(q) &= \sum_{n=0}^\infty \frac{q^n}{(q^{n+1};q)_{n+1}}
          =\sum_{n=0}^\infty \frac{q^{n}(q)_n}{(q)_{2n+1}}                \\
&=\sum_{n=0}^\infty
      \frac{1}{\aqprod{q^2}{q}{2n}}\cdot \frac{q^n(q)_n}{(1-q)}
=\frac{1}{(q)_\infty} \sum_{n=0}^\infty \beta_n \delta_n.
\end{align*}
By the Bailey Transform and \eqn{A2} we have
\begin{align}
&(q)_\infty \chi_1(q)
=\sum_{n=0}^\infty \beta_n \delta_n
= \sum_{n=0}^\infty \alpha_n \gamma_n
\label{eq:chi1id1}\\
&=\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j} q^{j(3j-1)/2 - m(15m-7)/2-1}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j+1} q^{j(3j-1)/2 - m(15m+7)/2-1}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+2}^\infty (-1)^{m+j+1}
      \left\{ q^{j(3j-1)/2 - m(15m+17)/2-3}\right.\nonumber\\
&\hphantom{XXXXXXXXXXXX}\left.+ q^{j(3j-1)/2 - m(15m+13)/2-2}\right\} (1+q^j)\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+7)/2 -1}(1+q^{j})\nonumber\\
& \qquad +
\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+13)/2 -2}(1+q^{j}),
\nonumber
\end{align}
by noting that
$$
-(-m-1)(15(-m-1)+17)/2-3 = -m(15m+13)/2-2.
$$
This completes the proof of Theorem \refthm{chi01}.

\section{Proof of Theorem \refthm{mock7}}
\label{sec:proofthm2}
To prove Theorem \refthm{mock7} we proceed as in Section \sect{proofthm1}.
This time we need Slater's Bailey pairs $A(6)$ and $A(8)$, and a variant
of her Bailey pair $A(7)$.

From \cite[Eq.(3.4),p.464]{Sl51} we have
\begin{align*}
&\frac{q^{n^2-n}}
       {(q)_{2n}}
= \sum_{r=-[(n+1)/3]}^{[n/3]}
\frac{(1-q^{6r+1})q^{3r^2-2r}}
     {(q)_{n+3r+1} (q)_{n-3r}}\\
&=
 \sum_{r=0}^{[n/3]} \frac{(1-q^{6r+1})q^{3r^2-2r}}
                                {(q)_{n-3r} (q)_{n+3r+1}}
+
 \sum_{r=1}^{[(n+1)/3]}
\frac{(1-q^{-6r+1})q^{3r^2+2r}}
     {(q)_{n+3r} (q)_{n+1-3r}}
\\
&=
 \sum_{r=1}^{[(n+1)/3]} \frac{q^{3r^2+2r}-q^{3r^2-4r+1}}
                                {(q)_{n-(3r-1)} (q)_{n+(3r-1)+1}}
+
 \sum_{r=0}^{[n/3]} \frac{q^{3r^2-2r}-q^{3r^2+4r+1}}
                                {(q)_{n-3r} (q)_{n+3r+1}},
\end{align*}
so that
\begin{align*}
&\frac{(1-q)q^{n^2-n}}
       {(q)_{2n}} \\
&=
 \sum_{r=1}^{[(n+1)/3]} \frac{q^{3r^2+2r}-q^{3r^2-4r+1}}
                                {(q)_{n-(3r-1)} \aqprod{q^2}{q}{n+(3r-1)}}
+
 \sum_{r=0}^{[n/3]} \frac{q^{3r^2-2r}-q^{3r^2+4r+1}}
                                {(q)_{n-3r}\aqprod{q^2}{q}{n+3r}}.
\end{align*}
This implies the following Bailey pair relative to $(q,q)$:
\begin{equation}
\beta_n = \frac{(1-q)q^{n^2-n}}{(q)_{2n}}
        ,\qquad
        \alpha_n =
        \left\{\begin{array}{ll}
                 q^{3m^2+2m}-q^{3m^2-4m+1}& \mbox{ if } n = 3m-1,
                \\
                q^{3m^2-2m}-q^{3m^2+4m+1}& \mbox{ if } n = 3m,
                \\
                0 & \mbox{ if } n = 3m+1.
        \end{array}\right.
\label{eq:A7star}
\end{equation}
We note that this Bailey pair was found by Warnaar \cite[p.375]{Wa03}
using a different method.
We have
\begin{align*}
\mathcal{F}_0(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{\aqprod{q^{n+1}}{q}{n}}\\
&=\sum_{n=0}^\infty
      \frac{(1-q)q^{n^2-n}}{(q)_{2n}}\cdot \frac{q^n(q)_n}{(1-q)}
=\frac{1}{(q)_\infty} \sum_{n=0}^\infty \beta_n \delta_n.
\end{align*}
Thus by the Bailey Transform and \eqn{A7star} we have
\begin{align}
&(q)_\infty \mathcal{F}_0(q)
= \sum_{n=0}^\infty \beta_n \delta_n
= \sum_{n=0}^\infty \alpha_n \gamma_n
\label{eq:F0id1}\\
&=\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j} q^{j(3j-1)/2 - m(21m-13)/2-1}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j+1} q^{j(3j-1)/2 - m(21m-1)/2}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j+1}
       q^{j(3j-1)/2 - m(21m+13)/2-1}
(1+q^j)\nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j}
       q^{j(3j-1)/2 - m(21m+1)/2}
(1+q^j)\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+13)/2-1 }(1+q^{j})\nonumber\\
& \qquad +
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j} q^{ j(3j-1)/2 - m(21m+1)/2}(1+q^{j}),
\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+13)/2-1 }\nonumber\\
&\hphantom{XXXXXXXXXXXXXX}(1+q^{j})(1-q^{6m+1}),
\nonumber
\end{align}
which is \eqn{F0id}.

To prove \eqn{F1id} we need Slater's \cite[p.463]{Sl51} A(8) Bailey pair relative to $(q,q)$:
\begin{equation}
\beta_n = \frac{q^{n^2+n}}{\aqprod{q^2}{q}{2n}}
        ,\qquad
        \alpha_n =
        \left\{\begin{array}{ll}
                 q^{3m^2-2m}& \mbox{ if } n = 3m-1,
                \\
                q^{3m^2+ 2m}& \mbox{ if } n = 3m,
                \\
                -q^{3m^2 + 4m + 1}-q^{3m^2+2m}& \mbox{ if } n = 3m+1.
        \end{array}\right.
\label{eq:A8}
\end{equation}
We have
\begin{align*}
\mathcal{F}_1(q) &= \sum_{n=1}^\infty \frac{q^{n^2}}
                                           {\aqprod{q^n}{q}{n}}
=q\sum_{n=0}^\infty \frac{q^{n^2+2n}}
                         {\aqprod{q^{n+1}}{q}{n+1}}\\
&=q\sum_{n=0}^\infty
      \frac{q^{n^2+n}}
           {\aqprod{q^2}{q}{2n}}\cdot \frac{q^n(q)_n}{(1-q)}
=\frac{q}{(q)_\infty} \sum_{n=0}^\infty \beta_n \delta_n.
\end{align*}

Thus by the Bailey Transform and \eqn{A8} we have
\begin{align}
&(q)_\infty \mathcal{F}_1(q)
= \sum_{n=0}^\infty \beta_n \delta_n
= \sum_{n=0}^\infty \alpha_n \gamma_n
\label{eq:F1id1}\\
&=\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j} q^{j(3j-1)/2 - m(21m-5)/2}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j+1} q^{j(3j-1)/2 - m(21m+5)/2}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+2}^\infty (-1)^{m+j+1} \left\{
       q^{j(3j-1)/2 - m(21m+19)/2-2} (1+q^j) \right.\nonumber\\
&\hphantom{XXXXXXXXX}\left.+ q^{j(3j-1)/2 - m(21m+23)/2-3} (1+q^j)\right\}
\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+5)/2 }(1+q^{j})\nonumber\\
& \qquad +
\sum_{j=2}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+19)/2-2}(1+q^{j}),
\nonumber
\end{align}
which is \eqn{F1id}.

To prove \eqn{F2id} we need Slater's \cite[p.463]{Sl51} A(6) Bailey pair relative to $(q,q)$:
\begin{equation}
\beta_n = \frac{q^{n^2}}{\aqprod{q^2}{q}{2n}}
        ,\qquad
        \alpha_n =
        \left\{\begin{array}{ll}
                 q^{3m^2+m}& \mbox{ if } n = 3m-1,
                \\
                q^{3m^2-m}& \mbox{ if } n = 3m,
                \\
                -q^{3m^2 + m}-q^{3m^2+5m + 2}& \mbox{ if } n = 3m+1,
        \end{array}\right.
\label{eq:A6}
\end{equation}
We have
$$
\mathcal{F}_2(q) = \sum_{n=1}^\infty \frac{q^{n^2+n}}
                                           {\aqprod{q^{n+1}}{q}{n+1}}
=\sum_{n=0}^\infty
      \frac{q^{n^2}}
           {\aqprod{q^2}{q}{2n}}\cdot \frac{q^n(q)_n}{(1-q)}
=\frac{1}{(q)_\infty} \sum_{n=0}^\infty \beta_n \delta_n.
$$

Thus by the Bailey Transform and \eqn{A6} we have
\begin{align}
&(q)_\infty \mathcal{F}_2(q)
= \sum_{n=0}^\infty \beta_n \delta_n
= \sum_{n=0}^\infty \alpha_n \gamma_n
\label{eq:F2id1}\\
&=\sum_{m=1}^\infty \sum_{j=3m}^\infty (-1)^{m+j} q^{j(3j-1)/2 - m(21m-11)/2-1}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+1}^\infty (-1)^{m+j+1} q^{j(3j-1)/2 - m(21m+11)/2-1}
 (1+q^j) \nonumber\\
& \qquad
+\sum_{m=0}^\infty \sum_{j=3m+2}^\infty (-1)^{m+j+1} \left\{
       q^{j(3j-1)/2 - m(21m+25)/2-4} (1+q^j)\right.\nonumber\\
&\hphantom{XXXXXXXXXXXX}\left. + q^{j(3j-1)/2 - m(21m+17)/2-2} (1+q^j)\right\}
\nonumber\\
&=
\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+11)/2-1 }(1+q^{j})\nonumber\\
& \qquad +
\sum_{j=2}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(21m+17)/2-2}(1+q^{j}),
\nonumber
\end{align}
which is \eqn{F2id}.
This completes the proof of Theorem \refthm{mock7}.

\section{Zagier's Mock Theta Function Identities and Related Results}
\label{sec:zagier}

In this section we write our double-series identities for the two fifth
order functions $\chi_0(q)$ and $\chi_1(q)$ and all three seventh
order functions $\mathcal{F}_j(q)$ ($j=0,1,2$) using Dirichlet characters.
This leads naturally to relations between the coefficients of these series
as in Theorems \refthm{chirels} and \refthm{mock7rels}.

As mentioned before Andrews \cite{An86} obtained indefinite theta series
identities
for all of Ramanujan's fifth order functions except $\chi_0(q)$
and $\chi_1(q)$. Using Andrews's results Zwegers \cite{Zw-thesis}
showed how to complete
all of Andrews's fifth order functions to weak harmonic Maass forms.
As noted by Watson \cite[pp.277-279]{Wa36b},
Ramanujan gave identities for $\chi_0(q)$
and $\chi_1(q)$ in terms of the other fifth order functions. Zagier
suggested  that indefinite theta function identities for $\chi_0(q)$ and $\chi_1(q)$
could be obtained from Ramanujan's results and Zwegers transformation
formulas, although he gave no details.
We state Zagier's results in a modified form in the following
\begin{thm}
\label{thm:zagchithm}
\begin{equation*}
(q)_\infty (2 - \chi_0(q))
=
\sum_{\substack{ 5\abs{b} < \abs{a} \\ a+b\equiv 2\pmod{4}
\\ a\equiv 2 \pmod{5}}}
(-1)^a \textnormal{sgn}(a) \leg{-3}{a^2-b^2}
  q^{\frac{1}{120}(a^2 - 5b^2) - \frac{1}{30}}
\end{equation*}
and
\begin{equation*}
(q)_\infty \chi_1(q)
=
\sum_{\substack{ 5\abs{b} < \abs{a} \\ a+b\equiv 2\pmod{4}
\\ a\equiv 4 \pmod{5}}}
(-1)^a \textnormal{sgn}(a) \leg{-3}{a^2-b^2}
  q^{\frac{1}{120}(a^2 - 5b^2) - \frac{19}{30}}
\end{equation*}
\end{thm}
\begin{rem}
Here $\leg{-3}{\cdot}$ is the Kronecker symbol, and is a Dirichlet character mod $3$.
\end{rem}

Our Theorem \refthm{chi01} seems to differ from Zagier's Theorem.
In contrast to Zagier's theorem which involves a character mod $3$ our version
involves the Dirichlet  character mod $60$:
$$
\chi_{60}(m) =
\begin{cases}
1 & \mbox{if $m\equiv 1$, $11$, $19$, $29 \pmod{60}$}\\
i & \mbox{if $m\equiv 7$, $13$, $17$, $23 \pmod{60}$}\\
-1 & \mbox{if $m\equiv 31$, $41$, $49$, $59 \pmod{60}$}\\
-i & \mbox{if $m\equiv 37$, $43$, $47$, $53 \pmod{60}$}
\end{cases}
$$

\begin{thm}
\label{thm:chi60}
\begin{equation}
\label{eq:chi0altid}
(q)_\infty (2 - \chi_0(q))
=
\sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 1,11 \pmod{30}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(5a^2 - b^2) - \frac{1}{30}}
\end{equation}
and
\begin{equation}
\label{eq:chi1altid}
(q)_\infty \chi_1(q)
= i\,
\sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv b\equiv 1\pmod{6}
\\ b\equiv \pm 2 \pmod{5}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(5a^2 - b^2) - \frac{19}{30}}
\end{equation}
\end{thm}

We find analogous identities for the seventh order functions.
Also Andrews \cite{An86} obtained indefinite theta series  identities
for these functions. Hickerson \cite[Theorem 2.0,p.666]{Hi88b} found
nice versions of Andrews identities, which he used to prove
his seventh order analogues of Ramanujan's mock theta conjectures \cite{An-Ga89}
for the fifth order functions. Our identities differ from Andrews's and Hickerson's
and appear to be new.
\begin{thm}
\label{thm:mock7alt}
\begin{equation}
\label{eq:F0altid}
(q)_\infty \mathcal{F}_0(q) =
\sum_{\substack{ 3\abs{b} < 7 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 1,13 \pmod{42}}}
\textnormal{sgn}(b) \leg{12}{a} \leg{12}{b}\leg{b}{7}
  q^{\frac{1}{168}(7a^2 - b^2) - \frac{1}{28}},
\end{equation}
\begin{equation}
\label{eq:F1altid}
(q)_\infty \mathcal{F}_1(q) =
-\sum_{\substack{ 3\abs{b} < 7 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 5,19 \pmod{42}}}
\textnormal{sgn}(b) \leg{12}{a} \leg{12}{b}\leg{b}{7}
  q^{\frac{1}{168}(7a^2 - b^2) + \frac{3}{28}},
\end{equation}
and
\begin{equation}
\label{eq:F2altid}
(q)_\infty \mathcal{F}_2(q) =
-\sum_{\substack{ 3\abs{b} < 7 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 11,17 \pmod{42}}}
\textnormal{sgn}(b) \leg{12}{a} \leg{12}{b}\leg{b}{7}
  q^{\frac{1}{168}(7a^2 - b^2) - \frac{9}{28}}.
\end{equation}
\end{thm}

We sketch the proof of \eqn{chi1altid}. Firstly we observe that
\begin{align*}
\tfrac{1}{2}j(3j\pm1) - \tfrac{1}{2}m(15m+7)-1
&= \tfrac{1}{120}\Parans{5(6j  \pm1)^2 - (30m+7)^2} - \tfrac{19}{30},\\
\tfrac{1}{2}j(3j\pm1) - \tfrac{1}{2}m(15m+13)-2
&= \tfrac{1}{120}\Parans{5(6j  \pm1)^2 - (30m+13)^2}- \tfrac{19}{30}.
\end{align*}
In the summations in equation \eqn{chi1altid}, we let $a=6(\pm j)+1$,
and $b=30m+r$, where $j\ge1$, $m\in\mathbb{Z}$, and $r=7$, $13$. We have
\begin{align*}
\leg{12}{a} &= \leg{12}{6(\pm j)+1} = (-1)^j,\\
i\chi_{60}(b) &= i\chi_{60}(30m+r)=(-1)^{m+1},\quad\mbox{and}\\
\textnormal{sgn}(b) &=\textnormal{sgn}(30m+r)=\textnormal{sgn}(m).
\end{align*}
Next we consider the inequalities for the variables in the summations.

\subsubsection*{Case 1} $m\ge0$ and $r=7$. Then we see that
$$
3\abs{b} < 5\abs{a} \Leftrightarrow 3m < j + \Parans{\frac{\pm5-21}{30}}
\Leftrightarrow 3m \le j-1.
$$
\subsubsection*{Case 2} $m<0$ and $r=7$. Then we see that
$$
3\abs{b} < 5\abs{a} \Leftrightarrow -j < 3m + \Parans{\frac{\pm5+21}{30}}
\Leftrightarrow -j \le 3m.
$$
\subsubsection*{Case 3} $m\ge0$ and $r=13$. Then we see that
$$
3\abs{b} < 5\abs{a} \Leftrightarrow 3m < j + \Parans{\frac{\pm5-39}{30}}
\Leftrightarrow 3m \le j-2.
$$
\subsubsection*{Case 4} $m<0$ and $r=13$. Then we see that
$$
3\abs{b} < 5\abs{a} \Leftrightarrow -j + \Parans{\frac{-39\pm5}{30}}< 3m
\Leftrightarrow -j-1 \le 3m.
$$
It follows that
\begin{align*}
&\sum_{j=1}^\infty
\sum_{\substack{-j \le 3m \le j-1}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+7)/2 -1}(1+q^{j})\\
&\qquad
=i\,
\sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv1\pmod{6}
\\ b\equiv 7 \pmod{30}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(5a^2 - b^2) - \frac{19}{30}},
\end{align*}
and
\begin{align*}
&\sum_{j=1}^\infty
\sum_{\substack{-j-1 \le 3m \le j-2}}
\textnormal{sgn}(m) (-1)^{m+j+1} q^{ j(3j-1)/2 - m(15m+13)/2 -2}(1+q^{j})\\
&\qquad
=i\,
\sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv1 \pmod{6}
\\ b\equiv 13 \pmod{30}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(5a^2 - b^2) - \frac{19}{30}}.
\end{align*}
Therefore we see that equation \eqn{chi01b} implies \eqn{chi1altid}.
The proof of the remaining parts of Theorems \refthm{chi60} and \refthm{mock7alt}
are analogous.

Theorems \refthm{chi60} and \refthm{mock7alt} imply simple relations between
the coefficients. We define the coefficients $C_0(n)$ and $C_1(n)$ by
\begin{align*}
\sum_{n=0}^\infty C_0(n) q^n &= (q)_\infty \,(2 - \chi_0(q)),\\
\sum_{n=0}^\infty C_1(n) q^n &= (q)_\infty \,\chi_1(q),
\end{align*}
define
\begin{equation}
\varepsilon_p =
\begin{cases}
-1 &\mbox{if $p\equiv 3 \pmod{10}$},\\
1 &\mbox{if $p\equiv 7 \pmod{10}$},
\end{cases}
\end{equation}
and for an integer $n$ and a prime $p$, define $\nu_p(n)$ to be the
exact power of $p$ dividing $n$.
\begin{thm}
\label{thm:chirels}
If $p>5$ is any prime congruent to $3$ or $7$ mod $10$, then
\begin{align}
C_0(n) &= 0  \qquad \mbox{if $\nu_p(30n+1)=1$},
\label{eq:chirel1}\\
C_0(p^2 n + \tfrac{1}{30}(19p^2-1)) &= - \varepsilon_p\, C_1(n)
\qquad \mbox{for $n \ge 0$},
\label{eq:chirel2}\\
C_1(n) &= 0  \qquad \mbox{if $\nu_p(30n+19)=1$},
\label{eq:chirel3}\\
C_1(p^2 n + \tfrac{1}{30}(p^2-19)) &=  \varepsilon_p\, C_0(n)
\qquad \mbox{for $n \ge 0$}.
\label{eq:chirel4}
\end{align}
\end{thm}
\begin{proof}
Suppose $p>5$ is any prime congruent to  $3$ or $7$ mod $10$. Then
$5$ is a quadratic nonresidue mod $p$.
Therefore $5a^2-b^2\equiv0\pmod{p}$ implies that $a\equiv b\equiv0\pmod{p}$
and \eqn{chirel1} clearly follows from \eqn{chi0altid}.
Similarly \eqn{chirel3} follows from \eqn{chi1altid}.

We suppose $a\equiv 1\pmod{6}$, $b\equiv1$, $11\pmod{30}$, $3\abs{b}< 5\abs{a}$,
and $a\equiv b\equiv0\pmod{p}$. Letting $a=pa'$, $b=pb'$ we have the following
table
\begin{center}
\begin{tabular}{c c c}
$p\pmod{30}$ & $a'\pmod{6}$ & $b'\pmod{30}$\\
$7$          & $1$          & $13,-7$ \\
$13$          & $1$          & $7,-13$ \\
$17$          & $-1$          & $-7,13$ \\
$23$          & $-1$          & $-13,7$
\end{tabular}
\end{center}
By considering the table and noting that the
summation term
$$
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(5a^2 - b^2) - \frac{1}{30}}
$$
is invariant under both $a\mapsto -a$ and $b\mapsto -b$ we see that
\begin{align*}
&\sum_{n=0}^\infty
C_0(p^2 n + \tfrac{1}{30}(19p^2-1))\,
q^{p^2 n + \tfrac{1}{30}(19p^2-1)}\\
&\qquad=
\sum_{\substack{ 3\abs{b'} < 5 \abs{a'} \\ a'\equiv 1\pmod{6}
\\ b'\equiv 7,13 \pmod{30}}}
\textnormal{sgn}(pb') \leg{12}{pa'} \chi_{60}(pb')
  q^{\frac{1}{120}(p^2(5(a')^2 - (b')^2) - \frac{1}{30}}
\\
&
\qquad=
\leg{12}{p} \chi_{60}(p) \sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 7,13 \pmod{30}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}(p^2(5a^2 - b^2)) - \frac{1}{30}}
\end{align*}

and

\begin{align*}
&\sum_{n=0}^\infty
C_0(p^2 n + \tfrac{1}{30}(19p^2-1))\,
q^{n}\\
&\qquad=
-i\,\varepsilon_p\,\sum_{\substack{ 3\abs{b} < 5 \abs{a} \\ a\equiv 1\pmod{6}
\\ b\equiv 7,13 \pmod{30}}}
\textnormal{sgn}(b) \leg{12}{a} \chi_{60}(b)
  q^{\frac{1}{120}((5a^2 - b^2) - \frac{19}{30}}\\
&\qquad = -\varepsilon_p  (q)_\infty \,\chi_1(q)
 = -\varepsilon_p \sum_{n=0}^\infty C_1(n) q^n,
\end{align*}
and \eqn{chirel2} follows.
The proof of \eqn{chirel3}--\eqn{chirel4} is analogous.
\end{proof}

In a similar fashion, Theorem \refthm{mock7alt} implies relations between the
coefficients of the seventh order mock theta functions. For $j=0$, $1$, $2$
we define $f_j(n)$ by
$$
\sum_{n=0}^\infty f_j(n) q^n = (q)_\infty \,\mathcal{F}_j(q).
$$
\begin{thm}
\label{thm:mock7rels}
Let $p$ be any odd for which $7$ is
a quadratic nonresidue mod $p$;i.e.\ $p\equiv\pm5, \pm 11$ or $\pm 13\pmod{28}$.
\begin{enumerate}
\item
Then
\begin{align*}
f_0(n) &= 0  \qquad \mbox{if $\nu_p(28n+1)=1$},\\
f_1(n) &= 0  \qquad \mbox{if $\nu_p(28n-3)=1$},\\
f_2(n) &= 0  \qquad \mbox{if $\nu_p(28n+9)=1$}.
\end{align*}
\item If $p\equiv \pm 5 \pmod{28}$ then
\begin{align*}
f_0(p^2n + \tfrac{1}{28}(9p^2-1)) &= \pm f_2(n),\\
f_1(p^2n + \tfrac{1}{28}(p^2+3)) &= \pm f_0(n),\\
f_2(p^2n + \tfrac{1}{28}(25p^2-9)) &= \mp f_1(n+1).
\end{align*}
\item If $p\equiv \pm 11 \pmod{28}$ then
\begin{align*}
f_0(p^2n + \tfrac{1}{28}(25p^2-1)) &= \mp f_1(n+1),\\
f_1(p^2n + \tfrac{1}{28}(9p^2+3)) &= \pm f_2(n),\\
f_2(p^2n + \tfrac{1}{28}(p^2-9)) &= \mp f_0(n).
\end{align*}
\item If $p\equiv \pm 13 \pmod{28}$ then
\begin{align*}
f_0(p^2n + \tfrac{1}{28}(p^2-1)) &= \mp f_0(n),\\
f_1(p^2n + \tfrac{1}{28}(25p^2+3)) &= \mp f_1(n+1),\\
f_2(p^2n + \tfrac{1}{28}(9p^2-9)) &= \mp f_2(n).
\end{align*}
\end{enumerate}
	\end{thm}

We omit the proof of Theorem \refthm{mock7rels}. The proof is analogous to that
of Theorem \refthm{chirels}.

\section{Concluding Remarks}
\label{sec:conclude}

In Theorems \refthm{chi60} and \refthm{mock7alt} we found new identities
for the fifth order mock theta functions $\chi_0(q)$, $\chi_1(q)$
and all three  seventh order mock theta functions
$\mathcal{F}_0(q)$, $\mathcal{F}_1(q)$, $\mathcal{F}_2(q)$,
in terms of Hecke-Rogers indefinite binary theta series.
This suggests the problem of relating these theorems directly to the results
of Zagier (Theorem \refthm{zagchithm}) for the fifth order functions,
and to the results of Andrews \cite[Theorem 13, pp.132--133]{An86}
and Hickerson \cite[Theorem 2.0,p.666]{Hi88b} for the seventh order
functions.

\subsection*{Acknowledgments}
I would like to thank Chris Jennings-Shaffer and Jeremy Lovejoy
for their  comments and suggestions. Also I would like to thank the
referee for corrections and suggestions.

redpajama_set_name: RedPajamaArXiv


Drexelbrook's Ultrasonic level transmitter, USonic, is part of a family of 2-wire, HART compatible, ultrasonic level transmitters with premium features in a durable and economical package. Drexelbrook has decades of experience with ultrasonic level measurements and the result is a unique ultrasonic level transmitter that can take on and simplify applications never previously thought possible for the ultrasonic technology. USonic also brings a number of unique features to the market that make using the USonic ultrasonic level transmitter a simple and cost effective solutions for liquids and slurries.
The USonic range of ultrasonic level transmitters include many standard features like the patent SMARTgain™ algorithm for auto profiling, PipeSoft for ferrules and underground tanks and industry leading accuracy and response time. The USonic Ultrasonic level transmitter is also IECEx approved for intrinsically safe Ex ia which makes it ideal for hazardous areas. The housing, keypad and display are also UV and weather rated for use outside without the need for sun or weather shielding.
A patent algorithm that automatically profiles the ultrasonic level transmitters surrounding environment to ignore obstructions like ladders and agitators that can interrupt the sonic signal. No more tedious tank mapping procedure. SMARTgain™ runs automatically without any user input required and does not require the vessel be emptied as future obstructions are automatically adjusted for. Another way Drexelbrook is making the Ultrasonic level experience easier and more reliable.
Ever wanted to use an ultrasonic level transmitter when a long and narrow ferrule or nozzle was involved? Well now you can. In the past long nozzles and ferrules that open up to a large area, like an underground diesel tank, would play havoc with ultrasonic level transmitters. The signal bounces of the walls of the pipe and scatter on exit meaning measurement past this the end is lost. This is where Drexelbrook unique PipeSoft comes into it's own by being able to make measurements off these scattered sound waves. The result is an Ultrasonic level transmitter that can be used for underground pits, tanks or vessels with ferrules as long as 7 meters.
Aside from being an accomplished Ultrasonic level transmitter. The USonic is also an excellent open channel flow meter. Drexelbrook's USonic comes complete with many common weir and flume functions as well as the ability to program your own geometry for optimum flow measurements.
Having a fast response time is a key advantage for an Ultrasonic level transmitter. The USonic family has a standard response time of 300 ms which is the fastest in the industry. Response time enables an ultrasonic level transmitter to respond quickly to changes in the process particularly important for filling, emptying or upset scenarios.
UV Rated display for full sun exposure. No shade required.
The USonic R ultrasonic level transmitter expands on the the features list of the USonic with options like dual sensors, relays output, modbus, data logging and additional programs for batch control and multiple tanks.


Also available as a combo (Rack plus a SureCan)!
Our SureCan Gas Can Rack secures your SureCan to your open or enclosed trailer with ease. Our SureCan Rack is fully lockable to secure your SureCan from theft or damage. Our SureCan rack features a cutout to allow you to easily fill your equipment without removing your SureCan from the rack. To remove your SureCan from the rack, simply pull the pull-pin on the side of the rack to release the spring loaded locking arm. This innovative design allows you effortless access to your SureCan at any time. To secure your SureCan rack simply push down on the locking arm, and run a padlock through the locking arm hole, this will prevent theft. Our rack also allows the SureCan to be mounted in both directions in the rack, preventing anyone from draining your fuel. Our SureCan rack will mount effortlessly to either your Open or Enclosed landscaping trailer.


text:
Q: How to create a DataFrame with Reddit API loop and manage the list I'm very new to Reddit API (PRAW/PSAW), Python, as well as programming in general. What I'm trying to do is get top submissions from certain subreddits within 6 months, then convert the list into a DataFrame and to CSV file later.
I want to:

*

*Get the length of the list

*Sort by date(epoch)

*Make a data frame out of this

What I tried so far:
list_submission = []
for submission in reddit.subreddit('bitcoin').top(limit=None):
    if submission.created_utc >=1569902400 and submission.created_utc <=1585627200:
        print(submission.created_utc, submission.title, submission.score, submission.id) # This seems to get me the data I want.
        len() # I want to check the length, but it doesn't work. It just gives me a row of zeroes.
        sorted(submission.created_utc) # This also doesn't work. It says 'float' object is not iterable.
                                       # I tried converting to int, but also didn't work.
pd.DataFrame(list_submission) # Also doesn't work.

So in brief,
I suppose making a data frame out of this can as well solve the first 2 problems, although I think being able to do that using the codes will be helpful when evaluating the list!

A: To answer the 3 parts of your question:

*

*To get the length of a list, you need to pass the list you want to evaluate to the len() method, so if you want to let's say find the length of list_submission, you would instead do len(list_submission). Right now you are basically trying to get the length of nothingness, so that is why you are seeing zeros.

*If the submission matches the requirements, you can append it to the list of submissions with list_submission.append(submission). Then after the for loop is complete, you can used sorted() to sort the entire list. You need to pass in the whole list plus the key you want to sort on, so it would look like sorted(list_submission, key=lambda submission: submission.created_utc). The reason you are getting an error is because you are passing in the wrong parameters.

*Your method for converting the list into a DataFrame should then work. You can use columns = ['created_utc', 'title', 'score', 'id'] to set the column names.

Final code will look something like the following:
list_submission = []
for submission in reddit.subreddit('bitcoin').top(limit=None):
    if submission.created_utc >= 1569902400 and submission.created_utc <= 1585627200:
        print(submission.created_utc, submission.title, submission.score, submission.id)
        list_submission.append(submission)
        print(len(list_submission))

sorted(list_submission, key=lambda submission: submission.created_utc)
pd.DataFrame(list_submission, columns = ['created_utc', 'title', 'score', 'id'])

redpajama_set_name: RedPajamaStackExchange


text:
You are viewing RINGO STARR
RINGO & HIS ALL STARR BAND AT BEACON THEATER, NY, USA
By	Beatles Magazine Posted on Wednesday, November 15, 2017 Thursday, November 16, 2017 In RINGO STARR 0 RINGO STARR	26
Ringo & His All Starr Band at Beacon Theater, New York – 15 Nov, 2017
RADIOHEAD'S PHIL SELWAY JOIN RINGO AT THE BACON THEATER
By	Beatles Magazine Posted on Wednesday, November 15, 2017 Monday, November 20, 2017 In RINGO STARR 0 RINGO STARR	26
Radiohead drummer Phil Selway joined Ringo and the All Starr Band for a performance of the Beatles' classic "With a Little Help From My Friends" at the Beacon Theater in New York City Wednesday.
Selway struck a characteristically stoic pose behind the kit, his steady playing style a sharp contrast to the flash of All Starr Band drummer Gregg Bissonette.The pair played in perfect tandem, providing an extra thump to the song's marching beat as Starr crooned the Sgt. Pepper's classic. While Starr ran off towards the end of the song, he quickly returned as Selway and the All Starr Band segued into a rendition of "Give Peace a Chance."
Ringo released his 19th solo album, Give More Love, in September.
RINGO & ALL STARR BAND AT MAYO PERFORMING ARTS CENTER, NJ USA
By	Beatles Magazine Posted on Tuesday, November 14, 2017 Thursday, November 16, 2017 In RINGO STARR 0 RINGO STARR	14
Ringo & His All Starr Band at Mayo Performing Arts Center, Morristown, NJ, USA 14 Nov, 2017
RINGO & HIS ALL STARR BAND AT CONSTANT CONVOCATION CENTER, VA
By	Beatles Magazine Posted on Sunday, November 12, 2017 Wednesday, November 15, 2017 In RINGO STARR 0 RINGO STARR	11
Ringo Starr & His All Starr Band at Constant Convocation Center – Old Dominion University, Norfolk, VA, USA 12 November, 2017
RINGO & HIS ALL STARR BAND AT FOX THEATER, ATLANTA
By	Beatles Magazine Posted on Saturday, November 11, 2017 Wednesday, November 15, 2017 In RINGO STARR 0 RINGO STARR	11
Ringo & His All Starr Band at Fox Theater, Atlanta, GA, USA 11 November, 2017
"RINGO" IS ONE OF 75 CLASSIC ALBUM PICKED TO CELEBRATE THE MILESTONE THAT IS CAPITOL RECORDS'75th BIRTHDAY
By	Beatles Magazine Posted on Saturday, November 11, 2017 In RINGO STARR 0 RINGO STARR	26
With his third solo album, Ringo, finally put his solo career in gear, showing that he had a lot more to offer than his eccentric first two solo albums, Sentimental Journey and Beaucoup Of Blues, both released in 1970. Issued three years later, on 2 November 1973, Ringo was a far more satisfying record, made on a big budget and featuring an amazing cast of backing musicians.
In its review of the album, Rolling Stone said, "This Ringo Starr album is the first to actually invoke The Beatles' aura." That was down to the fact that John Lennon, George Harrison and Paul McCartney each contributed as songwriters, singers, and instrumentalists on the album, across recording sessions that began in March 1973 and wrapped later that summer. Consequently, Ringo is the only solo Beatle album to feature all four of the Liverpudlians playing on one record. However, on no single track did all four appear together. Harrison played the guitars on the Lennon composition 'I'm The Greatest', with Lennon playing piano and singing harmony on a song he re-wrote for Starr and which was used as the album's opening track. Harrison also joined in on 'Sunshine Life For Me', 'Photograph' and 'You And Me (Babe)', co-writing the latter two.
In June 1973, Ringo flew to London, where Paul and Linda, joined in on the McCartney tune 'Six O'clock,' which had been written specifically for the album. With a tight structure and lyrical grace, it is a standout composition on the record. Paul also appeared on Ringo's cover of the 1960 Johnny Burnette No.1 hit 'You're Sixteen' (written by the Sherman Brothers), which provided the biggest single hit of the album. Nicky Hopkins, a session musician who appeared regularly with The Rolling Stones, provides some lively piano backing, and there is even a kazoo impression from McCartney. Starr had been able to persuade the latter to be involved in the project by telling him, "You don't want to be left out, do you?"
But it wasn't only the guests that made Ringo such a success: Starr advanced his own cause by co-writing two of the album's Top 10 singles, the No.1 'Photograph' and 'Oh My My', which had backing vocals from Motown star Martha Reeves. Ringo and Vini Poncia's 'Devil Woman' were just as good as the hits. Though Ringo's vocal range is not particularly wide, he sings with gusto throughout and his voice carries a certain pathos.
One of the highlights of Ringo is a version of master songwriter Randy Newman's composition 'Have You Seen My Baby'. Starr's version has real verve, helped by compelling boogie guitar from T.Rex main man Marc Bolan and fine honky-tonk piano from New Orleans legend James Booker. Though the album was recorded at Sunset Sound Studios in Los Angeles, Bolan's guitar was added as an overdub at A&M Studios.
Ringo's best and most consistent new studio album, Ringo represented both the drummer/singer's dramatic comeback and his commercial peak; it was only beaten to the top of the Billboard charts in November 1973 by Elton John's Goodbye Yellow Brick Road. The original 10-track 1973 album was reissued in 1991 as a 13-track CD, the bonus tracks including 'Early 1970', Starr's interesting perspective on The Beatles' break-up.
Ringo was produced by Richard Perry, who had worked with John's friend Harry Nilsson. Ringo said: "We met at a session for one of Harry's albums. I went down and played and Richard and I got to egging each other on about doing something together. We ended up at a club, and when we were leaving we promised we'd get together." Perry was a good choice as producer, and Nilsson returned the favour by singing backing vocals on 'You're Sixteen'.
Among the other leading guest musicians are Jimmy Calvert (guitar on five tracks), Steve Cropper (guitar), Billy Preston (piano), Jim Keltner (drums), Milt Holland (percussion), and The Band's Garth Hudson (accordion), Robbie Robertson and Levon Helm (mandolin).
The cover art was created by Tim Bruckner, who had been working as an apprentice jeweller in Beverly Hills when he met Perry and pitched to do some freelance album artwork. Bruckner was flown to London to show Starr some concept ideas and was hired by the singer. In 2015, the artist told Beatlesbible.com: "There are 26 portraits in the balcony. The rest are people I invented. Ringo's cover sidekick, the cherub, happened after I got home. Having met the man and spent some time with him, I understood how important humour was to him and his circle of friends. The cherub just seemed like a natural extension of that part of his character, funny and a little mischievous."
The Latin motto at the top of the sleeve – "Duit on mon dei" ("Do it on Monday") – was Nilsson's idea. "Harry wanted to make it a joke on a Latin motto. I think the only reason it got on the cover was they thought it was funny and asked that I include it," added Bruckner.
Ringo has a stellar cast; it's light-hearted, unpredictable and entertaining, and stands as a testament to Starr's ability as a musician outside of The Beatles.
Ringo is one of 75 classic albums picked to celebrate the milestone that is Capitol Records' 75th birthday. Buy the album here.
source:udiscovermusic


text:
Mike Zimmer to take his time replacing Vikings OC, DC
By Nick Shook
Published: Jan. 13, 2020 at 11:08 a.m.
Updated: Jan. 13, 2020 at 12:41 p.m.
Brady is 'open-minded' about where he could play
Titans DC Dean Pees retires after 16 NFL seasons
There are primarily two types of change in the NFL: change as a result of success, and change as a result of failure.
The Vikings are about to confront plenty of the former even after falling in the Divisional Round.
Mike Zimmer's staff will likely look considerably different in 2020 after the expected departures of offensive coordinator Kevin Stefanski and defensive coordinator George Edwards. Stefanski is set to take the Browns' head coaching job, while Edwards' contract expired at the conclusion of the 2019 season, which Zimmer described as "a situation where it's probably best." The Vikings are also parting ways with longtime defensive backs coach Jerry Gray, NFL Network's Tom Pelissero reported, per a source informed.
▶ Mahomes transcends hyperbole
▶ 49ers win with superb gameplan
▶ Packers still see open window
▶ Why this feels like Chiefs' time
Stefanski's replacement will be the hottest topic related to staff changes, as it's possible veteran coach Gary Kubiak, who helped guide Stefanski in reshaping Minnesota's offense in 2019, could take the job. Kubiak's son, Vikings quarterbacks coach Klint Kubiak, might also be a candidate for the job if he isn't lured to Cleveland by Stefanski. Zimmer declined to offer insight on what he and Gary Kubiak might have discussed regarding the job, adding the two will "discuss it a little bit more at another time."
"Right now, it's a little early so we're going through the process of all that and we'll determine that as we go," Zimmer said of possibly looking internally to find Stefanski's replacement. "So I haven't made any determinations yet. Obviously we've talked about it a little bit and we'll kind of go from there."
As for Edwards' replacement, Zimmer stressed a need to step away from football in general in order to process his team's disappointing postseason loss before evaluating and identifying candidates for the position.
That was the theme of Monday's media session with the coach, who was also noncommittal on a long-term plan for quarterback Kirk Cousins, but did say he felt Cousins was a better version of himself in 2019.
"I think Kirk [Cousins] played a lot better this year than he did the year before," Zimmer said. "I think obviously the scheme helped quite a bit and so I think, yeah, this may have been his best year he's played in the NFL."
While vague, Zimmer left a bread crumb or two for those trying to project who might be communicating in the quarterback's radio in 2020.
"It will be the same the system," Zimmer said of Minnesota's offensive future with Cousins. "Obviously he's going to have a new offensive coordinator -- his fifth one in five years I think he told me yesterday or at least what's in his ear on game day. There's nothing I can do about that. But I think it's important not just for Kirk but the entire offense to have the same system, the same calls, things like that so they come in here and it's not completely foreign to them."
That sure sounds like Zimmer would like to keep things consistent offensively, though we'll have to wait to see how he decides to fill the open coordinator position. As for his own long-term status, Zimmer is content with letting things play out.
"First of all -- like I said a couple of weeks ago when we were talking about this -- I love these players, this organization and the owners have been outstanding to me so I believe there will be conversations here in the near future and whatever happens, I'm happy with it," the coach said.
"Lions GM 'comfortable' with Stafford returnin..."
Lions GM 'comfortable' with Matt Stafford returning


On the 9th of April Andrey Sushentsov, head of the "Foreign Policy" agency and program director of Valdai Club participated in the Russian-Iranian dialogue. The event was held by international discussion club "Valdai" and Institute for Political and International Studies (IPIS, Tehran). More than 50 experts participated in the dialogue, including 15 leading Russian specialists in international relations, security and nuclear energy sector.


To communicate with students throughout the term outside of the classroom, you can post Announcements in your Blackboard course shell.
An Announcement can be used when you want to send a reminder to students; "Remember we'll be meeting in the Library tomorrow."
Click Announcements on the Course Menu.
Type a subject for your announcement (think of this like the subject line in an email).
Type your message for students in Message.
You can choose to have your announcement emailed to all your students in addition to posting it in Blackboard. Check Send a Copy of this Announcement Immediately to do so.
Note that sending an announcement from your course shell only sends the announcement to students in that section of the course. There is a way to send an announcement to students in multiple courses or sections at the same time using Qwickly.


I have never considered myself to be evangelical, as to do so risks confrontation and that is just not in me.
Another rule I assigned to simple living is to not engage in conversation with anyone seated next to me on a long flight. Why chance it? Some people don't know when to stop conversation. You are captive, and what does it matter anyway with a complete stranger you are not likely to encounter again?
For some reason, I have never been very good at following this rule and possibly a few others. In the process, I have met some of the most fascinating people, from a Frank Sinatra impersonator to the CFO of a major hotel chain. I was even invited to stay as a VIP Guest in his 4-star hotel on future business travels. The one passenger I recall with total clarity 30 years later is the self-proclaimed atheist traveling alone.
She boarded the plane late, and the door closed immediately behind her. The center seat was open, and I had the window seat as usual. She politely asked if she may sit next to me as she fastened her seat belt, still breathing heavily from her run to catch a nearly missed flight.
Ms. Rosenstein was on her way home from a Mothers Against Drunk Drivers Convention. She had been a keynote speaker as she often was. She told me about the fateful night her son was heading home from college, following his girlfriend, when she got a flat tire. He stopped to change her tire and was struck and killed by a drunk driver. Her pain was raw and real, and she wept as if it had happened yesterday, but it had been years. I wondered how many times she had told this story. How many times she had re-lived this pain in such detail?
As I do, I searched for words to comfort her, but clearly I had none. I mentioned God and faith, and it sounded like a canned platitude coming from a young man who knew little of either at the time.
She sobbed and told me that she did believe in God, but she was just so lost and angry. I told her that was okay, and she spoke of her pain as I listened. Our plane gently touched down. She looked into my eyes and we were both tearful but smiling. She hugged me and thanked me. I was certain she heard what she needed to hear as I said nothing further.
It was not of me, but through me. It doesn't matter if we don't know what to say. Sometimes it is okay if we just listen and let God do the talking.
When asked to speak on Stewardship, little did I know how the theme would make my head swim. I prayed to God and asked for help on how I should talk about Stewardship based on the theme, "Growing Thro..