If your Europe plans didn't pan out this summer, don't worry because you can still sneak away on the cheap later this year. Norwegian Airlines' fall sale has $89 flights to Europe that won't break the bank. If you're considering taking a trip across the pond, you'll need to book your tickets sooner rather than later. The discounted fares must be booked by early September. Pack your bags and get ready to fly, because these flight prices are so worth it.
Norwegian Airlines is a budget airline carrier, but these flight prices are almost too good to be true. In order to take advantage of Norwegian Airlines' fall flight sale, you must book your travel no later than Wednesday, Sept. 5. You won't be able to score any big sales in September or most of October, so keep your eyes peeled for a deal. The Norwegian Airlines fall flight sale is valid for travel between Wednesday, Oct. 31, 2018, to Sunday, March 31, 2019, per the airline's website.
If a trip to Edinburgh, Scotland has been on your travel bucket list, now is the time to make your flight arrangements. From November to March, you can fly from New York's JFK Airport to Edinburgh from $89 each way. There are roughly 3,257 miles in between the two cities. Flights to Europe can cost thousands of dollars, so the fact that you could fly roundtrip from NYC to Scotland for under $200 is almost hard to believe. If you've already checked out the castles Edinburgh has to offer, Ireland is just as beautiful. From JFK, you can fly to both Dublin and Shannon, Ireland for $99 each way any time from November to March. I know, March seems like forever from now. Don't forget that St. Patricks Day is March 17. There's no better place to celebrate than Ireland.
Flights aren't just good out of NYC. There are deals out of Los Angeles and San Francisco to many destinations across Europe. They don't exactly rival the $89 flight prices from New York, but they are still pretty affordable in the grand scheme of flying to Europe. If you live in Los Angeles, you can get away to Barcelona, London, and Madrid for $169 each way from November through March, according to Norwegian Airlines. Paris, Oslo, Rome, and Copenhagen aren't too far out of reach, either, at just $179 each way from November to March. Flying out of Oakland or San Francisco? You can save $30 each way. Fares to Barcelona, London, and Paris start at $149 for a one-way flight.
Europe may be a popular place to visit in the summertime, but there's something to be said for their awesome array of holiday markets. I'm always ready to pack my suitcase and head off to Europe, but most of the time it's too expensive. Knowing when and how to save big on airfare is half the battle of booking any trip. These flight deals from Norwegian Airlines are seriously tempting me to book right now. Remember, if you're thinking about booking a flight, make sure to do so by Wednesday, Sept. 5. That's when prices go up.


Art 210 / Digital Tools / Typography is the design, choice, and arrangement of printed letters in on a two-dimensional plane, in a sequence, or in motion graphics. You will learn and explore its methodologies through class exercises and assignments. As this course progresses, you will understand the cultural and aesthetic dimensions of the visible word. The knowledge of typography will play a substantial role in any future practice of effective design communication.
Prompt / The poster is one of the oldest forms of graphic design. The main purpose is catching someone's eye and advertising a product, service, or event. In this project you will be designing a few posters with the typography being the strong note.
Prompt / Digital media is a relativly new form compared to the printed book or poster. A designer working in the digital realm depends more on storytelling and mockups compared to final printed pieces. In this project you will be designing icons and prototypes for a forecasting app.
Prompt / For this last project we shift gears away from code, to a workflow you might use working with a team of developers. You will be designing a multi-page website of your choice along with all the necessary peripherals to communicate your idea clearly to both your developers and your clients.


Slowly but surely, Generation Xers have been taking over from Baby Boomers as the majority of parents in elementary and secondary education. In the early 1990s, Gen Xers began joining parent-teacher associations in the nation's elementary schools. Around 2005, they became the majority of middle school parents. By the fall of 2008, they took over as the predominant parents of high school seniors.
Gen-X parents and Boomer parents belong to two neighboring generations, each possessing its own location in history and its own peer personality. They are similar in some respects, but clearly different in others.
Throughout the 1990s, educators grew accustomed to "helicopter parents," Boomer parents of Millennials who sometimes are helpful, sometimes annoying, yet always hovering over their children and making noise.
Today, behold the era of the Gen-X "stealth-fighter parent." Stealth-fighter parents do not hover. They choose when and where they will attack. If the issue seems below their threshold of importance, they save their energy and let it go entirely. But if it crosses their threshold and shows up on their radar, they will strike — rapidly, in force and often with no warning.
Boomers always have cared deeply about the higher moral and civic goals of education, for the ultimate purpose (recall the '60s!) of creating a more ethical and socially conscious community. Gen-X moms and dads tend to be more interested in how the right school will create concrete opportunities for their own children.
Many Gen Xers believe they live in an individualistic world in which there is no common interest and people do best by looking out for their own interests. As voters, they are less sympathetic than Boomers to cross subsidies and state equalization formulas and more intent on preserving local funds for local purposes. As parents, they will be more exclusively focused on their own child than the good of the school. While Boomers might volunteer for a districtwide curriculum committee, Gen-X parents will prefer to be a class chaperone, which directly benefits (and allows them personally to monitor) their own child.
Boomer helicopter parents generally assume the rewards of school and college are vast but impossible to measure. But stealth-fighter parents are more likely to assume that anything immeasurable is untrustworthy. Back when Xers were graduating from K-12, one blue ribbon commission after another told them their schools had failed and that the passionate hopes of '60s reformers had miscarried. Now they want proof their children won't have the same problem.
This local, pragmatic, bottom-line perspective certainly contrasts with the more global, idealistic and aspirational perspective of Boomers. It has driven the rapid growth of parent-teacher organizations that opt out of any affiliation to the National Parent Teacher Association. According to many younger parents, the PTA is simply too large, too inflexible, too politically correct and too deferential to the educational establishment.
Skeptical of grandiose claims and worried about making ends meet economically, most Gen Xers are acutely sensitive to the prices they pay and the value they receive in return. As voters, they may doubt that a routine tax or fee is really worth whatever schools are buying with it. As parents, they comparison shop to make sure a school's reputation or brand is worth all the life costs they must incur (including mortgage payments and property taxes) for their children to attend.
When you talk generations, the first thing most people want to know is where they fit.
They are always looking for a discount or shortcut. While Boomers may brag about how much they paid for a BMW, Gen Xers are more likely to brag about how little they paid.
As a whole, Gen Xers feel comfortable with market outcomes — and most Gen-X parents expect schools to be run like customer-oriented businesses. As with other purchases and investments, Gen Xers believe their children's education should be a fair and open transaction with complete and accurate information and unconstrained consumer choice. They will evaluate the transaction on the basis of the value it appears to offer. If it doesn't offer the right value, they will take their business elsewhere — whether another school district, a charter school, a private school or home schooling. It is practically impossible to persuade most Gen-X parents they should relinquish their choice for the sake of some great public good.
Administrators long have complained about Boomer helicopter parents who argue endlessly with teachers and administrators whenever their child encounters a problem. Gen-X stealth-fighter parents are less likely to argue at great length and more likely to find a loophole, buck rank or go quickly to a confrontational posture. Some Xers may skip the discussion stage entirely and move immediately to a decisive action — suddenly filing a lawsuit, for example, or withdrawing their child from a school with no warning.
Employers already notice this difference with Gen Xers as employees. When they don't like their boss, they don't talk, they walk. As K-12 parents, they will walk with their child.
• Assume no trust. Market to them, spell out the rules and start relationships early. To win over Gen-X consumers, almost every industry — from autos to health insurance plans to financial and legal services — is trying to build trust by assuming no brand loyalty, by rejustifying product value from the ground up and by educating buyers about what they must do in order for the product to work as expected. K-12 schools will get along better with Gen X if they take a similar approach.
First, assume no trust. Given their generally positive or at least empowering experience with education institutions, Boomers generally have trusted the school bureaucracy to do right by their children. Not today's younger parents. Schools must gird themselves for Xers who feel they have no reason to trust the schools' competence — and take steps to justify their performance in every area, from physical safety to academic achievement.
Second, market to them. More business oriented than Boomers, Gen Xers are impressed by marketing skills that enable an institution to make its best case quickly and easily to busy consumers. Incredibly, public school districts routinely bury favorable findings about their performance in unreadable memos, while allowing the media to shape negative findings without rebuttal. Most districts would benefit by hiring a marketing consultant. To Gen Xers, who disproportionately fill the ranks of America's sales and marketing workforce, the inability to shape and pro-ject a clear message is itself a symptom of incompetence.
Third, spell out the rules. Gen Xers, accustomed to thriving in a high-option marketplace, tend to believe they have a right to anything not explicitly denied them in the contract. When introducing themselves to parents, K-12 districts should explicitly spell out the reciprocal obligations of parents and schools — and enumerate concretely where the school will take the lead (and parents should not interfere) and where parents must pitch in to ensure student success. Like many colleges, schools can frame this agreement as a contract or a covenant. While not legally binding, this document will catch the attention of these parents, who always want to know the rules of the game.
Finally, schools should make every effort to start their relationship with parents early and on their own terms before there is a problem. If a child ends up failing an exam or being suspended for misbehavior, the teacher or counselor could be an ally in whom the parent confides — or an enemy the parent fights tooth and nail. The first meeting often makes all the difference.
• Stress personal accountability and personal contribution. Compared to Boomers, Gen Xers are less likely to trust good intentions or a well-designed institutional process and more likely to trust bottom-line incentives and personal accountability. They don't want to see guidelines and flowcharts outlining a school's emergency response plan or curriculum acceleration plan. What they want to see is a real person who is concretely accountable for the outcome.
School leaders should stress a chain of personal accountability for anything that goes wrong, particularly in zero-tolerance areas like school safety. Designate a go-to person for everything from fire drills to hall monitoring to student counseling and mental health.
In their own market-oriented lives, Gen Xers are accustomed to bottom-line incentives ("win this contract and get a bonus; lose it and you're fired") and often are suspicious of institutions where individual accountability never seems to enter the picture.
Showcasing accountability will also help draw Gen-X parents into volunteering for tasks that don't involve their own children. Give Gen Xers "ownership" of the project. Make sure they understand they are helping real people, not just "the system." And find creative ways to reward them if it goes well, as many of these parents are entrepreneurial.
• Offer data, standards, transparency and return on investment. Gen-X parents want measurable standards for schools, teachers and students. They want to know how those standards are linked to career and life success. They want to see data measuring the achievement of those standards. And they expect transparency in all important deliberations about strategy.
Xer constituents will call on school boards to show in detail that each tax dollar serves a concrete purpose for their children's education. They are less likely than Boomers to accept credentials (of schools, administrators and teachers) at face value and will want access to data on everything from hiring practices to teacher evaluations.
Gen-X parents will be especially prone to pounce on issues that affect student success. Schools will face rising pressure to offer accurate and comprehensible information on all their evaluation systems — from consistency in grade point averages and Advanced Placement scores across schools and courses to procedures for assigning class levels and evaluating disabilities.
Above all, educators should collect data on what happens to students after graduation. Gen-X parents often are astonished to find out how little schools know about long-term student outcomes. After all, in the marketplace companies are judged, continuously and without mercy, according to how well their products measure up.
Gen-X parents will demand more than graduation and college enrollment rates. They will want to know the documented career outcomes and earnings capabilities of graduates five and 10 years down the road. They also may want to see data comparing long-term outcomes among an array of choices, such as "stopping out" of school for a year, home schooling, entering a career academy, loading up on AP courses, using community college as a stepping stone for a four-year degree and so on. Gen Xers want to advise their child in the same way they manage their finances: hands on, eyes open, all options on the table.
• Offer real-time service (the "Fed-Ex" test). Gen-X parents will apply the "FedEx" test to their children's schools, expecting the service to be cheerful, fast and efficient, with information and options in real time, online, 24/7. As educators are already beginning to note, those once-per-semester parent-teacher nights, snail-mail notifications and rotary phone tree messages no longer will cut it. If Gen-X parents can get instant, real-time information on something as trivial as a package, why should they stay in the dark about their child's academic performance?
School leaders should use digital technology to offer parents continuous access and include them in a tight cycle of intervention and redirection whenever their children hit an educational snag. Course management systems like Edline, which allow teachers to record and track each student's performance every day, also can be used to instantly share performance data with parents. Some years down the road (when fewer Baby Boomers are left to object), schools may even install real-time video monitoring systems that let parents tune in to whatever is happening in their child's classroom.
• Present your school as the best parent choice in a competitive market. Ever since their children first entered elementary school, Gen Xers have been the most vocal constituency for education policies that empower parent choice, including vouchers, magnet schools and home schooling. Even within the public school system, the share of parents who say they "chose" (rather than being "assigned to") their public school has grown steadily, from 11 percent in 1993 to 15 percent in 2003 to 18 percent in 2007.
Public school leaders who have grown accustomed to their role as default educators will no longer enjoy this luxury. Both public and private school leaders need to market their schools as top-notch options in a competitive education market and persuade parents they do indeed deliver the goods.
Gen Xers like being informed and energetic consumers. When Gen-X parents perceive they really are choosing a school, their enthusiasm and goodwill can be beneficial. When they perceive they don't have any real choice, however, educators should prepare for a strong backlash over any policy event that reinforces their sense of being "trapped" (such as a decision to close down or combine schools or to redraw district lines).
Boomers, while often complaining, usually end up adapting themselves to whatever the system dishes out. Gen Xers are more likely to fight tooth and nail and even "game the system" to send their child to the school they decide is best.
• Prepare for the modular "opt-out" consumer and the innovative high-tech competitor. Gen Xers like to compartmentalize, viewing every transaction as a menu-driven series of discrete and modular choices. Across America, Gen-X consumers are dealing out the middleperson, avoiding product packages and demanding every item be customized to their tastes. Why buy a whole CD when you can purchase just the one song you like on iTunes?
Similarly, when making educational choices for their children, they may wonder why they should sign on for the whole K-12 package when parts of it may not be a perfect fit for their special child.
Many will want to split the K-12 experience into its components and pick and choose exactly what they want for their children. If a high school student wants to take a course that is not offered, can he take it online or at a community college for credit? If a middle schooler is passionate about fencing, can she opt out of physical education to take lessons? If she wants to study marine biology, can she attend the career academy across town in the afternoons? If a student drops out, is there a customized, high-tech substitute to get that high school degree? If a certain school fails to facilitate this level of customization, Gen-X parents will take their business elsewhere.
As Gen-X parents generate a rising demand for a wide spectrum of educational choices and credential substitutes over the next couple of decades, other Gen Xers entering midlife are likely to rise up to meet this demand. Gen-X entrepreneurs will find ways to provide new options outside the traditional school setting, through auxiliary programs, online courses, career modules and home-schooling aids. Gen-X info-tech designers will steadily improve the efficacy, flexibility, interactivity and entertainment value of teaching tools.
Meanwhile, as Gen-X voters, executives and elected officials replace Boomers in top leadership roles, they may agree to relax the credentials and help legitimize these end runs around the system. K-12 schools can deal with these new options by working with them, adopting them or outcompeting them. They cannot simply ignore them.
In the coming era of accountability, K-12 leaders need to face up to the rising tide of Gen-X parents. Schools that figure it out, collect the right data and market themselves intelligently to this new generation of parents will be able to rebrand themselves for success in the decades ahead.
Neil Howe is president and co-founder of LifeCourse Associates in Great Falls, Va. E-mail: howe@lifecourse.com. He is the author of the AASA book Millennials & K-12 Schools (AASA members save 5% using coupon code AASA at checkout). He is also the author of the forthcoming book Millennials in the Workplace: Human Resource Strategies for a New Generation.
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A graduate of Interamerican University in San Juan, Puerto Rico and the University of Puerto Rico School of Medicine, Dr. Gil-Acosta completed his residency in internal medicine at Indiana University Medical Center in Indianapolis. He completed two clinical fellowships in pulmonary and critical care medicine at Emory University in Atlanta, and a postdoctoral research fellowship at Veteran's Affairs Medical Center in Decatur, Georgia. He is board certified in internal medicine, pulmonary medicine and critical care medicine. Dr. Gil-Acosta is bilingual in English and Spanish. He joined Oaklawn's active staff in December 2002.


The Sims 3 AmbitionsRequires The Sims 3 to play.
With The Sims 3 Ambitions, decide whether your Sim will be the brave town hero or cause loads of mischief.
Widen your Sim's horizons with a bunch of killer new career opportunities. With The Sims 3 Ambitions, decide whether your Sim will be the brave town hero or cause loads of mischief among their neighbors. Make your Sim save the day as a brave firefighter, change the town as a leading architect, or live on the edge as a tattoo artist. Will your Sim find success as a billionaire high-tech inventor or eke out a living as a bumbling mad scientist? Will they uncover secrets as a private investigator by dusting for prints or roughing up a suspect for information? Their future is entirely in your hands!
For the first time ever, you control your Sim's actions and interactions while they're on the job.
Choose how your Sim's career progresses - will they work for good or evil, pursue creative endeavors, rise to success as a mogul, or shirk responsibility as a slacker?
Your Sim's on-the-job choices now change their town and affect other Sims as well. Modify the town's structure as an architect or set neighborhood fashions as a stylist.
One of the most worth it expansion packs in my opinion! Adds so much to the game.
The sims 3 is all great in it self. But i just love this expansion. I GET TO WORK EVEN WHEM IM NOT AT WORK!!!


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Sports briefs for June 13, 2012
Sand Gnats game to help Goodwill
The Savannah Sand Gnats have discounted tickets for sale to tonight's home game against Charleston, S.C., as a fundraiser for area Goodwills.
Goodwill retail locations in Savannah, Pooler and Richmond Hill are selling the tickets for $5 apiece. The Sand Gnats will donate 50 cents for every person who enters Grayson Stadium with a ticket purchased through Goodwill.
If Goodwill sells 1,000 tickets and those 1,000 buyers come to the game, the Sand Gnats will donate an additional $500.
Also, the first 200 children ages 12 and younger who attend will receive a free Webkinz.
Grayson Stadium opens at 6 p.m. Game time is 7:05 p.m. For more information, call 354-6611, ext. 107.
LCHS football camp planned for June
Liberty County High School is hosting its annual football skills camp for ages 7-15 from 9 a.m.-11:30 a.m. June 25-28. The fee is $60 if registered and paid before June 20. The fee goes up to $70 after June 20. For more information, call coach Kirk Warner at 876-0210
Jiu Jitsu Training
Wolf Pack Jiu Jitsu training is open for ages 8 and older. Classes for ages 8-12 are 5:30-6:30 p.m. Mondays and Wednesdays; ages 13 and older are from 6:30-7:30 p.m. For more information, call Wolf Pack Jiu Jitsu at 368-1123.
Cornhole Classic benefit set
Coach's Corner in Savannah will host its first Cornhole Classic Tournament on Thursday, June 14, to benefit Senior Citizens Inc.The cost is $100 per team of two, with cash payouts going to the top five teams. All ages and skill levels are welcome. Entry to the tournament also includes a fish fry, all the beer you can drink and a concert by Liquid Ginger. The tournament begins at 6 p.m., followed by the fish fry at 7 and concert at 8.Tickets for non-entrants are $30 per person and include dinner, drinks and the concert. Teams are limited. Call 236-0363 or visit www.seniorcitizens-inc.org for more information.
Savannah Derby Devils doubleheader June 16
The Savannah Derby Devils will have a fundraiser and canned-good drive during their doubleheader at 5 p.m. Saturday, June 16 at the Savannah Civic Center.Those who bring at least one canned good item for donation to America's Second Harvest will receive $1 off their ticket. That night's raffle proceeds will go to America's Second Harvest of Coastal Georgia.Tickets are available through the Savannah Civic Center website at savannahcivic.com and box office. Admission ranges from $2 to $16 per ticket. Discount tickets are available for children, seniors, students, military and groups of 10 or more. The event includes live music from Portman's New Horizon's Band and an appearance by the Savannah Junior Roller Derby.For more information, visit us at savannahderby.com or at facebook.com/savannahderbydevils

redpajama_set_name: RedPajamaCommonCrawl


Have you already booked a table at your favourite restaurant to relish your favorite non-vegetarian dishes? Are you all geared up to cook a flavoursome meat dish to enjoy with your family?
Every 100 gram serving of real mutton has 21 grams of fat and LOTS of cholesterol.
It is a top contributor to heart disease and diabetes.
The meat you are consuming might be the highest contributor to carbon dioxide emissions.
In India, over 300 crore chickens and 8 crore goats are raised and slaughtered in horrific conditions.
These facts don't paint a pretty picture, but how do you satisfy your meat cravings?
The Good Dot dish I tasted was just like chicken curry.
Every diet chart shows you need to stop eating your red meat. Good Dot meat definitely seems like a healthier option.
Non-veg food preparation normally has a smoky or unique flavour that vegetarians miss out on. Most of us like the gravy but cannot have the meat part, so the hitch is always there. Good Dot provides this for which is really nice for vegetarians.
Good Dot, with the help of RCM, is bringing high quality and healthy vegetarian meat that is made from grain and plant proteins. Now you can enjoy taste and textures of real meat with vegetarian meat manufactured in a cruelty-free way.


Capital Letters to Start Sentences Year 1 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and application and reasoning resources. This pack is designed to work alongside our GPS Scheme of Work for Spring Block 3.
Capital Letters to Start Sentences Year 1 Teaching PowerPoint.
Capital Letters to Start Sentences Year 1 Varied Fluency with answers.
Capital Letters to Start Sentences Year 1 Application and Reasoning with answers.
Developing Questions to support using capital letters to start sentences where children mainly write the 10 easiest to form capital letters ABEHILOXPT.
Expected Questions to support using capital letters to start sentences where children mainly write the 10 easiest to form capital letters as well as the next 6 ABCDEFHILMOPSTW.
Greater Depth Questions to support using capital letters to start sentences where children mainly write the 10 hardest to form capital letters GJKNQRUVYZ.
Developing Explain whether the capital letter in the sentence has been used correctly where the 10 easiest to form capital letters are used. (ABEHILOXPT).
Expected Explain whether the capital letters in the sentences have been used correctly where the 10 easiest to form capital letters as well as the next six are used. (ABCDEFILMOPSTWX).
Greater Depth Explain whether the capital letters in the sentences have been used correctly where the 10 hardest to form capital letters are used. (GJKNQRUVYZ).
Developing Re-write the sentence using the correct capital letter where the 10 easiest to form capital letters are used. (ABEHILOXPT).
Expected Re-write the sentences using correct capital letters where the 10 easiest to form capital letters as well as the next six are used. (ABCDEFILMOPSTWX).
Greater Depth Re-write the sentences using correct capital letters where the 10 hardest to form capital letters are used. (GJKNQRUVYZ).
Developing Write a sentence to describe the picture using the correct capital letter. Word bank provided where all words for the sentence are given.
Expected Write a sentence to describe the picture using the correct capital letter. Word bank provided where some words for the sentence are given.
Greater Depth Write a sentence to describe the picture using the correct capital letters. No word bank provided.


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Die Kirche St. Marien Schmerzhafte Mutter (verkürzt auch: Katholische Kirche und Schmerzhafte Mutter) in Flensburg entstand 1898 bis 1900 im neugotischen Stil nach Plänen des Architekten Heinrich Flügel. Die römisch-katholische Pfarrkirche gehört zu den Kulturdenkmalen der Flensburger Innenstadt.

Geschichte
Mit der Reformation war Flensburg evangelisch geworden. 1528 waren die Mönche des Franziskanerklosters St. Katharinen vertrieben worden. Während des Deutsch-Dänischen Krieges engagierte sich der Malteserorden in Flensburg (vgl. Regierungshof (Flensburg) sowie Borgerforeningen). Im Anschluss des Krieges wurde auf Grundlage der in Deutschland herrschenden Religionsfreiheit in Flensburg eine katholische Gemeinde gegründet. Bischof Paulus von Osnabrück ernannte den ehemaligen Militärpfarrer Rudolf Rave zum ersten Seelsorger der neuen Gemeinde. Zunächst durfte die Gemeinde in der Heiliggeistkirche ihre Gottesdienste halten. Der Bischof gewährte der kleinen Gemeinde kurz darauf Mittel zum Ankauf mehrerer kleiner Häuser an der Ecke Heiligengeistgang/Nordergraben. Eines dieser Häuser diente als Pfarrwohnung. 1866 wurde eines der erworbenen Häuser, das an der Stelle der heutigen Kirche stand, zu einer katholischen Kapelle umgebaut und am 29. März des besagten Jahres geweiht. Die wirtschaftliche Entwicklung Flensburgs führte in der Folgezeit zum Anwachsen der Gemeinde auf 2000 Mitglieder, so dass die Kapelle sich schon zur Jahrhundertwende als zu klein für die gewachsene Gemeinde erwies. Zu Weihnachten 1897 war letztmals eine große Zahl der Gemeindeangehörigen gezwungen, dem Gottesdienst in der Kälte vor der Tür beizuwohnen. Aus diesem Grund sollte eine neue Kirche errichtet werden, für deren Finanzierung im Vorfeld tausende Briefe mit der Bitte um Unterstützung versandt und Kollekten gesammelt wurden.

1898 begannen unter der Leitung des Hamburger Baumeisters E. Breitschneider die Neubauarbeiten für die neue Kirche, die nach Plänen des Bremer Architekten Flügel entstand. Der Bau begann unter widrigen Umständen. Beim Ausschachten stießen die Arbeiter zu Beginn auf den Stadtgraben, der aus Morast und sumpfiger Erde bestand. Erst am 19. März 1899 konnte feierlich der Grundstein gelegt werden. Am 17./18. Januar 1900 fand die Erihe der neuen Kirche statt. Gefeiert wurde zu diesem Anlass im Nordischen Hof. Zur Weihe besaß die Kirche noch keine Turmspitze. Auf dem Turm war nur ein einfaches, flaches Dach gesetzt worden. Die Wölbung im Inneren sowie der Turmhelm wurden erst 1909 vollendet. 1922 schaffte es die Gemeinde, trotz der Inflationszeit nach dem Ersten Weltkrieg ein neues Pfarrhaus am Heiligengeistgang zu errichten. Während des Zweiten Weltkrieges wurde in der ehemaligen katholischen Schule, die sich direkt südlich der Kirche befand, eine Luftschutzschule für Luftschutzwarte eingerichtet. Das besagte Gebäude, das optisch das Hauptportal verdeckte, wurde nach dem Krieg abgerissen. Dort befindet sich heute ein Parkplatz.

1980/81 wurde auf der Nordseite der Kirche ein Gemeindezentrum angebaut. Gleichzeitig wurde die abrutschende Apsis der Kirche, die auf weichem Torf und Sand errichtet worden war, mittels eines neuen Fundamentes stabilisiert. Die entstandenen Schäden am Gewölbe wurden anschließend behoben. Zum 150-jährigen Jubiläum der Gemeindegründung erhielt die Kirche 2014 ein neues Kirchendach. Die Feierlichkeiten zum 150-jährigen Bestehen der Gemeinde wurden ökumenisch begangen.

Die Kirche fungiert als Pfarrkirche einer Großpfarrei mit ungefähr 12.000 Mitgliedern, die sich bis nach Kappeln ausdehnt. Zur Flensburger Kirchengemeinde werden 9.000 Mitglieder gerechnet. In Mürwik befindet sich mit der St.-Ansgar-Kirche eine zweite katholische Kirche im Stadtgebiet. Die St.-Michael-Kirche in Weiche wurde nach dem Abzug der Bundeswehr geschlossen. In den Vororten Harrislee und Glücksburg befinden sich zwei weitere katholische Kirchen (St. Laurentius (Glücksburg) und St. Anna).

Die katholische Kirche St. Marien Schmerzhafte Mutter besitzt heute die Adressen Nordergraben 36 und Heiligengeistgang 16/18. Vor der Kirche befindet sich eine Bushaltestelle, mit dem Stationsnamen "Katholische Kirche".

Architektur und Ausstattung
Die backsteinerne Hallenkirche wurde in der Formgebung der Neugotik errichtet. Sie befindet sich in einer städtebaulich hervorstechenden Hanglage auf der Westseite oberhalb der Flensburger Innenstadt, neben der kleinen Villenburg namens Burg Schöneck. Der polygonale Chor ist nach Norden ausgerichtet. An der westlichen Langseite führt die Straße Nordergraben entlang der Kirche. Im Südwesten des Baukörpers befindet sich der Kirchturm. Im Turm hängt ein dreistimmiges Geläut von Bronzeglocken mit folgender Schlagtonreihe: f' – g' – b'. In der Südseite des Turmes befindet sich das Hauptportal der Kirche. Ebenfalls auf der Südseite befindet sich eine große Fensterrosette. Im inneren ein Kreuzrippengewölbe. Von der Innenausstattung blieben die Kirchenbänke, hölzerne Heiligenfiguren und Bilder mit der Darstellung der Kreuzwegstationen erhalten. Zur Kirche gehört heute auch die kleine Josefskapelle, die vom Hofeingang erreichbar ist.

Orgel

Die heutige Orgel wurde im Jahr 2001 von der Werkstatt Michael Becker Orgelbau errichtet. Sie verfügt über 23 Register auf zwei Manualen und Pedal, die über Schleifladen mit mechanischer Traktur angespielt werden. Die Disposition ist wie folgt:

 Koppeln: II/I, I/P, II/P

Einzelnachweise

Weblinks

 Katholische Kirchengemeinde Flensburg

Katholische Kirche St. Marien Schmerzhafte Mutter
Flensburg
Kulturdenkmal in Flensburg
Erbaut in den 1900er Jahren
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redpajama_set_name: RedPajamaWikipedia


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Jim Denney: Would Walt Disney approve of 'Star Wars: Galaxy's Edge' theme park? The answer will surprise you
Anticipation builds for Disney's 'Star Wars' theme park attractions
Disney has released photos and videos of Star Wars: Galaxy's Edge, the first of two areas to open in Disneyland and Disney World.
Walt Disney died in December 1966, more than a decade before the first "Star Wars" movie was released. What would he think of Galaxy's Edge, Disney's new 14-acre "Star Wars"-themed land?
Galaxy's Edge opens at Disneyland in California on May 31, and at Disney's Hollywood Studios in Florida on August 29. As you enter, you step into another world, the planet Batuu. There you find the Black Spire Outpost, a battle-scarred waystation for bounty hunters and smugglers on the galactic fringe.
DISNEYLAND IS LIMITING STAR WARS: GALAXY'S EDGE VISITORS TO FOUR-HOUR WINDOWS
The centerpiece attraction is Millennium Falcon: Smuggler's Run, a simulator that puts you in the cockpit of Han Solo's souped-up Corellian freighter. As part of the Falcon's cockpit crew, you jump to hyperspace, steer the ship, and fire the weapons while smuggling dangerous cargo along the outer rim of the Galaxy.
There are also "Star Wars"-themed eateries, a build-your-own droid depot, and a build-your-own lightsaber workshop. A second thrill ride, Rise of the Resistance, opens this summer.
Walt Disney had a lot to say about the future — both humanity's future in space and Disneyland's future expansion. On Disneyland's opening day, July 17, 1955, he said, "To all who come to this happy place: Welcome. Disneyland is your land. Here age relives fond memories of the past, and here youth may savor the challenge and promise of the future."
He knew the future is always a moving target, and he wanted Disneyland to never stop aiming for it. "Disneyland will never be finished," he said. "It's something I can keep developing, keep plussing and adding to. It will be a living, breathing thing that will always keep changing."
Walt envisioned Disneyland as a place to stimulate young minds and fire up the imagination. "In Disneyland, people can actually take part in visible, touchable, moving, and dimensional fantasy. They can ride on it, fly with it, measure imagination with it, and glean information from it about the past, the present, and future."
There was a special place in Walt's heart for young people. He wanted future generations to believe their future is bright, that anything is possible.
He said, "To the youngsters of today, I say believe in the future, the world is getting better; there still is plenty of opportunity."
Though the "Star Wars" saga is set "a long time ago in a galaxy far, far away," it is really about the limitless possibilities of the future.
I once spoke with science-fantasy writer Ray Bradbury about the first time he met Walt Disney. At Christmastime in 1964, Ray spotted Walt walking through a Beverly Hills department store, his arms laden with gifts. Ray stopped Walt and introduced himself. Walt's eyes lit up and he said, "Ray Bradbury! I know your books!"
It's true — Walt knew Bradbury's "Fahrenheit 451" and "The Martian Chronicles" because he was a science fiction fan from way back. Walt had grown up on the classic science fiction tales of Jules Verne. That's why he made a lavish 1955 production of Verne's "20,000 Leagues Under the Sea."
Walt was also enthusiastic about humanity's future in space, which is why Tomorrowland is one of the four original lands of Disneyland. Walt himself conceived the wildly popular Space Mountain attraction, though he originally called it Space Port. The development of Space Mountain faltered for years after Walt's death, but it finally opened in January 1975 at Florida's Magic Kingdom Park.
"Star Wars" creator George Lucas was a confirmed Disney fan. Lucas once told Entertainment Weekly, "I was at Disneyland the second day it opened." He was eleven-years-old on July 18, 1955, and his favorite ride was the Tomorrowland Autopia. Today, attractions based on his "Star Wars" creations are among the most dazzling wonders of Disney theme parks.
Walt Disney and George Lucas are both innovative, imaginative geniuses who fought for their ideas and changed the entertainment landscape. Disney and Lucas impacted the global culture in ways too profound to measure.
They created entertainment that is optimistic, family-friendly, and loved by fans of all generations.
In a 2012 interview, Lucas said, "When I first made 'Star Wars,' everybody in Hollywood said, 'Well, this is a movie that Disney should have made.'"
"Star Wars" is a paradox, a fantasy future set in the distant past. That's why it's the perfect solution to a problem that long plagued Tomorrowland.
Walt talked about this problem shortly before his death in 1966: "When we opened Disneyland, outer space was Buck Rogers. I did put in a trip to the moon, and I got [NASA space scientist] Wernher von Braun to help me plan the thing. Since then has come Sputnik and then has come our great program in outer space. So I had to tear down my Tomorrowland that I built eleven years ago and rebuild it to keep pace."
Star Wars: Galaxy's Edge will open May 31, 2019, at Disneyland Park in Anaheim, California.
It's true. The futuristic attractions that dominated Tomorrowland in 1955 — the Rocket to the Moon and Space Station X-1 — have been gone for decades. They've been replaced by Space Mountain and the "Star Wars"-themed Star Tours space flight simulator. Tomorrowland makeovers tend to look dated within a decade or two, but "Star Wars" is ageless.
So, yes, Walt would definitely approve of Galaxy's Edge. As longtime Disney imagineer John Hench said, "Walt had one foot in the past and one foot in the future."
It's been more than fifty years since Walt left us, yet his optimistic, futuristic spirit lives on at Disneyland and the other Disney theme parks around the world. And yes, his spirit lives on in Star Wars: Galaxy's Edge.
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VINYL MASTERCUT RECORDS
Papeterie & Letterpress
What is SUPERSENSE?
SUPERSENSE
The Blue Screen of Death - There are just 16 steps down to hell 10"
White - €25.00 EUR Black - €25.00 EUR
The Blue Screen of Death play electronic-8bit-Punk. Because they can handle a computer better than a guitar. In 2008 the band was founded in Göttingen. Since then, its members are met with hostility or get beat up on the streets of the city - by "real" punks, whose parochial conception of the world they don't fit in. At the same time, they got everything what it takes to be a real punk band: dirt, volume, movement, energy and a hint of unpredictability. Sometimes the drumsticks fly off accidentally, another time the bass player runs against the trusses with his head and falls into the audience - similar to a baby giraffe, leaving the birth canal. Their first real record was a long time in the coming because some prick has stolen the laptop with the finished recordings on it. But now it's here! Should you need a musical connection point, there surely is the "Göttinger Schule" for starters. It's a breed of German-language, intelligent Hardcore which developed a particular sound and attitude in the mid 90s. The attitude stayed the same, but the sound was extended through electronoise. That's it! Well, not quite... The record sleeve was created by Gregor "the godfather of special packaging in conjunction with a vinyl record" Samsa and it's a giant floppy disk. So, there's only one thing left to say: Don't copy this floppy!
Let's get detailed:
Size 300 x 300 x 10 mm
Sleeve: plastic
Processing: steel / paper
Production Time: 26 Minutes per Copy
Edition: limited edition of 1000 copies 150 gram, black vinyl
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© 2020 SUPERSENSE


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Far Eastern Federal University's ALIEN ROV Revealed!
by Kevin Berry
The Far Eastern Federal University (FEFU) student team won the world championship in remotely operated vehicles at the recent 2012
MATE International ROV competition which was held in Orlando, FL.
The FEFU team — which goes by the name Primorye Coast — competed with more than
20 teams from different countries including the USA, China, India, Great Britain, Egypt, and others.
FEFU students have been taking part in these competitions since 2008. The team won first place for the first time in 2010. Primorye Coast consists of students with different specialties — from computer security to medical physics and interior design.
This year, the theme of the competition was the research of sunken World War II vessels. For the participants, there was a simulated event in which the fuel materials of the sunken vessels were an environmental threat that had to be neutralized. The teams were challenged with the development of methods for a secure fuel extraction.
The "Primorye Coast" company prepared a technical paper as part of the required submittal for the competition. This article summarizes that paper, with some editorial comments and rewrites to make the (over) 5,000 word document flow in this condensed format.
The ROV has six powerful thrusters that can provide a steady position while working with a variety of devices, or making video captures with four cameras. A special payload was installed on the vehicle, including a tank flushing device, a measuring tape for the determination of the ship's length, a metal detector to search for the debris of the vessel, a holder for the magnetic patches, and a manipulator for attaching the lift bag to the mast and also transplanting corals.
Total expense for the development of the vehicle was $10,420, while the total project cost — taking into account the materials and the cost of travel — was $40,133.
Initial Design Concepts
While developing the vehicle, the team decided to create something new — something not like their previous vehicles or industrial models, but capable to perform mission tasks.
They began with a classic brainstorming session. All team members became vehicle designers for a while. Team members proposed dozens of ideas and sketches (Figure 1). Sometimes heated discussions were held. "Will it float at all?" "How would it stand on the ground?" "Should the center of masses really be here?" and other more specialized questions were asked. The two best designs were simulated with Solid Works. The final choice is shown in Figure 2. The team notes that at this point, the most difficult task was still ahead: proving their design to their teachers and mentors. This design — a new concept unlike others the teachers were familiar with — had to be shown to be suitable for mission tasks. The mentors actually did not accept the design at first. After more dynamic discussions, all parties came to a compromise in details (but defended their initial concept).
The basis of the vehicle's design is the frame (Figure 3).


In this episode, Jeff takes his time in analyzing these two fine samples from Cornerstone Cellars. Get yourself a nice bottle of wine, and a big plate of food, before you watch this one… This one may take a while.
Wine Samples provided by Cornerstone Cellars.
What's your take on the Cabs of Napa Valley, Howell Mountain, and Cornerstone Cellars? Leave a comment, and let us know.


How To Write A Powerful Resume, download this wallpaper for free in HD resolution. How To Write A Powerful Resume was posted in August 27, 2018 at 5:27 am. This HD Wallpaper How To Write A Powerful Resume has viewed by 60 users.How To Write A Powerful Resume may related with how to write a powerful resume how to write a powerful resume objective how to write a powerful resume summary .


Stakeholders of the Peoples Democratic Party (PDP) in Essien Udim Local Government Area of Akwa Ibom State, have risen in strong condemnation of what they described as 'shameful and unethical' rampage of their sacked House of Assembly member, Mr. Nse Ntuen, who according to them has been illegally parading himself as Speaker of the State House of Assembly.
In their resolution, the stakeholders led by the political godfather of the district, Chief Michael Afangide, the chairman of the area, Raphael Isobara, serving and former political office holders, passed a vote of confidence on Governor Udom Emmanuel and Speaker of the House of Assembly, Barr. Onofiok Luke.
IF MY PEOPLE, WHICH ARE CALLED BY NAME, SHALL HUMBLE THEMSELVES, AND PRAY, AND SEEK MY FACE, AND TURN FROM THEIR WICKED WAYS; THEN I WILL HEAR FROM HEAVEN, AND WILL FORGIVE THEIR SIN, AND WILL HEAL THEIR LAND.
BUT I BELIEVE, NOTHING CAN GO WRONG WITH GOD THAT IS WITH US.


La Paz Chattanooga helped recruit and train over 75 bilingual volunteers to work with 14 Hamilton County public schools during the county-wide school registration day. Volunteers requested time off from their employers to help Chattanooga's Latino and Hispanic students enroll in school for the upcoming academic year.
"It's committed community partners like La Paz that go above and beyond to support our students and their families that will improve access and outcomes across Hamilton County," said Jared Bigham, executive director of Chattanooga 2.0.
Bilingual services are key in helping educators and students succeed, said officials. A recent survey conducted by La Paz found that educators and students in Hamilton County need greater access to culturally relevant services such as bilingual translation and interpretation services. Interpreters are needed at conferences, during school registration days and in general communications to families. These services allow parents to become active participants in their child's education by helping them understand their child's academic needs. Additionally, with additional training and support, teachers would be able to provide information on student academic progress and performance.
"The local Latino community is constantly seeking opportunities to learn English, but for many local Latino residents, their English proficiency is not yet good enough for them to fill out the paperwork on their own," explains Patrick Miles, La Paz Chattanooga's director of Community Engagement. "The process of filling out the mountain of papers required at school registration can be challenging for a native English speaker. So, for our Spanish speaking families, this process can be daunting. Just a little bit of assistance from caring community members goes a long way to show our city's collective efforts to help students in our area develop."
For more information regarding La Paz's community outreach efforts , please contact Patrick Miles, La Paz's director of Community Engagement and Administration, at pm@lapazchattanooga.org or call the La Paz office at 624-8414.


text:
Defining Marriage - Gay/LGBT News & Chat
Each week on Defining Marriage, hosts Matt Baume and James Morris chat about what's happening with marriage equality, featuring frequent digressions into pop culture, silly banter, and the jokes and quibbles that have kept them together as a couple for over a decade. The first eighteen episodes of the podcast contain the complete audiobook version of the book Defining Marriage, which traces the decades-long evolution of marriage through the personal stories of those who lived through it, featuring personal insights from the lives of Evan Wolfson, Dan Savage, Ken Mehlman, Dustin Lance Black, and many more.Read more »
Each week on Defining Marriage, hosts Matt Baume and James Morris chat about what's happening with marriage equality, featuring frequent digressions into pop culture, silly banter, and the jokes and quibbles that have kept them together as a couple for over a decade. The first eighteen episodes of the podcast contain the complete audiobook version of the book Defining Marriage, which traces the decades-long evolution of marriage through the personal stories of those who lived through it, featuring personal insights from the lives of Evan Wolfson, Dan Savage, Ken Mehlman, Dustin Lance Black, and many more.
@mattbaume
/mattbaume
Store-Bought Ham
Dec 24, 2018 · 54 minutes
It's a Christmas miracle! Defining Marriage is back, baby. And also finished. But James and Matthew aren't leaving you -- this is the start of something new and exciting and hopefully far less inappropriate. Join us for one last bumble, a round of what's-her-name, and a tease for where you can find us in the new year on our new upcoming podcast, Cozy Pants.
Do You Want to See My Foot
42 minutes | · Apr 2, 2018
Practice Man Full of Meat
38 minutes | · Dec 18, 2017
I Endorse Doug Funny
CLOUD, getReviews, 10.6ms


Posted September 13th, 2013 by Phil Bird & filed under Business, IT Support.
Asking the right questions can make selecting the right IT support provider for you and/or your business a much easier process. As an IT support business striving to provide a great service to our customers we love being asked tricky questions as we're confident we can answer them. Here is a list of the questions we think will help you sort the wheat from the chaff.
1. Can you provide me with some reference companies to speak to?
Everybody knows that there's no greater measure of a company's ability to deliver than through referral so always ask for a number of references. You don't have to call them all but if they can provide a few to select from it will instantly give you confidence.
Do they deliver what they promised?
Are they responsive and easy to get hold of?
Do they communicate well – letting you know what's happening with issues?
Do they stay within the projected budget?
IT support companies vary enormously but as they grow, their ability to offer more solutions or products increases, as does their knowledge base. Conversely, the "one man band" will struggle to provide a truly great service as there will always be times when they have to deal with another customer or they're ill or on holiday.
However, this doesn't mean you should necessarily choose for the biggest company you can as extremely large companies can sometimes be faceless and lack the ability to understand you and your requirements.
Just be confident the company can meet all your needs with the number of staff they have. You should weigh up the pros and cons.
3. Is your business insured and if so for how much?
Things can go wrong and you want to ensure the company you deal with is adequately insured for any eventuality. It should be at least a £1,000,000 professional indemnity package.
4. Do you offer different levels of support?
managed service support, where the support company actively manages your systems to reduce the chance of things going wrong. This usually involves signing an annual contract – see question 5.
5. Do your contracts tie me in for a specific period?
Most IT support companies provide contracts with a minimum time commitment or long cancellation periods. If they don't provide a good service you want the ability to stop using them so look for companies with a short minimum term and short cancellation period. Ideally 1 month for both.
6. What are your response times?
You should expect your support company to offer some sort of reasonable response time in the case of a problem with your IT network. This may vary depending on what level of support you are willing to pay for.
However, remember that a "guaranteed response time" is exactly that – it only guarantees that they won't ignore you and will start doing something with your problem.
Also, be wary of companies that guarantee to resolve problems within a certain period as this is invariably a hollow guarantee, because sometimes it can take much longer than expected to get to the root of problems.
7. Will we get a dedicated account manager?
It's good to have a high level single point of contact within the IT support company with who you can build a long term relationship with. They will get to know your business and your plans and so can work with you to ensure your IT accommodates these. It is reasonable to expect one or two permanent contacts that you can deal with.
8. Will I get a dedicated engineer?
It's unreasonable to expect you'll always deal with the same engineer, because most support companies allocate work to engineers best suited to each particular job. However, this is still a good question to ask – if they say 'yes', consider why this is. Do they only have a limited number of experienced staff?
9. When I phone, will I speak directly to a technician who can help?
Many IT support companies will operate call centres as the first point of contact, manned by people with limited or no technical knowledge. They will often take details of your issue and pass this onto a second tier of technicians. This may then be further escalated to a third tier. There is nothing more frustrating than being passed around so check that the company you are considering provides you with direct access to the technical people.
10. How can we log support calls with you?
When you have an IT problem, it's really frustrating when you can't get hold of the person you employ to fix it. Make sure your support company either has a dedicated number with someone always available to answer, or another facility such as an online form or email address that goes straight to the support team. Ideally you should have all three of these options.
11. Can we only call when we have a problem?
Does the IT support company you are considering only help with problems as they arise or do they offer additional support with common IT problems such as "How do I do this in Word?" Any additional assistance an IT support company can give to help your staff is clearly useful and makes the investment in the support even more viable.
12. Can I see the IT systems you use?
Would you use a dentist with bad teeth? Probably not! When using an IT support company it makes sense to see how they use technology to be efficient and provide a great service. If they don't have a CRM system, an effective email system and possibly other systems such as remote management and monitoring, then are they likely to be the best IT support company to deal with?
13. Do you supply hardware as well as support?
Even if you purchase all your computer equipment from a different supplier, it's still good to know that your support company can replace any faulty parts themselves.
And when it is time to replace your hardware, you may feel better knowing the people who are supporting it have recommended, supplied and installed it – giving you a single point of contact if something goes wrong.
14. How and when will I be charged?
IT Support company terms vary, but generally expect 15 to 30 day payment terms for one off ad-hoc work. Support contracts tend to be paid monthly, quarterly or annually and usually payable in advance.
15. What exactly do you support in your contract (and does it include peripherals such as printers, mobiles, etc)?
Most IT support companies will cover labour charges on any incident that relates to hardware or specified software and some also include mileage and travel time to and from your premises (if an on-site visit is required). However, support for custom software that was built by another company is unlikely to be included so be wary of any support company that claims "we support it all" – do they really?
Also, many support companies charge extra for devices that connect to the main computers such as printers, scanners, mobiles, etc. If, for example, you have a problem getting emails on your mobile, or even want it setting up to access your emails, will the IT support company help and is it included in the support or an extra?
16. What isn't covered in the support contract?
Although this is technically question 15 in reverse, the information it reveals can be very useful. The list could be very long but you need to know examples of things that you may be charged for. For example, if your server fails and needs replacing, will its installation be covered by the contract, or will it cost extra? Asking what the support contract includes may not necessarily bring out this detail.
17. Do you provide remote management and monitoring?
Many IT support companies still operate an on-site break-fix service and offer little in terms of remote management and monitoring, whilst others offer 24/7 monitoring and pro-actively take action (often remotely) to prevent problems before they cause major issues to your business. As they say, prevention is better than cure, and this is no truer than with IT systems.
For example, they may be able to spot that a disk drive is about to fail and then back up the data before it's too late. Remote management can also reduce the time it takes to fix things and cut any related travel costs.
18. Do they have testimonials and case studies from other satisfied customers?
There's nothing better than seeing real existing customers tell you what the support company has done for them and why they are happy with them. Just beware of testimonials that are attributed to vague people, such as Mr A. Smith, or that are completely anonymous – you need to check that these are real people and real customers.
19. Tell me about your partnership with other IT companies?
The best IT support companies will focus on maintenance and support of IT Infrastructures. They know what they are good at and stick to it.
To ensure you receive the best support they will take time to forge professional and trusted relationships with other IT companies that specialise in related fields such as cabling, telecoms, web design, etc. They aren't afraid to sub-contract or refer you to a trusted third party who can deliver the best in their field.
Most good IT support companies will also manage your relationship with your existing providers. This means you no longer need to telephone BT to spend time in a queue working out why your Broadband isn't working – your IT provider should manage that job for you.
Ask about how a "total" support service is provided and question any business that either says "That's not what we do" or "We do everything ourselves", as both extremes can be potentially risky.
20. What do you believe makes your service unique?


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redpajama_set_name: RedPajamaGithub


text:
What Is Tartrazine (E102) And Is It Vegan?
Tartrazine (E102) Chemical Structure
Tartrazine, also known as E102 in Europe, is an additive used in many foods to turn them a specific color, generally yellow. Its Vegan Status is: Definitely Vegan.
The reason Tartrazine/E102 is definitely vegan is because the only way it's produced is entirely synthetically. No Animal products go into the production whatsoever, nor do they have a single manufacturer that tested them on animals.
Is Tartrazine (E102) an Artificial Color?
Indisputably yes; it's an artificial color, and it's part of the Azo Dye group that many believe cause hyperactivity, particularly in children.
Wait, So What is Tartrazine (E102)?
As briefly touched on before, it's an artificial color which is used for dying foods yellow, or in some cases green when combined with Brilliant Blue for example. It's made out of various chemicals and is entirely synthetic. It generally comes in either a powder or liquid form that's added to foods, but also some soaps and medications.
Is Tartrazine (E102) Safe to consume?
It's been banned by multiple governments across the world and can cause mild allergic reactions in a microscopic portion of people, and it's not a health food or anything like that.
As for the Hyperactivity claim the evidence for it is questionable at best, as most of the studies that found it were self-reported by the child's parents, which has been proven to be completely inaccurate in other studies.
The bottom line is a little in your diet likely won't harm you, nor the animals, as E102 (Also known as Tartrazine) is 100% Vegan!


text:
Q: Identity with binomial coefficients and k^k In process of doing some computations on Hilbert schemes, I stumbled across the following identity, for $k \ge 2$:
$$
k^{k-3} = \frac{1}{2} \sum_{i=1}^{k-1} \binom{k-2}{i-1} i^{i-2} (k-i)^{k-i-2}
$$
which can be verified computationally for small values of $k$ (I did it for the first 500).
This is actually a special case of a bigger problem, so I'm hoping to get ideas about what methods might be applicable to prove this kind of identity.
I understand that there are good methods for automatically proving hypergeometric identities, but am I correct in thinking that this is not hypergeometric, due to the $k^{k-3}$ type terms?
EDIT
Here is a link to the bigger question that I had that this is a special case of: Counting some binary trees with lots of extra stucture

A: Let's denote $T(x)=\sum_{n\geq 1}\frac{n^{n-1}x^n}{n!}$ the exponential generating function of labeled rooted trees, and $U(x)=\sum_{n\geq 1}\frac{n^{n-2}x^n}{n!}$ the corresponding function for unrooted trees. We will use the fact that $xU'=T$ and $xe^T=T$, both of which can be derived algebraically or combinatorially. Your identity follows from the relation
$$2\left(T(x)-U(x)\right)=T^2$$
since the coefficient of $x^{k}$ on the left is $\frac{2k^{k-3}}{(k-2)!}$ and the corresponding coefficient on the right is $\sum_{i=1}^{k-1}\frac{i^{i-1}}{i!}\cdot\frac{(k-i)^{k-i-1}}{(k-i)!}$.
To get a quick proof of this notice that both sides only have terms of order $x^2$ or higher. Therefore it is enough to prove the derivative of this relation
$$2T'-2U'=2TT'$$
$$\iff xT'-xU'=xTT'\iff xT'(1-T)=T\iff T'(1-T)=e^T$$
$$\iff 1=\frac{d}{dx}\left(\frac{T}{e^T}\right).$$

I hope it's clear from above that in principle all such identities follow from the basic relation $xe^T=T$ and all its derivatives. I know it's usually common to give references to books or surveys, but I really like the video lectures by Zagier available here, under the title "Partitions, Modular Forms, and Moduli Spaces" (these are based on a forthcoming paper by Dubrovin, Zagier and Yang). In a series of talks he presents how quasimodular forms appear in enumerative geometry. However in the first couple of lectures he focuses on what he calls "Lambert space" (nonstandard terminology from what I can tell, coming from the closely related Lambert function) which is essentially the space of series that are linear combinations of $T$ and its derivatives. He shows, for example, how generating functions of maps or Hurwitz numbers belong to this Lambert space.

A: In my (first research) paper, A note on evaluation of Abel's sums, (that appeared second, in 1979) I described another way to evaluate Abel's sum of the form $$A_n(x,y;p,q)=\sum_{k=0}^n {{n} \choose {k}}(k+x)^{k+p}(n-k+y)^{n-k+q},$$ via (simpler) auxiliary sums defined by $$F_n(x,y;p,q)=\sum_{k=0}^n {{n} \choose {k}}(-1)^k (k+x)^{p}(n-k+y)^{q}.$$ The $F_n$ functions have simple evaluations in terms of difference operators. I was inspired by Riordan's book Combinatorial Identities and later corresponded with Riordan, and met him in 78.

A: Here are two proofs of your identity, which I have hinted at in my comments.
The identity
Let me first restate your identity with the letter $k$ renamed as $n$:

Theorem 1. Let $n\geq2$ be an integer. Then,
  $$
n^{n-3}=\dfrac{1}{2}\sum\limits_{i=1}^{n-1}\dbinom{n-2}{i-1}i^{i-2}\left(
n-i\right)  ^{n-i-2}.
$$

First proof: the Abel identity and its friends
The first proof of Theorem 1 will actually show a more general identity:

Theorem 2. Let $n$ be a nonnegative integer. In the polynomial ring
  $\mathbb{Z}\left[  x,y,z\right]  $, we have
  $$
\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(
n-k\right)  z\right)  ^{n-k-1}=\left(  x+y\right)  \left(  x+y-nz\right)
^{n-1}.
$$

Note that both sides of the equality in Theorem 2 are well-defined polynomials
in $\mathbb{Z}\left[  x,y,z\right]  $ (not just rational functions in
$\mathbb{Q}\left(  x,y,z\right)  $): Indeed,

*

*the only denominators that appear on the left hand side arise from the
$\left(  x-kz\right)  ^{k-1}$ term for $k=0$ and from the $\left(  y-\left(
n-k\right)  z\right)  ^{n-k-1}$ term for $k=n$; but both of them are cancelled
by the $xy$ factor appearing in each addend.

*the only denominator that appears on the right hand side arises from the
$\left(  x+y-nz\right)  ^{n-1}$ term for $n=0$; but this denominator is
cancelled by the $x+y$ factor.
Theorem 2 is a well-known fact in umbral calculus: It says that the sequence
of the Abel polynomials (the sequence $\left(  p_{0},p_{1},p_{2}
,\ldots\right)  $ whose $n$-th term is $p_{n}=x\left(  x-nz\right)  ^{n-1}$,
where $x$ is considered an indeterminate and $z$ a constant) is of binomial
type (i.e., it satisfies $p_{n}\left(  x+y\right)  =\sum\limits_{k=0}^{n}\dbinom
{n}{k}p_{k}\left(  x\right)  p_{n-k}\left(  y\right)  $). As such, it is
probably proven all across the literature. I also think that Theorem 2 is what
the equation (2) in Victor J. W. Guo, Bijective proofs of Gould-Mohanty's
and Raney-Mohanty's identities,
arXiv:1005.4258v1 is supposed to be (I say
"supposed" because equation (2) as it appears in Guo's preprint is false).
Let me derive Theorem 2 from an even better known identity, namely the Abel
identity (appearing as equation (1) in the above-cited paper by Victor J. W.
Guo):

Theorem 3.  Let $n$ be a nonnegative integer. In the polynomial ring
  $\mathbb{Z}\left[  x,y,z\right]  $, we have
  $$
\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(  x-kz\right)  ^{k-1}\left(  y+kz\right)
^{n-k}=\left(  x+y\right)  ^{n}.
$$

As before, both sides of the equality in Theorem 3 are well-defined
polynomials in $\mathbb{Z}\left[  x,y,z\right]  $.
The most well-known particular case of Theorem 3 is probably the one obtained
by substituting $-1$ for $z$:

Theorem 4. Let $n$ be a nonnegative integer. In the polynomial ring
  $\mathbb{Z}\left[  x,y\right]  $, we have
  $$
\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(  x+k\right)  ^{k-1}\left(  y-k\right)
^{n-k}=\left(  x+y\right)  ^{n}.
$$

Theorem 4 appears in many places, e.g. (with a boring proof) in my solution
to Problem 4 on the 6th QEDMO 2009 (scroll down to Theorem 4). Yes, I am that lazy at
reference hunting. $\square$
I am going to reverse the cart and derive Theorem 1 back from Theorem 4. The
first step is getting Theorem 3:
Proof of Theorem 3. Work in the field $\mathbb{Q}\left(  x,y,z\right)  $ of
rational functions in $x,y,z$. Substitute $\dfrac{x}{z}$ and $\dfrac{y}{z}$
for $x$ and $y$ in Theorem 4. The result is
$$
\sum\limits_{k=0}^{n}\dbinom{n}{k}\dfrac{x}{z}\left(  \dfrac{x}{z}-k\right)
^{k-1}\left(  \dfrac{y}{z}+k\right)  ^{n-k}=\left(  \dfrac{x}{z}+\dfrac{y}
{z}\right)  ^{n}.
$$
Multiplying this equality by $z^{n}$ and clearing denominators results in
$$
\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(  x-kz\right)  ^{k-1}\left(  y+kz\right)
^{n-k}=\left(  x+y\right)  ^{n}.
$$
Thus, Theorem 3 is proven. $\square$
Next, I shall derive Theorem 2 from Theorem 3:
Proof of Theorem 2. Substituting $y-nz$ for $y$ in Theorem 3 yields
$$
\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(  x-kz\right)  ^{k-1}\left(  y-nz+kz\right)
^{n-k}=\left(  x+y-nz\right)  ^{n}.
$$
Thus,
\begin{align}
\left(  x+y-nz\right)  ^{n}  & =\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(
x-kz\right)  ^{k-1}\left(  \underbrace{y-nz+kz}_{=y-\left(  n-k\right)
z}\right)  ^{n-k}\nonumber\\
\tag{1}
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(  x-kz\right)  ^{k-1}\left(  y-\left(
n-k\right)  z\right)  ^{n-k}.
\label{pf.t2.1}
\end{align}
Multiplying this equality by $y$, we find
\begin{align}
y\left(  x+y-nz\right)  ^{n}  & =y\sum\limits_{k=0}^{n}\dbinom{n}{k}x\left(
x-kz\right)  ^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k}\nonumber\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  ^{k-1}\underbrace{\left(
y-\left(  n-k\right)  z\right)  ^{n-k}}_{=\left(  y-\left(  n-k\right)
z\right)  \left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}}\nonumber\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(
n-k\right)  z\right)  \left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}
\nonumber\\
\tag{2}
& =\sum\limits_{k=0}^{n}\left(  y-\left(  n-k\right)  z\right)  \dbinom{n}
{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(  n-k\right)  z\right)
^{n-k-1}.
\label{pf.t2.2}
\end{align}
On the other hand, substituting $y$ and $x$ for $x$ and $y$ in \eqref{pf.t2.1},
we obtain
\begin{align*}
\left(  y+x-nz\right)  ^{n}  & =\sum\limits_{k=0}^{n}\dbinom{n}{k}y\left(
y-kz\right)  ^{k-1}\left(  x-\left(  n-k\right)  z\right)  ^{n-k}\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{n-k}y\left(  y-\left(  n-k\right)  z\right)
^{n-k-1}\left(  x-\left(  n-\left(  n-k\right)  \right)  z\right)  ^{n-\left(
n-k\right)  }
\end{align*}
(here, we have substituted $n-k$ for $k$ in the sum). Since $y+x=x+y$, this
rewrites as
\begin{align*}
& \left(  x+y-nz\right)  ^{n}\\
& =\sum\limits_{k=0}^{n}\underbrace{\dbinom{n}{n-k}}_{=\dbinom{n}{k}}y\left(
y-\left(  n-k\right)  z\right)  ^{n-k-1}\underbrace{\left(  x-\left(
n-\left(  n-k\right)  \right)  z\right)  ^{n-\left(  n-k\right)  }
}_{\substack{=\left(  x-kz\right)  ^{k}\\=\left(  x-kz\right)  \left(
x-kz\right)  ^{k-1}}}\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}y\left(  y-\left(  n-k\right)  z\right)
^{n-k-1}\left(  x-kz\right)  \left(  x-kz\right)  ^{k-1}\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}y\left(  x-kz\right)  \left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}.
\end{align*}
Multiplying this equality by $x$, we obtain
\begin{align*}
x\left(  x+y-nz\right)  ^{n}  & =x\sum\limits_{k=0}^{n}\dbinom{n}{k}y\left(
x-kz\right)  \left(  x-kz\right)  ^{k-1}\left(  y-\left(  n-k\right)
z\right)  ^{n-k-1}\\
& =\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  \left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}\\
& =\sum\limits_{k=0}^{n}\left(  x-kz\right)  \dbinom{n}{k}xy\left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}.
\end{align*}
Adding this equality to \eqref{pf.t2.2}, we find
\begin{align*}
& y\left(  x+y-nz\right)  ^{n}+x\left(  x+y-nz\right)  ^{n}\\
& =\sum\limits_{k=0}^{n}\left(  y-\left(  n-k\right)  z\right)  \dbinom{n}
{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(  n-k\right)  z\right)
^{n-k-1}\\
& \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k=0}^{n}\left(  x-kz\right)  \dbinom{n}
{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(  n-k\right)  z\right)
^{n-k-1}\\
& =\sum\limits_{k=0}^{n}\underbrace{\left(  \left(  y-\left(  n-k\right)  z\right)
+\left(  x-kz\right)  \right)  }_{=x+y-nz}\dbinom{n}{k}xy\left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}\\
& =\sum\limits_{k=0}^{n}\left(  x+y-nz\right)  \dbinom{n}{k}xy\left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}\\
& =\left(  x+y-nz\right)  \sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}.
\end{align*}
Thus,
\begin{align*}
& \left(  x+y-nz\right)  \sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)
^{k-1}\left(  y-\left(  n-k\right)  z\right)  ^{n-k-1}\\
& =y\left(  x+y-nz\right)  ^{n}+x\left(  x+y-nz\right)  ^{n}=\left(
y+x\right)  \left(  x+y-nz\right)  ^{n}\\
& =\left(  x+y\right)  \left(  x+y-nz\right)  ^{n}.
\end{align*}
Dividing this equality by $x+y-nz$, we obtain
$$
\sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left(  x-kz\right)  ^{k-1}\left(  y-\left(
n-k\right)  z\right)  ^{n-k-1}=\left(  x+y\right)  \left(  x+y-nz\right)
^{n-1}.
$$
This proves Theorem 2. $\square$
Finally, we can get Theorem 1:
First proof of Theorem 1. Theorem 2 (applied to $n-2$ instead of $n$) shows
that in the polynomial ring $\mathbb{Z}\left[  x,y,z\right]  $, we have
\begin{align*}
& \sum\limits_{k=0}^{n-2}\dbinom{n-2}{k}xy\left(  x-kz\right)  ^{k-1}\left(
y-\left(  \left(  n-2\right)  -k\right)  z\right)  ^{\left(  n-2\right)
-k-1}\\
& =\left(  x+y\right)  \left(  x+y-\left(  n-2\right)  z\right)  ^{n-3}.
\end{align*}
Substituting $1$, $1$ and $-1$ for $x$, $y$ and $z$ in this equality (and
making the obvious simplifications), we find
\begin{align*}
& \sum\limits_{k=0}^{n-2}\dbinom{n-2}{k}\left(  1+k\right)  ^{k-1}\left(  1+\left(
\left(  n-2\right)  -k\right)  \right)  ^{\left(  n-2\right)  -k-1}\\
& =2\left(  2+\left(  n-2\right)  \right)  ^{n-3}=2n^{n-3}.
\end{align*}
Hence,
\begin{align*}
2n^{n-3}  & =\sum\limits_{k=0}^{n-2}\dbinom{n-2}{k}\left(  1+k\right)  ^{k-1}\left(
1+\left(  \left(  n-2\right)  -k\right)  \right)  ^{\left(  n-2\right)
-k-1}\\
& =\sum\limits_{i=1}^{n-1}\dbinom{n-2}{i-1}\left(  1+\left(  i-1\right)  \right)
^{\left(  i-1\right)  -1}\left(  1+\left(  \left(  n-2\right)  -\left(
i-1\right)  \right)  \right)  ^{\left(  n-2\right)  -\left(  i-1\right)  -1}
\end{align*}
(here, we have substituted $i-1$ for $k$ in the sum). After some trivial
simplifications, this becomes
$$
2n^{n-3}=\sum\limits_{i=1}^{n-1}\dbinom{n-2}{i-1}i^{i-2}\left(  n-i\right)  ^{n-i-2}.
$$
Dividing this equality by $2$, we obtain the claim of Theorem 1. $\square$
Second proof: Cayley's formula and the symmetry of trees
Let me say right away that the above first proof of Theorem 1 is my favorite,
since it shows various rather general combinatorial identities in cascade. In
contrast, the following second proof is cute and memorable.
A tree shall here mean a simple undirected graph (i.e., a pair $\left(
V,E\right)  $ in which $V$ is a finite set and $E$ is a set of $2$-element
subsets of $V$) that is connected and has no cycles. Of course, various other
equivalent characterizations of trees exist. If $n\in\mathbb{N}$, then an
$n$-tree means a tree whose vertex set is $\left\{  1,2,\ldots,n\right\}  $.
Notice that there exist no $0$-trees, since a graph with no vertices is not
connected. If $G=\left(  V,E\right)  $ is a simple undirected graph, then the
edge set $E$ is denoted by $\operatorname*{E}\left(  G\right)  $.
The following theorem is one of the famous results of 19th century
combinatorics -- the Cayley
formula (actually
discovered by Borchardt in 1860):

Theorem 5. Let $n$ be a positive integer. Then, the number of all
  $n$-trees is $n^{n-2}$.

The similarity of the $n^{n-2}$ term here with the powers in Theorem 1
suggests a relation, and such indeed exists. First, let me derive a corollary
of Theorem 5:

Corollary 6. Let $n\geq2$ be an integer. Let $f$ be a $2$-element subset
  of $\left\{  1,2,\ldots,n\right\}  $. Then, the number of all $n$-trees $T$
  such that $f\in\operatorname*{E}\left(  T\right)  $ is $2n^{n-3}$.

Proof of Corollary 6. It is well-known that the number of edges in a tree is
always its number of vertices minus $1$. Thus, a tree with $n$ vertices must
have exactly $n-1$ edges. In other words, any $n$-tree must have exactly $n-1$ edges.
Let $F$ be the set of all $2$-element subsets of $\left\{  1,2,\ldots
,n\right\}  $. Then, for each $n$-tree $T$, we have $\operatorname*{E}\left(
T\right)  \subseteq F$ (that is, the edges of $T$ are elements of $F$). Also,
$f$ is a $2$-element subset of $\left\{  1,2,\ldots,n\right\}  $ and thus
belongs to $F$. Also, clearly, $\left\vert F\right\vert =\dbinom{n}{2}
=\dfrac{n\left(  n-1\right)  }{2}$.
Now, we claim that any $e\in F$ satisfies
\begin{align}
\left(  \text{the number of all }n\text{-trees }T\text{ such that }\left\{
1,2\right\}  \in\operatorname*{E}\left(  T\right)  \right)  \nonumber\\
\tag{3}
=\left(  \text{the number of all }n\text{-trees }T\text{ such that }
e\in\operatorname*{E}\left(  T\right)  \right)  .
\label{pf.c6.1}
\end{align}
[Proof of \eqref{pf.c6.1}: Let $e\in F$. Thus, $e$ is a $2$-element subset
of $\left\{  1,2,\ldots,n\right\}  $. In other words, $e=\left\{  x,y\right\}
$ for two distinct elements $x$ and $y$ of $\left\{  1,2,\ldots,n\right\}  $.
Consider these $x$ and $y$.
The elements $x$ and $y$ of $\left\{  1,2,\ldots,n\right\}  $ are distinct.
Thus, there exists some permutation $\sigma$ of $\left\{  1,2,\ldots
,n\right\}  $ such that $\sigma\left(  1\right)  =x$ and $\sigma\left(
2\right)  =y$. (For example, one can find such a permutation $\sigma$ by
setting its first two values $\sigma\left(  1\right)  $ and $\sigma\left(
2\right)  $ to be $x$ and $y$, and setting all its remaining $n-2$ values to
be the remaining $n-2$ elements of $\left\{  1,2,\ldots,n\right\}  $ in any
order.) Consider such a $\sigma$.
Recall that any permutation $\omega$ of $\left\{  1,2,\ldots,n\right\}  $ also
induces a permutation of the set $F$ of all $2$-element subsets of $\left\{
1,2,\ldots,n\right\}  $ (namely, it sends each $g\in F$ to $\omega\left( g \right)  =\left\{  \omega\left(  u\right)  \ \mid\ u\in g \right\}  $). In
particular, $\sigma$ induces a permutation of the set $F$. This permutation
satisfies $\sigma\left(  \left\{  1,2\right\}  \right)  =\left\{
\underbrace{\sigma\left(  1\right)  }_{=x},\underbrace{\sigma\left(  2\right)
}_{=y}\right\}  =\left\{  x,y\right\}  =e$.
If $T$ is an $n$-tree, and if $\omega$ is a permutation of $\left\{
1,2,\ldots,n\right\}  $, then $\omega\left(  T\right)  $ shall denote the
$n$-tree whose edges are the images of the edges of $T$ under $\omega$ (that
is, $\operatorname*{E}\left(  \omega\left(  T\right)  \right)  =\left\{
\omega\left(  e\right)  \ \mid\ e\in\operatorname*{E}\left(  T\right)
\right\}  $). It is easy to see that this $\omega\left(  T\right)  $ is indeed
an $n$-tree (indeed, it is just the $n$-tree $T$ with its vertices relabelled
by the map $\omega$). Hence, the map
$$
\left\{  n\text{-trees}\right\}  \rightarrow\left\{  n\text{-trees}\right\}
,\ T\mapsto\sigma\left(  T\right)
$$
is well-defined. Moreover, this map is a bijection (indeed, the map $\left\{
n\text{-trees}\right\}  \rightarrow\left\{  n\text{-trees}\right\}
,\ T\mapsto\sigma^{-1}\left(  T\right)  $ is easily seen to be inverse to it).
This bijection restricts to a bijection
\begin{align*}
\left\{  n\text{-trees }T\text{ such that }\left\{  1,2\right\}
\in\operatorname*{E}\left(  T\right)  \right\}    & \rightarrow\left\{
n\text{-trees }T\text{ such that }e\in\operatorname*{E}\left(  T\right)
\right\}  ,\\
T  & \mapsto\sigma\left(  T\right)
\end{align*}
(since $\sigma\left(  \left\{  1,2\right\}  \right)  =e$). Hence, the sets
$\left\{  n\text{-trees }T\text{ such that }\left\{  1,2\right\}
\in\operatorname*{E}\left(  T\right)  \right\}  $ and $\left\{  n\text{-trees
}T\text{ such that }e\in\operatorname*{E}\left(  T\right)  \right\}  $ are in
bijection. Thus, these two sets have the same size. In other words,
\eqref{pf.c6.1} holds.]
Now,
\begin{align*}
& \left(  \text{the number of all pairs }\left(  T,e\right)  \text{ where
}T\text{ is an }n\text{-tree and }e\in\operatorname*{E}\left(  T\right)
\right)  \\
& =\sum\limits_{T\text{ is an }n\text{-tree}}\underbrace{\left(  \text{the number of
all }e\in\operatorname*{E}\left(  T\right)  \right)  }_{\substack{=\left\vert
\operatorname*{E}\left(  T\right)  \right\vert =\left(  \text{the number of
all edges of }T\right)  =n-1\\\text{(since any }n\text{-tree must have exactly
}n-1\text{ edges)}}}\\
& =\sum\limits_{T\text{ is an }n\text{-tree}}\left(  n-1\right)  =\left(  n-1\right)
\cdot\underbrace{\left(  \text{the number of all }n\text{-trees}\right)
}_{\substack{=n^{n-2}\\\text{(by Theorem 5)}}}\\
& =\left(  n-1\right)  \cdot n^{n-2}.
\end{align*}
Hence,
\begin{align*}
& \left(  n-1\right)  \cdot n^{n-2}\\
& =\left(  \text{the number of all pairs }\left(  T,e\right)  \text{ where
}T\text{ is an }n\text{-tree and }e\in\operatorname*{E}\left(  T\right)
\right)  \\
& =\sum\limits_{e\in F}\underbrace{\left(  \text{the number of all }n\text{-trees
}T\text{ such that }e\in\operatorname*{E}\left(  T\right)  \right)
}_{\substack{=\left(  \text{the number of all }n\text{-trees }T\text{ such
that }\left\{  1,2\right\}  \in\operatorname*{E}\left(  T\right)  \right)
\\\text{(by \eqref{pf.c6.1})}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{E}\left(  T\right)
\subseteq F\text{ for each }n\text{-tree }T\right)  \\
& =\sum\limits_{e\in F}\left(  \text{the number of all }n\text{-trees }T\text{ such
that }\left\{  1,2\right\}  \in\operatorname*{E}\left(  T\right)  \right)  \\
& =\underbrace{\left\vert F\right\vert }_{=\dfrac{n\left(  n-1\right)  }{2}
} \cdot \underbrace{\left(  \text{the number
of all }n\text{-trees }T\text{ such that }\left\{  1,2\right\}  \in
\operatorname*{E}\left(  T\right)  \right)  }_{\substack{=\left(  \text{the
number of all }n\text{-trees }T\text{ such that }f\in\operatorname*{E}\left(
T\right)  \right)  \\\text{(by \eqref{pf.c6.1}, applied to }e=f\text{)}}}\\
& =\dfrac{n\left(  n-1\right)  }{2}\cdot\left(  \text{the number of all
}n\text{-trees }T\text{ such that }f\in\operatorname*{E}\left(  T\right)
\right)  .
\end{align*}
Solving this equality for $\left(  \text{the number of all }n\text{-trees
}T\text{ such that }f\in\operatorname*{E}\left(  T\right)  \right)  $, we
obtain
\begin{align*}
\left(  \text{the number of all }n\text{-trees }T\text{ such that }
f\in\operatorname*{E}\left(  T\right)  \right)    & =\left(  n-1\right)  \cdot
n^{n-2}/\dfrac{n\left(  n-1\right)  }{2}\\
& =2n^{n-3}.
\end{align*}
This proves Corollary 6. $\square$
Here is another trivial corollary of Theorem 5:

Corollary 7. Let $S$ be a finite nonempty set. Then, the number of all
  trees with vertex set $S$ is $\left\vert S\right\vert ^{\left\vert
S\right\vert -2}$.

Proof of Corollary 7. Let $n=\left\vert S\right\vert $. Then, there exists a
bijection $\left\{  1,2,\ldots,n\right\}  \rightarrow S$. We can use this
bijection to transform any $n$-tree into a tree with vertex set $S$ (by
relabelling its vertices) and vice versa. Hence, the number of all trees with
vertex set $S$ equals the number of all $n$-trees. But the latter number is
$n^{n-2}$ (by Theorem 5). Hence, the former number is also $n^{n-2}$. In other
words, the number of all trees with vertex set $S$ is $n^{n-2}=\left\vert
S\right\vert ^{\left\vert S\right\vert -2}$ (since $n=\left\vert S\right\vert
$). Corollary 7 is proven. $\square$
Now, we can finally reprove Theorem 1:
Second proof of Theorem 1. Corollary 6 (applied to $f=\left\{  1,2\right\}
$) shows that the number of all $n$-trees $T$ satisfying $\left\{
1,2\right\}  \in\operatorname*{E}\left(  T\right)  $ is $2n^{n-3}$.
Now, let me show a different way to compute this number.
Indeed, if we remove the edge $\left\{  1,2\right\}  $ from an $n$-tree $T$
satisfying $\left\{  1,2\right\}  \in\operatorname*{E}\left(  T\right)  $,
then the remaining graph splits into two connected components, one of which
contains the vertex $1$ while the other contains the vertex $2$. Both
connected components are trees; we denote them by $A_{T}$ and $B_{T}$,
respectively (so that $1$ is a vertex of $A_{T}$ and $2$ is a vertex of
$B_{T}$). Let $\mathcal{A}_{T}$ be the vertex set of the tree $A_{T}$, and let
$\mathcal{B}_{T}$ be the vertex set of the tree $B_{T}$. Then, $\mathcal{A}
_{T}$ and $\mathcal{B}_{T}$ are two disjoint subsets of $\left\{
1,2,\ldots,n\right\}  $ whose union is $\left\{  1,2,\ldots,n\right\}  $;
thus, $\mathcal{B}_{T}=\left\{  1,2,\ldots,n\right\}  \setminus\mathcal{A}
_{T}$. Note that $1\in\mathcal{A}_{T}$ (since $1$ is a vertex of $A_{T}$) and
$2\in\mathcal{B}_{T}$ (since $2$ is a vertex of $B_{T}$), thus $2\in
\mathcal{B}_{T}=\left\{  1,2,\ldots,n\right\}  \setminus\mathcal{A}_{T}$, thus
$2\notin\mathcal{A}_{T}$.
Obviously, an $n$-tree $T$ satisfying $\left\{  1,2\right\}  \in
\operatorname*{E}\left(  T\right)  $ is uniquely determined by the two trees
$A_{T}$ and $B_{T}$. Thus, any $n$-tree $T$ satisfying $\left\{  1,2\right\}
\in\operatorname*{E}\left(  T\right)  $ can be constructed by the following procedure:

*

*First, we decide how many elements the set $\mathcal{A}_{T}$ will have
(i.e., how many vertices $A_{T}$ will have). The number of elements of
$\mathcal{A}_{T}$ has to be an integer from $1$ to $n-1$ (in fact, it cannot
be $0$ (since we must have $1\in\mathcal{A}_{T}$); but it cannot be $n$ either
(since we must have $2\notin\mathcal{A}_{T}$)). We choose one such integer,
and denote it by $i$. We thus want the set $\mathcal{A}_{T}$ to have $i$ elements.

*Then, we choose the set $\mathcal{A}_{T}$ itself (i.e., we decide which
vertices $A_{T}$ will have). We already know that this set will have $i$
elements, and we know that $1$ will be one of them (since $1\in\mathcal{A}
_{T}$) but $2$ will be not (since $2\notin\mathcal{A}_{T}$); thus, we are left
to choose the remaining $i-1$ elements of $\mathcal{A}_{T}$ freely among the
$n-2$ numbers $3,4,\ldots,n$. There are $\dbinom{n-2}{i-1}$ ways to make this choice.

*Having chosen the set $\mathcal{A}_{T}$, we immediately know the set
$\mathcal{B}_{T}$: namely, $\mathcal{B}_{T}=\left\{  1,2,\ldots,n\right\}
\setminus\mathcal{A}_{T}$ (there are no choices to make here). Notice that
$\left\vert \mathcal{A}_{T}\right\vert =i$ (since the set $\mathcal{A}_{T}$
has $i$ elements), and from $\mathcal{B}_{T}=\left\{  1,2,\ldots,n\right\}
\setminus\mathcal{A}_{T}$ we obtain $\left\vert \mathcal{B}_{T}\right\vert
=\left\vert \left\{  1,2,\ldots,n\right\}  \setminus\mathcal{A}_{T}\right\vert
=\underbrace{\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert }
_{=n}-\underbrace{\left\vert \mathcal{A}_{T}\right\vert }_{=i}=n-i$. Note that
both sets $\mathcal{A}_{T}$ and $\mathcal{B}_{T}$ are nonempty (since
$1\in\mathcal{A}_{T}$ but $2\in\mathcal{B}_{T}$ (since $2\notin\mathcal{A}
_{T}$)).

*Now, we choose the tree $A_{T}$. This tree has to be a tree with vertex set
$\mathcal{A}_{T}$; the number of all such trees is $\left\vert \mathcal{A}
_{T}\right\vert ^{\left\vert \mathcal{A}_{T}\right\vert -2}$ (by Corollary 7,
applied to $S=\mathcal{A}_{T}$). In other words, the number of all such trees
is $i^{i-2}$ (since $\left\vert \mathcal{A}_{T}\right\vert =i$). Thus, there
are $i^{i-2}$ ways to choose $A_{T}$.

*Now, we choose the tree $B_{T}$. This tree has to be a tree with vertex set
$\mathcal{B}_{T}$; the number of all such trees is $\left\vert \mathcal{B}
_{T}\right\vert ^{\left\vert \mathcal{B}_{T}\right\vert -2}$ (by Corollary 7,
applied to $S=\mathcal{B}_{T}$). In other words, the number of all such trees
is $\left(  n-i\right)  ^{n-i-2}$ (since $\left\vert \mathcal{B}
_{T}\right\vert =n-i$). Thus, there are $\left(  n-i\right)  ^{n-i-2}$ ways to
choose $B_{T}$.

*Finally, we assemble the $n$-tree $T$ from $A_{T}$ and $B_{T}$ by adding
the edge $\left\{  1,2\right\}  $.
This procedure clearly yields any $n$-tree $T$ satisfying $\left\{
1,2\right\}  \in\operatorname*{E}\left(  T\right)  $ in a unique way. Thus,
the number of all $n$-trees $T$ satisfying $\left\{  1,2\right\}
\in\operatorname*{E}\left(  T\right)  $ is $\sum\limits_{i=1}^{n-1}\dbinom{n-2}
{i-1}i^{i-2}\left(  n-i\right)  ^{n-i-2}$ (because this is how many ways we
have to make choices in the above procedure). Comparing this with the fact
that this number is $2n^{n-3}$ (as shown above), we conclude that
$$
2n^{n-3}=\sum\limits_{i=1}^{n-1}\dbinom{n-2}{i-1}i^{i-2}\left(  n-i\right)  ^{n-i-2}.
$$
Dividing this equality by $2$, we obtain the claim of Theorem 1. $\square$

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