180 and 360 degree cameras continue to increase in popularity, with a number of manufacturers releasing new product in the past year. Sentry360, best known for their 360º line, has now announced two models of 180º cameras to compliment them. In this note, we overview these cameras, Sentry360's novel approach to 180º imaging, and compare to options such as Arecont's SurroundVision, low-cost panoramics, and wide angle lenses.
Instead of using multiple imagers or simply wide-angle lenses, as many cameras do, the FullSight180 series uses a single imager with panoramic lens, dewarping the view in the camera, and cropping to a custom resolution.
Maximum resolution is 2560x1800 (4.6MP) in the FS-IP4180 and 3840x2100 (8.2MP) in the 8180. Note that these resolutions are non-standard, not 4:3 or 16:9, and may not fit standard client viewing panes. We discuss this topic and associated challenges in our aspect ratio overview.
No client-side dewarping is required, as the video is transmitted dewarped from the camera.
H.264 and MJPEG streaming are supported, with maximum frame rates of 15 and 6 in the 4180 and 8180, respectively.
VMS integration via ONVIF is supported.
Aside from this custom dewarping, FullSight 180 models share most features with the FullSight 360 series. Users may see our overview of the Sentry360 line for more information.
The FS-IP4180 and FS-IP8180 are projected to be available in Q3 2012, with MSRP pricing of $1,295 and $2,395, respectively. Based on approximate dealer discounts, we expect street pricing of about $1,100 and $2,000 for these models.
Wide-angle lenses: Super wide angle lenses such as Theia's SY125 have become more popular options for covering wide fields of view, but generally top out at about 135º, as opposed to a true 180º FOV. Compared to using a Theia lens ($440 MSRP, $400+ online) with a 5MP camera such as the Arecont AV5115DN (~$550 online), Sentry360's FS-IP4180 is modestly more expensive, around $1,100 vs. $950-1,100. However, the FullSight 180 camera provides a full 180º FOV, and may be mounted flat to the wall, requiring no other hardware.
360º panoramic cameras: A number of lower-cost camera manufacturers, such as ACTi and Vivotek, have released 360º panoramic cameras in the past year. These cameras are generally priced below the FullSight 180, in the range of $700 or less online. Most of these cameras may be mounted flat to walls, similar to the FullSight series, and used in a 180º panorama mode. While some of these panoramic models require custom integration in the VMS client, many offer the ability to dewarp a 180º FOV in the camera, effectively the same as the FullSight 180 series.
Multi-imager 180º cameras: 180º cameras using multiple imagers have gained in popularity over the past 1-2 years, with Arecont's SurroundView line being especially popular. Arecont's 8MP SurroundView 180 sells for $1,300-1,400 online, about $600 less than the FS-IP8180. Arecont has a horizontal resolution advantage vs. the 8180 (6400 vs. 3840), but Sentry360 offers more vertical resolution (2100 vs. 1200). Users should keep this in mind, depending on what field of view is required. Lastly, some operators may find tracking subjects much simpler using the single-imager Sentry360 camera, instead of tracking across four separate imagers in the Arecont camera.
Ultimately, the FullSight 180 series takes some steps to simplify integration of wide angle cameras to VMSs. However, they come at a price premium above other options. Whether they are best suited for a user's site or not depends on the field of view required, VMS being used, and operator preferences, as discussed above, factors which should be carefully considered.


Hi, Cathy! To get to the Polynesian Resort from All Star Movies, there's the easy way and then there's the free way. Me? I'm pretty cheap, so let's start with the free option.
On Sun., Jan. 9, the Magic Kingdom opens at 9:00 a.m. That means that bus transportation from All Star Movies should start around 8:00 a.m. that morning. Take a bus to the Magic Kingdom. Then, instead of walking toward the park, keep walking up the ramp to the resort monorail and take it to the Poly.
If you are short on time, don't feel like walking and/or have $10-$15, take a cab straight to the Poly from your resort. You should try to depart All Star Movies by 8:30 a.m. to make your reservation. Bon appétit!


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washingtonpost.com > Education > Virginia Education
More Loudoun Schools Using 'Flex Periods'
By Ian Shapira
Washington Post Staff Writer
They sound like workout sessions at a gym, but "flex periods" are fast becoming a scheduling strategy among Northern Virginia high schools that want to offer students remediation or enrichment during the school day rather than before or after classes.
High schools in Loudoun, Fairfax and Prince William counties have been inserting these chunks of time -- from 40 to 90 minutes, depending on the school -- for several years, often to reduce after-school tutoring costs but also to raise achievement in the era of the federal No Child Left Behind law.
The program varies among schools, but the premise is similar: Between regular courses, students are assigned to a flex classroom to review material or work independently. Flex time can also be used for attending schoolwide events. And if a student needs help from a teacher in another part of the building, he or she can get a pass and visit the teacher during flex time.
Students at Stonewall Jackson High School near Manassas are in the second year of a flex program. Reactions have been mixed: Some students interviewed said the periods help them catch up on homework or review tough lessons with teachers, but others said the time is often exploited by students prone to goofing off or leaving school property. Schools are trying to crack down on the latter.
Annandale High School has used flex periods since the late 1990s. Principal John Ponton said the program is crucial for low-income students who hold after-school jobs and need extra help with schoolwork. This year, Annandale instituted guidelines requiring students and teachers to make arrangements via e-mail, 24 hours in advance, if a student wants to "flex out" of his or her assigned flex classroom to see a teacher for help.
"Some kids were just roaming around the building before, and they never got to the intended class," Ponton said. "There was too much traffic in the hallway."
Stonewall Principal Richard Nichols said the success of the program might help reduce budget costs.
"We offer tutoring two days a week after school in four core subject areas, for around $20,000 a year," Nichols said. "If we have to cut back next year on tutoring, we will at least have this program built in to support the kids."
Mel Riddile, an associate director at the National Association of Secondary School Principals, said he has noticed flex periods surfacing in schools nationwide in the past five to six years. Such periods have largely replaced study halls, which, he said, might have become largely extinct in the 1990s because their loose structures resulted in the loss of a lot precious academic time.
"People came to realize we needed to make the most of every minute," he said. "This is a big issue with student achievement. Some students need more time and don't learn at the same rate."
Some Stonewall Jackson students said it was sometimes hard to get approval from their assigned flex teacher to seek help in another course from another teacher. Others said they were grateful for the breather during the day.
A survey of Stonewall Jackson teachers indicated that staff members prefer the new system, with about 84 percent of teachers saying they agree that the program helps students learn. The majority of students also supported the program, but not to the same degree: About 70 percent of students said they think flex time helps them learn.
Stonewall Jackson senior Madison Whitehead, 17, said the program has given her more time to catch up on her English reading assignments. "I'm more focused at school than at home," she said. "Most of the time, I use flex to catch up."
© 2009 The Washington Post Company

redpajama_set_name: RedPajamaCommonCrawl


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Turkey widens purges to police after coup bid, Europe warns on rule of law
Mon, Jul 18, 2016 - 6:04 PM
Turkey suspended thousands of police officers on Monday, widening a purge of the armed forces and judiciary after a failed military coup, and raising concern among European allies that it was abandoning the rule of law.
[ISTANBUL] Turkey suspended thousands of police officers on Monday, widening a purge of the armed forces and judiciary after a failed military coup, and raising concern among European allies that it was abandoning the rule of law.
A senior security official told Reuters 8,000 police officers, including in the capital Ankara and the biggest city Istanbul, had been removed from their posts on suspicion of links to Friday's coup bid by a faction in the army.
Thirty regional governors and more than 50 high-ranking civil servants have also been dismissed, CNN Turk said.
Thousands of members of the armed forces, from foot soldiers to commanders, were rounded up on Sunday, some shown in photographs stripped to their underpants and handcuffed on the floors of police buses and a sports hall. Several thousand prosecutors and judges have also been removed.
More than 290 people were killed and around 1,400 wounded in the violence on Friday night, as soldiers commandeered tanks, attack helicopters and fighter jets in a bid to seize power, strafing parliament and the intelligence headquarters and trying to seize the main airport and bridges in Istanbul.
Turkish hackers target Greek government websites, stock exchange
President Recep Tayyip Erdogan on Sunday told crowds of supporters, called to the streets by the government and by mosques across the country, that parliament must consider their demands to apply the death penalty for the plotters.
"We cannot ignore this demand," he told a chanting crowd outside his house in Istanbul late on Sunday. "In democracies, whatever the people say has to happen."
He called on Turks to stay on the streets until Friday, and late into Sunday night his supporters thronged squares and streets, honking horns and waving flags.
The bloodshed shocked the nation of almost 80 million, where the army last used force to stage a successful coup more than 30 years ago, and shattered fragile confidence in the stability of a Nato member state already rocked by Islamic State suicide bombings and an insurgency by Kurdish militants.
EU foreign policy chief Federica Mogherini warned the Turkish government on Monday against taking steps that would damage the constitutional order.
"We were the first... during that tragic night to say that the legitimate institutions needed to be protected," she told reporters on arrival at an EU foreign ministers meeting, which was also to be attended by US Secretary of State John Kerry.
"We are the ones saying today rule of law has to be protected in the country," she said in Brussels. "There is no excuse for any steps that take the country away from that." Austrian Foreign Minister Sebastian Kurz said it would be unacceptable for Turkey to reintroduce the death penalty, which it abolished in 2004. Abolishing capital punishment was a prerequisite for talks with Turkey on membership of the European Union, to which it still aspires.
Turkey's pro-Kurdish HDP opposition, parliament's third largest party, said it would not support any government proposal to reintroduce the death penalty. The main CHP opposition said the response to the coup attempt must be conducted within the rule of law and that the plotters should face trial.
Turkish security forces are still searching for some of the soldiers involved in the coup bid in various cities and rural areas but there is no risk of a renewed bid to seize power, a senior security official told Reuters.
The official said Turkey's military command had been dealt "a heavy blow in terms of organisation" but was still functioning in coordination with the intelligence agency, police and the government. Some high-ranking military officials involved in the plot have fled abroad, he said.
Mr Erdogan has blamed US-based Muslim cleric Fethullah Gulen for orchestrating the attempted power grab. He has long accused the cleric of trying to create a "parallel state" within the courts, police, armed forces and media.
Mr Gulen, in turn, has said the coup attempt may have been staged, casting it as an excuse for Mr Erdogan to forge ahead with his purge of the cleric's supporters from state institutions.
The swift rounding up of judges and others indicated the government had prepared a list beforehand, the EU commissioner dealing with Turkey's membership bid, Johannes Hahn, said.
"I'm very concerned. It is exactly what we feared," he said in Brussels.
A Turkish official acknowledged that Mr Gulen's followers in the armed forces had been under investigation for some time, but denied that an arrest list had been prepared in advance.
"In our assessment, this group acted out of a sense of emergency when they realised that they were under investigation. There was a list of people who were suspected of conspiring to stage a coup," the official said. "There was no arrest list. There was a list of people suspected of planning a coup."
turkey coup


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Thai police fire tear gas at anti-goverment protesters in Bangkok
Student protester hit by tear gas on 16 October protest at Ratchaprasong Intersection, Bangkok. Photo: Milktea2020. CC BY-SA 4.0.
Thai police on Saturday used tear gas, rubber bullets and water cannons during a new massive anti-government demonstration in Bangkok.
The demonstration, called by the movement Redem (Restart Democracy), was scheduled to start at 14.00 at the Democracy Monument before marching to the Grand Palace, but was forced to change its route several times, which ended up spreading the riots throughout much of the city.
The group came out on Saturday with three demands: the unconditional resignation of the Prime Minister, General Prayut Chan Ocha, cuts to the military budget and more resources to combat the COVID-19 pandemic.
The demonstrations have left five people injured — two civilians and three police officers — and two arrested, according to sources from the Erawan Emergency Medical Center and the police, 'Bangkok Post' said.
A burning police vehicle is seen during clashes at a protest against what they call the government's failure in handling the coronavirus disease (COVID-19) outbreak, in Bangkok, Thailand, August 7, 2021. REUTERS/Shoon Naing pic.twitter.com/OYGBdORSrl
— Shoon Naing (@Shoon_Naing) August 7, 2021
"Rubber bullets and tear gas were used for crowd control. Our aim is to maintain order," explained police spokesman Col. Kritsana Pattanacharoen, before complaining that protesters threw several objects at officers, he added.
Bangkok has been the scene of numerous anti-government rallies since the second half of 2020.
The demonstrators are calling for the "abolition of Article 112", which refers to lèse-majesté laws and carries penalties of up to 15 years' imprisonment for those convicted of insulting the monarchy.
-Thailand News (TN)
Tags: Bangkok, Government, Police, Protest, Tear Gas
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Visitation: 4 to 7 p.m. Friday at Schilling Funeral Home, Sterling.
SERVICES: 10 a.m. Saturday at Community Baptist Church, Milledgeville, with the Rev. J. Kregg Farmer officiating.
COLETA, Ill. — Donald E. Hollaway, 71, of Coleta, died Monday at his home.
Burial will be in Oak Knoll Memorial Park.
He was employed as a crane operator with Northwestern Steel & Wire Co. until he retired.
He was born April 17, 1936, in Princeton to Thomas E. and Lila E. Forristall Hollaway. He married Helen Resel on Aug. 9, 1967, in Milledgeville, Ill.
Survivors include his wife; three daughters, Gayla (Carl) Krahl of Milan, Ill., Krista (Larry) Specht of Tampico, Ill., and LouAnn (Ed) Covell of Kasbeer, Ill.; three sons, Thomas Hollaway of Peach Tree City, Ga., Steve (Kathy) Bearsley of Albany, Ill., and Dana Hollaway of Sterling, Ill.; one sister, Debra Sippel of Peoria, Ill.; two brothers, Robert (Joyce) Hollaway of Sterling and Charles Hollaway of Ottawa; 16 grandchildren; and nine great-grandchildren.
A memorial has been established for Community Baptist Church and Hospice of the Rock River Valley.


Amitabh Bachchan has finally confirmed that he will star in Yash Raj's next movie to be directed by Vijay Krishna Acharya. The movie is tentatively titled 'Thug' and Hrithik Roshan has already been confirmed to play a lead role in it.
We have already informed you that Hrithik Roshan has already given his nod to the project. The stylish actor is rumored to play a high profile thief in the movie, though nothing is confirmed yet.
'Thug' will go on floor once Hrithik Roshan completes the shooting of Sanjay Gupta's 'Kaabil' which also star Yami Gautam. As per the last update, Thug will go on floor in the end of June.
Hrithik will be next seen in Ashutosh Gowariker's period drama Mohenjo Daro. The movie also stars Pooja Hegde and is slated to release on 12 Aug 2016. Amitabh Bachchan on another hand will be seen in TE3N releasing on 10 June 2016.


The Eugène Pesch Museum is located in the "Depienne House" on the former pithead in Lasauvage. Visitors can admire a unique collection of fossils and minerals. A reconstruction of a mining gallery evokes the difficult working conditions of miners in the early 20th century.


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This Week (2003 TV programme)
For other uses, see This Week.
Andrew Neil presenting the show
Current affairs, politics
Michael Portillo
Miranda Green
Liz Kendall
Jim Meacock[1]
Samir Shah
Juniper TV
Original network
Original release
16 January 2003 (2003-01-16) –
18 July 2019 (2019-07-18)
Despatch Box[2]
Brexitcast[3]
Daily Politics
Politics Live
This Week is a BBC One current affairs and politics TV programme, screened late on Thursday evenings. It was hosted by former Sunday Times editor Andrew Neil, with a panel of two commentators, one each from the right and left of the political spectrum. The show was introduced on 16 January 2003, as was the Daily Politics, after a major review of BBC political programmes. It replaced the nightly Despatch Box (1998–2002), for which Neil had been the sole presenter in its later years.[2]
In February 2019, following Neil's decision to step down as host, the BBC announced that This Week would end in July 2019.[4] The final episode aired on 18 July 2019, a live broadcast from Westminster Central Hall with an invited audience of political dignitaries and celebrities. Mick Hucknall of pop group Simply Red sang "Nobody Does it Better" to Andrew Neil and the 'Final Show' was closed by 'Quiet Man' covering "Make Luv" (sic). After September 2019, Neil went on to host The Andrew Neil Show on Wednesdays. This Week was initially replaced on Thursdays by the TV version of Brexitcast, a podcast about Brexit.
2 Recurring jokes
2.1 Guest commentators' nicknames
2.2 Other political nicknames
2.3 Election song
With a light-hearted tone compared with most political programming, This Week was "punchy, irreverent, satirical".[5] From start to finish, Michael Portillo was the regular Conservative commentator on the show and the Labour Party commentator until 2010 was Diane Abbott. The two were ostensibly an "odd couple" coming from different sides of the political spectrum (albeit with a long-standing friendship dating back to when both attended grammar school in Harrow, even having appeared in a production of Macbeth together).[6]
During her unsuccessful campaign in 2010 to lead the Labour Party and her subsequent tenure as Shadow Minister for Public Health, Abbott began making only occasional appearances, her ability to speak without constraint becoming noticeable. Eventually, her place was taken by another Labour MP, in rotation each week, always a backbencher, most often Alan Johnson. After returning to the backbenches in 2013, Abbott appeared on a fortnightly basis, alternating with Johnson. Since her appointment to Jeremy Corbyn's shadow cabinet in September 2015, she has not appeared. The tradition of Labour MPs alternating in the spot has continued.
Since 2013, Neil's golden retriever Miss Molly also frequently appeared on the show, often walking in front of the camera during shots or choosing to sleep next to guests.
This Week often appeared on the BBC Radio 4 Comedy Show Dead Ringers, where Andrew Neil interviewed Diane Abbott and then Labour leader Jeremy Corbyn often popped up.
Episode format
Shown directly following Question Time, This Week presented itself as a more laid-back companion to its predecessor, with episodes regularly opening with a summary of the week's main events in the form of a parody of a popular television series. At the beginning of each episode, Neil asked the two regular commentators for their "Moment of the Week", occasionally contributing his own "moment". The standard format consists of three segments, with a guest contributor featuring in each. The first featured a journalist or commentator who presents their "Take of the Week" in a short film before appearing in the studio to discuss their perspective further. The second segment was a light-hearted "Round-up of the Week" in and around Parliament, formerly presented by Mark Mardell, who left the show on becoming the BBC's Europe Editor in 2005. The "Round-up" segment was later presented by a rotation of writers and broadcasters. This was followed by a discussion between the hosts (often joined for this segment by Miranda Green) of the issues raised. The third main segment, "Spotlight", typically focused more on cultural topics and features a final guest.[7] For many years, there was often also a quiz at the end of the show, in which Neil took pleasure in demonstrating the commentators' ignorance of a range of topics, though this feature no longer appeared in later years.[8]
Though primarily a political discussion programme, This Week achieved notoriety for its humorous approach to current affairs. For example, during the 2005 General Election, the show's title sequence spoofed the recently re-released version of "Is This the Way to Amarillo" and its video featuring comedian Peter Kay.[9] In 2011, following the announcement that the techno band Underworld were to perform at the opening ceremony for the London Summer Olympics, the show's ending credits featured Neil and his guests raving to the band's music.[10] The clip subsequently went viral on YouTube, helping to increase the show's unconventional popularity amongst younger audiences.
In 2006, the programme won the Hansard Society Award for Opening Up Politics which was awarded at the Channel 4 Political Awards ceremony.[11]
During the general election campaign of 2010, This Week was broadcast on Monday nights in addition to its usual Thursday night slot, with contributors including Sarah Teather, Lynne Featherstone, Caroline Flint and James Purnell and the late Charles Kennedy.
Recurring jokes
In keeping with its comic style, This Week had several recurring jokes and nicknames. These included assertions that the show's viewers watched the show drinking Blue Nun, David Cameron watched the show in bed wearing his pyjamas, and that the cast regularly went to a nightclub after filming is completed, either Annabel's in Berkeley Square or Lou Lou's in Mayfair (with Charles Clarke providing the guests a minicab service, even when he was not appearing on the show). It was also frequently said that the show had a budget of almost zero, and had few or no regular viewers. While giving out the Twitter and Facebook handles for the show (in the process, creatively mangling the names of the social media sites), Neil also insisted that no comments posted by viewers would be read. In addition, every episode began with the words "Evenin' all" and ended with "That's your lot for this week" followed by "Nighty-night; don't let the [topical event or person] bite". The middle section was introduced with "Now, it's late; [topical event]-late." Following Diane Abbott's departure from the show, Neil would joke that her leadership bid and later appointment as Shadow Minister for Public Health were part of her "insatiable lust for power". On the rare occasions that she consequently made an appearance on the show, Neil introduced her by saying "And back by absolutely no public demand whatsoever...". In weeks where the Speaker of the House of Commons, John Bercow (or his spouse), had featured in the news, the end credits were frequently shown over a scene of the diminutive Speaker being ceremonially escorted into the House of Commons to the music and lyrics of Jimmy Dean's "Big Bad John".
Guest commentators' nicknames
Michael Portillo, regularly mocked by Andrew Neil for his garishly-designed shirts and his long-running railway series
"Itchy & Scratchy" – Diane Abbott and Michael Portillo
"#Choo-Choo" & "#SadManOnATrain" – Michael Portillo
"#baffled & #HackneyAbbott" – Diane Abbott
"#SadManOnTheLeft" – Alan Johnson
"#ShesLovely" - Miranda Green
"#ManInTheMiddle" or "Chat-Show Charlie" – the late Charles Kennedy
"#RivieraGigolo" – Alastair Campbell
"#FourPerCent" & "Liz Misérables" – Liz Kendall
"#DivisionBelle" – Caroline Flint
"#theLamminator" – David Lammy
"#LastKingofScotland" – Alex Salmond
"Vince the Cable" – Vince Cable
"Shirl the Girl" – Baroness Shirley Williams of Crosby
"#SadManinaSuit" – Jacob Rees-Mogg
"#JessWeCan" – Jess Phillips
"#SadScouserOnaSofa" – Esther McVey
"#SmoothOperator" – Chuka Umunna
"#NorthernSoul" – Lisa Nandy
"#MaroonedCamerooned" – Ed Vaizey
"#minesavindaloo" – Michael Dugher
"#glitterballs" – Ed Balls
"#OhItsNotEdBalls" – Chris Leslie
"#littlepinkbus" – Harriet Harman
"Molly the dog" – Molly Neil (Andrew Neil's dog)
"Scrubber" – Iris Bailey (Andrew Neil's dog)
Other political nicknames
"BoJo The Clown" – Boris Johnson, Prime Minister 2019–
"The Maybot" – Theresa May, Prime Minister 2016–19
"Call Me Dave" – David Cameron, Prime Minister 2010–16
"Nick Cleggover", "#NickFlixandChill" – Nick Clegg, Deputy Prime Minister 2010–15
"Spreadsheet Phil" – Philip Hammond, Chancellor 2016–19
"Boy George" – George Osborne, Chancellor 2010–16
"Not-So-Red Ed" – Ed Miliband, Labour Party leader 2010–15
"Her Maj" or "The Boss" – The Queen
"The Great Leader" – Gordon Brown, Prime Minister 2007–10
"Our Dear Leader" – BBC Directors-General Tony Hall 2013–19; Mark Thompson 2004–12
"Miss Trust" – BBC Trust
Jean-Claude "Mine's a triple Cognac, high five" Juncker – Jean-Claude Juncker, President of the European Commission 2014–19
"Bish" – Archbishop of Canterbury
"Jezza the Red" – Jeremy Corbyn, Labour Party leader 2015–20
"The Donald" – Donald Trump, President of the United States 2017–21
Election song
During general elections, the show started with the 'election song'. The 2005 election song was "Is this the way to Election Day?" (a spoof of "Is This the Way to Amarillo?" by Neil Sedaka and Howard Greenfield),[12] and for the 2010 election, the song was "We're Off to Find a PM" (a spoof of "We're Off to See the Wizard" by Harold Arlen).[13]
^ "BBC News – This Week theme music". BBC News. 20 November 2009. Retrieved 5 June 2010.
^ a b Cozens, Claire; Deans, Jason (20 September 2002). "The BBC: A new manifesto for viewers". The Guardian. London.
^ Sherwin, Adam (13 June 2019). "BBC Brexitcast podcast to become weekly TV show in place of Andrew Neil's This Week". inews.co.uk. Retrieved 8 July 2019.
^ Moore, Matthew (14 February 2019). "BBC to axe This Week as Andrew Neil steps down". The Times. Retrieved 15 February 2019.
^ "About This Week". BBC News. 4 September 2008.
^ "Andrew Neil". BBC News. 5 September 2008.
^ "This Week – About This Week – BBC One". BBC. Retrieved 7 July 2016.
^ "Andrew Neil raps for Tinchy Stryder in This Week's quiz". BBC News. Retrieved 8 July 2016.
^ "Amarillo becomes election soundtrack". BBC. 12 April 2005. Retrieved 8 July 2016.
^ "Underworld rave tribute credits from This Week panel". BBC News. Retrieved 8 July 2016.
^ "Opening Up Politics". BBC News. 11 April 2006.
^ (Is this the way to) Election Day? on YouTube
^ We're off to Find a PM on YouTube
BBC portal
Politics portal
United Kingdom portal
This Week at BBC Programmes
This Week on Twitter
This Week (2003) on IMDb
Buzzfeed behind-the-scenes article


An employer is obligated to pay a worker the same day—or, for the long-term employee, according to an agreed upon payment schedule—by both positive and negative Torah command: "B'yomo titein secharo (pay wages the same day) and "Lo salin" (do not delay payment overnight).
Must I pay with cash? Checks and credit cards are also an acceptable form of payment, if it's a common practice in that particular locale. This only applies to situations where checks are expected—paying a taxi driver via check, for example, would not be considered suitable.
Some poskim (halachic authorities) limit paying with a check to business hours—while banks are open; if the worker has to wait to cash the check until the next day, it may not be considered timely payment.
A bounced check violates "lo salin", since payment to the worker is delayed.
A post-dated check—and likewise a payment plan over a period of time—is acceptable payment if it was agreed upon in advance.
In the case of cash, payment is restricted to common legal tender, unless it is local custom to use certain foreign currency.


text:
POSITIVE NEWS FOR THE PACIFIC NORTHWEST
Shriners hospital helps girl get prosthetic leg
by Katrina Aman | Apr 2, 2015 | COMMUNITY | 0 comments
SPOKANE, Wash.–
The Shriners Hospital is helping a four-year-old girl get back on her feet after adding a prosthetic lab last year.
The new device will help Ruth Schrader, who was adopted from Haiti by her parents when her biological parents could not provide the medical assistance she needed.
Ruth has Amniotic Band Syndrome, which occurs when a child is entangled in the womb cutting off blood flow and development in the fetus. Her left leg was amputated when she was born and her right leg was severely curved.
Doctors are making her a prosthetic leg and are proud of how far she's come.
"She's just made so many strides from where she was with not being very mobile to now I would say she's pretty mobile," said Peter Springs, a prosthetist.
Springs is the only person in Spokane who focuses on pediatric prosthetic care. Before the new lab was built, the hospital had to send families elsewhere. Rush's family appreciates the blessing of it being near to their home.
"It's comforting to know she's going to get what she needs in life," said Ryan Schrader. "And that she has that support system of not just her family but there's a hospital and a care staff out there that are going to be taking care of her."
With the new lab, the hospital will be able to meet patients' needs more quickly. 2,000 patients annually is how many people the hospital sees annually. Upgrades cost the hospital $20,000.
Katrina Aman
Katrina Aman is an aspiring journalist who desires to be a person of positive influence. Particularly passionate about poverty alleviation and civil rights, she hopes her writing takes her where she can improve lives.
Eight-year-old Indiana boy takes initiative to save turtles crossing the road
Canadian woman wards off cougar with Metallica—gets call from James Hetfield
Rosie the Riveters recall WWII days at Flex Studios summer dance camp
Five-year-old girl rescued from Dublin beach by vacationing Virginians
Woman saves her twins after starting the abortion process
Project of OREGON RIGHT TO LIFE Copyright © 2015-2017


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Tais Awards (Czech, Slovakia & Overseas) Music
NOMINATED / WINNERS
TIE~BREAK
GREAT BOOK TAIS AWARDS
TAIS AWARDS & HARVEST PRIZE: description
NOMINATED ARTISTS FOR 13th TAIS AWARDS 2020
Úvodní stránka » ARCHIVE » Thomas Dybdahl
Thomas Dybdahl — What's Left Is Forever
— One of Norway's most popular and acclaimed singer/songwriters.
Born: 12 April 1979
Instruments: Vocals, Guitar, Bass, Drums/Perc., Harmonica, Hammond, Mandolin, Banjo, Piano, Harp
Location: Stavanger, Norway
Album release: September 16, 2013
Record Label: Petroleum/Sony Music/Strange Cargo~Universal
01. This Love Is Here To Stay (3:20)
02. Easy Tiger (4:09)
03. Running On Fumes (4:52)
04. Shine (3:46)
05. Soulsister (3:51)
06. Man On A Wire (3:20)
07. I Never Knew That What I Didn't Know Could Kill Me (4:43)
08. Interlude (1:47)
09. City Lights (3:13)
10. So Long (3:43)
11. The Sculptor (3:57)
12. But We Did (4:50)
13. This Next Wave Is A Big One (5:02)
Track 01: David Baerwald / Thomas Dybdahl / Larry Klein
Track 07: Thomas Dybdahl / Larry Klein / David Poe
Track 12: David Batteau / Thomas Dybdahl / Larry Klein
Tracks 02~5, 8~11, 13: Thomas Dybdahl
¬ David Baerwald Composer
¬ David Batteau Composer
¬ Thomas Dybdahl Composer
¬ Larry Klein Composer
¬ David Poe Composer
Website: http://www.thomasdybdahl.com/
MySpace: https://myspace.com/tdybdahl
YouTube: http://www.youtube.com/artist/thomas-dybdahl
Twitter: http://www.twitter.com/thomasdybdahl
Facebook: https://www.facebook.com/thomasdybdahlofficial?fref=ts
Scandinavia: Kenneth Anderson (kenande2@me.com)
Worldwide: Nick Mander (nickmander@blueyonder.co.uk)
Press: Norway: Kim Paulsen, kim@petroleumrecords.no
Artist Biography by Mark Deming
¬ One of Norway's most popular and acclaimed singer/songwriters, Thomas Dybdahl has gained a growing international following for his elegant and expressive songs of love and loss, which have earned him comparisons to Nick Drake and Jeff Buckley.
¬ Born in 1979, Dybdahl first caught the attention of music fans as a guitarist with the funk/acid jazz ensemble Quadraphonics, who had a following in their native Norway and eventually did some touring in Canada and the United States. Dybdahl had begun writing songs of his own with a folk~rock/alt-~ountry feel that didn't fit in with Quadraphonics, and in 2000 he cut a solo EP, Bird. The record received small sales and little press notice, and a second EP, 2001's John Wayne, fared no better. Dybdahl's luck changed for the better with his first full~length album, 2002's ...That Great October Sound; primarily written during a visit to New York City with Dybdahl playing most of the instruments, the album's leadoff track, "From Grace," was embraced by Norwegian radio and the album went gold. The follow~up, Stray Dogs, was equally successful, and in 2004 One Day You'll Dance for Me, New York City became Dybdahl's first album created specifically for release in the United States and Canada. (A Danish version was released a year later, and was also made available as part of a box set with ...That Great October Sound and Stray Dogs, called The October Trilogy.) An album by Dybdahl's side project, the National Bank, was also released in 2004. Dybdahl signed a deal with Rykodisc for the release of his fourth solo set, 2006's Science, and in the spring of 2007 he launched his first tour of the United States. Dybdahl returned in 2010 with a self~titled studio effort on PIAS as well as Waiting for That One Clear Moment on Universal. In 2011, Dybdahl released the career~spanning overview Songs on Decca. ~ Mark Deming
In french:
¬ Un singer songwriter Norvégien qui propose la un bel album varié d'indie pop folk soul. C'est son septieme opus, enregistré entre la Californie et la Norvege. A découvrir.
¬ ...that great October sound (2002)
¬ Stray Dogs (2003)
¬ One day you'll dance for me, New York City (2004), (Danish version 2005)
¬ Science (2006)
¬ Waiting for That One Clear Moment (2010)
¬ What's Left Is Forever (2013)
Compilation Albums:
¬ October Trilogy (2006) (box set feat. all 3 albums, the remastered DVD and a bonus CD)
¬ Thomas Dybdahl (2009)
¬ Songs (2011)
EPs/Singles:
¬ Bird EP (2000)
¬ John Wayne EP (2001)
¬ Rain down on Me (2003)
¬ A Lovestory (2004)
¬ Damn Heart (2006)
¬ Cecilia (2010)
Other releases:
¬ Quadrophonics — Q (1998)
¬ The National Bank — The National Bank (2004)
¬ The National Bank — Come On Over To The Other Side (2008)
¬ That Great October DVD (2003)
¬ That Great October DVD Digitally Remastered (2005)
¬ Live at Paradiso, Amsterdam DVD (2009)
Broken Social Scene — Live at Third Man Records
Deserta — Black Aura My Sun
Keeley Forsyth — Debris
John McLaughlin, Shankar Mahadevan, Zakir Hussain
Whyte Horses — Hard Times
David Cross & Peter Banks — Crossover
The Adobe Collective
Lucia Cadotsch — Speak Low
ALBUM COVERS XI.
Tais Awards & Harvest Prize
Strachovská 520, Pelhřimov, CZE
taisawards@yahoo.com
Copyright © 2020 Tais Awards & Harvest Prize | inPage - webové stránky, doména a webhosting snadno. | Mapa webu


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text:
\section{\textsf{Introduction}}

The Berezin transform was introduced and studied by Berezin \cite{Berezin75}
for certain classical symmetric domains in $\C^n$
and next extensively considered in the context of Bergman, Hardy and
Bargmann-Fock spaces \cite{Berger,Englis,HKZ}.
It is closely related to Topelitz operators \cite{Zhu} and is useful in the quantization theory and corresponding $\star$-product on a suitable algebras of functions as well as in the correspondence principle  \cite{Berezin75,Berger}.

This transform is defined as follows. Let $\Omega$ be a domain of $\C^n$
with a Borel measure $\mu$ and $\mathcal{H}$ a closed subspace of $L^2(\Omega,d\mu)$ consisting of continuous
functions and  possessing a reproducing kernel $K(\cdot,\cdot)$. Then, the Berezin symbol $\sigma(A) $ of a
bounded linear operator ${A}$ on $\mathcal{H}$ is the function on $\Omega$ given by
$\sigma({A})(x) :=\scal{{A}e_{x},e_{x}}$, where $e_{x}(\cdot) :=K(\cdot,x) K(x,x)^{-1/2} \in\mathcal{ H}$.
Thus the Toeplitz operator $T_f$ with symbol $f\in L^{\infty}(\Omega)$ is the operator defined on $\mathcal{H}$ by
$T_f(\varphi) =P(f\varphi)$; $\varphi \in \mathcal{H}$,
where $P$ is the orthogonal projection from $L^2(\Omega,d\mu)$ into $\mathcal{H}$.
The Berezin transform associated to $\mathcal{H}$ is then defined to be the positive
self-adjoint operator $\mathscr{B}:=\sigma T,$ which turns out to be a bounded
operator on $L^2(\Omega, K(x,x)d\mu(x))$.

In this paper, we deal with the Bergman unit ball $\Omega=\B$ in $(\C^n,\scal{\cdot,\cdot})$ as domain endowed with
its Lebesgue measure $d\mu$. Since the Berezin transform can be defined
provided that there is a given closed subspace of $L^2(\B,d\mu)$ possessing a
reproducing kernel, we are concerned here with the  $L^2$-eigenspaces
\begin{equation}\label{gBerg}
\gBerg =\left\{ \varphi \in L^2(\B,(1-|z|^2)^{-n-1}d\mu) ; \quad H_{\chp}\varphi
=\epsilon^{\chp,n}_m\varphi \right\} ,
\end{equation}
associated to the discrete spectrum
\begin{equation}
\epsilon^{\chp,n}_m=4\chp(2m+n)-4m(m+n); \qquad m=0,1,2,\cdots, [\chp-\frac n2],
\end{equation}
of the Schr\"{o}dinger operator with uniform magnetic field on $\B$ given by
\begin{equation}\label{MagnSchr0}
H_{\chp}=-4(1-|z|^2) \left\{
\sum_{j=1}^{n}( \delta _{ij}-z_{i}\overline{z_{j}}) \frac{\partial^2}{\partial z_{i}\partial \overline{z_{j}}}
+\chp\sum_{j=1}^{n}\left( z_{j}\frac{\partial }{\partial z_{j}}-\overline{z_{j}}\frac{\partial }{\partial \overline{z_{j}}}\right)
+\chp^2\right\} + 4\chp^2,
\end{equation}
provided that $\chp >n/2$. Above $[x]$ denotes the greatest integer not exceeding $x$.
Note that for $m=0$,  the space $\Berg$ reduces further to be isomorphic to the weighted Bergman space of holomorphic functions $g$ on $\B$ satisfying the growth condition
\begin{eqnarray}\label{GCBerg}
 \int_{\B} |g(z)|^2(1-|z|^2)^{2\chp -n-1}d\mu(z) <+\infty .
\end{eqnarray}
The associated Berezin transform can be expressed as
\begin{equation}
\mathscr{B}[\varphi] (z) = \frac{(2\chp-n)\Gamma(2\chp )}{\pi^n\Gamma(2\chp -n+1)}
\int_{\B}\left(\frac{(1-|z|^2)(1-|\xi|^2)}{|1-<z,\xi>|^2}\right)^{2\chp } \varphi
(\xi) \left( 1-|\xi|^2\right) ^{-\left( n+1\right) }d\mu(\xi) .  \label{B0}
\end{equation}
Moreover,  it can be written as a function of the Laplace-Beltrami operator $\Delta_{\B}$
(\cite[p.182]{peeter}) as
\begin{equation}\label{peetre}
 \mathscr{B}  =\frac{\left| \Gamma\left(\alpha+1+\frac{n}{2} +\frac i2 \sqrt{-\Delta_{\B}-n^2}\right) \right|^2}{\Gamma(\alpha+1) \Gamma(\alpha +n+1) } ,
\end{equation}
where $\Gamma(x)$ being the usual Gamma function
 and $\alpha =2\chp-n-1$ occurring in the measure weight of the Bergman space under consideration in \cite{peeter} by Peeter who derived \eqref{peetre} from a formula representing the Berezin transform on the Bergman ball as function of the $\Delta_{\B}$, which was first established by Berezin in \cite{Berezin2}.

Our aim is then to generalize \eqref{B0} and \eqref{peetre} to the case of the spaces in \eqref{gBerg}.
We precisely attach to each of these eigenspaces a Berezin transform $\Berm$ following the formalism described
above as
\begin{eqnarray}
\Berm [\varphi](z) &:&=\frac{\Gamma(n)m!(2(\chp-m)-n)\Gamma(2\chp-m)}{\pi^n \Gamma(n+m)\Gamma(2\chp-m-n+1)} \int_{\B}\left( \frac{(1-|z|^2)(1-|\xi|^2)}{|1-\scal{z,\xi}|^2}\right)
^{2(\chp-m) } \label{BerInt}\\
&&\qquad \times \left( P_{m}^{(n-1,2[\chp-m])}(1-2|\xi|^2) \right)^2  \varphi(\xi)( 1-|\xi|^2)^{-(n+1)}d\mu(\xi),
\nonumber
\end{eqnarray}
where $P_{m}^{(\alpha,\beta)}(\cdot)$ denotes the Jacobi polynomial \cite{GR}. Moreover, we have to
prove that such transform can also be expressed as a function of the Laplace-Beltrami operator
$\Delta_{\B}$ in terms of the $\bF$-sum as
\begin{align}
\Berm  =C^{\chp,n}_m &\sum\limits_{j=0}^{2m}   (-2)^{j} A_{j}
    \frac{\Gamma(n+j)\Gamma\left(2(\chp-m)-\frac 12\left(n-i\sqrt{-\Delta_{\B} -n^2}\right) \right) }
    {\Gamma\left( 2(\chp-m) +j + \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right)\right) }
     \label{***}
\\ &  \times {\bF }\left[ \begin{array}{c}
                 \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right), \, n+j, \, \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right) \\
                 (\chp-m) +j+\frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right), \, n
               \end{array} \bigg| 1
\right],
\nonumber
\end{align}
where
\begin{align}
C^{\chp,n}_m= \frac{(2(\chp-m)-n) \Gamma(n+m)}{m!\Gamma(2\chp -n-m+1) \Gamma(2\chp -n)}
\end{align}
and
\begin{align}
A_{j}=2^{-j}\sum_{p=\max \left( 0,j-m\right) }^{\min \left( m,j\right)
\frac{(m!)^2\Gamma(2\chp-m) \Gamma(2\chp-m+j-p)}{(j-p)!(m+p-j)!p!(m-p)!\Gamma(n+j-p) \Gamma(n+p) }.
\end{align}
In particular, when $m=0,$ the transform \eqref{BerInt} reduces to the well
known Berezin transform  $\mathscr{B} $ in \eqref{B0} on the unit ball \cite{Berezin75}
and the expression \eqref{***} becomes the formula \eqref{peetre}.

The paper is organized as follows. In Section 2, we fix some notations and
we review briefly some needed tools
on the bi-holomorphic maps of the ball, the spectral function of the
Laplace-Beltrami operator and the Fourier-Jacobi transform of this operator. In Section 3, we are dealing
with the $L^2$-spectral theory of the Schr\"{o}dinger operator with uniform  magnetic
field on the unit ball and we are concerned in particular with  the reproducing kernels of
the eigenspaces associated with the discrete spectrum. In Section 4, we
construct for each of these eigenspaces a Berezin transform that we express
 in terms of the bi-holomorphic maps of the ball. In Section 5, we give a
formula representing the constructed Berezin transform as function of the
Laplace-Beltrami operator.

\section{Some notations and tools}
Here we fix some notations and we recall some needed tools we will be using such as the bi-holomorphic maps of the ball,
the spectral function of the Laplace-Beltrami operator and the Fourier-Jacobi transform.

\subsection{ The Moebius group of $\B$.}

Let $\C^n$ be endowed with its standard inner product
$
\scal{z,w} =z_{1}\overline{w_{1}}+z_{1}\overline{w_{2}}+\cdots +z_{1}\overline{w_{n}}
$
for every $z=(z_1,\cdots, z_n)$, $w=(w_1,\cdots,w_n)$ in $\C^n$, so that
$
|z| =\sqrt{\scal{z,z} }=\sqrt{|z_{1}|^2+ \cdots +| z_{n}|^2}.
$
We denote by $\B=\left\{ z\in \C^n:|z| <1\right\}$ the unit ball of $\C^n$. The Moebius group of $\B$, denoted $Aut (\B)$, i.e.
the set of the biholomorphic mappings  $\varphi: \B\longrightarrow \B$ (called also the
autormorphisms of $\B$) acts transitively on $\B$. For fixed $a\in \B$, let $P_{a}$ be the orthogonal projection of
$\C^n$ onto the subspace $<a>$ generated by $a$ and let $Q_{a}=I-P_{a}$ be
the projection onto the orthogonal complement of $<a>$. To be quite explicit, we have
$P_{a}(z)= {\scal{z,a} a}/{\scal{a,a} }$ if $a\neq 0$ and $P_{0}(z)=0$.
For every $a\in \B$, set $s=(1-|a|^2)^{1/2}$ and define
\begin{equation}
\varphi _{a}(z) =\frac{a-P_{a}(z) - (1-|a|^2)^{1/2}Q_{a}(z) }{1-\scal{z,a}}.
\end{equation}
 Then, the maps $\varphi _{a}$ have the following properties \cite{rudin}:
(i)  $\varphi _{a}\left( 0\right) =a$ \ \ and \ \ $\varphi _{a}(a) =0$, (ii)  $\varphi _{a}^{\prime}(0) =-s^2P_a-sQ_a$ \ \ and \ \ $\varphi_{a}^{\prime}(a) =-\frac{1}{s^2}P_a-\frac{1}{s}Q_a$, (iii) $\varphi _{a}$ is an involution: $\varphi _{a}\left(\varphi _{a}(z) \right) =z$,
 (iv)  $\varphi _{a}: \overline{\B}\rightarrow \overline{\B}$ is an homeomorphism  and belongs to $Aut(\B)$, (v) The Jacobian $\frac{|Dw|}{|D\xi|}$  of the transformation $\xi =\varphi_{a}(w)$ is equal to
\begin{equation}\label{Jacobian}
\frac{\left| Dw\right| }{\left| D\xi \right| }=\left( \frac{1-\left|
a\right|^2}{\left| 1-\scal{a,\xi } \right|^2}\right) ^{n+1}
\end{equation}
and (vi) The identity
\begin{equation}\label{Inva}
1-\scal{\varphi _{a}(w) ,\varphi _{a}(w')}
=\frac{\left( 1-\scal{a,a} \right) \left(1-\scal{w,w'} \right) }{\left( 1-\scal{w,a} \right) (1-\scal{a,w'} )}
\end{equation}
holds for every $w,w'\in \overline{\B}$.

The elements of $Aut(\B)$ are the maps $\varphi_{a}$; $a\in\B$, and their unitary transformations.
 Indeed, if $\psi \in Aut(\B) $ and $a=\psi ^{-1}(0)$, then there is a unique $U$ in the unitary group $U(n)$ such that $\psi =U.\varphi _{a}$.
 Moreover, in view of \eqref{Inva}, $\psi$ satisfies the identity
\begin{align}\label{IdentityInv}
1-\scal{\psi (w) ,\psi (w') }
\frac{\left( 1-\scal{a,a} \right) \left( 1-\scal{w,w'} \right) }{\left( 1-\scal{w,a} \right)
(1-\scal{a,w'} )}; \qquad w,w'\in \overline{\B}.
\end{align}
For our purpose, we will be concerned with the automorphism
\begin{align}
\varphi _{z}(w) :=\frac{A_{z}w+z}{1+\scal{w,z} }, \label{varphiz}
\end{align}
where $A_{z}$ is the $n\times n$ matrix given by
\begin{align}
A_{z}:=(1-|z|^2) ^{\frac{1}{2}}I_{n}+\frac{zz^{\ast }}{1+\sqrt{1-|z|^2}}, \label{varphiAz}
\end{align}
$I_n$ being the identity matrix.

\subsection{The Laplace-Belrami operator. }

Recall that the Bergman kernel of the unit ball is given by
 \begin{equation}
K(z,w) =\frac{n!}{\pi ^{n}\left( 1-\scal{z,w} \right) ^{n+1}},
\end{equation}
so that one defines
 \begin{equation}
g_{ij}=\partial _{i}\overline{\partial }_{j}\left(\log K(z,z)\right) =\frac{n+
}{\left( 1-|z|^2\right)^2}\left( \left( 1-|z|
^2\right) \delta _{ij}-\overline{z}_{i}z_{j}\right),
 \end{equation}
where we have used the notation $\partial _{k}$ and $\overline{\partial }_{k}$ to mean
\begin{equation}
\partial _{k}=\frac{\partial }{\partial z_{k}}=\frac{1}{2}\left( \frac
\partial }{\partial x_{k}}-i\frac{\partial }{\partial y_{k}}\right) \mbox{ and } \ \ \overline{\partial }_{k}=\frac{\partial }{\partial \overline{z}_{k}}
\frac{1}{2}\left( \frac{\partial }{\partial x_{k}}+i\frac{\partial }
\partial y_{k}}\right)
\end{equation}
with $z_{k}=x_{k}+iy_{k}\in \C$.
Therefore, the Bergman metric on $\B$ defined through
\begin{align}
ds_{z}^2 := \sum\limits_{i,j} \partial _{i}\overline{\partial }_{j}\log K\left(K(z,z)\right) \xi _{i}\overline{\xi }_{j}
 =(1-|z|^2) ^{-2}\sum\limits_{i,j=1}^{n}\left((1-|z|^2) \delta _{i,j}+\overline{z_{i}}z_{j}\right) dz_{i}\otimes d\overline{z_{j}},
\end{align}
so that the Bergman ball $(\B,ds^2)$ carries a K\"ahlerian structure.
The corresponding invariant volume measure  is
\begin{equation} \label{vm}
dg(z) =K(z,z) dV(z) =\frac{n!}{\pi ^{n}}\frac{dV}{\left( 1-|z|^2\right)^{n+1}}
=(1-|z|^2) ^{-(n+1)} d\mu(z)
\end{equation}
 and the geodesic distance $d(z,w)$ reads
\begin{equation}\label{Dist}
\cosh^2\left( d(z,w) \right) =\frac{|1-\scal{z,w}|^2}{(1-|z|^2)(1-|w|^2)}.
\end{equation}
 The Laplace-Beltrami operator associated to $(\B,ds^2)$ is defined to be
\begin{equation} \label{LaplBeltrami}
\Delta_{\B}=4\sum\limits_{i,j}g^{ij}\partial _{j}\overline{\partial }_{i}
=-4(1-|z|^2)\sum\limits_{i,j}\left( \delta _{ij}-z_i\overline{z}_{j}\right) \partial_{i}\overline{\partial }_{j};
\end{equation}
$(g^{ij})$ being the inverse matrix of $(g_{ij})$.
Note that $\Delta_{\B}$ is invariant under the action of the  group $SU(n,1)$ and
its spectrum $\sigma(\Delta_{\B})$ in the Hilbert space $L^2(\B,(1-|z|^2)^{-n-1}d\mu)$
is purely continuous and equals to $]-\infty, -n^2]$.
Instead of $\Delta_{\B}$, we have to deal with the non-negative elliptic self-adjoint operator
$-L_n:= -\Delta_{\B} -n^2$ so that $\sigma(-L_n)=[0,+\infty[$. Hence, for a given suitable real valued function $f: \R \longrightarrow \R$,
 the operator $f\left( -L_{n}\right) $ is defined by
  \begin{align} \label{FctLn}
    f(-L_n)[\varphi](z) := \int_{\B} K_f(z,w) \varphi(w)(1-|w|^2)^{-n-1}d\mu(w),
    \end{align}
  whose the kernel function has the integral representation
\begin{align} \label{SpecKernel}
  K_f(z,w) = \int_{0}^{+\infty} \Psi(z,w;\lambda) f(\lambda) d\lambda,
\end{align}
 where the spectral kernel is given by (\cite[p.13]{Oueld}):
\begin{equation}
\Psi(z,w;\lambda) =   \frac{|\Gamma(\frac{n+i\lambda}2)|^4}{4\pi^{n+1}\Gamma(n)|\Gamma(i\lambda)|^2}
\, {\aF}\left[ \begin{array}{c} \frac{n+i\lambda }{2},\frac{n-i\lambda }{2}\\ n \end{array} \bigg | -\sinh^2 (d(z,w))\right] ,\label{BLBOueld}
\end{equation}
in terms of the ${_2\digamma_1}(\cdot)$-Gauss hypergeometric function \cite{GR}.

\subsection{The Fourier-Jacobi Transform. }

Let us recall  that the Jacobi functions are defined by
\begin{align}\label{JacFun}
\phi _{\lambda }^{(\alpha,\beta) }(t)
:=\aF \left[ \begin{array}{c} \frac{1}{2}\left( \alpha +\beta +1+i\lambda
\right) ,\frac{1}{2}\left( \alpha +\beta +1-i\lambda \right) \\ \alpha +1 \end{array} \bigg | -\sinh^2(|t|)\right] .
\end{align}
The particular case of $t=i\theta$ and $\lambda=i(2n+\alpha+\beta+1)$ leads to
\begin{eqnarray}
\phi _{i\left( 2n+\alpha +\beta +1\right) }^{(\alpha,\beta)
}\left( i\theta \right) &:=&\aF \left[ \begin{array}{c} -n,n+\alpha +\beta
+1-i\lambda \\ \alpha +1 \end{array} \bigg |\sin^2(|\theta|) \right] \\
&=&\frac{n!}{\left( \alpha +1\right) _{n}}P_{n}^{(\alpha,\beta) }\left( \cos (|2\theta|) \right),
\end{eqnarray}
 a normalized Jacobi polynomial. We have the following special cases
$\phi _{\lambda }^{\left( -\frac{1}{2},\frac{1}{2}\right) }(t)
=\cos (|\lambda t|)$ and $\phi _{\lambda }^{\left( 0,0\right) }(t) =P_{\frac{1}{2}\left(
i\lambda -1\right) }\left( \cosh (|2t|)\right)$,
where $P_{\nu }(\cdot)$ is the Legendre function.
Now, let us assume that $\alpha >-1$ and $\beta \in \Bbb{R\cup }i\R$, and set
\begin{align}\label{Deltaab}
\Delta _{\alpha ,\beta }(t) :=\left( 2\sinh (|t|)\right) ^{2\alpha+1}\left( 2\cosh(|t|)\right) ^{2\beta +1}.
\end{align}
Then the Jacobi function $\phi _{\lambda }^{(\alpha,\beta) }$ in \eqref{JacFun} is an even
C^{\infty }$-function on $\R$ satisfying the second order differential equation
 \begin{equation}
\left( \frac{d^2}{dt^2}+\frac{\Delta _{\alpha ,\beta }^{\prime }(t) }{\Delta _{\alpha ,\beta }(t) }\frac{d}{dt}+\lambda
^2+(\alpha+\beta+1)^2\right) \phi _{\lambda }^{(\alpha,\beta) }(t) =0
 \end{equation}
with $\phi _{\lambda }^{(\alpha,\beta)}(0) =1$, and is uniquely defined
by these properties.
Moreover, for $\alpha >-1$ and $\beta \in \left[ -\alpha -1,\alpha +1\right] \cup
\R$, we have the integral transform pair
\begin{align}
g(\lambda) &=\int\limits_{0}^{+\infty }f(t) \phi_{\lambda }^{(\alpha,\beta) }(t) \Delta _{\alpha,\beta }(t) dt \label{JacTransf}\\
f(t) &=(2\pi) ^{-1}\int_{0}^{+\infty }g(\lambda) \phi _{\lambda }^{(\alpha,\beta) }(t) \left| c_{\alpha ,\beta }(\lambda) \right|^{-2}d\lambda , \label{InvJacTransf}
\end{align}
where
\begin{align}
c_{\alpha ,\beta }(\lambda) :=\frac{2^{\alpha +\beta
+1-i\lambda }\Gamma(\alpha+1) \Gamma(i\lambda) }
\Gamma \left( \frac{1}{2}\left( \alpha +\beta +1+i\lambda \right) \right)
\Gamma \left( \frac{1}{2}\left( \alpha -\beta +1+i\lambda \right) \right) }. \label{ChandraT}
\end{align}
This establishes a one-to-one correspondence between the space of even $C^{\infty
}$-functions on $\R$ with compact support and its image under the
Fourier-cosine transform, as characterized by the classical Paley-Wiener
theorem. The mapping $f\mapsto g$, which is called the Fourier-Jacobi transform, can be extended
to an isometry of the Hilbert spaces between $L^2\left( \
_{+},\Delta _{\alpha ,\beta }(t) dt\right) $ and $L^2\left(
\R_{+},(2\pi) ^{-1}\left| c_{\alpha ,\beta }(\lambda) \right| ^{-2}d\lambda \right)$. Note that in the case of $\alpha >-1$ and
\left| \beta \right| >\alpha +1$, \eqref{JacTransf} remains valid except
that we have to add to the right hand side of the second formula in \eqref{InvJacTransf} the term
 \begin{equation}
\sum_{\lambda \in D_{\alpha ,\beta }}g(\lambda) d_{\alpha
,\beta }(\lambda),
 \end{equation}
where $D_{\alpha ,\beta }$ is a finite subset of $i\R$ and the weights
d_{\alpha ,\beta }(\lambda) $ can be given explicitly.
 For more information  on the Fourier-Jacobi transform see Koornwinder's paper \cite{Koornwinder}  and the references therein.

\section{The generalized Bergman spaces $\mathcal{A}_{m}^{\chp}\left( \Bbb{B}^{n}\right) $}

Associated to the K\"ahlerian manifold $(\B,ds^2)$, modeling the complex hyperbolic space
 of negative holomorphic sectional curvature, there is a canonical vector potential
\begin{eqnarray}
\theta(z)=- \sqrt{-1} (\partial -\bar \partial)\mbox{Log} (1-|z|^2).
\end{eqnarray}
Thus, the magnetic Schr\"odinger operator, describing a charged particle moving in $\B$
  under the action of the derived magnetic field $d\theta$, can be defined as
\begin{eqnarray}
H_{\chp}:=  ( d + \sqrt {-1}\chp  \theta  )^{*} ( d + \sqrt {-1}\chp \theta).
\end{eqnarray}
It appears then as the Laplace-Beltrami operator in \eqref{LaplBeltrami}
  perturbed by a first order differential operator. Its   expression
in the complex coordinates is given by (\cite{GhInJMP}):
\begin{eqnarray} \label{LanLikeHam}
H_{\chp} = -4\Big(1-|z|^2\Big)\set{\sum _{i,j=1}^n(\delta _{ij}- z_i{\bar z_j})\frac{\partial ^2}{\partial z_i
\partial {\bar z_j}} +\chp \sum _{j=1}^n\left(z_j\frac{\partial}
{\partial z_j}-{\bar z_j}\frac{\partial}{\partial{\bar z_j}}\right) +\chp^2} +4\chp^2 .
\end{eqnarray}
Different aspects of the spectral analysis of $H_{\chp}$ have been studied by many authors under different names and notations (like Maass Laplacians, Landau Hamiltonians, ... ), see for example \cite{zhang,ahern96,BouIn} for $n\geq 2$ and \cite{elstrodt,patterson} for $n=1$.
 For instance, note that $H_{\chp}$ is an elliptic densely defined operator on the Hilbert space  $L^2(\B,(1-|z|^2)^{-n-1}d\mu)$  and admits a unique self-adjoint realization denoted also by $H_{\chp}$.
Moreover, it is a known fact that the discrete spectrum of $H_{\chp}$ on $L^2(\B,(1-|z|^2)^{-n-1}d\mu)$ is not trivial whenever $\chp >n/2$ and consists of a finite number of infinitely degenerate eigenvalues of the form
\begin{eqnarray} \label{LandauLevels}
\epsilon^{\chp,n}_m:= 4\chp(2m+n)-4m(m+n)
\end{eqnarray}
for varying integer $m$ such that $0\leq m< \chp- n/2$. The corresponding $L^2$-eigenfunctions, i.e., the elements of
 the space $\gBerg$ in \eqref{gBerg}

can be written explicitly in terms of the Jacobi polynomials $P^{(\alpha,\beta)}_j(x)$ and the complex spherical harmonics
 $h^{p,q}(z,\bar z)$ \cite{folland,rudin}.  Namely, from \cite{zhang,GhInJMP} we check that the functions
\begin{eqnarray}
\psi^{\chp,m}_{p,q}(z)

:= (1-|z|^2)^{\chp-m} P^{(n+p+q-1,2[\chp-m]-n)}_{m-q}(1-2|z|^2) h^{p,q}(z,\bar z),
\label{L2eigenfunctions}
\end{eqnarray}
  constitute an orthogonal basis of the space in \eqref{gBerg}, for varying integers $p$ and $q$ such that $p=0,1,2, \cdots$, and  $q=0,1,2, \cdots ,m$, whose the square norms in $L^2(\B,(1-|z|^2)^{-n-1}d\mu)$ are given by
    \begin{eqnarray}
\norm{\psi^{\chp,m}_{p,q}}^
=\frac{\pi^n\Gamma(n+m+p)\Gamma(2\chp -n-m-q+1)}
{2n!(2[\chp-m]-n) (m-q)!\Gamma(2\chp -m+p)} . \label{L2normb}
\end{eqnarray}
Furthermore, the space $\gBerg$ is a reproducing kernel Hilbert space, i.e.,
for every fixed $z\in \B$, there exists a function $K^{\chp}_{m,z}$ belonging to this space
such that
 \begin{equation}
f(z) = \scal{f, K^{\chp}_{m,z} } = \int_{\B} K^{\chp,n}_{m}(z,w)f(w)(1-|z|^2)^{-n-1}d\mu(w)
 \end{equation}
 for every $f \in \gBerg$,  where we have set $K^{\chp,n}_{m}(z,w)=\overline{K^{\chp}_{m,z}(w)}$.
 Its expression is given in terms of the hyperbolic distance in \eqref{Dist} as
\begin{align}
K^{\chp,n}_{m}(z,w)=\gamma^{\chp,n}_m \left(\frac{1- \overline{\scal{z,w}}}
{1-\scal{z,w}}\right)^{\chp}&\left(\cosh(d(z,w))\right) ^{-2(\chp-m)}  \label{rkB2}\\ & \times   P_m^{(n-1,2[\chp-m]-n)}\left(1-2\tanh^2(d(z,w))\right),\nonumber
\end{align}
where
\begin{eqnarray}
\gamma^{\chp,n}_m= \frac{(2(\chp-m) -n)\Gamma(2\chp -m)}{\pi^n\Gamma(2\chp -m-n+1)}.\label{crkb}
\end{eqnarray}
For $m=0$, the $L^2$-eigenspace associated to the bottom
eigenvalue $\epsilon^{\chp,n}_0=4\chp n$  is isomorphic to the usual
weighted Bergman Hilbert space of holomorphic functions $g$ on $\B$ such that \eqref{GCBerg}
holds. The corresponding reproducing kernel reads
 \begin{eqnarray}
K^{\chp,n}_{0}(z,w)=\frac{(2\chp -n)\Gamma(2\chp )}{\pi^n\Gamma(2\chp -n+1)}
\Big(\cosh(d(z,w))\Big) ^{-2\chp }.
\end{eqnarray}

\begin{definition}
Assume that $\chp > n/2$ and let $m$ be an integer such that $0\leq m <\chp - n/2$. Then the space $\gBerg$ is
 called here a generalized Bergman space of index $m$.
\end{definition}

\section{Generalized Berezin transforms on the unit ball}
We adopt to the formalism presented in Section 1 to the case of the Bergman ball $\Omega=\B$ as domain of $\C^n$ and $\gBerg$ as a closed subspace  of $L^2(\B,(1-|z|^2)^{-n-1}d\mu )$. Then, the Berezin symbol of a bounded
operator  $A$ on $\gBerg$ is the function
 \begin{equation}
\sigma _{m}(A) (z) :=\scal{A\left( \widetilde
K_{m}}\right) _{z}, \left( \widetilde{K_{m}}\right) _{z}}
_{L^2(\B,(1-|z|^2)^{-n-1}d\mu )}; \qquad z \in \B.
 \end{equation}
Here $(\widetilde{K_{m}})_{z}$ denotes the normalized reproducing kernel with evaluation at $z$, that is
 \begin{equation}
(\widetilde{K_{m}})_{z}(w) =\frac{K^{\chp,n}_{m}(z,w) }{\sqrt
K^{\chp,n}_{m}(z,z) }},\qquad w\in \B,
 \end{equation}
where $K^{\chp,n}_{m}(z,w)$ is given in \eqref{rkB2}.
 Therefore, for given
 complex-valued function $\phi$ such that $\phi \widetilde{K_{m}} \in L^2(\B,(1-|z|^2)^{-n-1}d\mu )$,
 the Berezin transform of $\phi$ is defined to be the Berezin symbol of the Toeplitz operator $T_{m}(\phi ) $ with symbol $\phi $
 defined on $\gBerg $ by $T_{m}(\phi )[f]:=P_{m}[\phi f]$. That is,
\begin{align}
\Berm\left[ \phi \right] (z)  =\sigma _{m}(T_{m}(\phi))(z)
 =\scal{T_{m}(\phi )\left( \widetilde{K_{m}}\right) _{z},
\left( \widetilde{K_{m}}\right) _{z}} =\scal{P_{m}\left( \phi .\left( \widetilde{K_{m}}\right)
_{z}\right) , \left( \widetilde{K_{m}}\right) _{z}}. \label{BerTop}
\end{align}
By expressing \eqref{BerTop} as an integral,
we get
\begin{align}\label{Berezin1}
\Berm \left[ \phi \right] (z) =\int_{\B}B_{m}(z,w) \phi (w) (1-|w|^2)^{-n-1}d\mu (w),
\end{align}
where
\begin{align}
B_{m}(z,w) :=\frac{\left| K_{m}^{\chp,n}(z,w) \right|^
}{K_{m}^{\chp,n}(z,z) }, \qquad z,w\in \B.
\end{align}
Explicitly,
\begin{align}
B_{m}(z,w) = \frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m (\cosh d(z,w))^{-4(\chp-m)}\left( P_{m}^{\left( n-1,2[\chp-m] \right) }\left( 1-2\tanh^2(d(z,w))
\right) \right)^2,
\end{align}
where $\gamma^{\chp,n}_m$ is as in \eqref{crkb}.
We then state the following

\begin{definition} Let $m=0,1,2, \cdots, [\chp-\frac n2]$ and let $\gamma^{\chp,n}_m$ as in \eqref{crkb}. Then, the integral operator acting on $\phi\in L^{\infty }\left( \
\right) $  as
\begin{align}
\Berm [\phi] (z) =&\frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m    \int_{\B}(\cosh(d(z,w)))^{-4(\chp-m)}\label{GBT}\\
&\times
\left( P_{m}^{\left( n-1,2[\chp-m] \right) }\left( 1-2\tanh^2(d(z,w))
\right) \right)^2 \phi(w) \left( 1-|w|^2\right)^{-n-1}d\mu(w).\nonumber
\end{align}
 is called the generalized Berezin transform of index $m$.
\end{definition}

\begin{proposition} The transform $\Berm $ defined in \eqref{GBT} can be written as
\begin{align}
\Berm [\phi](z) &= \frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m \int_{\B}(1-|w|^2)^{2(\chp-m)}  \label{RHS}\\
&\times \left( P_{m}^{( n-1,2[\chp-m])}  ( 1-2|w|^2) \right)^2 \phi\left( \varphi _{z}(w) \right) (1-|w|^2)^{-n-1} d\mu(w), \nonumber
\end{align}
where $\varphi _{z}(w)$ is as in \eqref{varphiz}.
\end{proposition}

\noindent {\it Proof.} Fix $z\in \B$ and set $w=\varphi_{z}^{-1}(\xi) $, where $\varphi _{z}(w)$ is as in \eqref{varphiz}. We have
$\varphi _{z}\in Aut(\B)$ and $\varphi _{z}\left( 0\right) =z$. The quantity in the right hand side of \eqref{RHS} reads, in terms of $\xi $, as
\begin{eqnarray}
&& \frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m \left( 1-\left| \varphi _{z}^{-1}(\xi) \right|^2\right)^{2( \chp-m) }\label{expression} \\
&& \qquad \times
\left( P_{m}^{(n-1,2[\chp-m])}\left( 1-2\left|\varphi_{z}^{-1}(\xi)\right|^2\right) \right)^2 \phi(\xi)
\left(1-\left|\varphi_{z}^{-1}(\xi)\right|^2\right)^{-n-1} d\mu \left( \varphi _{z}^{-1}(\xi) \right). \nonumber
\end{eqnarray}
By writing down the identity \eqref{IdentityInv} for $\psi=\varphi_{z}^{-1}$ and $w=w'$, we get
 \begin{equation}
1-\left| \varphi _{z}^{-1}(\xi) \right|^2= \frac{(1-|z|^2)
\left( 1-|\xi|^2\right) }{\left| 1-\scal{\xi,z} \right|^2} =\cosh^{-2}(d(z,\xi)) = 1-\tanh^2(d(z,\xi))
. \end{equation}
Also, by making use of \eqref{Jacobian} giving the explicit expression of
the Jacobian of the transformation $\xi =\varphi _{z}(w) $ is given through \eqref{Jacobian}, we see that
\begin{align}
\left( 1-\left| \varphi _{z}^{-1}(\xi) \right|^2\right) ^{-n-1} d\mu \left( \varphi _{z}^{-1}(\xi) \right) &= \left( 1-\left| \varphi _{z}^{-1}(\xi) \right|
^2\right) ^{-n-1} \frac{\left|Dw\right| }{\left| D\xi \right| }d\mu (\xi)\nonumber \\
&= (1-|\xi|^2)^{-n-1}d\mu (\xi). \label{Jacobian}
\end{align}
Therefore, the expression in \eqref{expression} can also be written as
\begin{eqnarray}
&& \frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m \int_{\B} \left( \frac{(1-|z|^2) \left( 1-|\xi|^2\right) }{\left| 1-\scal{\xi ,z} \right|
^2}\right) ^{2(\chp-m)} \\ && \qquad \qquad\times \left( P_{m}^{\left( n-1,2[\chp-m] \right) }\left(
1-2\tanh^2(d(z,\xi)) \right) \right)^2 \phi(\xi) \left( 1-|\xi|^2\right) ^{-n-1}d\mu
(\xi) \nonumber
\end{eqnarray}
or equivalently as
\begin{align}
&\frac{m!\Gamma(n)}{\Gamma(m+n)}\gamma^{\chp,n}_m \int_{\
}\left( \cosh d(z,\xi) \right) ^{-4\left( \chp-m\right) }
\\ & \qquad \times \left(P_{m}^{\left( n-1,2[\chp-m] \right) }\left( 1-2\tanh^2(d(z,\xi))\right) \right)^2
\phi(\xi) \left( 1-|\xi|^2\right)^{-n-1}d\mu (\xi), \nonumber
\end{align}
which is exactly the expression of $\Berm $ as defined in \eqref{GBT}. \fin

\section{$\Berm $ as function of the Laplace-Beltrami operator}

Our task here is to express the generalized Berezin transform $\Berm $
 as a function of the Laplace-Beltrami operator $\Delta_{\B}$. Precisely, we establish the following result.

\begin{theorem}
Let $m=0,1,2, \cdots, [\chp-\frac n2]$. Then, the transform $\Berm $ defined in \eqref{GBT}
can be expressed in terms of the Laplace-Beltarmi operator $\Delta_{\B}$ as
\begin{align*}
\Berm  &=C^{\chp,n}_m \sum\limits_{j=0}^{2m}   (-2)^{j} A_{j}
    \frac{\Gamma(n+j)\Gamma\left(2(\chp-m)-\frac 12\left(n-i\sqrt{-\Delta_{\B} -n^2}\right) \right) }
    {\Gamma\left( 2(\chp-m) +j + \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right)\right) }

\\ &  \times  {\bF }\left[ \begin{array}{c}
                 \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right), \, n+j, \, \frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right) \\
                 (\chp-m) +j+\frac 12\left(n+i\sqrt{-\Delta_{\B} -n^2}\right), \, n
               \end{array} \bigg| 1
\right], \nonumber
\end{align*}
where
\begin{align*}
C^{\chp,n}_m&= \frac{(2(\chp-m)-n) \Gamma(n+m)}{m!\Gamma(2\chp -n-m+1) \Gamma(2\chp -n)} \\
A_{j}&=2^{-j}\sum_{p=\max \left( 0,j-m\right) }^{\min \left( m,j\right)
\frac{(m!)^2\Gamma(2\chp-m) \Gamma(2\chp-m+j-p)}{(j-p)!(m+p-j)!p!(m-p)!\Gamma(n+j-p) \Gamma(n+p) }.
\end{align*}
\end{theorem}

\noindent{\bf Proof.}
On one hand, the transform $\Berm $ is the integral operator
\begin{align}\label{GBT2}
\Berm \left[ \varphi \right] (z) =\int_{\B}B_{m}(z,w) \varphi (w)  (1-|w|^2)^{-n-1}d\mu (w)
\end{align}
with the kernel function
\begin{align}\label{KerGBT}
B_{m}(z,w) = \frac{{m!}\Gamma(n)}{\Gamma(n+m)} \gamma^{\chp,n}_m
(\cosh (\rho))^{-4(\chp-m)} \left(P_{m}^{\left( n-1,2\left[ \chp-m\right] \right) }\left( 1-2\tanh^2 (\rho)\right) \right)^2,
\end{align}
where $\rho=d(z,w)$ and $\gamma^{\chp,n}_m$ is as in \eqref{crkb}.
On the other hand, from \eqref{FctLn} combined with \eqref{SpecKernel}, we obtain
\begin{align}
\label{FctLn2}
f(-\Delta_{\B} -n^2)[\varphi](z) = \int_{\B} \left(\int_{0}^{+\infty} \Psi(z,w;\lambda) f(\lambda) d\lambda \right) \varphi(w)  (1-|w|^2)^{-n-1}d\mu(w),
\end{align}
$\Psi(z,w;\lambda)$ being the spectral function given in \eqref{BLBOueld}.
Now, by equating \eqref{GBT2} and \eqref{FctLn2} we get
\begin{equation}
\int_{0}^{+\infty }\Psi(z,w;\lambda) f(\lambda) d\lambda = B_{m}(z,w)   \label{BLB10}
\end{equation}
by uniqueness of the kernels.
Explicitly,
\begin{align}
\int_{0}^{+\infty }
& \frac{\left|\Gamma\left(\frac{n+i\lambda}{2}\right)\right|^4}{\left| \Gamma(i\lambda) \right|^2}\ \
 \aF \left[ \begin{array}{c} \frac{n+i\lambda }{2},\frac{n-i\lambda }{2}\\ n \end{array} \bigg| -\sinh^2(\rho) \right] f(\lambda) d\lambda =  h(\rho),
\label{BLB15}
\end{align}
 where $h$ is used to represents the function $4\pi^{n+1}{m!}\Gamma(n) B_{m}(z,w)$, i.e.,
 \begin{align}
  h(\rho):=
 \frac{4\pi^{n+1}{m!}\Gamma^2(n)}{ \Gamma(n+m)}{\gamma^{\chp,n}_m}\left( \cosh (\rho) \right) ^{-4(\chp-m) } \left( P_{m}^{\left( n-1,2(\chp-m) -n\right)}(1-2\tanh^2(\rho)) \right)^2.  \label{h}
\end{align}
 To  determine the function $f$ we need to invert the integral equation \eqref{BLB15}. For this, we make appeal to the Fourier-Jacobi transform
 \begin{align}
 h\in L^2\left( \R_{+},\Delta _{\alpha ,\beta }(t) dt\right)  \longmapsto  g\in L^2\left( \R_{+},(2\pi) ^{-1}\left|c_{\alpha ,\beta }(\lambda) \right| ^{-2}d\lambda \right)
\end{align}
 as defined in \eqref{JacTransf} by
\begin{equation}
g(\lambda) =\int_{0}^{+\infty }h(t) \phi_{\lambda }^{(\alpha,\beta) }(t) \Delta _{\alpha
.\beta }(t) dt. \label{BLB17}
\end{equation}
Its inverse is given by
\begin{equation}
h(t) =\frac{1}{2\pi }\int_{0}^{+\infty }g(\lambda) \phi _{\lambda }^{(\alpha,\beta) }(t)
\left| c_{\alpha ,\beta }(\lambda) \right| ^{-2}d\lambda .
\label{BLB19}
\end{equation}
For $\alpha =n-1$, $\beta =0$ and  $t=\rho $, the involved quantities
  $\phi_{\lambda}^{(\alpha,\beta)}(t)$,
 $\Delta_{\alpha,\beta}(t)$ and $c_{\alpha ,\beta}(\lambda)$ given respectively by \eqref{JacFun}, \eqref{Deltaab}
 and \eqref{ChandraT} read
 \begin{align}
&\phi _{\lambda }^{(n-1,0) }(\rho) =\aF \left[ \begin{array}{c}
\frac{n+i\lambda}{2} ,\frac{1}{2}(n-i\lambda)
\\ n \end{array} \bigg | -\sinh^2(\rho) \right]   \label{C1}
\\
& \Delta_{n-1,0}(\rho) = 2 \cosh(\rho)(\sinh(\rho))^{2n-1} \label{C2}\\
&c_{n-1,0}(\lambda) =\frac{2^{n-i\lambda }\Gamma(n)
\Gamma(i\lambda) }{\Gamma^2\left( \frac{n+i\lambda}{2} \right)}.  \label{C3}
\end{align}
Therefore, \eqref{BLB15}  can be rewritten in terms of $\phi_{\lambda}^{(n-1,0)}(\rho)$ and
 $c_{n-1,0}(\lambda)$  as
\begin{equation}
\frac{1}{2\pi }\int_{0}^{+\infty }f(\lambda) \phi _{\lambda
}^{(n-1,0) }(\rho) \left| c_{n-1,0}(\lambda) \right|^{-2}d\lambda
=\frac{1}{2\pi  2^{2n}\Gamma^2(n) } h(\rho) . \label{BLB28}
\end{equation}
In view of the transform \eqref{BLB17}, the function $f$ takes the form
\begin{align}
f(\lambda) =\frac{1}{2 \pi 2^{2n}  \Gamma^2(n) }
\int_{0}^{+\infty} h(\rho) \phi_{\lambda }^{(n-1,0)}(\rho) \Delta_{n-1,0}(\rho) d\rho   . \label{BLB33}
\end{align}
 Substitution of $h(\rho)$, $\phi_{\lambda }^{(n-1,0)}(\rho)$ and
 $\Delta_{n-1,0}(\rho)$ by their explicit expressions, as given by \eqref{h}, \eqref{C1} ad \eqref{C2} respectively, yields
\begin{align}
f(\lambda) = &
\frac{2\pi^{n}m!}{\Gamma(n+m)} \gamma^{\chp,n}_m\int_{0}^{+\infty }
{\aF}\left[ \begin{array}{c}  \frac{n+i\lambda}{2},\frac{n-i\lambda}{2}\\n \end{array} \bigg |-\sinh^2(\rho) \right] \label{flamb}
\\ & \times  (\sinh(\rho))^{2n-1}\left( \cosh(\rho) \right) ^{-4(\chp-m)
+1} \left( P_{m}^{(n-1,2(\chp-m)-n)}(1-2\tanh^2(\rho)) \right)^2d\rho . \nonumber
\end{align}
By the change of variable $x=\sinh^2(\rho)$, Equation \eqref{flamb} becomes
\begin{align}
f(\lambda
=\frac{\pi^{n}m!}{ \Gamma(n+m)} \gamma^{\chp,n}_m& \int_{0}^{+\infty }{\aF} \left[ \begin{array}{c} \frac
n+i\lambda }{2},\frac{n-i\lambda }{2}\\n \end{array} \bigg | -x\right] \label{BLBBack}
\\  & \times x^{n-1}(1+x)^{-2(\chp-m)} \left( P_{m}^{\left( n-1,2(\chp-m) -n\right)
}\left( \frac{1-x}{1+x}\right) \right)^2dx .\nonumber
\end{align}
Next, we adopt a formula of the product of two Jacobi polynomials \cite[Eq (10), p.648]{TT} to write that
 \begin{equation}
\left( P_{m}^{(n-1,2(\chp-m)-n) }\left( \frac{1-x}{1+x}\right) \right)^2
=\frac{\left( \Gamma(n+m) \right)^2}{(m!)^2\left( \Gamma(2\chp -m) \right)^2}
\sum\limits_{j=0}^{2m}(-1) ^{j}A_{j}\left( 1-\frac{1-x}{1+x}\right) ^{j}, \label{JacProd}
 \end{equation}
where
 \begin{equation}\label{Aj}
A_{j}=2^{-j}\sum_{p=\max \left( 0,j-m\right) }^{\min \left( m,j\right)
\frac{(m!)^2\Gamma(2\chp-m) \Gamma(2\chp-m+j-p)}
{(j-p)!(m+p-j)!p!(m-p)!\Gamma(n+j-p) \Gamma(n+p) }.
\end{equation}
Inserting \eqref{JacProd} into \eqref{BLBBack}, we arrive at
\begin{align}
f(\lambda)
 \pi^{n}\gamma^{\chp,n}_m \Gamma(n+m)  \sum\limits_{j=0}^{2m}\left( -2\right) ^{j}A_{j} I_{n,j}(\lambda),
\end{align}
where
\begin{align}
I_{n,j}(\lambda) :=\int_{0}^{+\infty }\ \
\aF\left[ \begin{array}{c} \frac{n+i\lambda }{2},\frac{n-i\lambda }{2}\\ n \end{array} \bigg | -x\right]
x^{n-1+j}\left( 1+x\right) ^{-2(\chp-m) -j}dx.
\end{align}
Using the integral representation for the $\aF$-sum \cite[p.1005]{GR}:
\begin{align}
{\aF}\left[ \begin{array}{c} \alpha ,\beta \\ \gamma \end{array} \Big | z\right]  =\frac{1}{\mathcal{B}\left( \beta ,\gamma -\beta \right) }
\int_{0}^{1}t^{\beta -1}(1-t)^{\gamma -\beta -1}
                                      (1-tz)^{-\alpha }dt, \quad
                                      \Re\left( \gamma \right) >\Re \left( \beta \right) >0 ,
\end{align}
where $\mathcal{B}(\cdot,\cdot)$ is the Beta function, for the special case of
 $ \alpha =\frac{n+i\lambda }{2}$, $\beta =\frac{n-i\lambda }{2}$, $\gamma =n$ and $z=-x$, it follows
\begin{align}
I_{n,j}(\lambda) & = \int_{0}^{\infty }\left[ \frac{\Gamma(n) }{\left| \Gamma
\left( \frac{n+i\lambda }{2}\right) \right|^2}\int_{0}^{1}t^{\frac
n-i\lambda }{2}-1}(1-t) ^{\frac{n+i\lambda }{2}-1}\left(
1+tx\right) ^{-\frac{\left( n+i\lambda \right) }{2}}dt\right]
x^{n-1+j}\left( 1+x\right) ^{-2\left( \chp-m\right) -j}dx \nonumber\\
&=\frac{\Gamma(n) }{|\Gamma(\frac{n+i\lambda }{2})|^2}
\int_{0}^{1} t^{\frac{n-i\lambda }{2}-1}(1-t)^{\frac{n+i\lambda }{2}-1}
Q_{\chp,m,j,\lambda }^{n}(t)  dt , \label{IntI1}
\end{align}
where
\begin{align}
Q_{\chp,m,j,\lambda }^{n}(t) = \int_{0}^{\infty }\left( 1+tx\right)^{-\frac{( n+i\lambda) }{2}}
              x^{n-1+j}(1+x)^{-2(\chp-m) -j}dx . \label{IntQj0}
\end{align}
Making use of the identity \cite[p.317]{GR}:
\begin{align}
\int_{0}^{\infty }x^{\lambda_{0}-1}(1+x)^a(1+\alpha x)^{b}dx =
\mathcal{B}(\lambda_{0},-(a+b)-\lambda_{0}) \aF \left[ \begin{array}{c} -b ,\lambda_{0}\\ -(a+b) \end{array} \Big | 1-\alpha \right]
\end{align}
requiring that $|\arg(\alpha)| <\pi$ and $-\Re(a+b) >\Re(\lambda_{0})>0$ that are both fulfilled when $\lambda _{0}=n+j$,
$a =-2\left( \chp-m\right) -j$, $b=-(n+i\lambda)/2$ and $\alpha =t$, we get
\begin{align}
Q_{\chp,m,j,\lambda }^{n}(t) =\mathcal{B}\left( n+j,\frac{i\lambda -n}{2}+2\left( \chp-m\right) \right) \ \
\aF\left[ \begin{array}{c} \frac{n+i\lambda }{2},n+j\\ \frac{n+i\lambda }{2}+2\left( \chp-m\right) +j\end{array} \bigg| 1-t\right]. \label{IntQj}
\end{align}
By inserting \eqref{IntQj} in  \eqref{IntI1}, we see that
\begin{align}
I_{n,j}(\lambda) &=\frac{\Gamma(n)}{\left| \Gamma\left(\frac{n+i\lambda}{2}\right) \right|^2}
  \mathcal{B}\left( n+j,
\frac{i\lambda -n}{2}+2(\chp-m) \right) \\ &\times \int_{0}^{1}t^{\frac
n-i\lambda }{2}-1}(1-t) ^{\frac{n+i\lambda }{2}-1} {\aF}
\left[ \begin{array}{c} \frac{n+i\lambda }{2},n+j\\ \frac{n+i\lambda }{2} +2(\chp-m) +j\end{array} \bigg |1-t\right] dt.\nonumber
\end{align}
The change of the variable $1-t=u$ and the use of the formula \cite[p.813]{GR}:
 \begin{equation}
\int_{0}^{1}u^{a-1}(1-u)^{b-1}\aF \left[ \begin{array}{c} \alpha ,\beta \\ \gamma \end{array} \Big | u\right] du =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\ \
{\bF }\left[ \begin{array}{c} \alpha ,\beta ,a \\ \gamma , a+b \end{array} \Big| 1 \right],
 \end{equation}
$\Re(a) >0$, $\Re(b) >0$ and $\Re(b+\gamma -\alpha -\beta) >0$ for
$a = \frac{n+i\lambda }{2}$, $b =\frac{n-i\lambda }{2}$, $\alpha =\frac{n+i\lambda
}{2},\beta =n+j$ and $\gamma =\frac{n+i\lambda }{2}+2\left( \chp-m\right) +j$, yield
\begin{align}
I_{n,j}(\lambda) &=B\left( n+j,-\frac{n}{2}+\frac{i\lambda }{2}+2(\chp-m) \right) \label{Ij}
\\ & \qquad \times {\bF }\left[ \begin{array}{c} \frac{n+i\lambda }{2},n+j,\frac{n+i\lambda }{2}\\
\frac{n+i\lambda }{2}+2(\chp-m) +j,n\end{array} \Big| 1 \right]. \nonumber
\end{align}
 Finally, we arrive at
\begin{align}
f(\lambda) &=
\frac{(2(\chp-m) -n) \Gamma(n+m)
}{m!\Gamma(2\chp-(n+m)+1) \Gamma(2\chp-m) }\sum\limits_{j=0}^{2m}(-2)^{j} \mathcal{B}\left( n+j,\frac{i\lambda-n}{2}+2(\chp-m) \right)
A_{j} \nonumber
\\ & \qquad \times {\bF }\left[ \begin{array}{c} \frac{n+i\lambda }{2},n+j,\frac{
n+i\lambda }{2}\\ \frac{n+i\lambda }{2}+2(\chp-m) +j,n \end{array} \Big | 1\right].\label{fm}
\end{align}
Replacing $\lambda$ by $\sqrt{-\Delta_{\B}-n^2}$ and expressing the Beta function in terms of Gamma functions we obtain the announced formula. \fin

\section{ The case $m=0$}

In this case, the transform $\Berezin  $ turns out to be the well known Berezin transform $\mathscr B$
related to the Bergman ball of $\Bbb{C}^{n}$
\begin{equation}
\mathscr B \left[ \varphi \right] (z) = \frac{(2\chp -n)\Gamma(2\chp )}{\pi^n\Gamma(2\chp -n+1)}
\int_{\B}\left(\frac{(1-|z|^2)(1-|\xi|^2)}{|1-<z,\xi>|^2}\right)^{2\chp } \varphi
(\xi) \left( 1-|\xi|^2\right) ^{-\left( n+1\right)
}d\mu(\xi)   \label{BLB3}
\end{equation}
with the kernel function
 \begin{equation}
B(z,w) = \frac{(2\chp -n)\Gamma(2\chp )}{\pi^n\Gamma(2\chp -n+1)}
\left(\frac{(1-|z|^2)(1-|\xi|^2)}{|1-<z,\xi>|^2}\right)^{2\chp }.
 \end{equation}
Based on a formula due to Berezin \cite[p.377]{Berezin2}, a formula  expressing $\mathscr B  $ as a function of the Laplace-Beltrami
operator $\Delta_{\B}$ has been derived by  Peeter \cite[p.182]{peeter} as
 \begin{equation}\label{Lambdaa}
\mathscr B  =\frac{\left| \Gamma\left(\alpha+1+\frac{n}{2} +\frac i2 \sqrt{-\Delta_{\B}-n^2} \right)
\right|^2}{\Gamma(\alpha+1) \Gamma \left( \alpha +n+1\right) }.
 \end{equation}
 There $\alpha =\frac{1}{\hbar }-n-1$ occurs in the measure weight of the Bergman
space considered in \cite{peeter}. The result we obtain recovers the Peeter's one by taking $m=0$ and $\alpha=2\chp-n-1$ (i.e., $\frac{1}{\hbar }=2\chp$). Indeed, we establish the following

\begin{proposition}
 For $m=0$, the function in \eqref{fm} reduces to
 \begin{equation}
f_{0}(\lambda) =\frac{\left| \Gamma(2\chp -\frac{n-i\lambda}{2}) \right|^2}{\Gamma(2\chp-n) \Gamma(2\chp) }.
 \end{equation}
\end{proposition}

\noindent {\bf Proof.}
For $m=0$, the expression in \eqref{fm} reads
\begin{align}
f_0(\lambda) &= \frac{(2\chp-n)\Gamma(n)}{\Gamma(2\chp-n+1)\Gamma(2\chp)}
                \frac{\Gamma(n)\Gamma\left(\frac{i\lambda-n}{2}+2\chp \right)}{\Gamma\left(2\chp+\frac{n+i\lambda}{2} \right)}
                A_{0} \\
                &\qquad  \times {\bF }\left[ \begin{array}{c} \frac{n+i\lambda }{2},n,\frac{n+i\lambda }{2}\\ \frac{n+i\lambda }{2}+2\chp,n \end{array} \Big | 1\right].\nonumber
\end{align}
Now, according to the expression of $A_0$ given through \eqref{Aj} as
\begin{equation}
A_0=\left(\frac{\Gamma(2\chp)}{\Gamma(n)}\right)^2
\end{equation}
 together with the equality between the hypergeometric series $\bF$ and $\aF$:
\begin{equation}
\bF\left[\begin{array}{c} a,b,c\\ d,e \end{array} \Big | x\right]=\aF\left[ \begin{array}{c} a,b\\ d \end{array} \Big |x \right]
\end{equation}
in the particular case when $c=e$, it follows that
\begin{align}
f_0(\lambda) = \frac{(2\chp-n)\Gamma(2\chp)}{\Gamma(2\chp-n+1)}
                \frac{\Gamma\left(2\chp +\frac{i\lambda-n}{2}\right)}{\Gamma\left(2\chp+\frac{n+i\lambda}{2} \right)}
                \, \,\aF\left[ \begin{array}{c} \frac{n+i\lambda }{2},\frac{n+i\lambda }{2}\\ \frac{n+i\lambda }{2}+2\chp
                 \end{array} \bigg | 1\right] .\nonumber
\end{align}
Next, by applying the Gauss theorem \cite[p.1008]{GR}
 \begin{equation}
\aF\left[ \begin{array}{c} \alpha ,\beta \\ \gamma \end{array} \Big | 1\right] =
\frac{\Gamma(\gamma)\Gamma(\gamma -\alpha -\beta)}
{\Gamma(\gamma-\alpha)\Gamma(\gamma -\beta) },
 \end{equation}
whenever $\Re(\gamma-\alpha-\beta) >0$, $\gamma \neq 0,-1,-2, \cdots$, we arrive at
\begin{align*}
f_0(\lambda)  = \frac{(2\chp-n)  |\Gamma(2\chp-\frac{n-i\lambda}{2})|^2 }{\Gamma(2\chp-n+1)\Gamma(2\chp)}
                .\nonumber
\end{align*}
Making use of the identity $\Gamma(x+1)=x\Gamma(x)$ leads to
\begin{align*}
f_{0}(\lambda) =\frac{\left| \Gamma \left( 2\chp -\frac{n-i\lambda}{2}\right)
\right|^2}{\Gamma \left( 2\chp -n\right) \Gamma \left( 2\chp \right) }.
\end{align*}
This ends the proof. \fin

As consequence, replacing $\lambda$ by $\sqrt{-\Delta_{\B}-n^2}$ in the result of the previous proposition yields get \eqref{Lambdaa}.

redpajama_set_name: RedPajamaArXiv


text:
Yuan Jiajun, Party secretary of Zhejiang province, presided over the first meeting of the province's leadership team for the 19th Asian Games and the 4th Asian Para Games on Jan 7.
The Asian Games will be held in the provincial capital of Hangzhou in September. Wang Hao, acting governor of Zhejiang, and other officials of the provincial government attended the meeting.
At the meeting, Liu Xin, mayor of Hangzhou and the vice-chairman of the organizing committee for the 19th Asian Games, gave a briefing about the progress made by the city in preparing for the Asian Games and the Asian Para Games.
Officials from relevant departments also gave presentations about publicity, security, and pandemic control situations in relations to the Asian Games Hangzhou 2022.
Yuan acknowledged the provincial government's progress in the preparation work, but also pointed out existing difficulties and problems in this regard. Yuan said officials should be particularly meticulous so as to ensure that the Asian Games Hangzhou 2022 will fully demonstrate Zhejiang's characteristics.
Editor:Huang Yan
Zhejiang authorities discuss preparation work for Asian Games Hangzhou 2022 public html


To encourage and assist the academic progress of fraternity brothers.
The Fraternity of Delta Omega Epsilon (ΔΩΕ; official name Delta Omega Epsilon Incorporated) was founded on April 15, 1985 at the campus ofThe College of Staten Island. Delta Omega Epsilon was the first fraternity to be chartered at the school. They were a co-ed Greek Organization until it was believed a sister sorority should be broken off called Sigma Omega Epsilon. Delta Omega Epsilon was established to form a brotherhood, something that reaches beyond that of a friendship. On Feb 10th, 2011 Delta Omega Epsilon officially became incorporated.
The official symbol of the fraternity is the Phoenix and the Dolphin. The colors of the fraternity are black and silver.
To provide free tutorial and counseling to fraternity brothers.
To give a sense of brotherhood and while attending School.
To provide students with an opportunity for a business like experience in a fraternity corporation.
To expand the idea of multiculturalism on college campuses.
To learn and build on ideas on professionalism & advancement in the community.


Got a question or comment about Health in Sao Tome And Principe? Ask a Travel Expert!
Hepatitis A, typhoid, yellow fever, measles-mumps-rubella, and tetanus are all recommended as pre-travel immunisations. Cholera vaccinations are not generally necessary for travel to Sao Tome and Principe. Schistosomiasis may be acquired through contact with contaminated water, and travellers are advised to avoid swimming, wading, or rafting in bodies of fresh water, such as lakes, ponds, streams, or rivers.
Malaria is not as common a concern as it once was in Sao Tome, however visitors are still advised to take precautions, including insect repellent, netting, and antimalarial tablets. Either mefloquine (Lariam), atovaquone/proguanil (Malarone), or doxycycline may be taken. It is recommended to wear long sleeves, long pants, hats, and closed shoes in forested areas to prevent tick bites.
Medical facilities in Sao Tome and Principe are extremely limited. The main healthcare facility is the Ayres Menezes Hospital, located on Sao Tome. Be advised that most doctors and hospitals will expect cash payment, regardless of whether you have travel health insurance. Serious medical problems will require air evacuation to a country with state-of-the-art medical facilities. It is also recommended that travellers carry a personal supply of basic first aid and antibiotics.


I couldn't tell you when I first held a paintbrush but I CAN tell you I was young! I'm an animal lover with a border collie mix rescue pup, Brody. That's him on the left. He was my very first pet portrait. Home watercolors stemmed from a painting of our first house as newlyweds. And now that I have a private pilot husband, I paint an increasing number of planes!
My son was born in 2017 and after painting on the side as a hobby for years I decided to go for it, leaving my job in computer software with an amazing company who remains dear to my heart. I am living the dream watching Jacob grow and painting away during naptime.
The wonderful journey that led me to my husband, Hunter, took us from our southern roots across the country to Boise, Idaho. As much as we miss our family and friends, we love the adventure and the beautiful scenery around us (talk about art inspiring). I thank God for all He has given me and a talent I can share with others!


text:
<?php

namespace Baker\RecipeBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class RecipeType extends AbstractType
{
        /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('title')
            ->add('author')
            ->add('content', 'textarea', array(
                'attr' => array(
                    'class' => 'tinymce',
                    'data-theme' => 'bbcode'
                )
            ))
            ->add('file')
            ->add('tags')
        ;
    }

    /**
     * @param OptionsResolverInterface $resolver
     */
    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'Baker\RecipeBundle\Entity\Recipe'
        ));
    }

    /**
     * @return string
     */
    public function getName()
    {
        return 'baker_recipebundle_recipe';
    }
}

redpajama_set_name: RedPajamaGithub


2- The applicant must be a high school graduate that belongs to the Ministry of Education in Kurdistan Region. He/she needs to provide a copy of his/her official transcript certified by the Directorate of Education.
3- The applicant needs to pass a standard medical test with specific conditions for each college and institute.
4- The applicant should not be over 35 for morning studies and 40 for evening studies and 45 for parallel studies. Applicants who have a final grade point average of more than 80 percent are not subjected to age conditions.
1- The applicants high school certificate with his/her Grade Point Average (GPA).
2- A copy of Iraqi Nationality Card.
3- A copy of Identity Card.
6- A medical test form.
7- Eight passport sized pictures.
Declaration that the given information and related documents are true and correct.


Authorities in Cuba have decided to bring an end to a ban on the individual import of domestic appliances.
According to a report by the Guardian newspaper, a ban on items such as microwaves and air conditioners was introduced in 2005 during an energy crisis. Following power shortages and eventual blackouts, president Fidel Castro announced restrictions on the sale and import of domestic appliances in a bid to reduce pressure on the grid.
In 2011, current president Raul Castro introduced economic reforms which allowed for local sales of domestic appliances to resume, although the government has continued to stress the need for care to be taken to avoid placing excessive pressure on the nation's energy system.
Under the new rules, Cubans can bring two appliances into the country as long as they're for non-commercial purposes. The list of approved items includes toasters, irons, water heaters, microwaves under 2,000 watts, ovens under 1,500 watts and certain air conditioners.
At the end of last year, 3,000 Cubans living in a town to the southeast of Havana were given microwave ovens as part of a government programme.