text:
Record Year for PortMiami
PortMiami has announced a record cruise and cargo year.
The port experienced an increase of 5.8 per cent in containerized cargo movements for its 2018 Fiscal Year with a total of 1,084,000 (twenty-foot equivalent units) and an increase of 4.7 per cent in cruise visitors with a total of 5,592,000 passengers during the 12 month period ending September 30, 2018, according to a press release.
On the cruise side, PortMiami officials attribute the growth to expanded business opportunities, according to a statement. n 2018, Royal Caribbean International added sailings to the Empress of the Seas and welcomed the newly renovated Mariner of the Seas. Oceania Cruises and Disney Cruise Line extended their winter seasons.
The port also welcomed three new cruise lines: Viking Ocean Cruises, Victory Cruise Lines, and Seabourn. Additionally, it welcomed Carnival Cruise Line's Carnival Magic and the new Carnival Horizon. MSC also introduced its new-build, the MSC Seaside, to the port's line-up.
"I am proud to say that today PortMiami is a full-service global hub handling some of the largest vessels in the industry," said Miami-Dade County Mayor Carlos A. Gimenez. "We're starting to see the results of more than $1 billion of completed infrastructure and a port team that gets the job done. I would like to thank the entire PortMiami team for their hard work and dedication. The Port's continued growth would not be possible without them."
"I want to thank all of PortMiami's dedicated partners including the cruise and cargo lines, the International Longshoremen's Associations' Locals 1416 and 1922, Miami-Dade Police and Fire Departments, the Transportation Security Administration, U.S. Customs and Border Protection Miami, U.S. Coast Guard, the Biscayne Bay Pilots among so many others," said Chairwoman of the Economic Development and Tourism Committee Rebeca Sosa. "New facilities, berths and terminals are in the works to accommodate the industry's new generation of vessels. The future is only getting brighter at our seaport."
Posted in Uncategorized	| Tagged cruise visitors, MSC Seaside, NCL Escape, Port of Miami, PortMiami, Record year, royal caribbean international | Leave a comment
8 June, 2018 by wetravel2u
MSC Cruises chief warns yields could drop amid mass growth
Cruise line yields could fall in the future due to the volume of new ships entering the market in the next decade, according to the boss of MSC Cruises.
A total of 106 ships are expected to launch between 2017-26, including 12 alone from MSC Cruises. Other mainstream cruise brands such as Royal Caribbean International has six ships on its order books and Norwegian Cruise Line has seven. 48 expedition and luxury ships are also planned.
Asked whether occupancy levels were at risk of falling from so much growth, chief executive Gianni Onorato told delegates at the ITT Conference in Sicily: "No, we are not crazy. But in difficult times it can have an impact on yields more than occupancy. So I think this will be more of an issue than occupancy levels."
Onorato said the slowdown in cruise bookings last year in the UK was due to a "lack of capacity" in 2017 and said he was very "optimistic" about the British market going forward.
He said the biggest challenge facing cruise lines was future proofing ships which are built to last 20 years.
"Knowing what guests want in 2040 when you're building them now is very difficult to predict so we need to have a flexible mindset and be able to follow guests' needs (in terms of technology) but at the same time maintain human contact," Onorato said.
Posted in Uncategorized	| Tagged British market, Carnival Paradise, cruise market, Harmony of the Seas, mass growth, MSC Cruises, MSC Seaview, NCL Bliss, royal caribbean international | Leave a comment
4 February, 2018 by wetravel2u
New Features for Mariner of the Seas in Short Cruise Market
Royal Caribbean International today announced the details of a $90 million refurbishment on the Mariner of the Seas, as she moves into the Caribbean short cruise market and is redeployed from the Asia and China market.
Service starts from Miami with a three- and four-day cruise program in June, with the ship sailing to Nassau and CocoCay.
"At Royal Caribbean, we believe in the constant pursuit of adventure and that every weekend is an opportunity to experience something new and live life to its fullest," said Michael Bayley, President and CEO, Royal Caribbean International. "With the new Mariner of the Seas, we invite travellers to weekend like they really mean it and make the most of a quick getaway than they even thought was possible."
Among the highlights is the Sky Pad, a brand new virtual reality, bungee trampoline experience on the Mariner of the Seas.
Guests will strap in and put on a virtual reality headset, which the company said will " transport them to another time and planet to bounce over moon craters or compete in intergalactic games. The out-of-this-world experience is for guests of all ages, and also can be enjoyed without virtual reality headsets, for those who prefer to look out over the ocean as they leap toward the sky."
Other additions include The Perfect Storm, a duo of racing waterslides called Cyclone and Typhoon, and a glow-in-the-dark laser tag experience in Studio B, where friends and family can team up to play "Battle for Planet Z."
Mariner will offer Puzzle Break: The Observatorium, a new escape room experience.
The company is also adding the FlowRide surfing simulator to the ship.
New food and beverage options include Jamie's Italian, Starbucks and Izumi Hibachi & Sushi.
Posted in Uncategorized	| Tagged Battle for Planet Z, Caribbean cruise, Jamie's Italian, Mariner of the Seas, royal caribbean cruise, royal caribbean international, Short Cruise Market, Sky Pad, Starbucks, The Perfect Storm | Leave a comment


Exclusive Dunes Court Sale - $50 per night in the month of April when you stay two nights or more at the Dunes Court Building.
Cancellation Policy: Reservations cancelled from the time of booking to 72 hours (3 days) prior to arrival will forfeit a $25.00 service fee. Reservations cancelled within 72 hours prior to arrival will forfeit one night's room and tax.
Fun abounds at the casual Dunes Court at the northern tip of the famous Ocean City Boardwalk where all accommodations overlook the outdoor pool, each with semi-private balcony or patio, 48" HDTV, two double comfort beds, complimentary Comcast fiber WiFi and a fresh new hip look. Enjoy the comfort of this airy extra-large guest room offering two double beds microwave, refrigerator and in-room safe. This guest room is spacious and provides ample closet space and a relaxing seating area. Located at the corner of Baltimore Avenue and 27th Street, convenient check can be found at the Dunes Court on-site office with complimentary coffee and ice available offered daily. At the Dunes Court you will find ample off-street parking, outdoor pool and daily housekeeping service allowing for a fully relaxing stay. Enhance your stay and take advantage of the amenities of the Dunes Hotel at 28th Street and Baltimore Avenue, such as daily Afternoon Tea from 3pm – 4pm, an indoor/outdoor pool and deck, restaurant, bar, verandah, complimentary morning coffee and sing-alongs on scheduled evenings. This guest room also contains a refurbished bathroom with refreshing bath amenities. Additional features include an alarm clock radio, black out curtains, iron & ironing board and outdoor seating area.


text:
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Searching For Tony Jaa The Hottest Martial Arts Movie Star Since Jackie Chan And Jet Li Part 1
back kick
back kicks
boxing hook
November, 2013,
Just when most moviegoers were ready to abandon all hope that a fresh face would ever appear in martial arts cinema, we got Tony Jaa, star of 2003's Ong-Bak. As an added bonus, he brought with him a deadly new fighting style. In the blink of an eye, the sacred Thai art of pounding a person senseless with the knees and elbows was introduced to the world. Aside from having the most incredible fight scenes ever and showing us Bangkok rather than Hong Kong, Ong-Bak is an important movie for two reasons. It was the first major film to feature muay Thai and the first Thai movie to have wide distribution in the States — all thanks to a high-flying martial artist from the jungles of Southeast Asia. And now Tony Jaa is making the leap from Bangkok to Hollywood thanks to his recent casting in the upcoming Fast & Furious 7, helmed by Saw director James Wan and starring Dwayne Johnson, Jason Statham, Vin Diesel, Michelle Rodriguez, Paul Walker and veteran actor Kurt Russell. Watch a video of Vin Diesel and Tony Jaa training:
TONY JAA VIDEO Tony Jaa and Vin Diesel Train for Fights in the Upcoming Fast & Furious 7
— From the Fast & Furious 7 Fan Channel (unofficial)
And now on to the search for Tony Jaa ...
Tony Jaa's Hometown
Walking down Silom, one of Bangkok's main shopping areas, I see Tony Jaa posters and DVDs of Tony Jaa movies everywhere. Even in remote villages, children wear T-shirts bearing photos of Jaa Pnom, as he's called in Thailand. The locals know that he's been an action hero for 15 years — which prompts my muay Thai instructor to ask, "Why did it take you guys so long to realize how great he is?"
Take your muay thai training to a new level with our FREE download! Master Toddy's MMA-Tested Muay Thai Techniques
Being a movie star, Tony Jaa lives in the capital, but my first stop is a village in Surin province, seven hours north of Bangkok. Here, Tony Jaa was born and given the name Pnom Yeerum, which he later changed to Jaa Pnom. At the bus station, everyone knows Tony Jaa and loves to talk about him. They're proud of the local boy who made it big in America, and they're grateful that he hasn't forgotten his roots. They tell stories of his frequent visits to the region and the money he donates to schools and temples. A driver agrees to hire out his bus for $25 a day, and he says he'll take me to see Tony Jaa's house. After driving for an hour, we arrive at a beautiful, two-story home surrounded by a brick wall. The driver says this is the mansion that Tony Jaa had built for his parents. "But they don't answer when people ring the bell," he adds. "Too many reporters come to bother them." He motions for me to get back on the bus, and I begin suspecting that I've been had. He could have driven to any big house, pointed at it and said it was Tony Jaa's place. I discuss the matter with two Cambodian monks, friends who elected to tag along to translate, when an elderly man and woman come out to greet us. They're Tony Jaa's parents.
Searching for Tony Jaa: Meeting His Parents
"We don't usually answer the door, but we saw the monks," the mother says, bowing. As we enter the compound, the father gives me a sly look: "Say 'thank you' to your friends — you would never have gotten in here if it wasn't for them." Mr. and Mrs. Jaa and I sit barefoot and cross-legged on the floor, drinking ice tea. The monks perch on a bamboo platform above us. Being taller than most Thais, I have to make sure my head isn't higher than the monks' — I don't want to offend the Jaas, who are known to be very religious. The parents speak at length with the monks. They're curious to know how a foreigner came to be friends with the Cambodians. The monks explain that we met in Cambodia several years ago while I was studying the martial arts there. The father signals me to start the interview. I compliment them on the beauty of their house. It's strange to see a mansion beside all the tiny two-room affairs. "After he became famous, Tony bought this house for us," the mother says with pride. "It cost 10 million baht." "Where did you live before?" I ask. "In that house," she says, pointing. Upon entering the compound, I noticed a dilapidated wooden shack and wondered why they kept it on the grounds. Now, I understand. "Every day, we look out the window and remember where we came from," Mrs. Jaa says.
Tony Jaa Growing Up
"Tony was always playing sports," the father says. "He liked all sports, including basketball, but he especially loved any kind of martial art. When he was 5, he fell in love with martial arts. He watched Bruce Lee movies and copied the moves. He said that when he grew up, he'd be famous. We never doubted him. Tony taught himself martial arts as a kid. He wanted to be like [martial arts actor] Panna Rittikrai."
Your FREE guide to Bruce Lee movies is here! Our Bruce Lee Movies List: Little-Known Trivia From Bruce Lee's Pictures
As a teenager, Tony Jaa studied taekwondo and other arts, winning several competitions. His father, a former boxer, also taught him muay Thai. When he was older, he earned a scholarship to study at a sports university. Then he came back and studied with Panna Rittikrai, who helped him get his first work in the movies. "He's 175 centimeters (5 feet 9 inches) tall and weighs just over 50 kilograms (110 pounds)," the father says. "He's tall and strong and good looking, with big muscles," his mother adds. "And women go crazy for him." "Even foreign ladies want to date him," his father quips with a naughty grin.
The Celebrity Status of Tony Jaa vs. His True Roots
Tony Jaa is nearly as big a celebrity in neighboring Cambodia as he is in Thailand. Surin province, his home, was actually part of Cambodia until 250 years ago. As a result, 70 percent of the people here have Khmer blood and speak the Khmer language. The Cambodians all say that Tony Jaa is a Khmer and that the martial art he uses in his films is Khmer bokator. We've been speaking Thai, but now the conversation switches to Khmer, which Mr. and Mrs. Jaa speak perfectly. Hoping to settle the argument once and for all, I ask if Tony Jaa is Khmer or Thai. The father shocks us by answering, "We are Kuy!" The Kuy (Suoy in the Thai language) are a tribe, an ethnic minority that came to Thailand 400 years ago, migrating from India, down through Cambodia and into Thailand. They brought with them their elephants, which the Thai king quickly realized had excellent military applications. He gave the Kuy people Thai citizenship and made them the royal elephant handlers. Today, no matter where you go in Thailand, if you see elephants, the handlers are Kuy. "I was born in Ban Taklang," the father says, referring to the famous elephant village. "We still keep elephants to this day." Suddenly, Tony Jaa's second film, Tom Yum Goon (The Protector in the United States), makes sense. The whole movie revolves around the theft of an elephant from Thailand. Once again, Tony Jaa was showing that he's true to his roots.
Tony Jaa's Elephant Obsession
Our next stop is Ban Taklang, where, amid a sea of elephants, we meet Kun Da, an aged Kuy man who knew Jaa and his family when they lived in the village. "I was born in this village," Kun Da says. "My grandparents and all my ancestors were born here, too. I started living with elephants when I was a boy. Now I am 60 or 70 years old — I don't know for sure — and I live with the elephants. There are some elephants who are older than me. "I speak Khmer, Thai and Kuy. Kuy doesn't have letters, but I learned it from my parents at home. I know Jaa Pnom. He lived here before. His parents were born here. Then he was born here. Then they went to live near the mountain." In the village, when a boy is 1 or 2, he's given a baby elephant, and the two grow up together. Elephants have a life span similar to human beings, often living to be 60 years old. The locals give respect to elephants the same way they do to people of the same age. When they get older, the elephants are consulted on family decisions. If a Kuy wants to marry, build a house or move to a new town, he has to ask permission from the elephants. At a religious shrine, dedicated to the god of the elephant handlers, I meet some child monks. They're all excited because a Buddhist holiday is almost here, a day when people bring food, clothing and money to the monks. Tony Jaa will come home to preside over the festival, the kids say. Nearby, we enter the forest monastery where Tony Jaa's father was once a monk. The temple is dedicated to the elephants. There are altars with elephant bones and the entire skeleton of a baby pachyderm. Set into stone grottoes and covered in vines are massive statues of the Buddha, the elephant god Ganesh and the monkey god Hanuman. Hanuman is respected throughout Thailand and considered the most intelligent of the animal gods. He practices muay boran, the ancient art Tony Jaa studied, the moves of which are based on the simian. To be continued ...
About the Author: Antonio Graceffo is a freelance writer currently based in Asia. His book, Warrior Odyssey: The Travels of a Martial Artist Through Asia, is Antonio Graceffo's record of where culture, communication and martial arts meet during his decadelong travels through nine countries, including Taiwan, Vietnam, Laos, Cambodia and the Philippines.
Join Antonio Graceffo for a trip around the world in these exciting martial arts travelogue videos:
Martial Arts in Southeast Asia: Thailand
Martial Arts in Southeast Asia: Taiwan (part 1)
Martial Arts in Southeast Asia: Vietnam (part 1)
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redpajama_set_name: RedPajamaCommonCrawl


text:
Timeline of the helicopter
Helicopter blades
By: Greg, Abi, Blake
Feb, 5 2015
Greg: Historical part of the science research paper
Paul Cornu, a French scientist invented the first helicopter in 1907 but it
was unsuccessful because it was only in the air for 20 seconds and only
went about 1 metre in the air. However, it was Igor Sikorski who
invented the first successful helicopter, which were early models of the
VS-300 that successfully flew into the air in 1939.
- Paul Cornu's unsuccessful helicopter in 1907
Igor Sikorsky's VS-300 from September 1939.
From 1907-1922, there were small helicopters being developed by a
few people. From 1923-1933, Juan de la Cierva invented the world's
first practical rotorcraft called the Autogiros. From 1934-1940, France
and Germany developed helicopters and added more effective
mechanisms for lift, propulsion, stability and control. During World War
II, from 1941-1945, 6 countries used the helicopter like Germany,
United States, Great Britain, Canada, Soviet Union and Japan, revealing
the use of the helicopter to the world. The Germans had 27 models;
Great Britain 8 models, United States 7 models and Canada and Japan
only had 2 models.
After World War II, from 1946-1949 was a period of struggle for
helicopters therefore helicopter manufacturers were getting less
business and they were making less money because many people
started using aeroplanes. From 1950-1954 there were wars in Korea as
well as the British and French colonies, which was a period of success
for helicopters due to the fact that they were very useful for taking off
and landing without runways. This was useful to transport soldiers from
one place to the next, because these countries didn't have many roads
and almost no runways. From 1955-1961 there was a big improvement
with the turbine engines for helicopters. From 1962-1972 helicopters
were used in the Vietnam War, which forever changed the way people
thought of fighting wars and changed the way the people thought about
the way that the public thought about helicopters.
From 1973-1989, commercial operations became more profitable
because it's demand and supply, there were a lot of helicopters and
pilots after the Vietnam War, and a demand, much of it from oil
companies because of the oil crisis. Nowadays helicopters are faster,
more reliable and more comfortable because of the advanced
technology that we have, leading to more and more new features like
anti-torque features, better engines etc…
Abi: Social perspective of the research paper
Helicopters have many different purposes like in military
operations for example. They were used in Korea, Vietnam, Afghanistan
and Ethiopia. They patrolled and rescued people in the air and at sea.
They also evacuated the injured people and transported equipment.
They're also used to attack. These types of helicopters can have
weapons such as auto cannons, machine-guns, rockets, missiles and
more. Lots of ideas came from this, for example some hospitals now
have landing pads for the helicopter ambulances to land on. People
rescued from catastrophes like earthquakes and floods usually are
evacuated on these helicopters. An advantage of using helicopters
instead of planes is that they can take-off and land vertically instead of
having to drive on a runway.
Helicopters are also used on construction sites to lift heavy
equipment for example steel beams and also sometimes to install
television towers. They are also cheaper than using a crane. And they
are also a safe way of getting up t high floors or places on the
construction site. Transportation by helicopter is getting more popular.
They're being used as taxis and for sight seeing. Companies also use
them for flying their staff and clients places.
Police now use them to prevent traffic jams and to find criminals
they're chasing. They're also beneficial to the United States Forest
Service in fighting forest fires; water is poured from hoses carried in the
craft. Helicopters are also useful for photography and film, to take aerial
action shots, and get different angles, for example in for the News.
Blake: Technological perspective of the research paper
In 1480, Leonardo da Vinci designed a sketch aircraft that held a
vertically mounted screw. He could not figure out how to make the
screws to turn fast enough to allow his aircraft to fly, because he did not
understand the physics of how the torque works. Torque when an
object is pushed or pulled to make it rotate. The torque would make the
aircraft turn while the screw stayed without moving. Despite all of this,
there were no motors at the time to spin the screw. In 1877, due to the
lack of motors, Forlanini created a helicopter powered by steam that
was able to be lifted 15m off the ground for about one minute. In the
20th century, Mumford and Bréguet each created helicopters powered
by a gasoline engine and this engine enabled it to rise above the ground
because of the spinning blades. In their own experiments, they were
never able to lift the helicopter more than 7.5m off the ground. In 1939,
a Russian engineer Igor Sikorsy flew the first single blade helicopter.
Because he flew 70km/h this solved his problem of torque. In 1942, the
US army used Sikorsy's helicopters in the second world war.
Greg's resources:
http://www.helicopter-history.org/
Rotary Wing History. 2002. AHS Helicopter History Committee.
http://school.eb.com/levels/middle/article/277062
"Igor Sikorsky." Britannica School. Encyclopedia Britannica,
Inc., 2014. Web. 4 Dec. 2014.
Abi's resources:
"Helicopter" (no author) Britannica school middle 2015
Encyclopedia Britannica, inc.
Blake's resources:
"AVIATION: REACHING FOR THE SKY"
Author: Don Berliner
The Oliver Press,INC.
Practice Problems 10/10 Test
29A_Early Helicopters
Informative Speech Topics from http://www.best-speech
Helicopter Operation in Severe Environmental Conditions


text:
Q: PHP SOAP returning NULL Good morning,
I am setting up a PHP application to communicate with a SOAP web service but have run into an issue. When calling the "GetIntradaySnapshotForCommodity" function on the service, PHP is returning NULL. However, when I inspect the "__getLastResponse()" content, a valid XML response is being returned.
The following is my code for the SOAP interaction.
$client = new SoapClient('liffecommoditiesservice.wsdl', array('trace' => TRUE));

$result = $client->GetIntradaySnapshotForCommodity(array(
    'securityToken' => 'XXX',
    'commoditiySymbol' => 'XT'
));

var_dump($result); //This returns NULL

var_dump($client->__getLastRequest()); //This response is included below
var_dump($client->__getLastResponse()); //This response is included below

$client->__getLastRequest()
<?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://liffecommoditieswebservice.if5.com/"><SOAP-ENV:Body><ns1:GetIntradaySnapshotForCommodity><ns1:securityToken>XXX</ns1:securityToken><ns1:commoditiySymbol>XT</ns1:commoditiySymbol></ns1:GetIntradaySnapshotForCommodity></SOAP-ENV:Body></SOAP-ENV:Envelope>

$client->__getLastResponse()
<?xml version="1.0" encoding="utf-8"?><soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"><soap:Body><ArrayOfIntradaySnapshot xmlns="http://liffecommoditieswebservice.if5.com/"><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Jan-13</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>191 </Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>192.95 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Jan-14</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>161.5 </Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>165.75 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Jul-13</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>-</Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>201.3 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Mar-13</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>-</Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>195.2 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Mar-14</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>-</Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>168.1 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>May-13</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>195 </LastTradePrice><LastTradeVolume>30 </LastTradeVolume><LastTradeTime>08:57:17 </LastTradeTime><Bid>194.25 </Bid><Ask>199 </Ask><High>195 </High><Low>195 </Low> <SettlementPrice>198 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>30 </CurrentCumulativeVolume><PriceChange>-3</PriceChange> <PercentagePriceChange>-1.52%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>May-14</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>-</Bid><Ask>-</Ask><High>-</High><Low>-</Low> <SettlementPrice>172.25 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>-</CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Nov-12</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>190 </LastTradePrice><LastTradeVolume>20 </LastTradeVolume><LastTradeTime>09:20:13 </LastTradeTime><Bid>190 </Bid><Ask>190.5 </Ask><High>191 </High><Low>189 </Low> <SettlementPrice>191.95 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>129 </CurrentCumulativeVolume><PriceChange>-1.94999999999999</PriceChange> <PercentagePriceChange>-1.02%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Nov-13</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>161 </Bid><Ask>167 </Ask><High>-</High><Low>-</Low> <SettlementPrice>165.25 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>0 </CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot><IntradaySnapshot><CommoditySymbol>XT</CommoditySymbol><DeliveryMonth>Nov-14</DeliveryMonth><TimeStamp>01-Aug-2012 10:48:01</TimeStamp><LastTradePrice>-</LastTradePrice><LastTradeVolume>-</LastTradeVolume><LastTradeTime>-</LastTradeTime><Bid>-</Bid><Ask>164 </Ask><High>-</High><Low>-</Low> <SettlementPrice>164 </SettlementPrice><OpenInterest>-</OpenInterest><CurrentCumulativeVolume>0 </CurrentCumulativeVolume><PriceChange>0</PriceChange> <PercentagePriceChange>0%</PercentagePriceChange></IntradaySnapshot></ArrayOfIntradaySnapshot></soap:Body></soap:Envelope>

$client->__getFunctions()
array(4) {
    [0]=>
        string(101) "GetEODSnapshotForCommodityResponse GetEODSnapshotForCommodity(GetEODSnapshotForCommodity $parameters)"
    [1]=>
        string(116) "GetIntradaySnapshotForCommodityResponse GetIntradaySnapshotForCommodity(GetIntradaySnapshotForCommodity $parameters)"
    [2]=>
        string(101) "GetEODSnapshotForCommodityResponse GetEODSnapshotForCommodity(GetEODSnapshotForCommodity $parameters)"
    [3]=>
        string(116) "GetIntradaySnapshotForCommodityResponse GetIntradaySnapshotForCommodity(GetIntradaySnapshotForCommodity $parameters)"
}

A: For anybody that is interested, this issue appears to have been caused by a bug in the PHP PEAR library whereby a perfectly valid SOAP response was not being converted to an object properly.
I ended up using the SOAP PEAR library instead which worked like a dream.

redpajama_set_name: RedPajamaStackExchange


Monday marked the 75th anniversary of the National Labor Relations Act – also known as the Wagner Act – one of the lesser known, but key components of President Franklin Roosevelt's New Deal. In addition to Social Security, Unemployment Insurance, a Federal minimum wage and laws regulating child labor – all controversial concepts at the time that we now take for granted as basic elements of fairness – the New Deal included the National Labor Relations Act (NLRA) which protected workers' rights to join or form unions and engage in collective bargaining.
But President Roosevelt and his Labor Secretary, Frances Perkins, understood the importance of giving workers bargaining power and saw the NLRA as central to his strategy to rebuild the economy. The famous economist John Maynard Keynes would later write a letter to FDR praising his efforts to restore the American economy and emphasizing the expansion of collective bargaining as an essential effort to the economic recovery.
Collective bargaining helped create our middle class. Working people were able to share in the gains of their productivity and labor and management together forged creative solutions to create the powerful engine of the American economy we all are proud of.
In order to rebuild the middle class today, we need to level the playing field for all working people and update our labor laws to fit the 21st century workplace. That's why the President and I support the Employee Free Choice Act – which would update the NLRA so workers can form unions if they choose to without fear or pressure. In addition, millions of workers are not covered by the NLRA including public sector workers, farm workers, domestic workers, and more – so other laws, like the Public Safety Cooperation Act would ensure that firefighters and other public servants have a voice on the job, too.
Some people say that given the state of the economy, we can't afford unions right now. They've got it backwards.
Workers are facing unprecedented challenges and they need the voice on the job that unions provide. Leveling the playing field is a win for workers and employers. Most employers believe that investing in their workforce makes good business sense, but businesses that play by the rules and respect workers rights should not have to compete unfairly with those who do not.


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Exactly one child sits in each chair.
No boy sits next to another boy.
Ivan sits next to and east of the fourth child in the row.
Sylvia sits east of Ivan.
Frank sits next to Ruby.
What is the maximum possible number of different pairs of chairs in which Frank and Ruby could sit?
Both Harry and Joel sit east of Frank.
Both Harry and Ruby sit east of Frank.
Both Harry and Joel sit west of Frank.
Both Harry and Ruby sit west of Frank.
Both Joel and Ruby sit east of Frank.
If Thelma sits next to Ivan, and if Frank sits next to Thelma, which one of the following statements could be false?
Both Frank and Ivan sit east of Ruby.
Both Frank and Ruby sit west of Thelma.
Both Frank and Sylvia sit east of Ruby.
Both Frank and Thelma sit west of Sylvia.
Both Frank and Ruby sit west of Joel.
If Frank does not sit next to any child who sits next to Ivan, which one of the following statements could be true?
Harry sits west of Frank.
Joel sits west of Ivan.
Ruby sits west of Frank.
Thelma sits west of Frank.
Thelma sits west of Ruby.
If Frank sits east of Ruby, which one of the following pairs of children CANNOT sit next to each other?
1) Which children must, can, or cannot sit east and west of which other children?
2) Which children sit next to which other children?
The Initial Setup: Since the east-west thing seems to play a big part in the questions, there's no harm in jotting down a W and E to remind yourself at a glance which children are east of and west of each other. Get the boys and girls down on the page, throw in seven chairs, and you're ready to see how the seating arrangement will be determined.
1) Talk about strict. No fun for these kids; one kid per chair. Just a loophole closer that you most likely assumed anyway.
3) Here's the most concrete rule of the bunch, one that you may have jumped to right off the bat: It requires a touch of interpretation, but essentially means that Ivan is 5th. Chairs 5, 6, and 7 are the chairs east of chair 4, and the only one of those next to 4 is 5.
4) "East of" in this set means to the right, so this boils down to our usual sequencing notation I . . . S. Perhaps you thought this through in accordance with the deduction stated above. We'll hold off until Step 4 and deal with it in Key Deductions below.
5) FR or RF must appear somewhere in the ordering.
The major challenge of Q. 1 is figuring out exactly what they're looking for, so a careful reading is in order. First of all, what is a "pair of chairs in which Frank and Ruby can sit"? Let's check it out: We see from our sketch that Frank can be in chair 1 with Ruby in chair 2, so chairs 1 and 2 together count as one acceptable pair for this couple linked by Rule 5. Keeping Ruby in 2, Frank can move to 3 without a hitch, so there's a second pair, chairs 2 and 3. But Ruby can also be in chair 4 when Frank's in 3, so 3 and 4 together counts as our third pair. Any others? Nope, so three it is, choice (C).
Next we're to find a statement that must be false, which means that each wrong choice will either be true or possibly true. Notice how all the choices involve Frank, which is nice because we know plenty about Frank's prospects. In fact, simply recognizing that Frank can be in the first chair allows us to kill choices (A), (B), and (E), since if Frank were to take chair 1, everyone would be east of him. Upon analysis, we see that (C) is the impossibility we seek: We deduced up front that Frank must sit in either chair 1 or 3. But even if we slide Frank down as far east as possible—that is, chair 3—there's no way to fit both free agents Harry and Joel west of him, considering that either Ruby or Thelma must take chair 2. (D) is possible: One of our boy floaters, in this case Harry, can certainly hang out with Ruby on the west side of Frank—H in 1, R in 2, and F in 3 would do the trick. But we can't get both Harry and Joel west of Frank. As mentioned above, one of those guys has to be in the easternmost chair 7.
Good solid info is offered in this one, so here's a perfect time to use your pencil—get it all down in a new sketch. Ivan's in chair 5, so if Thelma's next to him, she must be in chair 4, considering that we've already slotted Sylvia for chair 6. The stem also tells us that Frank is next to Thelma, so put F in chair 3, which leaves R for chair 2. Free agents Harry and Joel will float between chairs 1 and 7. We're looking for something that could be false, and we can shorten our work by scanning the choices for Harry and Joel; after all, they're the only ones not fully restricted here. Only Joel appears, and way down in choice (E). Lo and behold, Joel's elusive nature does in fact make (E) the correct choice. (E) would be true if Joel's in chair 7, but would be false if Joel's in chair 1.
If you're not convinced yet as to the importance of the Frank deduction, here's another question that falls pretty easily thanks to the fact that Frank must be in 1 or 3. If Frank took chair 1, he wouldn't be east of Ruby as the stem demands; so Frank must sit in chair 3 with Ruby in chair 2. That again lands Thelma in chair 4 (familiar territory for her), which again leaves 1 and 7 for Harry and Joel to fight over. Again, everyone's placed but those two. From this ordering, we can see quite readily that the only pair that cannot be next to each other is the duo of Ivan and Ruby—Frank and Thelma come between them. Choice (A) contains folks that are next to each other, while the others contain pairs that could be, depending on where Harry and Joel sit.


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Home Business News 'Everybody lives happily till next blackout' quips energy expert
'Everybody lives happily till next blackout' quips energy expert
26 April 2016 11:06 am - 0 - {{hitsCtrl.values.hits}}
Both the current and past governments have swept post-blackout reports and recommendations which may have averted successive blackouts under the rug, which has resulted in an ongoing crisis situation, one of the country's leading power and energy consultants said. "The reports have gone to someone's WPB.
Everyone knows what WPB is; it's the waste paper basket," Dr. Tilak Siyambalapitiya said at a seminar on the electricity crisis themed 'Is there light at the end of the tunnel', held at the Ceylon Chamber of Commerce yesterday. None of the Public Utilities Commission of Sri Lanka (PUCSL) or Ceylon Electricity Board (CEB) present at the discussion refuted the claims. Dr. Siyambalapitiya said that the report following the 2009 blackout, which outlined 37 recommendations and was written by him as part of the investigative committee, was never given any serious thought, or was published by the PUCSL.
"Everybody lives happily till the next blackout. The 2015 September blackout report was ignored till after the 2016 blackout, when people looked around, asking 'where is that 2015 report'," Dr. Siyambalapitiya added. CEB Chairman Anura Wijayapala, who outlined some challenges for the future, went on to say that he was confident that the state-run power generator will overcome any adversity.
"The PUCSL last week wrote to the CEB and the CEB said that all is well and that everything that can be done has been done. The PUCSL didn't ask for any timeframe for action and the CEB didn't give one," Dr. Siyambalapitiya said. However, it is interesting to note that when the CEB requested additional power requisition this week, the PUCSL shot down the proposal. Dr. Siyambalapitiya noted that in the US, when the last major blackout happened in 2003, there were seminars, research and public criticism for months, all of which is still available to the public today, unlike in Sri Lanka, where no public discussion has taken place, in addition to the shortcomings in report publication. He noted that the current crisis is due to a mix of shortcomings in reliability, accountability and capacity. "Up to a certain measure, it's all of it. Is there light at the end of the tunnel?
The indications are that there is no such light," he said. He said that there were no reliability indices or modern security analysis published, which are done in real time in other countries, where power generation company officials are fined for unreliable power supply and the proceeds are distributed among the customers.
Dr. Siyambalapitiya said that Sri Lanka cannot keep comparing itself to India anymore either. "In some states in India, they are achieving 24/7 power. Even in India, there's only a 2 percent power outage. Now we shout and say India's power outage is this, and we should do 10 times better," he said. He added that the CEB has not been accountable under the Electricity Act of 2009, not publishing reports on transmission performance standards, distribution performance standards or the grid codes for the past four years, except just once.
He noted that while there is no significant crisis in capacity yet, it will definitely come in 2018, since the commissioning of the Sampur Power Plant has been delayed till 2021. Wijayapala agreed, saying that the power demand is expected to increase by 5 percent each year and only micro projects are expected to come online till 2021. "I told the Prime Minister that the demand is expected to increase by 5 percent. He got a bit angry and said that we should cater to a 10 percent increase or more, because the government will be bringing in a lot of industries," he said. However, Dr. Siyambalapitiya said that the authorities should create comprehensive, holistic plans to satisfy the 5 percent increase before looking any further. Pix by Pradeep Pathirana
Litro Chairman reappointed several hours after removal
Son of arrested doctor makes controversial statement against police


Miura Iron headcovers in carbon fiber. With magnet enclosures on each cover, dropped headcovers are greatly minimized. The magnets attracts itself to the iron head. You can easily lay your headcover on the ground while you take your shot. Tap your iron head on the cover and you can pick it up without bending down. This is undoubtedly the best iron headcover around.


Corey Graves is the World Champion Of Love (at least that's what he wants to be known as from now on). He isn't shy about his opinions at all and it shows in his commentary style. If you thought that Graves is unglued on RAW and SmackDown Live, it's nothing compared to his Twitter.
Graves clapped back at a fan who was condemning his commentary style. The biggest issue some fans had was the fact that the WWE commentator took a direct shot at government assistance.
"Well, 'Blunt Boss,' clearly you're setting the world on fire. Maybe you could take over?
There was a slew of hate in Corey Graves' direction after that. Graves tried to explain his comment, but it might have been too late.
You can check out some of the tweets he received below as well as a couple of his replies. It sounds like he doesn't care and figures that this kind of thing will get him a couple extra zeros on his paycheck when it is all said and done.


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'We Don't Hate Jews,' Claim Activists Who Want to Eliminate Jewish State During Orwellian Panel
An anti-Israel group hosted an Orwellian 'anti-Semitism panel' that featured speakers who promote boycotts of the Jewish state and spread modern day blood libels.
Outrage Erupts at Anti-Semitism Panel Featuring Noted Anti-Semites
'Well-known anti-Semites have no place defining anti-Semitism,' commented the pro-Israel progressive group Zioness.
New Video Rebuts Smear that US-Israel Police Exchanges Lead to Death of Black Americans
The Investigative Project on Terrorism's two-part film disproves the 'anti-Semitic, intersectional' narrative about the Israel Police force.
Opinion: What to Teach (and Not Teach) Your Children about Israel
The pandemic has become an educational disaster, but as the Seth Rogen kerfuffle reminds us, it's also an opportunity to inform and inspire kids about Jewish life.
300 Percent Spike in Harassment of Jews at Universities with 'Anti-Zionist' Activity
A new report shows that Jewish students' complaints of Israel-related harassment have regularly been dismissed by the Department of Education and ignored by university administrators.
WATCH: Roger Waters Sings About Destroying Israel
'We'll take back the land, from the Jordan River to the [Mediterranean] Sea,' anti-Israel activist Roger Waters sang.
Exposed: Left-wing Jewish Group Manages Facebook Page from Lebanon!
The anti-Israel organization recently scrubbed its Facebook account of Lebanon-based page manager.
WATCH: 'Jewish Voice for Peace' Caught Peddling Anti-Israel Lies
More lies about Israel exposed and dispelled.
New Report: Chicago-Area Colleges 'New Hub' of Anti-Israel Activism
According to a new report published by the Israel On Campus Coalition (ICC) on Tuesday, 'Anti-Israel activity declined or remained constant in every region of the country except the Midwest, where campuses experienced a sharp increase in anti-Israel activism.'
Spineless UC Davis Faculty Capitulates to Violent Anti-Israel Students
Pro-Israel campus activists are furious at the University of California, Davis administration over its weak response to violent and disruptive activities by anti-Israel groups on campus.
Fake 'New York Times' Newspaper Distributed by Activists to Bash Israel
Anti-Israel activists distributed a fake edition of The New York Times across the Big Apple on Tuesday morning.
Report Shows 30% Increase in Anti-Israel Activity on US Campuses
Anti-Israel and anti-Semitic activity on US campuses is becoming more aggressive and vicious, a report by the ADL shows.


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Tucson to continue every-other week recycling program
22February20
Tucson will keep its curbside recycling pickup at every other week, making the move more than four months after launching a pilot program to account for rising costs amid a change in the global recycling market.
The City Council unanimously voted to put the change into law Wednesday after Carlos De La Torre, the city's director of environmental and general services, said the pilot program that began Sept. 30 was a success.
"We're seeing that essentially we're no longer picking up too much air and we're basically picking up real recyclables," he said.
De La Torre said the program specifically found that more than 80% of blue bins were at least half full on the every-other-week schedule, while the tonnage per load increased by 35%, compared to the previous weekly pickups.
That coincided with a reduction in fuel, maintenance and overtime costs.
The city has received about 1,400 calls since the change went into effect, the majority regarding the pickup schedule, which designates "A" and "B" areas to stagger pickup times, he said.
The city also received nearly 750 requests for a secondary recycling bin; 315 of the requests were fulfilled, 66 were denied and the remainder are pending.
He noted that Tucson has made about $32 a ton on recycling this fiscal year so far, while it cost about $100 a ton, resulting in a net loss; that's compared to $94 and $30 just five years ago.
One of the biggest costs, he said, is contamination. Tucson is being charged about $33,000 a month — or $400,000 a year — in fees because of non-recyclables being placed into the blue bins, a list that included items such as diapers, paint and excess lumber.
That fee would be eliminated should the city reduce the contamination number in half, he said.
The city's contamination rate of about ª of total recyclables is higher than the rate of peer recycling cities, including El Paso, Austin, Mesa, Seattle and Denver, but lower than Albuquerque and Phoenix, he said.
De La Torre said Tucson has a three-strike program to remove offenders from the program, but it is not being enforced.
The council's vote included a recommendation to enforce the three-strike program.
"The way I see it is that people who are contaminating our recycling are costing us $400,000 a year," Mayor Regina Romero said. "What I'm hearing from you is that you want to charge all of us for what a certain percentage users are doing to contaminate our recycling. I don't like that."
Tucson's decision to cut down recycling pickups came as China and other nations opted to no longer accept many recyclables from the United States, forcing municipalities around the country to seek ways to reduce rising recycling costs.
Local officials opted in August to implement the every-other-week recycling program, estimating at the time that it would save the city roughly $1.4 million a year.
The city also took $2 million from the hotel bed surcharge to cover increased recycling processing costs.
The city expects to have a total savings estimate at the six-month mark.
De La Torre's presentation also included a recommendation to stop glass recycling, noting that the practice makes very little revenue and has less environmental impact compared to other recyclables, such as metals and paper.
The council rejected the recommendation, instead asking De La Torre to come up with a specific replacement plan, including modifications to the city's contract with Republic Services and whether to set up glass drop-off sites around the city.
Vice Mayor Paul Cunningham said he wants that discussion to be part of a larger scale effort to "rethink how we're doing recycling."
"I want to be the trend setter, not the trend follower," he said. "We have some opportunity here."
Cunningham offered his support for another pilot program, that of fellow Council member Steve Kozachik, who has started crushing glass bottles and turning them into sand in the garage of his Ward 6 office. Kozachik has previously pitched using that sand for city roads, sandbags and sidewalks.
"If we're one of those cities that's milling even 10% of our new road materials internally from our Environmental Services department to Transportation department, that's going to make us unique. That's going to make Tucson special," Cunningham said.
Kozachik, who passed around a mason jar of the sand during Wednesday's meeting, said "the community has bought into this notion of finding a secondary market."
He estimated that he's crushed 4,000 gallons of bottles — provided to him from local businesses — into 500 gallons of sand, which he's used to fill sandbags and for concrete for the sidewalk in front of his office.
He told the Arizona Daily Star he's sent a cost estimate for a larger-scale crusher for the city, which would cost about $40,000.
"There's a market and a way to do this," Kozachik said. "We need to think collaboratively to figure out how to do it without breaking the bank."
This article was originally published by Justin Sayers, tucson.com
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Soffits, fascias and new guttering installation in St. Albans, Hertfordshire
November 22, 2016 6:34 pm Published by Summit Cladding
Summit Cladding was recently contacted by our customers, who live in St Albans, Hertfordshire. They moved into their detached property five years ago and had noticed the deteriorating condition of structural timber around their investment. They had replaced the original windows four years ago and realised that uPVC was a great alternative to the continuous maintenance required by wooden fascias and soffits. One of our client's concerns was how the work was going to be done. They had a conservatory at the rear of their address and a garage at the side with a tiled roof that made access challenging. Summit Cladding arranged an on-site meeting to discuss the logistics of the installation with the client. The solution to the access... View Article
Door canopy installation in Barnet, north London
Summit were recently contacted by a client who lives in Barnet, north London. The client was very specific in her requirements and had lived at the property for over ten years. Having completely modernised the interior, she wanted to give the house an outside makeover, giving it the kerb appeal it deserved. After several site visits, Summit Cladding came up with an ideal solution that our client loved. We suggested constructing a pitched roof door canopy over the front door. This would alter the flat look of the front of the address and create more depth. To start, we hand-carved two wooden gallows brackets that would support the new timber frame. This we combined with a plain tile exactly the... View Article
James Hardie cladding installation in north London
November 4, 2016 4:34 pm Published by Summit Cladding
Summit Cladding were contacted recently by clients we had worked for previously, who live in Enfield, north London. We had previously replaced all the original wooden fascias and guttering at their house six years. The address did not grant easy access to it's roofline, which was causing issues and scaffolding was always necessary to gain access. The wooden soffits and fascias were beginning to rot as the timber really needed painting every three years, which had been difficult for the client to achieve. The second issue facing them was the original render on the house was now starting to crack. Although the house had been constructed in the nineties, it unfortunately was starting to look tired and dated. With years... View Article


A new study suggests that long-term exposure to secondhand smoke during childhood increases the risk of chronic obstructive pulmonary disease death in adulthood.
Childhood exposure to passive smoking or secondhand smoke may increase chronic obstructive pulmonary disease and death risk in adulthood.
The diabetes epidemic in Guatemala is worse than previously thought: more than 25 percent of its indigenous people, who make up 60 percent of the population, suffer from type 2 diabetes or pre-diabetes.
Swedish adults with type 2 diabetes with five selected risk factor variables within the target range had only marginally higher risks for death, stroke and myocardial infarction vs. the overall population without diabetes, although excess risk for hospitalization for heart failure persisted, according to findings published in The New England Journal of Medicine.
Men and women who gain weight in the 5 to 7 years after quitting smoking experience a short-term, increased risk for type 2 diabetes vs. current smokers and those who do not gain weight after smoking cessation, according to findings published in The New England Journal of Medicine.


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« Catholica comes to tea
Come back for Christmas »
Denied Sunday Mass again
By Church Mouse | 6 December, 2008 - 10:46 pm | 1 January, 2020 Our pastors
At the prayers of the faithful this morning, Len started with the following:
I came across some words on the internet the other day:
"Everyone has a place in the church, every person without exception should be able to feel at home and never rejected"
They were on a banner outside St Mary's church in South Brisbane.
Some might describe these as words of "pretend christianity", but they were actually spoken by Pope Benedict XVI.
I pray for a church where positive words like these are put into action.
He got as far as "pretend christianity" (a phrase borrowed from the sermon to which the congregation had just been subjected), when the celebrant, Pelle, tried to terminate the prayer and stop further offerings. Len attempted to finish his prayer.
The two Neocat clerics then threatened to terminate the Mass, and when several other members of the congregation ignored their bullying tactics by offering their own prayers, they scurried from the altar to the sacristy.
Several disappointed parishioners tried to reason with them, only to be berated by the same old Neocat mantra:
You have no respect for the liturgy
You have no respect for the Roman Catholic Church
And a new one from Mendes:
You treat this as your club – this is not your club
This is my church
who then made out that he was going to lock the church, but changed his mind, and the "priests" headed for the church door with their small entourage.
Len followed, trying to reason with them, only to have the door slammed in his face by the Neocat woman, M. Fortunately he suffered only a glancing blow to the side of the head. The assault was witnessed by several of those present.
The community, sans Neocats, regathered and said the Lord's prayer.
Jack suggests that The Hollow Men, by T. S. Eliot offers a detached, subjective description that might be applied to our pastors:
We are the hollow men
We are the stuffed men
Leaning together
Headpiece filled with straw. Alas!
Our dried voices, when
We whisper together
Are quiet and meaningless
As wind in dry grass
Or rats' feet over broken glass
In our dry cellar
Shape without form, shade without colour,
Paralysed force, gesture without motion;
Those who have crossed
With direct eyes, to death's other Kingdom
Remember us—if at all—not as lost
Violent souls, but only
As the hollow men
The stuffed men.
The St Vincent's community lives with the constant threat of having their Masses terminated. The notice discussed in Update from February this year appears in every Sunday's church bulletin.
From the General Instruction for the Roman Missal:
The Prayer of the Faithful
69. In the General Intercessions or the Prayer of the Faithful, the people respond in a certain way to the word of God which they have welcomed in faith and, exercising the office of their baptismal priesthood, offer prayers to God for the salvation of all. It is fitting that such a prayer be included, as a rule, in Masses celebrated with a congregation, so that petitions will be offered for the holy Church, for civil authorities, for those weighed down by various needs, for all men and women, and for the salvation of the whole world.
70. As a rule, the series of intentions is to be:
a. for the needs of the Church;
b. for public authorities and the salvation of the whole world;
c. for those burdened by any kind of difficulty;
d. for the local community.
Nevertheless, in a particular celebration, such as Confirmation, Marriage, or a Funeral, the series of intentions may reflect more closely the particular occasion.
71. It is for the priest celebrant to direct this prayer from the chair. He himself begins it with a brief introduction, by which he invites the faithful to pray, and likewise he concludes it with a prayer. The intentions announced should be sober, be composed freely but prudently, and be succinct, and they should express the prayer of the entire community.
The intentions are announced from the ambo or from another suitable place, by the deacon or by a cantor, a lector, or one of the lay faithful. The people, however, stand and give expression to their prayer either by an invocation said together after each intention or by praying in silence.


The Smoke More Red Meat Special!
Another week passes and both of us successfully avoid a one way trip to Fishtown. So get ready for nine great songs from Suicide Generation, Musta Paraati, Neville & Sugary Staple, The Cavemen, Restless, Bloodcocks UK, Puppy And The Hand Jobs, Fuzzbox and The Dirtiest.
Poetry Corner, Tony has your Facebook comments, Mild Quirkiness, Happy Birthday Ann K, south of Fishtown, last week, Tony's been ill, smoke more weed, Paul went to Oslo, super expensive red wine, the worst reason ever for a failed MOT, From the Vaults, Tony's International Gig Guide, PGOTW, Sumac/Sumac, Crash Induction, this week, work and MOT fun, Polyfest, no Izzatwat this week, NTNS, Tony has your Podchaser reviews and a reminder of the stations which feature us.
Song 8: Fuzzbox - WGAF... AWGUI!


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Subida al cielo es una película de comedia mexicana de 1952 dirigida por Luis Buñuel y protagonizada por Lilia Prado.

El origen de esta película está en el poeta español exiliado Manuel Altolaguirre, amigo de Buñuel, ya desde la época de la Residencia de Estudiantes de Madrid. Tras un accidentado viaje con su esposa, una millonaria cubana, propuso a Buñuel financiar una película que narrara de modo estrafalario acontecimientos curiosos que suceden en un viaje, siempre y cuando Buñuel aceptara dirigirla.

Argumento

En una ciudad de la costa del estado de Guerrero, Oliverio, quien se acaba de casar, se ve obligado a regresar en pleno viaje de novios porque su madre se está muriendo. En su lecho de muerte, la madre pide a su hijo que busque a un letrado para que redacte su testamento. Oliverio viajará en autobús para cumplir el deseo de su madre moribunda. Durante el viaje sucederán muchos imprevistos y Oliverio tendrá que sortear los coqueteos de una mujer.

Comentarios, crítica y algunos datos
Tras su sonado éxito con Los olvidados, Luis Buñuel se tomó la libertad de realizar a su libre imaginación un tema que le propuso su amigo Manuel Altolaguirre, con la financiación de su esposa. La temática, tan trivial como puede ser un viaje en autobús, se torna insólita. Buñuel, que se caracterizó por realizar filmes con actores de no mucho renombre, realizó una interesante cinta donde cada personaje, pese a su simplicidad, realizaba un buen desempeño con diálogos chispeantes. Sin duda las aportaciones que hicieran el guionista Juan de la Cabada y la joven guerrerense Lilia Solano fueron de gran ayuda para darle frescura a la cinta. En el filme los villanos (si puede hablarse de villanos en los filmes de Buñuel) son los hermanos de Oliverio, dos ambiciosos e hipócritas que pretenden quedarse con la mejor parte de la herencia de su madre, que agoniza. Pero Buñuel no cayó en dulzuras y así, la recién casada Albina, pese a su abnegación, confiesa que hubiera querido pasar su luna de miel en la casa que la familia posee en la Ciudad de México. Oliverio, por su parte, es un joven con el complejo de Edipo, que ve en Albina a su madre, pero a quien no desea sexualmente, sino a la sensual y provocativa Raquel (Lilia Prado). Otros personajes llamativos son: el candidato a diputado que, pese a dar lisonjeros discursos, quiere resolver todo a balazos; un chofer que ama a su madre y que es a la vez un irresponsable; un cojo metiche y burlón que ya adelanta la corte de mendigos de Viridiana; una mujer a quien se le ocurre parir a media subida de un monte; una madre que batalla con sus dos hijos; un anciano de la época porfiriana que quiere que le devuelvan sus tierras, y hasta un vendedor de gallinas desfilan en esta cinta.

Sin embargo, la película vale principalmente por el humor erótico en clave surrealista que tanto cultivó Buñuel, como lo fue el sueño de Oliverio, en donde el protagonista lanza al agua a la sensual Raquel (que intercambia su rostro con Albina), luego se encuentra con Raquel en el camión, solos (un anhelo que él niega revelar), y devuelve a su madre una interminable mondadura de manzana, que simboliza un cordón umbilical y que significa que él se libera de ella y se entrega a Raquel, pero un tropel de ovejas y cabras irrumpen en el camión y Oliverio regresa con Albina, mientras cuatro de los pasajeros se han convertido en músicos. Un hombre inexplicablemente cruza con una oveja negra y la hermosa y abnegada Albina se convierte en la sensual Raquel, quien ríe burlonamente y a continuación emerge de una laguna, coqueta, semejando una criatura que sale de la placenta. Antes de que Oliverio despierte, Raquel, quien apenas unos minutos invitó a Oliverio a nadar en una poza, y le ha ofrecido una mondadura de manzana y la manzana misma que éste mordió, le pregunta con la mayor coquetería posible: "¿Estabas pensando en mí?". En la ida, el camión es escenario de un nacimiento; en el regreso, acarrea con la niña muerta, quien un día anterior había sacado con una par de bueyes al camión atascado en un río. Al final, el aún casto Oliverio cede ante Raquel en el puerto de montaña llamado "Subida al cielo"; después de ello, Raquel se muestra indiferente hacia un Oliverio sonriente, otrora arisco ante la seductora mujer. Al regresar a su pueblo y encontrar a su madre muerta, Oliverio recuerda a Raquel dando un mordisco a una manzana que enseguida escupe para decir burlonamente: "Lo que quería ya lo tuve", por lo que decide que tiene que engañar a sus hermanos para que estos no se salgan con la suya. Así, surge la tétrica pregunta de si son más válidas las cruces que hizo viva que las huellas digitales muerta.

La cinta fue la primera colaboración de Lilia Prado con Buñuel, con quien participaría en dos cintas más: Abismos de pasión y La ilusión viaja en tranvía; asimismo, sería la segunda participación de Roberto Cobo con el cineasta, cuando ya previamente había participado en Los olvidados.

Premios

Entre otros 16 premios internacionales, recibió en Francia el Premio de la crítica a la mejor película de vanguardia del Festival de Cannes (1952). En México, sin embargo, sólo obtuvo 6 nominaciones.

 1952, Festival de Cannes, Edición 6.º, Nominada, Gran Premio
 1953, Premio Ariel - Academia Mexicana de Artes y Ciencias Cinematográficas, Edición 8.º, Nominada, Ariel de Oro a la Mejor Película
 1953, Premio Ariel - Academia Mexicana de Artes y Ciencias Cinematográficas, Edición 8.º, Nominada, Ariel de Plata a Mejor Dirección
 1953, Premio Ariel - Academia Mexicana de Artes y Ciencias Cinematográficas, Edición 8.º, Nominada, Ariel de Plata a Mejor Coactuación Masculina
 1953, Premio Ariel - Academia Mexicana de Artes y Ciencias Cinematográficas, Edición 8.º, Nominada, Ariel de Plata a Mejor Actor de Cuadro
 1953, Premio Ariel - Academia Mexicana de Artes y Ciencias Cinematográficas, Edición 8.º, Nominada, Ariel de Plata a Mejor Escenografía

Bibliografía
 González, Reynaldo. Manuel Altolaguirre, Poeta también en el cine
 Ficha técnica en la página del cine mexicano del ITESM

Ligas externas

Películas de Roberto Cobo
Películas de Lilia Prado
Películas dramáticas de México
Road movies de México

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

Let $p,\ q \in (1,\infty)$ be any constants.
We define $\sin_{p,q}{x}$ by
the inverse function of
$$\sin_{p,q}^{-1}{x}:=\int_0^x \frac{dt}{(1-t^q)^{1/p}}, \quad 0 \leq x \leq 1,$$
and
\begin{equation}
\label{eq:pipq}
\pi_{p,q}:=2\sin_{p,q}^{-1}{1}=2\int_0^1 \frac{dt}{(1-t^q)^{1/p}}
=\frac2q B\left(\frac{1}{p^*},\frac1q\right),
\end{equation}
where $p^*:=p/(p-1)$ and $B$ denotes the beta function.
The function $\sin_{p,q}{x}$ is increasing in $[0,\pi_{p,q}/2]$ onto $[0,1]$.
We extend it to $(\pi_{p,q}/2,\pi_{p,q}]$ by $\sin_{p,q}{(\pi_{p,q}-x)}$
and to the whole real line $\mathbb{R}$ as the odd $2\pi_{p,q}$-periodic
continuation of the function.
Since $\sin_{p,q}{x} \in C^1(\mathbb{R})$,
we also define $\cos_{p,q}{x}$ by $\cos_{p,q}{x}:=(\sin_{p,q}{x})'$.
Then, it follows that
$$|\cos_{p,q}{x}|^p+|\sin_{p,q}{x}|^q=1.$$
In case $p=q=2$, it is obvious that $\sin_{p,q}{x},\ \cos_{p,q}{x}$
and $\pi_{p,q}$ are reduced to the ordinary $\sin{x},\ \cos{x}$ and $\pi$,
respectively.
This is a reason why these functions and the constant are called
\textit{generalized trigonometric functions} (with parameter $(p,q)$)
and \textit{the generalized $\pi$}, respectively.

Dr\'{a}bek and Man\'{a}sevich \cite{DM} introduced
the generalized trigonometric functions with two parameters
to study an inhomogeneous eigenvalue problem of $p$-Laplacian.
They gave a closed form of
solutions $(\lambda,u)$ of the eigenvalue problem
$$-(|u'|^{p-2}u')'=\lambda |u|^{q-2}u, \quad u(0)=u(L)=0.$$
Indeed, for any $n=1,2,\ldots$, there exists a curve of
solutions $(\lambda_{n,R},u_{n,R})$
with a parameter $R \in \mathbb{R} \setminus \{0\}$ such that
\begin{gather}
\lambda_{n,R}=\frac{q}{p^*}\left(\frac{n\pi_{p,q}}{L}\right)^p |R|^{p-q},
\label{eq:ev}\\
u_{n,R}(x)=R\sin_{p,q}\left(\frac{n\pi_{p,q}}{L}x\right)
\label{eq:ef}
\end{gather}
(see also \cite{T}). Conversely, there exists no other solution of the
eigenvalue problem. Thus, the generalized trigonometric functions
play important roles to study problems of the $p$-Laplacian.

It is of interest to know whether the generalized trigonometric functions
have multiple-angle formulas unless $p=q=2$.
A few multiple-angle formulas seem to be known.
Actually, in case $2p=q=4$, the function $\sin_{p,q}{x}=\sin_{2,4}{x}$
coincides with the lemniscate sine function $\slem{x}$,
whose inverse function is defined as
$$\slem^{-1}{x}:=\int_0^x \frac{dt}{\sqrt{1-t^4}}.$$
Furthermore, $\pi_{2,4}$ is equal to the lemniscate constant
$\varpi:=2\slem^{-1}{1}=2.6220\cdots$.
Concerning $\slem{x}$ and $\varpi$,
we refer to the reader to \cite[p.\,81]{PS}, \cite{To} and \cite[\S{22.8}]{WW}.
Since $\slem{x}$ has the multiple-angle formula
\begin{equation}
\label{eq:maflem}
\slem{(2x)}=\frac{2\slem{x}\sqrt{1-\slem^4{x}}}
{1+\slem^4{x}},\quad 0 \leq x \leq \frac{\varpi}{2},
\end{equation}
we see that
$$\sin_{2,4}{(2x)}=\frac{2\sin_{2,4}{x}\cos_{2,4}{x}}
{1+\sin_{2,4}^4{x}},\quad 0 \leq x \leq \frac{\pi_{2,4}}{2}.$$
Also in case $p^*=q=4$, it is possible to show that
$\sin_{p,q}{x}=\sin_{4/3,4}{x}$ can be expressed in terms of
the Jacobian elliptic function,
whose multiple-angle formula yields
\begin{equation}
\label{eq:EGL}
\sin_{4/3,4}{(2x)}=\frac{2\sin_{4/3,4}{x}\cos^{1/3}_{4/3,4}{x}}
{\sqrt{1+4\sin_{4/3,4}^4{x}\cos_{4/3,4}^{4/3}{x}}}
\quad 0 \leq x <\frac{\pi_{4/3,4}}{4}.
\end{equation}
The formula \eqref{eq:EGL} was investigated by Edmunds, Gurka
and Lang \cite[Proposition 3.4]{EGL}.
They also proved an addition theorem for $\sin_{4/3,4}{x}$
involving \eqref{eq:EGL}.

In this paper, we will present multiple-angle formulas
which are established between
two kinds of the generalized trigonometric functions
with parameters $(2,p)$ and $(p^*,p)$.

\begin{thm}
\label{thm:main}
For $p \in (1,\infty)$ and $x \in [0,2^{-2/p}\pi_{2,p}]
=[0,\pi_{p^*,p}/2]$, we have
\begin{equation}
\label{eq:main1}
\sin_{2,p}{(2^{2/p} x)}=2^{2/p} \sin_{p^*,p}x \cos_{p^*,p}^{p^*-1}x
\end{equation}
and
\begin{align}
\cos_{2,p}{(2^{2/p} x)}&=\cos_{p^*,p}^{p^*}{x}-\sin_{p^*,p}^p{x} \notag \\
&=1-2\sin_{p^*,p}^p{x}=2\cos_{p^*,p}^{p^*}{x}-1. \label{eq:main2}
\end{align}
Moreover, for $x \in \mathbb{R}$, we have
\begin{equation}
\label{eq:formula2}
\sin_{2,p}{(2^{2/p} x)}=2^{2/p} \sin_{p^*,p}{x}|\cos_{p^*,p}{x}|^{p^*-2}\cos_{p^*,p}{x}
\end{equation}
and
\begin{align}
\cos_{2,p}{(2^{2/p} x)}
&=|\cos_{p^*,p}{x}|^{p^*}-|\sin_{p^*,p}{x}|^p \notag \\
&=1-2|\sin_{p^*,p}{x}|^p=2|\cos_{p^*,p}{x}|^{p^*}-1.\label{eq:formula}
\end{align}
\end{thm}

In Theorem \ref{thm:main}, the fact
\begin{equation}
\label{eq:key}
\frac{\pi_{2,p}}{2^{2/p}}=\frac{\pi_{p^*,p}}{2}
\end{equation}
is the special case $n=2$ of the following identity.

\begin{thm}
\label{cor:pi}
Let $2 \leq n<p+1$. Then
$$\pi_{\frac{p}{p-1},p} \pi_{\frac{p}{p-2},p} \cdots \pi_{\frac{p}{p-n+1},p}
=n^{1-n/p}\pi_{\frac{n}{n-1},p}\pi_{\frac{n}{n-2},p}\cdots \pi_{\frac{n}{1},p}.$$
\end{thm}

We give a series expansion of $\pi_{p^*,p}$
as a counterpart of the Gregory-Leibniz series for $\pi$.
It is worth pointing out that $\pi_{p^*,p}$ is the area
enclosed by the $p$-circle $|x|^p+|y|^p=1$
(see \cite{LE}).

\begin{thm}
\label{prop:gls}
\begin{align*}
\frac{\pi_{p^*,p}}{4}
&=\sum_{n=0}^\infty\frac{(2/p)_n}{n!}\frac{(-1)^n}{pn+1} \\
&=1-\frac{2}{p(p+1)}+\frac{2+p}{p^2(2p+1)}-\frac{(2+p)(2+2p)}{3p^3(3p+1)}+\cdots,
\end{align*}
where $(a)_n:=\Gamma(a+n)/\Gamma(a)=a(a+1)(a+2)\cdots(a+n-1)$
and $\Gamma$ denotes the gamma function.
\end{thm}

We will apply Theorems \ref{thm:main}--\ref{prop:gls}
to the following problems (I)--(V).

(I) \textit{An alternative proof of \eqref{eq:EGL}}.
It should be noted that
the multiple-angle formula \eqref{eq:main1} in Theorem \ref{thm:main}
allows \eqref{eq:EGL}
to be rewritten in terms of the lemniscate function $\slem{x}=\sin_{2,4}{x}$:
\begin{equation*}
\sin_{4/3,4}{(2x)}=\frac{\sqrt{2}\slem{(\sqrt{2}x)}}
{\sqrt{1+\slem^4{(\sqrt{2}x)}}},
\quad 0 \leq x <\frac{\pi_{4/3,4}}{4}=\frac{\varpi}{2\sqrt{2}},
\end{equation*}
where the last equality above follows from \eqref{eq:key} with $\pi_{2,4}=\varpi$.
This indicates that it is possible to obtain \eqref{eq:EGL} from
the multiple-angle formula \eqref{eq:maflem} for the lemniscate function.

(II) \textit{A relation between eigenvalue problems of the $p$-Laplacian
and that of the Laplacian}.
Let $u$ be a function with $(n-1)$-zeros in $(0,L)$
satisfying
$$-(|u'|^{p-2}u')'=\lambda |u|^{p^*-2}u, \quad u(0)=u(L)=0$$
for some $\lambda>0$.
Similarly, let $v$ be a function with $n$-zeros in $(0,L)$
satisfying
$$-(|v'|^{p-2}v')'=\mu |v|^{p^*-2}v, \quad v'(0)=v'(L)=0$$
for some $\mu>0$.
Then, by Theorem \ref{thm:main},
we can show that the product $w=uv$ is a function
with $(2n-1)$-zeros in $(0,L)$ satisfying
$$-w''=2p^*(\lambda \mu)^{1/p} |w|^{p^*-2}w,\quad w(0)=w(L)=0.$$
The curious fact is the consequence of a straightforward calculation with
\eqref{eq:ev},\ \eqref{eq:ef},\ \eqref{eq:formula2} and \eqref{eq:key}.
Such a relation between the eigenvalue problems of the $p$-Laplacian
and that of the Laplacian may be known.
However, we can not find a literature proving it,
while the assertion in case $p=2$ is trivial because
$$w=\sin{\left(\frac{n\pi}{L}x\right)}\cos{\left(\frac{n\pi}{L}x\right)}
=\frac12 \sin{\left(\frac{2n\pi}{L}x\right)}.$$

(III) \textit{A pendulum-type equation with the $p$-Laplacian}.
We give a closed form of
solutions of the pendulum-type equation
$$-(|\theta'|^{p-2}\theta')'
=\lambda^p |\sin_{2,p}{\theta}|^{p-2}\sin_{2,p}{\theta}.$$
In case $p=2$, this equation is the ordinary pendulum equation
$-\theta''=\lambda^2 \sin{\theta}$ and it is well known that the solutions
can be expressed in terms of
the Jacobian elliptic function.
We will obtain an expression of the solution for
the pendulum-type equation above by using our special functions
involving a generalization of the Jacobian elliptic function in \cite{T,T2}.
There are studies of other (forced) pendulum-type equations with $p$-Laplacian
versus $\sin{\theta}$ in \cite{M}; versus $\sin_{p,p}{\theta}$ in \cite{AMN},
for the purpose of finding periodic solutions.

(IV) \textit{Catalan-type constants}.
Catalan's constant, which occasionally appears in estimates in combinatorics,
is defined by
$$G=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=0.9159\cdots.$$
We can find a lot of representation of $G$ in \cite{Bra};
for a typical example,
\begin{equation}
\label{eq:Gs}
\frac12\int_0^{\pi/2} \frac{x}{\sin{x}}\,dx=G.
\end{equation}
The multiple-angle formula \eqref{eq:main1} gives a generalization of \eqref{eq:Gs} as
\begin{equation}
\label{eq:Gp1}
\frac{1}{2^{2/p}}\int_0^{\pi_{2,p}/2} \frac{x}{\sin_{2,p}{x}}\,dx
=\sum_{n=0}^\infty \frac{(2/p)_n}{n!} \frac{(-1)^n}{(pn+1)^2}.
\end{equation}
In case $p=2$,
the formula \eqref{eq:Gp1} coincides
with \eqref{eq:Gs}. Moreover, for $p=4$ we obtain the interesting formula
$$\frac{1}{\sqrt{2}} \int_0^{\varpi/2} \frac{x}{\slem{x}}\,dx
=\sum_{n=0}^\infty \frac{(1/2)_n}{n!} \frac{(-1)^n}{(4n+1)^2}.$$

(V) \textit{Series expansions of the lemniscate constant $\varpi$}.
The lemniscate constant $\varpi$ has the formula (\cite[Theorem 5]{To}):
\begin{equation*}
\frac{\varpi}{2}
=1+\frac{1}{10}+\frac{1}{24}+\frac{5}{208}+\cdots
+\frac{(2n-1)!!}{(2n)!!}\frac{1}{4n+1}+\cdots,
\end{equation*}
where $(-1)!!:=1$.
For this, using Theorem \ref{prop:gls} with \eqref{eq:key}, we will obtain
$$\frac{\varpi}{2\sqrt{2}}
=1-\frac{1}{10}+\frac{1}{24}-\frac{5}{208}+\cdots
+\frac{(2n-1)!!}{(2n)!!}\frac{(-1)^n}{4n+1}+\cdots,$$
which does not appear in Todd \cite{To} and seems to be new.
We will also produce some other formulas of $\varpi$.

This paper is organized as follows.
Section 2 is devoted to the proofs of Theorems \ref{thm:main}--\ref{prop:gls}.
In Section 3, we deal with the above-mentioned problems (I)--(V).

\section{The multiple-angle formulas}
\label{sec:def}

Let $p,\ q \in (1,\infty)$ and $x \in (0,\pi_{p,q}/2)$.
It is easy to see that
\begin{gather*}
\cos_{p,q}^p{x}+\sin_{p,q}^q{x}=1, \\
(\sin_{p,q}{x})'=\cos_{p,q}{x}, \quad
(\cos_{p,q}{x})'=-\frac{q}{p}\sin_{p,q}^{q-1}{x}\cos_{p,q}^{2-p}{x}, \\
(\cos_{p,q}^{p-1}{x})'=-\frac{q}{p^*}\sin_{p,q}^{q-1}{x}.
\end{gather*}
If we extend to these formulas for any $x \in \mathbb{R}$, then
the last one, for example, corresponds to
\begin{equation}
(|\cos_{p,q}{x}|^{p-2}\cos_{p,q}{x})'
=-\frac{q}{p^*}|\sin_{p,q}{x}|^{q-2}\sin_{p,q}{x}. \label{eq:diffsin}
\end{equation}

In a particular case,
\begin{gather}
\cos_{p^*,p}^{p^*}{x}+\sin_{p^*,p}^p{x}=1, \label{eq:pyta} \\
(\sin_{p^*,p}{x})'=\cos_{p^*,p}{x}, \quad
(\cos_{p^*,p}{x})'=-(p-1)\sin_{p^*,p}^{p-1}{x}\cos_{p^*,p}^{2-p^*}{x}, \notag \\
(\cos_{p^*,p}^{p^*-1}{x})'=-\sin_{p^*,p}^{p-1}{x}. \notag
\end{gather}
From the last one and the differentiation of inverse functions,
$$(\cos_{p^*,p}^{p^*-1})^{-1}{(y)}=\int_y^1 \frac{dt}{(1-t^p)^{1/p^*}}, \quad 0 \leq y \leq 1,$$
hence
$$\sin_{p^*,p}^{-1}{y}+(\cos_{p^*,p}^{p^*-1})^{-1}{(y)}=\frac{\pi_{p^*,p}}{2}.$$
Therefore, for $x \in [0,\pi_{p^*,p}/2]$
\begin{gather}
\sin_{p^*,p}\left(\frac{\pi_{p^*,p}}{2}-x\right)=\cos_{p^*,p}^{p^*-1}{x}, \label{eq:sym} \\
\cos_{p^*,p}^{p^*-1}\left(\frac{\pi_{p^*,p}}{2}-x\right)=\sin_{p^*,p}{x}.
\label{eq:sym2}
\end{gather}

Throughout this paper, the following function is useful:
$$\tau_{p,q}(x):=\frac{\sin_{p,q}{x}}{|\cos_{p,q}{x}|^{q/p-1}\cos_{p,q}{x}}, \quad
x \neq \frac{2n+1}{2}\pi_{p,q},\ n \in \mathbb{Z}.$$
Then, it follows immediately from \eqref{eq:sym} and \eqref{eq:pyta} that

\begin{lem}
\label{lem:pi4}
For $x \in (0,\pi_{p^*,p}/2)$, $\tau_{p^*,p}{(x)}=1$ implies
$x=\pi_{p^*,p}/4$. Moreover,
$\sin_{p^*,p}^{-1}{(2^{-1/p})}=\cos_{p^*,p}^{-1}{(2^{-1/p^*})}
=\pi_{p^*,p}/4$.
\end{lem}

Let us prove the multiple-angle formulas in Theorem \ref{thm:main}.

\begin{proof}[Proof of Theorem \ref{thm:main}]
Let $x \in [0,\pi_{p^*,p}/4]$. Then, $y=\sin_{p^*,p}{x} \in [0,2^{-1/p}]$ by Lemma \ref{lem:pi4}.
Setting $t^p=(1-(1-s^p)^{1/2})/2$ in
$$\sin_{p^*,p}^{-1}{y}=\int_0^y \frac{dt}{(1-t^p)^{1/p^*}},$$
we have
\begin{align*}
\sin_{p^*,p}^{-1}{y}
&=\int_0^{y(4(1-y^p))^{1/p}}
\dfrac{\dfrac{2^{-1-1/p}s^{p-1}}{(1-s^p)^{1/2}(1-(1-s^p)^{1/2})^{1-1/p}}\,ds}
{2^{-1+1/p} (1+(1-s^p)^{1/2})^{1-1/p}}\\
&=2^{-2/p}
\int_0^{y(4(1-y^p))^{1/p}}
\frac{ds}{(1-s^p)^{1/2}};
\end{align*}
that is,
\begin{equation}
\label{eq:pe}
\sin_{p^*,p}^{-1}{y}=2^{-2/p} \sin_{2,p}^{-1}{(y(4(1-y^p)^{1/p})}.
\end{equation}
Hence we obtain
$$\sin_{2,p}{(2^{2/p} x)}=2^{2/p} \sin_{p^*,p}{x}\cos_{p^*,p}^{p^*-1}{x},$$
and \eqref{eq:main1} is proved.
In particular, letting $y=2^{-1/p}$ in \eqref{eq:pe} and using Lemma \ref{lem:pi4},
we get
$$\frac{\pi_{p^*,p}}{4}=2^{-2/p} \sin_{2,p}^{-1}{1}
=\frac{\pi_{2,p}}{2^{1+2/p}},$$
which implies \eqref{eq:key}.

Next, let $x \in (\pi_{p^*,p}/4,\pi_{p^*,p}/2]$
and $y:=\pi_{p^*,p}/2-x \in [0,\pi_{p^*,p}/4)$.
By the symmetry properties
\eqref{eq:sym} and \eqref{eq:sym2}, we obtain
$$2^{2/p}\sin_{p^*,p}{x}\cos_{p^*,p}^{p^*-1}{x}
=2^{2/p}\cos_{p^*,p}^{p^*-1}{y}\sin_{p^*,p}{y}.$$
According to the argument above,
the right-hand side is identical to $\sin_{2,p}{(2^{2/p}y)}$.
Moreover, \eqref{eq:key} gives
$$\sin_{2,p}{(2^{2/p}y)}=\sin_{2,p}{(\pi_{2,p}-2^{2/p}x)}
=\sin_{2,p}{(2^{2/p}x)}.$$

The formula \eqref{eq:main2} is deduced from differentiating both sides
of \eqref{eq:main1}. Moreover, \eqref{eq:formula2} and \eqref{eq:formula}
come from the periodicities of the functions,
\end{proof}

\begin{proof}[Proof of Theorem \ref{cor:pi}]
Setting $x=1/n$ and $y=1/p$ in the formula of the beta function
(see \cite[\S{12.15}, Example]{WW})
$$B(nx,ny)=\frac{1}{n^{ny}} \frac{\prod_{k=0}^{n-1}
B\left(x+k/n,y\right)}{\prod_{k=1}^{n-1}B(ky,y)}, \quad n \geq 2,\ x,\ y>0,$$
we have
$$\frac{p}{n}=\frac{1}{n^{n/p}}
\frac{\prod_{k=1}^nB(k/n,1/p)}{\prod_{k=1}^{n-1}B(k/p,1/p)}.$$
Hence,
$$\prod_{k=1}^{n-1} B\left(\frac{k}{p},\frac1p\right)=n^{1-n/p}
\prod_{k=1}^{n-1}B\left(\frac{k}{n},\frac1p\right), \quad n \geq 2.$$
From \eqref{eq:pipq}, this is rewritten as
$$\prod_{k=1}^{n-1} \pi_{(p/k)^*,p}=n^{1-n/p}
\prod_{k=1}^{n-1}\pi_{(n/k)^*,p}, \quad
2 \leq n <p+1,$$
which is precisely the assertion of the theorem.
\end{proof}

\begin{rem}
Taking $n=2$ in Theorem \ref{cor:pi}, we have the relation \eqref{eq:key}
between $\pi_{p^*,p}$ and $\pi_{2,p}$.
In fact,
\eqref{eq:key} is equivalent to the duplication formula of the gamma function
(see \cite[\S{12.15}, Corollary]{WW})
$$\Gamma (2x)=\frac{2^{2x-1}}{\sqrt{\pi}} \Gamma(x)\Gamma\left(x+\frac12\right).$$
\end{rem}

\begin{proof}[Proof of Theorem \ref{prop:gls}]
Let $x \in (0,1)$. Differentiating the inverse function of $\tau_{p^*,p}{(x)}$, we have
$$\tau_{p^*,p}^{-1}{(x)}=\int_0^x \frac{dt}{(1+t^p)^{2/p}}.$$
Hence
\begin{align}
\label{eq:tan}
\tau_{p^*,p}^{-1}{(x)}
=\int_0^x \sum_{n=0}^\infty \binom{-2/p}{n} t^{pn}\, dt
=x \sum_{n=0}^\infty \frac{(2/p)_n}{n!}\frac{(-x^p)^n}{pn+1}.
\end{align}
By Abel's continuity theorem \cite[\S3.71]{WW},
it is sufficient to show that the right-hand side of \eqref{eq:tan} converges
at $x=1$; i.e., the series
$$\sum_{n=0}^\infty \frac{(2/p)_n}{n!}\frac{(-1)^n}{pn+1}
=:\sum_{n=0}^\infty (-1)^na_n$$
converges. This is an alternating series and $\{a_n\}$ is decreasing because
$$0 \leq \frac{a_{n+1}}{a_n}=\frac{(2/p+n)(pn+1)}{(n+1)(pn+p+1)}
=\frac{(n+1/p)(pn+2)}{(n+1)(pn+p+1)}<1.$$
Moreover, $\{a_n\}$ converges to $0$ as $n \to \infty$. Indeed,
Euler's formula for the gamma function \cite[\S12.11, Example]{WW} gives
\begin{align*}
\lim_{n \to \infty} a_n
&=\lim_{n \to \infty}
\frac{2/p (2/p+1)(2/p+2)\cdots (2/p+n-1)}{(n-1)!n^{2/p}}
\frac{n^{2/p}}{n(pn+1)}\\
&=\frac{1}{\Gamma(2/p)}\cdot 0=0.
\end{align*}
Therefore, the series above converges to $\tau_{p^*,p}^{-1}(1)$
(see for instance \cite[\S 2.31, Corollary (ii)]{WW}).
From Lemma \ref{lem:pi4}, we concludes the theorem.
\end{proof}

\begin{rem}
Combining \eqref{eq:formula2} and \eqref{eq:formula},
we can assert that $\tau_{2,p}$ and $\tau_{p^*,p}$ satisfy the multiple-angle formula
$$\tau_{2,p}{(2^{2/p}x)}=\frac{2^{2/p}\tau_{p^*,p}{(x)}}{1-|\tau_{p^*,p}{(x)}|^p},$$
which coincides with that of the tangent function if $p=2$.
\end{rem}

\section{Applications}
\label{sec:appl}

\subsection{An alternative proof of \eqref{eq:EGL}}

Let us give
an alternative proof of the multiple-angle formula
of $\sin_{4/3,4}{x}$:
\begin{equation}
\sin_{4/3,4}{(2x)}=\frac{2\sin_{4/3,4}{x}\cos_{4/3,4}^{1/3}{x}}
{\sqrt{1+4\sin_{4/3,4}^4{x}\cos_{4/3,4}^{4/3}{x}}},
\quad 0 \leq x <\frac{\pi_{4/3,4}}{4}, \tag{\ref{eq:EGL}}
\end{equation}
which was discovered by Edmunds, Gurka and Lang \cite{EGL}.

Recall that $\sin_{2,4}{x}$ is equal to the
lemniscate function $\slem{x}$.
Applying \eqref{eq:main1} in case $p=4$
with $x$ replaced by $2x \in [0,\pi_{4/3,4}/2)$, we get
\begin{equation}
\label{eq:cubic}
\slem{(2\sqrt{2}x)}=\sqrt{2}\sin_{4/3,4}{(2x)}(1-\sin_{4/3,4}^4{(2x)})^{1/4}.
\end{equation}

First, we consider the case
$$0 \leq x <\frac{\pi_{4/3,4}}{8}=\frac{\varpi}{4\sqrt{2}}.$$
Then, since $0 \leq 2\sin_{4/3,4}^4{(2x)}<1$ by Lemma \ref{lem:pi4},
the equation \eqref{eq:cubic} gives
$$2\sin_{4/3,4}^4{(2x)}
=1-\sqrt{1-\slem^4{(2\sqrt{2}x)}}.$$
Set $S=S(x):=\slem{(\sqrt{2}x)}$. Using the multiple-angle formula \eqref{eq:maflem}
for the lemniscate function, we have
\begin{align*}
2\sin_{4/3,4}^4{(2x)}
&=1-\sqrt{1-\left(\frac{2S\sqrt{1-S^4}}{1+S^4}\right)^4}\\
&=1-\frac{\sqrt{1-12S^4+38S^8-12S^{12}+S^{16}}}{(1+S^4)^2}\\
&=1-\frac{|1-6S^4+S^8|}{(1+S^4)^2}.
\end{align*}
Since $0 \leq S <\slem{(\varpi/4)}=(3-2\sqrt{2})^{1/4}$,
evaluated by \eqref{eq:maflem},
we see that $1-6S^4+S^8 \geq 0$.
Thus,
\begin{equation}
\label{eq:case1}
2\sin_{4/3,4}^4{(2x)}
=1-\frac{1-6S^4+S^8}{(1+S^4)^2}=\frac{8S^4}{(1+S^4)^2}.
\end{equation}
Therefore, by \eqref{eq:main1},
$$\sin_{4/3,4}{(2x)}=\frac{\sqrt{2}S}{\sqrt{1+S^4}}
=\frac{2\sin_{4/3,4}{x}\cos_{4/3,4}^{1/3}{x}}
{\sqrt{1+4\sin_{4/3,4}^4{x}\cos_{4/3,4}^{4/3}{x}}}.$$

In the remaining case
$$\frac{\varpi}{4\sqrt{2}}=\frac{\pi_{4/3,4}}{8}
\leq x <\frac{\pi_{4/3,4}}{4}=\frac{\varpi}{2\sqrt{2}},$$
it follows easily that $1 \leq 2\sin_{4/3,4}^4{(2x)}<2$ and $1-6S^4+S^8 \leq 0$,
hence we obtain \eqref{eq:case1} again.
The proof of \eqref{eq:EGL} is complete.

\subsection{A relation between eigenvalue problems of the $p$-Laplacian and that of
the Laplacian}

\begin{thm}
Let $n \in \mathbb{N}$ and $p \in (1,\infty)$.
Let $u$ be an eigenfunction with $(n-1)$-zeros in $(0,L)$
for an eigenvalue $\lambda>0$ of the eigenvalue problem
\begin{equation}
\label{eq:dirichlet}
-(|u'|^{p-2}u')'=\lambda |u|^{p^*-2}u, \quad u(0)=u(L)=0,
\end{equation}
and $v$ an eigenfunction with $n$-zeros in $(0,L)$
for an eigenvalue $\mu>0$ of the eigenvalue problem
\begin{equation}
\label{eq:neumann}
-(|v'|^{p-2}v')'=\mu |v|^{p^*-2}v, \quad v'(0)=v'(L)=0.
\end{equation}
Then, the product $w=uv$ is an
eigenfunction for the eigenvalue $\xi=2p^* (\lambda \mu)^{1/p}$
with $(2n-1)$-zeros in $(0,L)$ of the eigenvalue problem
\begin{equation}
\label{eq:laplace}
-w''=\xi |w|^{p^*-2}w,\quad w(0)=w(L)=0.
\end{equation}
\end{thm}

\begin{proof}
By \eqref{eq:ev} and \eqref{eq:ef},
the solution $(\lambda,u)$
of \eqref{eq:dirichlet} can be expressed
as follows:
\begin{gather*}
\lambda=\left(\frac{n\pi_{p,p^*}}{L}\right)^p|R|^{p-p^*},\\
u(x)=R\sin_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right), \quad R \neq 0.
\end{gather*}
Similarly, by the symmetry \eqref{eq:sym},
the solution $(\mu,v)$ of \eqref{eq:neumann}
is represented as
\begin{gather*}
\mu=\left(\frac{n\pi_{p,p^*}}{L}\right)^p|Q|^{p-p^*},\\
v(x)=Q\left|\cos_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right)\right|^{p-2}
\cos_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right), \quad Q \neq 0.
\end{gather*}
Applying \eqref{eq:formula2} in Theorem \ref{thm:main}
and \eqref{eq:key}
to the product $w=uv$, we have
\begin{align*}
w(x)
&=RQ\sin_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right)
\left|\cos_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right)\right|^{p-2}
\cos_{p,p^*}\left(\frac{n\pi_{p,p^*}}{L}x\right)\\
&=2^{-2/p^*}RQ\sin_{2,p^*}\left(\frac{2n\pi_{2,p^*}}{L}x\right),
\end{align*}
which belongs to $C^2(\mathbb{R})$ and has $(2n-1)$-zeros in $(0,L)$.
Therefore, by \eqref{eq:diffsin} with $p=2$,
a direct calculation shows
\begin{align}
\label{eq:w''}
w''&=-p^*2^{1-2/p^*} \left(\frac{n\pi_{2,p^*}}{L}\right)^2 RQ
\left|\sin_{2,p^*}\left(\frac{n\pi_{2,p^*}}{L}x\right)\right|^{p^*-2}
\sin_{2,p^*}\left(\frac{n\pi_{2,p^*}}{L}x\right) \notag \\
&=-p^*2^{3-4/p^*}\left(\frac{n\pi_{2,p^*}}{L}\right)^2
|RQ|^{2-p^*}|w|^{p^*-2}w.
\end{align}
On the other hand, \eqref{eq:key} gives
\begin{equation}
\label{eq:lm}
(\lambda\mu)^{1/p}
=2^{2-4/p^*}\left(\frac{n\pi_{2,p^*}}{L}\right)^2|RQ|^{2-p^*}.
\end{equation}
Combining \eqref{eq:w''} and \eqref{eq:lm}, we obtain \eqref{eq:laplace}.
\end{proof}

\subsection{A pendulum-type equation}

We give an expression of the solution of the following initial value problem:
\begin{equation}
\label{eq:p-pendulum}
-(|\theta'|^{p-2}\theta')'
=\lambda^p |\sin_{2,p}{\theta}|^{p-2}\sin_{2,p}{\theta}, \quad
\theta(0)=0, \ \theta'(0)=\omega_0.
\end{equation}

For $p,\ q \in (1,\infty)$ and $k \in [0,1)$ we define
$\am_{p,q}{(x,k)}$ by the inverse function of
$$\am_{p,q}^{-1}{(x,k)}:=\int_0^x \frac{d\theta}{(1-k^q|\sin_{p,q}{\theta}|^q)^{1/p^*}},
\quad -\infty<x<\infty,$$
in particular,
\begin{equation}
\label{eq:kpq}
K_{p,q}(k):=\am_{p,q}^{-1}\left(\frac{\pi_{p,q}}{2},k\right)
=\int_0^{\pi_{p,q}/2} \frac{d\theta}{(1-k^q\sin_{p,q}^q{\theta})^{1/p^*}}.
\end{equation}
Then, the function
$$\sn_{p,q}{(x,k)}:=\sin_{p,q}{(\am_{p,q}{(x,k)})}$$
is a $4K_{p,q}(k)$-periodic odd function in $\mathbb{R}$.
In case $p=q=2$, the functions $\am_{p,q},\ K_{p,q}$ and $\sn_{p,q}$
coincide with the amplitude function, the complete elliptic integral
of the first kind and the Jacobian elliptic function, respectively
(see \cite[\S 2.8]{PS} for them).
It is easy to see that
$K_{p,q}(0)=\pi_{p,q}/2,\
\sn_{p,q}(x,0)=\sin_{p,q}{x},\
\lim_{k \to 1-0}K_{p,q}(k)=\infty$
and
$\lim_{k \to 1-0}\sn_{p,q}(x,k)=\tanh_q{x}$,
defined as the inverse function of
$$\tanh_q^{-1}{x}=\int_0^x \frac{dt}{1-|t|^q},\quad -1<x<1.$$
See \cite{T2} for details of the generalized Jacobian elliptic function.
Another type of generalization of the Jacobian elliptic function
also appears in a bifurcation problem of the $p$-Laplacian; see \cite{T}.

\begin{thm}
Let $p \in (1,\infty)$ and $\lambda,\ \omega_0 \in (0,\infty)$,
and set $k:=\omega_0/(2^{2/p} \lambda)$.
Then, the solution of \eqref{eq:p-pendulum}
is as follows:
\begin{enumerate}
\item Case $k<1$.
\begin{equation*}
\theta (t)=2^{2/p} \sin_{p^*,p}^{-1}{\left(k
\sn_{p,p} \left(\lambda t,k\right)\right)},
\end{equation*}
which is a $4K_{p,p}(k)/\lambda$-periodic function and
$$\max_{0 \leq t<\infty}|\theta (t)|=2^{2/p} \sin_{p^*,p}^{-1}{k}.$$
\item Case $k=1$.
\begin{equation*}
\theta (t)=2^{2/p} \sin_{p^*,p}^{-1}{(\tanh_p{(\lambda t)})},
\end{equation*}
which is a strictly monotone increasing function and
$$\lim_{t \to \infty}\theta (t)=2^{2/p-1}\pi_{p^*,p}=\pi_{2,p}.$$
\item Case $k>1$.
\begin{equation*}
\theta (t)=2^{2/p} \am_{p^*,p}{\left(k\lambda t,\frac1k \right)},
\end{equation*}
which is a strictly monotone increasing function and
$$\lim_{t \to \infty}\theta (t)=\infty.$$
\end{enumerate}
\end{thm}

\begin{proof}

By the standard argument (for example, \cite[Proposition 2.1]{DM} or
\cite[Theorem 3.1]{LE}),
we can show that there exists a unique global
solution of \eqref{eq:p-pendulum}.

Let $\theta$ be the solution of \eqref{eq:p-pendulum}
and $T:=\inf\{t:\theta'(t)=0\}$.
On the interval $(0,T)$, $\theta$ satisfies $\theta(t)>0$ and $\theta'(t)>0$.
Then, using \eqref{eq:diffsin} with $p=2$, we obtain
\begin{equation}
\label{eq:ee}
\frac1p \theta'(t)^p-\frac1p \omega_0^p
=\frac{2\lambda^p}{p} \cos_{2,p}\theta(t)-\frac{2\lambda^p}{p}.
\end{equation}
From \eqref{eq:formula} in Theorem \ref{thm:main} we have
\begin{equation}
\label{eq:mafc}
\cos_{2,p}\theta(t)
=1-2|\sin_{p^*,p}{(2^{-2/p}\theta(t))}|^p.
\end{equation}
Combining \eqref{eq:ee} and \eqref{eq:mafc} we obtain
$$\theta'(t)^p
=4\lambda^p(k^p-|\sin_{p^*,p}{(2^{-2/p}\theta(t))}|^p),$$
where $k=\omega_0/(2^{2/p} \lambda)$, hence, for $t \in [0,T]$,
\begin{equation}
\label{eq:t}
t=\frac{1}{2^{2/p} \lambda}
\int_0^{\theta(t)} \frac{d\theta}{(k^p-|\sin_{p^*,p}{(2^{-2/p}\theta)}|^p)^{1/p}}.
\end{equation}

(i) Case $k<1$.
We can find $\alpha \in (0,2^{2/p-1}\pi_{p^*,p})$ such that
$k=\sin_{p^*,p}(2^{-2/p}\alpha)$ and $T=\theta^{-1}{(\alpha)}$.
Letting $\sin_{p^*,p}{(2^{-2/p}\theta)}=k\sin_{p,p}{\vp}$
in \eqref{eq:t}, we obtain
\begin{equation*}
t=\frac{1}{\lambda}
\int_0^{\vp(t)} \frac{d\vp}{(1-k^p|\sin_{p,p}{\vp}|^p)^{1/p^*}},
\end{equation*}
which implies
$$\vp(t)=\am_{p,p}(\lambda t,k).$$
Therefore,
$$\theta(t)=2^{2/p} \sin_{p^*,p}^{-1}{(k\sn_{p,p}(\lambda t,k))}.$$
We have thus found the unique solution $\theta$ of \eqref{eq:p-pendulum} in $[0,T]$.
However, in view of the periodicity properties of $\sn_{p,p}$, this function $\theta$
is actually the unique global solution of \eqref{eq:p-pendulum},
which is periodic of $4T=4K_{p,p}(k)/\lambda$ and
whose maximum value is $\alpha=2^{2/p} \sin_{p^*,p}^{-1}{k}$.

(ii) Case $k=1$. In this case, letting $\sin_{p^*,p}(2^{-2/p}\theta)=x$
in \eqref{eq:t}, we obtain
$$t=\frac{1}{\lambda} \int_0^{x(t)} \frac{dx}{1-|x|^p}=\frac{1}{\lambda}\tanh_p^{-1}{x(t)}.$$
Therefore,
$$\theta(t)=2^{2/p} \sin_{p^*,p}^{-1}{(\tanh_p{(\lambda t)})}$$
and $T=\infty$. Moreover, by \eqref{eq:key}
$$\lim_{t \to \infty}\theta(t)
=2^{2/p} \sin_{p^*,p}^{-1}{1}=2^{2/p-1} \pi_{p^*,p}= \pi_{2,p}.$$

(iii) Case $k>1$. In this case, \eqref{eq:t} becomes
\begin{align*}
t
&=\frac{1}{2^{2/p} k \lambda}
\int_0^{\theta(t)} \frac{d\theta}{(1-k^{-p}|\sin_{p^*,p}{(2^{-2/p}\theta)}|^p)^{1/p}}\\
&=\frac{1}{k \lambda}
\int_0^{\vp(t)} \frac{d\vp}{(1-k^{-p}|\sin_{p^*,p}{\vp}|^p)^{1/p}}\\
&=\frac{1}{k \lambda}
\am_{p^*,p}^{-1}\left(\vp(t),\frac1k\right).
\end{align*}
Therefore,
$$\theta(t)=2^{2/p} \am_{p^*,p}{\left(k\lambda t,\frac1k\right)}$$
and $T=\infty$. It is obvious that $\lim_{t \to \infty}\theta(t)=\infty$.
\end{proof}

\begin{rem}
The solution $\theta (t)$ in (ii) does not attain $\pi_{2,p}$ for any finite $t$,
while equations with $p$-Laplacian sometimes have flat-core
solutions (cf. \cite{T}).
\end{rem}

\subsection{Catalan-type constants}

We define
$$G_p:=\sum_{n=0}^\infty \frac{(2/p)_n}{n!}\frac{(-1)^n}{(pn+1)^2}.$$
It is clear that $G_2=G$, i.e. Catalan's constant described in Introduction.

\begin{thm}
\label{thm:Gp}
Let $p \in (1,\infty)$, then
\begin{equation}
\label{eq:Gp}
G_p
=\frac{1}{2^{2/p}} \int_0^{\pi_{2,p}/2} \frac{x}{\sin_{2,p}{x}}\,dx
=\frac{1}{2^{2/p}} \int_0^1 K_{2,p}(k)\,dk,
\end{equation}
where $K_{2,p}(k)$ is defined by \eqref{eq:kpq}.
\end{thm}

\begin{proof}
By \eqref{eq:tan},
$$\int_0^1 \frac{\tau_{p^*,p}^{-1}{(x)}}{x}\,dx
=\sum_{n=0}^\infty \frac{(2/p)_n(-1)^n}{n!} \int_0^1 \frac{x^{pn}}{pn+1}\,dx
=G_p.$$
On the other hand, letting $\tau_{p^*,p}^{-1}{(x)}=2^{-2/p}y$,  we obtain
\begin{align*}
\int_0^1 \frac{\tau_{p^*,p}^{-1}{(x)}}{x}\,dx
&=\frac{1}{2^{4/p}}\int_0^{2^{2/p-2}\pi_{p^*,p}}
\frac{y}{\sin_{p^*,p}{(2^{-2/p}y)}\cos_{p^*,p}^{p^*-1}{(2^{-2/p}y)}}\,dy\\
&=\frac{1}{2^{2/p}} \int_0^{\pi_{2,p}/2} \frac{y}{\sin_{2,p}{y}}\,dy.
\end{align*}
Here, we have used \eqref{eq:key} and \eqref{eq:main1}.
This shows the first equality in \eqref{eq:Gp}.

The second equality in \eqref{eq:Gp} follows from Fubini's theorem that
\begin{align*}
\int_0^1 K_{2,p}(k)\,dk
&=\int_0^{\pi_{2,p}/2} \int_0^1
\frac{1}{(1-k^p\sin_{2,p}^p{x})^{1/2}}\,dk\,dx\\
&=\int_0^{\pi_{2,p}/2} \frac{1}{\sin_{2,p}{x}}
\int_0^{\sin_{2,p}{x}} \frac{1}{(1-t^p)^{1/2}}\, dt\, dx\\
&=\int_0^{\pi_{2,p}/2} \frac{\sin_{2,p}^{-1}{(\sin_{2,p}{x})}}
{\sin_{2,p}{x}}\,dx\\
&=\int_0^{\pi_{2,p}/2} \frac{x}
{\sin_{2,p}{x}}\,dx.
\end{align*}
The proof is accomplished.
\end{proof}

By Theorem \ref{thm:Gp} with $p=4$, we obtain

\begin{cor}
$$\frac{1}{\sqrt{2}} \int_0^{\varpi/2} \frac{x}{\slem{x}}\,dx
=\sum_{n=0}^\infty \frac{(1/2)_n}{n!} \frac{(-1)^n}{(4n+1)^2}.$$
\end{cor}

\begin{rem}
In a similar way to \cite[\S 3.2]{BE} or the last paragraph of \cite[\S 2.1]{LE},
we can also obtain the formula
\begin{equation}
\label{eq:Cp}
\int_0^{\pi_{2,p}/2} \frac{x}{\sin_{2,p}{x}}\,dx
=\frac{\pi_{2,p}}{2}\sum_{n=0}^\infty
\frac{(1/2)_n (1/p)_n}{(1/2+1/p)_n n!} \frac{1}{pn+1}=:\frac{\pi_{2,p}}{2}C_p.
\end{equation}
Therefore, from \eqref{eq:Gp}, \eqref{eq:Cp} and \eqref{eq:key},
\begin{equation}
\label{eq:pip*}
\frac{\pi_{p^*,p}}{4}=\frac{\pi_{2,p}}{2^{2/p+1}}
=\frac{G_p}{C_p}
=\frac{\sum_{n=0}^\infty \frac{(2/p)_n}{n!}\frac{(-1)^n}{(pn+1)^2}}
{\sum_{n=0}^\infty \frac{(1/2)_n (1/p)_n}{(1/2+1/p)_n n!} \frac{1}{pn+1}},
\end{equation}
particulary,
$$\frac{\pi}{4}=\frac{\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}}
{\sum_{n=0}^\infty \left(\frac{(1/2)_n}{n!}\right)^2 \frac{1}{2n+1}}.$$
\end{rem}

\subsection{Series expansions of the lemniscate constant $\varpi$}

The series of Proposition \ref{prop:gls} for $p=2$ is nothing but the Gregory-Leibniz series.
Letting $p=4$ and using \eqref{eq:key}, we have the expansion series
for the lemniscate constant $\varpi$:
\begin{equation}
\label{eq:lpi1}
\frac{\varpi}{2\sqrt{2}}
=1-\frac{1}{10}+\frac{1}{24}-\frac{5}{208}+\cdots
+\frac{(2n-1)!!}{(2n)!!}\frac{(-1)^n}{4n+1}+\cdots.
\end{equation}
On the other hand, there is the similar series to this in
\cite[Theorem 5]{To}:
\begin{equation}
\label{eq:lpi2}
\frac{\varpi}{2}
=1+\frac{1}{10}+\frac{1}{24}+\frac{5}{208}+\cdots
+\frac{(2n-1)!!}{(2n)!!}\frac{1}{4n+1}+\cdots.
\end{equation}
Combining \eqref{eq:lpi1} and \eqref{eq:lpi2}, we obtain
\begin{align*}
\frac{2+\sqrt{2}}{8}\varpi
&=1+\frac{1}{24}+\cdots
+\frac{(4n-1)!!}{(4n)!!}\frac{1}{8n+1}+\cdots,\\
\frac{2-\sqrt{2}}{8}\varpi
&=\frac{1}{10}+\frac{5}{208}+\cdots
+\frac{(4n+1)!!}{(4n+2)!!}\frac{1}{8n+5}+\cdots.
\end{align*}
Finally, letting $p=4$ in \eqref{eq:pip*} we obtain
$$\frac{\varpi}{2\sqrt{2}}=
\frac{\sum_{n=0}^\infty \frac{(1/2)_n}{n!}\frac{(-1)^n}{(4n+1)^2}}
{\sum_{n=0}^\infty \frac{(1/2)_n (1/4)_n}{(3/4)_n n!} \frac{1}{4n+1}}.$$

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