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Category Archives: Senegal
Crossing Borders, Mali, Senegal, Travel
Africa Overland
August 12, 2011 John	4 Comments
Waiting for the bus at the Mali-Senegal border
Crossing into Mali from Senegal, you immediately know you've transitioned into a poorer country. The amenities change. Everywhere in Mali are half-finished concrete buildings – the remnants of some great idea on the part of the government or an NGO or whomever, for which the money ran out. Senegal's by no means a developed country, but you get the sense that there's an entrepreneurial, if misguided, spirit present in the towns and cities: Senegal is moving forward – maybe at the expense of some of its charm and West African-ness – but it's moving forward nonetheless.
The apprentice collecting fares from the front seat in a bush taxi in Mali
Both border controls were a breeze. On the Senegalese side, the officer thanked us for visiting his country, stamped our passports without hassle, and wished us safe travels with a kindness utterly absent in the way I've seen American immigration officials berate and belittle foreigners.
The Malian side too was little trouble. An intimidating gendarme cracked into a teddy bear when we told him we'd been Peace Corps volunteers and especially when we started asking him how to say a few words in Bambara.
The hassle came on the road, especially in the first few hundred kilometers, where frequent checkpoints turned the 800-900 km – a long distance by any measure – into a marathon day. Our visa receipts got us a simple nod of acceptance from the police, but they used the opportunity to extract a little bit of pocket money from the Malians and other Africans on board the bus.
Anne-Claire in the back of a pickup in Mali
Maybe the Senegalese officials are just better paid and don't need the extra income from bribes. Even traveling through the heavily militarized Casamance region in southern Senegal, where rebel separatist sentiments necessitate (apparently) the tanks that roll through the towns with soldiers manning the gun turrets and the frequent stops on the roadway – though they're so cursory, I can't imagine they're ever effective in rooting out insurgents – even there, little money seemed to change hands.
The land too changed almost immediately after crossing the Senegal River, which separates Kidira in Senegal from Diboli in Mali. Anne-Claire and I often joke that when they drew the border between California and Oregon, the California folks just drove north until the sun stopped shining, then took a few steps back and planted their stakes. Of course, there's quite a bit of rain in northern California, but the four times I've driven across the border, our little joke has held water – sun on the California side, rain or snow in Oregon.
The same idea seems to have been put in place by the folks who drew the Senegal-Mali border. Right over the border, the verdant fields swarming with birds give way to the dusty hardpan of the true Sahel. Mango trees heavy with fruit are replaced by scrappier neem, acacia and shea trees, and the baobabs, in comparison to their grotesquely huge and heavily branched cousins in Senegal, look spindly and stunted.
Nigerien bush taxis (van in the background) always seem to get more for their money
The bleakness is oddly comforting in a way, as I can feel we're getting closer to Niger. You can hear it in the cadence of speech and see it in the well-worn dress and thatched roofs and mud huts melting in the rainy season.
Even the trundling bus has the comfort of familiarity, swerving from shoulder to shoulder to avoid the slightest pockmark in the tarmac – West African bus drivers will gladly run down donkey carts and cyclists, and they seem to see dogs and livestock as little more than videogame obstacles: best not to hit them, but not at the cost of speed. But present a driver with an inch-deep pothole, perhaps a foot in diameter, and he'll tiptoe his overloaded rig across it as gently as if he were balancing stacks of dishes on the roof.
As green as it gets – driving through Niger on the way to Kollo
Travel is a perennial challenge here. The air conditioners in the buses stopped working long ago, but of course the windows don't open. The only air comes from the vented escape hatches in the roof, and then only when the bus is moving. And, in a region that's produced the likes of Yousou N'Dour, Ali Farke Touré and Salif Keita, bus drivers seem to love with a cult-like obsession the same way that some people love campy B movies in the States the clangy, scream-laden, over-instrumented, over-amplified music of less adept artists that's always too loud.
Still, I wouldn't trade this way of seeing West Africa, even with all its challenges. It's almost as if jigsawing our way through the region's roads, we're unlocking the secrets this part of the world holds. It's only like this you can see how the harshness of the land forges tough-as-nails people, tempered by relentless hospitality and an eagerness to laugh that's unlike anywhere else.
It's an area of the world I was proud to call home for two years, and I'm excited to be back.
bush taxiDakarferryGambiaGambia RiverMaliNigerSenegalZiguinchor
Crossing Borders, Senegal, Travel
August 9, 2011 John	4 Comments
Watching the sunset on the beach, we could have easily been on a tropical island somewhere. Lightning flashed far in the distance, and the heavy moist air from the day's teasing rainstorms settled over the beach, leaving us glistening with sweat in the last light of day.
But this wasn't just anywhere. This was Senegal. Cows shared the sand with us, and down from where we'd just walked, smoke bellowed from the skyscraping palm trees, as women dried the day's catch that had just come in on the massive wooden boats.
We'd arrived in Ziguinchor to visit Haroon, one of Anne-Claire's former classmates. Our second day, he took us to stay at a work colleague's house on the coast, up near the border with Gambia. Cheikh, the colleague, runs an NGO called Water and Sanitation for West Africa. Over his decades-long career, he's worked with the World Health Organization, the UN, and now USAID, currently focused on improving sanitation in villages through strategies like hand washing and building latrines – in short, combining behavior change with infrastructure improvements.
Here on the outskirts of Kafountine, Cheikh's built an oasis of calm with bungalows for visiting friends, solar-powered electricity, and a pack of friendly dogs named after world leaders, Tony Blair, Angela Merkel and Vladimir Putin among them. (Nicholas Sarkozy died recently after being bitten by a cobra.) Cheikh has planted trees on the land with his small personal staff, and both nights we stayed there, we enjoyed walks down to the beach to watch the sunset. The second evening, we watched as a flotilla of massive fishing boats and their crews of 25-30 men each hauled the day's harvest through the surf.
We ate well, enjoying dishes of fish with fried potatoes, rice and fish sauce, and chicken served with fried bananas. Always accompanying the meal were Cheikh's stories, detailing his adventures in development. We spent a long time Saturday evening discussing the failures of massive, one-size-fits-all projects and the need to tailor solutions to the cultures they're designed to work within.
It's heartening to see a well-educated African, who's lived and worked all over the world, focus his talents on improving things for other Africans and to see that he doesn't just accept poverty for his continent. One of my biggest takeaways from Peace Corps is that as outsiders, we'll never affect real positive change leading the charge in development. We just don't have the knowledge of the culture, the vested interest and connections to local communities that takes a lifetime to acquire. Outsider-led projects, in my opinion, do little but muck up the situation even further. Best case, we should be supporting folks like Cheikh and taking direction from them as to how we can best help those around us.
Africaanimalsbeach cowsCasamancedevelopmenthuman capitalNGOsSenegalWestern aidZiguinchor
Crossing Borders, Senegal, Travel, Video
Virtual Bush Taxi Ride
Here's a short video Anne-Claire took on our ride through Senegal and the Gambia. The roads in parts of Senegal are notoriously bad, even for West Africa. Drivers, to save their cars from gaping holes in the road that would devastate a vehicle that hasn't had shocks for decades, swing wildly from one side of the road to the other. There's an elaborate system of using turning signals to indicate to oncoming traffic which way you're going to pass, though I still haven't figured it out. I'm not sure they have either, which might be why road accident fatalities are 8-10 times higher in the developing world than elsewhere. Often, the smoothest track is with one set of wheels on the roadway and the other on the less-worn dirt path running alongside.
I'm sharing this now, after we've left West Africa, so there's no need to worry. We traveled some 3500 kilometers on roads in Senegal, Mali and Niger (roughly the equivalent of driving from California to Ohio), all thankfully without incident.
Africabush taxivideo
Crossing Borders, Senegal, The Gambia, Travel
Crossing the Gambia
We left the soft cradle of the diplomatic family's house early in the morning and took to the already-muggy streets at 5:30 a.m. After Anne-Claire argued the price of our luggage down from the exorbitant price the driver first demanded, we got in a bush taxi headed for Ziguinchor in southern Senegal near the border with Guinea-Bissau.
Five and a half hours straight in the 'sept-place' (seven-seater station wagon, excluding the driver) were about all my seat bones could take; the airflow to our seats in the third row had stopped, creating a suffocating, sauna-like atmosphere that pushed my claustrophobia to the limit and nearly sent me in a panicked launch toward the cracked windows. A commonality over much of West Africa we've found is that people often prefer to be sweaty rather than dusty, so they're often loath to roll down the windows in all but the worst heat. It was a good reminder of what real heat is. I looked vaguely like I'd been through a rainstorm, and I was intimately aware of every sweat gland on my body. I'd forgotten how much my eye sockets can sweat.
When we arrived at the Gambia border, I thought it would be a welcome reprieve. First, it was just an opportunity to get out of the car. We had to get an exit stamp in our passports from Senegal (easy and free), then pay a small 'entrance fee' (read: bribe?) to get into the Gambia without a proper visa.
Second, the country's namesake river holds a powerful place in my memory, as this was where the story of 'Roots' by Alex Haley began. The film drove my curiosity about Africa when I first saw it at 11 years old, and reading the book years later in my first months at post in Peace Corps deepened my understanding of the rhythm of village life.
The real fun started as we got close to the Gambia River. Defying all logic, our driver sped past the miles-long stretches of cars, buses and trucks waiting for the ferry to cross the river. Word has it that the Gambian government refuses to build a bridge for fear of losing the bustling commerce that passengers forced to wait days generate in the thin sliver of land that forms this odd country.
While Anne-Claire and I had egg sandwiches, the driver took our contributions to bribe the ferry operators into letting us move up in the line (our fellow passengers contributed as well). Hot, cranky and tired, we decried the entropic disarray as people fought for the few places on one of the two slow, decrepit boats that left each bank of the river at most once an hour.
On reflection, though, this is Africa – things operate differently that the way we know, and in the end there's a method hidden amongst the madness. It may not look like what we're used to or can even comprehend, like the way news in a village always seems to spread like wildfire: I'd get home to my Peace Corps house in Niger, say hello to my host family, and not 10 minutes later, folks from the far side of the village a 15-minute walk away would arrive just to greet me because they'd heard I was home.
I suppose that's part of what makes traveling interesting. If everything looked the same, the impetus to light out wouldn't be so strong.
Africabush taxiferryGambia RiverRoots
Sometimes traveling opens up a different sort of experience than you'd expect to have. And sometimes it happens to be one of things you'll remember most. Our stop in Derry, Northern Ireland, fits this description – what we'd originally intended to be just a stopover became a memorable visit with new friends that opened up a window to the history of the area.
Our first day in Dakar ended with one of those evenings. We arrived early Tuesday morning from Casablanca. I'm not sure if it was because it was 1:30 in the morning, but the immigration line was more efficient than most we've had to wait in. Anne-Claire had left the poles for our tent strapped to the outside of her bag, and they were missing when it came down the carousel. When she asked a guard if this was 'just something we should expect traveling here,' he looked offended and said it absolutely was not. Though we didn't mean to offend him, it was impressive to see someone taking pride in the fact that this isn't just another middling developing country where that sort of stuff is par for the course.
Sure enough, the poles eventually came down the beltway. After winding our way out to the taxi stand, we hailed one and worked out a price, and he took us to our hotel in the center of the city. We hadn't realize just how centrally located it was, so it was nice to step out the next morning (OK, not really the morning – around 12:30 pm) and be right in the center of bustling Dakar.
Anne-Claire and I both noticed how little attention we got, especially compared to what we'd been used to in Morocco. Even though we stick out here quite a bit more, most folks leave us alone, and the few that are interested in our business – for cell phone cards mostly – are just as likely to proffer their goods to a Senegalese.
The helpful clerk at our hotel recommended a restaurant for lunch called Chez Loutcha a couple blocks away that serves Cape Verdean and Senegalese dishes. The food was terrific – I had Dibi, sort of steak frites with spiced lamb instead of beef. The prices are a bit higher then we're used to from Niamey, but the bustling café full of Senegalese and ex-pats alike enjoying their lunch hour was evidence that Senegal does not sit at the bottom of most economic and development indicators (as Niger does).
We ran errands around Dakar in the afternoon, sent a few souvenirs from Morocco home to the State, and had a Peace Corps-esque indulgent ice cream at a fancy café on the big square called Place de l'Independence. Our real treat in Dakar came in meeting a friend of a friend who is a Foreign Service officer. She and her husband were kind enough to let us stay with their family for a couple of nights. They live in a gorgeous ex- pat house and have three charming children who were eager to have visitors.
After nearly a month and a half on the road, a couple of nights that felt like we were in America were a welcome break, similar to the times in Peace Corps that we spent with the missionary family in Maradi who took in us strays on holidays, or with our country director and his wife when they'd host us for dinners at their house.
We shared a delicious dinner with the family and a current Senegalese Peace Corps volunteer who they'd also taken in for the night. The kids are whip-smart, speak French and English, and have devised all sorts of games to keep themselves entertained. We enjoyed their company so much that we came back early from a visit to Ile de Goré to swim with them the next afternoon.
Ile de Goré is a small island off the coast of Dakar that played a small role in the slave trade and now houses a slave museum and memorial. On it are some excellent examples of colorful French colonial architecture, as well as a small beach. We might have been a little road weary for such a touristy spot, as we didn't have a lot of patience for the people offering their guide services or pedaling paintings.
Overall, we enjoyed our time in Dakar. The city has character to be sure. You're just as likely to see a stunning woman dressed like she just stepped off the plane from Paris as you are to see a family that's come in from the bush to make a better life for themselves. You'll find French coffee shop chains and United Colors of Benetton, but like so many other African cities, Dakar has become a victim of its own success. The area it now occupies is huge, and the strain of so many people has stretched its electrical power far beyond its capacity. We were told there's a 200-Megawatt shortfall (not sure what this would mean in the States, but it sounds big). For comparison, the entire country of Niger requires about 200 Megawatts of electricity.
Africaelectricityforeign serviceNigerSenegal

redpajama_set_name: RedPajamaCommonCrawl


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a writing blog
← the immortal story (1968)
the cradle will rock →
f for fake (1974)
by Anthony Lee Collins Posted on February 4, 1999
"I am a charlatan."
(Orson Welles)
Welles' films are mostly quite somber in tone. There is little humor, except in Falstaff, but Welles had another side that came out in his magic act, and in interviews. He was a trickster, a ham, a gleeful illusionist. This side dominates this film, which is more or less a documentary (a film essay he called it, which he hoped would catch on as a new genre of filmmaking).
He set out to make a documentary about Elmyr de Houry, one of the greatest art forgers in history, whose paintings hang in museums throughout the world. Actually, Welles saw a documentary about Elmyr on French television and bought up all the footage, including outtakes, in order to expand it into a feature film. But the unpredictable element was that the expert on Elmyr in the documentary was Clifford Irving, and in the middle of the making of Welles' movie it was revealed that Irving was quite a forger and illusionist himself (he wrote a fake Howard Hughes autobiography which he sold for a tidy sum, and then the whole scam was discovered).
At that point, the film became much more of a meditation on illusion, trickery, and authorship in general, especially since Welles claimed that he had originally intended to make Citizen Kane about Howard Hughes.
But, as he said, if you told the story of Hughes' life in a film, nobody would believe it.
He also performs a bit of trickery on the audience, which is revealed at the end of the film. I won't give it away.
return to the orson welles page
This entry was posted in Movie Reviews and tagged orson welles. Bookmark the permalink.
Stevie One: an adventure
the town hall mystery (part fifteen)
the town hall mystery (part fourteen)
the town hall mystery (part thirteen)
writers, and writing "rules"
the town hall mystery (part twelve)
the town hall mystery (part eleven)
the town hall mystery (part ten)
the town hall mystery (part nine)
thomas carnacki
the town hall mystery (part eight)
dare to betray the book
the town hall mystery (part seven)
there is a point (and it's labeled)
journals, with a bullet
Millennium Trilogy
Pre-Blog
Seven Sentence Sunday
Storytelling Lessons
The Ten Pillars


These tea leaves are dried with jasmine flower petals to absorb the flavor.
Two Leaves takes these big green tea leaves and blends them delicately with Italian red orange for a memorable yet not overpowering citrus flavor.
Two Leaves calls this organic rooibos tea (say "roy-bus") "African Sunset" because red tea is grown in South Africa, where it is cherished for its spicy, herbal flavor.
These are big, black, organic tea leaves grown in Assam, India, where hot days and cool nights bring out a rich, full flavor.
Delicious little flowers - that's how they think of their Chamomile.
These organic, golden orange pekoe tea leaves are grown on the rolling hills of Darjeeling, India.
This tea is entirely caffeine free and is a great after-dinner mint. A zing for your taste buds, rich aroma for your nose and a gift for your psyche.
There's a reason this is Two Leave's best selling green tea - it has a sweet, light flavor you want in a cuppa' green tea, without being bitter.


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William Pfaff: The Pundit Who Hated Militarism & War
"Nowhere in the elite establishment newspaper sphere could you find regular, sustained, and well-informed criticism of an aggressive and overly militarized American foreign policy."--Scott McConnell.
The Great Deal David Petraeus Got Shows a Double Standard at Work
It's another example of how our pretense of equal justice for all is simply untrue.
Kent State: After 45 Years We Need a Serious Look at What happened and Why
"There has never been a completely satisfactory explanation for why the Guard fired."
Obama's Dangerous Legacy Could Include a War in Russia's Backyard
The U.S. is now an active player in regions Russia views as an integral part of its sphere of influence.
The Neocons: First in War, Last in Peace
"American Jewish right-wingers [like Irving Kristol] are almost without exception partisans of the Likud policy of de facto annexation of the West Bank."
Why Three Jailed Pacifists Took on the Empire and What We've Learned Because They Broke the Law.
"Michael is trying to save lives. Your life."
After Obama's Executive Order Against Venezuela, We'll Be Hearing "Yanqui, Go Home" Once Again
"Whoever looks at the map and says that Venezuela could be a threat to the U.S. has to be out of his mind."
Are You Ready To Fight Putin's Russia?
There's only one way out of the Ukrainian mess and that's patient, deliberate, nonviolent diplomacy. The alternative could be a nuclear war.
It's SUPER BOWL Time!!!!
The Super Bowl is designed to remind us that the USA, the NFL, and the 1% team owners are one and the same.
Who Knows Eric Fair?
The recent revelation about unaccountable Americans using torture on prisoners have again raised the question of morality and legality.
Happy Holidays, President Obama! Harding Pardoned Debs So Why Not Pardon Snowden and Manning Too?
President Barack Obama should pardon Snowden and Manning just as Warren Harding pardoned Eugene Debs
Deja vu, Afghans: We're Ba-a-a-a-ck!
Obama is becoming more hawkish and his antiwar base isn't welcome
Are Iraq, Afghanistan and Ukraine, etc. Worth the Life of a Single U.S. Soldier or Marine?
A new and non-interventionist foreign policy?
Putin, Putin, Putin & More Putin
Whatever the eventual outcome in Ukraine, Russian interests will have to be atken into account.
How Much Freedom for Journalists?
Veteran Washington-based reporters complain about the Obama Administration's restrictions on press freedom
Resisting Evil
Patrick Henry, a Catholic scholar, proves that Jews did resist the Nazis, physically, spiritually,. violenty and non-violently
Hiroshima & Nagasaki 69 Years Later
Was the atomic attack on Japan justified? Is the mass killing of civilians by nuclear bombs a war crime?
Remembering Al Campanis: Was Justice Done?
Al Campanis made one stupid remark and was then banned from baseball. Was justice done?
Stalin & American Myths
Why and how Stalin badly hurt Russia and why Hiroshima did not end WWII.
War Crimes: Who Arre Accuntable?
The trial of Japanese General Yamashita and why he did not deserve to hang.


And I don't know what an Emmylou Harris album has with Sherry Fiester's fascinating book , Enemy of the Truth. The book is a forensic anyalyst's look at some of the known aspects of the JFK murder. And the recommendation is a "country" album. Huh?
PS, note the uncomfortable abbreviation of Enemy of the Truth's subtitle.
I loved watching Beany & Cecil cartoons as a kid. I guess you could say I was part of their "second wave" of fandom, because I ONLY knew them from cartoons.
As a kid I had no idea that they started out as puppets.
One luxury of my childhood, though, was that for my birthday I got an official Beany-Copter.
Now, do you see that hollow space in the center of the propeller? The propellers originally had caps that fit over this little cavity. This small hollow space was to put secret messages into.
You set the propeller onto the top of the hat and turned it clockwise two or three times. There was a ratcheting spring inside, and you could hear it click-pop every quarter-turn or so. Then you put your Beany-Copter on your head and pulled the chin-string. This released the propeller, which zoomed into the air and probably landed on your neighbor's roof, never to be seen again until the next windy day.
I can still remember the zip-click-whir sound made when you had your Beany-copter on and yanked the string to let the propeller go!
And, yes, children, it was typical to have commercials that ran for one full minutes, not ten or fifteen seconds.


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ATZORI Alberto's profile
ATZORI Alberto
Department of Agriculture, University of Sassari , Sassari, Italy
Agricultural sustainability, Animal nutrition modelling, Cattle production, Mathematical modelling, Ruminant nutrition, Small ruminants
Modelling the impact of the macroalgae Asparagopsis taxiformis on rumen microbial fermentation and methane production
Rafael Muñoz-Tamayo , Juana C. Chagas, Mohammad Ramin, Sophie J. Krizsan 10.1101/2020.11.09.374330
Understanding the mechanisms behind natural bioactive compounds that can potentially reduce methane production in anaerobic conditions. A case study of Asparagopsis taxiformis
Recommended by Luis Tedeschi based on reviews by Alberto Atzori, Henk van Lingen and 2 anonymous reviewers
Naturally occurring compounds that can reduce methane production in anaerobic conditions have been studied for quite some time as feasible approaches to mitigate methane production in ruminant animals, especially those of commercial importance. Asparagopsis taxiformis (red algae) and Dictyota bartayresii (brown algae) are effective inhibitors of methane synthesis under in vitro anaerobic fermentation systems (Machado et al., 2014) likely because of their high concentration of secondary metabolites that are toxic to the typical rumen microbiota, including protozoa. In addition to phytoplankton (Palmer and Reason, 2009), Asparagopsis contains a high concentration of haloform compounds (e.g., bromoform, chloroform) while Dictyota has a high concentration of isoprenoid terpenes. Despite the economic and biological impact of diverse phytochemicals on reducing methane production in ruminant animals (Tedeschi et al., 2021), haloform compounds' environmental impact and safety, in particular, are still unclear. In the present study, Munõz-Tamayo and collaborators (2021) listed relevant literature about the impact of A. taxiformis on ruminal methane production.
Concurrent to the understanding of mechanisms and biology behind the reduction of ruminal methane, mathematical models can lead the way to enhance the effectiveness of feeding A. taxiformis under commercial applications. Thus, in the present study, Munõz-Tamayo and collaborators (2021) sought to develop a mathematical model to understand the rumen fermentation changes in vitro experimentation containing extract of A. taxiformis by adapting a previously documented model by Muñoz-Tamayo et al. (2016).
Modeling development, calibration, and evaluation steps should be independent of each other, requiring complete, distinct, and separate databases (Tedeschi, 2006). However, in rare circumstances where such requirements cannot be met because data availability is scarce, the cross-validation technique, when possible, should be considered to assess data dispersion's effects on model adequacy. In other situations, clear reasoning for failing to do so must be addressed in the paper. In the present paper, Munõz-Tamayo and collaborators (2021) explained the limitations in their modeling efforts were primarily due to the lack of ideal data: "experiments with simultaneous dynamic data of bromoform, volatile fatty acids, hydrogen, and methane." Regardless of the availability of ideal data, improvements in the conceptual model are warranted to include amino acids and branched-chain fatty acids fermentation dynamics in the rumen and the fluctuations in ruminal pH.
Machado L, Magnusson M, Paul NA, Nys R de, Tomkins N (2014) Effects of Marine and Freshwater Macroalgae on In Vitro Total Gas and Methane Production. PLOS ONE, 9, e85289. https://doi.org/10.1371/journal.pone.0085289
Muñoz-Tamayo R, Chagas JC, Ramin M, Krizsan SJ (2021) Modelling the impact of the macroalgae Asparagopsis taxiformis on rumen microbial fermentation and methane production. bioRxiv, 2020.11.09.374330, ver. 4 peer-reviewed and recommended by PCI Animal Science. https://doi.org/10.1101/2020.11.09.374330
Muñoz-Tamayo R, Giger-Reverdin S, Sauvant D (2016) Mechanistic modelling of in vitro fermentation and methane production by rumen microbiota. Animal Feed Science and Technology, 220, 1–21. https://doi.org/10.1016/j.anifeedsci.2016.07.005
Palmer CJ, Reason CJ (2009) Relationships of surface bromoform concentrations with mixed layer depth and salinity in the tropical oceans. Global Biogeochemical Cycles, 23. https://doi.org/10.1029/2008GB003338
Tedeschi LO (2006) Assessment of the adequacy of mathematical models. Agricultural Systems, 89, 225–247. https://doi.org/10.1016/j.agsy.2005.11.004
Tedeschi LO, Muir JP, Naumann HD, Norris AB, Ramírez-Restrepo CA, Mertens-Talcott SU (2021) Nutritional Aspects of Ecologically Relevant Phytochemicals in Ruminant Production. Frontiers in Veterinary Science, 8. https://doi.org/10.3389/fvets.2021.628445
Lactation curve model with explicit representation of perturbations as a phenotyping tool for dairy livestock precision farming.
Ben Abdelkrim Ahmed, Puillet Laurence, Gomes Pierre, Martin Olivier 10.1101/661249
Developing smart fitting algorithms to identify random perturbations in time-series data
Recommended by Luis Tedeschi based on reviews by Alberto Atzori, Jennifer Spencer and 1 anonymous reviewer
The ability to adequately characterize the lactation curve of livestock is important not only to ensure proper nutrition of the lactating animal but, among many other benefits, it can assist in diagnosing the incidence of diseases, predicting the quantity of milk production, and comparing animals within the herd for managerial strategies such as culling. Eventually, such smart fitting algorithms can lead to improved genetic selection of more productive animals after genetic-unrelated noises are removed from the data, systematically.
The manuscript by Ben Abdelkrim et al. developed and explained an algorithm to detect perturbations in lactation curves of dairy goats. Researchers have been interested in accurately describing lactation curves since the early-1960s. Johansson [1] proposed a nonlinear decay function, Nelder [2] described an inverse polynomial, and Wood [3] proposed the incomplete gamma function to describe milk production of dairy cows. Unfortunately, many of the lactating animals that yielded lactation curves that did not comply with the typical, expected pattern of milk production were usually discarded and, until then, efforts to address this lack of adherence were not conducted. The recommended manuscript provides a different perspective in which rather than discarding the lactation profile, one can model the perturbations of the lactation curve as an attempt to identify possible problems (e.g., mastitis) and minimize their occurrence. Such an algorithm is important to separate females that show resilient attributes from those females that show sustainable attributes, as per existing definitions proposed by Tedeschi et al. [4].
The recommended manuscript proposes the Perturbed Lactation Model to explicitly account for multiple perturbations in the time-series milk production in dairy goats. When perturbations occur in biological processes, a typical negative impact is observed in the animal's response, but on rare occasions, positive impacts can occur. In this case, the animal responds positively to the perturbation (i.e., responsive), and as a result, there is an increase in their output when compared to unperturbed animals. The recommended manuscript only considered negative impacts due to perturbations in the lactation curve of dairy goats. Future modifications should include positive responses due to perturbations. In this case, animals would be "positively responsive" to perturbations, and examples of such behavior include feed intake and growth curves.
[1] Johansson, I. (1961). Genetic Aspects of Dairy Cattle Breeding. University of Illinois Press, Urbana, IL.
[2] Nelder, J. A. (1966). Inverse polynomials, a useful group of multi-factor response functions. Biometrics. 22 (1):128-141. doi: 10.2307/2528220
[3] Wood, P. D. P. (1967). Algebraic model of the lactation curve in cattle. Nature. 216 (5111):164-165. doi: 10.1038/216164a0
[4] Tedeschi, L. O., J. P. Muir, D. G. Riley, and D. G. Fox. (2015). The role of ruminant animals in sustainable livestock intensification programs. Int. J. Sustainable Dev. World Ecol. 22 (5):452-465. doi: 10.1080/13504509.2015.1075441


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Q: How to wait for appendChild() to update and style the DOM, before trying to measure the new element's width I'm appending a DOM element like this:
this.store.state.runtime.UIWrap.appendChild(newElement)

When I immediately try to measure the new element's width I get 2.
I tried:

*

*setTimeout()

*double nested window.requestAnimationFrame()

*MutationObserver
The above works very unreliably, like 50% of the time. Only when I set a large timeout like 500ms it worked.
This happens only on mobile.
This is the workaround that I'm using, but it's ugly:
function getWidthFromStyle(el) {
  return parseFloat(getComputedStyle(el, null).width.replace('px', ''))
}
function getWidthFromBoundingClientRect(el) {
  return el.getBoundingClientRect().width
}

console.log(getWidthFromBoundingClientRect(newElement))

while (getWidthFromBoundingClientRect(newElement) < 50) {
  await new Promise(r => setTimeout(r, 500))
  console.log(getWidthFromBoundingClientRect(newElement))
}

I tried with both functions getWidthFromStyle() and getWidthFromBoundingClientRect() - no difference. The width gets calculated properly after a couple of cycles.
I also tried using MutationObserver without success.
Is there a way to know when the DOM is fully updated and styled before I try to measure an element's width/height?
P.S. I'm not using any framework. this.store.state.runtime... is my own implementation of a Store, similar to Vue.
EDIT: The size of the element depended on an image inside it and I was trying to measure the element before the image had loaded. Silly.

A: it can done with MutationObserver.
doesn't this method solve your problem?

const div = document.querySelector("div");
const span = document.querySelector("span");

const observer = new MutationObserver(function () {
  console.log("new width", size());
});

observer.observe(div, { subtree: true, childList: true });

function addElem() {
  setTimeout(() => {
    const newSpan = document.createElement("span");
    newSpan.innerHTML = "second";
    div.appendChild(newSpan);
    console.log("element added");
  }, 3000);
}

function size() {
  return div.getBoundingClientRect().width;
}

console.log("old width", size());
addElem();
div {
  display: inline-block;
  border: 1px dashed;
}
span {
  background: gold;
}

  one

A: You can use something like this:
export function waitElement(elementId) {
  return new Promise((resolve, reject) => {
    const element = document.getElementById(elementId);
    if (element) {
      resolve(element);
    } else {
      let tries = 10;
      const interval = setInterval(() => {
        const element = document.getElementById(elementId);
        if (element) {
          clearInterval(interval);
          resolve(element);
        }
        if (tries-- < 0) {
          clearInterval(interval);
          reject(new Error("Element not found"));
        }
      }, 100);
    }
  });
}

redpajama_set_name: RedPajamaStackExchange


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text:
16:9 Ratio Notebook Panels To Become Mainstream In H2
By DigiTimes
With panel makers pushing 16:9 ratio panels in 2008, Acer has decided to open up all its notebook product lines to be based on the ratio, while other notebook vendors are readying to push at least 14" and 15.6" lines in the third quarter this year. Notebook makers predict the proportion of 16:9 models will increase 10% per quarter during in the second half of 2008.
More here at Digitimes.
DigiTimes
nvalhalla
Shouldn't it be 16:10? 1920x1200 and such?
gm0n3y
Yeah, 16:9 is an HDTV format, 16:10 is the computer format, but perhaps that's the point? They want people to watch movies on them I guess. Just don't try to play a game designed for 16:10.
I wonder how long it'll take EA to support the new format.
frozenlead
I haven't seen a 4:3 screen in a while...I was under the impression this was already really accepted as standard.
thehelo
"I haven't seen a 4:3 screen in a while...I was under the impression this was already really accepted as standard."
the standard for pc screens is 16:10. 16:9 is what hdtv's are.


An order to buy a specified product at or below a specified price, or to sell at or above a specified price.
When market price is equal to or more than the stop loss price, your buy order will be sent to market.
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Why can't I change or delete orders on Hang Seng Index (HSI), Mini-Hang Seng Index (MHI), H-shares index (HHI) and Mini H-shares index (MCH) during the price-opening session?
There are regulations by HKEX on order placing of HSI, MHI, HHI and MCH futures during the pre-opening session (8:45am-9:15am and 12:30pm-1:00pm).
During the pre-opening periods of 8:45am-9:11am and 12:30pm-12:56pm, the exchange accepts any adding, changing or deleting of auction orders or limit orders.
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The following table displays the actions that can be done during the trading day.
"" indicates the actions that can be performed at a particular time period.
How long do the order instructions for trading local futures remain valid for?
All the order instructions for local futures trading are valid in both T and T+1 sessions, that is, they are valid until the close of T+1 session..
Is there any price limit up/down mechanism in the After-Hours Futures Trading Session (T+1)?
Yes. HKEX has set the percentage for the price limit up/down at 5% in the After-Hours Trading Session(T+1).
What is the meaning of 5% price limit up/down Mechanism?
It means no sell order of price below 95% and no buy order of price above 105% of the last traded price of the spot month contract in the T Session will be allowed. Trading (for all contract months) will be allowed only within the price limit range during the T+1 session. Moreover, trading in the T session will not be subject to this price limit up/down mechanism.
You cannot place buy order above 21,000 or sell order below 19,000 for all contract months.
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How the available withdraw balance is being calculated?
What is the sequence for closing out position?
What are the last trading date and the settlement method for HK index futures?
The lasting trading date for HK index futures is two working days before month end. The settlement method is cash settled. The official settlement price is determined by HKEX, which is an average of index prices taken at five minute intervals over the last trading day. The final settlement price can be viewed on the HKEX website.
What is the close out deadline for physical delivery contracts?
For relative long positions, the close out deadline is either 3 trading days before the First Notice Day or Last Trading Day, whichever is earlier.
For relative short positions, the close out deadline is 3 trading days before the Last Trading Day.


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text:
\section{Preliminaries and statements of main results}\label{csw-sec1}
\subsection{Notations}
For $n\geq2$, let $\mathbb{R}^{n}
$ denote the usual real vector space of dimension $n$. Sometimes it is convenient to
identify each point $x=(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}$ with an $n\times 1$ column matrix
so that
$$x=\left(\begin{array}{cccc}
x_{1} \\
\vdots \\
 x_{n}
\end{array}\right).
$$
For $w=(w_{1},\ldots,w_{n})$ and $x\in\mathbb{R}^{n}$, we define the
Euclidean inner product $\langle \cdot ,\cdot \rangle$ by
$\langle x,w\rangle=x_{1}w_{1}+\cdots+x_{n}w_{n}
$
so that the Euclidean length of $x$ is defined by
$$|x|=\langle x,x\rangle^{1/2}=(|x_{1}|^{2}+\cdots+|x_{n}|^{2})^{1/2}.
$$
Denote a ball in $\mathbb{R}^{n}$ with center $w\in\mathbb{R}^{n}$
and radius $r$ by
$$\mathbb{B}^{n}(w,r)=\{x\in\mathbb{R}^{n}:\, |x-w|<r\}.
$$
In particular, $\mathbb{B}^{n}$ and $\mathbb{S}^{n-1}$ denote the unit ball
$\mathbb{B}^{n}(0,1)$  and the unit sphere in $\mathbb{R}^{n}$, respectively.
Set $\mathbb{D}=\mathbb{B}^2$, the open unit
disk in the complex plane $\mathbb{C}\cong \mathbb{R}^{2}$. For
$k\in\mathbb{N}_0:=\mathbb{N}\cup\{0\}$ and $m\in\mathbb{N}_0$, we
denote by $\mathcal{C}^{k}(\Omega_{1},\Omega_{2})$ the set of all
$k$-times continuously differentiable functions from $\Omega_{1}$
into $\Omega_{2}$, where $\Omega_{1}$ and $\Omega_{2}$ are subsets
of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$, respectively. In
particular, let
$\mathcal{C}(\Omega_{1},\Omega_{2}):=\mathcal{C}^{0}(\Omega_{1},\Omega_{2})$,
the set of all continuous functions of $\Omega_{1}$ into
$\Omega_{2}$. For
$f=(f_{1},\ldots,f_{m})\in\mathcal{C}^{1}(\Omega_{1},\Omega_{2})$,
we denote the derivative $D_{f}$ of $f$ by
$$D_{f}=\left(\begin{array}{cccc}
\displaystyle D_{1}f_{1}\; \cdots\;
 D_{n}f_{1}\\[4mm]
\vdots\;\; \;\;\cdots\;\;\;\;\vdots \\[2mm]
 \displaystyle
D_{1}f_{m}\; \cdots\;
 D_{n}f_{m}
\end{array}\right), \quad D_{j}f_{i}(x)=\frac{\partial f_{i}(x)}{\partial x_j}.
$$
In particular, if $n=m$, the Jacobian of $f$ is defined by
$J_{f}=\det D_{f}$ and the Laplacian of
$f\in\mathcal{C}^{1}(\Omega_{1},\Omega_{2})$ is defined by
$$\Delta f=\sum_{k=1}^{n}D_{kk}f.
$$
For an $m\times n$ matrix $A$, the operator norm of $A$ is defined by
$$|A|=\sup_{x\neq 0}\frac{|Ax|}{|x|}=\max\{|A\theta|:\,
\theta\in\mathbb{S}^{n-1}\}
$$
where $m\geq1$ and $n\geq2$.

\subsection{Poisson equation and Schwarz lemma}
For $x,y\in\mathbb{R}^{n}\backslash\{0\}$, we define $x'=x/|x|$, $y'=y/|y|$ and let
$$[x,y]:=\left|y|x|-x'\right|=\left|x|y|-y'\right|.
$$
Also, for $x,y\in\mathbb{B}^{n}$ with $x\neq y$ and $|x|+|y|\neq0$,
we use $G(x,y)$ to denote the {\it Green function}:
\begin{equation}\label{eq-ex0}
G(x,y)=c_{n}\left(\frac{1}{|x-y|^{n-2}}-\frac{1}{[x,y]^{n-2}}\right),
\end{equation} where $c_{n}=1/[(n-2)\omega_{n-1}]$ and
$\omega_{n-1}=2\pi^{\frac{n}{2}}/\Gamma\big(\frac{n}{2}\big)$
denotes the {\it Hausdorff measure} of $\mathbb{S}^{n-1}$. The {\it
Poisson kernel} $P:\,\mathbb{B}^{n}\times
\mathbb{S}^{n-1}\rightarrow {\mathbb R}$ is defined by
$$P(x,\zeta)=\frac{1-|x|^{2}}{|x-\zeta|^{n}}.
$$
We write
$$\nabla =\left ( \frac{\partial }{\partial x_1}, \ldots, \frac{\partial }{\partial x_n} \right )
$$
and for a vector valued function $f=(f_1, \ldots, f_m)$, we define
the directional derivative $\frac{\partial f}{\partial
\overrightarrow{n}}$ by a componentwise interpretation:
$$ \frac{\partial f}{\partial \overrightarrow{n}}(x)=\left ( \langle \nabla f_1(x), x'\rangle, \ldots, \ldots, \langle \nabla f_m(x), x'\rangle \right ).
$$
For a given bounded integrable function $\psi:~\mathbb{S}^{n-1}\rightarrow\mathbb{R}^{m}$ and
$g\in\mathcal{C}(\mathbb{B}^{n},\mathbb{R}^{m})$, the solution of the {\it Poisson equation}
\begin{equation}\label{eq-1}
\begin{cases} \displaystyle \Delta f=g
& \mbox{ in } \mathbb{B}^{n}\\
\displaystyle f=\psi &\mbox{ in }\, \mathbb{S}^{n-1}
\end{cases}
\end{equation} is given by \begin{equation}\label{eq-p}
f(x)=\mathcal{P}_{\psi}(x)-\mathcal{G}_{g}(x), \end{equation} where
\begin{equation}\label{eq-p1a}
\mathcal{P}_{\psi}(x)=\int_{\mathbb{S}^{n-1}}P(x,\zeta)\psi(\zeta)d\sigma(\zeta)
~\mbox{ and }~\mathcal{G}_{g}(x)=
\int_{\mathbb{B}^{n}}G(x,y)g(y)dV(y) \end{equation} for $x\in\mathbb{B}^{n}$.
Here $d\sigma$ denotes the normalized Lebesgue surface measure on
$\mathbb{S}^{n-1}$ and $dV$ is the Lebesgue volume measure on
$\mathbb{B}^{n}$. It is well known that if $\psi$ and $g$ are
continuous in $\mathbb{S}^{n-1}$ and in $\overline{\mathbb{B}^{n}}$,
respectively, then $f=\mathcal{P}_{\psi}-\mathcal{G}_{g}$ has a
continuous extension $\tilde{f}$ to the boundary, and
$\tilde{f}=\psi$ in $\mathbb{S}^{n-1}$ (see \cite[p.~118--120]{Ho}
or \cite{K4,K3,K1}).

The classical Schwarz lemma states that a holomorphic function $f$
from $\mathbb{D}$ into itself with $f(0)=0$ satisfies
$|f(z)|\leq|z|$ for all $z\in\mathbb{D}$. It is well-known that the
Schwarz lemma has become a crucial theme in a lot of branches of
mathematical research for more than a hundred years to date. For
$n\geq3$,  the classical Schwarz lemma of harmonic
mappings in $\mathbb{B}^{n}$ infers that if $f$ is a
harmonic mapping of $\mathbb{B}^{n}$ into itself satisfying
$f(0)=0,$ then
$$|f(x)|\leq U(rN),
$$
where $r=|x|$, $N=(0,\ldots,0,1)$ and $U$ is
a harmonic function of $\mathbb{B}^{n}$ into $[-1,1]$ defined by
$$U(x)=\mathcal{P}_{(\mathcal{X}_{S^{+}}-\mathcal{X}_{S^{-}})}(x).
$$
Here $\mathcal{X}$ is the indicator function,
$S^{+}=\{x=(x_{1},\ldots,x_{n})\in\mathbb{S}^{n-1}:~x_{n}\geq0\}$
and
$S^{-}=\{x=(x_{1},\ldots,x_{n})\in\mathbb{S}^{n-1}:~x_{n}\leq0\}$
(see \cite{ABR}). For the case $n=2,$ we refer to \cite{CK,CV,He,
K0}.
In \cite{K5}, Kalaj proved the following result for harmonic mappings $f$ of $\mathbb{B}^{n}$
into itself:
\begin{equation}\label{eq-K}
\left|f(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}f(0)\right|\leq U(|x|N).
\end{equation}
\subsection{Main results and Remarks}
The first aim of the paper is to extend the result \eqref{eq-K} to mappings satisfying the Poisson equation.
More precisely, we shall prove the following.

\begin{thm}\label{thm-1}
Let $n\geq3$, $m\geq1$ and $g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$. If
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{m})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{m})$
satisfies $\Delta f=g$, then, for $x\in\overline{\mathbb{B}^{n}}$,
\begin{equation}\label{eq-thm1}
\left|f(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}\mathcal{P}_{f}(0)\right|
\leq\|\mathcal{P}_{f}\|_{\infty}U(|x|N)
+\frac{\|g\|_{\infty}}{2n}(1-|x|^{2}),\end{equation} where
$$\mathcal{P}_{f}(x)=\int_{\mathbb{S}^{n-1}}P(x,\zeta)f(\zeta)d\sigma(\zeta),~\|f\|_{\infty}=
\sup_{x\in\mathbb{B}^{n}}|f(x)|~\mbox{and}~\|g\|_{\infty}=\sup_{x\in\mathbb{B}^{n}}|g(x)|.
$$
If we choose $g(x)=(-2nM,0,\ldots,0)$ and
$f(x)=(M(1-|x|^{2}),0,\ldots,0)$ for
$x\in\overline{\mathbb{B}^{n}}$, then
 the inequality
{\rm(\ref{eq-thm1})} is sharp in $\mathbb{S}^{n-1}\cup\{0\}$, where
$M>0$ is a constant.
\end{thm}

By Theorem \ref{thm-1}, we give an explicit estimate as follows.

\begin{cor}\label{cor-1}
Let $n\geq3$, $m\geq1$ and $g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$. If
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{m})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{m})$
satisfies $\Delta f=g$, then, for $x\in\overline{\mathbb{B}^{n}}$,
\begin{equation}\label{eq-thm1a}
\left|f(x)-\frac{1-|x|}{(1+|x|)^{n-1}}\mathcal{P}_{f}(0)\right|
\leq\|\mathcal{P}_{f}\|_{\infty}\left[1-\frac{1-|x|}{(1+|x|)^{n-1}}\right]
+\frac{\|g\|_{\infty}}{2n}(1-|x|^{2}).
\end{equation}
If we choose $g(x)=(-2nM,0,\ldots,0)$ and $f(x)=(M(1-|x|^{2}),0,\ldots,0)$ for
$x\in\overline{\mathbb{B}^{n}}$, then  the inequality {\rm(\ref{eq-thm1a})} is sharp
in $\mathbb{S}^{n-1}\cup\{0\}$, where $M>0$ is a constant.
\end{cor}

There is a classical Schwarz lemma at the boundary, which is as
follows and the same may be obtained from standard texts. See for instance,
\cite{G} or \cite[p.~249, Corollary~6.62]{pon}.

\begin{Thm
\label{Thm-B}
Let $f$ be a holomorphic function from
$\mathbb{D}$ into itself. If $f$ is holomorphic at $z=1$ with
$f(0)=0$ and $f(1)=1$, then $f'(1)\geq1$. Moreover, the inequality
is sharp.
\end{Thm}

Theorem \Ref{Thm-B} has been generalized in various forms. For
example, Krantz \cite{Kra} explored many versions of the Schwarz
lemma at the boundary point of a domain in $\mathbb{C}$, and
reviewed several results of many authors. A natural response to
these results is to ask whether we can extend the Schwarz lemma at
the boundary to higher dimensional cases. This form of the Schwarz lemma has attracted much
attention, see \cite{BK,LWT,LT} for holomorphic functions, and see
\cite{ABR,Bu, K5,K6,MM} for harmonic functions. In the following, by
using Theorem \ref{thm-1}, we establish a Schwarz lemma at the
boundary for mappings satisfying Poisson's equation, which is also a
generalization of Theorem \Ref{Thm-B}.

\begin{thm}\label{thm-2} For $n\geq3$, $m\geq1$ and a given
$g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$, let
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{m})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{m})$
be a mapping of $\mathbb{B}^{n}$ into itself satisfying $\Delta f=g,$
where
$$\|g\|_{\infty}<\frac{nA_{n}}{\big(1+\frac{1}{2^{\frac{n}{2}}}\big)} ~\mbox{ and }~ A_{n}=\frac{n!\big[1+n-(n-2)F\big(\frac{1}{2},1;\frac{n+3}{2};-1\big)\big]}{2^{\frac{3n}{2}}
\Gamma\big(\frac{n+1}{2}\big)\Gamma\big(\frac{n+3}{2}\big)}.
$$
If $f(0)=0$ and $\lim_{r\rightarrow1^{-}}|f(r\zeta)|=1$ for some $\zeta\in\mathbb{S}^{n-1}$, then
\begin{equation}\label{eq-Sch}
\liminf_{r\rightarrow1^{-}}\frac{|f(\zeta)-f(r\zeta)|}{1-r}\geq
A_{n}-\frac{\|g\|_{\infty}}{n}\left(1+\frac{1}{2^{\frac{n}{2}}}\right).\end{equation}
In particular, if $\|g\|_{\infty}=0$, then the estimate of
{\rm(\ref{eq-Sch})} is sharp.
\end{thm}

The definition of the classical hypergeometric function $F(a,b;c;z)$ is given in Section \ref{csw-sec2}.
Next, we state a Schwarz-Pick type lemma, which is a generalization
of \cite[Theorem 1.3]{K6}, \cite[Theorem 2.12]{MM} and
\cite[Corollary 2.2]{K5}.

\begin{thm}\label{thm-3}
Let $n\geq3$ and $m\geq1$. If
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{m})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{m})$
satisfies $\Delta f=g$ for a given $g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$, then
$$|D_{f}(x)|\leq\frac{\|\mathcal{P}_{f}\|_{\infty}}{1-r}\sup_{\gamma>0}C(\gamma,r)+
\frac{n}{n+1}\|g\|_{\infty} ~\mbox{ for $x\in\mathbb{B}^{n}$},
$$
where $r=|x|$ and
$$C(\gamma,r)=\frac{4\omega_{n-3}}{\omega_{n-1}}
\frac{2^{n-1}}{(1+r)^{n-1}}\frac{1}{\sqrt{1+\gamma^{2}}}\int_{0}^{1}\frac{\Psi_{r}(\gamma
t)+\Psi_{r}(-\gamma t)}{\sqrt{(1-t^{2})^{4-n}}}dt<\infty.
$$
Here
$$\Psi_{r}(\gamma)=\int_{0}^{\frac{\gamma+\sqrt{\gamma^{2}+1-\alpha^{2}(r)}}{1-\alpha(r)}}
\frac{n-\beta(r)+n\gamma
\rho-\beta(r)\rho^{2}}{(1+\rho^{2})^{\frac{n}{2}+1}(1+\tau^{2}(r)\rho^{2})^{\frac{n}{2}-1}}\rho^{n-2}d\rho,
$$
$$\tau(r)=\frac{1-r}{1+r},~\alpha(r)=\frac{r(n-2)}{n}~\mbox{and}~\beta(r)=\frac{[n-(n-2)r]}{2}.
$$
\end{thm}

\begin{rem}
We observe that if $n=4$ in Theorem \ref{thm-3}, then
$\sup_{\gamma>0}C(\gamma,r)$ can be replaced by
$$\frac{\left[r\sqrt{4-r^{2}}(2+r^{2})+4(1-r^{2})\arctan\big(r\frac{\sqrt{4-r^{2}}}{r^{2}-2}\big)\right]}{\pi(1+r)r^{3}},
$$
where $r=|x|$ (see \cite[Theorem 1.3]{K6}). In particular, using
(\ref{eq-K}), we can obtain an explicit estimate for
$|D_{\mathcal{P}_{f}}(x)|$. But it is not going to yield a better
bound than the bound stated in Theorem \ref{thm-3}. Indeed, for any fixed
$x\in\mathbb{B}^{n}$, consider the function
$$\nu(y)=\mathcal{P}_{f}\big(x+(1-|x|)y\big).
$$
Now, if we apply (\ref{eq-K}) to $\nu/\|\mathcal{P}_{f}\|_{\infty}$,
then we deduce that
$$\left|\frac{\mathcal{P}_{f}\big(x+(1-|x|)y\big)-\mathcal{P}_{f}(x)}{|y|}-
\frac{\left[(1-|y|^{2})(1+|y|^{2})^{-n/2}-1\right]}{|y|}\mathcal{P}_{f}(x)\right|\leq\|\mathcal{P}_{f}\|_{\infty}\frac{U(|y|N)}{|y|},
$$
which, together with the fact that
$$\lim_{t\rightarrow0^{+}}\frac{(1-t^{2})(1+t^{2})^{-n/2}-1}{t}=0,
$$
implies \begin{equation}\label{eq18c}
(1-|x|)|D_{\mathcal{P}_{f}}(x)|\leq\|\mathcal{P}_{f}\|_{\infty}\frac{\partial
U(rN)}{\partial
r}\Big|_{r=0}=\frac{2\omega_{n-1}}{V(\mathbb{B}^{n})}\|\mathcal{P}_{f}\|_{\infty}=2n\|\mathcal{P}_{f}\|_{\infty},\end{equation}
where $V(\mathbb{B}^{n})$ is the volume of $\mathbb{B}^{n}$.
Finally, by  (\ref{eq18c}) and (\ref{eq-15c}),   we conclude that
\begin{equation}\label{eq-19cp}|D_{f}(x)|\leq|D_{\mathcal{P}_{f}}(x)|+|D_{\mathcal{G}_{g}}(x)|\leq\frac{2n\|\mathcal{P}_{f}\|_{\infty}}{1-|x|}+
\frac{n}{n+1}\|g\|_{\infty} ~\mbox{ for $x\in\mathbb{B}^{n}$} \end{equation} as
desired. \hfill $\Box$ \end{rem}

There exists a number of articles dealing with Landau type theorems
in geometric function theory and, for general class of functions
without some additional condition(s), there is no Landau's theorem.
See for example \cite{CK,CMPW,CP,CPW-211,CPW-2011,CPW,CPW-2015,CV,W}
and the related references therein. In our next result, we use
Theorems \ref{thm-1} and \ref{thm-3} to establish a Landau type
theorem for mappings satisfying the Poisson equation.

For $n\geq3$ and a given
$g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{n})$, we use
$\mathcal{F}_{g}$ to denote the set of all mappings
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{n})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{n})$
satisfying $\Delta f=g$ and $|f(0)|=J_{f}(0)-1=0$. Let
$\mathcal{F}_{g}^{M}$ be the set of all mappings
$f\in\mathcal{F}_{g}$ satisfying $\|f\|_{\infty}+\|g\|_{\infty}\leq
M$, where  $M$ is a positive constant.

\begin{thm}\label{thm-4}
For $n\geq3$ and a given $g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{n})$, let
$f\in\mathcal{F}_{g}^{M}$.  Then there is a positive constant $r_{0}$ depending only on $M$ and $g$ such that
$\mathbb{B}^{n}(0,r_{0})\subset f(\mathbb{B}^{n}).$
\end{thm}

\begin{rem}
We wish to point out that the Landau type theorem fails for the
family $\mathcal{F}_{g}$ without appropriate additional
condition(s). For example, consider $g(x)=(0,\ldots,0, 2n/3)$ and
$f_{k}(x)=(kx_{1},x_{2}/k, x_{3},\ldots,x_{n-1},|x|^{2}/3+x_{n})$
for $k=\{1,2,\ldots\}$, where $n\geq3$ and
$x=(x_{1},\ldots,x_{n})\in\mathbb{B}^{n}$. It is easy to see that
each $f_{k}$ is univalent and $|f_{k}(0)|=J_{f_{k}}(0)-1=0$.
Furthermore, each  $f_{k}(\mathbb{B}^{n})$ contains no ball with
radius bigger than $1/k$. Hence, there does not exist an absolute
constant $r_{0}>0$ which can work for all $k\in\{1,2,\ldots\}$, such
that $\mathbb{B}(0,r_{0}) $ is contained in the range
$f_{k}(\mathbb{B}^{n})$. Although Theorem \ref{thm-4} provides
existence of the Landau-Bloch constant for
$f\in\mathcal{F}_{g}^{M}$, an explicit estimate on the Landau-Bloch
constant is not established.
\end{rem}

The proofs of Theorems \ref{thm-1}, \ref{thm-2}, \ref{thm-3} and Corollary \ref{cor-1} will be
presented in Section \ref{csw-sec2}. Moreover, the proof of Theorem \ref{thm-4} will be
given in Section \ref{csw-sec3}.

\section{The Schwarz Lemmas for mappings satisfying Poisson's equation}\label{csw-sec2}

\subsection{M\"obius Transformations of the Unit Ball}\label{sbcsw-sec2.1}

For $x\in\mathbb{B}^{n}$, the {\it M\"obius transformation} in
$\mathbb{B}^{n}$ is defined by \begin{equation}\label{eq-ex1}
\phi_{x}(y)=\frac{|x-y|^{2}x-(1-|x|^{2})(y-x)}{[x,y]^{2}},~y\in\mathbb{B}^{n}.
\end{equation} The set of isometries of the hyperbolic unit ball is a {\it
Kleinian subgroup} of all M\"obius transformations of the extended
spaces $\mathbb{R}^{n}\cup\{\infty\}$ onto itself. In the following,
we make use of the {\it automorphism group} ${\operatorname{Aut}}(\mathbb{B}^{n})$
consisting of all M\"obius transformations of the unit ball
$\mathbb{B}^{n}$ onto itself. We recall the following facts from
\cite{Bea}: For $x\in\mathbb{B}^{n}$ and
$\phi_{x}\in{\operatorname{Aut}}(\mathbb{B}^{n})$, we have $\phi_{x}(0)=x$,
$\phi_{x}(x)=0$, $\phi_{x}(\phi_{x}(y))=y \in\mathbb{B}^{n}$,
\begin{equation}\label{II}
|\phi_{x}(y)|=\frac{|x-y|}{[x,y]}, ~1-|\phi_{x}(y)|^{2}=\frac{(1-|x|^{2})(1-|y|^{2})}{[x,y]^{2}}
\end{equation}
and
\begin{equation}\label{III}
|J_{\phi_{x}}(y)|=\frac{(1-|x|^{2})^{n}}{[x,y]^{2n}}.
\end{equation}

\subsection{Gauss Hypergeometric Functions}\label{sbcsw-sec2.2}

For $a, b, c\in\mathbb{R}$ with $c\neq0, -1, -2, \ldots,$ the {\it
hypergeometric} function is defined by the power series in the variable $x$
$$F(a,b;c;x)=\sum_{k=0}^{\infty}\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{x^{k}}{k!},~|x|<1.
$$
Here $(a)_{0}=1$,  $(a)_{k}=a(a+1)\cdots(a+k-1)$ for $k=1, 2, \ldots$, and
generally $(a)_{k}=\Gamma(a+k)/\Gamma(a)$ is the {\it Pochhammer} symbol,
where $\Gamma$ is the {\it Gamma function}. In particular, for $a, b, c>0$ and
$a+b<c$, we have (cf. \cite{PBM})
$$F(a,b;c;1)=\lim_{x\rightarrow1}
F(a,b;c;x)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}<\infty.
$$

The following result is useful in showing one of our main
results of the paper.

\begin{prop}{\rm (\cite{K2}~$\mbox{or}$~\cite[2.5.16(43)]{PBM})}\label{pro-1}
For $\lambda_{1}>1$ and $\lambda_{2}>0$, we have
$$\int_{0}^{\pi}\frac{\sin^{\lambda_{1}-1}t}{(1+r^{2}-2r\cos t)^{\lambda_{2}}}dt=
\mathbf{B}\left(\frac{\lambda_{1}}{2},\frac{1}{2}\right)
F\big(\lambda_{2},\lambda_{2}+\frac{1-\lambda_{1}}{2};\frac{1+\lambda_{1}}{2};r^{2}\big),
$$
where $\mathbf{B}(.,.)$ denotes the beta function and $r\in[0,1)$.
\end{prop}

\subsection{Proofs}
\subsection*{Proof of Theorem \ref{thm-1}}
Let $n\geq3$.
For $x,y\in\mathbb{B}^{n}$ with $x\neq y$ and $|x|+|y|\neq0$, by
\eqref{II}, we have
\begin{eqnarray}\label{eq-2}
\left|\frac{1}{|x-y|^{n-2}}-\frac{1}{[x,y]^{n-2}}\right|&=&\frac{1}{|x-y|^{n-2}}\left|1-\frac{|x-y|^{n-2}}{[x,y]^{n-2}}\right|\\
\nonumber&=&\frac{1}{|x-\phi_{x}(z)|^{n-2}}\left(1-|z|^{n-2}\right),
\end{eqnarray}
where $\phi_{x}\in{\operatorname{Aut}}(\mathbb{B}^{n})$ and $z=\phi_{x}(y)$. By \eqref{eq-ex1}, direct calculation shows that
\begin{eqnarray*}
x-\phi_{x}(z)=\frac{x[x,z]^{2}-|x-z|^{2}x+(1-|x|^{2})(z-x)}{[x,z]^{2}}
=\frac{(z-x|z|^{2})(1-|x|^{2})}{[x,z]^{2}},
\end{eqnarray*}
which gives \begin{equation}\label{eq-3}
|x-\phi_{x}(z)|=\frac{|z|(1-|x|^{2})}{[x,z]}. \end{equation} By (\ref{eq-2})
and (\ref{eq-3}), we obtain \begin{equation}\label{eq-4}
\left|\frac{1}{|x-y|^{n-2}}-\frac{1}{[x,y]^{n-2}}\right|=\frac{[x,z]^{n-2}(1-|z|^{n-2})}{|z|^{n-2}(1-|x|^{2})^{n-2}}.
\end{equation} Now, let
$g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$ be
given. Then, by (\ref{eq-p}) with $f$ in place of $\psi$, we have
$f(x)=\mathcal{P}_{f}(x)-\mathcal{G}_{g}(x), $ where
$\mathcal{P}_{f}$ and $\mathcal{G}_{g}$ are defined by
\eqref{eq-p1a}.
It follows that
\begin{eqnarray}\label{eq-5}
\nonumber|\mathcal{G}_{g}(x)|&\leq&\int_{\mathbb{B}^{n}}|G(x,y)g(y)|dV(y)\\
\nonumber
&\leq&c_{n}\|g\|_{\infty}\int_{\mathbb{B}^{n}}\left|\frac{1}{|x-y|^{n-2}}-\frac{1}{[x,y]^{n-2}}\right|dV(y),~\mbox{ by \eqref{eq-ex0}},\\
\nonumber &=&c_{n}\|g\|_{\infty}\int_{\mathbb{B}^{n}}
\frac{[x,z]^{n-2}(1-|z|^{n-2})}{|z|^{n-2}(1-|x|^{2})^{n-2}}\frac{(1-|x|^{2})^{n}}{[x,z]^{2n}}dV(z),~\mbox{ by   \eqref{II} and (\ref{eq-4})},\\
\nonumber
&=&c_{n}\|g\|_{\infty}(1-|x|^{2})^{2}\int_{\mathbb{B}^{n}}\frac{(1-|z|^{n-2})|z|^{n+2}}{|z|^{n-2}\big|x|z|^{2}-z\big|^{2+n}}dV(z)\\
&=&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{n-2}\int_{0}^{1}dr\int_{\mathbb{S}^{n-1}}\frac{r(1-r^{n-2})}{|rx-\zeta|^{2+n}}d\sigma(\zeta).
\end{eqnarray}

Using the polar coordinates and Proposition \ref{pro-1}, we obtain
\begin{eqnarray}\label{eq-6}\nonumber
\int_{\mathbb{S}^{n-1}}\frac{d\sigma(\zeta)}{|rx-\zeta|^{2+n}}&=&\frac{1}{\int_{0}^{\pi}\sin^{n-2}tdt}\int_{0}^{\pi}
\frac{\sin^{n-2}tdt}{\left(1+r^{2}|x|^{2}-2r|x|\cos
t\right)^{\frac{n+2}{2}}}dt\\ \nonumber
&=&\frac{\Gamma\big(\frac{n}{2}\big)}{\sqrt{\pi}\Gamma\big(\frac{n-1}{2}\big)}\cdot
\frac{\sqrt{\pi}\Gamma\big(\frac{n-1}{2}\big)}{\Gamma\big(\frac{n}{2}\big)}F\Big(\frac{n+2}{2},2;\frac{n}{2};r^{2}|x|^{2}\Big)\\
&=& F\Big(\frac{n+2}{2},2;\frac{n}{2};r^{2}|x|^{2}\Big).
\end{eqnarray}
By (\ref{eq-5}) and (\ref{eq-6}), we get
\begin{eqnarray}\label{eq-7}\nonumber
|\mathcal{G}_{g}(x)|&\leq&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{n-2}
\int_{0}^{1}F\Big(\frac{n+2}{2},2;\frac{n}{2};r^{2}|x|^{2}\Big)
r(1-r^{n-2})dr\\ \nonumber
&=&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{n-2}\int_{0}^{1}\sum_{k=0}^{\infty}
\frac{\Gamma\big(\frac{n}{2}\big)\Gamma\big(\frac{n}{2}+k+1\big)
\Gamma(2+k)}{\Gamma\big(\frac{n}{2}+1\big)\Gamma\big(\frac{n}{2}+k\big)\Gamma(2)}
\frac{r^{2k+1}(1-r^{n-2})|x|^{2k}}{k!}\\ \nonumber
&=&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{n-2}\int_{0}^{1}
\sum_{k=0}^{\infty}\frac{2}{n}\big(k+\frac{n}{2}\big)(k+1)r^{2k+1}(1-r^{n-2})|x|^{2k}dr\\
\nonumber
&=&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{n-2}\sum_{k=0}^{\infty}
\frac{2}{n}\big(k+\frac{n}{2}\big)(k+1)\frac{(n-2)}{2(k+1)(2k+n)}|x|^{2k}\\
\nonumber
&=&\frac{\|g\|_{\infty}(1-|x|^{2})^{2}}{2n}\sum_{k=0}^{\infty}|x|^{2k}\\
&=&\frac{\|g\|_{\infty}(1-|x|^{2})}{2n}.
\end{eqnarray}

For $\rho\in(0,1)$, let $F(x)=\mathcal{P}_{f}(\rho x)$, $x\in\mathbb{B}^{n}$. Then
\begin{equation}\label{eq-ex2}
F(x)=\int_{\mathbb{S}^{n-1}}P(x,\zeta)F(\zeta)d\sigma(\zeta),
\end{equation}
which, together with (\ref{eq-K}), yields that
\begin{equation}\label{eq-10}
\left|F(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}F(0)\right|
\leq\|\mathcal{P}_{f}\|_{\infty}U(|x|N).
\end{equation}
Applying (\ref{eq-10}), we see that
\begin{eqnarray}\label{eq-11}\nonumber
\left|\mathcal{P}_{f}(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}\mathcal{P}_{f}(0)
\right|&=&\lim_{\rho\rightarrow1^{-}}\left|F(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}F(0)\right|\\
&\leq& \|\mathcal{P}_{f}\|_{\infty}U(|x|N).\end{eqnarray} Hence, by
(\ref{eq-7}) and (\ref{eq-11}), we conclude that
\begin{eqnarray*}
\left|f(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}\mathcal{P}_{f}(0)\right|&=&\left|\mathcal{P}_{f}(x)-
\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}\mathcal{P}_{f}(0)-\mathcal{G}_{g}(x)\right|\\
&\leq&\left|\mathcal{P}_{f}(x)-\frac{1-|x|^{2}}{(1+|x|^{2})^{\frac{n}{2}}}\mathcal{P}_{f}(0)\right|+|\mathcal{G}_{g}(x)|\\
&\leq&\|\mathcal{P}_{f}\|_{\infty}U(|x|N)+\frac{\|g\|_{\infty}}{2n}(1-|x|^{2}).
\end{eqnarray*}

Now we prove the sharpness part. For
$x\in\overline{\mathbb{B}^{n}}$, let $g(x)=(-2nM,0,\ldots,0)$ and
$f(x)=(M(1-|x|^{2}),0,\ldots,0)$, where $M$ is a positive constant.
If $x\in\mathbb{S}^{n-1}$, then the optimality of (\ref{eq-thm1}) is
obvious. On the other hand, if $x=0$, then
$$\|\mathcal{P}_{f}\|_{\infty}=0~\mbox{and}~M=|f(0)-\mathcal{P}_{f}(0)|=|f(0)|
= \frac{\|g\|_{\infty}}{2n}=M.
$$
The proof of the theorem is complete.
\qed

\subsection*{Proof of Corollary \ref{cor-1}}
For $\rho\in(0,1)$, let $F(x)=\mathcal{P}_{f}(\rho x)$,
$x\in\mathbb{B}^{n}$. Then, following the proof of Theorem \ref{thm-1} and considering $F$ described by \eqref{eq-ex2},
we deduce that
\begin{eqnarray}
\nonumber
\left|F(x)-\frac{1-|x|}{(1+|x|)^{n-1}}F(0)\right|&=&
\left|\int_{\mathbb{S}^{n-1}}\Big[P(x,\zeta)-\frac{1-|x|}{(1+|x|)^{n-1}}\Big]F(\zeta)d\sigma(\zeta)\right|\\
\nonumber
&\leq&\int_{\mathbb{S}^{n-1}}\Big[P(x,\zeta)-\frac{1-|x|}{(1+|x|)^{n-1}}\Big]|F(\zeta)|d\sigma(\zeta)\\
&\leq&
\|\mathcal{P}_{f}\|_{\infty}\left[1-\frac{1-|x|}{(1+|x|)^{n-1}}\right]
\end{eqnarray*}
and therefore,
\begin{eqnarray}
\nonumber
\left|\mathcal{P}_{f}(x)-\frac{1-|x|}{(1+|x|)^{n-1}}\mathcal{P}_{f}(0)
\right|&=&\lim_{\rho\rightarrow1^{-}}\left|F(x)-\frac{1-|x|}{(1+|x|)^{n-1}}F(0)\right|\\
&\leq& \|\mathcal{P}_{f}\|_{\infty}\left[1-\frac{1-|x|}{(1+|x|)^{n-1}}\right]
\end{eqnarray*}
which by  Theorem \ref{thm-1} leads to
\begin{eqnarray*}
\left|f(x)-\frac{1-|x|}{(1+|x|)^{n-1}}\mathcal{P}_{f}(0)\right|&=&\left|\mathcal{P}_{f}(x)-\frac{1-|x|}{(1+|x|)^{n-1}}\mathcal{P}_{f}(0)-\mathcal{G}_{g}(x)\right|\\
&\leq&\left|\mathcal{P}_{f}(x)-\frac{1-|x|}{(1+|x|)^{n-1}}\mathcal{P}_{f}(0)\right|+|\mathcal{G}_{g}(x)|\\
&\leq&\|\mathcal{P}_{f}\|_{\infty}\left[1-\frac{1-|x|}{(1+|x|)^{n-1}}\right]
+\frac{\|g\|_{\infty}}{2n}(1-|x|^{2}).
\end{eqnarray*}
The proof of the corollary is complete.
 \qed

\begin{Lem}{\rm (\cite[Lemma 2.3]{K5})}\label{Lem-A}
For $r\in[0,1]$, let $\varphi(r)=\frac{\partial U(rN)}{\partial r}$.
Then $\varphi(r)$ is decreasing on $r\in[0,1]$ and
$$\varphi(r)\geq \left . \frac{\partial U(rN)}{\partial r} \right |_{r=1}=\varphi(1)=A_{n},
$$
where $A_{n}$ is the same as in
Theorem {\rm\ref{thm-2}}.
\end{Lem}

\subsection*{Proof of Theorem \ref{thm-2}}
For $x\in\mathbb{B}^{n}$, there is a $\rho\in(r,1)$ such that
\begin{equation}\label{eq-12c} \frac{1-U(rN)}{1-|x|}=\frac{\partial U(\rho
N)}{\partial r}, \end{equation} where $r=|x|$. Now, for $n\geq3$ and a given
$g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$, by
(\ref{eq-p}) with $f$ in place of $\psi$, we have
$$f(x)=\mathcal{P}_{f}(x)-\mathcal{G}_{g}(x),
$$
where
$\mathcal{P}_{f}$ and $\mathcal{G}_{g}$ are defined as in \eqref{eq-p1a}. Since
$f(0)=\mathcal{P}_{f}(0)-\mathcal{G}_{g}(0)=0$, by (\ref{eq-7}),
Theorem \ref{thm-1} and the assumptions, we see that
\begin{eqnarray}\label{eq-13c}\nonumber
|f(\zeta)-f(r\zeta)|&=&\left|f(\zeta)+\mathcal{P}_{f}(0)\frac{1-r^{2}}{(1+r^{2})^{\frac{n}{2}}}-
\mathcal{G}_{g}(0)\frac{1-r^{2}}{(1+r^{2})^{\frac{n}{2}}}-f(r\zeta)\right|\\
\nonumber &\geq&1-
\left|f(r\zeta)-\mathcal{P}_{f}(0)\frac{1-r^{2}}{(1+r^{2})^{\frac{n}{2}}}\right|-|\mathcal{G}_{g}(0)|\frac{1-r^{2}}{(1+r^{2})^{\frac{n}{2}}}\\
&\geq&1-U(rN)-\frac{\|g\|_{\infty}}{2n}(1-r^{2})-\frac{\|g\|_{\infty}}{2n}\frac{1-r^{2}}{(1+r^{2})^{\frac{n}{2}}},
\end{eqnarray}
where $r\in[0,1)$. Finally, by (\ref{eq-12c}), (\ref{eq-13c}) and Lemma \Ref{Lem-A}, there is a $\rho\in(r,1)$ such that
\begin{eqnarray*}
\frac{|f(\zeta)-f(r\zeta)|}{1-r}
&\geq&\frac{1-U(rN)}{1-r}-\frac{\|g\|_{\infty}}{2n}(1+r)-\frac{\|g\|_{\infty}}{2n}\frac{1+r}{(1+r^{2})^{\frac{n}{2}}}\\
&\geq&\frac{\partial U(\rho N)}{\partial r}-\frac{\|g\|_{\infty}}{2n}(1+r)\left (1+\frac{1}{(1+r^{2})^{\frac{n}{2}}}\right )\\
&\geq&A_{n}-\frac{\|g\|_{\infty}}{2n}(1+r)\left (1+\frac{1}{(1+r^{2})^{\frac{n}{2}}}\right ),
\end{eqnarray*}
which gives \eqref{eq-Sch}.
The sharpness part easily follows from \cite[Theorem 2.5]{K5}. The
proof of the theorem is complete. \qed

\begin{Thm}{\rm (\cite[Theorem 2.12]{MM})}\label{Lem-MM}
Let $u$ be a bounded harmonic function from $\mathbb{B}^{n}$ into
$\mathbb{R}$, where $n\geq3$. Then, for $x\in\mathbb{B}^{n}$,
$$|\nabla
u(x)|\leq\frac{\|u\|_{\infty}}{1-|x|}\sup_{\gamma>0}C(\gamma,|x|),$$
where $C(\gamma,|x|)$ is defined in Theorem {\rm\ref{thm-3}}.
\end{Thm}

\subsection*{Proof of Theorem \ref{thm-3}}
Let $n\geq3$ and $g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$ be given. As before, by
(\ref{eq-p}), we have
$$f(x)=\mathcal{P}_{f}(x)-\mathcal{G}_{g}(x),
$$
where $\mathcal{P}_{f}$ and $\mathcal{G}_{g}$ are defined as in
\eqref{eq-p1a}.
Set $\mathcal{G}_{g}=(\mathcal{G}_{g,1},\ldots,\mathcal{G}_{g,m})$
and $g=(g_{1},\ldots,g_{m})$. For $k\in\{1,2,\ldots,m\}$, we let
$$I_{k}=\int_{\mathbb{B}^{n}}\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right||g_{k}(y)|dV(y).
$$
If we apply Cauchy-Schwarz's inequality and \cite[Theorem 2.1]{K2}, it follows that
\begin{eqnarray}\label{eq16c}
I_{k}^{2}&\leq&
\int_{\mathbb{B}^{n}}\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right|dV(y)\\
\nonumber &&\times\bigg[\int_{\mathbb{B}^{n}}
\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right||g_{k}(y)|^{2}dV(y)\bigg]\\
\nonumber &\leq&
\frac{2n\pi^{\frac{n}{2}}}{(n+1)\Gamma\big(\frac{n}{2}\big)}\int_{\mathbb{B}^{n}}
\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right||g_{k}(y)|^{2}dV(y),\end{eqnarray}
which yields that
\begin{eqnarray}\label{eq17c}|\nabla
\mathcal{G}_{g,k}(x)|^{2}&\leq&\frac{I_{k}^{2}}{\omega_{n-1}^{2}}\nonumber \\
&\leq&\frac{n}{(n+1)\omega_{n-1}}\int_{\mathbb{B}^{n}}
\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right||g_{k}(y)|^{2}dV(y).
\end{eqnarray}
Then, using (\ref{eq17c}) and \cite[Theorem 2.1]{K2}, we obtain
\begin{eqnarray}\label{eq-15c}\nonumber
|D_{\mathcal{G}_{g}}(x)|&=&\sup_{\theta\in\mathbb{S}^{n-1}}\left(\sum_{k=1}^{m}|\langle\nabla
\mathcal{G}_{g,k}(x),\theta\rangle|^{2}\right)^{\frac{1}{2}}\\
\nonumber &\leq&\left(\sum_{k=1}^{m}|\nabla
\mathcal{G}_{g,k}(x)|^{2}\right)^{\frac{1}{2}}\\ \nonumber
&\leq&\bigg[\frac{n}{(n+1)\omega_{n-1}}\int_{\mathbb{B}^{n}}
\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right|\sum_{k=1}^{m}|g_{k}(y)|^{2}dV(y)\bigg]^{\frac{1}{2}}\\
\nonumber&\leq&\left[\frac{n}{(n+1)\omega_{n-1}}\right]^{\frac{1}{2}}\|g\|_{\infty}\left(\int_{\mathbb{B}^{n}}
\left|\frac{x-y}{|x-y|^{n}}-\frac{|y|^{2}x-y}{[x,y]^{n}}\right|dV(y)\right)^{\frac{1}{2}}\\
\nonumber
&\leq&\left[\frac{n}{(n+1)\omega_{n-1}}\right]^{\frac{1}{2}}\left(\frac{n\omega_{n-1}}{n+1}\right)^{\frac{1}{2}}\|g\|_{\infty}\\
&=&\frac{n}{n+1}\|g\|_{\infty}.
\end{eqnarray}

Now we estimate $|D_{\mathcal{P}_{f}}|$. For $x\in\mathbb{B}^{n}$, we may let
$\mathcal{P}_{f}(x)=(\mathcal{P}_{f,1}(x),\ldots,\mathcal{P}_{f,m}(x)).$ Then, for any
$\theta\in\mathbb{S}^{n-1}$ and $k\in\{1,2,\ldots,m\}$, by
Cauchy-Schwarz's inequality, we have
\begin{eqnarray*}
\left|\langle\nabla
\mathcal{P}_{f,k}(x),\theta\rangle\right|^{2}&=&\left|\int_{\mathbb{S}^{n-1}}\langle\nabla
P(x,\zeta),\theta\rangle\mathcal{P}_{f,k}(\zeta)d\sigma(\zeta)\right|^{2}\\
 &\leq&\left(\int_{\mathbb{S}^{n-1}}\big|\langle\nabla
P(x,\zeta),\theta\rangle\big||\mathcal{P}_{f,k}(\zeta)|d\sigma(\zeta)\right)^{2}\\
&\leq&\delta(x,\theta)\int_{\mathbb{S}^{n-1}}\big|\langle\nabla
P(x,\zeta),\theta\rangle\big||\mathcal{P}_{f,k}(\zeta)|^{2}d\sigma(\zeta),
\end{eqnarray*}
which gives that
\begin{eqnarray}\label{eq-cp1}\nonumber
 \left(\sum_{k=1}^{m}\big|\langle\nabla
\mathcal{P}_{f,k}(x),\theta\rangle\big|^{2}\right)^{\frac{1}{2}}&\leq&(\delta(x,\theta))^{\frac{1}{2}}
\left(\int_{\mathbb{S}^{n-1}}\big|\langle\nabla
P(x,\zeta),\theta\rangle\big|\sum_{k=1}^{m}|\mathcal{P}_{f,k}(\zeta)|^{2}d\sigma(\zeta)\right)^{\frac{1}{2}}\\
&\leq&\delta(x,\theta)\|\mathcal{P}_{f}\|_{\infty},
\end{eqnarray}
where
$$\delta(x,\theta)=\int_{\mathbb{S}^{n-1}}\big|\langle\nabla P(x,\zeta),\theta\rangle\big|d\sigma(\zeta).
$$
Applying (\ref{eq-cp1}), \cite[Lemma 2.3]{MM} and Theorem \Ref{Lem-MM}, we
see that, for $x\in\mathbb{B}^{n}$,
\begin{eqnarray}\label{eq-cp2}
\nonumber|D_{\mathcal{P}_{f}}(x)|&=&\sup_{\theta\in\mathbb{S}^{n-1}}\left(\sum_{k=1}^{m}\big|\langle\nabla
\mathcal{P}_{f,k}(x),\theta\rangle\big|^{2}\right)^{\frac{1}{2}}\\
\nonumber &\leq&\|\mathcal{P}_{f}\|_{\infty}
\sup_{\theta\in\mathbb{S}^{n-1}}\delta(x,\theta)\\
&\leq&
\frac{\|\mathcal{P}_{f}\|_{\infty}}{1-|x|}\sup_{\gamma>0}C(\gamma,|x|)
\end{eqnarray}
By   (\ref{eq-15c}) and (\ref{eq-cp2}),   we conclude that
$$|D_{f}(x)|\leq|D_{\mathcal{P}_{f}}(x)|+|D_{\mathcal{G}_{g}}(x)|
\leq\frac{\|\mathcal{P}_{f}\|_{\infty}}{1-|x|}\sup_{\gamma>0}C(\gamma,|x|)+ \frac{n}{n+1}\|g\|_{\infty}, ~~x\in\mathbb{B}^{n}.
$$
The proof the theorem is complete. \qed

\section{An application of the Schwarz Lemma}\label{csw-sec3}

Let $f:\,\overline{\Omega}\to
\mathbb{R}^n$ be a differentiable mapping and $x$ be a regular value
of $f$, where $x\notin f(\partial\Omega)$ and $\Omega\subset
\mathbb{R}^n$ is a bounded domain. Then the degree
$\deg(f,\Omega,x)$ is defined by the formula (cf. \cite{Ll,V})
$$\deg(f,\Omega,x):=\sum_{y\in f^{-1}(x)\cap\Omega}\mbox{ sign} \big(\det J_{f} (y)\big).
$$

\begin{Lem}\label{Lem-X}
The $\deg(f,\Omega,x)$ satisfies the following properties {\rm (cf.
\cite[p.~125-129]{RR}):}
\begin{enumerate}
\item[(a)] If $x\in\mathbb{R}^ n \backslash f(\partial D)$ and $\deg(f,\overline{\Omega},x)\neq 0$, then there exists an $w\in\Omega$ such that $f(w)=x$.
\item[(b)]\label{(II)} If $D$ is a domain with $\overline{D}\subset\Omega$ and $x\in \mathbb{R}^ n \backslash f(\partial D)$,
then $\deg(f,D,x)$ is a constant on each component of $\mathbb{R}^
n \backslash f(\partial D)$.
\end{enumerate}
\end{Lem}

\begin{lem}\label{lem-L}
Let $n\geq3$ and $m\geq1$. For a given
$g\in\mathcal{C}(\overline{\mathbb{B}^{n}},\mathbb{R}^{m})$, if
$f\in\mathcal{C}^{2}(\mathbb{B}^{n},\mathbb{R}^{m})\cap\mathcal{C}(\mathbb{S}^{n-1},\mathbb{R}^{m})$
satisfies $\Delta f=g$ and $\|f\|_{\infty}+\|g\|_{\infty}\leq M$ for some constant $M>0$,
then there is a constant $L>0$ such that
$$|f(x_{1})-f(x_{2})|\leq L|x_{1}-x_{2}| ~\mbox{ for all
$x_{1},x_{2}\in\overline{\mathbb{B}^{n}(x_{0},\rho_{0})}$,}
$$
where $x_{0}\in\mathbb{B}^{n}$ and
$\rho_{0}\in(0,1-|x_{0}|)$ are some constants.
\end{lem}
\begin{pf} By Theorem \ref{thm-3} or (\ref{eq-19cp}), for all
$x\in\overline{\mathbb{B}^{n}(x_{0},\rho_{0})}$, we have
$$|D_{f}(x)|\leq\frac{\|\mathcal{P}_{f}\|_{\infty}}{1-|x|}\sup_{\gamma>0}C(\gamma,|x|)+
\frac{n}{n+1}\|g\|_{\infty}\leq\frac{2nM}{1-\rho_{0}-|x_{0}|}+\frac{n}{n+1}M:=L,
$$
which implies, for all $x_{1},x_{2}\in\overline{\mathbb{B}^{n}(x_{0},\rho_{0})}$,
$$|f(x_{1})-f(x_{2})|\leq\int_{[x_{1},x_{2}]}|D_{f}(x)|\, |dx|\leq L\int_{[x_{1},x_{2}]}|dx|=L|x_{1}-x_{2}|,$$
where $[x_{1},x_{2}]$ is the segment from $x_{1}$ to $x_{2}$ (or
$x_{2}$ to $x_{1}$) with the endpoints $x_{1}$ and $x_{2}$. \end{pf}

\subsection*{Proof of Theorem \ref{thm-4}}

We prove the theorem by the method of contradiction. Suppose that
the result is not true. Then there is a sequence $\{a_k\}$ of
positive real numbers, and a sequence of functions $\{f_k\}$ with
$f_{k}\in\mathcal{F}_{g}^{M}$, such that $\{a_k\}$ tends to $0$ and
$a_k \notin f_k(\mathbb{B}^{n})$ for $k\in\{1,2,\ldots\}$. By
Theorem \ref{thm-1}, Lemma \ref{lem-L} and Arzel${\rm \grave{a}}$-Ascoli's theorem,
we know that there is a subsequence $\{f_{k}^{\ast}\}$ of $\{f_{k}\}$ which converges
uniformly on a compact subset of $\mathbb{B}^{n}$ to a function
$f^{\ast} $. For each $k$, it is easy to see that the function
$h_k=f_{k}^{\ast}-f_{1}^{\ast}$ is harmonic. As a consequence, the
sequence $\{h_k\}$ converges uniformly on compact subsets of
$\mathbb{B}^{n}$ to $f^{\ast}-f_{1}^{\ast}$ and therefore, the
partial derivatives of $f_{k}^{\ast}$ converge uniformly on compact
subsets of $\mathbb{B}^{n}$ to the partial derivatives of $
f^{\ast}$. In particular, $f_{k}^{\ast} (0) \rightarrow f^{\ast}(0)$
and $J_{f_{k}^{\ast}} (0) \rightarrow J_{f^{\ast}} (0)$ as
$k\rightarrow\infty$, which imply that $f^{\ast} \in
\mathcal{F}_{g}^{M}$. Since $J_{f^{\ast}} (0)-1=|f^{\ast}(0)|=0,$
there are $r_0\in(0,1)$ and $c_1
> 0$ such that $J_{f^{\ast}} >0$ on $
\overline{\mathbb{B}^{n}(0,r_{0})}$,
$f^{\ast}(\mathbb{B}^{n}(0,r_{0})) \supset
\overline{\mathbb{B}^{n}(0,c_1)}$ and $|f^{\ast}(x)| \geq c_1$ for
$x \in
\partial\mathbb{B}^{n}(0,r_{0})$.

Now, we let $c_2=c_1/2$, $\mathbb{B}_{r_{0}}=\mathbb{B}^{n}(0,r_{0})$ and
$\mathbb{B}_{c_{2}}= \mathbb{B}^{n}(0,c_2) $. Then there is a $k_0$
such that $|f^{\ast}_{k}(x)| \geq c_2$ for $k \geq k_0$ and
$J_{f^{\ast}_k}
>0$ on $\overline{\mathbb{B}_{r_{0}}}$. Since $\deg(f^{\ast}_k,\mathbb{B}_{r_{0}},0)\geq 1$,
by Lemma \Ref{Lem-X}, we see that, for $y \in \mathbb{B}_{c_{2}}$
and $k \geq k_0$, $\deg(f^{\ast}_k,\mathbb{B}_{r_{0}},y)\geq 1$.
Hence, for $k \geq k_0$, $f^{\ast}_k(\mathbb{B}_{r_{0}})\supset
\mathbb{B}_{c_{2}}$ which contradicts our assumption.
The proof of the theorem is complete.
\qed

\bigskip

{\bf Acknowledgements:}  This research was partly supported by the
National Natural Science Foundation of China ( No. 11571216 and No.
11401184),  the Science and Technology Plan Project of Hunan
Province (No. 2016TP1020) and the Construct Program of the Key
Discipline in Hunan Province. The second author is currently on leave from IIT Madras.

\subsection*{Conflict of Interests}
The authors declare that there is no conflict of interests regarding the publication of this paper.

redpajama_set_name: RedPajamaArXiv


text:
UPDATE: 26 candidates file for Congressional seat
Jannette Pippin Daily News Staff
Mar 8, 2019 at 4:30 PM Mar 8, 2019 at 6:08 PM
The following candidates have filed to run in the upcoming special election for the vacant North Carolina District 3 Congressional seat previously held by the late Walter Jones. The filing period ended Friday.
Richard Bew
Ike Johnson
Dana E. Outlaw
Ernest T. Reeves
Gregory Humphrey
Kevin Baiko
Paul Beaumont
Graham Boyd
Celeste Cairns
Gary Ceres
Chimer Davis Clark Jr.
Don Cox
Francis X. De Luca
Phil Law
Jeff Moore
Greg Murphy
Michele Nix
Mike Payment
Joan Perry
Eric Rouse
Phil Shepard
Michael Speciale
Shannon Wilson Bray
Greg Holt
As a result of the random drawing held by the N.C. Board of Elections to determine the order in which candidates names will appear on the ballot, 3rd Congressional District candidates whose last names begin with the letter "Z" will appear first in contests in the primary, second primary (if necessary) and general election. A coin flip determined that additional candidates will follow in reverse alphabetical order: Z, Y, X, W, V, … A.
The ballot order selection process was streamed live on Facebook and is available on the State Board's Facebook page, https://www.facebook.com/NCSBE/.
The bid to fill the vacant Congressional seat previously held by the late Walter Jones will be a highly contested one.
As of 4 p.m. Friday 21 candidates had filed for the special election that has been scheduled to fill North Carolina's 3rd District Congressional seat for the remainder of Jones' unexpired term. Jones died Feb. 10 just months after being elected to a 13th term in the U.S. House of Representatives.
As the week-long filing period closed Friday, and the list of candidates include 15 Republicans, four Democrats, one libertarian, and one candidate from the Constitution Party.
The packed field of Republican candidates will mean a highly contested primary to determine the party's nominee to proceed to the general election and bid to maintain a Republican hold on the seat. The Democrats will also have a primary.
The special election is scheduled to be held July 9, with the primary election on April 30. If a second primary is required, it will be held July 9 and the general election on Sept. 10.
According to the State Board of Elections, the candidate who receives the second highest number of votes may demand a second primary if no candidate receives more than 30 percent of the votes. The top two vote-getters would be on the ballot for the second primary.
Democrats filing included: Ike Johnson of Jacksonville, Dana E. Outlaw of New Bern, Ernest T. Reeves of Greenville, and Richard Bew of Newport.
Republican candidates include: Phil Law of Jacksonville, Phil Shepard of Jacksonville, Michele Nix of Kinston, Michael Speciale of New Bern, Greg Murphy of Greenville; Gary Ceres of Greenville, Chimer Davis Clark Jr. of Morehead City, Graham Boyd of Wake Forest, Celest Cairns of Emerald Isle, Paul Beaumont of Shawboro, Jeff Moore of Raleigh, Joan Perry of Kinston, Kevin Baiko of Moyock, Francis De Luca of Cary and Eric Rouse of Kinston.
Shannon Bray, of Apex, filed as a Libertarian candidate and Greg Holt, of New Bern, filed as a candidate for the Constitution Party.
Reporter Jannette Pippin can be reached at 910-382-2557 or Jannette.Pippin@JDNews.com.


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text:
> Words of wine
Words of wine
A AOC - Acidity - Alcohol -
B BIVC - Bacteria - Balance - Bud - Budburst
C Caillottes - Cask - Caudalie - Cellar -
D Decanting - De-earthing - Disbudding
E Empyreumatic -
F Fault (of Sancerre) - Flowering
G Geology -
H Harsh - Hoeing
I INAO -
K Kieselguhr - Kimmeridgien
L Label - Leaf - Lees
M Maceration - Malolactic fermentation - Marc - Marls
N Neck-label - Noble rot - Nose
O Oak cask - Oenology - Opening out of the vine buds
P Palate - Pedology - Peppery
Q Quality -
R Racking - Refractometer - Residual sugars
S SICAVAC - "Saint Paul" - "Saint Vincent" - Selection - Sensory analysis
T Tannin - Tartar - Taste (sense of) - Tasting
V Vine - Vine branch - Vineyard - Vintage
W White soils - Wine - Wine grower - Wine making method
Y Yeasts - Yield
A.O.C. - Appellation d'Origine Contrôlée. Created in 1935 and administrated by the professionals within the framework of the French National Institute for the Labels of Origin (I.N.A.O), A.O.C labels guarantee the origin, the authenticity and the quality of the product. It exists in France for more than 400 A.O.C wines.
Acidity - Generic term which qualifies the fresh taste of a wine and contributes to its balance.
Alcohol - Main constituent of the wine after water. It comes from the fermentation of sugars by yeasts.
BIVC - Bureau Interprofessionnel des Vins du Centre. Interprofessional house for the wines of the region Centre. Created in 1994, the B.I.V.C groups together the wine growers and the wine merchants of the region. It permits the economic follow-up of the vineyards as well as the qualitative follow-up of wines after their launching on the market. It is also in charge of the promotion of wines and public relations within France and abroad.
Bacteria - Microorganisms present in wine which belong to different species,some favorable, some not, to its development. Lactic bacteria permit a second obligatory fermentation for red wines (the malolactic fermentation).
Balance - Harmony between the various elements from the first nose to the finish.
Bud - Shoot which grows at the start of the leaves. The buds will produce grapes.
Budburst - The time when the buds develop and come out of the bud after the winter dormancy. In the vineyards of Centre-Loire, budburst takes place during the month of April.
Caillottes - Term that refers to the chalky-calcareous soils of the Sancerre appellation. They cover 35% of the appellation surface area. The wines produced on these soils generally develop faster and have very developed floral notes.
Cask - Wooden container invented by Gauls. The size is different according to the regions.
Caudalie - Unit of measure for the finish in the mouth of a wine length of after-taste. One caudalie corresponds to one second.
Cellar - Term that designates the place where the vinification is made and where wine growers age and preserve their wines.
Decanting - It consists in pouring the wine into a carafe to separate it from the deposits (mainly for old wines). It also aerates wines with a closed nose.
De-earthing - The process of removing excess soil covering the vine plant.
Disbudding - The removal of some of the buds borne by the vine. It permits a better control of the yields. The exposure of the grapes to the sun is also improved.
Empyreumatic - Aromas of roasting with a slight toasted taste (toasted bread, smoke, tobacco etc).
Fault (of Sancerre) - Caused by a seismic movement, this north-south oriented fault that goes through the village of Sancerre, created the bed of the river Loire. The great diversity of terroirs (in Sancerre and the Coteaux du Giennois) is due to this fault.
Flowering - Stage of the vine growth cycle that determines the quantity of grapes. The date of the blossoming may indicate the date of grape harvest.
Geology - Science describing soils and subsoils. Geology influences the vine growth.
Harsh - Term used for an extremely astringent wine.
Hoeing - Land work that does not require chemical weed-killer.
I.N.A.O - (Institut National des Appellations d'Origine). French National Institute for the Labels of Origin (A.O.C and A.O.V.D.Q.S). Created in 1935, this professional organization is in charge of administrating and granting the labels of origin. The chairman is appointed by the Ministry of Agriculture on the proposal of the winemakers.
Kieselguhr - Powder of diatoms stemming from siliceous rock, used for the filtering of wines.
Kimmeridgien - Geological stratum of the Jurassic era (secondary era). It corresponds to the white soils, where white wines, particularly good for long-term cellaring, are produced.
Label - Created towards 1860, the label is used to notify the consumer of compulsory comments (name of the appellation, capacity, alcohol content of the wine, volume, surname, address and occupation of the bottler ). The vintage, the awards won and the grape varieties are optional comments.
Leaf - Thanks to photosynthesis, this part of the vine feeds the plant and especially produces the sugars of the grape.
Lees - They are formed by the sedimentation of (dead) yeasts after the fermentation. They are usually eliminated during the rackings.
Maceration - Operation that consists in bringing into contact the juice and the solid matters (grape skins) in order to obtain colour (for red wines). The skin contact is a variant, sometimes used to increase the aromatic content of white wines.
Malolactic fermentation - Transformation of malic acid into lactic acid. It naturally makes the wine less acid. In Centre-Loire, only red wines do this fermentation.
Marc - Residue of the white grape pressing. Solid matters remaining after the devatting of red wines.
Marls - Chalky-clayey soils that contain between 65 and 95% of clays. It mainly represents the white soils situated in the geological stratum called kimmeridgien.
Neck-label - Label on the neck of the bottle with the mention of the vintage.
Noble rot - Beneficial form of the Botrytis Cinerea development that leads to a concentration of sugars and to a modification of the organoleptic characters.
Nose - Organ of the sense of smell which is essential for the perception of a taste. It is also used to define aromas (aroma of citrus fruits etc).
Oak cask - A wooden receptacle intended to keep the wine during the fermentation and/or the maturing. In Centre-Loire, barrels are used for red wines.
Oenology - The science of the wine.
Opening out of the vine buds - When the buds grow and open out after winter. In the Centre-Loire vineyards it occurs during April.
Palate - Upper part of the mouth which perceives tactile and thermic sensations.
Pedology - Study of the soils and their chemical, physical and organic characters as well as their evolution and their properties.
Peppery - Spicy character found in some red and rosé wines.
Quality - Main characters of a grape, a must or a wine…
Racking - It consists in transferring the wine to separate it from the lees settled at the bottom of a vat or a barrel or to aerate it.
Refractometer - A device used to measure the density of a liquid by the use of light refraction. This tool is used to follow the evolution of the grapes maturity.
Residual sugars - Sugars, no transformed into alcohol during the fermentation, present in the wine.
SICAVAC - (Service Interprofessionnel de Conseil en Agronomie, Vinification et Analyses du Centre). Interprofessional department of advice in agronomy, wine making, analyses of the region Centre. Professional technical centre created in 1994, specialized in oenological/wine studies and advice.
"Saint Paul" - Patron saint of farmers. He is celebrated in the villages of Menetou-Salon and Morogues.
"Saint Vincent" - Patron saint of wine growers. Celebrated the 22nd of January, it gives rise to a great reunion among wine growers for one or more days.
Selection - Selection of the highest quality grapes during the harvest, avoiding the rotten or unripe grapes.
Sensory analysis - Very precise description of all the feelings perceived during tasting.
Tannin - Substance contained in the grapes and that brings a feeling of astringency to the wine.
Tartar - Crystalline deposit left by a wine after the end of its fermentation.
Taste (sense of) - Sense that permits the perception of flavours. The main 4 tastes are: sweetness, acidity, bitterness and saltiness.
Tasting - Sensory analysis of the wine's qualities. Three stages are necessary: the eye, the nose, the mouth.
Vine - The vine is a liana in a natural state. Most of the wine produced in the world comes from the Vitis Vinifera species.
Vine branch - Green shoot of the vine. The vine branch becomes the vine shoot when it turns into wood.
Vineyard - All the vines of a wine estate or an appellation.
Vintage - Year of a wine harvest. Its mention is not obligatory but when it is mentioned on the bottle, it means that 100% of the wine has to come from this vintage.
White soils - Soil made of clay and limestone, found in the western part of the Sancerre vineyard. Called ''terres blanches".
Wine - Alcoholic drink obtained from the partial or total fermentation of the grape.
Wine grower - Person who grows his own grape vine and vinifies his grapes.
Wine making method - Operations that go with the transformation of grapes into wine.
Yeasts - Unicellular organism permitting the fermentation of wines.
Yield - Production of a vine per hectare. In the Centre-Loire vineyards, the maximum yields for white wines represent 65 hectolitres per hectare; for rosé wines, 63 hectolitres per hectare and for red wines, 59 hectolitres per hectare. The control of yields is the first quality factor of a wine.


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text:
Watt se poate referi la următorii autori care au denumit cel puțin o specie:

 David Allan Poe Watt
 George Watt
 John Charles Watt

redpajama_set_name: RedPajamaWikipedia


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text:
UFE is Coming to Brooklyn! 1 day, 2 events
October 10, 2011 — mlapham
Join United for a Fair Economy and Responsible Wealth on Sunday, October 30th for two events in Brooklyn, NY. Please RSVP for one or both of these events to Kathy Lique at [email protected] or 617-423-2148 x121 to get directions and tell us how many will attend with you.
How Do We Coddle the Super-Wealthy?
September 30, 2011 — Shannon M.
What is #OccupyWallStreet & How Can You Join?
September 28, 2011 — Martin Bourqui
After visiting the Occupy Wall Street protests on their seventh day yesterday I feel like I need to put in writing — and in longer than 140 character status updates! — what I see and hear about what's going on here and the way in which these protests are being perceived, both by me and by those on the right and the left.
VIDEO: Mike Lapham vs. CNBC Teleposse
September 23, 2011 — Maz
Responsible Wealth director, Mike Lapham was invited onto CNBC's "The Kudlow Report" to discuss Obama's recent tax proposal, including "The Buffett Plan" to raise taxes on millionaires. In the show's typical bullying fashion, Lapham was posed with opposition not just from two guests, but also from the host and "moderator" himself, Larry Kudlow.
VIDEO: Scott Brown's Party, Crashed!
Hundreds of Bay State activists rolled up on Scott Brown's $1000 per plate fundraiser at a harborside restaurant in Boston to scold the Senator for consistently voting against the interests of working families.


and sent to Catch 22 Solutions.
For answers, contact me, Cherie-Lynn. My contact info is on the right.
Catch 22 Solutions is a program of Agora Employment Essentials | The Government of Canada has contributed funding to this initiative.