text:
Q: R Shiny How to access slider min & max values in server code Is there a way to reference the minimum & maximum values of a slider within the server.R portion of a Shiny app?
For instance, given the following definition within ui.R:
sliderInput("slider1", "", min = 0, max = 100, value = 50)

how do I determine that the preset minimum is 0, and the maximum  is 100?
Thank you.

A: It depends on what you want to do with the min and max values when you access them.  I assume that you don't want to just read them, since you would already know their fixed values.  If you want to be able to manipulate them, you could try defining the slider in server.R using renderUI.  Then you could set the min and max parameters to variables, which could be manipulated elsewhere.  Below is an example of that.
ui.R
library(shiny)
shinyUI(fluidPage(
  titlePanel("Access the min and max of a slider"),
  sidebarLayout(
    sidebarPanel(uiOutput("SliderWidget")),
    mainPanel()
  )
))

server.R
library(shiny)
shinyServer(function(input, output) {
  SlideMax = 100
  SlideMin = 0
  output$SliderWidget <- renderUI({
    sliderInput("Slider1","",min = SlideMin,max = SlideMax,value = 50)
  })
})

Of course, your question actually asks how to determine the preset values.  I'm not sure I understand that.  I think that the preset values would be static, so there would be no need to determine them.

A: I've now found that one can have a third file as part of a shiny app, called global.R.
So, for the use case above I could define the min and max parameters as constants in the global.R file or dynamically create the slider as per Paul's suggestion.


text:
Shoshana Damari (ur. 31 marca 1923 w Dhamar (Jemen), zm. 14 lutego 2006 w Tel Awiwie) – izraelska wokalistka, nazywana Królową żydowskiej piosenki.

Do Izraela przyjechała w 1925 r., a jej rodzina osiedliła się w Riszon le-Cijjon. W wieku 14 lat zaczęła śpiewać w izraelskim radiu. Zasłynęła z wykonań wielu hebrajskich piosenek.

Najbardziej znane utwory wykonywane przez Shoshanę Damari to: Mah Tovu, Migdalor, Or.

Linki zewnętrzne
 Biografia na stronie Jewish Virtual Library (ang.)

Urodzeni w 1923
Zmarli w 2006
Izraelskie wokalistki
Ludzie urodzeni w Zamarze

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text:
Hacker Fund
Mentor Spotlight
Introducing Hacker Fund
08 February 2016 on News
The Mentor Army
Student: "I was told to find the guy in the cat suit if I needed a mentor."
Me: "Hellllllllllyeah!"
Hacker Fund is a 501(c)(3) public charity that is dedicated to providing resources for computer science education. But how did we go from idea to a squad of mentors in animal onesies around the country?
It all started with a Facebook post in Hackathon Hackers.
I was sitting on the flight from HackGT and kept remembering how distraught collegiate hacker Anthony Lobko was when he realized he couldn't raise enough funds to attend CalHacks. Here's a student who built a hologram in 36 hours and wants to keep learning in the hackathon environment but can't make it to his next event because of a lack of funds. It was painful.
I felt compelled to ask the community for their opinion. So I posted this into a popular Facebook group, Hackathon Hackers:
That post went viral in the group and the very next day, I bought the domain hacker.fund to begin the journey. At that time, I was consulting local LA startups about growth-hacking alongside my cofounders Justin Ezor (Zoro) & Shahan Khan, and was also finishing my sixth month as Coaching Program Director of Major League Hacking. When I proposed this idea of Hacker Fund to Shahan & Zoro, they were immediately interested in applying to YC with the idea.
We didn't get in. That's when we made our first pivot.
We noticed that there was no support for hackathons at the high school level. But while I was at Major League Hacking, I noticed that high school students were building projects that made it to Top 3 at the collegiate competitions.
Why not try to organize a high school hackathon?
Shahan and Zoro joined me at a meeting with the school board of my alma mater, Beverly Hills High School. During that conversation, the Chief Academic Officer walked into the room and interrupted our hackathon pitch with the request of a lifetime:
Can you consult our school district and assist the "Intro to Programming" teachers with delivering relevant computer science education?
It was at that moment that our team realized that Hacker Fund could be much more than just a hackathon organization, so we incorporated Hacker Fund as a nonprofit.
We received our 501(c)(3) tax exemption in June 2015 and built a strong founding Board of Directors to join us, comprised of vocal figures from the hackathon community: Daniel Eckert, Ari Vaniderstine, Cassidy Williams, David Mace, Connor Crowley, and Ishmael Philip. We had friends at Google, Microsoft, Apple, Snapchat, and Tinder who all wanted to volunteer because they remembered a time in which they wanted a mentor.
We felt unstoppable.
After inspiring one educator, the waterfall of introductions came pouring down on us and we made it to one of the largest high school districts in the nation, Los Angeles Unified School District. We began to receive questions from teachers like:
"How do we connect to the industry?"
"Can you bring an engineer from a famous tech company into our classroom?"
"Will you teach our kids after-school?"
"Can you organize a hackathon for our school?"
"Do you have summer opportunities?"
"Is there a place students can go outside of school to continue learning?"
All of those questions became projects...and we did it all.
We organized events for K-12 students with the Office of Los Angeles Mayor Eric Garcetti and the i.am.angel Foundation.
We supported HSHacks, the largest high school hackathon in history, checking in 1000+ students who wanted to learn computer science.
We staffed the Harold Robinson Foundation's summer camps with our mentors.
We partnered with Rhubarb Studios to create an enrichment center for after-school programs in the tallest building West of the Mississippi River, the US Bank Tower in Los Angeles.
But the epiphany came when we received this question from an administrator who was directing curriculum standards in her school district:
"How do we teach like you?"
BOOM. That was it.
It wasn't our curriculum — students can get it online.
It wasn't our hardware — teachers can buy it online.
It was our mentors — we are friends with them, online.
We realized that we were friends with top professionals from the tech industry. Thank you, hackathons.
Now, it's 2016 and we are on a mentour — a cross-country quest to build mentor communities that can change the face of computer science education. We are recruiting, training, and staffing mentors who want to give back to the community by inspiring the next generation of technologists.
If you're interested in becoming a mentor and inspiring students to believe that they will hack, click here to apply.
Justin Brezhnev
Annual Report FY15
Dear Mentors, Donors, Teachers, Students, & Parents - In our last year of operations, we engaged with 150 different…
Hacker Fund © 2021

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text:
Attempts On Her Life – May 2007
May11 by sheilajtheatre
By: Martin Crimp
Directed by: Katie Mitchell + company
Venue: Lyttelton Theatre
This was dreadful. Not all the way through, but it's an hour and three quarters I'll never have back again.
The performance style was based on modern media. The actors didn't play specific characters, instead they morphed in and out of various roles, as well as moving cameras and lights around, filming other actors, then becoming the focus of the cameras themselves. There was a huge screen lowered down so the audience could see what was being filmed, intercut with footage shot previously and with pictures layered and superimposed. All very technical, but to what end?
The general idea seemed to be to look at the role of women in our media-driven society, and particularly issues around women committing suicide. There was no specific woman – it's any woman. There's a good section looking at the use of women as sexual objects of desire in advertising, in this case, advertising a car. The advert (in Russian?) was translated into English, so we could get the humour. At the end, the usual caveats are scrolled across the screen, and the combination of these over pictures of a sexy woman, make it clear that we're not meant to read the words – it's the advertising equivalent of small print.
Another good part was the Abba imitation – the style is as for one of their Eighties' hits, but the words are much tougher. The police interrogation sketch didn't work so well for me – there have been so many comedy send-ups, never mind Life On Mars, that I found most of it just boring. There was one good line, though, when the coppers are pushing this guy to sign his statement, and he says he hasn't got a pen.
Apart from that, I enjoyed the Newsnight Review sketch, with recognisable imitations of regular participants, e.g. Germaine Greer. Otherwise, I could barely get through the turgid stuff that was passing for a theatrical performance. No criticism of the actors is intended, even though they participated in the staging. I just didn't find this performance style remotely engaging, in fact, quite the reverse. The use of cameras, the screen, mikes for the actors, etc., meant the whole piece was distanced from the audience – we might as well have watched a film, and the actors might as well have been acting in an empty theatre for all the exchange that was going on between us.
The opening section showed a bit of promise. The prison doors of the stage curtain creaked open to reveal a vast open space, filled with the cameras, etc that took such a central role later on. All the cast are milling around, and finally come forward to talk through some ideas about a woman, like a group of creatives at an advertising agency. There are a few good lines, but mostly, it's a jumble, and not at all clear where it's going. However, I stuck with it (unlike one gentleman behind us), and, sadly, was disappointed. The chorus line effect was repeated at the end, only in an even more incoherent fashion, though as I'd pretty much lost interest by this time, I really didn't care.
With no characters, plot or anything resembling a play taking place on stage, it was impossible to get involved in these performances or any of the issues raised. The distancing effects previously mentioned added to that, and I actually felt disrespected as an audience member, and increasingly irrelevant. For the first time, I chose not to applaud at the end. I will go a long way to avoid seeing anything this banal again.
This entry was posted in 3/10, Drama, Movement, Theatre and tagged Katie Mitchell, Lyttelton Theatre, Martin Crimp, NT.
← Office Suite – May 2007
A Midsummer Night's Dream – May 2007 →


Battle Rifles are full-powered, full-length rifles that use either an integral or detachable magazine. Battle Rifles are available to the Assault, Support and Recon classes.
A battle rifle has two main definitions that exist, given the term has existed prior to World War 2 with different contexts. The term originated from the United States.
The post-WW2 definition of a battle rifle - the one used in present day and most commonly used - is: a full-length rifle that fires a full-powered rifle cartridge (e.g. 7.62x51mm NATO), is capable of select-fire and is magazine fed. The term was used to differentiate rifles that fired intermediate rifle cartridges such as the German Sturmgewehr 44 from weapons that fired full-powered rifle rounds. Examples of such battle rifles are the Belgian FN FAL. However, the term can also be applied to weapons made before 1945, such as the German FG 42, Soviet AVS-36 and American M1918 BAR.
Prior to 1945, the term was mainly used to describe weapons such as the American M1 Garand, German Gewehr 43 and Soviet SVT-40, with the definition roughly being: an automatic full-length rifle that fires a full-powered cartridge. This definition is still used today, mostly by those in the United States. The term "battle rifle" was also used often as a synonym to a service or infantry-rifle before the 20th century.
"Battle rifle" as a term is not commonly used throughout the world. European nations tend to classify weapons that would fall under the 'battle rifle' moniker as assault rifles, such as Norway or Austria. The term is mainly used in western, English-speaking nations. Most nations do not use the term officially in their military, although due to popular culture, the term has become more well-known in other countries in both their armed forces and members of the public. The U.S still remains the predominant user of the term.
During the late 1960s and onwards, the concept of the assault rifle became far more appealing to armed forces; double the ammunition for the same weight, higher velocity and less recoil were such factors. Battle rifles such as the M14 and FAL gradually were withdrawn from service during the late 20th century in favor of rifles such as the M16, Steyr AUG and L85. However, battle rifles remained in service in limited capacities or remained the standard-issue rifle, such as the German G3 that remained in service with the West German Bundeswher up until the early 1990s. Battle rifles are still in large usage today, however. The Belgian FN SCAR-H made by FN Herstal - a modern successor to the older FAL, the American Mk 14 - a modernized M14 or the German HK417 see front-line service as either an alternative to the usual assault rifle, offering more stopping power, or as Designated Marksman Rifles.
Much like Light Machine Guns, the traits and attributes of a battle rifle have them perform best in mid-range combat. Like LMGs, Battle Rifles deal high damage, have a much more slow and controllable Rate of Fire (RoF) and posses an extremely high recoil compared to other weapons. Unlike LMGs, Battle Rifles possess a very low amount of ammunition, but mostly have a quicker reload time and lighter weight, making them much more capable on standing on their own compared to the LMG, which acts more as a support weapon.
While the large ammunition pool of the LMG allows for continuous firing and mistakes, Battle Rifles have a much lower amount of ammunition, making accuracy and trigger control extremely important when using these weapons. Also, since the low ammunition pool will force anyone using these weapons to reload often, it is much more preferable to fight on one-on-one engagements with Battle Rifles, rather than engaging with multiple enemies than is expected from LMGs. Still, due to a low RoF but high damage, it is best to use this weapon in mid to long range engagements, though it is still recommended to be careful and not fire automatically unless if in Close Quarters Combat (CQC) or in closer mid-range combat.
With the exception of the FAL 50.00, Battle Rifles do not possess any alternative aiming mode, making visual and gun recoil much more challenging to work with. As a side note, the lack of a more open aiming mode further disallows Battle Rifles to perform favorably in CQC. Still, as recoil can be negated to an extent with specific attachments, and skilled trigger control will be used more in longer engagements when using these weapons, one can easily work around Battle Rifle recoil.
While a lot like Light Machine Guns, Battle Rifles possess special characteristics that differentiate themselves from LMGs and give them a completely different usage. Battle Rifles, with their high damage and controllable RoF, these weapons outperform in mid to long range combat than most other weapons. While possessing a low ammunition pool and rather lengthy reload, Battle Rifles can still hold better by themselves than LMGs, mostly due to a quicker reload time; it is still recommended to engage in one-on-one engagements due to make up for the lack of ammunition thereof. Just be careful not to try engage in CQC.


text:
Lisa Chun, HAWAIIUSA
Christina O'Connor
Some members of HawaiiUSA's Pearl City branch Polar Plunge team: (back) Nate Brown, Natasha Daley, Jaci Miyashiro, (front) Vera Kim-Todd and Lisa Chun. Photo courtesy HawaiiUSA FCU
On April 1, daring volunteers from around the island will jump into a pool filled with 10 tons of ice. While this may sound like a torturous task, the participants are braving the cold in the name of Special Olympics Hawaii, which provides year-round activities for children and adults with disabilities.
Polar Plunge raises money and awareness for Special Olympics athletes. Among the participants is presenting sponsor HawaiiUSA Federal Credit Union, which has been volunteering with Special Olympics since 2007. HawaiiUSA AVP-Community & Business Development Lisa Chun says that the company's employees are eager to help at Special Olympics events.
"Every year, our employees always are all waiting to help," Chun says of the company's annual involvement with the Summer Games.
As the presenting sponsor of this year's Polar Plunge, HawaiiUSA is working to raise money for Special Olympics. To reach its financial goals, HawaiiUSA is selling Polar Plunge icons at all of its branches, conducting internal fundraising campaigns and collecting donations.
HawaiiUSA employees also have a friendly competition going: The staff has been divided into teams for fundraising, and the losing team has to get in the water. Plus, a group of 20 HawaiiUSA employees – including Chun – has volunteered to take the plunge whether they win or lose.
"I'm very impressed with the Special Olympics," Chun says. "There are so many athletes, and Special Olympics only has seven full-time employees. They work so hard, and they impact so many people.
"Since we have been volunteering since 2007, we have gotten to know a lot of the athletes, coaches and parents," Chun adds. "The (athletes) always come show us the medal that they won. You can't put a price tag on that."
In addition to its work with Special Olympics, HawaiiUSA also is involved with a number of other volunteer efforts. Chun explains that giving back to the community is an integral part of the company's mission, and it strives to help bring recognition to a variety of nonprofit organizations. This year, the company also will work with March of Dimes, Making Strides Against Breast Cancer, Shriners Hospital and Lanakila Pacific, in addition to continuing its work at the Special Olympics Summer Games.
Polar Plunge takes place on the lawn at Waterfront Plaza beginning at 10 a.m. The event also will feature a costume contest, food and entertainment. For more information or to register to plunge, visit specialolympicshawaii.org.
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HAWAIIUSALisa Chun
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text:
Walid Shoebat Former Muslim Brotherhood
Member Now Peace Activist
Articles and Press Releases
Evidence: The Story of Walid Shoebat. It Really Checks Out.
Comments from Walid Shoebat
Obamas' Wahhabist Fundraising Empire
Innocence of Muslims
CNN Smear Campaign
The CNN Hatchet Job
CNN Reporter Drew Griffin is a Proven Liar
⇒ NavigateAbout Me – Articles and Press Releases – Evidence: The Story of Walid Shoebat. It Really Checks Out. – Comments from Walid Shoebat Important Information – Obamas' Wahhabist Fundraising Empire – Innocence of Muslims – Huma Abedin – Palestinian Justice – CNN Smear Campaign – – The CNN Hatchet Job – – I Confess – – CNN Reporter Drew Griffin is a Proven Liar Donate Contact
You are here: Home › Benghazi Select Committee › Featured › General › IkhwanLeaks › THE FIVE DAMNING LETTERS Ignored By The Obama Administration and Hillary Clinton Regarding Muslim Brotherhood Infiltration
THE FIVE DAMNING LETTERS Ignored By The Obama Administration and Hillary Clinton Regarding Muslim Brotherhood Infiltration
By Shoebat Foundation on June 7, 2014 in Benghazi Select Committee, Featured, General, IkhwanLeaks
By Walid Shoebat and Ben Barrack
**SHOEBAT EXCLUSIVE**
The Benghazi attacks and subsequent coverup could have been prevented. The Obama administration, to include everyone who sided with him, are responsible for not heeding the words of five U.S. Congressmen who sent five separate letters to various agencies and departments exactly 90 days earlier.
On June 13th 2012, five congressmen sent some very serious warning letters regarding Muslim Brotherhood infiltration, one of which was sent to the Deputy Inspector General (IG) of the State Department, led by then Secretary of State Hillary Clinton.
The letters caused so much controversy that nothing was done about them. The congressmen contacted and consulted with Walid Shoebat to etch some details regarding such Muslim Brotherhood infiltration.
Today, resurrecting these letters (discussed later) makes an incredible argument for bringing the Obama Administration, the media, and the sharp tongues of some stupid politicians to answer some tough questions.
[CLICK ON IMAGE BELOW TO VIEW 42 PAGES WORTH OF DETAILED LETTERS REGARDING THIS SUBJECT]
Letter about Muslim Brotherhood infiltration sent 90 days before Benghazi.
One of those Congressmen who sent the letters is Rep. Lynn Westmoreland (R-GA) who sits on the newly formed House Select Committee on Benghazi, chaired by Rep. Trey Gowdy.
(Note: due to Westmoreland's role on the Select Committee, we will refer to these five letters as being from him as he is the only signatory who was assigned to the committee).
Each Inspector's General (IG) office that received these letters was given 90 days to respond.
Due to the political controversy caused by the letters' addressing Muslim infiltration, none responded or even cared. It gets even worse. No one even paid attention that those 90 days expired ON THE DAY of the Benghazi attacks.
The red line was crossed by these Obama administration agencies and departments when they ignored that warning and the deadline.
And not only were there no consequences but due to the attacks and the smear campaigns against the five courageous congressmen, neither Westmoreland nor any of the other signatories kept the issue alive.
Westmoreland, who also sits on the House Intelligence Committee, needs to pick up the ball where it was left off, especially since the House Select Committee on Benghazi is compromised by the appointment of Philip Kiko as the Executive Director, which is in itself an extremely warranted concern as explained by Shoebat.com [here and here].
Rep. Lynn Westmoreland: Cannot deny he knows.
Of the five letters Westmoreland signed on June 13, 2012, one was sent to the Deputy Inspector General (IG) of the State Department, led by then Secretary of State Hillary Clinton (who was cc'd on the letter). One of the other letters was sent to the IG of the Office of Director of National Intelligence (ODNI), which presides over the Central Intelligence Agency (CIA), led at the time by James Clapper and David Petraeus respectively. Perhaps more than any other, State and CIA are the two entities most central to and impacted by the Benghazi attacks.
So what do the Benghazi attacks have to do with Muslim Brotherhood infiltration? And what type of infiltration did these letters discuss?
Westmoreland's letter to Deputy IG Howard Geisel includes the following excerpts regarding Muslim Brotherhood infiltration of the State Department:
The State Department and, in several cases, the specific direction of the Secretary of State, have taken actions recently that have been enormously favorable to the Muslim Brotherhood and its interests.
One of those actions identified by Westmoreland alleges that a series of meetings with the Organization of Islamic Cooperation (OIC) involved attempts to do EXACTLY what blaming the video for the Benghazi attacks was supposed to do. The letter sent 90 days before the Benghazi attacks stated that these meetings…
…are now known as "the Istanbul Process" and are aimed at finding ways to accommodate the OIC's demands for restrictions on freedom of expression guaranteed by the U.S. Constitution, so as to preclude "blasphemy" against Islam and its adherents.
As Shoebat.com has clearly and extensively demonstrated, attempts to blame the video for the Benghazi attacks were specifically designed to do exactly what Westmoreland suggested the OIC was attempting to accomplish.
Then Secretary of State Hillary Clinton with Rashad Hussain in 2010.
The most prominent Muslim figure to work at the State Department is stealth jihadist Rashad Hussain, who is the U.S. envoy to… the OIC.
In fact, our "Ironclad" Report includes a screenshot of the YouTube channel belonging to the man who produced the very anti-Muhammad video the Obama administration itself attempted to blame for the attacks. The only two men featured on the "Feed" page of Nakoula Basseley Nakoula (aka Sam Bacile) – who contacted us earlier this year to persuade us of his good intentions – were the same exact two men at the heart of the campaign in Egypt, Wisam Abdul Waris (top) and Nader Bakkar (bottom). These men sought to blame the video for riots across the Middle East for the purpose of silencing free speech:
YouTube Channel of anti-Muhammad video filmmaker.
Chillingly, the June 13th letter to Geisel also included the following excerpt after Westmoreland enumerated five specific concerns, including the one about the OIC's efforts to suppress speech mentioned above:
We believe these actions and policies are deeply problematic. They may even pose security risks for this nation, its people and interests.
Westmoreland was proven right on the very day that the deadline for a response to his letter had expired. He was proven right in the subsequent days as well, when the administration intentionally parroted the lie that the anti-Muhammad video was responsible for the attacks. That lie was perpetuated in the Arab world for one reason – to lobby for non-Muslim governments to enact and enforce laws that made criticism of Islam a crime. The Obama administration helped to push this narrative every time it blamed the video; Obama inexplicably did so himself a full two weeks after the attacks (Exhibit J in our "Ironclad" Report).
Kirkpatrick: Pushed the anti-Muhammad video lie more than a year after Benghazi attacks.
Perhaps even more shockingly, the New York Times Cairo Bureau Chief David Kirkpatrick pushed the narrative more than one year later, on December 28, 2013 in a fictional missive masquerading as journalism, as Shoebat.com reported. Wrote Kirkpatrick:
Months of investigation by The New York Times, centered on extensive interviews with Libyans in Benghazi who had direct knowledge of the attack there and its context, turned up no evidence that Al Qaeda or other international terrorist groups had any role in the assault. The attack was led, instead, by fighters who had benefited directly from NATO's extensive air power and logistics support during the uprising against Colonel Qaddafi. And contrary to claims by some members of Congress, it was fueled in large part by anger at an American-made video denigrating Islam.
In response to the Kirkpatrick article's lies, Westmoreland said he thought the Times was…
"…trying to absolve [Clinton] from the lack of security that was sent over there, the number of requests for security that was turned down."
In reality, all Westmoreland had to do is to stop heeding to political pressure, read his own letter to learn what the initial reason for the video narrative was about. The very Muslim Brotherhood infiltration that Westmoreland sounded alarm bells over is why the narrative of the Obama administration matched the agenda of the Muslim Brotherhood in the Middle East.
Westmoreland knows this!
Besides Rashad Hussein, two other Muslim Brotherhood agents who work at the State Department are Farah Pandith and Salam al-Marayati. In 2009, Pandith was sworn in – on the Qur'an – as the U.S. Representative to Muslim Communities by Secretary Clinton, as Shoebat.com reported.
Hillary swears in Pandith on Qur'an in 2009.
Perhaps more relevant to Westmoreland's letters as they relate to State Department infiltration is Pandith's appearance at an Islamic Eid dinner just two days after the Benghazi attacks, inside the Ben Franklin room. Joining Pandith besides Hillary was Ali Sulaiman Aujali, then Libya's Ambassador to the U.S., as Shoebat.com reported.
Aujali has a history that includes speaking at Muslim Brotherhood front groups. In 2011, as reported by Shoebat.com, he spoke at the ISNA convention and was introduced by the Executive Director of CAIR-Chicago, Ahmed Rehab:
Sulaiman Aujali and Ahmed Rehab at 2011 ISNA Convention.
Is Pandith far from being involved? No. In 2006, Pandith joined a Muslim Brotherhood agent who was also the recipient of the 'smoking gun' Benghazi email sent by a White House Deputy three days after the attacks (detailed below). Accompanying her was a stealth Muslim Brotherhood agent named Yahya Basha, who is closely connected to a very large stealth jihadist fish, the chairman of the very notorious Dar al-Hijrah mosque, Bassam Estwani, as Shoebat.com has revealed.
As for Muslim Public Affairs Council (MPAC) founder and Executive Director Salam al-Marayati, he has a history that includes time as the State Department's envoy to the Organization for Security and Co-operation in Europe (OSCE).
As Shoebat.com revealed, one of the figures on the distribution list of what's been billed as the 'smoking gun' email that ties the anti-Muhammad video narrative to the White House, is a Muslim Brotherhood agent. Mehdi K. Alhassani received the email sent by White House Deputy Ben Rhodes on September 14, 2012 that instructed Susan Rice to lie during her appearance on the September 16th Sunday talk shows.
Alhassani: National Security Council Staff, Executive Office of the President
As a Muslim Brotherhood infiltrator, Alhassani would almost necessarily have the same agenda as Wisam Abdul Waris and Nader Bakkar, mentioned above. Blaming the video was about censoring speech critical of Islam.
Alhassani served as President of the Muslim Students Association (MSA) – a Muslim Brotherhood front group – chapter at George Washington University (GWU) and has extensive ties to Muslim Brotherhood groups and individuals. In 2006, he joined Pandith and Basha on European Muslim Brotherhood goodwill tour. Alhassani was on the receiving end of this email in his capacity as an employee of the White House. More specifically, Alhassani was / is Special Assistant, Office of the Chief of Staff, National Security Council Staff, Executive Office of the President [EOP]. Here is a screen shot of the email read by Alhassani three days after the Benghazi attacks but not shown to the American people for well more than one year later:
Why did Alhassani receive 'Smoking gun' email?
More than 90 days earlier, in his June 13, 2012 letter to I. Charles McCullough III, IG for the ODNI, Westmoreland provided examples that represent cause for concern about Muslim Brotherhood infiltration of the Intelligence Community. One of those concerns was the demonstrably false testimony of the Director of National Intelligence (DNI) James Clapper, which included Clapper's assertion that the Muslim Brotherhood is "largely secular" and "has eschewed" violence. Westmoreland then wrote:
These sorts of activities strike us as deeply problematic with respect to the national interest and security, and are within the scope of your duties under Section 2 of the Inspector General Act of 1978, as amended, "to conduct and supervise audits and investigations relating to the programs and operations of the [Office of the Director of National Intelligence] and… to provide a means for keeping the head of the establishment and the Congress fully and currently informed about problems and deficiences relating to the administration of such programs and operations and the necessity for and progress of corrective action."
As was the case with the State Department's IG office, McCullough did not respond to Westmoreland's letter despite it being brought to his attention that the head of the ODNI, Clapper said this more than a year earlier, as the Arab Spring was unfolding:
John Brennan, the man who would replace David Petraeus as CIA Director one day after Obama's 2012 re-election, is also cause for great concern. In his prior position as Homeland Security Advisor, Brennan received a letter from a woman named Farhana Khera, Executive Director and President of Muslim Advocates, a group with extensive connections to the Muslim Brotherhood. Khera herself is featured on the website of the Muslim Brotherhood's largest front group in the U.S., as Shoebat.com reported.
In her letter to Brennan, co-signed by dozens of Muslim organizations, Khera urged him to create an "inter-agency task force led by the White House" with the directive that it "purge" counter-terrorism training materials the Muslim Brotherhood front groups found offensive. Shockingly, Brennan responded to the letter in the affirmative, granting Khera most of what she requested.
CIA Director Brennan
Based on what Westmoreland has demonstrated that he knows, he should already know that the Executive Director of the Committee on which he now sits – Philip Kiko – is registered to lobby on behalf of Muslim Advocates. In 2013, as Shoebat.com reported, Kiko was employed by the Smith-Free Group that counts the Leadership Conference on Civil Rights (LCCR) among its clients. Muslim Advocates is listed as a member of LCCR.
According to the Insider's Guide to Key Committee Staff of the U.S. Congress in 2010, Kiko:
…served as chief of staff on the House Judiciary Committee, where he helped craft the USA Patriot Act and pushed for its first two extensions.
According to Khera's own bio:
Ms. Khera was Counsel to the U.S. Senate Judiciary Committee, Subcommittee on the Constitution, Civil Rights, and Property Rights. In the Senate, she worked for six years directly for Senator Russell D. Feingold (D-WI), the Chairman of the Constitution Subcommittee. Ms. Khera focused substantially on the Patriot Act, racial and religious profiling, and other civil liberties issues raised by the government's anti-terrorism policies after September 11, 2001.
The Patriot Act was clearly watered down to avoid identifying the face of America's Muslim Brotherhood enemies. Khera's letter to Brennan illustrates exactly how this kind of thing is done. As an individual who "helped craft" the Patriot Act, Kiko was part of a compromised or worse, corrupted process. As such, the Benghazi investigation is compromised by having such an individual as its Executive Director.
In conclusion, the blood of four dead Americans including the silence on all the excuses, is on the hands of all who sided with the Obama administration or even dismissed the warnings of those five congressmen. Westmoreland should take the opportunity to resurrect these concerns if he is serious about getting to the bottom of what happened at those U.S. installations in Benghazi on September 11, 2012. If he does not, he risks being complicit in a coverup because unlike the other Select Committee members, Westmoreland's letters demonstrate he does not have plausible deniability.
It makes no difference how well-intentioned or pure Westmoreland or Committee chair Gowdy may be. If this committee is compromised by the likes of someone who has a stake in doing just that, based on what he's done in the past, every member of this committee risks having to do the same thing in the future based on what they will have done in the past.
Rep. Lynn Westmoreland (far right) is the one Committee member who cannot plead ignorance.
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text:
\section{Introduction}

Given a pair of points in the closure of the unit disk ${\mathbb B}^2\,,$ there exists a unique hyperbolic geodesic line joining these two points.
Hyperbolic lines are simply sets of the form $C \cap \mathbb{B}^2$, where $C$ is a circle perpendicular to the unit circle, or a Euclidean diameter of
$\mathbb{B}^2\,.$ For a quadruple of four points
$\{a,b,c,d\}$ in the closure of the unit disk, we can draw these hyperbolic lines
joining each of the four pairs of points $\{a,b\},$ $\{b,c\},$ $\{c,d\},$ and $\{d,a\}\,.$ If these hyperbolic lines bound a domain $D \subset {\mathbb B}^2$  such that the points $\{a,b,c,d\}$ are in the positive order on the boundary of the domain, then we say
that the quadruple of points $\{a,b,c,d\}$ determines a hyperbolic quadrilateral $Q(a,b,c,d)\,$ and that the points $a,b,c,d$ are its vertices.
A hyperbolic quadrilateral with angles equal to $\pi/2, \pi/2, \pi/2, \phi\,(0\leq \phi < \pi/2)\,,$ is
called a hyperbolic {\em Lambert} quadrilateral \cite[p. 156]{be}, see Figure 1. Observe that one of the vertices of a Lambert quadrilateral may be on the unit circle, in which case the angle at that vertex is $\phi=0\,.$

In this paper, we study bounds for the product and the sum of
two hyperbolic distances between the opposite sides of hyperbolic Lambert quadrilaterals in the unit disk. Also, we consider the same product expression for the images of these hyperbolic Lambert quadrilaterals under quasiconformal mappings from the unit disk onto itself. In particular, we obtain similar results for ideal hyperbolic quadrilaterals, i.e., in the case when all the vertices are on the unit circle and all the angles are zero.
This follows, because an ideal hyperbolic quadrilateral can be subdivided
into four Lambert quadrilaterals.

For the formulation of our main results we introduce some notation --
further notation will be given below in Section 2.
Let $J^*[a,b]$ be the hyperbolic geodesic line with end points $a\,,b\in\partial\mathbb{B}^2,$
and let $J[a,b]$ be the hyperbolic geodesic segment joining $a$ and $b$ when $a,b \in {\mathbb B}^2\,,$ or the hyperbolic geodesic ray when one of the two points $a\,,b$ is on $\partial\mathbb{B}^2$.

Given two nonempty subsets $A, B$ of $ {\mathbb{B}^2}$ (or of the upper
 half plane $ \mathbb{H}^2\,$),
let $d_\rho(A,B)$ denote the hyperbolic distance between them,  defined as
 $$d_\rho(A,B)=\inf_{{x\in A\atop y\in B}}\rho(x,y)\,,$$
where $\rho(x,y)$ stands for the hyperbolic distance \eqref{th} (or \eqref{cosh} in the case $ A, B\subset\mathbb{H}^2$).

We now formulate our main results.

\begin{theorem}\label{Lamdd}
Let $Q (v_a\,,v_b\,,v_c\,,v_d)$ be a hyperbolic Lambert quadrilateral in $\mathbb{B}^2$ and let the quadruple of interior angles
$(\frac{\pi}{2}\,,\frac{\pi}{2}\,,\phi\,,\frac{\pi}{2})$, $\phi\in[0, \pi/2)\,,$ correspond to the quadruple $(v_a\,,v_b\,,v_c\,,v_d)$ of vertices. Let
$d_1=d_\rho(J[v_a,v_d]\,,J[v_b,v_c])\,,$  $d_2=d_\rho(J[v_a,v_b]\,,J[v_c,v_d])$ (see Figure \ref{Lamb}), and
let $L={\rm th} \rho(v_a,v_c)\in(0,1]$.
Then
\begin{eqnarray*}
d_1d_2\leq \left({\rm arth}\left(\frac{\sqrt 2}{2}L\right)\right)^2 \,.
\end{eqnarray*}
The equality holds if and only if $v_c$ is on the bisector of the interior angle at $v_a$.
\end{theorem}

\medskip
\begin{figure}[h]
\centering
\includegraphics[width=7cm]{lambfig1.pdf}
\caption{\label{Lamb} A hyperbolic Lambert quadrilateral in $\mathbb{B}^2$.}
\end{figure}
\medskip

\begin{theorem}\label{Lamdad}
Let $Q(v_a\,,v_b\,,v_c\,,v_d)$\,, $d_1$, $d_2$ and $L$ be as in Theorem \ref{Lamdd}.
Let $m= \sqrt{(2-L^2)(3L^2-2)}$, $r_0=\sqrt{\frac{1-m/L^2}{2}}$ and $r'_0=\sqrt{1-r^2_0}$.

(1) If $0<L\leq\sqrt{\frac 23}$, then
\begin{eqnarray*}
{\rm arth}\,L<d_1+d_2\leq {\rm arth}\left(\frac{2\sqrt 2L}{2+L^2}\right).
\end{eqnarray*}
The equality holds in the right-hand side if and only if $v_c$ is on the bisector of the interior angle at $v_a$.

(2) If $\sqrt{\frac 23}<L<\sqrt{2(\sqrt 2-1)}$, then
\begin{eqnarray*}
{\rm arth}\,L<d_1+d_2\leq {\rm arth}\left(\frac{L(r_0+r'_0)}{1+L^2r_0r'_0}\right).
\end{eqnarray*}
The equality holds in the right-hand side if and only if the interior angle between $J[v_a,v_b]$ and $J[v_a,v_c]$ is $\arccos r_0$ or $\arccos r'_0$.

(3) If $\sqrt{2(\sqrt 2-1)}\leq L<1$, then
\begin{eqnarray*}
{\rm arth}\left(\frac{2\sqrt 2L}{2+L^2}\right)\leq d_1+d_2\leq {\rm arth}\left(\frac{L(r_0+r'_0)}{1+L^2r_0r'_0}\right).
\end{eqnarray*}
The equality holds in the left-hand side if and only if $v_c$ is on the bisector of the interior angle at $v_a$.
The equality holds in the right-hand side if and only if the interior angle between $J[v_a,v_b]$ and $J[v_a,v_c]$ is $\arccos r_0$ or $\arccos r'_0$.

(4) If $L=1$, then
\begin{eqnarray*}
d_1+d_2\geq {\rm arth}\left(\frac{2\sqrt 2}{3}\right)\,.
\end{eqnarray*}
The equality holds if and only if $v_c$ is on the bisector of the interior angle at $v_a$.
\end{theorem}

In a Lambert quadrilateral, the angle $\phi$ is related to the lengths
$d_1$, $d_2$ of the sides "opposite" to it as follows
\cite[Theorem 7.17.1]{be}:
$${\rm sh}\, d_1{\rm sh}\, d_2=\cos\phi.$$
See also the recent paper of A. F. Beardon and D. Minda \cite[Lemma 5]{bm}.
The proof of Theorem \ref{Lamdd} yields the following corollary, which provides a connection between $d_1$, $d_2$ and $L={\rm th} \rho(v_a,v_c)$.

\begin{corollary} Let $L$, $d_1$ and $d_2$ be as in Theorem \ref{Lamdd}. Then
\begin{eqnarray*}
{\rm th}^2\,d_1+{\rm th}^2\,d_2=L^2.
\end{eqnarray*}
\end{corollary}

By Theorem \ref{Lamdd} and Theorem \ref{Lamdad}, we obtain the following corollary which deals with the ideal hyperbolic quadrilaterals.

\begin{corollary}\label{dd}
Let $Q(a,b,c,d)$ be an ideal hyperbolic quadrilateral in $\mathbb{B}^2$. Let $d_1=d_\rho(J^*[a,d],J^*[b,c])$ and
$d_2=d_\rho(J^*[a,b], J^*[c,d])$ (see Figure \ref{thdd}).
Then
\begin{eqnarray*}
d_1d_2\leq \left(2\log(\sqrt{2}+1)\right)^2
\end{eqnarray*}
and
\begin{eqnarray*}
d_1+d_2\geq 4\log(\sqrt{2}+1)\,.
\end{eqnarray*}
In both cases the equalities hold if and only if $|a,b,c,d|=2$.
\end{corollary}

\medskip
\begin{figure}[h]
\centering
\includegraphics[width=7cm]{lambfig2.pdf}
\caption{\label{thdd} An ideal hyperbolic quadrilateral in $\mathbb{B}^2$.}
\end{figure}
\medskip

\begin{remark}
$|a,b,c,d|=2$ means that there exists a M\"obius transformation $f$ such that $f(a)=1$,$f(b)=i$,$f(c)=-1$,$f(d)=-i$, see \eqref{crossratio}.
\end{remark}

\begin{theorem}\label{thqcLam}
Let $f: \mathbb{B}^2\rightarrow\mathbb{B}^2$ be a $K$-quasiconformal mapping with $f \mathbb{B}^2= \mathbb{B}^2$
and let $Q(v_a, v_b, v_c, v_d)$, $d_1\,, d_2$, $L$ be as in Theorem \ref{Lamdd}. Let $A(K)$ be as in Lemma \ref{leqc} and $f_L(r)$ be as in Lemma \ref{lecr}(1) by taking $c=L$.
Denote $D_1=d_\rho(f(J[v_a,v_d]),f(J[v_b,v_c]))$ and
$D_2=d_\rho(f(J[v_a,v_b]), f(J[v_c,v_d]))\,.$

(1) If $0<L\leq \frac{e^2-1}{e^2+1}\approx0.761594$, then
$$D_1D_2\leq A(K)^2 \left({\rm arth}\left(\frac{\sqrt 2}{2}L\right)\right)^{2/K} .$$

(2) If $\frac{e^2-1}{e^2+1}<L\leq1$, then let $r_L=\frac 1L\frac{e^2-1}{e^2+1}\approx\frac{0.761594}{L}$ and $$M_L=\frac{f_L(\sqrt{1-r^2_L})}{f_L(r_L)}>1.$$
Let $r_L(K)$ be the unique solution $r$ to the equation $K f_L(r)=f_L(\sqrt{1-r^2})$ with $r_L<r<1$.
Further, define
$$T(x,L)={\rm arth}(L x)\left({\rm arth}\left(L\sqrt{1- x^2}\right)\right)^{1/K}\,,\,0<x<1.$$
Then
\begin{eqnarray*}
D_1D_2\leq A(K)^2\max\left\{T(r_L(K),L),\,\, \left({\rm arth}\left(\frac{\sqrt 2}{2}L\right)\right)^{2/K} \right\}
\end{eqnarray*}
if $K> M_L$, and
\begin{eqnarray*}
D_1D_2\leq A(K)^2\max\left\{T(r_L,L),\,\, \left({\rm arth}\left(\frac{\sqrt 2}{2}L\right)\right)^{2/K} \right\}
\end{eqnarray*}
if $1\leq K\leq M_L$.
\end{theorem}

\begin{corollary}\label{thqc}
Let $f: \mathbb{B}^2\rightarrow\mathbb{B}^2$ be a $K$-quasiconformal mapping with $f \mathbb{B}^2= \mathbb{B}^2\,$
and let $Q(a,b,c,d)$, $d_1, d_2$ be as in Corollary \ref{dd}. Let $A(K)$ be as in Lemma \ref{leqc} and $f_1(r)$ be as in Lemma \ref{lecr}(1)
by taking $c=1$. Denote $D_1=d_\rho(f(J^*[a,d]), f(J^*[b,c]))$ and
$D_2=d_\rho(f(J^*[a,b]),f(J^*[c,d])).$
 Further denote $r_1=\frac{2 \sqrt e}{e+1}\approx0.886819$ and
$$M_1=\frac{(e-1)(\log(\sqrt e+1)-\log(\sqrt e-1))}{ \sqrt e}\approx 1.46618\,$$
and
define $r_1(K)$ to be the unique solution $r$ to the equation $K f_1(r)=f_1(\sqrt{1-r^2})$ with $r_1<r<1$.
With the notation
$$T(x)={\rm arth}(x)\left({\rm arth}(\sqrt{1- x^2})\right)^{1/K}\,,\,0<x<1\,,$$
we have
\begin{eqnarray*}
D_1D_2\leq A(K)^2\max\left\{2^{1+1/K}T(r_1(K)),\,\, \left(2\log(\sqrt{2}+1)\right)^2 \right\}
\end{eqnarray*}
if $K> M_1$, and
\begin{eqnarray*}
D_1D_2\leq A(K)^2\max\left\{2^{1+1/K}T(r_1),\,\, \left(2\log(\sqrt{2}+1)\right)^2\right\}
\end{eqnarray*}
if $1\leq K\leq M_1$.
\end{corollary}

\section{Preliminaries}

It is assumed that the reader is familiar with basic definitions of geometric
function theory and quasiconformal mapping theory, see e.g. \cite{be,v}.
We recall here some basic information on hyperbolic geometry \cite{be}.

The chordal metric is defined by
\begin{equation}\label{q}
\left\{\begin{array}{ll}
q(x,y)=\frac{|x-y|}{\sqrt{1+|x|^2}\sqrt{1+|y|^2}},&\,\,\, x\,,y\neq\infty,\\
q(x,\infty)=\frac{1}{\sqrt{1+|x|^2}},&\,\,\, x\neq\infty,
\end{array}\right.
\end{equation}
for $x,y\in\overline{\mathbb{R}^2}$.

For an ordered quadruple $a,b,c,d$ of distinct points in $\overline{\mathbb{R}^2}$ we define the absolute ratio by
$$|a,b,c,d|=\frac{q(a,c)q(b,d)}{q(a,b)q(c,d)}.$$
It follows from (\ref{q}) that for distinct points $a,b,c,d\in \mathbb{R}^2$
\begin{equation}\label{crossratio}
|a,b,c,d|=\frac{|a-c||b-d|}{|a-b||c-d|}.
\end{equation}
The most important property of the absolute ratio is M\"obius invariance, see \cite[Theorem 3.2.7]{be}, i.e., if $f$ is a M\"obius transformation, then
$$|f(a),f(b),f(c),f(d)|=|a,b,c,d|,$$
for all distinct $a,b,c,d\in\overline{\mathbb{R}^2}$.

For a domain $G\subsetneq \mathbb{R}^2$ and a continuous weight function $w: G\rightarrow(0,\infty)\,,$  we define the weighted length of a rectifiable curve $\gamma\subset G$ to be
$$\ell_w(\gamma)=\int_{\gamma}w(z)|dz|$$
and the weighted distance between two points $x,y \in G $ by
$$d_w(x,y)=\inf_{\gamma}\ell_w(\gamma),$$
where the infimum is taken over all rectifiable curves in $G$ joining $x$ and $y$ ($x=(x_1,x_2),\,y=(y_1,y_2)$). It is easy to see that $d_w$ defines a metric on $G$ and $(G,d_w)$ is a metric space. We say that a curve $\gamma: [0,1]\rightarrow G$ is a geodesic joining $\gamma(0)$ and $\gamma(1)$ if for all $t\in (0,1)$, we have
$$d_w(\gamma(0),\gamma(1))=d_w(\gamma(0),\gamma(t))+d_w(\gamma(t),\gamma(1)).$$
The hyperbolic distance in $\mathbb{H}^2$ and $\mathbb{B}^2$ is defined in terms of the weight functions
$w_{\mathbb{H}^2}(x)=1/{x_2}$ and  $w_{\mathbb{B}^2}(x)=2/{(1-|x|^2)}\,,$ resp.  We also have the corresponding explicit formulas
\begin{equation}\label{cosh}
\cosh\rho_{\mathbb{H}^2}(x,y)=1+\frac{|x-y|^2}{2x_2y_2}
\end{equation}
for all $x,y\in \mathbb{H}^2$  \cite[p.35]{be}, and
\begin{equation}\label{th}
{\rm th}\frac{\rho_{\mathbb{B}^2}(x,y)}{2}=\frac{|x-y|}{\sqrt{|x-y|^2+(1-|x|^2)(1-|y|^2)}}
\end{equation}
for all $x,y\in \mathbb{B}^2$  \cite[p.40]{be}. In particular, for $t\in(0,1)$,
\begin{equation}\label{arth}
\rho_{\mathbb{B}^2}(0,t e_1)=\log\frac{1+t}{1-t}=2{\rm arth} t.
\end{equation}

There is a third equivalent way to express the hyperbolic distances.
Let $G\in\{\mathbb{H}^2,\mathbb{B}^2\}$, $x,y\in{G}$ and let $L$ be an arc of a circle
perpendicular to $\partial G$ with $x,y\in L$ and let
$\{x_*,y_*\}=L\cap\partial G$, the points being labelled so that $x_*, x, y, y_*$ occur in this order on $L$. Then by \cite[(7.2.6)]{be}
\begin{equation}\label{rho}
\rho_G(x,y)=\sup\{\log|a,x,y,b|:a,b\in\partial G\}=\log|x_*,x,y,y_*|.
\end{equation}
We will omit the subscript $G$ if it is clear from the context.
The hyperbolic distance is invariant under M\"obius transformations of $G$ onto $G'$ for $G,\,G'\in\{\mathbb{H}^2,\mathbb{B}^2\}$.

Hyperbolic geodesics are  arcs of circles which are orthogonal to the boundary of the domain. More precisely, for $a,b\in \mathbb{B}^2$ (or $\mathbb{H}^2)$, the hyperbolic geodesic segment joining $a$ to $b$ is an arc of a circle orthogonal to $S^1$ (or $\partial \mathbb{H}^2)$. In a limiting case the points $a$ and $b$ are located on a Euclidean line through $0$ (or located on a normal of $\partial \mathbb{H}^2$), see \cite{be}.  Therefore, the points $x_*$ and $y_*$ are the end points of the hyperbolic geodesic. For any two distinct points the hyperbolic geodesic segment is unique (see Figure \ref{h2} and \ref{b2}). For basic facts about hyperbolic geometry we refer the interested reader to \cite{a}, \cite{be} and \cite{kl}.

\medskip
\begin{figure}[h]
\begin{minipage}[t]{0.45\linewidth}
\centering
\includegraphics[width=8cm]{lambfig3.pdf}
\caption{\label{h2} Hyperbolic geodesic segments in $\mathbb{H}^2$.}
\end{minipage}
\hfill
\hspace{1cm}
\begin{minipage}[t]{0.45\linewidth}
\centering
\includegraphics[width=6cm]{lambfig4.pdf}
\caption{\label{b2} Hyperbolic geodesic segments in $\mathbb{B}^2$.}
\end{minipage}
\end{figure}
\medskip

By \cite[Exercise 1.1.27]{k} and \cite[Lemma 2.2]{kv}, for  $x\,,y \,\in \mathbb{R}^2\setminus\{0\}$ such that $0,x,y$ are noncollinear,
the circle $S^1(a,r_a)$ containing $x, y$ is orthogonal to the unit circle, where
\begin{equation}\label{orar}
a =i\frac{y(1+|x|^2)-x(1+|y|^2)}{2(x_2y_1-x_1y_2)}\,\,\,\, {\rm and}\, \,\,\,r_a=\frac {|x-y|\big|x|y|^2-y\big|}{2|y||x_1y_2-x_2y_1|}\,.
\end{equation}

For $r,s\in(0,+\infty)$, the \emph{H\"older mean of order $p$} is defined by
$$H_p(r,s)=\left(\frac{r^p+s^p}{2}\right)^{1/p} \quad\mbox{for}\quad{p\neq 0},
\quad H_0(r,s)=\sqrt{r\,s}.$$
For $p=1$, we get the arithmetic mean $A=H_1$; for $p=0$, the
geometric mean $G=H_0$; and for $p=-1$, the harmonic mean $H=H_{-1}$.
It is well-known that $H_p(r,s)$ is continuous and increasing with respect to $p$.
Many interesting properties of H\"older means are given in \cite{bu} and \cite{hlp}.

A function $f:I\to J$ is called {\it $H_{p,q}$-convex (concave)} if it satisfies
$$f(H_p(r,s))\leq(\geq)H_q(f(r),f(s))$$
for all $r,s\in I$, and  {\it strictly $H_{p,q}$-convex (concave)} if the
inequality is strict except for $r=s$. For $H_{p,q}$-convexity
of some special functions the reader is referred to \cite{avv2, ba, wzj1, wzj2, zwc}.

Some other notation is also needed in the paper. Let $[a,b]$ be the Euclidean segment with end points $a$ and $b$.
Let $X, Y$ be the real axis and imaginary axis, resp.
Let ${\rm Arc}(abc)$ be the circular arc with end points $a,c$ and through $b$, and ${\rm SArc}(ac)$ be semicircle with end point $a,c$.

\medskip

The next lemma, so-called {\em
monotone form of l'H${\rm \hat{o}}$pital's rule}, has found recently numerous applications in proving inequalities. See the extensive bibliography of \cite{avz}.

\begin{lemma} \label{lhr}{\rm \cite[Theorem 1.25]{avv1}}
For $-\infty<a<b<\infty$, let $f,\,g: [a,b]\rightarrow \mathbb{R}$ be continuous on $[a,b]$, and be differentiable on $(a,b)$, and let $g'(x)\neq 0$ on $(a,b)$. If $f'(x)/g'(x)$ is increasing (deceasing) on $(a,b)$, then so are
\begin{eqnarray*}
\frac{f(x)-f(a)}{g(x)-g(a)}\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\frac{f(x)-f(b)}{g(x)-g(b)}.
\end{eqnarray*}
If $f'(x)/g'(x)$ is strictly monotone, then the monotonicity in the conclusion is also strict.
\end{lemma}

From now on we let $r'=\sqrt{1-r^2}$ for $0<r<1$.

\begin{lemma}\label{lecr}
Let $c\in(0,1]$ and $r\in(0,1)$.

(1) The function $f_c(r)\equiv\frac{1-(c r')^2}{r{\rm arth}(c r)}$ is strictly decreasing and concave with range $(0,1)$ if $c=1$, and strictly decreasing with range $(0,\infty)$ if $0<c<1$.

(2) The function $F_c(r)\equiv{\rm arth}(c r) {\rm arth}(c r')$ is strictly increasing on $(0,\frac{\sqrt{2}}{2}]$ and strictly decreasing on $[\frac{\sqrt{2}}{2}, 1)$ with maximum value $({\rm arth}(\frac{\sqrt{2}}{2} c))^2$.
\end{lemma}

\begin{proof}

(1) If $c=1$, then $f_1(r)=\frac{r}{{\rm arth}\,r}$. By differentiation,
$$f'_1(r)=\frac{h_{11}(r)}{h_{12}(r)},$$
where $h_{11}(r)={\rm arth}\, r-\frac{r}{r'^2}$ and $h_{12}(r)=({\rm arth}\, r)^2$. It is easy to see that $h_{11}(0^+)=h_{12}(0^+)=0$. Then
\begin{eqnarray*}
\frac{h'_{11}(r)}{h'_{12}(r)}=-\frac{h_{13}(r)}{h_{14}(r)},
\end{eqnarray*}
where $h_{13}(r)=(\frac{r}{r'})^2$ and $h_{14}(r)={\rm arth}\, r$. Then $h_{13}(0^+)=h_{14}(0^+)=0$. By differentiation, we have
\begin{eqnarray*}
\frac{h'_{13}(r)}{h'_{14}(r)}=\frac{2r}{r'^2},
\end{eqnarray*}
which is strictly increasing. Hence by Lemma \ref{lhr}, $\frac{h'_{11}(r)}{h'_{12}(r)}$ is strictly decreasing and so is $f'_1$
with $f'_1(r)<f'_1(0^+)=0$. Therefore, $f_1$ is strictly decreasing and concave on $(0,1)$. The limiting value $f_1(1^-)=0$ is clear and $f_1(0^+)=1$ by l'H${\rm \hat{o}}$pital's Rule.

If $0<c<1$, then
$$f_c(r)=f_1(cr)h(r),$$
where $h(r)=\frac{1-(cr')^2}{cr^2}$. By differentiation,
\begin{eqnarray*}
h'(r)=-\frac{2(1-c^2)}{cr^3}<0.
\end{eqnarray*}
Therefore, $f_c$ is strictly decreasing. The limiting values are clear.
\medskip

(2) By differentiation,
$$F'_c(r)=\frac{c}{r'f_c(r)}\left(\frac{f_c(r)}{f_c(r')}-1\right).$$
$\frac{f_c(r)}{f_c(r')}$ is strictly decreasing from $(0,1)$ onto $(0,\infty)$ by (1) and $F'_c(\frac{\sqrt{2}}{2})=0$. Then
$F_c$ is strictly increasing on $(0,\frac{\sqrt{2}}{2}]$ and strictly decreasing on $[\frac{\sqrt{2}}{2}, 1)$ with maximum value $F_c(\frac{\sqrt{2}}{2})$.
\end{proof}

\begin{lemma}\label{lecra}
Let $c\in(0,1]$, $r\in(0,1)$, $m=\sqrt{(2-c^2)(3c^2-2)}$ and $r_0=\sqrt{\frac{1-m/c^2}{2}}$. Let
$$G_c(r)\equiv{\rm arth}(c r)+{\rm arth}(c r').$$

(1) If $0<c\leq\sqrt{\frac 23}$, then the range of $G_c$ is $({\rm arth}\, c, {\rm arth}(\frac{2\sqrt 2 c}{2+c^2})]$.

(2) If $\sqrt{\frac 23}<c<\sqrt{2(\sqrt 2-1)}$, then the range of $G_c$ is $({\rm arth}\, c, {\rm arth} (\frac{c (r_0+r'_0)}{1+c^2r_0r'_0})]$.

(3) If $\sqrt{2(\sqrt 2-1)}\leq c<1$, then the range of $G_c$ is $[{\rm arth} (\frac{2\sqrt 2 c}{2+c^2}), {\rm arth} (\frac{c (r_0+r'_0)}{1+c^2r_0r'_0})]$.

(4) If $c=1$, then the range of $G_c$ is $[{\rm arth}(\frac{2\sqrt{2}}{3}), \infty)$.

\end{lemma}

\begin{proof}
It is clear that the limiting values
$$
G_c(0^+)=G_c(1^-)=
\left\{\begin{array}{ll}
{\rm arth}\, c,&\,\,\,0<c<1,\\
\infty,&\,\,\,c=1.
\end{array}\right.
$$
Let
$$g_c(r)={\rm th}G_c(r)=\frac{c(r+r')}{1+c^2 r r'}\,.$$
By differentiation, we have
$$
r'(1+c^2r r')^2 g'_c(r)=c(r'-r)(1-c^2-c^2 r r').
$$
Making substitution of $x=r^2$, we get
\begin{eqnarray}\label{cx}
& &1-c^2-c^2 r r'=0\\
&\Leftrightarrow&
c^4x^2-c^4x+(1-c^2)^2=0\,.\nonumber
\end{eqnarray}
Therefore, equation (\ref{cx}) has no root if $0<c^2< \frac 23$ or $c=1$ ,
only one root $\frac{\sqrt{2}}{2}$ if $c^2=\frac 23$,
and two different roots $r_0\,,r'_0$ if $\frac 23<c^2<1$. It is obvious that $r_0\in (0,\frac{\sqrt 2}{2})$.

It is easy to see that $c=g_c(0)<g_c(\frac{\sqrt{2}}{2})=\frac{2\sqrt 2}{2+c^2}c$ if and only if $c^2<2(\sqrt 2-1)\approx0.828427$. If $\sqrt{\frac 23}<c<1$, $g_c$ is increasing on $(0,r_0)$ , $(\frac{\sqrt{2}}{2}, r'_0)$ and decreasing on $(r_0, \frac{\sqrt{2}}{2})$, $(r'_0,1)$.

Now we get the following conclusions.

(1) If $0<c\leq\sqrt{\frac 23}$, $g'_c(r)=0$ is equivalent to $r=\frac{\sqrt{2}}{2}$.
Then
$${\rm arth}\, c={\rm arth} (g_c(0))<G_c(r)\leq {\rm arth} (g_c({\sqrt{2}}/{2}))={\rm arth}\left(\frac{2\sqrt 2 c}{2+c^2}\right).$$

(2) If $\sqrt{\frac 23}<c<\sqrt{2(\sqrt 2-1)}$, then
 $${\rm arth}\, c={\rm arth} (g_c(0))<G_c(r)\leq{\rm arth} (g_c(r_0))={\rm arth}\left(\frac{c (r_0+r'_0)}{1+c^2r_0r'_0}\right).$$

(3) If $\sqrt{2(\sqrt 2-1)}\leq c<1$, then
$${\rm arth} \left(\frac{2\sqrt 2 c}{2+c^2}\right)={\rm arth} (g_c({\sqrt{2}}/{2}))\leq G_c(r)\leq{\rm arth} (g_c(r_0))={\rm arth}\left(\frac{c (r_0+r'_0)}{1+c^2r_0r'_0}\right).$$

(4) If $c=1$, then
$$G_c(r)\geq {\rm arth}(g_1({\sqrt{2}}/{2}))={\rm arth}({2\sqrt{2}}/{3}).$$

This completes the proof.
\end{proof}

\begin{lemma}\label{le1} Let $r\in(0,1)$.

(1) The function $h_1(r)\equiv \frac{r'}{{\rm arth}\,r'}$ is strictly increasing and concave with range $(0,1)$.

(2) The function $h(r)\equiv \frac{r}{{\rm arth}\,r}+\frac{r'}{{\rm arth}\,r'}$ is strictly increasing on $(0,\frac{\sqrt 2}{2}]$, strictly decreasing on  $[\frac{\sqrt 2}{2},1)$, and concave on $(0,1)$ with range $(1, \frac{\sqrt2}{\log(\sqrt2+1)}]$.
\end{lemma}

\begin{proof}

(1) The monotonicity and the limiting values of $h_1$ can be easily obtained  by Lemma \ref{lecr}(1).

Now we prove the concavity of $h_1$. By differentiation,
$$h'_1(r)=\frac{r'-r^2{\rm arth}\,r'}{rr'({\rm arth}\,r')^2}$$
and
\begin{eqnarray}\label{le13}
h''_1(r)r^2r'^3({\rm arth}\,r')^3=2r'^2-r'{\rm arth}\,r'-r^2({\rm arth}\,r')^2\equiv\psi_1(r').
\end{eqnarray}
Then $\psi_1(r)=2r^2-r{\rm arth}\,r-r'^2({\rm arth}\,r)^2$ and by differentiation
\begin{eqnarray*}
\psi'_1(r)=r\left(4-\psi_2(r)\right),
\end{eqnarray*}
where $\psi_2(r)=\frac{1}{r'^2}+3\frac{{\rm arth}\,r}{r}-2({\rm arth}\,r)^2$.
Since
\begin{eqnarray*}
{\rm arth}\,r=\frac12\log\frac{1+r}{1-r}=\sum_{n=0}^{\infty}\frac{r^{2n+1}}{2n+1},
\end{eqnarray*}
we have
\begin{eqnarray*}
r^2r'^4\psi'_2(r)
&=&(r^4+2r^2-3){\rm arth}\,r-r^3+3r\nonumber\\
&=&\sum_{n=0}^{\infty}\frac{r^{2n+5}}{2n+1}+2\sum_{n=1}^{\infty}\frac{r^{2n+3}}{2n+1}-3\sum_{n=2}^{\infty}\frac{r^{2n+1}}{2n+1}\\
&=&\sum_{n=2}^{\infty}\frac{16(n-1)}{(2n-3)(2n-1)(2n+1)}r^{2n+1}>0.
\end{eqnarray*}
Hence $\psi_2$ is increasing with $\psi_2(0^+)=4$ and  $\psi_1$ is decreasing with $\psi_1(0^+)=0$. Therefore by (\ref{le13}), $h''_1$ is negative  and $h'_1$ is decreasing. Then $h_1$ is concave on $(0,1)$.

\medskip
(2) By differentiation,
$$h'(r)=f'_1(r)+h'_1(r)=f'_1(r)-\frac{r}{r'}f_1'(r'),$$
where $f_1(r)=\frac{r}{{\rm arth}\,r}=h_1(r')$.
It is easy to see that $h'(\frac{\sqrt2}{2})=0$. Therefore, $h$ is strictly increasing on $(0,\frac{\sqrt 2}{2}]$ and strictly decreasing on  $[\frac{\sqrt 2}{2},1)$. By (1) and Lemma \ref{lecr}(1), $h'$ is strictly decreasing and $h(0^+)=h(1^-)=1$. Hence $h$ is concave and
$$1<h(r)\leq h({\sqrt2}/{2})=\frac{\sqrt2}{\log(\sqrt2+1)}.$$

This completes the proof.
\end{proof}

\begin{lemma}\label{le2}
Let $p\in\mathbb{R}$, $r\in(0,1)$, and $C=1-\frac{\log(\sqrt{2}+1)}{\sqrt2}\approx 0.376775 $. Let
$$g(r)=\frac{r}{r'}\left(\frac{{\rm arth}\,r}{{\rm arth}\,r'}\right)^{p-1}.$$

(1) $g$ is strictly decreasing if $p\leq 0$ and strictly increasing if $p\geq C$.

(2) If $p\in(0,C)$, then there exists exactly one point $r_0\in(0,\frac{\sqrt2}{2})$ such that $g$ is increasing on $(0,r_0)$, $(r'_0,1)$ and decreasing on $(r_0,r'_0)$.

(3) If $p\in(0,C)$, then $g(0^+)=0$ and $g(1^-)=\infty$.
\end{lemma}

\begin{proof}
(1) By logarithmic differentiation,
\begin{eqnarray*}
\frac{rr'^2{\rm arth}\,r{\rm arth}\,r'}{r{\rm arth}\,r'+r'{\rm arth}\,r}\cdot\frac{g'(r)}{g(r)}&=&p-1+\frac{{\rm arth}\,r{\rm arth}\,r'}{r{\rm arth}\,r'+r'{\rm arth}\,r}\\
&=&p-\left(1-\frac{1}{h(r)}\right),
\end{eqnarray*}
where $h(r)$ is as in Lemma \ref{le1}(2). Since
$$0<1-\frac{1}{h(r)}\leq C,$$
we see that
$g$ is strictly increasing if $p\geq C$ and decreasing if $p\leq 0$.

\medskip
(2) If $p\in(0,C)$, then there exists exactly one point $r_0\in(0,\frac{\sqrt2}{2})$ such that $g'(r_0)=g'(r'_0)=0$ because $1-\frac{1}{h(r)}$ is increasing on $(0,\frac{\sqrt2}{2})$ and decreasing on $(\frac{\sqrt2}{2},1)$ by Lemma \ref{le1}(2). Therefore, $g'>0$ if $r\in(0,r_0)\cup(r'_0,1)$ and $g'<0$ if $r\in(r_0,r'_0)$.  Hence $g$ is increasing on $(0,r_0)$, $(r'_0,1)$ and decreasing on $(r_0,r'_0)$.

\medskip
(3) Since $0<p<C<1$ and
\begin{eqnarray*}
\lim\limits_{r\rightarrow 0^+}\frac{({\rm arth}\,r')^{1-p}}{r^{-p}}
=\lim\limits_{r\rightarrow 0^+}\frac{1-p}{p}\cdot\frac{r^p}{r'({\rm arth}\,r')^p}
=0,
\end{eqnarray*}
we have
\begin{eqnarray*}
\lim\limits_{r\rightarrow 0^+}g(r)=\lim\limits_{r\rightarrow 0^+}\left(\frac{r}{{\rm arth}\,r}\right)^{1-p}\cdot\left(\frac{1}{r'}\right)\cdot\frac{({\rm arth}\,r')^{1-p}}{r^{-p}}=0
\end{eqnarray*}
and
\begin{eqnarray*}
\lim\limits_{r\rightarrow 1^-}g(r)=\lim\limits_{r\rightarrow 1^-}\left(\frac{{\rm arth}\,r'}{r'}\right)^{1-p}\cdot r\cdot\frac{r'^{-p}}{({\rm arth}\,r)^{1-p}}=\infty.
\end{eqnarray*}
This completes the proof.
\end{proof}

\begin{theorem}\label{th1}
Let $C$ be as in Lemma \ref{le2}. Then for all $r\in(0,1)$,
\begin{eqnarray}\label{th11}
H_p({\rm arth}\,r,{\rm arth}\,r')\leq {\rm arth}\left(\frac{\sqrt2}{2}\right)
\end{eqnarray}
holds if and only if $p\leq 0$, and
\begin{eqnarray}\label{th12}
H_p({\rm arth}\,r,{\rm arth}\,r')\geq {\rm arth}\left(\frac{\sqrt2}{2}\right)
\end{eqnarray}
holds if and only if $p\geq C$.
The equalities hold if and only if $r=r'=\frac{\sqrt2}{2}$ and all inequalities are sharp in both cases.
\end{theorem}
\begin{proof}
We immediately obtain the inequality (\ref{th11}) if $p=0$ by Lemma \ref{lecr}(2). Therefore, it suffices to discuss the case $p\neq 0$.

Let
$$f(r)=\frac 1p\log\frac{({\rm arth}\,r)^p+({\rm arth}\,r')^p}{2}\,,\,\,p\neq 0.$$
By differentiation,
\begin{eqnarray*}
f'(r)=\frac{({\rm arth}\,r')^{p-1}}{rr'\left(({\rm arth}\,r)^p+({\rm arth}\,r')^p\right)}\left(g(r)-1\right),
\end{eqnarray*}
where $g(r)$ is as in Lemma \ref{le2} and it is easy to see that $g(\frac{\sqrt2}{2})=1$.

{\it Case 1.} $p<0$.
$f$ is strictly increasing on $(0,\frac{\sqrt 2}{2})$ and strictly decreasing on $(\frac{\sqrt 2}{2},1)$ by Lemma \ref{le2}(1).
Therefore,  $f(r)\leq f(\frac{\sqrt2}{2})$ for all $p<0$.

{\it Case 2.} $p\geq C$.
By Lemma \ref{le2}(1), $f$ is strictly decreasing on $(0,\frac{\sqrt 2}{2})$ and strictly increasing on $(\frac{\sqrt 2}{2},1)$, and hence $f(r)\geq f(\frac{\sqrt2}{2})$.

{\it Case 3.} $p\in(0,C)$.
By Lemma \ref{le2}(2), there exists exactly one point $r_1\in(0,r_0)$ such that $g(r_1)=g(r'_1)=1$, where $r_0\in(0,\frac{\sqrt2}{2})$ is as in Lemma \ref{le2}(2). Then $f$ is strictly decreasing on $(0,r_1)$, $(\frac{\sqrt2}{2},r'_1)$ and strictly increasing on $(r_1,\frac{\sqrt2}{2})$ , $(r'_1,1)$. Thus,
$$f(r_1)=f(r'_1)<f({\sqrt2}/{2}).$$
Since $f(0^+)=f(1^-)=\infty$, there exists $r_2\in(0,r_1)\cup(r'_1,1)$  such that
$$f({\sqrt2}/{2})<f(r_2).$$
Therefore, neither (\ref{th11}) nor (\ref{th12}) holds for all $r\in(0,1)$.

This completes the proof of Theorem \ref{th1}.
\end{proof}

\begin{lemma}\label{t1l1}
Let $r\in(0,1)$.

(1) The function $f(r)\equiv\frac{r'^4{\rm arth}\,r-r(1+r^2)}{r'^2\left((1+r^2){\rm arth}\, r-r\right)}$ is strictly decreasing with range $(-\infty, -2)$.

(2) For $p \in \mathbb{R}$ define
$$h_p(r)\equiv1+p r'^2\frac{{\rm arth}\, r}{r}-(1+r^2)\frac{{\rm arth}\, r}{r}.$$

(i) If $p\geq-2$, then the range of $h_p$ is $(-\infty,p)$.

(ii) If $p<-2$, then the range of $h_p$ is $(-\infty, C(p)]$,
where $C(p)\equiv\sup\limits_{0<r<1}{h_p(r)}\in (p\,,-1)$ with $\lim\limits_{p\to-2}C(p)=-2$ and $\lim\limits_{p\to-\infty}C(p)=-\infty$.
\end{lemma}

\begin{proof}
(1) Let $f_1(r)= r'^2{\rm arth} r-\frac{r(1+r^2)}{r'^2}$ and $f_2(r)=(1+r^2){\rm arth} r-r$, then $f_1(0^+)=f_2(0^+)=0$.
By differentiation, we have
$$
\frac{f'_1(r)}{f'_2(r)}=-1-\frac{2r}{r'^2(r+r'^2{\rm arth}\,r)}=-1-\frac{2}{r'^2\left(1+\frac{f_3(r)}{f_4(r)}\right)},
$$
where $f_3(r)={\rm arth}\,r$ and $f_4(r)=\frac{r}{r'^2}$. It is easy to see that $f_3(0^+)=f_4(0^+)=0$, then
$$
\frac{f'_3(r)}{f'_4(r)}=\frac{r'^2}{1+r^2},
$$
which is strictly decreasing. Hence by Lemma \ref{lhr}, $\frac{f'_1(r)}{f'_2(r)}$ is strictly decreasing and so is $f$ with $f(0^+)=-2$. Since $(1+r^2){\rm arth} r-r>0$ and by l'H$\rm \hat{o}$pital's Rule,
\begin{eqnarray}\label{r'arth}
\lim_{r\rightarrow1^-}\frac{{\rm arth} r}{r'^{-2}}=\lim_{r\rightarrow1^-}\frac{r'^2}{2r}=0,
\end{eqnarray}
we get $f(1^-)=-\infty$.

\medskip
(2) By Lemma \ref{le1}(1)  and (\ref{r'arth}), it is easy to see that $h_p(0^+)=p$ and $h_p(1^-)=-\infty$.

Next by differentiation, we have
$$
h'_p(r)=\frac 1r\left((1+r^2)\frac{{\rm arth}\, r}{r}-1\right)\left(f(r)-p\right),
$$
where $f(r)$ is as in (1).

If $p\geq-2$, and by (1), we see that $p>f(r)$, which implies that  $h_p$ is strictly decreasing and hence $h_p(r)<p$.

If $p<-2$, since the range of $f$ is $(-\infty,-2)$, we see that
there exists exactly one point $r_0\in(0,1)$ such that $p=f(r_0)$. Then $h_p$ is increasing on $(0, r_0)$ and decreasing on $(r_0, 1)$. Since
$$
h_p(r)=-1+2\left(1-\frac{{\rm arth}\,r}{r}\right)+(p+1)r'^2\frac{{\rm arth}\,r}{r}<-1,
$$
by the continuity of $h_p$, there is a continuous function
$$C(p)\equiv\sup\limits_{0<r<1}{h_p(r)}$$
with $p<C(p)<-1$,\,$\lim\limits_{p\to-2}C(p)=-2$ and $\lim\limits_{p\to-\infty}C(p)=-\infty$.
\end{proof}

\begin{lemma}\label{t1l2}
Let $p\,,q\in\mathbb{R}$, $r\in(0,1)$, and let $C(p)$ be as in Lemma \ref{t1l1}(2). Let
$$g_{p,q}(r)\equiv\frac{{\rm arth}^{q-1}r}{r^{p-1}r'^2}.$$
(1) If $p\geq-2$, then $g_{p,q}$ is strictly increasing for each $q\geq{p}$, and $g_{p,q}$ is not monotone for any $q<p$.\\
(2) If $p<-2$, then $g_{p,q}$ is strictly increasing for each $q\geq{C(p)}$, and $g_{p,q}$ is not monotone for any $q<C(p)$.
\end{lemma}
\begin{proof} By logarithmic differentiation in $r$,
$$\frac{g'_{p,q}(r)}{g_{p,q}(r)}=\frac{1}{r'^2{\rm arth}\, r}(q-h_p(r)),$$
where $h_p(r)$ is as in Lemma \ref{t1l1}(2). Hence the results immediately follow from Lemma \ref{t1l1}(2).
\end{proof}

\medskip

The following theorem studies the $H_{p,q}$-convexity of ${\rm arth}$.

\begin{theorem}\label{ath1}
The inverse hyperbolic tangent function ${\rm arth}$  is strictly $H_{p,q}$-convex on $(0,1)$ if and only if $(p,q)\in{D_1}\cup{D_2}$,
where $$D_1=\{(p,q)|-2\leq{p}<+\infty,\, p\leq q<+\infty\},$$
$$D_2=\{(p,q)|-\infty<p<-2,\,C(p)\leq q<+\infty\},$$
and $C(p)$ is a continuous function as in Lemma \ref{t1l1}(2).
There are no values of $p$ and $q$ for which ${\rm arth}$ is $H_{p,q}$-concave on the whole interval $(0,1)$.
\end{theorem}

\begin{proof}  The proof is divided into the following four cases.

{\it Case 1.} $p\neq0$ and $q\neq0$.
We may suppose that $0<x\leq y<1$. Define
$$F(x,y)={\rm arth}^{q}\left(H_p(x,y)\right)-\frac{{\rm arth}^{q}x+{\rm arth}^{q}y}{2}.$$
Let $t=H_p(x,y)$, then $\frac{\partial t}{\partial x}=\frac 12(\frac xt)^{p-1}$. If $x<y$, we see that $t>x$.
By differentiation, we have
$$\frac{\partial F}{\partial x}=\frac{q}{2}x^{p-1}\left(\frac{{\rm arth}^{q-1}t}{t^{p-1}t'^2}-\frac{{\rm arth}^{q-1}x}{x^{p-1}x'^2}\right).$$

{\it Case 1.1.} $p\geq-2$, $q\geq{p}$, and $pq\neq0$.
By Lemma \ref{t1l2}(1), $\frac{\partial{F}}{\partial{x}}<0$ if $q<0$ and $\frac{\partial{F}}{\partial{x}}>0$ if $q>0$.
Then $F(x,y)$ is strictly decreasing and $F(x,y)\geq F(y,y)=0$ if $q<0$, and $F(x,y)$ is strictly increasing and $F(x,y)\leq F(y,y)=0$ if $q>0$.
Hence we have
$${\rm arth}(H_p(x,y))\leq H_q({\rm arth}\, x,{\rm arth}\, y)$$
with equality if and only if $x=y$.

In conclusion, ${\rm arth}$ is strictly $H_{p,q}$-convex on $(0,1)$ for $(p,q)\in\{(p,q)|-2{\leq}p<0, p\leq q<0\}\cup\{(p,q)|-2{\leq}p<0, q>0\}\cup\{(p,q)|0<p<+\infty,q\geq p\}$.

{\it Case 1.2.} $p\geq-2$, $q<p$, and $pq\neq0$.
By Lemma \ref{t1l2}(1), with an argument similar to Case 1.1, it is easy to see that
${\rm arth}$ is neither $H_{p,q}$-concave nor $H_{p,q}$-convex on the whole interval $(0,1)$.

{\it Case 1.3.} $p<-2$, $q\geq{C(p)}$, and $pq\neq0$.
By Lemma \ref{t1l2}(2), $\frac{\partial{F}}{\partial{x}}<0$ if $q<0$ and $\frac{\partial{F}}{\partial{x}}>0$ if $q>0$.
Then $F(x,y)$ is strictly decreasing and $F(x,y)\geq F(y,y)=0$ if $q<0$, and $F(x,y)$ is strictly increasing and $F(x,y)\leq F(y,y)=0$ if $q>0$.
Hence we have
$${\rm arth}(H_p(x,y))\leq H_q({\rm arth}\, x,{\rm arth}\, y)$$
with equality if and only if $x=y$.

In conclusion, ${\rm arth}$ is strictly $H_{p,q}$-convex on $(0,1)$ for $(p,q)\in\{(p,q)|p<-2, C(p)\leq q<0\}\cup\{(p,q)|p<-2, q>0\}$.

{\it Case 1.4.} $p<-2$, $q<{C(p)}$, and $pq\neq0$.
By Lemma \ref{t1l2}(2), with an argument similar to Case 1.3, it is easy to see that
${\rm arth}$ is neither $H_{p,q}$-concave nor $H_{p,q}$-convex on the whole interval $(0,1)$.

{\it Case 2.} $p\neq0$ and $q=0$.
For $0<x\leq y<1$, let
$$F(x,y)=\frac{{\rm arth}^2(H_p(x,y))}{{\rm arth}\,{x}\,{\rm arth}\,{y}},$$
 and $t=H_p(x,y)$. If $x<y$, we see that $t>x$.
By logarithmic differentiation, we obtain
$$\frac1{F}\frac{\partial F}{\partial x}=x^{p-1}\left(\frac{({\rm arth}\,{t})^{-1}}{t^{p-1}t'^2}-\frac{({\rm arth}\,{x})^{-1}}{x^{p-1}x'^2}\right).$$

{\it Case 2.1.} $-2\leq{p}<0$ and $q=0>p$. By Lemma \ref{t1l2}(1), we have $\frac{\partial F}{\partial x}>0$ and $F(x,y)\leq F(y,y)=1$. Hence we have
$${\rm arth}(H_p(x,y))\leq\sqrt{{\rm arth}\,{x}\,{\rm arth}\,{y}}$$
with equality if and only if $x=y$.

In conclusion, ${\rm arth}$ is strictly $H_{p,q}$-convex on $(0,1)$ for $(p,q)\in\{(p,q)|-2\leq{p}<0, q=0\}$.

{\it Case 2.2.} $p>0$ and $q=0<p$.  By Lemma \ref{t1l2}(1), with an argument similar to Case 2.1, it is easy to see that
${\rm arth}$ is neither $H_{p,q}$-concave nor $H_{p,q}$-convex on the whole interval $(0,1)$.

{\it Case 2.3.} $p<-2$. We have $q=0\geq C(p)$, and by Lemma \ref{t1l2}(2), with an argument similar to Case 2.1, it is easy to see that
${\rm arth}$ is $H_{p,q}$-convex on $(0,1)$ for $(p,q)\in\{(p,q)|p<-2, q=0\}$.

{\it Case 3.} $p=0$ and $q\neq0$.
For $0<x\leq y<1$, let
$$F(x,y)={\rm arth}^q(\sqrt{xy})-\frac{{\rm arth}^qx+{\rm arth}^qy}{2},$$
 and $t=\sqrt{xy}$. If $x<y$, we have that $t>x$. By differentiation, we obtain
$$\frac{\partial F}{\partial x}=\frac{q}{2x}\left(\frac{{\rm arth}^{q-1}t}{t^{-1}t'^2}-\frac{{\rm arth}^{q-1}x}{x^{-1}x'^2}\right).$$

{\it Case 3.1.} $q>p=0$. By Lemma \ref{t1l2}(1), we have $\frac{\partial F}{\partial x}>0$ and $F(x,y)\leq F(y,y)=0$. Hence we have
$${\rm arth}(\sqrt{xy})\leq H_q({\rm arth}\,{x},\,{\rm arth}\,{y})$$
with equality if and only if $x=y$.

In conclusion, ${\rm arth}$ is strictly $H_{p,q}$-convex on $(0,1)$ for $(p,q)\in\{(p,q)|p=0, q>0\}$.

{\it Case 3.2.} $q<p=0$. By Lemma \ref{t1l2}(1), with an argument similar to Case 3.1, it is easy to see that
${\rm arth}$ is neither $H_{p,q}$-concave nor $H_{p,q}$-convex on the whole interval $(0,1)$.

{\it Case 4.} $p=q=0$. By Case 1.1, for all $x\,,y\in(0,1)$, we have
$${\rm arth}(H_p(x,y))\leq H_p({\rm arth}\,{x},\,{\rm arth}\,{y}),\quad\mbox{for}\quad{p\geq-2}\quad\mbox{and}\quad p\neq0.$$
By the continuity of $H_p$ in $p$ and ${\rm arth}$ in $x$, we have
$${\rm arth}(H_0(x,y))\leq H_0({\rm arth}\,{x},{\rm arth}\,{y}).$$
In conclusion, ${\rm arth}$ is strictly $H_{0,0}$-convex on $(0,1)$.

This completes the proof of Theorem \ref{ath1}.
\end{proof}

\medskip

Setting $p=1=q$ in Theorem \ref{ath1}, we easily obtain the convexity of ${\rm arth}$.

\begin{corollary}
The inverse hyperbolic tangent function ${\rm arth}$ is strictly convex on $(0,1)$.
\end{corollary}
\medskip
By (\ref{arth}), Theorem \ref{ath1} has a simple application to the hyperbolic metric.

\begin{corollary}
Let $z\in S^1(H_p(|x|,|y|))$. Then for all $x,y\in\mathbb{B}^2\setminus \{0\}$ and
$p\geq-2$
$$\rho(0,z)\leq H_p(\rho(0,x),\rho(0,y))$$
with equality if and only if $|x|=|y|$.
\end{corollary}

\section{Some Propositions for Hyperbolic Metric}

Next we give some geometric propositions for the hyperbolic metric.

\begin{proposition}
Let $c,d$ be arbitrary two points on the unit circle such that $0\,,c\,,d$ are noncollinear. Let $b\in[c,d]\cap \mathbb{B}^2$,
 $\{a\}=[0,b]\cap J^*[c,d]$, and let $s$ be the midpoint of
the Euclidean segment $[c,d]$. Then\\
(1) $a$ is the hyperbolic midpoint of the hyperbolic segment $J[0,b]$;\\
(2) $s$ is on the hyperbolic circle $S_{\rho}(a,\rho(0,a))=\{z|\rho(z,a)=\rho(0,a)\}$.
\end{proposition}

\begin{figure}[h]
\begin{minipage}[t]{0.45\linewidth}
\centering
\includegraphics[width=6cm]{lambfig5.pdf}
\end{minipage}
\caption{The point $a$ is the hyperbolic midpoint of $J[0,b]\,.$ The circular arc $(cad)$ is orthogonal to the unit circle. Here $s$ is the Euclidean midpoint of the chord $[c,d]$. \label{abcd}}
\end{figure}

\begin{proof}
Without loss of generality, we may assume $c=e^{i\alpha}$ and $d=e^{-i\alpha}$, where $0<\alpha<\pi/2$.

(1) The point $a$ is the hyperbolic midpoint of the hyperbolic segment $J[0,b]$
\begin{eqnarray}\label{rhoab}
\Leftrightarrow \rho(0,b)=2\rho(0,a) \Leftrightarrow \log\frac{1+|b|}{1-|b|}=2\log\frac{1+|a|}{1-|a|} \Leftrightarrow |b|=\frac{2|a|}{1+|a|^2}.
\end{eqnarray}
Let $b=|b|e^{i\beta}$( $-\frac{\pi}{2}<\beta<\frac{\pi}{2}$).
Then by the orthogonality of $J^*[c,d]$ and the unit circle
$$
s=\cos\alpha=|b|\cos\beta=|b|\frac{|a|^2+|w|^2-r^2}{2|a| |w|},
$$
where $w=1/\cos\alpha$ and $r=\sqrt{|w|^2-1}$ are the center and the radius, resp., of the circle containing $c\,,d$ and orthogonal to $\partial\mathbb{B}^2$.
Therefore, the last equality in (\ref{rhoab}) holds and $a$ is the hyperbolic midpoint
of $J[0,b]$.

\medskip
(2) It is easy to see that $[0,b]$ is also the Euclidean diameter of
$S_{\rho}(a,\rho(0,a))$ by geometric observation and (1). Therefore,
$s$ is on the hyperbolic circle $ S_{\rho}(a,\rho(0,a))$, see Figure \ref{abcd}.
\end{proof}

\begin{proposition}\label{a}
Let $J_1=J^*[e^{i\alpha},-e^{-i\alpha}]$,
$J_2=J^*[-e^{i\alpha},e^{-i\alpha}]$ be two hyperbolic geodesics in $\mathbb{B}^2$,
$0<\alpha<\pi/2$. Let $\{t e_2\}=J_1\cap Y$, see Figure \ref{a1}. Then
\begin{eqnarray}\label{ad}
d_\rho(J_1,J_2)=\rho(-t e_2, t e_2)=2\log\frac{1+t}{1-t}.
\end{eqnarray}
\end{proposition}

\begin{proof}
Let $g: \mathbb{B}^2\rightarrow \mathbb{H}^2$ be a M\"obius  transformation which satisfies $g(i)=\infty$ and $g(-i)=0$. Then $g([-te_2, te_2])=[g(-te_2),g(te_2)]$ which is on the ray emanating from $0$ and perpendicular to $\partial {\mathbb{H}}^2$.
By the orthogonality of $J_i$($i=1,2$) and $Y$, we get
$$J^*_1=g(J_1)={\rm Arc}(g(-e^{-i\alpha})g(te_2)g(e^{i\alpha})) \,\ ,\,\
J^*_2=g(J_2)={\rm Arc}(g(-e^{i\alpha})g(-te_2)g(e^{-i\alpha})).$$
Here $g(-e^{-i\alpha})=-g(e^{i\alpha})$ and $g(-e^{i\alpha})=-g(e^{-i\alpha})$, see Figure \ref{a2}.
For every rectifiable arc $\gamma_{xy}$, $x\in J^*_1$ and $y\in J^*_2$, by geometric observation we get
$$\int_{\gamma_{xy}}\frac{1}{d(z,\partial \mathbb{H}^2)} |dz|\geq\int_{[g(-te_2),g(te_2)]}\frac{1}{d(z,\partial \mathbb{H}^2)} |dz|,$$
and hence
\begin{equation}\label{adg}
d_\rho(J^*_1, J^*_2)=\rho(g(-t e_2),g(t e_2)).
\end{equation}
Since the hyperbolic distance is invariant under M\"obius transformations,
by (\ref{adg}) we get
\begin{eqnarray*}
d_\rho(J_1,J_2)=\rho(-t e_2, t e_2)=2\log\frac{1+t}{1-t}.
\end{eqnarray*}
\end{proof}
\begin{figure}[h]
\begin{minipage}[t]{0.45\linewidth}
\centering
\includegraphics[width=6cm]{lambfig6.pdf}
\caption{\label{a1}}
\end{minipage}
\hfill
\hspace{1cm}
\begin{minipage}[t]{0.45\linewidth}
\centering
\includegraphics[width=7.8cm]{lambfig7.pdf}
\caption{\label{a2}}
\end{minipage}
\caption*{The M\"obius transformation $g$ maps the unit disk in Figure \ref{a1} onto the upper half plane in Figure \ref{a2}.}
\end{figure}

Let $J_3=J^*[e^{i\alpha},e^{-i\alpha}]$ and $J_4=J^*[-e^{-i\alpha},-e^{i\alpha}]$. Let $\{s e_1\}=J_3\cap X$. Since the hyperbolic distance is invariant under rotations, by Proposition \ref{a} we also obtain
\begin{equation}\label{j3j4}
d_\rho(J_3,J_4)=\rho(-se_1, se_1).
\end{equation}

By the proof of Proposition \ref{a}, $J^*_3=g(J_3)={\rm SArc}(g(e^{i\alpha})g(e^{-i\alpha}))$ and $J^*_4=g(J_4)={\rm SArc}(g(-e^{-i\alpha})g(-e^{i\alpha}))$.  Then
$g$ maps ${\rm Arc}((-i)(s e_1)i)$ to the ray emanating from $0$ and tangent to $J^*_3$ at the point $g(se_1)$.
Similarly, $g$ maps ${\rm Arc}((-i)(-s e_1)i)$ to the ray emanating from $0$ and tangent to $J^*_4$ at the point $g(-se_1)$.
Then we have
\begin{equation}\label{j3j4g}
d_\rho(J^*_3, J^*_4)=\rho(g(-se_1),g(se_1)).
\end{equation}

\begin{remark}
By (\ref{ad}) and (\ref{j3j4}), we find the two distances between the opposite sides for the hyperbolic quadrilaterals with four vertices $e^{i\alpha},\,-e^{-i\alpha},\,-e^{i\alpha},\,e^{-i\alpha}$ on the unit circle and the counterpart for the upper half plane by (\ref{adg}) and (\ref{j3j4g}).
\end{remark}

\section{Proof of Main Results }

\begin{proof}[Proof of Theorem \ref{Lamdd}]
Since the hyperbolic distance is M\"obius invariant, we may assume that $v_a=0$, $v_b$ is on $X$, $v_d$ is on $Y$ and $v_c=t e^{i\theta}$, $0<t\leq1$ and $0<\theta<\frac{\pi}{2}$ (see Figure \ref{Lamb}).
Then by (\ref{orar}) the circle $S^1(b,r_b)$ through $v_c$ and $\overline{v_c}$ is orthogonal to $\partial\mathbb{B}^2$, where
$$
b=\frac{1+t^2}{2t\cos\theta}\,\,\,\,\,{\rm and}\,\,\,\,r_b=\frac{\sqrt{(1+t^2)^2-4t^2\cos^2\theta}}{2t\cos\theta}.
$$
By Proposition \ref{a}, we get
$$
d_1=\rho(0,v_b)=\log\frac{1+(b-r_b)}{1-(b-r_b)}={\rm arth}\left(\frac{2t}{1+t^2}\cos\theta\right).
$$
Similarly, we get
$$
d_2=\rho(0,v_d)={\rm arth}\left(\frac{2t}{1+t^2}\sin\theta\right).
$$
Then
$$d_1d_2={\rm arth}(Lr){\rm arth}(Lr'),$$
where $L=\frac{2t}{1+t^2}={\rm th}\rho(0,v_c)\in(0,1]$ and $r=\cos\theta\in(0,1)$. By Lemma \ref{lecr}(2), we have
$$d_1d_2\leq \left({\rm arth}\left(\frac{\sqrt 2}{2}L\right)\right)^2.$$

This completes the proof of Theorem \ref{Lamdd}.
\end{proof}

\medskip

\begin{proof}[Proof of Theorem \ref{Lamdad}]
By the proof of Theorem \ref{Lamdd}, we have
$$d_1+d_2={\rm arth}(Lr)+{\rm arth}(Lr'),$$
where $L\in(0,1]$ and $r\in(0,1)$.
Then by Lemma \ref{lecra}, the desired conclusion follows.
\end{proof}

\medskip

\begin{proof}[Proof of Corollary \ref{dd}]
There exists a M\"obius transformation $g$ which maps $a,b,c,d$ to
$e^{i\alpha}$, $-e^{-i\alpha}$, $-e^{i\alpha}$, $e^{-i\alpha}$, resp., where $\alpha=\arccos\sqrt{1/|a,b,c,d|}\in(0\,,\pi/2)$, see Figure \ref{a1}.
By Proposition \ref{a} and the proof of Theorem \ref{Lamdd}, we have
\begin{equation}\label{d2}
d_1=2 {\rm arth}\,r\,\,\,\,{\rm and}\,\,\,\,d_2=2 {\rm arth}\,r'
\end{equation}
where $r=\cos\alpha$.

Therefore, by Theorem \ref{Lamdd} and Theorem \ref{Lamdad}, we have
$$d_1d_2=4 ({\rm arth}\,r)({\rm arth}\,r')\leq \left(2 {\rm arth}({\sqrt{2}}/{2})\right)^2$$
and
$$d_1+d_2=2 ({\rm arth}\,r+{\rm arth}\,r')\geq 4{\rm arth}({\sqrt{2}}/{2}).$$
The equalities hold if and only if $\alpha=\frac{\pi}{4}$, namely, $|a,b,c,d|=\frac{1}{\cos^2\alpha}=2$.

This completes the proof of Corollary \ref{dd}.
\end{proof}

Let $G\,,G'$ be domains in $\overline{\mathbb{R}^n}$ and let $f: G\rightarrow G'$ be a homeomorphism. Then $f$ is $K$-quasiconformal if
$$M(\Gamma)/K\le M(f\Gamma)\le K M(\Gamma)$$
for every curve family $\Gamma$ in $G$, where $M(\Gamma)$ is the modulus of $\Gamma$, see \cite[10.9]{v}.

For $r \in (0,1)$ and $K \ge 1$, we define the distortion function
 $$
  \varphi_{K}(r) = \mu^{-1}(\mu(r)/K)\,,
$$
where $\mu(r)$ is the modulus of the planar Gr\"otzsch ring, see \cite[Exercise 5.61]{v}.

\begin{lemma}\label{leqc}{\rm  \cite[Theorem 1.10] {bv} }
Let $f: \mathbb{B}^2\rightarrow\mathbb{B}^2$ be a $K$-quasiconformal mapping with $f \mathbb{B}^2= \mathbb{B}^2$, and let $\rho$ be the hyperbolic metric of $\mathbb{B}^2$. Then
$$\rho(f(x),f(y))\leq A(K)\max\{\rho(x,y),\rho(x,y)^{1/K}\}$$
for all $x,y\in \mathbb{B}^2$, where $A(K)=2 {\rm arth} (\varphi_K({\rm th} \frac 12))$ and
$$K\leq u(K-1)+1\leq\log({\rm ch} (K {\rm arch}(e)))\leq A(K)\leq v(K-1)+K$$
with  $u= {\rm arch}(e){\rm th} ( {\rm arch}(e))>1.5412$ and $v=\log(2(1+\sqrt{1-{1}/{e^2}}))<1.3507$. In particular,
$A(1)=1$.
\end{lemma}

\medskip

\begin{proof}[Proof of Theorem \ref{thqcLam}]
In the same way as in the proof of Theorem \ref{Lamdd}, we still assume that $v_a=0$, $v_b$ is on $X$, $v_d$ is on $Y$ and $v_c=t e^{i\theta}$, $0<t\leq1$ and $0<\theta<\frac{\pi}{2}$ (see Figure \ref{Lamb}). Then
$$d_1=\rho(0,v_b)={\rm arth}(Lr)\,\,\,{\rm and} \,\,\,d_2=\rho(0,v_d)={\rm arth}(Lr'),$$
where $0<L\leq1$ and $0<r<1$.

By Lemma \ref{leqc}, we have
\begin{eqnarray*}
D_1D_2&\leq&\rho(f(0),f(v_b))\rho(f(0),f(v_d))\\
&\leq& A(K)^2\max\{\rho(0,v_b),\rho(0,v_b)^{1/K}\}\cdot\max\{\rho(0,v_d),\rho(0,v_d)^{1/K}\}\\
&=& A(K)^2\max\{d_1,d_1^{1/K}\}\cdot\max\{d_2,d_2^{1/K}\}.
\end{eqnarray*}

(1) $0<L\leq \frac{e^2-1}{e^2+1}\approx0.761594$.
This implies that $d_1<1$ and $d_2<1$. Then by Theorem \ref{Lamdd},
$$D_1D_2\leq A(K)^2(d_1d_2)^{1/K}\leq A(K)^2 \left({\rm arth}\left(\frac{\sqrt{2}}{2}L\right)\right)^{2/K}.$$

(2) $\frac{e^2-1}{e^2+1}<L\leq1$.

{\it Case 1.} $\frac{0.761594}{L}\approx\frac 1L\frac{e^2-1}{e^2+1}= r_L<r<1$.
This implies that $d_1>1$ and $d_2<1$. Then
$$
d_1d_2^{1/K}={\rm arth}(L r)({\rm arth}(L r'))^{1/K}\equiv F_{L,K}(r).
$$

By logarithmic differentiation, we have
\begin{eqnarray}\label{c22}
\frac{F'_{L,K}(r)}{F_{L,K}(r)}=\frac{L r}{r'(1-(L r')^2){\rm arth} (L r')}\left(\frac{f_L(r)}{f_L(r')}-\frac 1K\right),
\end{eqnarray}
where $f_L(r)=\frac{1-(L r')^2}{r\,{\rm arth}(L r)}$. By Lemma \ref{lecr}(1), $\frac{f_L(r)}{f_L(r')}$ is strictly decreasing from
$(r_L, 1)$ onto $(0, \frac {1}{M_L})$. Here $M_L=\frac{f_L(r'_L)}{f_L(r_L)}>1$ since $r_L\approx\frac{0.761594}{L}\geq 0.761594>0.707107\approx\frac{\sqrt 2}{2}$.

{\it Case 1.1} $1\leq K\leq M_L$.
By (\ref{c22}), we have $F'_{L,K}(r)\leq 0$ and hence $F_{L,K}(r)$ is strictly decreasing on $(r_L,1)$.
Therefore,
\begin{equation}\label{qc11}
d_1d_2^{1/K}\leq F_{L,K}(r_L)={\rm arth}(L r_L)({\rm arth}(L r'_L))^{1/K}.
\end{equation}

{\it Case 1.2} $K>M_L$.
There exists exactly one point $r_L(K)\in(r_L, 1)$ such that $\frac{f_L(r_L(K))}{f_L(\sqrt{1-r_L(K)^2})}=\frac 1K$. Then $F_{L,K}$ is strictly increasing on $(r_L, r_L(K))$ and strictly decreasing on $(r_L(K),1)$. Therefore,
\begin{eqnarray}\label{qc22}
d_1d_2^{1/K}\leq F_{L,K}(r_L(K))={\rm arth}\left(L r_L(K)\right)\left({\rm arth}\left(L\sqrt{1- r_L(K)^2}\right)\right)^{1/K}.
\end{eqnarray}

{\it Case 2.} $\sqrt{1-r^2_L}<r<r_L$.
This implies that  $d_1<1$ and $d_2<1$. Since $\frac{\sqrt 2}{2}\approx0.707107\in (\sqrt{1-r^2_L},r_L)$ , by Theorem \ref{Lamdd} we have
\begin{equation}\label{qc33}
(d_1d_2)^{1/K}\leq \left({\rm arth}\left(\frac{\sqrt{2}}{2}L\right)\right)^{2/K}.
\end{equation}

{\it Case 3.} $0<r<\sqrt{1-r^2_L}$.
This implies that  $d_1<1$ and $d_2>1$.
Putting $p=r'$, we have $r_L<p<1$ and
\begin{eqnarray*}
d_1^{1/K} d_2={\rm arth}(L p)({\rm arth}(L p'))^{1/K}.
\end{eqnarray*}
Hence Case 3 is the same as Case 1.

Therefore, by (\ref{qc11}) and (\ref{qc33}), we have
if $1\leq K\leq M_L$,
$$D_1D_2\leq A(K)^2\max\left\{
{\rm arth}(L r_L)\left({\rm arth}\left(L \sqrt{1-r^2_L}\right)\right)^{1/K}\,,\left({\rm arth}\left(\frac{\sqrt{2}}{2}L\right)\right)^{2/K}\right\}.$$
And
by (\ref{qc22}) and (\ref{qc33}), we have
if $K> M_L$,
$$D_1D_2\leq A(K)^2\max\left\{
{\rm arth}(L r_L(K))\left({\rm arth}\left(L\sqrt{1- r_L(K)^2}\right)\right)^{1/K}\,, \left({\rm arth}\left(\frac{\sqrt{2}}{2}L\right)\right)^{2/K}\right\}.$$

This completes the proof of Theorem \ref{thqcLam}.
\end{proof}

\medskip

\begin{proof}[Proof of Corollary \ref{thqc}]
First,  let $g$ be the same as in the proof of Corollary \ref{dd}. Let $\{s e_1\}=g(J^*[a,d])\cap X$, $\{t e_2\}=g(J^*[a,b])\cap Y$ and denote
$$z_1=g^{-1}(se_1),\,\,z_2=g^{-1}(te_2),\,\,z_3=g^{-1}(-se_1),\,\,z_4=g^{-1}(-te_2).$$

Since the hyperbolic distance is  M\"obius invariant, by the proof of Corollary \ref{dd}, we get
$$d_1=\rho(z_1,z_3)=2 {\rm arth}\, r\,\,\,{\rm and} \,\,\,d_2=\rho(z_2,z_4)=2 {\rm arth}\, r',\,\,\,0<r<1.$$

Then by Lemma \ref{leqc}, we have
\begin{eqnarray*}
D_1D_2&\leq&\rho(f(z_1),f(z_3))\rho(f(z_2),f(z_4))\\
&\leq& A(K)^2\max\{\rho(z_1,z_3),\rho(z_1,z_3)^{1/K}\}\cdot\max\{\rho(z_2,z_4),\rho(z_2,z_4)^{1/K}\}\\
&=& A(K)^2\max\{d_1,d_1^{1/K}\}\cdot\max\{d_2,d_2^{1/K}\}.
\end{eqnarray*}

{\it Case 1.} $0.886819\approx \frac{2 \sqrt e}{e+1}= r_1<r<1$.
This implies that  $d_1>1$ and $d_2<1$. Then
\begin{eqnarray*}
d_1 d_2^{1/K}=2^{1+1/K} {\rm arth}\, r ({\rm arth}\, r')^{1/K}\equiv2^{1+1/K}F_{1,K}(r).
\end{eqnarray*}
Let $M_1=\frac{f_1(r'_1)}{f_1(r_1)}$, where $f_1(r)=\frac{r}{{\rm arth} r}$ . By the proof of Case 1 in Theorem \ref{thqcLam}, we have the following conclusions.

{\it Case 1.1} $1\leq K\leq M_1$.
\begin{equation}\label{qc1}
d_1d_2^{1/K}\leq 2^{1+1/K} F_{1,K}(r_1)=2^{1+1/K} {\rm arth}\,r_1({\rm arth}\,r'_1)^{1/K}.
\end{equation}

{\it Case 1.2} $K>M_1$.
There exists exactly one point $r_1(K)\in(r_1, 1)$ such that $\frac{f_1(r_1(K))}{f_1(\sqrt{1-r_1(K)^2})}=\frac 1K$. Therefore,
\begin{equation}\label{qc2}
d_1 d_2^{1/K}\leq 2^{1+1/K} F_{1,K}(r_1(K))= 2^{1+1/K}{\rm arth}\left(r_1(K)\right) \left({\rm arth}\left(\sqrt{1-r_1(K)^2}\right)\right)^{1/K}.
\end{equation}

{\it Case 2.} $0.462117\approx\frac{e-1}{e+1}=\sqrt{1-r^2_1}<r<r_1\approx0.886819$.
This implies that  $d_1>1$ and $d_2>1$.
Since $\frac{\sqrt 2}{2}\approx0.707107\in (\sqrt{1-r^2_1},r_1)$ , by Corollary \ref{dd} we have
\begin{eqnarray}\label{qc3}
d_1d_2\leq\left(2 \log(\sqrt{2}+1)\right)^2.
\end{eqnarray}

{\it Case 3.} $0<r<\sqrt{1-r^2_1}\approx 0.462117$.
This implies that  $d_1<1$ and $d_2>1$. Putting $p=r'$, we have $r_1<p<1$ and
\begin{eqnarray*}
d_1^{1/K} d_2=2^{1+1/K} {\rm arth}\,p ({\rm arth}\, p')^{1/K}.
\end{eqnarray*}
Hence Case 3 is the same as Case 2.

Therefore, by (\ref{qc1}) and (\ref{qc3}), we have
if $1\leq K\leq M_1$,
$$D_1D_2\leq  A(K)^2\max\left\{2^{1+1/K} {\rm arth}\,r_1({\rm arth}\,r'_1)^{1/K}, \left(2\log(\sqrt{2}+1)\right)^2\right\}.$$
And
by (\ref{qc2}) and (\ref{qc3}), we have
if $K> M_1$,
$$D_1D_2\leq A(K)^2\max\left\{2^{1+1/K}{\rm arth}\left(r_1(K)\right) \left({\rm arth}\left(\sqrt{1-r_1(K)^2}\right)\right)^{1/K}, \left(2\log(\sqrt{2}+1)\right)^2\right\}.$$

This completes the proof of Corollary \ref{thqc}.
\end{proof}

\medskip

\subsection*{Acknowledgments}
The research of Matti Vuorinen was supported by the Academy of Finland,
Project 2600066611. The research of Gendi Wang was supported by CIMO
of Finland, Grant TM-10-7364.

redpajama_set_name: RedPajamaArXiv


This apartment, located in Les Trois-Îlets, is ideal for 4 vacationers. It offers 2 bedrooms, a furnished terrace and WiFi.
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Artificial nutrition and hydration, also called ANH or tube feeding, may be offered to someone who is unable to swallow food and liquids without choking. In most instances, liquid nutrition is delivered through a tube that has been surgically inserted into the stomach or intestines, bypassing the normal digestive process that begins in the mouth. ANH can be helpful to a patient whose condition can improve or be reversed, or whose ability to swallow can be safely restored such as when surgery affects the digestive system, or if there is a blockage that prevents normal digestion.
But studies show that ANH does not extend life in patients who are near death or whose disease is so advanced they are unlikely to improve. In Alzheimer's disease and dementia patients in particular, studies show that tube feeding can actually cause harm. Complications from tube feeding can include pain, bloating and aspiration pneumonia, and may decrease quality of life as well as life span. Additionally, dementia patients may need to be physically restrained or sedated by medicine to keep them from dislodging the feeding tube. As a result of becoming immobile, they are at much greater risk of developing painful pressure ulcers.
The patient or his/her legally-authorized surrogate decision-maker has the right to decide whether to withhold or withdraw ANH based on medical information concerning the patient. Medical ethicists draw no distinction between withholding and withdrawing ANH.
But the issue of nutrition and feeding is complicated and can be fraught with significant emotional burdens. Providing food and the act of feeding symbolize love and nurturing and are associated with improving health and sustaining life. Thus, it is difficult sometimes for family members to understand how tube feeding could be harmful.
Although families often are concerned that hospices will not accept a patient with a feeding tube, this is rarely the case. Hospices generally agree to enroll such patients but will likely try to educate them and/or family or surrogate about the benefits and burdens of ANH. If or when a decision is made to discontinue ANH, the feeding tube usually can just be capped and allowed to remain in place.
In 2012, Hospice Foundation of America brought together a panel of clinical experts to educate other clinicians about the dilemmas that sometimes arise around tube feeding and to correct misinformation about artificial nutrition and hydration. You can watch a video clip of part of their discussion on this page.
Do all patients benefit from artificial nutrition and hydration?
Artificial Nutrition and Hydration at the End of Life explores medical, legal, ethical issues, and communication barriers that surround ANH. It will emphasize the need for end-of-life provider organizations to have clear and transparent policies on ANH and to offer training to staff and education for families to minimize family misunderstanding and discord as well as moral distress and anguish of staff.


text:
Spider Solitaire 1 suit
Play Spider Solitaire 1 suit for free!
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1 suit Spider Solitaire is the first of this series of card games, which involves cards of only one suit (any of the following: ♥, ♠ , ♣ , ♦). The game got its name due to the fact that the columns with cards can be associated with the paws of a spider. In the classic version, the following card denominations are involved: milking, three, four, five, six, seven, eight, nine, ten, jack, queen, king, ace. At the same time, the youngest of them is considered the "Ace", and the oldest is the "King".
The playing field consists of ten columns, in four of which at the very beginning there are five cards with the back up and one facing the player and in six of which there are four cards with the back up and one with the front side. At the bottom right is a handout deck from which you can take additional cards until they run out.
The goal of the game
The main goal of the One Suit Spider Solitaire game is to add up all possible sequences of cards ordered by face value in descending order from king to ace. For example, ♠K, ♠Q, ♠J, ♠10, ♠9, ♠8, ♠7, ♠6, ♠5, ♠4, ♠3, ♠2, ♠A. Each fully assembled chain is immediately removed from the playing field. The game is considered won as soon as all the cards are removed from the game table.
How to increase the probability of winning?
Let's look at the basic tips that will help you in this:
Try to move all possible cards in the columns at the very beginning of the game, without using additional cards from the deck.
Try to collect the correct decreasing sequences as quickly as possible, as with further moves it will be increasingly difficult to move them to the right places.
Stick to the strategy of evenly revealing closed cards in columns, so as not to create long disordered chains that become difficult to disassemble and require even more additional moves to parse them.
It is not necessary to deal all the cards at once from an additional deck. If you have taken cards from it, then try to perform the maximum possible number of useful moves first and only then use it again.
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"New game" is a button for starting a new layout that differs from the current one.
"Start over" is a button by clicking on which you can repeat the current batch again from the very beginning.
"Cancel move" - cancels one or more previously performed actions on the playing field.
"Hint" is a button for getting possible options for moves. When pressed, a visual demonstration of the move occurs.
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Home News Tenants on benefits struggle to find homes in the private sector
Tenants on benefits struggle to find homes in the private sector
St Peter's Grove is popular with buy-to-let landlords (file pic)
Fewer landlords are letting their properties to people on Universal Credit and other benefits, according to the Canterbury Housing Advice Centre.
Over the last year, the charity helped 1,265 households regarding 1,844 calls for assistance — with the majority of these people being "vulnerably housed in insecure accommodation".
The difficulties in finding rented accommodation and the precariousness of private sector tenants are two of the themes that underlie the newly-released annual report.
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The review, covering the year to 31 March, shows that nearly a quarter (24%) of Canterbury Housing Advice Centre (CHAC) clients are 45 or over, a third (32%) are unemployed, a quarter (25%) are unemployed and 37% are disabled.
Commenting on the way the local market is changing for renters, the CHAC trustees say: "It is increasingly difficult to find private rented accommodation if you are receiving any form of benefits.
"In the past landlords and estate agents were sceptical of certain benefits. Increasingly landlords and estate agents apply a simplified rule against any benefits claimants, making it very difficult for many to rent privately."
In the past, uncertain housing has been seen as a problem for the younger sectors of society who might be changing job or moving location before settling down.
But with pensioners accounting for 6% of CHAC clients, and those in the 45-pension age category accounting for 18 per cent, the charity suggests, through its statistics, that older age groups are now also significantly at risk of losing their homes.
The report includes a recent case study involving a 74-year old man whose Canterbury landlord decided to sell up after 15 years.
The pensioner was issued with a possession notice. CHAC managed to negotiate more time with the landlord, secured Housing Benefit for the man and helped him be re-housed in a one-bedroom flat in a sheltered housing association flat.
Commenting on the case, CHAC said: "This case illustrates a real problem with assured shorthold tenancies –the tenant's security is at the whim of the landlord notwithstanding paying rent, age, or any other vulnerabilities."
CHAC helped people stay on in their homes in 30% of cases involving threatened homelessness.
For the other clients, the advisers helped them move elsewhere. Prevention of housing problems also saves money for the Council and taxpayers. CHAC estimates it has saved £318,756 in 2017/18 for the taxpayer — or £873 a day.
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