Download "UNT System Campus VPN Guide"
3 Introduction This is a guide on the different ways to connect to the University of North Texas System Campus VPN. There are several different methods of connecting to the VPN that this guide will discuss. Each method accomplishes the same goal just in a different way, which allows the users more flexibility. This flexibility is important since we have a wide range of users that have different needs and different hardware requirements. The UNT System Campus VPN is a device that will allow you to connect remotely to on-campus resources. This will allow employees and students of the University of North Texas to work from off campus using resources they otherwise couldn t access. The connection from the user machine to the Campus VPN is an encrypted connection which allows us to securely allow access to resources we otherwise wouldn t allow. As stated above there are three different methods to connect to the Campus VPN. Each method is available for all employees and students of the University of North Texas. There may be a preferred method that your network manager or teacher would like you to use, so it s always best to discuss the issue with them first. The first method is the web portal. This is a SSL web page that acts as a proxy server to the on-campus resources. This will allow access to the on-campus resources from machines that you might not want or need to install a client. This way uses SSL and Java to accomplish this goal. The second method is the AnyConnect client. This is an ActiveX or Java client which uses SSL protocols to setup an encrypted connection to the Campus VPN. The AnyConnect client gives the user a UNT IP address making their machine logically part of the network. Only traffic going to the University of North Texas will use the encrypted tunnel. All other traffic will not be encrypted and use your normal internet provider connection. The final method is the IPSec client. This is a client that uses the IPSec protocols to connect to the Campus VPN. The official IPSec client is downloadable from: You will need to use your EUID and password to download it. Any third party IPSec client that uses the standard IPSec protocols should work, however they may have problems which may not be supportable. Just like the AnyConnect client this method will also give the user a UNT IP address. It will also only use the encrypted tunnel for traffic going to the UNT like the AnyConnect client does. Just like the AnyConnect client all other traffic will use your normal internet provider connection.
4 SSL Web Portal The web portal is the easiest way to connect to the Campus VPN without having to install a client on your machine. Just point a web browser to: vpn.unt.edu You will have to accept the certificates first and you can setup a permanent exception so you only have to do this once. To login you will need to enter in your EUID and password. After that the computer usage policy will appear. You will have to accept the policy before you can continue on to the web portal home page.
5 Once you accept the computer usage policy you will be taken to the web portal home page. From this page you can go to the different areas of the web portal by using the menu to the left or the pull down menu show below. The first menu item is for web applications. This is where you can use the Campus VPN as a web proxy. On the right side of the page you will see instructions on how to use this function. Just enter in a web address in the address bar and hit enter or the browse button.
6 The Campus VPN will then browse to the webpage. Once you are at the webpage you will notice two strange things. The first thing is the address in the URL bar. You will notice that it has the Campus VPN address first then the webpage where you are. This is because you are using the Campus VPN as a proxy, piping the webpage through the Campus VPN. The second thing you will notice is the strange menu bar on the top right side of the screen. From left to right this menu will allow you to switch the menu to the other side, enter in a new address, go back to the web portal home page, and logoff the Campus VPN. If you hit the new address button it will pop up a window that you can enter in a new web address. The next menu item is the browse network page. This page will allow you to browse network shares and files.
7 Enter in the network share location into the address bar and hit enter or the browse button. This will bring up the network login screen shown below. Your network manager or teacher may have to give you access to the network share you are trying to access. Once you have access enter in your username and password for that network share. After you login, you will be able to browse the network share and access the files. Below is what the page will look like without any network folders or files. The third menu item is the AnyConnect page. From this page you will install the AnyConnect client on your machine. We will discuss the AnyConnect client and the installation later on in this guide.
8 The next menu item is the Telnet/SSH page. This page will allow you to telnet or SSH to servers on campus. Just like all the other pages, just enter in the server address you are trying to SSH or telnet to in the address bar. After you enter in the server address and hit enter a java applet will show up on page. Once it connects to the server you will get a login screen shown below. Enter in your username and password for that server and you will be logged in like normal.
9 The next menu item the VNC connection page. This page allows you to use the Campus VPN as a VNC proxy. Once you enter in the address of the machine you are trying to VNC to you will be taken to a new page. The new page uses a Java program called TightVNC as shown below. Once TightVNC connects to your VNC machine you will need to login as normal.
10 The final menu item is for terminal services usually called remote desktop. This page allows you to remote desktop to machines on campus that would otherwise be blocked from the outside. After you enter in the address of the machine you want to remote desktop to, a page like the one below will appear. This page is showing that it s trying to connect to the machine you have entered.
11 Once connected, the Campus VPN will pop up java window showing your remote desktop connection to the machine. Below is an example of the remote desktop window. That is all the features of the Campus VPN web portal. The web portal is the easiest and fastest way to connect to on campus resources without having to worry about installing any software. It provides a secure encrypted connection from your machine to the resources you are accessing.
12 Installing AnyConnect VPNClient As discussed earlier the AnyConnect client is an ActiveX or Java client that uses the SSL protocols to make an encrypted connection from your machine to the Campus VPN. To install the client you need to login to the Campus VPN web portal. If you need help logging in, please read the above section titled SSL Web Portal. 1. Select AnyConnect on the left navigation pane. Then, click Start AnyConnect.
13 2. An installation window will appear. If Java is installed on your machine, you may follow the automatic installation prompts. 3. If the automatic web install fails for any reason, click on the Windows 7/Vista/64/XP link. You may either install directly or save the installation file.
14 4. Launch the AnyConnect installation. Click Next. 5. Click on the button for I accept the terms in the License Agreement. Then, click Next.
15 6. Click Install. 7. It will take a few seconds for the installation to finish. Then, click Finish.
16 Connecting AnyConnect VPN client 1. Select the Cisco AnyConnect Secure Mobility Client from the Metro or Start menu. 2. Enter vpn.unt.edu in the text box. Then, click Connect. 3. Enter your EUID and EUID Password. Then, click OK.
18 IPSec Client Configuration The IPSec client connects to the Campus VPN using the IPSec protocols. To download the IPSec client you need to go to: There you will need to enter in your EUID and password and select the client for your OS. 1. Launch VPN Client from either the Metro or the Start menu. 2. Click New.
19 3. Configure the following items. All items are case sensitive. Then, click Save. Host: vpn.unt.edu Name: General Password: untvpnaccess 4. Highlight the new connection entry and click Connect.
20 5. Enter your EUID and EUID Password. 6. Click Continue to agree to the UNT Terms and Conditions of Service. Once you click continue you will be connected to the Campus VPN. Just like the AnyConnect client your machine now has logical connection to the University of North Texas network. You will notice an icon in the bottom right hand of your task bar. This is the IPSec client icon and right clicking it will bring up the menu below. From there you can disconnect, show statistics or bring up the connection entry list again. As stated above, once you want to log out of the Campus VPN you can bring up the menu above or bring up the connection entry list and click the disconnect button. The next time you need to connect to the Campus VPN just run the IPSec client and everything will be saved and ready to go.
21 Apple OS X Configuration This series of steps applies to Apple OS X Snow Leopard, Lion, Mountain Lion, or Mavericks. You will need an active Internet connection and administrator credentials to access the Campus VPN. 1. Access System Preferences by choosing it from the Apple menu or by opening it from your Dock.
22 2. Click Network. 3. Click on the plus sign (+) in the bottom left corner to add a new connection.
24 6. Configure the following items. Then, click OK. Shared Secret (case sensitive): untvpnaccess Group Name (case sensitive): General 7. Click Connect.
25 8. A prompt will appear. Configure the following items. Then, click OK. Account Name: EUID Password: EUID Password 9. Read the UNT Terms of Service. Click OK if you understand and agree to the Terms of Service.
26 Android Configuration Most Android devices can also support a VPN connection using an app. While there are a variety of apps which can effectively connect to the UNT System Campus VPN, this guide continues to use Cisco AnyConnect. To download and install the Cisco AnyConnect app, go to the Google Play Store and find the appropriate app for the device. =en Cisco AnyConnect ICS is a free app and requires Android or later. Not all manufactures of Android devices support ICS. Cisco offers alternative versions for some Samsung devices and rooted devices. 1. Launch the Cisco AnyConnect Secure Mobility Client app. 2. Tap Add New VPN Connection.
27 3. Configure the following items. Description: UNT System Campus VPN Server Address: vpn.unt.edu 4. Tap Done at the bottom of the screen. 5. Tap on the UNT System Campus VPN button.
29 8. Tap Accept if you agree to the Terms of Service and finalize your VPN connection. 9. The main button will now appear green to confirm an active VPN connection. 10. To close the connection, swipe the green tab to the left.
30 Apple ios Configuration This guide details setting up the UNT System Campus VPN using the built in ios VPN functionality. Alternatively, you may use the Cisco AnyConnect Secure Mobility Client app available for free in the Apple App Store. 1. Tap Settings on your iphone / ipad / ipod Touch.
31 2. Tap General 3. Tap VPN (some devices will have Network listed in the General menu, then tap VPN).
32 4. Tap Add VPN Configuration 5. Select IPSec from the options at the top of the screen.
Mac: VPN and RemoteApp Detailed Guide Update: February 2016 What is RemoteApp? RemoteApp is similar to Remote Desktop in that it enables you to access programs using a Mac computer and using the MFDA VPN.
Print How to: Set up TTUnet VPN on your Mac in OS X Snow Leopard, Lion, Mountain Lion, or Mavericks Details: 1) Access System Preferences by choosing it from the Apple menu or by opening it from your Dock.


Creating an enterprisewide "culture of security" is almost always listed as a top priority for experienced security and technology leaders in the public and private sectors.
GCN: What's the biggest challenge left?
LOHRMANN: Continuing to work on the culture, to help people understand how important security is at an individual level. ... Helping people understand the impact of their actions, I think that's the biggest challenge.
Fast-forward more than a decade, And I believe transforming the security culture still remains our greatest challenge as we head toward 2020. But how can we get to this elusive "culture of security" while balancing the cost, benefits and many other business priorities we face? As we think about people, processes and technology, what can we do to enable people and reduce risk over time?
Can "Just in Time" Training Help?
"It is walking down to the desk of a more experienced co-worker to ask for a solution when you get stuck on a project. It is looking up Wikipedia when you come across a novel concept during your browsing sessions. It is calling up mom when you want advice on a recipe.
Just-in-time learning is having access to knowledge just when you need it. It is not having to wait till the public library opens or you can catch hold of a subject matter expert.
The articles goes on to list many benefits of just-in-time learning, such as how it enhances worker productivity and speeds up the learning process.
Taking the "just in time" concept further, several companies are advocating the use of these techniques for enterprise security awareness training. The general concept that I am seeing is to provide a very basic compliance-focused training for the majority of people, and to enforce much more specific training for the select few who they identify as needing it most, since they violate some security policy, and/or do something inappropriate, such as clicking on a simulated phishing link.
"The first time Joe Employee saves a document to an unapproved cloud storage site (for example), he gets a system-generated pop-up that directs him to company policy on the use of cloud storage. Problem solved, 70% of the time — but not always.
Other companies are now offering related "education triggers" or "teachable moments" that are targeted at those who violate security policy or need the training the most because they do something wrong. These approaches claim to identify, focus training, (and where necessary) get rid of the bad apples and focus on those in the organization who (despite being non-malicious) pose the greatest risk.
But other industry leaders are not in favor of this "just in time" security training approach. They say this practice is like watering down the soup at your favorite restaurant. Yes, it may produce an immediate cost-saving benefit, but is it being penny wise and pound foolish in the end? They insist: With technology and threats moving so fast, don't all staff need constantly refreshed, relevant, focused security awareness training?
This blog from Info Security Magazine offers another specific contrarian argument. Kai Roer, founder and CEO of CLTRe, writes, "Blaming people for not handling poor technology correctly is — in my opinion — simply wrong."
While there are certainly some benefits to fear-based (or carrot-stick-based) approaches that send you off to training if you fail, others ask where the carrot is for staff in this model. Most parents understand the need for a mix of rewards and punishments, but this is all about punishments.
Myth: Crisis is a powerful impetus for change.
Reality: Ninety percent of patients who've had coronary bypasses don't sustain changes in the unhealthy lifestyles that worsen their severe heart disease and greatly threaten their lives.
Myth: Change is motivated by fear.
Reality: It's too easy for people to go into denial of the bad things that might happen to them. Compelling, positive visions of the future are a much stronger inspiration for change.
Other training experts I have spoken with say that fear can certainly help, if done in the appropriate balance with rewards. They have seen the classic "carrot and stick" approach work well in security training.
Back in 2014, I wrote this article on how to change the security culture in government. Of course, training is only one piece. The most important thing senior leaders must do is lead from the front. Second, all leaders must constantly communicate the vision for excellence, the process for getting there and the sense of urgency necessary by all.
I also pointed to this fascinating research on what motivates us at work beyond carrots and sticks. If you haven't seen this before, I urge you to watch the video.
Spoiler Alert: For those you don't want to watch the YouTube video, the main points are that research shows that what best motivates us is a sense of purpose in work, self-direction and mastery of a subject.
Or, in other terms, we need to offer compelling content that is intriguing and teaches people what they don't already know about security in sticky ways, in order to change behaviors and motivate people.
Content is still king, and I also believe that brief, frequent and focused content works best with gamification or game-based learning. I have even suggested to CSO Magazine that we need to make security awareness training more about culture change with a potential name change.
And while I do think that "just in time" security training may be able to help select organizations in a very limited context (as a supplemental approach), I have a more fundamental concern with this trend, if it is front and center.
Beyond views of the awareness training, the security team's reputation can suffer. In this type of enterprise culture, the security team members are the bad guys — or "Dr. No" who might pull you over or get you fired. My blog followers are familiar with the seven reasons that security pros fail — and what you can do about it. Bottom line: You DON'T WANT TO GO THERE!
I was one of the first security pros to say: Be a security enabler — not a disabler. However, I learned this lesson the hard way, after I was almost fired for opposing WiFi deployments in my earlier days as Michigan CISO. Nevertheless, that negative experience set our team on a new course that won us awards for our WiFi implementation and our overall cybersecurity program. We saw that powerful role that each employee must play in protecting the enterprise from cyberthreats.
In a healthy security culture, all front-line staff are proactively well trained on information and physical security, know what to do (and not do), where to report incidents, when to ask for help, who to contact and how to work together effectively. Staff have a good relationship with the security team — because the cyberpros are helpful. There is not an "us vs. them" problem.
I recognize that some in the security industry will disagree with me on this blog. But I hope we can agree on this: We need to be passionately building (or rebuilding) enterprise cultures that put security at the top of the priority list.
Just as model parents and teachers train their children by demonstrating, encouraging, motivating and challenging in fun, positive ways, much more than disciplining them, we must do the same to build healthy security cultures that endure. We want the staff to say "thank you!" They will, if we offer helpful security lessons that are intriguing, thoughtful and memorable.
And please don't make end-user security awareness training mainly a punishment for doing something wrong.


text:
Q: Why do these foreach and for loops create different outputs? $value = 1234;
$number_array = str_split($value);

for($i = 0; $i < sizeof($number_array); $i++) {
  $number_array[$i] *= 10;
}
// output: 10, 20, 30, 40.

foreach($number_array as &$value) {
  $value *= 10;
}
// output: 10, 20, 30, 30.

Can someone explain why the for loop gives the desired result, but the foreach doesn't?

A: Because  for loop changes the values of   $number_array
do like this will give same result
    $value = 1234;
    $number_array = str_split($value);

    foreach($number_array as $value) {
       $value *= 10;
       echo $value;
       // output: 10, 20, 30, 40.
    }

    for($i = 0; $i < sizeof($number_array); $i++) {
      $number_array[$i] *= 10;
    }
    // output: 10, 20, 30, 40.
    print_r($number_array);

A: Here is the result with var_dump :
for($i = 0; $i < sizeof($number_array); $i++) {
  $number_array[$i] *= 10;
  var_dump($number_array[$i]);
}
// output: int(10) int(20) int(30) int(40)

foreach($number_array as &$value) {
  $value *= 10;
  var_dump($value);
}
// output: int(100) int(200) int(300) int(400)

The output is different because you assigned to each variable their multiplication by 10 in the first loop and you do it again in the second.
If you call them individually with the same values, they will do the same output.

A: This two loops have exactly same result if you call it separately. But in first loop you are changing $number_array. Before first loop it's value is [1,2,3,4] and after it is [10,20,30,40].
take a look:
$value = 1234;
$number_array = str_split($value);

for($i = 0; $i < sizeof($number_array); $i++) {
  $number_array[$i] *= 10;
}

var_dump($number_array); // 10, 20, 30, 40

foreach($number_array as &$value) {
  $value *= 10;
}

P.S if you want to get same result for each loop, change your code like that:
$value = 1234;
$number_array = str_split($value);

for($i = 0; $i < sizeof($number_array); $i++) {
   $x = $number_array[$i];
   $x  *= 10;

    echo $x."
";
}
echo "---
";

foreach($number_array as &$value) {
  $value *= 10;
    echo $value."
";
}

Note: don't change values into array, but write them in $x variable and change it

A: It turned out the loop I was using to create the string was causing the incorrect output because I used $value again in the foreach loop of the string.
$value = 1234;
$number_array = str_split($value);

foreach($number_array as &$value) {
  $value *= 10;
}
foreach($number as $value) {
  $answr .= $value . " ";
}
// $answr = 10 20 30 30

Once I had changed the loop to $num to test, the correct output was printed.
foreach($number as $num) {
  $answr .= $num . " ";
} //$answr: 10 20 30 40

An unsual quirk, that I haven't quite got my head around yet, but I'm assuming it's because $value was still set as a var during the second foreach loop.
Thank you for all answers and I apologise for not posting the whole code snippet and causing confusion.

redpajama_set_name: RedPajamaStackExchange


text:
TELLER OF TALES
CAROLYN SAUNDERS was swept away at birth to spend her first months of babyhood sleeping in a cardboard box in a cupboard somewhere near Nellis AFB in Nevada, while her fighter pilot father did fighter pilot things like saving the world and playing Dead Ants. Laying in that dark cupboard, listening to the gentle snores of her older siblings who got to sleep in the real bedroom, she had lots of time to reflect on her young life and decided she already had some pretty awesome stories to tell, if only she could learn how to talk. Life after that was an exciting whirlwind (note to the grammar police: she knows there is no such thing as a boring whirlwind) of new bases, new countries, and new cultures, as she followed her mum and dad to such exotic postings as Germany, France and Chatham, New Brunswick, popping into England often enough to remember some kind of roots and replenish her supply of sticky toffee pudding. She wrote her first novel at age 10, a self-illustrated, self-bound (staples!) tome called Copper, Dog of the North, that rocked the literary world at her elementary school. Heady with success and the joy of storytelling, she never looked back, eventually parlaying her talent into a career as a newspaper reporter before being seduced by movie magic.
CAROLYN'S WORK
Carolyn first feature film, The Wasting, had a limited Canadian theatrical release after
a prize-winning festival run that included Best Feature at Vittorio Veneto. The
coming-of-age ghost story starred Lauren McQueen and Alexz Johnson. She is currently in pre-prep on her second feature, Island West, a female-led sci-fi set in a dystopian near-future, which was a finalist in the Meryl Streep/Nicole Kidman Writer's Lab. Her recently completed short film, Sundown, stars Art Hindle and Brendan Flynn.
Sometimes Carolyn is happy to just write, as with the upcoming Amazon series Ted Bundy: Falling for a Killer, and sometimes she is happy to write funny, as with the comedy web-series Tokens. Carolyn is also a story editor who most recently worked on the Richard Dreyfuss film Astronaut. She is an alumna of the prestigious Women in the Director's Chair (WIDC) and greatly appreciates all the help and support they have given.
While her focus is on writing and directing feature films, Carolyn takes forays into television documentary, where she has written and/or creative produced series such as the Gemini-nominated Mighty Ships (Discovery), Yukon Gold (History), the long-running Ghostly Encounters, Outlaw Tech, Fatal Vows, and Discovery's hit series Forensic Factor. She also has extensive experience developing doc series. The years spent pretending to love German food and trespassing in the potato fields of European farmers gave her an excellent story sense and a knack for finding the drama.
Though her dark beginnings have caused her to favour comedy, Carolyn works in many genres and formats. She has story edited and consulted on countless feature scripts and documentaries, in addition to being a script analyst for 13 years for the Harold Greenberg Fund.
Bonus fact: Carolyn is a citizen of both the UK and Canada, and thus can write almost anywhere, including squished into a Ryanair seat, if the situation calls for it. Also, she is mad for dogs and ice cream.
WAGGER OF TAILS
ENDZONE spent his first year of life being a puppy, possibly sleeping in a box, in all likelihood not anywhere near Nellis AFB. At the age of one he went to live with Carolyn, where he spent the next 13 years chasing sticks into the water, knocking things off the coffee table with his tail, patiently enduring psychotic cat attacks, immensely enjoying car rides and worming his way into her cold hard heart, until she felt compelled to name her company after him so he could live forever. Endzone's motto is "swim softly and carry a big stick."

redpajama_set_name: RedPajamaCommonCrawl


Astronaut Tom Jones: When we reached orbit on Sept. 30, '94, the Taklamakan was a major landmark and science target for Space Radar Lab 2. Our radar targets were the alluvial fans and dune fields along the southern margin of the Taklamakan, where the Silk Road oases hosted caravans and travelers on an ancient trade route. Note the alluvial fans and vegetation in foreground, fed by streams from the Altyn-Tagh mountains bordering the desert on the south. Our radar images found traces of Silk Road irrigation channels and sand-covered villages.
NASA: This south-looking view shows most of the west end of snow-dusted ranges on the Tibetan Plateau. A major fault line separates the plateau from the low-lying Takla Makan Desert (foreground). The darker areas along two rivers (foreground) make up one of the largest agricultural regions in the Takla Makan Desert. The hazy atmosphere over India (top) contrasts with the thinner, clear air over the plateau. The Vale of Kashmir in northern India is the prominent valley within the first wall of the Himalayan Mountains.


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On March 2, 1962 the Philadelphia Warriors, who would later go on to be the Golden State Warriors, played the New York Knicks in Hershey, Pennsylvania. The Warriors won the contest 169-147 in what was then a record for most combined points in an NBA game. That feat was overshadowed by perhaps the single greatest performance ever by an NBA player, Wilt Chamberlain scoring 100 points. "Wilt The Stilt" shot 36 for 63 from the field and went 28 for 32 from the charity stripe to amass his 100 points. Due to Chamberlain's scoring prowess the accomplishment was overlooked; however the game would eventually go down as Chamberlain's signature moment.
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text:
Printed from https://www.righttoknow.org.au/request/in_regards_to_an_address_to_the?utm_campaign=alaveteli-experiments-87&utm_content=sidebar_similar_requests&utm_medium=link&utm_source=right+to+know on January 31, 2023 05:18
In regards to an Address to the National Equity in Higher Education Policy Forum
P. Gale made this Freedom of Information request to Department of Industry, Science, Energy & Resources
P. Gale
Department of Industry, Science, Energy & Resources
The request was successful.
P. Gale June 21, 2013
To: The FOI Contact Officer
Department of Industry ... and Tertiary Education
Dear FOI Contact Officer,
This is a request under the FOI Act for documents.
On your department's web-site, you have a speech the Minister for Higher Education and Skills (The Hon Sharon Bird MP) recently made. The speech is called "Address to the National Equity in Higher Education Policy Forum". The web page says the minister gave the speech on April 30th. In the speech, she says that it was at Sydney University. I request documents (details below) in regards to the forum.
Please interpret "document" to mean "documents" where necessary or appropriate (for reasons, see http://bit.ly/Yu3jzG) and to include electronic documents as appropriate. Please interpret "apply" "application", "request" as the same.
Documents I request:
1. I would like a copy of the agenda for the conference.
2. If forum attendees got an "attendee package" (or whatever it was really called) from the department, I would like a copy of each document in the package.
3. I request a copy of any presentations (Power Point slides etc.)that attendees got shown.
4. I request a copy of any briefing the department gave to the minister for the speech at the forum.
In regards to item 4, I limit my request to the final version of briefing notes provided for the minister no earlier than 21 days preceding the forum (ie: on or after Monday, April 8th 2013) with the intention of preparing the minister for questions that she may have been expected to receive from attendees at the forum. If more than one such document exists, perhaps you could briefly describe their nature. If the minister was not given any briefing notes, please let me know and then consider the request for that item to be withdrawn.
P. Gale (Mrs)
June 21st, 2013.
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text:
Pope names Filipino archbishop as envoy to Israel and Cyprus
Archbishop Adolfo Tito Yllana, the new apostolic nuncio to Israel and Cyprus and apostolic delegate to Jerusalem and Palestine. SCREENSHOT/ARCHDIOCESE OF BRISBANE
By CBCP News
Pope Francis has appointed a Filipino archbishop to carry out an important mission in the Holy Land, the birthplace of Christianity.
In a statement on Thursday night, the Vatican announced that the new apostolic nuncio to Israel and Cyprus is Archbishop Adolfo Tito Yllana.
At the same time, the pope named the 73-year-old as apostolic delegate to Jerusalem and Palestine.
Prior to his appointment, Yllana has been serving as apostolic nuncio to Australia since 2015.
He succeeds Archbishop Leopoldo Girelli, who was appointed Apostolic Nuncio to India in March.
The prelate has represented the Holy See on four continents: Africa, Asia, Europe and Oceania.
Born in Naga City on Feb. 6, 1948, Yllana was ordained priest for the Archdiocese of Caceres on March 19, 1972.
He later earned a doctorate in civil and canon law, and entered the Ecclesiastical Academy, the Holy See's school of diplomacy.
After his studies, he entered the diplomatic service of the Holy See, serving successively at Pontifical Representations in Ghana, Sri Lanka, Turkey, Lebanon, Hungary and Taiwan.
In December 2001, St. John Paul II appointed him as apostolic nuncio to Papua New Guinea and consecrated him bishop on January 6, 2002 in St. Peter's Basilica.
Subsequently, he entrusted him with the leadership of the nunciature in the Solomon Islands.
In 2006, Pope Benedict XVI appointed him as apostolic nuncio to Pakistan and apostolic nuncio to the Democratic Republic of Congo in 2010.
In February 2015, Pope Francis appointed him as apostolic nuncio to Australia.
Yllana is one of the three Filipino apostolic nuncios who are active in the service. The others are Archbishop Bernardito Auza, the nuncio to Spain, and Archbishop Francisco Padilla, the nuncio to Guatemala.
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text:
\section{Introduction}

In this paper we will investigate semiclassical variations of the Stieltjes-Wigert polynomials, the $q$-Laguerre polynomials and
the little $q$-Laguerre polynomials. We are mainly interested in the recurrence coefficients of the monic polynomials in the three term
recurrence relation
\begin{equation}  \label{TTR}
   xP_n(x) = P_{n+1}(x) + b_n P_n(x) + a_n^2 P_{n-1}(x), \qquad n \geq 0,
\end{equation}
where $P_0=1$ and $P_{-1}=0$. There are two reasons for combining these three families in one paper. First of all, the Stieltjes-Wigert
polynomials are really just a special instance of the $q$-Laguerre polynomials and there is also a connection between the $q$-Laguerre polynomials
and the little $q$-Laguerre polynomials (changing $q$ to $1/q$). The second reason is that we use the same technique to prove the results for these three families of
recurrence coefficients, which results in very similar computations.

In the first three subsections we will briefly recall the Stieltjes-Wigert, $q$-Laguerre and little $q$-Laguerre polynomials. In \S \ref{section:main}
we will introduce the semiclassical variation of these three families and we will formulate our main results.

\subsection{Stieltjes-Wigert polynomials}
Stieltjes introduced in
\cite[\S 56]{Stieltjes} the family of positive weight functions
\[ \frac{1}{\sqrt{\pi}} x^{-\log x}[1+\lambda \sin (2\pi\log x)], \qquad x \in [0,\infty) ,\quad -1 \leq \lambda \leq 1, \]
and showed that the moments exist and are independent of $\lambda$:
\[   \frac{1}{\sqrt{\pi}}\int_0^\infty x^n x^{-\log x}[1+\lambda \sin (2\pi\log x)]\, dx = e^{(n+1)^2/4}.  \]
We denote the weight function with $\lambda=0$ by
$$
w_1(x) = \frac{1}{\sqrt{\pi}} \exp(-\log^2(x)), \qquad x \in [0,\infty),
$$
and Stieltjes' main contribution is that this weight function is not uniquely determined by its moments, since there
are other positive measures on $[0,\infty)$ with the same moments: the measure with weight function $w_1$ corresponds to
an \textit{indeterminate moment problem}. Wigert \cite{Wigert} introduced the more general weight function
\begin{equation}\label{sw-weight}
w_k(x) = \frac{k}{\sqrt{\pi}} \exp(-k^2\log^2(x)), \qquad  x \in [0,\infty),\,\, k>0,
\end{equation}
for which the moments are given by
$$
\mu_n = \int_0^\infty x^n w_k(x)\, dx = q^{-\frac{(n+1)^2}{2}},
$$
where $q=\exp(-1/(2k^2))$. This weight function also induces a positive measure on $[0,\infty)$ with an
indeterminate Stieltjes moment problem (\cite[Section 2.6]{Chihara}).
Another measure with the same moments is given by Askey \cite{Askey}: he
considers weight functions of the form
\begin{equation}\label{askey-weight}
\tilde{w}_\alpha(x) = \frac{x^\alpha}{(-x;q)_\infty(-q/x;q)_\infty},
\qquad x >0,\ 0 < q < 1.
\end{equation}
Here $(x;q)_\infty$ is the $q$-Pochhammer symbol
\[  (x;q)_\infty = \prod_{k=0}^\infty(1-xq^k). \]
The moments $m_n$ associated with this weight function satisfy
$$
\frac{m_n}{m_0}= q^{-n(\alpha+1)-n(n-1)/2}.
$$
This can be compared with the moments of the Stieltjes-Wigert weight $\mu_n/\mu_0=q^{-n^2/2-n}$, and hence for
$\alpha=1/2$ Askey's weight function has the same moments as the Stieltjes-Wigert weight\footnote{in \cite{Askey} there is an unfortunate misprint
in (4.14) and (4.15) which should be $d(n) = q^{-n^2/2-n-1/2}$ and $d(n)/d(0) = q^{-n^2/2-n}$. Then $\gamma=3/2$ and in (4.16) the $x^{-5/2}$ should
be $x^{1/2}$.}. In particular this means that
the orthogonal polynomials for the weight function (\ref{askey-weight}) with $\alpha=1/2$ coincide with the Stieltjes-Wigert polynomials, which are
orthogonal with respect to $w_k$. Observe that Askey's weight function has a factor $x^\alpha$ so that (\ref{askey-weight})
is in fact a one-parameter family of weight functions with explicitly known moments. This is implicitly also the case for the Stieltjes-Wigert
weight (\ref{sw-weight}), which has the scaling property
\[    w_k(q^\beta x) = q^{\beta^2/2} x^\beta w_k(x), \]
where we used $\log q = -1/(2k^2)$. Chihara \cite{Chihara2} also gives a number of discrete measures with the same moments
as the Stieltjes-Wigert weight. A detailed study of various solutions of the moment problem for the Stieltjes-Wigert weight,
including N-extremal solutions, can be found in the work of Christiansen \cite{Jacob1}.
The orthonormal Stieltjes-Wigert polynomials are given by
\[  P_n(x) = (-1)^n \frac{q^{(2n+1)/4}}{\sqrt{(q;q)_n}}  \sum_{k=0}^n \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}} q^{k^2} (-q^{1/2}x)^k, \]
\cite[Eq. (11)]{Wigert}, \cite[Eq. (2.3) on p.~173]{Chihara}, \cite[Eq. (2.7.4) on p.~33]{Szego}.
In terms of  basic hypergeometric functions \cite{GasperRahman}, the monic Stieltjes-Wigert polynomials are
\begin{equation}  \label{Pn}
P_n(x) = (-1)^n q^{-n^2-\frac{n}{2}} {}_1\phi_1 \left(\begin{array}{c} q^{-n} \\ 0 \end{array} ;q , -q^{n+3/2}x \right).
\end{equation}
Note that the polynomials in \cite[Eq. (3.27.1) on p.~116]{KoekoekSwarttouw} \cite[Eq. (14.27.1) on p.~544]{KoekoekLeskySwarttouw}
\[    S_n(x) = \frac{1}{(q;q)_n}{}_1\phi_1 \left( \begin{array}{c} q^{-n} \\ 0 \end{array} ; q , -q^{n+1}x \right) \]
are orthogonal for the weight function $x^{-1/2} w_k(x)$ (or Askey's weight function $\tilde{w}_0(x)$)\footnote{The remark in \cite[p.~117]{KoekoekSwarttouw} and \cite[p.~546--547]{KoekoekLeskySwarttouw} is missing this factor $x^{-1/2}$ in the weight function.}.
The monic Stieltjes-Wigert polynomials satisfy the three-term recurrence relation \eqref{TTR}
with coefficients $b_n = q^{-2n-3/2}\left(1+q-q^{n+1}\right)$ and $a_n^2 = q^{-4n}(1-q^n)$. The monic orthogonal polynomials
for the weight $w_k(q^\beta x)=q^{\beta^2/2} x^\beta w_k(x)$ are $Q_n(x) = q^{-n\beta} P_n(q^\beta x)$ and satisfy
\[   xQ_n(x) = Q_{n+1}(x) + b_nq^{-\beta} Q_n(x) + a_n^2q^{-2\beta} Q_{n-1}(x), \]
with $a_n^2$ and $b_n$ the recurrence coefficients of the Stieltjes-Wigert polynomials $P_n$.
The case $\beta=-1$ gives the log-normal distribution, which is well known in statistics (the distribution of $Y= e^X$, where
$X$ has a standard normal distribution). The case $\beta=\alpha-1/2$ gives a weight function with the same moments
as Askey's weight function $\tilde{w}_\alpha$ in \eqref{askey-weight}.

\subsection{$q$-Laguerre polynomials}  \label{sec:1.2}
Chihara \cite[pp.~173--174]{Chihara} \cite{Chihara1} \cite{Chihara3} introduced a generalization of the
Stieltjes-Wigert weight function:
\begin{equation}\label{chihara-weight}
\rho_0(x)=  \bigl(\frac{-p}{\sqrt{q}x};q\bigr)_\infty w_k(x),
\qquad x \in [0,\infty),\,\,\, p \in [0,1).
\end{equation}
The monic orthogonal polynomials with respect to \eqref{chihara-weight} are
\[    S_n(x;p,q) = (-1)^n q^{-n(2n+1)/2} (p;q)_n \sum_{k=0}^n \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}} q^{k^2} \frac{(-q^{1/2}x)^k}{(p;q)_k}, \]
or, in basic hypergeometric form,
\begin{equation} \label{Pn-p}
   S_n(x;p,q) = (-1)^n q^{-n(2n+1)/2} (p;q)_n\ {}_2\phi_1 \left( \begin{array}{c} q^{-n} \\ p \end{array} ; q, -q^{n+\frac32}x \right),
\end{equation}
which, for $p=0$, reduces to \eqref{Pn}. These orthogonal polynomials are nowadays known as $q$-Laguerre polynomials
\[   L_n^{(\alpha)}(x;q) = \frac{(q^{\alpha+1};q)_n}{(q;q)_n}
  {}_1\phi_1 \left( \begin{array}{c} q^{-n} \\ q^{\alpha+1} \end{array} ; q, -q^{n+\alpha+1} x \right), \]
\cite[Eq. (3.21.1) on p.~108]{KoekoekSwarttouw}, \cite[Eq. (14.21.1) on p.~522]{KoekoekLeskySwarttouw} and were also investigated by Moak \cite{Moak}.
Indeed, if $p=q^{\alpha+1}$ then $S_n(x;p,q) = C_n L_n^{(\alpha)}(xq^{3/2}/p;q)$, where $C_n$ is a normalizing constant.
The $q$-Laguerre polynomials were also investigated by Askey \cite{Askey2}, Ismail and Rahman \cite{IsmailRahman} and Christiansen \cite{Jacob2}.
Chihara found the recurrence coefficients for the monic orthogonal
polynomials $S_n(x;p,q)$ with respect to the weight \eqref{chihara-weight}:
\begin{align*}
b_n &= q^{-n-3/2}\left(-p-q+(1+q)q^{-n} \right), \\
a_n^2 &= q^{-4n}\left( 1-q^n\right)\left( 1-pq^{n-1}\right),
\end{align*}
which for $p=0$ also reduces to the Stieltjes-Wigert case.
The moment problem for these $q$-Laguerre polynomials is also indeterminate and in fact one has the orthogonality relation
\[  \int_0^\infty L_n^{(\alpha)}(x;q)L_m^{(\alpha)}(x;q) \frac{x^\alpha}{(-x;q)_\infty}\, dx = 0, \qquad n \neq m, \]
which uses a somewhat simpler weight function than \eqref{chihara-weight}.
Christiansen investigated the various solutions of the moment problem for $q$-Laguerre polynomials and one of the weights
he found is
\[   v(x) =  \frac{(-q/x;q)_\infty}{(-q^{\alpha+1-c}x)_\infty (-q^{-\alpha+c}/x;q)_\infty} x^{c-1}  \]
\cite[Eq. (3.0.6) on p.~9]{Jacob2}.
Indeed $v(xq^{3/2}x/p)$ corresponds to the weight $(-p/(\sqrt{q}x);q)_\infty \tilde{w}_{1/2}$ when $c=3/2$ and hence has the same moments as
the weight $\rho_0$ in \eqref{chihara-weight}.

\subsection{Little $q$-Laguerre polynomials}
The monic little $q$-Laguerre polynomials are given by
\[  P_n(x|q) = (-1)^n q^{\binom{n}{2}} (q^{\alpha+1};q)_n\ {}_2\phi_1 \left( \begin{array}{c} q^{-n}, 0 \\ q^{\alpha+1} \end{array} ; q,qx \right), \]
and their orthogonality relation is
\[  \sum_{k=0}^\infty P_n(q^k|q) P_m(q^k|q) q^k \frac{q^{k\alpha}}{(q;q)_k} = 0, \qquad m\neq n, \]
(\cite[\S 14.20]{KoekoekLeskySwarttouw}). This can also be written as
\[   \int_0^1 P_n(x|q)P_m(x|q) x^\alpha (qx;q)_\infty \, d_qx = 0, \qquad m\neq n, \]
where Jackson's $q$-integral is
\begin{equation} \label{q-Jackson}
  \int_0^1 f(x)\, d_qx = (1-q) \sum_{k=0}^\infty q^k f(q^k).
\end{equation}
We can therefore say that the little $q$-Laguerre polynomials are orthogonal on the $q$-lattice $\{q^k \ | \ k=0,1,2,\ldots\}$ for the
weight function $w(x) = x^\alpha (qx;q)_\infty$. The little $q$-Laguerre polynomials are intimately related to the $q$-Laguerre polynomials:
\[   L_n^{(\alpha)}(x;q) = (-1)^n P_n(-x|q^{-1}), \]
and from this we easily find the recurrence coefficients from those of the $q$-Laguerre polynomials by changing $q$ to $1/q$ and
changing the sign in $b_n$:
\begin{align*}
b_n &= q^{n} \bigl( 1+q^\alpha - q^{n+\alpha}(1+q) \bigr), \\
a_n^2 &= q^{2n+\alpha-1} (1-q^n)(1-q^{n+\alpha}) .
\end{align*}
The moment problem for the little $q$-Laguerre polynomials is determinate.

\subsection{Semiclassical extensions: main results} \label{section:main}

In this paper we will consider semiclassical extensions of the
Stieltjes-Wigert, $q$-Laguerre and little $q$-Laguerre weights by replacing the variable $x$ by $x^2$ and the parameter $q$ by $q^2$, but still restricting the weight to the positive real axis. This is very similar to the first semiclassical extension of the Hermite polynomials
(with weight $w(x) = e^{-x^2}$ on $\mathbb{R}$) to orthogonal polynomials for the Freud weight $w(x) = e^{-x^4}$ on $\mathbb{R}$, which leads to
recurrence coefficients satisfying the discrete Painlev\'e I equation (d-$\textrm{P}_{\textrm{I}}$), see e.g., \cite{VanAssche} or the semiclassical extension of the Laguerre polynomials (with weight $w(x)=x^\alpha e^{-x}$ on $[0,\infty))$ to $w(x)=x^\alpha e^{-x^2+tx}$ on $[0,\infty)$, which leads to an asymmetric discrete Painlev\'e IV equation
(d-$\textrm{P}_{\textrm{IV}}$), see \cite{BoelenVA}.
Our main interest is in the recurrence coefficients of the corresponding orthogonal polynomials. We will show that they are related to solutions
of the $q$-discrete Painlev\'e III ($q$-$\textrm{P}_{\textrm{III}}$) and Painlev\'e V ($q$-$\textrm{P}_{\textrm{V}}$) equations.

After this paper was submitted, we learned that Ormerod  considered simple generalizations of the Big $q$-Laguerre polynomials
and their relation with $q$-$\textrm{P}_{\textrm{V}}$ in \cite[\S 5]{Ormerod}. His results suggest that it would be of interest to investigate semi-classical extensions
of the Big $q$-Laguerre polynomials as well.

In Section \ref{section:Stieltjes-Wigert} we will prove the following result:
\begin{theorem}  \label{thm:1}
The recurrence coefficients of the orthogonal polynomials for the
semiclassical Stieltjes-Wigert weight
$$
w(x) = \frac{x^\alpha}{(-x^2;q^2)_\infty(-q^2/x^2;q^2)_\infty},
\qquad x \in [0,\infty),\  \alpha \in \mathbb{R}
$$
can be found by
\begin{align*}
  q^{2n+\alpha}b_n^2x_n &= x_{n+1}+q^{2n+2\alpha}x_{n-1}\left(x_n+q^{-n-\alpha}\right)^2+2\left(x_n+q^{-\alpha}\right)
\intertext{and}
  a_n^2 &= q^{1-n}x_n+q^{-2n-\alpha+1},
\end{align*}
where $x_n$ satisfies the $q$-discrete Painlev\'e III equation
\begin{equation}  \label{qPIII-thm1}
x_{n-1}x_{n+1} = \frac{(x_n+q^{-\alpha})^2}{(q^{n+\alpha}x_n+1)^2}.
\end{equation}
\end{theorem}

In Section \ref{section:q-Laguerre} we will prove
\begin{theorem} \label{thm:2}
The recurrence coefficients of the orthogonal polynomials for the
semiclassical $q$-Laguerre weight
$$
w(x)= \frac{x^\alpha
(-p/x^2;q^2)_\infty}{(-x^2;q^2)_\infty(-q^2/x^2;q^2)_\infty},\qquad x\in [0,\infty),\ p\in [0,q^{-\alpha}), \alpha \geq 0
$$
can be found by
\begin{align*}
  b_n^2q^{2n+\alpha}z_n^2 &=
  z_nz_{n+1}-1+q^{2n+2\alpha}\left(\sqrt{pq^{-2-\alpha}}z_n+q^{-n-\alpha}\right)^2\left(z_nz_{n-1}-1\right)\\
 & \qquad + 2\left(z_n+\sqrt{q^{2-\alpha}/p}\right)\left(z_n+\sqrt{pq^{\alpha-2}}\right),
\intertext{and}
  a_n^2 &= q^{-n+1}z_n\sqrt{pq^{-2-\alpha}}+q^{-2n-\alpha+1},
\end{align*}
where $z_n$ satisfies the $q$-discrete Painlev\'e V equation
$$
(z_nz_{n-1} -1)(z_nz_{n+1}-1)=
\frac{\left(z_n+\sqrt{q^{2-\alpha}/p}\right)^2\left(z_n+\sqrt{pq^{\alpha-2}}\right)^2}
{\left(q^{n+\alpha/2-1}\sqrt{p}z_n+1\right)^2}.
$$
\end{theorem}
Note that this result reduces to Theorem \ref{thm:1} if we take $x_n=z_n\sqrt{p}q^{-1-\alpha/2}$ and let $p \to 0$.
The result for $p=q^2$ was already obtained in \cite[Thm.~1.3]{BoelenVA}.

In Section \ref{section:little-q-Laguerre} we will prove
\begin{theorem}  \label{thm:3}
The recurrence coefficients of the orthogonal polynomials  for the weight
\[   w(x) = x^\alpha (q^2x^2;q^2)_\infty, \qquad \alpha >0  \]
on the positive exponential lattice $\{q^k \ | \ k=0,1,2,3,\ldots\}$ are given by
\[      a_n^2 = q^{n+\alpha/2-1}(x_n - q^{n+\alpha/2}), \qquad n \geq 1,  \]
and
\begin{multline*}  b_n^2 x_n^2q^{-2n-\alpha} = 1 - x_nx_{n+1}-q^{-2n}(x_nx_{n-1}-1)(x_nq^{-\alpha/2}-q_n)^2 \\
     -\ 2(x_n-q^{\alpha/2})(x_n-q^{-\alpha/2}), \qquad n \geq 1,
\end{multline*}
with $b_0^2=1-q^{\alpha/2}x_1$, where $x_n$ satisfies the $q$-discrete Painlev\'e V equation
\begin{equation}   \label{qPV-3}
 (x_nx_{n-1}-1)(x_nx_{n+1}-1) = \frac{q^{2n+\alpha}(x_n-q^{\alpha/2})^2(x_n-q^{-\alpha/2})^2}{(x_n-q^{n+\alpha/2})^2},
\end{equation}
with initial values $x_0=q^{\alpha/2}$ and $x_1=q^{-\alpha/2}(1-\mu_1^2/\mu_0^2)$, where
$\mu_0$ and $\mu_1$ are the first two moments of the weight $w$.
\end{theorem}

\section{Ingredients of the proofs}

\subsection{Discrete Painlev\'e equations}
Our Theorems \ref{thm:1}--\ref{thm:3} express the recurrence coefficients as solutions of $q$-discrete Painlev\'e equations.
These equations  are second order, nonlinear difference
equations which have a continuous Painlev\'e equation as a
continuous limit. They pass an integrability test called singularity
confinement \cite{GrammaticosRamaniPapageorgiou}. This integrability detector is the discrete analogue of the
Painlev\'e property for differential equations.
For an overview of discrete Painlev\'e equations, see
\cite{GrammaticosRamani2}. Three types of discrete Painlev\'e equations are
distinguished: those of additive nature (the independent variable
$n$ entering through $\alpha n+\beta$) are usually denoted by d-P or
$\delta$-P, those of multiplicative nature, where the independent
variable $n$ enters the equation in an exponential way, which are denoted by $q$-P, and an elliptic
equation. Sakai gave a classification of discrete Painlev\'e equations in terms of affine Weyl groups, starting from the
exceptional Weyl group $E_8$ (Sakai \cite{Sakai}, see also \cite[\S 6, p. 293--300]{GrammaticosRamani2}).
In this paper we are dealing with $q$-discrete Painlev\'e III ($q$-$\textrm{P}_{\textrm{III}}$)
\begin{equation}  \label{qPIII}
     x_{n+1}x_{n-1} = \frac{(x_n+a)(x_n+b)}{(1+cq^nx_n)(1+dq^nx_n)},
\end{equation}
and $q$-discrete Painlev\'e V ($q$-$\textrm{P}_{\textrm{V}}$)
\begin{equation}   \label{qPV}
    (x_{n+1}x_n-1)(x_nx_{n-1}-1) = cdq^{2n} \frac{(x_n-a)(x_n-1/a)(x_n-b)(x_n-1/b)}{(x_n-cq^n)(x_n-dq^n)} ,
\end{equation}
\cite[p.~269]{GrammaticosRamani2}, \cite[p.~720]{VanAssche}. We will find special cases of these $q$-discrete Painlev\'e equations with
$a=b$ and $c=d$.

\subsection{Technique of ladder operators}
In what follows we will be interested in retrieving the recurrence
coefficients of semiclassical extensions of the Stieltjes-Wigert polynomials, the $q$-Laguerre polynomials and the little $q$-Laguerre
polynomials. We will use the technique of ladder operators for $q$-orthogonal
polynomials which was introduced by Chen and Ismail in \cite{ChenIsmail}. They consider monic polynomials orthogonal with
respect to an absolutely continuous measure, defined through a
weight function $w$, on the positive real axis. The potential $u$ is
defined as
\begin{equation}\label{potential}
u(x) = -\frac{D_{q^{-1}}w(x)}{w(x)}
\end{equation}
where $D_q$ is the $q$-analogue of the difference operator,
$$
(D_qf)(x) = \begin{cases}
   \displaystyle \frac{f(x)-f(qx)}{x(1-q)}  &\textrm{if $x \neq 0$},\\
   f'(0)                      & \textrm{if } x=0.
             \end{cases}
$$
The main result involves two entities
\begin{equation}\label{Aqhalfline}
A_n(x)  =  \gamma_n^2 \int_0^\infty \frac{u(qx)-u(y)}{qx-y}
P_n(y)P_n(y/q)w(y)\, dy
\end{equation}
\begin{equation}\label{Bqhalfline}
B_n(x) = \gamma_{n-1}^2 \int_0^\infty
\frac{u(qx)-u(y)}{qx-y}P_n(y)P_{n-1}(y/q)w(y)\, dy ,
\end{equation}
where
\[   \frac{1}{\gamma_n^2} = \int_0^\infty P_n^2(x)w(x)\, dx, \]
which appear in the lowering relation \cite[Theorem 1.1]{ChenIsmail}
$$
D_qP_n(x) = a_n^2 A_nP_{n-1}(x) - B_nP_n(x).
$$
We can make use of the following identities \cite[Eqs. (1.15)--(1.16)]{ChenIsmail}
\begin{equation}
\int_0^\infty u(y) P_n(y)P_n(y/q)w(y)\, dy = 0
\end{equation}
and
\begin{equation}
\gamma_n^2 \int_0^\infty u(y)P_{n+1}(y)P_n(y/q)w(y)\, dy =
\frac{1-q^{n+1}}{1-q}q.
\end{equation}
Furthermore, the quantities $A_n$ and $B_n$ satisfy two compatibility relations
\begin{equation}\label{rel1}
B_{n+1}(x) +B_n(x) = (x-b_n)A_n(x) + x(q-1)\sum_{j=0}^nA_j(x)-
u(qx),
\end{equation}
\begin{equation}\label{rel2}
1+(x-b_n)B_{n+1}(x) - (qx-b_n)B_n(x) =
a_{n+1}^2A_{n+1}(x) - a_n^2A_{n-1}(x).
\end{equation}
These equations will enable us to find expressions for the
recurrence coefficients $b_n, a_n^2$.

Ismail \cite{Ismail} shows that the ladder operators also work for orthogonal polynomials on the exponential lattice $\{q^n\, | \, n =0,1,2,\ldots\}$,
in particular for the $q$-Jackson integral \eqref{q-Jackson}. He shows that the same formulas hold for the expressions
\begin{equation}\label{Aq^n}
A_n(x)  =  \gamma_n^2 \int_0^1 \frac{u(qx)-u(y)}{qx-y}
P_n(y)P_n(y/q)w(y)\, d_qy
\end{equation}
and
\begin{equation}\label{Bq^n}
B_n(x) = \gamma_{n-1}^2 \int_0^1
\frac{u(qx)-u(y)}{qx-y}P_n(y)P_{n-1}(y/q)w(y)\, d_qy .
\end{equation}

\section{Semiclassical Stieltjes-Wigert weight} \label{section:Stieltjes-Wigert}

In this section we will prove Theorem \ref{thm:1} using the compatibility relations for the ladder operators.
For the weight
\[   w(x)= \frac{x^\alpha}{(-x^2;q^2)_\infty (-q^2/x^2;q^2)_\infty}, \qquad x \in [0,\infty), \]
we have
\begin{equation} \label{Pearson1}
   w(x/q)=q^{2-\alpha} w(x)/x^2,
\end{equation}
and hence the potential, defined in \eqref{potential}, is given by
\[   u(x) = \frac{q}{1-q} \left( \frac{1}{x} - \frac{q^{2-\alpha}}{x^2} \right), \]
so that
\[  \frac{u(qx)-u(y)}{qx-y} = -\frac{u(y)}{qx} + \frac{q^{1-\alpha}}{(1-q)x^2y^2} + \frac{q^{-\alpha}}{(1-q)x^3y} . \]
The formula \eqref{Aqhalfline} then becomes
\[   A_n(x) = \frac{T_n}{(1-q)x^3} + \frac{\gamma_n^2 q^{1-\alpha}}{(1-q)x^2} \int_0^\infty P_n(y)P_n(y/q) \frac{w(y)}{y^2}\, dy, \]
with
\[  T_n = \gamma_n^2 q^{-\alpha} \int_0^\infty P_n(y)P_n(y/q) \frac{w(y)}{y}\, dy , \]
The relation \eqref{Pearson1} gives
\begin{align*}
   \int_0^\infty P_n(y)P_n(y/q) \frac{w(y)}{y^2}\, dy &= q^{\alpha-2} \int_0^\infty P_n(y)P_n(y/q)w(y/q)\, dy  \\
             &=  q^{\alpha-1} \int_0^\infty P_n(qx)P_n(x)w(x)\, dx \\
             &= \frac{q^{n+\alpha-1}}{\gamma_n^2},
\end{align*}
where we have used $P_n(qx) = q^nP_n(x) + \cdots$.  Hence we have that $A_n$ is the rational function
\begin{equation}  \label{Aq1}
   A_n(x) = \frac{T_n}{(1-q)x^3} + \frac{q^n}{(1-q)x^2}.
\end{equation}
If we write $P_n(x) = x^n + \delta_n x^{n-1} + \cdots$, then \eqref{Pearson1} and a calculation similar as before give
\begin{equation}  \label{T}
   T_n = q^{n-1}(\delta_n - q \delta_{n+1}).
\end{equation}
 From this we easily find
\[   \sum_{j=0}^n T_j = -q^n\delta_{n+1}.  \]
The formula \eqref{Bqhalfline} gives the rational function
\begin{equation}  \label{Bq1}
   B_n(x) = - \frac{1-q^n}{(1-q)x} + \frac{r_n}{(1-q)x^2} + \frac{t_n}{(1-q)x^3},
\end{equation}
with
\begin{align*}
   r_n &= q^{1-\alpha} \gamma_{n-1}^2 \int_0^\infty P_n(y)P_{n-1}(y/q) \frac{w(y)}{y^2}\, dy, \\
   t_n &= q^{-\alpha} \gamma_{n-1}^2  \int_0^\infty P_n(y)P_{n-1}(y/q) \frac{w(y)}{y}\, dy,
\end{align*}
where $r_0=t_0=0$. Using $P_n(y)=y^n + \delta_ny^{n-1} + \cdots$ and \eqref{Pearson1} one finds
\begin{equation}  \label{r}
  r_n = q^{n-1}(1-q) \delta_n.
\end{equation}
Inserting \eqref{Aq1} and \eqref{Bq1} into the compatibility relation \eqref{rel1} and comparing powers of $x$ gives
\begin{subequations}
\begin{align}
   t_{n+1}+t_n &= -b_nT_n + q^{-\alpha},  \label{tT}  \\
   r_{n+1}+r_n &= -b_nq^n + T_n + (q-1) \sum_{j=0}^n T_j,  \label{rT}
\end{align}
\end{subequations}
and the second compatibility relation \eqref{rel2} gives
\begin{subequations}
\begin{align}
     -b_nt_{n+1}+b_n t_n &= a_{n+1}^2 T_{n+1} -a_n^2 T_{n-1}, \label{tTT} \\
     t_{n+1}-qt_n-b_nr_{n+1} + b_n r_n &= a_{n+1}^2q^{n+1} -a_n^2 q^{n-1},  \label{tr}  \\
      b_nq^n(1-q) + r_{n+1} -qr_n &= 0.   \label{rr}
\end{align}
\end{subequations}
The equations \eqref{rT} and \eqref{rr} give $b_n = \delta_n-\delta_{n+1}$, but this relation is well known and
follows by comparing the coefficients of $x^{n}$ in the three term recurrence relation \eqref{TTR}.
When multiplying \eqref{tTT} by $T_n$  and using \eqref{tT}, one finds
\[     (t_{n+1}+t_n-q^{-\alpha})(t_{n+1}-t_n) = a_{n+1}^2 T_{n+1}T_n - a_n^2 T_{n}T_{n-1}. \]
Summing over $n$ gives
\begin{equation}  \label{TT}
   a_n^2T_nT_{n-1} = t_n(t_n-q^{-\alpha}).
\end{equation}
 From \eqref{r} and \eqref{T} one has
\[   r_{n+1}-r_n = (1-q)(q^n\delta_{n+1} - q^{n-1}\delta_n) = -(1-q)T_n. \]
Inserting this in \eqref{tr} gives
\[  t_{n+1}-qt_n + (1-q)b_nT_n = a_{n+1}^2 q^{n+1} - a_n^2 q^{n-1} . \]
Use \eqref{tT} to obtain
\[   q^{-\alpha}(1-q) - t_n + qt_{n+1} = a_{n+1}^2 q^{n+1}-a_n^{2} q^{n-1}.  \]
Multiply this by $q^n$, then summing over $n$ gives
\begin{equation} \label{at}
  a_n^2 q^{2n-1}  = q^{-\alpha}(1-q^n) + q^n t_n,
\end{equation}
which expresses $t_n$ in terms of $a_n^2$. We use this to get rid of $t_n$ and $t_{n+1}$ in  \eqref{tTT}:
\[  a_{n+1}^2 (T_{n+1} + q^n b_n) - a_n^2 (T_{n-1}+q^{n-1} b_n)  = q^{-n-1-\alpha} b_n(1-q). \]
Replacing $b_n$ and $T_{n\pm 1}$ by their expressions involving $\delta_n$ gives
\[ b_n(1-q)q^{-\alpha} = q^{2n+1} a_{n+1}^2 (\delta_n - q\delta_{n+2}) - q^{2n-1} a_n^2 (\delta_{n-1}-q\delta_{n+1}). \]
Summing over $n$ gives
\begin{equation}  \label{ba}
 q^{2n+\alpha-1} a_n^2 (\delta_{n-1}-q\delta_{n+1}) = (1-q) \sum_{j=0}^{n-1} b_j = -(1-q) \delta_n.
\end{equation}
 From \eqref{rT} one finds
\[ q^nb_n = T_n - q^{n-1}(1-q)\delta_n, \]
and this gives for \eqref{ba}
\[  b_n q^n(1-q^{2n+\alpha-1}a_n^2) = T_n + q^{2n+\alpha} b_n T_{n-1}.  \]
We can use this expression in \eqref{tT} to find
\[   (t_{n+1}+t_n-q^{-\alpha})(1-q^{2n+\alpha-1}a_n^2) q^n = -T_n^2-q^{2n+\alpha} a_n^2 T_nT_{n-1}.  \]
The last term in this expression can be replaced using \eqref{TT}, and by using \eqref{at} one has
\[  T_n^2 = q^{2n+\alpha}(t_n-q^{-\alpha})(t_{n+1}-q^{-\alpha}).  \]
We can use this in \eqref{TT} and together with \eqref{at} we find an expression in $t_n$, $t_{n-1}$ and $t_{n+1}$ only
\[  (q^{-\alpha}(1-q^n)+q^nt_n)^2q^{2\alpha}(t_{n+1}-q^{-\alpha})(t_{n-1}-q^{-\alpha}) (t_n-q^{-\alpha})^2 = t_n(t_n-q^{-\alpha}), \]
where we can divide both sides by $(t_n-q^{-\alpha})^2$ to obtain
\[ t_n^2 = q^{2\alpha}\bigl(q^{-\alpha}(1-q^n)+q^nt_n\bigr)^2(t_{n+1}-q^{-\alpha})(t_{n-1}-q^{-\alpha}). \]
If we define $x_n = t_n-q^{-\alpha}$ then we get \eqref{qPIII-thm1} from Theorem \ref{thm:1}. This is $q$-$\textrm{P}_{\textrm{III}}$
\eqref{qPIII} with $a=b=q^{-\alpha}$ and $c=d=q^{\alpha}$. The recurrence coefficients are expressed in terms of $x_n$ in the following way:
\begin{align*}
    q^{2n-1}a_n^2 &= q^nx_n+q^{-\alpha}, \qquad n \geq 0 \\
    q^{2n+\alpha} b_n^2 x_n &= x_{n+1} + q^{2n+2\alpha}x_{n-1}(x_n+q^{-n-\alpha})^2 + 2(x_n+q^{-\alpha}), \qquad n \geq 1,
\end{align*}
where $x_0=-q^{-\alpha}$ and $x_1= b_0^2$ and $b_n$ is to be taken positive.   This proves Theorem \ref{thm:1}.

\section{Semiclassical $q$-Laguerre weight} \label{section:q-Laguerre}
In this section we will look at a semiclassical variation on the
weight functions \eqref{askey-weight} and \eqref{chihara-weight}, namely
$$
w(x)= \frac{x^\alpha(-p/x^2;q)_\infty}{(-x^2;q^2)_\infty(-q^2/x^2;q^2)_\infty},
\qquad x\in [0,\infty), p\in [0,q^{-\alpha}), \alpha \geq 0.
$$
The case $p=q^2$ is a semiclassical generalization of the
$q$-Laguerre weight $x^\alpha /(-x;q)_\infty$ described in \S \ref{sec:1.2} and \cite{BoelenVA}.

We denote the sequence of monic orthogonal polynomials by
$(P_n)_{n\geq 0}$. When we compute $w(x/q)$ then we find
\begin{equation}\label{breuk}
\frac{w(x)}{p+x^2}= \frac{w(x/q)}{q^{2-\alpha}}.
\end{equation}
The potential for this weight $w$ is
$$
u(x) = \frac{q}{1-q}\left(\frac{1}{x}-\frac{q^{2-\alpha}}{x(p+x^2)}
\right),
$$
from which we find
$$
\frac{u(qx)-u(y)}{qx-y} =
-\frac{1}{qx}u(y)+\frac{q^{3-\alpha}}{1-q}\frac{1}{(p+q^2x^2)(p+y^2)}+\frac{q^{2-\alpha}}{1-q}\frac{y}{x(p+q^2x^2)(p+y^2)}.
$$
We now look for expressions for the functions $A_n$ and $B_n$ of the lowering relation. For $A_n$ we have
\begin{equation}
A_n(x) =
\frac{q^2}{1-q}\frac{T_n}{x(p+q^2x^2)}+\frac{q^{n+2}}{1-q}\frac{1}{p+q^2x^2},
\end{equation}
with
$$
T_n =  q^{-\alpha} \gamma_n^2 \int_0^\infty
yP_n(y)P_n(y/q)\frac{w(y)}{p+y^2}\, dy ,
$$
while
\begin{equation}
B_n(x) =
-\frac{1-q^n}{1-q}\frac{1}{x}+\frac{q^2}{1-q}\frac{r_n}{p+q^2x^2}+\frac{q^2}{1-q}\frac{t_n}{x(p+x^2)},
\end{equation}
with
$$
r_n = q^{1-\alpha} \gamma_{n-1}^2 \int_0^\infty
P_n(y)P_{n-1}(y/q)\frac{w(y)}{p+y^2}\, dy
$$
and
$$
t_n = q^{-\alpha}\gamma_{n-1}^2 \int_0^\infty
yP_n(y)P_{n-1}(y/q)\frac{w(y)}{p+y^2}\, dy.
$$
It is clear that $r_0=t_0=0$ as $P_{-1}=0$. Bearing in mind that we
write the polynomials $P_n$ as
$$
P_n(x) = x^n + \delta_n x^{n-1} +  \cdots,
$$
it's easy to see that
\begin{equation}  \label{T}
T_n = q^{n-1}\left(\delta_n-q \delta_{n+1} \right)
\end{equation}
and
\begin{equation}\label{r}
r_n = q^{n-1}(1-q)\delta_n.
\end{equation}
We used \eqref{breuk} in these computations and the substitution
$u=y/q$ in the integrals and we wrote $uP_n(qu)$ and $P_n(qu)$ in
their Fourier expansion to obtain these results. Using the newly found
expression for $T_n$, we can easily compute the following sum which
will appear in the compatibility relations:
$$
\sum_{j=0}^n T_j = -q^n\delta_{n+1}.
$$
Multiplying the first compatibility relation \eqref{rel1} by
$q^{-2}(1-q)x(p+x^2$) and comparing coefficients of powers of $x$,
we get two relations
\begin{subequations}
\begin{align}\label{10}
(t_{n+1}+t_n)+pq^{-2}(-1+q^n+q^{n+1}) &= -b_n T_n +q^{-\alpha}, \\
r_{n+1}+r_n &= -b_nq^n+T_n + (q-1)\sum_{j=0}^nT_j.  \label{11}
\end{align}
\end{subequations}

From the second compatibility relation \eqref{rel2} we get in the
same way the following three relations:
\begin{subequations}
\begin{align}\label{20}
-b_n(t_{n+1}-t_n)+b_nq^n(1-q)pq^{-2} &=
a_{n+1}^2T_{n+1}-a_n^2T_{n-1}, \\
 \label{21}
t_{n+1}-qt_n -b_nr_{n+1}+b_nr_n &=
a_{n+1}^2q^{n+1}-a_n^2q^{n-1}, \\
 \label{22}
b_n(1-q^{n+1})+r_{n+1}-b_n(1-q^n)-qr_n &=0.
\end{align}
\end{subequations}
The equations \eqref{11} and \eqref{22} are trivial in the light of
\eqref{T} and \eqref{r} as
$$
b_n = \delta_n-\delta_{n+1}.
$$
We elaborate on \eqref{21} by noticing that
\begin{align*}
r_{n+1}-r_n &= (1-q)\left( q^n\delta_{n+1}-q^{n-1}\delta_n \right)\\
     &=  -(1-q)T_n.
\end{align*}
Inserting this in \eqref{21} we obtain
$$
t_{n+1}-qt_n + (1-q)b_nT_n = a_{n+1}^2q^{n+1}-a_n^2q^{n-1}.
$$
We use \eqref{10} to eliminate $b_nT_n$ and we obtain
$$
q^{-\alpha}(1-q) - t_n +qt_{n+1}+(1-q)pq^{-2}(1-q^n-q^{n+1}) =
a_{n+1}^2q^{n+1}-a_n^2q^{n-1}.
$$
This equation has the integrating factor $q^n$:
$$
q^{n-\alpha}(1-q)+q^{n+1}t_{n+1}-q^t_n
+(1-q)pq^{-2}(q^n-q^{2n}(1+q))= a_{n+1}^2q^{2n+1}-a_n^2q^{2n-1}.
$$
Summing over $n$ (a telescopic sum) we get
\begin{equation}\label{21'}
a_n^2q^{2n-1}=\left(q^{-\alpha}+pq^{-2}\right)(1-q^n)-pq^{-2}(1-q^{2n})+q^nt_n,
\end{equation}
which expresses $t_n$ in terms of $a_n^2$.

The right hand side of \eqref{20} has the integrating factor $T_n$. When
multiplying this identity by $T_n$, the factor $b_nT_n$
appears twice on the left hand side. We use \eqref{10} to eliminate
this factor when it accompanies the term $t_{n+1}-t_n$ and \eqref{21}
when it accompanies the term $pq^n$. Then we get
\begin{align*}
a_{n+1}^2T_{n+1}T_n -a_n^2T_nT_{n-1} &=
\Bigl(t_{n+1}+t_n-q^{-\alpha}+p(q^{-2}-1+q^n+q^{n+1})\Bigr) (t_{n+1}-t_n)\\
 & \quad  +\ pq^{-2}q^n\Bigl( a_{n+1}^2q^{n+1}-a_n^2q^{n-1}-t_{n+1}+qt_n \Bigr).
\end{align*}
Using \eqref{21'} in the factor of $pq^{-2+n}$ we get
\begin{align*}
a_{n+1}^2T_{n+1}T_n -a_n^2T_nt_{n-1} &= t_{n+1}^2-t_n^2 -
(t_{n+1}-t_n)(q^{-\alpha}+pq^{-2})+ 2pq^{-2}q^{n+1}t_{n+1} \\
 & \quad  -\ 2pq^{-2}q^nt_n + pq^{-2} (q^{-\alpha}+pq^{-2})(1-q)q^n-p^2q^{-4}
q^{2n}(1-q^2).
\end{align*}
Summing over $n$ (telescopic sum) and taking into account that
$t_0=a_0^2=0$, we get
\begin{align}\label{20'}
  a_n^2T_nT_{n-1} &= t_n(t_n-q^{-\alpha}-pq^{-2}+2pq^{-2}q^n)   \nonumber \\
& \quad   +\ pq^{-2}(q^{-\alpha}+pq^{-2})(1-q^n)-p^2q^{-4}(1-q^{2n}).
\end{align}
We  now use \eqref{21'} to eliminate $t_n$ and $t_{n+1}$ in \eqref{20}
and obtain
$$
a_{n+1}^2 ( T_{n+1} +q^n b_n ) -a_n^2 ( T_{n-1}+q^{n-1}b_n ) = q^{-n-1}b_n(1-q).
$$
Replacing $b_n$ and $T_n$ in the left hand side by their
expressions in $\delta_n$, we get
$$
b_n(1-q)q^{-\alpha} =
q^{2n+1}a_{n+1}^2(\delta_n-q\delta_{n+2})-q^{2n-1}a_n^2(\delta_{n-1}-q\delta_{n+1}).
$$
Taking a telescopic sum, we obtain
\begin{equation}\label{p_1}
q^{2n+\alpha-1}a_n^2\left(\delta_{n-1} -q\delta_{n+1} \right) =
(1-q)\sum_{j=0}^{n-1} b_j = -(1-q)\delta_n.
\end{equation}
We can now look at $b_n$:
$$
q^nb_n = T_n -q^{n-1}(1-q)\delta_n.
$$
Replacing $\delta_n$ using \eqref{p_1}, we get
\begin{align*}
q^n b_n &=  T_n + q^{3n+\alpha-2}a_n^2 ( \delta_{n-1}-q\delta_{n+1} )  \\
 &=  T_n + q^{3n+\alpha-1}a_n^2b_n+q^{2n+\alpha}a_n^2T_{n-1}.
\end{align*}
Collecting all terms with $b_n$ then gives
\begin{equation}\label{alpha}
b_nq^n ( 1-q^{2n+\alpha-1}a_n^2 )=T_n +q^{2n+\alpha}a_n^2 T_{n-1}  .
\end{equation}
We can use this expression in \eqref{10}
$$
\Bigl( t_{n+1}+t_n-q^{-\alpha}+pq^{-2}(-1+q^n+q^{n+1}) \Bigr) (1-q^{2n+\alpha-1}a_n^2)q^n
= -T_n^2-q^{2n+\alpha}a_n^2T_nT_{n-1}.
$$
The last term on the right hand side can be replaced using
\eqref{20'}. We then get an expression for $T_n^2$ in terms of $t_n$:
\begin{align*}    \label{beta}
T_n^2 &= -q^{2n +\alpha}\Big( t_n(q^{-\alpha}+pq^{-2}-pq^{-2}q^{n+1})+t_{n+1}(-tn+q^{-\alpha} +
pq^{-2}-pq^{-2}q^n)\nonumber\\
 &\quad   +\ pq^{-2}q^{n-\alpha+1}-p^2q^{-4}(1-q^n-q^{n+1}+q^{2n+1})
  -q^{-\alpha}(q^{-\alpha}+pq^{-2}-pq^{-2}q^n) \Big) .
\end{align*}
This expression can be simplified by using the substitution $y_n = t_n
-q^{-\alpha}-pq^{-2}+pq^{-2}q^n$, giving
\begin{equation}\label{T_n^2}
T_n^2 = q^{2n+\alpha}(y_ny_{n+1}-pq^{-2}q^{-\alpha}).
\end{equation}
We can use this expression in \eqref{20'}, which after squaring becomes
$$
 a_n^4T_n^2 T_{n-1}^2 = \left( y_n +q^{-\alpha} \right)^2 \left(
y_n +pq^{-2} \right)^2 ,
$$
and together with \eqref{21'} we are left with an expression in
$y_{n-1},y_n$ and $y_{n+1}$ only:
$$
(y_ny_{n-1}-pq^{-2}q^{-\alpha})(y_ny_{n+1}-pq^{-2}q^{-\alpha}) =
\frac{(y_n+q^{-\alpha})^2 (y_n+pq^{-2})^2}{(q^{n+\alpha}y_n+1)^2}.
$$
If we now use the substitution $z_n = y_n/\sqrt{pq^{-2-\alpha}}$
(assuming $p \neq 0$), we get
$$
(z_nz_{n-1} -1)(z_nz_{n+1}-1)=
\frac{\left(z_n+\sqrt{q^{2-\alpha}/p}\right)^2\left(z_n+\sqrt{pq^{\alpha-2}}\right)^2}
{\left(q^{n+\alpha/2-1}\sqrt{p}z_n+1\right)^2},
$$
which is an instance of $q$-P$_{\rm V}$  \eqref{qPV}
with $a=b=c=d=-\sqrt{q^{2-\alpha}/p}$ but with $q$ replaced by $1/q$.
The initial conditions are
$$
z_0 =-\sqrt{q^{2-\alpha}/p}, \hspace{3mm} z_1 =
\frac{\mu_2\mu_0-\mu_1^2-\mu_0^2q^{-\alpha-1}}{\mu_0^2\sqrt{pq^{-\alpha-2}}},
$$
where $\mu_0$ and $\mu_1$ are the first two moments of the weight $w$. Using
\eqref{21'}, \eqref{alpha} and \eqref{beta}, we can express the
recurrence coefficients $b_n, a_n^2$ in terms of $z_n$:
\begin{multline*}
b_n^2q^{2n+\alpha}z_n^2  =
z_nz_{n+1}-1+q^{2n+2\alpha}\left(\sqrt{pq^{-2-\alpha}}z_n+q^{-n-\alpha}\right)^2\left(z_nz_{n-1}-1\right)\\
+\ 2\left(z_n+\sqrt{q^{2-\alpha}/p}\right)\left(z_n+\sqrt{pq^{\alpha-2}}\right),
\end{multline*}
and
$$
a_n^2 =q^{-n+1}z_n\sqrt{pq^{-2-\alpha}}+q^{-2n-\alpha+1}.
$$
This concludes the proof of Theorem \ref{thm:2}.

\section{Semiclassical little $q$-Laguerre polynomials} \label{section:little-q-Laguerre}

We consider the weight
\[   w(x) = x^\alpha (q^2x^2;q^2)_\infty, \qquad \alpha > 0, \]
on the exponential lattice $\{q^n\, | \, n=0,1,2,\ldots\}$ and consider monic orthogonal polynomials satisfying
\[  \int_0^1 P_n(x)P_m(x) w(x)\, d_qx = \delta_{n,m}/\gamma_n^2, \]
for the $q$-Jackson integral \eqref{q-Jackson}. The potential $u$ satisfies
\[    u(qx) = - \frac{D_qw(x)}{w(qx)} = \frac{1}{1-q} \left( \frac{1-q^{-\alpha}}{x} + q^{-\alpha+2}x \right), \]
which leads to
\[   \frac{u(qx)-u(y)}{qx-y} = \frac{1}{1-q} \left( q^{-\alpha+1} - \frac{1-q^{-\alpha}}{xy} \right). \]
Observe that $w(0)=w(1/q)=0$ which implies that all the boundary terms in the expressions for $A_n$ and $B_n$ in \cite{Ismail} vanish.
The function $A_n$ in \eqref{Aq^n} is
\begin{equation*}
 A_n(x) = \gamma_n^2 \frac{q^{-\alpha}-1}{(1-q)x} \int_0^1 P_n(y)P_n(y/q) \frac{w(y)}{y}\, d_qy
    + \gamma_n^2 \frac{q^{1-\alpha}}{1-q} \int_0^1 P_n(y)P_n(y/q) w(y)\, d_qy .
\end{equation*}
It is easy to see that the last integral is $q^{-n}/\gamma_n^2$. We then have
\begin{equation}  \label{Aq3}
   A_n(x) = \frac{R_n}{(1-q)x} + \frac{q^{-n-\alpha+1}}{1-q},
\end{equation}
where
\[  R_n = \gamma_n^2 (q^{-\alpha}-1) \int_0^1 P_n(y)P_n(y/q) \frac{w(y)}{y}\, d_qy.  \]
In a similar we find that $B_n$ in \eqref{Bq^n} is
\begin{equation}  \label{Bq3}
   B_n(x) = \frac{r_n}{(1-q)x},
\end{equation}
with
\[  r_n = \gamma_{n-1}^2 (q^{-\alpha}-1) \int_0^1 P_n(y)P_{n-1}(y/q) \frac{w(y)}{y}\, d_qy . \]
If we insert \eqref{Aq1}--\eqref{Bq1} into the compatibility relations \eqref{rel1}--\eqref{rel2} and equate the powers of $x$, then
\begin{subequations}
\begin{align}
    r_{n+1}+r_n &= - b_n R_n - (1-q^{-\alpha}), \label{rR} \\
    b_nq^{-n-\alpha+1} &= R_n + (q-1) \sum_{j=0}^n R_j, \label{bR}
\end{align}
\end{subequations}
and
\begin{subequations}
\begin{align}
   a_{n+1}^2 R_{n+1} - a_n^2 R_{n-1}  &= -b_n (r_{n+1}-r_n), \label{aRbr} \\
   q^{-n-\alpha} a_{n+1}^2 - q^{-n-\alpha+2}a_n^2 &= r_{n+1}-qr_n+1-q .   \label{ar}
\end{align}
\end{subequations}
We will use these four equations to find expressions for the recurrence coefficients $a_n^2,b_n$ of the orthogonal polynomials.
Multiplying \eqref{ar} by $q^{-n-2}$ and summing over $n$ gives
\begin{equation}  \label{a2r}
    a_n^2 = q^{n+\alpha-1} (r_n+1-q^n),
\end{equation}
with $a_0=r_0=0$. Multiplying \eqref{aRbr} by $R_n$, eliminating $-b_nR_n$ using \eqref{rR}, and summing over $n$ gives
\begin{equation}  \label{aRR}
   a_n^2R_nR_{n-1} = r_n(r_n+1-q^{-\alpha}).
\end{equation}
Next, we use \eqref{a2r} in \eqref{aRbr} and replace $b_n$ by using \eqref{bR} to find
\begin{multline*}
  (1-q^{-n-1})(q^{n+2}R_{n+1}+q^{n+1}R_n) - (1-q^{-n})(q^{n+1}R_n+q^nR_{n-1}) + R_n(1-q) \\
  = r_{n+1} \left(qR_{n+1} + R_n - (1-q) \sum_{j=0}^n R_j \right)
     - r_n \left( qR_n + R_{n-1} - (1-q) \sum_{j=0}^{n-1} R_j \right).
\end{multline*}
Summing over $n$ (telescoping sum) gives
\[  r_n \left( qR_n + R_{n-1} - (1-q) \sum_{j=0}^{n-1} R_j \right) = (1-q) \sum_{j=0}^{n-1} + (1-q^{-n})(q^{n+1}R_n+q^nR_{n-1}). \]
Now use \eqref{bR} to get
\begin{equation}  \label{bRR}
  b_n(1+r_n)q^{-\alpha-n+1} = q^{n+1}R_n - R_{n-1}(r_n+1-q^n).
\end{equation}
We use this expression for $b_n$ in \eqref{rR}
\[ q^{-\alpha-n+1} (1+r_n)(r_{n+1}+r_n+1-q^{-\alpha}) = -q^{n+1}R_n^2 + R_nR_{n-1}(r_n+1-q^n). \]
Combining \eqref{a2r} and \eqref{aRR}  we get
\begin{equation}  \label{R^2}
   R_n^2 = -q^{-2n-\alpha}\Bigl( (r_n+1)(r_{n+1}+1)-q^{-\alpha} \Bigr).
\end{equation}
We can use this expression for $R_n^2$ together with \eqref{a2r} in \eqref{aRR} to find
\[  r_n^2(r_n+1-q^{-\alpha})^2 = q^{-2n}(r_n+1-q^n)^2 \Bigl( (r_n+1)(r_{n+1}+1)-q^{-\alpha} \Bigr)
           \Bigl( (r_n+1)(r_{n-1}+1)-q^{-\alpha} \Bigr). \]
Now define $x_n = q^{\alpha/2}(r_n+1)$, then we find \eqref{qPV-3}, which is $q$-$\textrm{P}_{\textrm{V}}$ \eqref{qPV}
with $a=b=c=d=q^{\alpha/2}$. The initial values are $x_0=q^{\alpha/2}$ and $x_1 = q^{-\alpha/2} (1-b_0^2)$, where
$b_0=\mu_1/\mu_0$ is the ratio of the first two moments of $w$. Using \eqref{a2r} we have
\[   a_n^2 = q^{n+\alpha-1} (x_nq^{-\alpha/2}-q^n), \]
and $b_n$ can be found using \eqref{aRR}--\eqref{R^2} by
\[   b_n^2x_n^2q^{-2n-\alpha} = 1-x_nx_{n+1}-q^{-2n}(x_nx_{n-1}-1)(x_nq^{-\alpha/2}-q^n)^2
 -2(x_n-q^{\alpha/2})(x_n-q^{-\alpha/2}). \]
This proves Theorem \ref{thm:3}.

\section{Acknowledgements}
This research is part of the PhD dissertation \cite{Boelen} of Lies Boelen. We acknowledge support of the Belgian Interuniversity Attraction Poles Programme P7/18, FWO research grant G.0934.13, and KU Leuven research grant OT/12/073.

redpajama_set_name: RedPajamaArXiv


Adapted from Great Food Fast, a great little cookbook from the kitchens of Martha Stewart Living. Try to find asparagus spears that are straight, and of uniform width. Makes one tart that serves 4-6 people for lunch, with a side salad or bowl of soup.
Preheat oven to 400°F. On a floured surface, open the sheet of pastry into a square. Trim uneven edges. Place the pastry on a baking sheet. With a sharp knife score the dough one inch in from the edges on all four sides, to form a border. Using a fork, pierce the dough inside the markings at half-inch intervals. Bake until golden, about 12-13 minutes.
Remove the pastry from the oven, and sprinkle the cheese inside the border. Trim the bottoms of the asparagus spears to fit crosswise inside the tart shell; arrange in a single layer over the cheese. Brush with oil, and season with salt and pepper. Bake 18-20 minutes, or until spears are tender. Serve warm.


text:
Polish Journal of Public Health
2015 | 125 | 2 | 65-71
Rubella outbreak in Poland in 2013 – an analysis of surveillance data at the national and province level
Dorota Mrozek-Budzyn ,	Renata Majewska ,	Agnieszka Kiełtyka ,	Małgorzata Augustyniak
Introduction. Poland is a member of the WHO European Region where a complete eradication of measles and rubella is planned to be finished by 2015. Poland accounted for 99% of all reported rubella cases in 27 EU/EEA countries in 2013. It is a good time to evaluate whether the established Polish vaccination strategy was sufficient to reach the goal of rubella elimination in the near future. Aim. The aim of this study was to analyze the epidemiology of rubella in Poland when the disease outbreak took place in 2013, to determine the reasons of that situation and to find the solution for future rubella elimination strategies. Material and methods. To analyze the epidemiology of rubella in Poland during the disease outbreak in 2013 the authors used rubella surveillance data collected by the Provincial and National Notifiable Disease Reporting System in 2004-2013. The information at the provincial level derived from one of the 16 provinces (Malopolska). The data on MMR vaccination coverage in 2003-2012 derived from the National Surveillance System. The percentages of rubella cases and vaccine coverage between Poland with Malopolska province were compared. Results. The outbreak started in late 2012 and continued through 2013, when 38548 rubella cases (incidence rate 100.1/ 100 000) were notified. Geographically, rubella cases were reported from the entire country, with the highest incidence rate in Malopolska province (254.9/100 000). Only 5 cases from Malopolska and 120 in whole country were laboratory confirmed, the remaining 99.7% were reported solely on the basis of clinical signs. The vaccination coverage was not sufficient to protect the population against rubella outbreak in Poland, especially among adolescents and young adult males. Conclusions. The strengthening of routine immunization program and implementation of some additional vaccination campaigns in young adults as well as laboratory confirmation of all suspected cases are the challenges that will have to be met to eliminate rubella in Poland
rubella epidemiological situation Poland
Dorota Mrozek-Budzyn
Department of Epidemiology, Chair of Epidemiology and Preventive Medicine, Jagiellonian University Medical College, Krakow, Poland / The member of Polish National Verification Committee for Measles and Rubella Elimination, dorotamrozek@tlen.pl
Department of Epidemiology, Chair of Epidemiology and Preventive Medicine, Jagiellonian University Medical College, Krakow, Poland
Agnieszka Kiełtyka
Małgorzata Augustyniak
1. World Health Organization. Eliminating measles and rubella. Framework for the verification process in the WHO European Region. World Health Organization Regional Office for Europe, Copenhagen, Denmark; 2012. [http://www.euro.who.int/__data/assets/pdf_file/0005/156776/e96153-Eng-final-version.pdf]
2. Pan American Health Organization. Plan of Action of the documentation and verification of measles, rubella, and congenital rubella syndrome elimination in the region of the Americas. Washington DC: Pan American Health Organization; 2011.
3. Castillo-Solórzano C, Marsigli C, Bravo-Alcántara P, et al. Elimination of rubella and congenital rubella syndrome in the Americas. J Infect Dis. 2011;204:S571-S578.[WoS]
4. Pasławska A, Mrozek-Budzyn D. Is measles elimination possible in WHO European Region up to 2015? Przegl Epidemiol. 2013;67:451-4.
5. National Institute of Public Health - National Institute of Hygiene. Vaccinations in Poland in 2012. [http://www.pzh.gov.pl/oldpage/epimeld/2012/Sz_2012.pdf]
6. Mrozek-Budzyn D. The evolution of Polish Immunization Schedule during the last 10 years. Przegl Epidemiol. 2012;66:107-12.
7. European Commission. Commission decision of 19 March 2002 laying down case definitions for reporting communicable diseases to the Community network under Decision No 2119/98/EC of the European Parliament and of the Council. Official Journal of the European Union. Luxembourg: Publications Office of the European Union. Communities. 3.4.2002:L 86/44. [http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2002:086:0044:0062:EN:PDF]
8. National Institute of Public Health - National Institute of Hygiene. Infectious diseases and poisoning in Poland 2004-2013.[http://www. pzh.gov.pl/]
9. World Health Organization. Rubella vaccines: WHO position paper. WHO Weekly epidemiological report. 2011;86:301-16. [http://www.who.int/wer/2011/wer8629.pdf 7]
10. National Institute of Public Health - National Institute of Hygiene. Zachorowania na wybrane choroby zakaźne w Polsce od 1 stycznia do 15 sierpnia 2014 r. oraz w porównywalnym okresie 2013 r. Zakład Epidemiologii NIZP-PZH. [http://www.pzh.gov.pl/oldpage/epimeld/2014/INF_14_08A.pdf]
11. European Center for Disease Control. Surveillance report. Measles and rubella monitoring. Stockholm: ECDC; 2014.
12. Janta D, Stanescu A, Lupulescu E, et al. Ongoing rubella outbreak among adolescents in Salaj, Romania, September 2011-January 2012. Euro Surveill. 2012;17(7):1-4
13. Centers for Disease Control. Nationwide rubella epidemic - Japan, 2013. MMWR. 2013;62:457-62.
14. Hopkins DR. Disease eradication. N Engl J Med. 2013;368:54-63.
15. World Health Organization. Global measles and rubella strategic plan: 2012-2020. Switzerland: World Health Organization; 2012. [http:// www.measlesrubellainitiative.org/wp-content/uploads/2013/06/Measles-Rubella-Strategic-Plan.pdf].
16. World Health Organization. Eliminating measles and rubella. Framework for the verification process in the WHO European Region. Geneva: World Health Organization 2012. [http://www.euro.who.int/__data/assets/pdf_file/0005/156776/e96153-Eng-final-version.pdf].
17. Casiday R. Uncertainty, decision-making and trust: lessons from the MMR controversy. Community Pract. 2006;79:354-7.
18. Brown KF, Kroll JS, Hudson MJ, et al. Factors underlying parental decisions about combination childhood vaccinations including MMR: a systematic review. Vaccine. 2010;28:4235-48.[PubMed]
19. Usonis V, Anca I, André F, et al. Rubella revisited: where are we on the road to disease elimination in Central Europe? Vaccine. 2011;29:9141-7.[WoS]
20. Edmunds WJ, van de Heijden OG, Eerola M, Gay NJ. Modelling rubella in Eur Epidemiol Infect. 2000;125:617-34.
21. Vandermeulen C, Roelants M, Theeten H, et al. Vaccination coverage and sociodemographic determinants of measles-mumps-rubella vaccination in three different age groups. Eur J Pediatr. 2008;167:1161-8.
22. McGreevy D. Risks and benefits of the single versus the triple MMR vaccine: how can health professionals reassure parents? J R Soc Promot Health. 2005;125:84-6.
10.1515/pjph-2015-0026
bwmeta1.element.-psjd-doi-10_1515_pjph-2015-0026


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text:
Flygdrakar (Draco) är ett ödlesläkte i familjen agamer bland kräldjuren. De förekommer i Sydostasien.

Utseende och ekologi

Flygdrakar har en spenslig, smal kropp som är täckt med tegelliknande fjäll. De har en lång stjärt. Den främsta egenskapen är en halvmånformig, uttänjbar hud, som hålls utspänd av fem till sju starkt förlängda revbenspar och som spänns ut som segel och som används för att djuret ska kunna glidflyga. Tack vare denna kan flygdrakar, som uppehåller sig i trädens toppar, göra stora språng, ända till cirka 30 meters längd. Det sker bara i riktning uppifrån och nedåt.

De praktfulla färgerna, som utmärker flera arter, finns troligtvis på grund av mimikry. Deras föda består huvudsakligen av små insekter, bland annat myror.

Systematik
Flygdrakarnas systematik är inte helt klarlagd. De utgör en artrik grupp, kanske "representerande en anmärkningsvärd radiering". Uppfattningarna om antalet arter varierar mellan över 40 och cirka 25. Vissa problem med att gruppera de filippinska varianterna i volans-gruppen talar för att just dessa är resultaten av en mycket hastig radiering, där alltså många arter skulle ha uppstått under en mycket kort tidrymd.

Arter enligt Catalogue of Life i bokstavsordning:
 Draco affinis
 Draco biaro
 Draco bimaculatus
 Draco blanfordii
 Draco caerulhians
 Draco cornutus
 Draco cristatellus
 Draco cyanopterus
 Draco dussumieri
 Draco fimbriatus
 Draco guentheri
 Draco haematopogon
 Draco jareckii
 Draco lineatus
 Draco maculatus
 Draco maximus
 Draco melanopogon
 Draco mindanensis
 Draco norvillii
 Draco obscurus
 Draco ornatus
 Draco palawanensis
 Draco quadrasi
 Draco quinquefasciatus
 Draco taeniopterus

Källor

Noter

Agamer

redpajama_set_name: RedPajamaWikipedia


text:
Locamagic.biz Blogging Football Scores Things To Know Before You Buy
Football Scores Things To Know Before You Buy
August 17, 2020August 17, 2020 Howdy HHowdy H 0 Comments
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text:
Jazmín Delgado Flores
Sector 5: 19th Century Literature of the Standard Major
Junior Research Seminar Requirement of the Standard Major
Hailed as a masterpiece of American fiction, Herman Melville's Moby-Dick; or, The Whale (1851) is a genre-defying work that pulls epic, romantic, dramatic, scientific, and historiographic forms into its literary vortex. The cosmic scope and metaphysical complexity of this text have enthralled, and sometimes left stranded, many an intrepid reader. Members of this research seminar, however, will likely avert the latter fate by harnessing their collective interpretive skills in a semester-long study of this endlessly rewarding novel, one that has become a key touchstone for writers, artists, philosophers, and political thinkers alike. Guided by their own close-readings of Moby-Dick and selected contemporaneous texts, students will consider the historical and cultural contexts in which the novel was written--the rise of industrial capitalism, the national debate over slavery, U.S. expansionist efforts across the globe, and so on--as well as engage with the novel's own speculations about the nature of agency, freedom, and meaning-making. At once a rigorous and irreverent meditation on the relationship between literature and knowledge, Moby-Dick will serve both as inspiration and point of departure for students' own critical explorations in and beyond the major.
The Junior Research Seminar is designed to involve students in the kinds of research that the discipline of literary studies currently demands, including: working with primary sources and archival materials; reviewing the critical literature; using online databases of historical newspapers, periodicals, and other cultural materials; exploring relevant contexts in literary, linguistic, and cultural history; studying the etymological history and changing meanings of words; experimenting with new methods of computational analysis of texts; and other methodologies. The course typically involves a few main texts that are studied intensively from a variety of approaches. Research exercises throughout the semester will enable and culminate in a final project: either a scholarly essay of 10-15 pages or a creative project. In either case, the final project must emerge out of each student's intensive, independent research agenda.