text:
Home > Staff > STA-ADMIN > 4
Staff Scholarship
Outside Literary Studies: Black Criticism And The University
Andy Hines, Swarthmore CollegeFollow
This striking contribution to Black literary studies examines the practices of Black writers in the mid-twentieth century to revise our understanding of the institutionalization of literary studies in America. Andy Hines uncovers a vibrant history of interpretive resistance to university-based New Criticism by Black writers of the American left. These include well-known figures such as Langston Hughes and Lorraine Hansberry as well as still underappreciated writers like Melvin B. Tolson and Doxey Wilkerson. In their critical practice, these and other Black writers levied their critique from "outside" venues: behind the closed doors of the Senate Permanent Subcommittee on Investigations, in the classroom at a communist labor school under FBI surveillance, and in a host of journals. From these vantages, Black writers not only called out the racist assumptions of the New Criticism, but also defined Black literary and interpretive practices to support communist and other radical world-making efforts in the mid-twentieth century. Hines's book thus offers a number of urgent contributions to literary studies: it spotlights a canon of Black literary texts that belong to an important era of anti-racist struggle, and it fills in the pre-history of the rise of Black studies and of ongoing Black dissent against the neoliberal university.
Black Studies, Literature and Literary Criticism, American and Canadian Literature
Andy Hines. (2022). "Outside Literary Studies: Black Criticism And The University". Outside Literary Studies: Black Criticism And The University.
https://works.swarthmore.edu/sta-admin/4
Find in Tripod


PCSchool has now created an export format that meets the new ACARA (Australian Curriculum, Assessment and Reporting Authority) requirements.
Student & Family origin and background information has been entered and Family DETYA addresses have been verified as per the Government Reporting tech tip.
NOTE: At present 'Visa SubClass Number' and 'Arrival in Australia Date' is not stored in a standard field in the database so must be added manually to the csv file once the bulk of the data has been exported.
If there are any critical errors they can be seen in the Error Log and will need to be corrected before the file can be Completed / created.
Upon selecting 'Complete' the file creation to prompt the ACARA Key School Data screen to appear. You will need to ensure you enter the correct information for your school.
Report formats include the Full ACARA export or a smaller SBD ACARA export depending on your requirements.
NOTE AGAIN: At present 'Visa SubClass Number' and 'Arrival in Australia Date' is not stored in a standard field in the database so must be added manually to the csv file once the bulk of the data has been exported.
If you have any problems or questions please contact the PCSchool Help Desk (07 4939 5995 – or – helpdesk@pcschool.net) for assistance.
This entry was posted in File Imports / Exports, Reporting to Government and tagged ACARA, Classroom Management, Data Capture, pcschool, school administration software, School Administration System, student management Software, Student Management System by studentmanagementsoftware. Bookmark the permalink.


text:
Q: Postgresql's Schemas and Background Jobs I'm running a Rails 3.2 app on Phusion Passenger, Nginx, and Postgres 9.1.4. I've moved my multi-tenant app to Postgresql's schemas (in development), following the method described in this post:
Since this approach seemingly sets the schema search path globally (via handle_subdomain, see below) for each request, how does this affect background jobs (via Resque)?
class ApplicationController < ActionController::Base
  before_filter :handle_subdomain

  def handle_subdomain
    if @tenant = Tenant.find_by_subdomain(request.subdomain)
      PgTools.set_search_path @tenant.id
    else
      PgTools.restore_default_search_path
    end
  end
end

What happens if I then set the search path in a Resque background job that takes a decent amount of time (say, deleting a bunch of Paperclip images that are hosted on S3)?  Could this potentially interfere with a request to the app, which sets the path in handle_subdomain?
Should I patch ActiveRecord to hard code the schema search path?  E.G. select * from "1"."users" ("1" being the schema path)
Thank you for any help.

A: If ActiveRecord is just issuing "SET search_path = x" commands then that is setting the value per-session, so you should be fine.
In fact, if you set it inside an explicit transaction it will be rolled back at the end of the transaction if the transaction gets rolled back*.

*

*clarified following comment from araqnid

redpajama_set_name: RedPajamaStackExchange


Dashboard gives your business a branded website, iOS app, and Android app, e-commerce, chat, social media, and business management tools, all for one low monthly subscription fee. You control every aspect of your business from one central location. Updates show up immediately. Menu changes, appointment reschedules, inventory updates, all in realtime.
Dashboard is a turnkey subscription product that gives you a website, smartphone apps for iOS and Android, and a powerful set of tools that let you control everything that matters to your business: sales, support, scheduling, content, social media, and more.
What comes with the subscription fee?
You get a custom design for your website and apps, complete control over the content included in your website and apps, and an administrative tools for you and your staff that give you control over everything your customers and visitors see.
How much does Dashboard cost?
There are three monthly subscription levels: Small, Medium, and Large. Depending on your needs, you can get started for as low as $199 per month.
Give us a call at 1-800-504-0569 to schedule a live demo. We'll be happy to show you everything that's possible with Dashboard.
How long does it take for my website and apps to launch?
Depending on the features you need and your package level, usually between 1-3 weeks if design revisions stay within 3 rounds.
Right now! Just click Subscribe to get started.


Our social media division is proud to offer content creation & posting for businesses of all sizes who don't have time, the staff or the inclination to post on their own.
Our NYC based Social Media agency will write & delivery the perfect amount of social content, at the right time, for each platform.
Our dedicated team will research/audit your company and create custom content for promotion, which is sent out 5 days a week.
Need help picking the best solution for your business?
Once you sign up for a plan, you will be asked to add your profiles to our Management Platform.
Upon completion of adding the Social Profiles, you will be asked to fill out the business profiler, so we have all of the nitty-gritty information about your business.
To make a modification to your plan, contact us. Support is given with 1 business day.
You will be billed every month, the same date of initial sign-up.
At this moment, we only accept credit card.
Each month, we automatically renew via the billing details on file. If there's an issue with payment, we will notify you immediately. If no action has been taken after THREE DAYS, we will cancel & suspend services.
All payments are process through Stripe, known to be among the most secure & safe payment protocols.


text:
From hobby to landing a spot on the U.S. Figure Skating Team
"Representing the Skating Club of Boston, please welcome now, Kevin Shum!" says the announcer at the 2018 U.S. Collegiate National Championship in Adrian, Michigan. Shum, wearing a sparkly dark-blue-and-black jumpsuit, begins his free skate performance to "The Sound of Silence" by Disturbed.
According to Shum, a senior majoring in computer science with a concentration in theater arts, life is about discovering and pursuing passions. Shum is a two-time collegiate champion figure skater and has traveled across the nation and around the world with Team USA's world junior team. Despite competitive figure skating's intense environment, the freedom of skating without restrictions or boundaries has sustained Shum's passion for the sport and has even contributed to his success at MIT.
Shum grew up in the San Francisco Bay Area and first got involved with figure skating at the age six. What started as a hobby quickly evolved into a passion. At age 10, Shum performed in his first skating competition. "I love the big arenas. I love the bright lights. I love the big audience," Shum says. "It's a pretty surreal feeling."
Shum is also passionate about his work at MIT. At MIT, Shum has participated in an Undergraduate Research Opportunity Program (UROP) through MIT's Media Lab. In addition, Shum is an active student blogger for the Office of Admissions, where he writes freely about his MIT experience, internships, projects, travels, and skating. He especially enjoyed writing posts about the time he spent in Germany while participating in MIT's International Science and Technology Initiatives Global Teaching Labs (MISTI GTL) during 2018's Independent Activities Period (IAP). While there, Shum taught computer science to high school students, and shortly after he traveled to Zurich, Switzerland, for the inaugural MIT EECS x ETH exchange program.
Recently, Shum worked on a project for the class Engineering Interactive Technologies (6.810). He and a partner applied concepts related to adaptive sports to the process of learning to ride a skateboard. Shum's skateboard detects a rider's skill level by monitoring how often the board wobbles, and it shrinks or expands in length to fit the rider's abilities.
Each semester, Shum trains at the Cronin Skating Rink in Revere, Massachusetts, up to six days a week for two hours a day. Training includes both on- and off-ice preparation: practicing jumps and footwork on the ice, body conditioning and strengthening at the gym, physical therapy, and a lot of stretching. During IAP, Shum will compete in the 2019 U.S. Figure Skating Championships in Detroit, Michigan, skating to Frank Sinatra's rendition of "Luck Be a Lady" from the musical Guys and Dolls.
His extraordinary commitment to skating has taught Shum to manage and prioritize his time working in the classroom, doing research, and completing projects; applying learning from one passion to another. "If there's something I want to accomplish, I know I just need to put in the work, put in those hours. Work smart, work hard, and get it done," he says, "I know I only have 24 hours in a day. It has really forced me to really be intentional about what I spend my time on."
Written by Kailey Tse-Harlow, Video by Stephanie Tran
Have a question about this article?
Contact the Division of Student Life's Communications Office
dsl-comm@mit.edu
MIT celebrates a year of excellence at the 2019 Awards Convocation
On May 13, over 300 members of the MIT community gathered in the Samberg Center (E52) to celebrate outstanding achievements by students, student groups, faculty, and community members at the annual MIT Awards Convocation.
Diversity, Inclusion, Undergraduate, Student Groups, Leadership, Public Service, PKG Center, Awards, Traditions
3Q: Chancellor Barnhart and Dean Nelson on Room-Assignment and Move-In Design Process
Senior leaders talk about concerns that prompted their collaboration with students and the student-generated ideas that will be implemented this fall to improve the room-assignment and move-in process for everyone.
Undergraduate, Housing

redpajama_set_name: RedPajamaCommonCrawl


Save up to 25% Off with these current icon meals coupon code, free iconmeals.com promo code and other discount voucher. There are 19 iconmeals.com coupons available in June 2018.
Founded in 2013 by a group of bodybuilders, ICON Meals, Inc. addresses the demand for freshly cooked, nutritious meals, especially by the health-conscious.
ICON MEALS Promo Codes for May 2018. Today: 37 Active ICON Meals Promo Codes, Free Shipping Deals & Site-Wide Coupons. Enter Code at Iconmeals.com — Save 20% Off Your Next Order at ICON Meals (Site-wide).
Icon Meals shares 12 coupon codes and promo codes. Get 30% off discount and save money online.
Choose ICON Meals for healthy meal delivery and customized meal plans today!
Get 16 Icon Meals coupon codes and promo codes at CouponBirds. Click to enjoy the latest deals and coupons of Icon Meals and save up to 20% when making purchase at checkout.
3/31/2016 · Don't forget to use the SHIBBOLETH coupon code at check out.
Up to 15% off Icon Meals at Icon Meals. 6 Icon Meals promo codes and coupons for June 2018. Offers end soon!
Save up to 25% with these current Iconmeals coupons for June 2018. The latest iconmeals.com coupon codes at CouponFollow.


Mystgrix I like this EP although I am a bit disappointed of Mall Grabs remix, as I would have expected a more individual take on Bob Turner, which is a track that in my opinion can really be messed around with a lot. Still, its probably personal taste. Favorite track: Bob Turner (Original Mix).
Printed cover and inner sleeve!
Includes unlimited streaming of 2 via the free Bandcamp app, plus high-quality download in MP3, FLAC and more.
Get all 28 AVA. RECORDS releases available on Bandcamp and save 40%.
Includes unlimited streaming via the free Bandcamp app, plus high-quality downloads of Unda Pressure, Kunststoff (Bonus), TSC 1, Hallo Welt, Recklinghausen EP, Kunststoff, In Case You Don't Know What To Play, I'll Be In Paris On Sunday If You're About?, and 20 more. , and , .
Damiano's 2nd EP on DVE - including a remix by Mall Grab. Distributed by Rubadub.


These are the EZ2 RESULT December 3, 2018 (Monday) updated after each draw today. The prizes and result summary for PCSO EZ2 can also be found below. NEXT: See the EZ2 result history here for the past 30 days as well as the latest Swertres Result as of today.
We appreciate any feedback or comment concerning these EZ2 lotto result for December 3, 2018. Our comment box can also be found at the bottom of this post. You can now access here the Swertres result December 3, 2018.
Note: Your EZ-2 number combination must be in exact order to win the prize Php 4,000.00.
9PM UPDATE: The PCSO lotto results December 3, 2018 are now ready for checking.
Please also read the updated PCSO payout chart for more information on other PCSO Prizes.
If you win any of the EZ2 prizes above, please proceed to this page to see how to claim your prize.
This post is about lotto result dec 3 2018, ez2 result dec 3 2018, ez2 result dec 4 2018 and much more!


Recently, I selected 21 of my most popular columns and edited them into a book called A Light Shines in Bury. I'm selling it to raise money for St Nicholas Hospice, here in Bury.
The people in my book are folk you may know, not because they're bishops, councillors, or mayors (though Patrick Chung, former Mayor of St Edmundsbury, is named).
They're shop owners, sanitation workers, waiters, buskers, men-and-women in the street. Basically your neighbours and mine.
I was delighted when this paper and St Nicholas Hospice got on board, giving the book publicity. Happily, many individuals and businesses agreed to sell copies, too.
Here's a big thanks to David Lynch of David's Barbers, George and Annette Stanford of the Really-Rather-Good coffee and tea shop, the Rev Robert Green of Garland Street Baptist Church, Robert Todd, businessman, Richard Summers, man- about-town (how else can you describe him? His CV is five pages long!) John Young, talented local musician, and last, but not least, Carla Carlisle, owner of Wyken Vineyard, and Steve and Joanna Knight, co-owners of Reallyverynice Art Gallery.
I saved them for last because in addition to selling the books, they very generously reached into their pockets and gave their own gifts of money, too. Merci!
While I'm at it, I don't know the individual personally, but when I collected the sales receipts and money from the Stanfords, they told me a customer came in and, while not buying the book, asked if she could give a gift towards supporting the charity.
The lady promptly chipped in 70p. I was so touched by this. Talk about the gift of the Widow's Mite (Luke 21:1-4)! I'm pleased to say, so far, we've raised over £400 and 70p.
"A Light Shines in Bury" is still available at St Nicholas Hospice reception at the West Suffolk Hospital, as well as at the Bury Free Press reception. Priced at £8, proceeds all go to the hospice. Hurry and buy a few for friends while supplies last. If you read this column regularly, it's no secret. Bury St Edmunds is like oxygen in my lungs, inspiring my work the way Paris, London or New York inspired many other writers. I even hatched my Bury-based novel The Chronicles of Johnny Pilgrim over coffee in town.
Oh, I'm not putting myself in the same league as Andrew Marr, Jeremy Clarkson or Jilly Cooper. Nevertheless, give me coffee, a dry spot in the sun, and a bran muffin, and the writing ideas come to me like fleas on a barnyard cat.
That said, I'd be remiss if I didn't admit how the waiters and waitresses who serve us help me out. The friendlier they are, the more material I find about which to write.
No kidding. A cup of joe and a grin is good science. Did you know the guys and gals who serve us our java are every bit as effective as GPs and psychiatrists when it comes to our mental and physical health?
According to many independent studies, drinking coffee in moderation helps prevent diabetes, Parkinson's disease, liver disease, cancer, and heart disease.
So where do the servers come in? Other studies show that being around smiling people combats sickness and depression, because friendly people produce oxytocin and vasopressin – two hormones in your body stimulating feelings of generosity, love and health.
Since this is so, our servers ought to earn as much as their colleagues in the NHS. As they don't, today, after you've had your friendly cup of cheer, dig deeply and give your server a big tip. Meantime, drink more coffee, smile more and prosper!


text:
\section{Introduction}
Let $F(x, y)$ be a binary form with real coefficients of degree $n \geq 3$ and non-zero discriminant $D_F$. The set
$$
\{(x, y) \in \mathbb R^2 \colon |F(x, y)| \leq 1\},
$$
which represents the collection of all points bounded by the curve $|F(x, y)| = 1$, is called the \emph{fundamental region} of $F$. It was proved by Mahler \cite{mahler33} that the area $A_F$ of the fundamental region of $F$ is finite.

In what follows, we restrict our attention to binary forms $F$ having integer coefficients. The quantity $A_F$, which can be evaluated via the formula \mbox{\cite[Section 3]{bean94}}
\begin{equation} \label{eq:AF-formula}
A_F = \int\limits_{-\infty}^{+\infty}\frac{dx}{|F(x, 1)|^{2/n}},
\end{equation}
plays a significant role in the analysis of certain Diophantine equations and inequalities associated with $F$. Mahler \cite{mahler33} proved that, for a positive \mbox{integer $h$}, the number of integer solutions $Z_F(h)$ to the \emph{Thue inequality} $|F(x, y)| \leq h$ satisfies
$$
\left|Z_F(h) - A_Fh^{2/n}\right| \ll_F h^{1/(n - 1)}.
$$
In 2019, Stewart and Xiao \cite{stewart-xiao} proved that the number of integers of absolute value at most $h$ which are represented by the form $F$ is asymptotic to $C_Fh^{2/n}$, where $C_F$ is a positive number which depends on $F$ and is a rational multiple of $A_F$.

The values $A_F$ for certain binary forms $F$ are intimately connected to the values of the \emph{beta function}
\begin{equation} \label{eq:beta-reg}
B(x, y) = \int\limits_0^{1}t^{x - 1}(1 - t)^{y - 1}dt,
\end{equation}
where $x$ and $y$ are complex numbers with positive real parts. An overview of important properties of $B(x, y)$ is given in Section \ref{sec:beta}. For a matrix $M = \left(\begin{smallmatrix}a & b\\c & d\end{smallmatrix}\right)$, with real coefficients, define
$$
F_M(x, y) = F(ax + by, cx + dy).
$$
Two forms $F$ and $G$ are said to be \emph{equivalent under} $\operatorname{GL}_2(\mathbb R)$ if $G = F_M$ for some $M \in \operatorname{GL}_2(\mathbb R)$. Analogously, we define the equivalence of forms under $\operatorname{GL}_2(\mathbb Z)$. In 1994, Bean \mbox{\cite[Corollary 1]{bean94}} proved that $A_F = 3B\left(\frac{1}{3}, \frac{1}{3}\right) = 15.8997\ldots$ when $F$ is a binary form that is equivalent under $\operatorname{GL}_2(\mathbb Z)$ to $xy(x - y)$. Furthermore, he proved that this value is the largest among all binary forms with integer coefficients, degree at least three and non-zero discriminant.

It was proved by Stewart and Xiao \cite[Corollary 1.3]{stewart-xiao} that if $a, b$ are fixed non-zero integers and $F_n(x, y) = ax^n + by^n$ is a binomial form of degree $n \geq 3$, then
\begin{equation} \label{eq:AFn}
A_{F_n} = \begin{cases}
\frac{1}{n|ab|^{1/n}}\left(2B\left(\frac{1}{n},1-\frac{2}{n}\right) +B\left(\frac{1}{n},\frac{1}{n}\right)\right) & \text{if $n$ is odd,}\\
\frac{2}{n|ab|^{1/n}}B\left(\frac{1}{n},\frac{1}{n}\right) & \text{if $ab$ is positive and $n$ is even,}\\
\frac{4}{n|ab|^{1/n}}B\left(\frac{1}{n},1-\frac{2}{n}\right) & \text{if $ab$ is negative and $n$ is even.}
\end{cases}
\end{equation}
In Section \ref{sec:binomial-form} we prove that $\lim\limits_{n \rightarrow \infty} A_{F_n} = 4$.

In 2019, Fouvry and Waldschmidt \cite[Th\'eor\`eme 1.5]{fouvry-waldschmidt} estimated $A_{\Phi_n}$ for cyclotomic binary forms $\Phi_n$ (note that a positive integer $n$ no longer refers to the degree of the form). It is a consequence of their result that for any $\varepsilon > 0$ there exists $n_0 = n_0(\varepsilon)$ such that for all $n \geq n_0$ the inequalities
$$
\left(2 - n^{-1 + \varepsilon}\right)^2 < A_{\Phi_n} < \left(2 + n^{-1 + \varepsilon}\right)^2
$$
are satisfied. Consequently, $\lim\limits_{n \rightarrow \infty} A_{\Phi_n} = 4$.

In this article we estimate $A_F$ for four families of binary forms. Let $\Psi_n(x)$ and $\Pi_n(x)$ denote the minimal polynomials of $2\cos\left(\frac{2\pi}{n}\right)$ and $2\sin\left(\frac{2\pi}{n}\right)$, respectively. The first two families that we are interested in are $\Psi_n(x, y)$ and $\Pi_n(x, y)$, which are homogenizations of $\Psi_n(x)$ and $\Pi_n(x)$, respectively. By \cite[Lemma]{watkins-zeitlin},
\begin{equation} \label{eq:Psin-formula}
\Psi_n(x, y) = \prod\limits_{\substack{1 \leq k < \frac{n}{2}\\\gcd(k, n) = 1}}\left(x - 2\cos\left(\frac{2\pi k}{n}\right)y\right).
\end{equation}
Further, since $\sin\left(\frac{2\pi}{n}\right) = \cos\left(\frac{2\pi(n - 4)}{4n}\right)$, it follows from (\ref{eq:Psin-formula}) that $\sin\left(\frac{2\pi}{n}\right)$ is an algebraic conjugate of $\cos\left(\frac{2\pi}{c(n)}\right)$, where $c(n)$ is the denominator of $\frac{n - 4}{4n}$ (in lowest terms). Consequently,
$$
\Pi_n(x, y) = \Psi_{c(n)}(x, y).
$$
See Figure \ref{fig:Psi13Pi9} for the graphs of $|\Psi_{13}(x, y)| = 1$ and $|\Pi_9(x, y)| = 1$. Note that in both cases the fundamental regions are \emph{not} compact.

\begin{figure}[t!]
\centering
\includegraphics[width=170pt]{psi13.png}
\includegraphics[width=170pt]{pi9.png}
\caption{Graphs of $|\Psi_{13}(x, y)| = 1$ (left) and $|\Pi_9(x, y)| = 1$ (right)}
\label{fig:Psi13Pi9}
\end{figure}

Next, let $T_n(x)$ and $U_n(x)$ denote Chebyshev polynomials of the first and second kinds, respectively. The other two families that we are interested in are $T_n(x, y)$ and $U_n(x, y)$, which are homogenizations of $T_n(x)$ and $U_n(x)$, respectively. It is known \cite{mason-handscomb} that
$$
T_n(x, y) = 2^{n - 1}\prod\limits_{k = 0}^{n-1}\left(x - \cos\left(\frac{(2k + 1)\pi}{2n}\right)y\right)
$$
and
$$
U_n(x, y) = 2^n\prod\limits_{k = 1}^n\left(x - \cos\left(\frac{k\pi}{n + 1}\right)y\right).
$$
See Figure \ref{fig:T6U6} for the graphs of $|T_6(x, y)| = 1$ and $|U_6(x, y)| = 1$. Note that in both cases the fundamental regions are \emph{not} compact.
It is also known that the binary forms $T_n(x, y)$ and $U_n(x, y)$ both have integer coefficients \cite{mason-handscomb}.

For a positive integer $n$, let $d(n)$ denote the number of its positive divisors. Let $\varphi(n)$ denote the Euler's totient function. Our results are summarized in Theorem \ref{thm:main3}, Corollary \ref{cor:main3}, Theorem \ref{thm:main1} and Theorem \ref{thm:main2}.

\newpage

\begin{thm} \label{thm:main3}
Let $n$ be a positive integer such that $\varphi(n) \geq 6$. Let $\Psi_n(x, y)$ denote the homogenization of the minimal polynomial of $2\cos\left(\frac{2\pi}{n}\right)$. Then
\small
\begin{align} \label{eq:APsin-bounds}
A_{\Psi_n} & > \frac{16}{3}\exp\left(-\frac{2d(n)\log n}{\varphi(n)}\right)\\
A_{\Psi_n} & < 2^{\frac{4}{\varphi(n)}}\exp\left(\frac{2d(n)^2\log n}{\varphi(n)}\right) \times\notag\\
& \times \left[\frac{16}{3} + 2^{1-\frac{4}{\varphi(n)}}B\left(\frac{1}{2},\frac{1}{2} - \frac{2}{\varphi(n)}\right) - 2\pi + \left(\frac{2}{n}B\left(\frac{1}{n}, 1 - \frac{4}{\varphi(n)}\right) - 2\right) - \left(\frac{2}{n}B\left(\frac{3}{n}, 1 - \frac{4}{\varphi(n)}\right) - \frac{2}{3}\right)\right].\notag
\end{align}
\normalsize
Furthermore, $\lim\limits_{n \rightarrow \infty} A_{\Psi_n} = \frac{16}{3}$.
\end{thm}

\begin{figure}[t!]
\centering
\includegraphics[width=170pt]{t6.png}
\includegraphics[width=170pt]{u6.png}
\caption{Graphs of $|T_6(x, y)| = 1$ (left) and $|U_6(x, y)| = 1$ (right)}
\label{fig:T6U6}
\end{figure}

\begin{cor} \label{cor:main3}
Let $n$ be a positive integer. Let $\Pi_n(x, y)$ denote the homogenization of the minimal polynomial of $2\sin\left(\frac{2\pi}{n}\right)$. Then $\Pi_n(x, y) = \Psi_{c(n)}(x, y)$, where
$$
c(n) =
\begin{cases}
4n & \text{if $n$ is odd,}\\
2n & \text{if $n \equiv 2 \pmod 4$,}\\
n & \text{if $n \equiv 0 \pmod 8$,}\\
n/2 & \text{if $n \equiv 12 \pmod{16}$,}\\
n/4 & \text{if $n \equiv 4 \pmod{16}$.}
\end{cases}
$$
Furthermore, $\lim\limits_{n \rightarrow \infty} A_{\Pi_n} = \frac{16}{3}$.
\end{cor}

\begin{thm} \label{thm:main1}
Let $n$ be a positive integer such that $n \geq 3$. Let $T_n(x, y)$ denote the homogenization of the Chebyshev polynomial of the first kind $T_n(x)$. Then
\begin{equation} \label{eq:ATn-bounds}
\frac{8}{3} < A_{T_n} < \frac{8}{3} + \frac{2}{3}\left(\sqrt[n]{4} - 1\right) + B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) - \pi
\end{equation}
Furthermore, $\lim\limits_{n \rightarrow \infty} A_{T_n} = \frac{8}{3}$.
\end{thm}

\begin{thm} \label{thm:main2}
Let $n$ be a positive integer such that $n \geq 3$. Let $U_n(x, y)$ denote the homogenization of the Chebyshev polynomial of the second kind $U_n(x)$. Then
\begin{align} \label{eq:AUn-bounds}
A_{U_n} & > \frac{8}{3} + \left(B\left(1 + \frac{1}{n}, \frac{1}{2}\right) - 2\right) + \frac{2}{3}\left((n + 1)^{-2/n} - 1\right)\\
A_{U_n} & < \frac{8}{3} + \left(B\left(1 + \frac{1}{n}, \frac{1}{2}\right) - 2\right) + B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) - \pi\notag
\end{align}
Furthermore, $\lim\limits_{n \rightarrow \infty} A_{U_n} = \frac{8}{3}$.
\end{thm}

Theorem \ref{thm:main3} follows from Lemma \ref{lem:psi-area}, which we establish in Section \ref{sec:Psi}. The proof of Corollary \ref{cor:main3} is outlined in Section \ref{sec:pi}. Theorem \ref{thm:main1} follows from Lemma \ref{lem:chebyshev1-area}, which we establish in Section \ref{sec:Tn}. Theorem \ref{thm:main2} follows from Lemma \ref{lem:chebyshev2-area}, which we establish in Section \ref{sec:Un}.

Our last result, which we present in Section \ref{sec:bean}, concerns the quantity
$$
Q(F) = |D_F|^{1/n(n - 1)}A_F
$$
associated to a binary form $F$ of degree $n \geq 3$ and non-zero discriminant $D_F$. It was demonstrated by Bean \cite{bean94} that the quantities $A_F$ and $Q(F)$ remain invariant for all binary forms that are equivalent under $\operatorname{GL}_2(\mathbb Z)$ to $F$. Unlike $A_F$, the quantity $Q(F)$ also remains invariant for all binary forms that, up to multiplication by a complex number, are equivalent under $\operatorname{GL}_2(\mathbb R)$ to $F$. An important conjecture about $Q(F)$ was formulated by Bean \mbox{\cite[Conjecture 1]{bean95}}. Let
$$
F_n^*(x, y) = \prod\limits_{k = 1}^n\left(\sin\left(\frac{k\pi}{n}\right)x - \cos\left(\frac{k\pi}{n}\right)y\right).
$$

\begin{conj} \label{conj:bean}
The maximum value $M_n$ of $Q(F)$ over the forms $F$ with complex coefficients of degree $n$ with discriminant $D_F \neq 0$ is attained precisely when $F$ is a form which, up to multiplication by a complex number, is equivalent under $\operatorname{GL}_2(\mathbb R)$ to the form $F_n^*$. That is, $M_n = Q(F_n^*)$. Moreover, the limit of the sequence $\{M_n\}$ is $2\pi$.
\end{conj}

It was proved by Bean and Laugesen \cite[Theorem 1]{bean-laugesen} that
\begin{equation} \label{eq:QFn}
Q(F_n^*) = 2^{1 - 2/n}n^{1/(n - 1)}B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right)
\end{equation}
and that $\lim\limits_{n \rightarrow \infty} Q(F_n^*) = 2\pi$. Let $\nu_2(n)$ denote the $2$-adic order of $n$. In \cite{mosunov} it was proved by the author that the binary form
$
S_n(x, y) = 2^{n - 1 - \nu_2(n)}F_n^*(x, y)
$
has integer coefficients and that the greatest common divisor of its coefficients is equal to one.\footnote{Note that in \cite[Proposition 1]{mosunov} where the coefficients of $F_n^*$ and $S_n$ are determined there should be no $Y$ in front of summation symbols.} See Figure \ref{fig:S6} for the graph of $|S_6(x, y)| = 1$. Just like the graph of $|F_n^*(x, y)| = 1$, the graph of $|S_n(x, y)| = 1$ is invariant under rotation by any integer multiple of $\frac{\pi}{n}$. In \cite{mosunov} it was also demonstrated that
$$
A_{S_n} = 4^{\nu_2(n)/n}B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right),
$$
and as a consequence that $\lim\limits_{n \rightarrow \infty}A_{S_n} = \pi$. Since $Q(F_n^*)$ remains invariant under multiplication by a non-zero complex number, we see that $Q(S_n) = Q(2^{n - 1 - \nu_2(n)}F_n^*) = Q(F_n^*)$. In Section \ref{sec:bean} we will prove that $Q(T_n) \leq Q(S_n)$ and $Q(U_n) \leq Q(S_n)$ for every integer $n \geq 3$. We will also prove that
$$
\lim\limits_{n \rightarrow \infty} Q(\Psi_n) = \lim\limits_{n \rightarrow \infty} Q(\Pi_n) = \lim\limits_{n \rightarrow \infty} Q(T_n) = \lim\limits_{n \rightarrow \infty} Q(U_n) = \frac{16}{3}.
$$

\begin{figure}[t!]
\centering
\includegraphics[width=200pt]{s6.png}
\caption{Graph of $|S_6(x, y)| = 1$}
\label{fig:S6}
\end{figure}

\begin{table}[b]
\centering
\begin{tabular}{c | c | c | c | c | c}
$n$ & $Q(\Psi_n)$ & $Q(\Pi_n)$ & $Q(S_n)$ & $Q(T_n)$ & $Q(U_n)$\\
\hline
$3$ & $\infty$ & $\infty$ & $15.8997$ & $15.8997$ & $15.8997$\\
$4$ & $\infty$ & $\infty$ & $11.7726$ & $11.4798$ & $11.5438$\\
$5$ & $\infty$ & $10.8953$ & $10.3228$ & $9.85622$ & $9.94897$\\
$6$ & $\infty$ & $\infty$ & $9.55526$ & $8.97702$ & $9.08227$\\
$7$ & $15.8997$ & $8.5577$ & $9.06881$ & $8.41412$ & $8.52393$\\
$8$ & $\infty$ & $\infty$ & $8.72772$ & $8.01814$ & $8.12848$\\
$9$ & $15.8997$ & $9.00056$ & $8.47265$ & $7.72218$ & $7.83097$
\end{tabular}
\caption{Values of $Q(\Psi_n)$, $Q(\Pi_n)$, $Q(S_n)$, $Q(T_n)$ and $Q(U_n)$ for $n = 3, 4, \ldots, 9$.}
\label{tab:table2}
\end{table}

\section{Beta Function and Its Properties} \label{sec:beta}

For a complex number $z$ with positive real part, let
\begin{equation} \label{eq:gamma}
\Gamma(z) = \int\limits_0^\infty t^{z - 1}e^{-t}dt
\end{equation}
denote the \emph{gamma function}. This function satisfies the functional equation
\begin{equation} \label{eq:functional-equation}
\Gamma(z + 1) = z\Gamma(z).
\end{equation}
The gamma function is analytic, with simple pole at $x = 0$. Its residue at $x = 0$ is equal to $1$, i.e.,
\begin{equation} \label{eq:gamma-residue}
\operatorname{Res}(\Gamma(x), 0) = \lim\limits_{x \rightarrow 0}x\Gamma(x) = 1.
\end{equation}

The beta function (\ref{eq:beta-reg}) can be expressed in terms of the gamma function by means of the relation
\begin{equation} \label{eq:beta-gamma}
B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}.
\end{equation}
If in (\ref{eq:beta-reg}) we make the change of variables $t = \sin^2 \theta$, then we obtain a trigonometric form of the beta function:
\begin{equation} \label{eq:beta}
B(x, y) = 2\int\limits_0^{\pi/2}(\sin \theta)^{2x - 1}(\cos\theta)^{2y - 1}d\theta.
\end{equation}
This representation of $B(x, y)$ is used in the derivation of the formula for $A_{F_n^*}$ \cite{mosunov}. In Lemmas \ref{lem:psi-area}, \ref{lem:chebyshev1-area} and \ref{lem:chebyshev2-area} we will use it to obtain upper and lower bounds on $A_{\Psi_n}$, $A_{T_n}$ and $A_{U_n}$.

Finally, we would like to remark that the values of beta function can be easily computed numerically, as this function is present in various mathematical software, such as Maple (\texttt{Beta(x, y)}), Mathematica  (\texttt{Beta(x, y)}) and MATLAB (\texttt{beta(Z, W)}).

\section{The Limit of $A_{F_n}$ for $F_n(x, y) = ax^n + by^n$} \label{sec:binomial-form}

In this section we prove the following result.

\begin{prop} \label{prop:binomial-forms}
Let $a$, $b$ and $n$ be non-zero integers with $n \geq 3$ and let $F_n(x, y) = ax^n + by^n$. Then $\lim\limits_{n \rightarrow \infty} A_{F_n} = 4$.
\end{prop}

\begin{proof}
In view of (\ref{eq:AFn}) and $\lim\limits_{n \rightarrow \infty}|ab|^{1/n} = 1$, the result will follow once we show that
$$
\lim\limits_{n \rightarrow \infty}\frac{1}{n}B\left(\frac{1}{n}, 1 - \frac{2}{n}\right) = 1 \quad \text{and} \quad
\lim\limits_{n \rightarrow \infty}\frac{1}{n}B\left(\frac{1}{n}, \frac{1}{n}\right) = 2.
$$
To see that this is the case, first note that the identities
$$
B\left(\frac{1}{n}, 1 - \frac{2}{n}\right) = \frac{\Gamma(1/n)\Gamma(1 - 2/n)}{\Gamma(1 - 1/n)} \quad \text{and} \quad B\left(\frac{1}{n}, \frac{1}{n}\right) = \frac{\Gamma(1/n)^2}{\Gamma(2/n)}
$$
follow directly from (\ref{eq:beta-gamma}). It follows from (\ref{eq:gamma-residue}) that
$$
\lim\limits_{n \rightarrow \infty} \frac{\Gamma(1/n)}{n} = 1 \quad \text{and} \quad \lim\limits_{n \rightarrow \infty} \frac{\Gamma(2/n)}{n} = \frac{1}{2}.
$$
Since $\Gamma(1) = 1$, we see that
\begin{align*}
\lim\limits_{n \rightarrow \infty}\frac{1}{n}B\left(\frac{1}{n}, 1 - \frac{2}{n}\right)
 = \lim\limits_{n \rightarrow \infty}\frac{\Gamma(1/n)\Gamma(1 - 2/n)}{n\Gamma(1 - 1/n)}
 = \lim\limits_{n \rightarrow \infty} \frac{\Gamma(1/n)}{n}
 = 1
\end{align*}
and
\begin{align*}
\lim\limits_{n \rightarrow \infty}\frac{1}{n}B\left(\frac{1}{n}, \frac{1}{n}\right)
 = \lim\limits_{n \rightarrow \infty}\frac{\Gamma(1/n)^2}{n\Gamma(2/n)}
 = \frac{\lim\limits_{n \rightarrow \infty}\left(\frac{\Gamma(1/n)}{n}\right)^2}{\lim\limits_{n \rightarrow \infty}\frac{\Gamma(2/n)}{n}}
 = 2.
\end{align*}
\end{proof}

\section{Upper and Lower Bounds on $A_{\Psi_n}$} \label{sec:Psi}

In this section we prove the following result.

\begin{lem} \label{lem:psi-area}
For any $n \in \mathbb N$ and $\alpha \in \mathbb R$ such that $\varphi(n) \geq 4$ and $2/\varphi(n) < \alpha < 1$,
\small
\begin{align*}
\int\limits_{-\infty}^{+\infty}\frac{dx}{|\Psi_n(x)|^\alpha} & > \left(4 + \frac{16}{\varphi(n)^2\alpha^2 - 4}\right)\exp\left(-\frac{d(n) \log n}{2}\alpha\right)\\
\int\limits_{-\infty}^{+\infty}\frac{dx}{|\Psi_n(x)|^\alpha} & < 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right), \times\\
& \times \left[4 + 2^{1-\alpha}B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - 2\pi + \frac{2}{n}B\left(\frac{\varphi(n)\alpha - 2}{2n}, 1 - \alpha\right) - \frac{2}{n}B\left(\frac{\varphi(n)\alpha + 2}{2n}, 1 - \alpha\right)\right].
\end{align*}
\normalsize
\end{lem}

Let us see why Theorem \ref{thm:main3} directly follows from Lemma \ref{lem:psi-area}.

\begin{proof}[Proof of Theorem \ref{thm:main3}]
It follows from (\ref{eq:Psin-formula}) that the degree of $\Psi_n$ is equal to $\varphi(n)/2$. Since the area $A_{\Psi_n}$ bounded by the curve $|\Psi_n(x, y)| = 1$ can be computed via (\ref{eq:AF-formula}), the inequalities (\ref{eq:APsin-bounds}) follow from Lemma \ref{lem:psi-area} once we take $\alpha = 4/\varphi(n)$.

The fact that $\lim\limits_{n \rightarrow \infty} A_{\Psi_n} = \frac{16}{3}$ is a consequence of the following:
$$
\lim\limits_{n \rightarrow \infty} B\left(\frac{1}{2}, \frac{1}{2} - \frac{2}{\varphi(n)}\right) = B\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{\Gamma(1/2)^2}{\Gamma(1)} = \pi,
$$
$$
\lim\limits_{n \rightarrow \infty} \frac{2}{n}B\left(\frac{1}{n}, 1 - \frac{4}{\varphi(n)}\right) = \lim\limits_{n \rightarrow \infty} \frac{2}{n}\Gamma\left(\frac{1}{n}\right) = 2,
$$
$$
\lim\limits_{n \rightarrow \infty} \frac{2}{n}B\left(\frac{3}{n}, 1 - \frac{4}{\varphi(n)}\right) = \lim\limits_{n \rightarrow \infty} \frac{2}{n}\Gamma\left(\frac{3}{n}\right) = \frac{2}{3}.
$$
Further, we have
$$
\lim\limits_{n \rightarrow \infty} \frac{2d(n)\log n}{\varphi(n)} = 0 \quad \text{and} \quad \lim\limits_{n \rightarrow \infty} \frac{2d(n)^2\log n}{\varphi(n)} = 0.
$$
These two limits follow from the fact that, for every integer $n > 2$,
\begin{equation} \label{eq:d}
d(n) \leq n^{\frac{1.067}{\log\log n}}
\end{equation}
and
\begin{equation} \label{eq:phi-lower-bound}
\varphi(n) > \frac{n}{e^\gamma \log\log n + \frac{3}{\log \log n}}.
\end{equation}
Here $\gamma \approx 0.5772$ denotes the Euler-Mascheroni constant. The upper bound on $d(n)$ was derived by Nicolas and Robin \cite{nicolas-robin}, while the lower bound on $\varphi(n)$ can be found in the paper of Rosser and Schoenfeld \cite[Theorem 15]{rosser-schoenfeld}.
\end{proof}

Before proving Lemma \ref{lem:psi-area}, we need to establish three lemmas.

\begin{lem} \label{lem:psi}
Let $n$ be an integer such that $n \geq 3$. Then, for every \mbox{real number $\theta$},
\begin{equation} \label{eq:psi-cos}
\Psi_n(\pm 2\cos \theta) = (\pm 1)^{\frac{\varphi(n)}{2}}e^{-\frac{\varphi(n)}{2}\theta i}\Phi_n\left(\pm e^{\theta i}\right)
\end{equation}
\begin{equation} \label{eq:psi-cosh}
\Psi_n(\pm 2\cosh \theta) = (\pm 1)^{\frac{\varphi(n)}{2}}e^{\frac{\varphi(n)}{2}\theta}\Phi_n\left(\pm e^{-\theta}\right).
\end{equation}
\end{lem}

\begin{proof}
According to Lehmer \cite{lehmer},
$$
\Psi_n(x + x^{-1}) = x^{-\varphi(n)/2}\Phi_n(x).
$$
Putting $x = e^{\theta i}$, we obtain the identity (\ref{eq:psi-cos}), where the plus signs are chosen everywhere. Replacing $\theta$ with $\pi + \theta$, $\theta i$ and $\pi + \theta i$ we obtain the identity (\ref{eq:psi-cos}) where the minus signs are chosen everywhere, as well as the two identities in (\ref{eq:psi-cosh}).
\end{proof}

For a polynomial $f(x)$, let $L(f)$ denote the sum of the absolute values of coefficients of $f(x)$. The following lemma can be found in \cite[Lemme 4.1]{fouvry-waldschmidt}.

\begin{lem} \label{lem:L-bound}
For every integer $n > 1$, $L(\Phi_n) \leq n^{\frac{d(n)}{2}}$.
\end{lem}

\begin{lem} \label{lem:prod-L-bound}
For every positive integer $n$,
$$
\prod\limits_{\substack{m \mid n\\m \neq 1}}L(\Phi_m) \leq \exp\left(\frac{d(n)^2\log n}{2}\right).
$$
\end{lem}

\begin{proof}
By Lemma \ref{lem:L-bound},
\begin{align*}
\prod\limits_{\substack{m \mid n\\m \neq 1}}L(\Phi_m)
& \leq \prod\limits_{m \mid n}m^{\frac{d(m)}{2}}\\
& = \exp\left(\frac{1}{2}\sum\limits_{m \mid n}d(m)\log m\right)\\
& \leq \exp\left(\frac{d(n)\log n}{2}\sum\limits_{m \mid n}1\right)\\
& = \exp\left(\frac{d(n)^2\log n}{2}\right).
\end{align*}
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:psi-area}]
Note that
\small
\begin{align*}
\int\limits_{-\infty}^{+\infty}\frac{dx}{|\Psi_n(x)|^{\alpha}}
& = \int\limits_{-2}^2\frac{dx}{|\Psi_n(x)|^{\alpha}} + \int\limits_{2}^{+\infty}\frac{dx}{|\Psi_n(x)|^{\alpha}} + \int\limits_{-\infty}^{-2}\frac{dx}{|\Psi_n(x)|^{\alpha}}\\
& = -\int\limits_{0}^\pi\frac{d(2\cos\theta)}{|\Psi_n(2\cos \theta)|^{\alpha}} + \int\limits_{0}^{+\infty}\frac{d(2\cosh\theta)}{|\Psi_n(2\cosh\theta)|^{\alpha}} + \int\limits_{-\infty}^{0}\frac{d(-2\cosh\theta)}{|\Psi_n(-2\cosh\theta)|^{\alpha}}\\
& = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Psi_n(2\cos \theta)|^{\alpha}} + 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{|\Psi_n(2\cosh\theta)|^{\alpha}} + 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{|\Psi_n(-2\cosh\theta)|^{\alpha}}\\
& = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Phi_n(e^{\theta i})|^{\alpha}} + 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(e^{-\theta})^{\alpha}} + 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(-e^{-\theta})^{\alpha}},
\end{align*}
\normalsize
where the last equality follows from Lemma \ref{lem:psi}. Let
\small
$$
C_1 = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Phi_n(e^{\theta i})|^{\alpha}}, \quad C_2 = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(e^{-\theta})^{\alpha}}, \quad C_3 = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(-e^{-\theta})^{\alpha}}.
$$
\normalsize
We will now derive upper and lower bounds on $C_1$, $C_2$ and $C_3$.

We begin with the derivation of a lower bound. By Lemma \ref{lem:L-bound},
$$
|\Phi_n(z)| \leq L(\Phi_n)\max\{1, |z|\}^{\varphi(n)} \leq n^{\frac{d(n)}{2}}
$$
for all $z \in \mathbb C$ such that $|z| \leq 1$. In view of this,
\begin{align*}
C_1
& = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Phi_n(e^{\theta i})|^{\alpha}}\\
& \geq 2n^{-\frac{d(n)}{2}\alpha}\int\limits_{0}^{\pi}\sin\theta d\theta\\
& = 4\exp\left(-\frac{d(n) \log n}{2}\alpha\right)
\end{align*}
and
\begin{align*}
C_2
& = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n\left(e^{-\theta}\right)^{\alpha}}\\
& \geq 2n^{-\frac{d(n)}{2}\alpha}\int\limits_0^{+\infty}e^{-\frac{\varphi(n)}{2}\alpha\theta}\sinh\theta d\theta\\
& = 2n^{-\frac{d(n)}{2}\alpha}\frac{1}{\left(\frac{\varphi(n)}{2}\alpha\right)^2 - 1}\\
& =  \frac{8}{\varphi(n)^2\alpha^2 - 4}\exp\left(-\frac{d(n) \log n}{2}\alpha\right).
\end{align*}
Similarly,
$$
C_3 \geq \frac{8}{\varphi(n)^2\alpha^2 - 4}\exp\left(-\frac{d(n) \log n}{2}\alpha\right).
$$
We conclude that
$$
\int\limits_{-\infty}^{+\infty}\frac{dx}{|\Psi_n(x)|^{\alpha}} = C_1 + C_2 + C_3 \geq \left(4 + \frac{16}{\varphi(n)^2\alpha^2 - 4}\right)\exp\left(-\frac{d(n) \log n}{2}\alpha\right).
$$

To derive an upper bound, we will make use of the identity
$$
z^n - 1 = \prod\limits_{m \mid n}\Phi_m(z).
$$
Notice that
$$
\Phi_n(z) = (z^n - 1) \prod\limits_{\substack{m \mid n\\ m \neq n}}\Phi_m(z)^{-1}.
$$
Thus for every $z \in \mathbb C$ such that $|z| < 1$ we have
\begin{align*}
|\Phi_n(z)|^{-1}
& = \frac{1}{|z^n - 1|}\prod\limits_{\substack{m \mid n\\ m \neq n}}|\Phi_m(z)|\\
& = \frac{|z - 1|}{|z^n - 1|}\prod\limits_{\substack{m \mid n\\ 1 < m < n}}|\Phi_m(z)|\\
& \leq \frac{|z - 1|}{|z^n - 1|}\prod\limits_{\substack{m \mid n\\m \neq 1}}L(\Phi_m)\\
& < \frac{2}{|z^n - 1|}\exp\left(\frac{d(n)^2\log n}{2}\right),
\end{align*}
where the last inequality follows from Lemma \ref{lem:prod-L-bound}. In view of this,
\begin{align*}
C_1
& = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Phi_n(e^{\theta i})|^{\alpha}}\\
& \leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\int\limits_0^{\pi}\frac{\sin\theta d\theta}{|e^{n\theta i} - 1|^{\alpha}}\\
& = 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(\int\limits_0^{\pi}\sin\theta d\theta + \int\limits_0^{\pi}\sin\theta\left(|e^{n\theta i} - 1|^{-\alpha} - 1\right) d\theta\right)\\
&\leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(2 + \int\limits_0^{\pi}\left(\frac{1}{|e^{n\theta i} - 1|^{\alpha}} - 1\right) d\theta\right)\\
& = 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(2 + \int\limits_0^{\pi}\frac{d\theta}{|e^{n\theta i} - 1|^{\alpha}} - \pi\right).
\end{align*}
It remains to estimate the integral $\int_0^{\pi}\frac{d\theta}{|e^{n\theta i} - 1|^{\alpha}}$. Note that
$$
|e^{n\theta i} - 1|^2 = \left(\cos(n\theta) - 1\right)^2 + \sin^2(n\theta) = 2 - 2\cos(n\theta) = 4\sin^2\left(\frac{n\theta}{2}\right).
$$
Therefore,
\begin{align*}
\int_0^{\pi}\frac{d\theta}{|e^{n\theta i} - 1|^{\alpha}}
& = 2^{-\alpha}\int\limits_0^{\pi}\left|\sin\left(\frac{n\theta}{2}\right)\right|^{-\alpha}d\theta\\
& = \frac{2^{1-\alpha}}{n}\int\limits_0^{n\pi/2}\left|\sin \theta\right|^{-\alpha}d\theta\\
& = 2^{-\alpha}\cdot 2\int\limits_0^{\pi/2}\left(\sin \theta\right)^{-\alpha}d\theta\\
& = 2^{-\alpha}B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right).
\end{align*}
We conclude that
\begin{align*}
C_1
& = 2\int\limits_{0}^{\pi}\frac{\sin\theta d\theta}{|\Phi_n(e^{\theta i})|^{\alpha}}\\
& \leq 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(4 + 2^{1-\alpha}B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - 2\pi\right).
\end{align*}

Next, we derive an upper bound on $C_2$:
\begin{align*}
C_2
& = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(e^{-\theta})^{\alpha}}\\
& \leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\int\limits_0^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}(1 - e^{-n\theta})^{\alpha}}\\
& = 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\frac{1}{2n}\int\limits_0^1\left(t^{\frac{\varphi(n)\alpha - 2}{2n} - 1} - t^{\frac{\varphi(n)\alpha + 2}{2n} - 1}\right)(1 - t)^{-\alpha} dt\\
& = 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(\frac{1}{n}B\left(\frac{\varphi(n)\alpha - 2}{2n}, 1 - \alpha\right) - \frac{1}{n}B\left(\frac{\varphi(n)\alpha + 2}{2n}, 1 - \alpha\right)\right),
\end{align*}
where the second-to-last equality follows from the change of variables $t = e^{-n\theta}$.

Finally, we derive an upper bound on $C_3$. We consider two cases.

\emph{Case 1.} Suppose that $n$ is even. Then
\begin{align*}
C_3 & = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(-e^{-\theta})^{\alpha}}\\
& \leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\int\limits_0^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\left(1 - (-e^{-\theta})^n\right)^{\alpha}}\\
 & = 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\int\limits_0^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\left(1 - e^{-n\theta}\right)^{\alpha}}\\
& = 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(\frac{1}{n}B\left(\frac{\varphi(n)\alpha - 2}{2n}, 1 - \alpha\right) - \frac{1}{n}B\left(\frac{\varphi(n)\alpha + 2}{2n}, 1 - \alpha\right)\right),
\end{align*}
where the last equality was derived from the estimate for $C_2$.

\emph{Case 2.} Suppose that $n$ is odd. Then for any real number $x$ we have $\Phi_n(-x) = \Phi_{2n}(x)$, so
\begin{align*}
C_3 & = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_n(-e^{-\theta})^{\alpha}}\\
& = 2\int\limits_{0}^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\Phi_{2n}(e^{-\theta})^{\alpha}}\\
& \leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right) \int\limits_0^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\left(1 - e^{-2n\theta}\right)^{\alpha}}\\
& \leq 2^{1 + \alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right) \int\limits_0^{+\infty}\frac{\sinh\theta d\theta}{e^{(\varphi(n)/2)\alpha\theta}\left(1 - e^{-n\theta}\right)^{\alpha}}\\
& = 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right)\left(\frac{1}{n}B\left(\frac{\varphi(n)\alpha - 2}{2n}, 1 - \alpha\right) - \frac{1}{n}B\left(\frac{\varphi(n)\alpha + 2}{2n}, 1 - \alpha\right)\right),
\end{align*}
where the last equality was derived from the estimate for $C_2$.

We conclude that
\small
\begin{align*}
\int\limits_{-\infty}^{+\infty}\frac{dx}{|\Psi_n(x)|^{\alpha}}
& = C_1 + C_2 + C_3\\
& \leq 2^{\alpha}\exp\left(\frac{d(n)^2\log n}{2}\alpha\right) \times\\
& \times \left(4 + 2^{1-\alpha}B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - 2\pi + \frac{2}{n}B\left(\frac{\varphi(n)\alpha - 2}{2n}, 1 - \alpha\right) - \frac{2}{n}B\left(\frac{\varphi(n)\alpha + 2}{2n}, 1 - \alpha\right)\right).
\end{align*}
\normalsize
\end{proof}

\section{Proof of Corollary \ref{cor:main3}} \label{sec:pi}

In view of the identity $\sin\left(\frac{2\pi}{n}\right) = \cos\left(\frac{2\pi(n - 4)}{4n}\right)$, we can follow the argument of Niven \cite[III.4]{niven} and consider the following five cases.
\begin{enumerate}[(1)]
\item If $n$ is odd, then the fraction $\frac{n - 4}{4n}$ is in lowest terms. Thus $2\cos\left(\frac{2\pi(n - 4)}{4n}\right)$ is conjugate to $2\cos\left(\frac{2\pi}{4n}\right)$ and $c(n) = 4n$.

\item If $n \equiv 2 \pmod 4$, then $\frac{n - 4}{4n}$ reduces to a fraction with denominator $2n$. Thus $2\cos\left(\frac{2\pi(n - 4)}{4n}\right)$ is conjugate to $2\cos\left(\frac{2\pi}{2n}\right)$ and $c(n) = 2n$.

\item If $n \equiv 0 \pmod 8$, then $\frac{n - 4}{4n}$ reduces to a fraction with denominator $n$. Thus $2\cos\left(\frac{2\pi(n - 4)}{4n}\right)$ is conjugate to $2\cos\left(\frac{2\pi}{n}\right)$ and $c(n) = n$.

\item If $n \equiv 12 \pmod{16}$, then $\frac{n-4}{4n}$ reduces to a fraction with denominator $n/2$. Thus $2\cos\left(\frac{2\pi(n - 4)}{4n}\right)$ is conjugate to $2\cos\left(\frac{2\pi}{n/2}\right)$ and $c(n) = n/2$.

\item If $n \equiv 4 \pmod{16}$, then $\frac{n - 4}{4n}$ reduces to a fraction with denominator $n/4$. Thus $2\cos\left(\frac{2\pi(n - 4)}{4n}\right)$ is conjugate to $2\cos\left(\frac{2\pi}{n/4}\right)$ and $c(n) = n/4$.
\end{enumerate}
A combination of this result with Theorem \ref{thm:main3} yields \mbox{$\lim\limits_{n \rightarrow \infty} A_{\Pi_n} = \lim\limits_{n \rightarrow \infty} A_{\Psi_n} = \frac{16}{3}$}.

\section{Upper and Lower Bounds on $A_{T_n}$} \label{sec:Tn}

In this section we prove the following result.

\begin{lem} \label{lem:chebyshev1-area}
For any $n \in \mathbb N$ and $\alpha \in \mathbb R$ such that $n \geq 3$ and $2/n \leq \alpha < 1$,
$$
2 + \frac{2}{n^2\alpha^2 - 1} < \int_{-\infty}^{+\infty}\frac{dx}{|T_n(x)|^\alpha} < 2 + \frac{2^{1 + \alpha}}{n^2\alpha^2 - 1} + B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi.
$$
\end{lem}

Let us see why Theorem \ref{thm:main1} directly follows from Lemma \ref{lem:chebyshev1-area}.

\begin{proof}[Proof of Theorem \ref{thm:main1}]
Since for an arbitrary binary form $F$ of positive degree $n$ the area bounded by the curve $|F(x, y)| = 1$ can be computed via (\ref{eq:AF-formula}), the inequalities (\ref{eq:ATn-bounds}) follow from Lemma \ref{lem:chebyshev1-area} if we take $\alpha = 2/n$.

To see that $\lim\limits_{n \rightarrow \infty} A_{T_n} = \frac{8}{3}$, it is sufficient to demonstrate that the limit of $B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right)$ as $n$ approaches infinity is equal to $\pi$. Recall that the gamma function (\ref{eq:gamma}) satisfies $\Gamma(1) = 1$ and $\Gamma(1/2) = \sqrt{\pi}$. In view of (\ref{eq:beta-gamma}), we see that
$$
\lim\limits_{n \rightarrow \infty}B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) = B\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{\Gamma(1/2)^2}{\Gamma(1)} = \pi.
$$
\end{proof}

We will now turn our attention to the proof of Lemma \ref{lem:chebyshev1-area}.

\begin{proof}[Proof of Lemma \ref{lem:chebyshev1-area}]
Since the function $|T_n(x)|$ is even, we find that
$$
\int_{-\infty}^{+\infty}\frac{dx}{|T_n(x)|^\alpha} = C_1 + C_2,
$$
where
$$
C_1 = 2\int\limits_{1}^{+\infty}\frac{dx}{T_n(x)^\alpha}, \,\,\, C_2 = 2\int\limits_{0}^{1}\frac{dx}{|T_n(x)|^\alpha},
$$
We will now derive upper and lower bounds on $C_1$ and $C_2$.

First, we determine upper and lower bounds on $C_1$. Recall that
$$
T_n(x) = \frac{(x + \sqrt{x^2 - 1})^n + (x - \sqrt{x^2 - 1})^n}{2}
$$
for all $x$ such that $|x| \geq 1$. Hence
\begin{equation} \label{eq:Tn-bound}
\frac{(x + \sqrt{x^2 - 1})^n}{2} < T_n(x) < (x + \sqrt{x^2 - 1})^n
\end{equation}
for all $x > 1$. Using the lower bound in (\ref{eq:Tn-bound}), as well as the change of variables $x = \csc \theta$, we find that
\begin{align*}
C_1
& = 2\int\limits_{1}^{+\infty}\frac{dx}{T_n(x)^\alpha}\\
& < 2^{\alpha + 1}\int\limits_1^{+\infty}\frac{dx}{(x + \sqrt{x^2 - 1})^{n\alpha}}\\
& = 2^{\alpha + 1}\int\limits_0^{\pi/2}\frac{(\cos \theta)(\sin \theta)^{n \alpha - 2}d\theta}{(1 + \cos \theta)^{n\alpha}}\\
& = 2^{\alpha + 1}\left.\frac{(\sin \theta)^{n\alpha - 1}(n\alpha\cos \theta + 1)}{(n^2\alpha^2 - 1)(1 + \cos \theta)^{n\alpha}}\right|_{0}^{\pi/2}\\
& = \frac{2^{\alpha + 1}}{n^2\alpha^2 - 1}
\end{align*}
A lower bound on $C_1$ can be obtained analogously. As a result, we get
$$
\frac{2}{n^2\alpha^2 - 1} < C_1 < \frac{2^{1 + \alpha}}{n^2\alpha^2 - 1}.
$$

Next, we determine upper and lower bounds on $C_2$. We apply the substitution $x = \cos(\theta/n)$ and make use of the identity $\cos \theta = T_n(\cos(\theta/n))$ as follows:\footnote{The author is grateful to Fedor Petrov for recognizing the third equality. See \url{https://mathoverflow.net/questions/345820/an-integral-of-sinx-cosnx-2-n-from-pi-to-pi}.}
\begin{align*}
C_2
& = 2\int\limits_0^1\frac{dx}{|T_n(x)|^\alpha}\\
& = \frac{2}{n}\int\limits_0^{\pi n/2}\frac{\sin(\theta/n)d\theta}{|\cos \theta|^\alpha}\\
& = \frac{2}{n}\int\limits_0^{\pi n/2}\sin(\theta/n)d\theta + \varepsilon_n,
\end{align*}
where
$$
\varepsilon_n = \frac{2}{n}\int\limits_0^{\pi n/2}\sin(\theta/n)\left(|\cos \theta|^{-\alpha} - 1\right)d\theta.
$$
Note that $\varepsilon_n > 0$. Since
$$
\frac{2}{n}\int\limits_0^{\pi n/2}\sin(\theta/n)d\theta = 2\int\limits_0^{\pi/2}\sin\theta d\theta = 2,
$$
we have
$$
C_2 = \frac{2}{n}\int\limits_0^{\pi n/2}\sin(\theta/n)d\theta + \varepsilon_n > 2.
$$

It remains to find an upper bound on $C_2$. Note that
\begin{align*}
\varepsilon_n & = \frac{2}{n}\int\limits_0^{\pi n/2}\sin(\theta/n)\left(|\cos \theta|^{-\alpha} - 1\right)d\theta\\
& \leq \frac{2}{n}\int\limits_0^{\pi n/2}|\cos \theta|^{-\alpha}d\theta - \pi\\
& = 2\int\limits_0^{\pi/2}(\cos \theta)^{-\alpha}d\theta - \pi\\
& = B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi,
\end{align*}
where the last equality follows from (\ref{eq:beta}). We conclude that
$$
2 < C_2 \leq 2 + B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi.
$$
\end{proof}

\section{Upper and Lower Bounds on $A_{U_n}$} \label{sec:Un}

In this section we prove the following result.

\begin{lem} \label{lem:chebyshev2-area}
For any $n \in \mathbb N$ and $\alpha \in \mathbb R$ such that $n \geq 3$ and $2/n \leq \alpha < 1$,
\begin{align*}
\int\limits_{-\infty}^{+\infty}\frac{dx}{|U_n(x)|^\alpha} & > B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right) + \frac{2}{n^2\alpha^2 - 1}(n + 1)^{-\alpha}\\
\int\limits_{-\infty}^{+\infty}\frac{dx}{|U_n(x)|^\alpha} & < B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right) + \frac{2}{n^2\alpha^2 - 1} + B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi\notag
\end{align*}
\end{lem}

Let us see why Theorem \ref{thm:main2} directly follows from Lemma \ref{lem:chebyshev2-area}.

\begin{proof}[Proof of Theorem \ref{thm:main2}]
Since for an arbitrary binary form $F$ of positive degree $n$ the area bounded by the curve $|F(x, y)| = 1$ can be computed via (\ref{eq:AF-formula}), the inequalities (\ref{eq:AUn-bounds}) follow from Lemma \ref{lem:chebyshev2-area} if we take $\alpha = 2/n$.

To see that $\lim\limits_{n \rightarrow \infty} A_{U_n} = \frac{8}{3}$, it is sufficient to demonstrate that the limits of $B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right)$ and $B\left(1 + \frac{1}{n}, \frac{1}{2}\right)$ as $n$ approaches infinity are equal to $\pi$ and $2$, respectively. Recall that the gamma function (\ref{eq:gamma}) satisfies $\Gamma(1) = 1$ and $\Gamma(1/2) = \sqrt{\pi}$. In view of (\ref{eq:beta-gamma}), we see that
$$
\lim\limits_{n \rightarrow \infty}B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) = B\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{\Gamma(1/2)^2}{\Gamma(1)} = \pi
$$
and
$$
\lim\limits_{n \rightarrow \infty}B\left(1 + \frac{1}{n}, \frac{1}{2}\right) = B\left(1, \frac{1}{2}\right) = \frac{\Gamma(1)\Gamma(1/2)}{\Gamma(3/2)} = \frac{\Gamma(1)\Gamma(1/2)}{(1/2)\Gamma(1/2)} = 2,
$$
where the second-to-last equality follows from the functional equation (\ref{eq:functional-equation}).
\end{proof}

We will now turn our attention to the proof of Lemma \ref{lem:chebyshev2-area}.

\begin{proof}[Proof of Lemma \ref{lem:chebyshev2-area}]
Since the function $|U_n(x)|$ is even, we find that
$$
\int_{-\infty}^{+\infty}\frac{dx}{|U_n(x)|^\alpha} = C_1 + C_2,
$$
where
$$
C_1 = 2\int\limits_{1}^{+\infty}\frac{dx}{U_n(x)^{\alpha}}, \,\,\, C_2 = 2\int\limits_{0}^{1}\frac{dx}{|U_n(x)|^{\alpha}}.
$$
We will now derive upper and lower bounds on $C_1$ and $C_2$.

Put $C_1 = 2\int\limits_{1}^{+\infty}\frac{dx}{U_n(x)^{\alpha}}, \,\,\, C_2 = 2\int\limits_{0}^{1}\frac{dx}{|U_n(x)|^{\alpha}}$, so that $\int\limits_{-\infty}^{+\infty}\frac{dx}{|U_n(x)|^\alpha} = C_1 + C_2$.

First, we determine upper and lower bounds on $C_1$. Recall that
$$
U_n(x) = \frac{\left(x + \sqrt{x^2 - 1}\right)^{n + 1} - \left(x - \sqrt{x^2 - 1}\right)^{n + 1}}{2\sqrt{x^2 - 1}}
$$
for all $x$ such that $|x| \geq 1$.
Thus for every $\theta \in [0, \pi/2]$ we have
\begin{align*}
(\sin \theta)^nU_n(\csc \theta)
& =
\frac{(1 + \cos \theta)^{n + 1} - (1 - \cos \theta)^{n + 1}}{2\cos \theta}\\
& =
\frac{\left(2\cos^2\frac{\theta}{2}\right)^{n + 1} - \left(2\sin^2\frac{\theta}{2}\right)^{n + 1}}{2\cos \theta}\\
& = 2^n\sum\limits_{k = 0}^n\left(\sin^2\frac{\theta}{2}\right)^k\left(\cos^2\frac{\theta}{2}\right)^{n - k}.
\end{align*}
Since
$$
\left(\cos^2\frac{\theta}{2}\right)^n < \sum\limits_{k = 0}^n\left(\sin^2\frac{\theta}{2}\right)^k\left(\cos^2\frac{\theta}{2}\right)^{n - k} < (n + 1)\left(\cos^2\frac{\theta}{2}\right)^n
$$
for every $\theta \in (0, \pi/2]$, we find that
\begin{equation} \label{eq:Un-bound}
\left(1 + \cos \theta\right)^n < (\sin \theta)^nU_n(\csc\theta) < (n + 1)\left(1 + \cos \theta\right)^n
\end{equation}
for every $\theta \in (0, \pi/2]$.

 Using the change of variables $x = \csc \theta$, as well as the lower bound in (\ref{eq:Un-bound}), we find that
\begin{align*}
C_1
& = 2\int\limits_1^{+\infty}\frac{dx}{U_n(x)^\alpha}\\
& = 2\int\limits_0^{\pi/2}\frac{\cos\theta d\theta}{(\sin \theta)^2U_n(\csc\theta)^\alpha}\\
& < 2\int\limits_0^{\pi/2}\frac{\cos\theta (\sin \theta)^{n\alpha - 2} d\theta}{(1 + \cos\theta)^{n\alpha}}\\
& = 2\left.\frac{(\sin \theta)^{n\alpha - 1}(n\alpha\cos \theta + 1)}{(n^2\alpha^2 - 1)(1 + \cos \theta)^{n\alpha}}\right|_{0}^{\pi/2}\\
& = \frac{2}{n^2\alpha^2 - 1}.
\end{align*}
A lower bound on $C_1$ can be obtained analogously. As a result, we get
$$
\frac{2}{n^2\alpha^2 - 1}(n + 1)^{-\alpha} < C_1 <  \frac{2}{n^2\alpha^2 - 1}.
$$

It remains to determine upper and lower bounds on $C_2$. We apply the substitution $x = \cos\left(\theta/(n + 1)\right)$ and make use of the identity
$$
U_n\left(\cos\left(\frac{\theta}{n + 1}\right)\right) = \frac{\sin \theta}{\sin\left(\frac{\theta}{n + 1}\right)}
$$
as follows:
\begin{align*}
C_2
& = 2\int\limits_0^1\frac{dx}{|U_n(x)|^\alpha}\\
& = \frac{2}{n + 1}\int\limits_0^{\pi (n + 1)/2}\sin\left(\frac{\theta}{n + 1}\right)^{1 + \alpha}|\sin \theta|^{-\alpha}d\theta\\
& = \frac{2}{n + 1}\int\limits_0^{\pi (n + 1)/2}\sin\left(\frac{\theta}{n + 1}\right)^{1 + \alpha}d\theta + \varepsilon_n\\
& = 2\int\limits_0^{\pi/2}(\sin\theta)^{1 + \alpha} d\theta + \varepsilon_n\\
& = B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right) + \varepsilon_n,
\end{align*}
where the last equality follows from (\ref{eq:beta}) and
$$
\varepsilon_n = \frac{2}{n + 1}\int\limits_0^{\pi(n + 1)/2}\sin\left(\frac{\theta}{n + 1}\right)^{1 + \alpha}\left(|\sin \theta|^{-\alpha} - 1\right)d\theta.
$$
Since $\varepsilon_n > 0$, we obtain the lower bound $C_2 > B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right)$.

To find an upper bound on $C_2$, note that
\begin{align*}
\varepsilon_n
& \leq \frac{2}{n + 1}\int\limits_0^{\pi(n + 1)/2}\left(|\sin \theta|^{-\alpha} - 1\right)d\theta\\
& = 2\int\limits_0^{\pi/2}(\sin \theta)^{-\alpha} - \pi\\
& = B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi,
\end{align*}
where the last equality follows from (\ref{eq:beta}). We conclude that
$$
B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right) < C_2 \leq B\left(\frac{2 + \alpha}{2}, \frac{1}{2}\right) + B\left(\frac{1 - \alpha}{2}, \frac{1}{2}\right) - \pi.
$$
\end{proof}

\section{Bean's Conjecture} \label{sec:bean}

We conclude this article with the proof of Proposition \ref{prop:bean}, which provides further theoretical evidence in support of Conjecture \ref{conj:bean}.

\begin{prop} \label{prop:bean}
For any integer $n \geq 3$, $Q(T_n) \leq Q(S_n)$ and $Q(U_n) \leq Q(S_n)$. Furthermore,
$$
\lim\limits_{n \rightarrow \infty} Q(\Psi_n) = \lim\limits_{n \rightarrow \infty} Q(\Pi_n)  = \lim\limits_{n \rightarrow \infty} Q(T_n) = \lim\limits_{n \rightarrow \infty} Q(U_n) = \frac{16}{3}.
$$
\end{prop}

\begin{table}[t]
\centering
\begin{tabular}{c | c | c | c | c | c | c | c | c | c | c}
$n$ & $D_{\Psi_n}$ & $A_{\Psi_n}$ & $D_{\Pi_n}$ & $A_{\Pi_n}$ & $D_{S_n}$ & $A_{S_n}$ & $D_{T_n}$ & $A_{T_n}$ & $D_{U_n}$ & $A_{U_n}$\\
\hline
$3$ & 1 & $\infty$ & $2^23$ & $\infty$ & $2^23^3$ & $7.28585$ & $2^4 3^3$ & $5.78286$ & $2^{11}$ & $4.46217$\\
$4$ & $1$ & $\infty$ & $1$ & $\infty$ & $2^2$ & $10.4882$ & $2^{17}$ & $4.30008$ & $2^{16}5^2$ & $3.50332$\\
$5$ & $5$ & $\infty$ & $2^45^3$ & $5.78302$ & $2^{12}5^5$ & $4.55444$ & $2^{16}5^{5}$ & $3.78568$ & $2^{28}3^3$ & $3.19719$\\
$6$ & $1$ & $\infty$ & $2^23$ & $\infty$ & $2^{16}3^6$ & $5.29992$ & $2^{31}3^6$ & $3.52082$ & $2^{36}7^4$ & $3.04985$\\
$7$ & $7^2$ & $8.31171$ & $2^67^5$ & $5.38644$ & $2^{30}7^7$ & $3.99650$ & $2^{36}7^7$ & $3.35841$ & $2^{64}$ & $2.96434$\\
$8$ & $2^3$ & $\infty$ & $2^3$ & $\infty$ & $2^{24}$ & $6.48467$ & $2^{73}$ & $3.24832$ & $2^{64}3^{12}$ & $2.90894$\\
$9$ & $3^4$ & $7.64379$ & $2^63^9$ & $5.63543$ & $2^{56}3^{18}$ & $3.75495$ & $2^{64}3^{18}$ & $3.16867$ & $2^{88}5^7$ & $2.87035$
\end{tabular}
\caption{Invariants associated with $\Psi_n$, $\Pi_n$, $S_n$, $T_n$ and $U_n$ for $n = 3, 4, \ldots, 9$.}
\label{tab:table1}
\end{table}

\begin{proof}
Since $Q(F_n^*) = Q(S_n)$, we prove our statement with $F_n^*$ in place of $S_n$.

Let $n$ be an integer such that $n \geq 3$. According to Tran \cite{tran}, the discriminants of $T_n$ and $U_n$ are given by
\begin{equation} \label{eq:discriminants}
D_{T_n} = 2^{(n - 1)^2}n^n \quad \text{and} \quad D_{U_n} = 2^{n^2}(n + 1)^{n - 2}.
\end{equation}
Along with Theorems \ref{thm:main1} and \ref{thm:main2}, these equalities yield the upper bounds
\small
\begin{equation} \label{eq:QTn-bound}
2^{-\frac{n - 1}{n}}n^{-\frac{1}{n - 1}}Q(T_n) < \frac{8}{3} + \frac{2}{3}\left(\sqrt[n]{4} - 1\right) + B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) - \pi
\end{equation}
\normalsize
and
\small
\begin{equation} \label{eq:QUn-bound}
2^{-\frac{n}{n - 1}}(n + 1)^{-\frac{n - 2}{n(n - 1)}} Q(U_n) < \frac{8}{3} + \left(B\left(1 + \frac{1}{n}, \frac{1}{2}\right) - 2\right) + B\left(\frac{1}{2} - \frac{1}{n}, \frac{1}{2}\right) - \pi
\end{equation}
\normalsize

Now, the values of $Q(F_n^*) = Q(S_n)$, $Q(T_n)$ and $Q(U_n)$ for $n = 3, 4, \ldots, 9$ are given in \mbox{Table \ref{tab:table2}}. The integrals associated with $A_{S_n}$, $A_{T_n}$ and $A_{U_n}$ were approximated with Mathematica. Notice that $Q(T_3) = Q(U_3) = Q(S_3) = 3B\left(\frac{1}{3}, \frac{1}{3}\right)$, because $S_3$, $T_3$ and $U_3$ are all equivalent under $\operatorname{GL}_2(\mathbb R)$ to \mbox{$xy(x - y)$}. From Table \ref{tab:table1} it is clear that the statement of \mbox{Corollary \ref{prop:bean}} holds for the aforementioned values of $n$.

For $n \geq 10$, the inequalities $Q(T_n) \leq Q(S_n)$ and $Q(U_n) \leq Q(S_n)$ can be easily verified by combining the formula (\ref{eq:QFn}) with the upper bounds (\ref{eq:QTn-bound}) and (\ref{eq:QUn-bound}).

The formulas for $D_{T_n}$ and $D_{U_n}$ given in (\ref{eq:discriminants}) imply
$$
\lim\limits_{n \rightarrow \infty}D_{T_n}^{1/n(n - 1)} = \lim\limits_{n \rightarrow \infty}D_{U_n}^{1/n(n - 1)} = 2.
$$
Combining this with the fact that $\lim\limits_{n \rightarrow \infty} A_{T_n} = \lim\limits_{n \rightarrow \infty} A_{U_n} = \frac{8}{3}$, we find that
$$
\lim\limits_{n \rightarrow \infty} Q(T_n) = \lim\limits_{n \rightarrow \infty} D_{T_n}^{\frac{1}{n(n - 1)}}A_{T_n} = \left(\lim\limits_{n \rightarrow \infty} D_{T_n}^{\frac{1}{n(n - 1)}}\right)\cdot\left(\lim\limits_{n \rightarrow \infty} A_{T_n}\right) = \frac{16}{3}.
$$
The same calculation applies with $Q(U_n)$ in place of $Q(T_n)$.

Next, we prove that $\lim\limits_{n \rightarrow \infty}Q(\Psi_n) = \frac{16}{3}$. For a positive integer $n$, let $\omega(n)$ denote the number of distinct prime divisors of $n$, and let \mbox{$n = p_1^{e_1} \cdots p_{\omega(n)}^{e_{\omega(n)}}$} be the prime factorization of $n$. According to Liang \cite{liang}, the discriminant $D_{K_n}$ of the number field $K_n = \mathbb Q\left(2\cos\left(\frac{2\pi}{n}\right)\right)$ can be computed according to the formula
\begin{equation} \label{eq:Dn}
D_{K_n} =
\begin{cases}
2^{(m-1)2^{m - 2} - 1} & \textrm{if $n = 2^m$, $m > 2$,}\\
p^{(mp^m - (m+1)p^{m-1} - 1)/2} & \textrm{if $n = p^m$ or $2p^m$, $p > 2$ prime,}\\
\left(\prod_{i = 1}^{\omega(n)}p_i^{e_i - 1/(p_i - 1)}\right)^{\frac{\varphi(n)}{2}} & \textrm{if $\omega(n) > 1, n \neq 2p^m$.}
\end{cases}
\end{equation}
It was also established by Liang that the ring of integers $\mathcal O_{K_n}$ of $K_n$ has a power integral basis, i.e., $\mathcal O_{K_n} = \mathbb Z\left[2\cos\left(\frac{2\pi}{n}\right)\right]$. Consequently, the discriminant $D_{\Psi_n}$ of $\Psi_n$ is equal to $D_{K_n}$. Since $1 \leq D_{\Psi_n} \leq n^{\varphi(n)/2}$, it follows from (\ref{eq:phi-lower-bound}) and Theorem \ref{thm:main3} that
$$
\lim\limits_{n \rightarrow \infty} Q(\Psi_n) = \frac{16}{3}.
$$
In view of Corollary \ref{cor:main3}, this result also implies \mbox{$\lim\limits_{n \rightarrow \infty}Q(\Pi_n) = \lim\limits_{n \rightarrow \infty}Q(\Psi_n) = \frac{16}{3}$}.
\end{proof}

\section*{Acknowledgements}

The author is grateful to Prof.\ Cameron L.\ Stewart for his numerous suggestions, to Patrick Naylor for many productive conversations, and to the anonymous reviewer for their valuable recommendations.

redpajama_set_name: RedPajamaArXiv


text:
European Cars on Display at Saratoga Auto Museum
"The British are coming!" says Jean Hoffman, a trustee at the Saratoga Automobile Museum (SAM).
Ms. Hoffman says the nationally renowned museum will host its 4th Annual "Invitational at Saratoga" May 18 – 20 on the grounds of the Saratoga Spa State Park. According to Hoffman, the gathering of exquisite automobiles from around the world is quickly becoming a premier East Coast automobile event.
First held in 2009 as a supplement to SAM's longstanding Spring Auto Show, the "Invitational" immediately became a popular kick-off event for the Spa City's summer season.
Just like the thoroughbred horses which come to the city in July, horsepower and speed are the buzzwords of the "Invitational."
The three-day weekend begins with a $45 per person cocktail reception at 6 pm on Friday at SAM. Members of the public are invited to meet special guest Wayne Carini of Chasing Classic Cars fame. Carini will get the weekend underway as Friday's featured speaker. According to a recent article written by Don Weberg, Carini is one of those guys that pretty much any car guy or gal can relate to. He's constantly on the quest for the rarest, most sought after, most unique cars ever built, cars that he personally loves, and more to the point, cars that his clients are looking for.
Saturday's line-up of Euro cars has historically attracted over 200 widely recognized examples of British, French, German, Italian and other European makes and models. According to Hoffman, anyone who owns or drives a European made automobile is invited to enter their vehicle in the show. "Saturday is not by invitation only…only Sunday," she explained.
The "Euro Experience on the Field" will be followed by a fund-raising dinner and auction at the Gideon Putnam Resort in Saratoga Springs.
The auction will feature items such as one-day of driving with the Sports Car Drivers Association School at Lime Rock Park or your chance to bid on a package which includes two tickets to the Tonight Show, a photograph in the Green Room with Jay Leno and a tour of Jay's car collection in Burbank, California.
Featured guest speaker will be noted actor Ed Herrmann, best known for his Emmy-nominated portrayals of Franklin D. Roosevelt on television, and to younger generations for his role as Richard Gilmore in Gilmore Girls and as a ubiquitous narrator for historical programs on the History Channel. Tickets are available for the 6 pm Saturday event at a cost of $150 per person. Contact Becky Earls at 518-587-1932 ext. 16.
Rounding out the weekend's activities, the Sunday "Invitational" should provide attendees with an up-close-and-personal look at cars from around the US and beyond. The "Invitational", as the name implies, is open only to select collectors with rare classic and historic automobiles.
Over 100 extraordinary vehicles and vintage motorcycles will be on display Sunday. The featured marque for 2012 is Cadillac, one of America's most highly respected names in automobile manufacturing. Reportedly at least 20 vintage Caddies will be featured during the "Invitational".
Saturday's and Sunday's events are open from 10 am to 3 pm each day and the cost of admission to the grounds is $15 and children under 12 are admitted free. No pets are allowed.
As an added bonus, the Museum itself will welcome visitors without charge with paid admission to either the Euro Show on Saturday or the "Invitational" on Sunday.
According to Hoffman, all proceeds from the weekend events will benefit the Museum's Educational programs and the Make-A-Wish Foundation of Northeast New York.


Montana Department of Revenue Administers more than 30 state taxes and fees, including income taxes, natural resource taxes, corporation taxes and miscellaneous taxes.
All Montana State Taxes (Summary & Expanded List from CPAide) Click here for a handy summary of all Montana taxes and a larger list of links to Montana taxes and other tax help and forms for each Montana tax.
Montana Judicial Info Courts for Montana, including Workers' Comp, Limited, Water, District and Supreme.
Montana Legislative Info The Montana Legislature is a bicameral legislature, composed of the Senate and the House of Representatives.
Montana State Home Page Big Sky Country.
Montana Statutes Montana Code, annotated 2014. See Title XV for Taxation.
Montana Tax Forms for Businesses The most popular Montana business-related income tax forms, worksheets and instructions.
Montana Tax Forms for Individuals The most popular Montana individual income tax forms, worksheets and instructions.
Montana: Voluntary Disclosure Montana's Voluntary Disclosure Program allows qualified entities and individuals to disclose their tax liabilities voluntarily and settle their tax obligations in a Voluntary Disclosure Agreement (VDA).


text:
ESL Voices
The Truth About Santa Claus Revealed!
Guest Plans Guidelines
Beginners: How to Find ESL Info on the Web
Tutorial: Refining Web Searches for ESL Topics
Reading Strategies and Lesson Plans
Writing Strategies and Activities
Modes of Writing
Speaking Strategies and Lesson Plans
Listening Strategies and Lesson Plans
Vocabulary Lesson Plans
Grammar Strategies and Activities
Charts and Organizers
Significant Educational Articles
ESL Acronyms and Abbreviations
Business Writing: Resumes & Cover Letters
Checking, Savings, and Credit Card Applications
Current Business Articles with Activities
News Articles List
Business Newspapers
Vocabulary Review
Basic Listening Practice: NPR & VOA
Stephen Colbert Listening Videos
Classic Short Stories
CLASSIC LESSON PLANS
Answer Key for Classic Lesson Plans
Contemporary Novels
Contemporary Novels List
Ghost & Horror
Guest Plans
Vocabulary Strategies and Lesson Plans
Grammer Strategies and Activities
Busines Writing: Resumes & Cover Letters
Business Newpapers
Classics: Lesson Plans
Adventrue
"Don't tell the kids, but we've got Santa Claus all wrong. Countless Christmas songs tell us that Santa is basically the Judge Judy of juveniles. He decides who's been naughty or nice, and doles out presents or punishments accordingly. But historians say Kris Kringle was originally created to keep 'adults', not children, off the naughty list. Being crafty codgers, we ducked Santa's surveillance, turning the spotlight on kids and dramatically changing Christmas celebrations." D. Burke, CNN
ESL Voices Lesson Plan for this post with Answer Key
Santa in the famous Macy's Day Parade in New York City, U.S.A. dailymail
Excerpt: The real story behind Santa Claus, By Daniel Burke, CNN
"How did we achieve this very important historical victory? Picture this: It's the early 1800s, and America's Christian leaders — most of whom were Protestant Reformation-types — had banned religious celebrations of Christmas as unscriptural and paganish.
In Tokyo Japan Santas from the Harley Santa Club deliver toys in a campaign against child abuse. CNN
But people still wanted to party. Because, why not? It was midwinter, the crops were harvested and sailors were waiting for better weather to disembark. So, on December 25, working-class stiffs got fall-down drunk and stumbled around cities looking for stuff to loot. Imagine Black Friday, spring break and New Year's Eve — then smash them together like sumo wrestlers full of saki. That was Christmas in the early 1800s.
Colorful Santa in Italy. CNN
A bunch of blue-blood New Yorkers decided all this fun must stop. They wanted to domesticate Christmas, bring it indoors, and focus it on children, says Gerry Bowler, author of Santa Claus: A History and professor of history at the University of Manitoba in Canada. These grinches, who formed the Saint Nicholas Society of New York, would change the world with two little poems. Yep. Poems.
Father Frost (equivalent of Santa) rests on a high-rise building in Kemerovo Russia
But let's back up for a minute. When the Dutch came to the New World in the 1600s, they brought a fellow from folklore named Sinterklaas with them, Bowler says. Sinterklaas, who wore a red bishop's miter and a snowy white beard…Despite being a bishop, this Nick was a bit of a bad boy. An archaeologist who dug up his bones in 2005 found that Nicholas had a broken nose, perhaps a result of the persistent persecution of Christians around that time… Or could it have been Christian-on-Christian violence?
Volunteer Santas in Seoul, Korea. CNN
Early icons of Nicholas depict him without bishop's garb, a subtle suggestion that he had been demoted, possibly for fisticuffs…Thankfully, St. Nicholas, was known for more than brawling. He also had a reputation for giving gifts and protecting children…While staying at an inn, Nicholas discovered three dismembered children in pickle barrels. He reassembled and resurrected the briny kids and punished the guilty innkeeper. These deeds, along with his everyman persona, (he wasn't a martyr or hermit like so many other model Christians of the time), made Nicholas the greatest male saint of the Middle Ages.
Santa in the U.S. toovia.com
Bowler says that St. Nick's feast day, December 6, (the day he supposedly died) was celebrated across Europe for hundreds of years, often by giving gifts to children…About a decade later, in 1821, an anonymous poem called The Children's Friend, featured a magical figure called 'Santeclaus,' who drove a reindeer-led sleigh full of "rewards" and filled obedient children's stockings with little presents. Building on that, an Episcopalian scholar named Clement Clarke Moore wrote a poem for his big brood called A Visit From St. Nicholas. It's now better known as The Night Before Christmas."
ESL Voices Lesson Plan for this post
NOTE: Lessons can also be used with native English speakers.
Level: Intermediate – Advanced
Language Skills: Reading, writing, and speaking. Vocabulary and grammar activities are included.
Time: Approximately 2 hours.
Materials: Student handout (from this lesson) and access to news article.
Objective: Students will read and discuss the article with a focus on improving reading comprehension and learning new vocabulary. At the end of the lesson students will express their personal views on the topic through group work and writing.
I. Pre-Reading Activities
Predictions: Analyzing headings and photos
Directions: Have students examine the titles of the post and of the actual article. After they examine the photos, ask students to create a list of words and ideas that they think might be related to this article.
Pre-reading Organizer By Scholastic
II. While Reading Tasks
Word Inference
Directions: Students are to infer the meanings of the words in bold taken from the article. They may use a dictionary, thesaurus, and Word Chart for assistance.
How did we achieve this very important historical victory?
They wanted to domesticate Christmas.
They brought a fellow from folklore named Sinterklaas.
An archaeologist dug up his bones in 2005.
The meeting in 325 formed the first consensus on Christian doctrine.
There was a suggestion that he had been demoted possibly for fisticuffs.
St. Nicholas, was known for more than brawling.
He reassembled and resurrected the briny kids.
In some early depictions, Santa Claus looks like an overgrown elf.
By the early 1900s, Santa became benevolent grandfather that we all know and love.
Vocabulary Cluster By Learnnc.org
True /False/NA-Statements
Directions: Review the following statements from the reading. If a statement is true they mark it T. If the statement is not applicable, they mark it NA. If the statement is false they mark it F and provide the correct answer.
Kris Kringle was originally created children.
In the early 1800s, America's Christian leaders encouraged celebrations of Christmas.
From this article we learn that in midwinter, the crops were harvested.
Back then on December 25, the working-class donated money to the poor.
A group of blue-blood New Yorkers applauded the working class for their donations.
They wanted to domesticate Christmas, bring it indoors, and focus it on children.
When the English came to the New World in the 1600s, they brought a fellow named Sinterklaas.
Sinterklaas had five children of his own.
Sinterklaas wore a red bishop's miter and had a snowy white beard.
An archaeologist who dug up his bones in 2005 found that Nicholas had a broken nose.
Grammar Focus
Using Adjectives to describe pictures
Directions: Have students choose a picture from the article and write a descriptive paragraph using adjectives.
Discussion/Writing Exercise
Directions: Place students in groups and have them answer the following questions. Afterwards, have the groups share their thoughts as a class. To reinforce the ideas, students can write an essay on one of the following discussion topics.
1. The following three statements were taken from the article. Rephrase each statement in your own words, then discuss the meaning with the members of your group.
"These deeds, along with his everyman persona, (he wasn't a martyr or hermit like so many other model Christians of the time), made Nicholas the greatest male saint of the Middle Ages…One measure of his popularity is the long list of people, places, churches and Christian groups that list St. Nick as their patron….St. Nick's feast day, December 6, (the day he supposedly died) was celebrated across Europe for hundreds of years, often by giving gifts to children."
"Some countries… such as the Netherlands, kept alive traditions associated with Sinterklaas. And it was these customs that 19th century New Yorkers wanted to revive. As they sought to make Christmas more family friendly, the Saint Nicholas Society found the perfect front man in their namesake, who, after all, was known for being nice to children."
"The real goal was getting drunks off the street, remember? Now they could do that by turning Christmas into a family event when children — who had it pretty rough back then — would receive gifts for good behavior. Drawing on the Dutch legends about Sinterklaas, the American author Washington Irving wrote a series of sketches featuring St. Nicholas soaring high above New York houses, smoking a pipe and delivering presents to well-behaved children."
3-2-1-Writing
Directions: Allow students 5 minutes to write down three new ideas they've learned about the origins of Santa Claus from the reading, two things they did not understand in the reading, and one thing they would like to know that the article did not mention. Review the responses as a class.
Category: Celebrations | Tags: Santa Claus
Listening exercises with
Stephen Colbert, NPR , & VOA.
Get Email Alerts for Lesson Plans
How To Print Lesson Plans
Feeds : RSS Atom
Resource Sites:
ESL and Native Americans:
Talking Feather
Tribalpedia
Books. (3)
Children/Teens (12)
Guest Plan (4)
Health Issues (27)
Marine Animals (2)
| Terms of Use | ©2021 Amerindian7
Powered by WordPress using a customized Rundown and jquery mmenu.
Site Design and Content: Amerindian7 | Implementation: Applied Logic Systems


text:
Home/Tech-Verse/OnePlus 10 Pro Set to Launch in January with Snapdragon 8 Gen 1 – Specifications, Geekbench
OnePlus has released a new teaser trailer for the OnePlus 10 Pro, marking the company's first formal announcement of the forthcoming flagship. Let's see what the teaser video has to say.
OnePlus has released a new teaser trailer for the OnePlus 10 Pro, marking the company's first formal announcement of the forthcoming flagship. The short film shows a look of the device's new and updated camera module, which has been rumored for quite some time. It also shows when the OnePlus 10 Pro will be available for purchase.
On the Chinese social media network Weibo, OnePlus has posted an official teaser video. The teaser video also discloses the launch date for the OnePlus 10 Pro, as pointed out by tipster Ishan Agarwal in a previous tweet. As a result, we know the new OnePlus flagship will be unveiled on January 11, 2022.
Image Source-NewsBytes
This is a significant departure from OnePlus' regular March launch schedule for its flagship phones. Because it missed out on the T-moniker devices in 2021, the company will release the phones early next year. An early launch would also help OnePlus compete with flagship smartphones from other OEMs, such as the Xiaomi 12 that was recently released and the Galaxy S22 Ultra that will be released later this month.
First Official look at the OnePlus 10 Pro from a new teaser video posted on Weibo!
Looks cool, kinda fresh. Thoughts? #OnePlus10Pro #OnePlus pic.twitter.com/dVtnNl4e1o
— Ishan Agarwal (@ishanagarwal24) December 30, 2021
For the time being, though, the OnePlus launch will be limited to China. It's unclear when the OnePlus 10 Pro will be available worldwide, while previous sources have suggested a March release date. This will be in keeping with how the new OnePlus flagship smartphones are usually released.
OnePlus 10 Pro Specifications-
A 6.7-inch QHD+ (3216×1,440 pixels) AMOLED screen with a 120Hz refresh rate and LTPO 2.0 technology is expected to be included in the OnePlus 10 Pro. It will be powered by a Qualcomm Snapdragon 8 Gen 1 processor and might have up to 12GB of RAM and 256GB of storage.
A 5,000mAh battery may power the phone, with support for 80W rapid charging. The OnePlus 10 Pro is said to have a squarish camera module with a triple-lens configuration and an LED flash, which will be located towards one end of the back panel. Hasselblad will power this camera system, which could have a 48-megapixel primary lens.
Image Source-samacharnagari. in
OnePlus 10 Pro Geekbench Listing
With model number NE2210, the OnePlus 10 Pro visited the Geekbench 5 database. In the single-core test, it received 976 points. In the Geekbench 5 test, however, the multi-core round resulted in a score of 3469 points. An octa-core CPU with 1 + 3 + 4 core configuration, up to 3.0GHz frequency, and codename taro is also revealed in the listing. The Qualcomm Snapdragon 8 Gen 1 processor will be used in the forthcoming OnePlus flagship, according to these reports. The Snapdragon 8 Gen 1 is made using Samsung's 4nm technology and has an Adreno 730 GPU.
The OnePlus 10 Pro will have 12GB of RAM, according to the Geekbench listing. However, we can anticipate OnePlus offering the gadget in a variety of RAM configurations. Android 12 will be pre-installed on the handset, according to reports. The gadget will come with a ColorOS 12 skin on top in China.
Also Read: 7 Biggest Tech Fails of 2021
7 Best Apps for Chromecast with Google TV
7 Best Free PDF Editors to use in 2022
BlackBerry phones will die on January 4, Once a Loved Monarch in Mobile World
Apple Music expands Chinese music reservoir via Tencent deal
Prime Gaming is giving away EA Games Over the Next Few Months
Pingback: BlackBerry Phones Will Die On January 4, Once A Loved Monarch In Mobile World | TechiDroid
Sony has filed a patent for the PlayStation Mobile Controller
Apple Releases iOS 16 Beta 4 To Registered Developers: Let's See What's New
Xiaomi Mi 11 Ultra Review: Beast from Xiaomi


I live in Maryland where blue crabs and crab cakes are a staple. I learned to eat blue crabs at a very early age. Knowing how to strip the shell of a blue crab and eat the succulent meat inside in under 5 minutes was a childhood ritual. On hot summer days, my father used to drive us down to the DC Wharf to grab a bushel of live blue crabs. We would bring the crabs home and steam them in beer. My father would occasionally let a few crabs crawl on the floor to scare the cat. Great memories!
In Maryland, blue crabs are typically in season from May until October. When they are out of season, I grab some fresh lump crab meat at my local grocery store and fry up some crab cakes. I use a very simple "lazy keto" recipe to create a high fat, low carb keto meal that can be enjoyed in less than 20 minutes.
The internal temperature of a cooked crab cake should be 155 degrees F on an instant-read thermometer.
Mix everything in a bowl, shape the mixture in mini crab cakes and fry them on each side for 6-8 minutes.
Crab meat is naturally low in fat and carbs. To add a bit of fat back in, fry the mini crab cakes in uncured bacon grease or garlic flavored olive oil. Very tasty!


The Marriott Portsmouth hotel required updating and a substantial overhaul of the entire building, with the intention of revitalising its appearance and improving its thermal and acoustic performance externally and providing a full internal refurbishment. This will be achieved through the recladding of the hotel's bedroom tower on all sides. The tower features high quality materials chosen to enliven the building, making it a distinctive beacon on the main route into Portsmouth. Folded aluminium-composite cassette panels were chosen to enliven the utilitarian building through both colour and shifting light and shade. This is intended to vary in appearance both on approach and when traveling past on large adjacent roads. The entrance canopies have been stripped back to slender carefully lit frames.


Q: I'm analytical about why refunds for accuse fabricated on a acclaim agenda about don't appearance up on your annual for seven to 10 "business days," while the aboriginal allegation shows up anon back the allegation is made, no amount what day of the anniversary it is. In both cases, the bell-ringer processes the affairs electronically on the aforementioned terminal.
A: It's barefaced that this would leave a lot of bodies abrading their active or alike frustrated.
The accessible one: Merchants don't accept as abundant action to action a acquittance as they do to action a acquirement that will put money in their pockets.
The additional reason: It's a bit of an illusion. Back you accomplish a acquirement and your agenda is swiped, it sends a acting allotment or authority to your bank. The transaction will appearance up as a "pending" transaction about immediately. But the transaction that was accustomed absolutely doesn't absolutely action for conceivably a day or two or three.
With a refund, there is no acting allotment that shows up and needs to be completed. It's all one transaction. And it processes aural a day or two or three afterwards the merchant initiates it.
Something absorbing that's array of accompanying to this: At some rental car companies, you accept to put bottomward a drop of conceivably $100 or $150 in accession to your rental fee. Back you acknowledgment the car, the drop is about refunded aural minutes. Clearly, the rental car companies apperceive this is article that's important to customers. If they don't acquittance it immediately, barter may go to rental car companies that do action refunds anon or don't allegation a drop at all.
Other merchants could accelerate refunds if they wanted.
Q: My bedmate anesthetized abroad beforehand this year. He had a coffer acclaim agenda in his name, and several retail acclaim cards also. Should I abutting these cards? And is there article I should do with his acclaim reports? We accept had a benumb on anniversary of the acclaim bureaus.
A: What you should do depends on whether you accept acclaim cards in your own name, or whether any of your husband's cards had contributed balances.
If you don't accept any cards in your name, I wouldn't automatically abolish all of them. You appetite to accomplish abiding you accept a MasterCard or Visa or Discover or American Express agenda in your own name. If your bedmate had a antithesis on any cards, it may be accessible to get your name added to the account, so the coffer will accept burning abundance that the debts will be repaid quickly.
If there are retail cards that you use for coupons or discounts, such as a Macy's or a Kohl's or a gasoline card, you may appetite to accumulate it and get yourself added or get your own card.
If there are cards that you accept no absorption in using, by all means, you should abolish the accounts so you don't accept to anguish about them.
To the additional allotment of your question, you absolutely should acquaint one of the three above acclaim bureaus. That one will acquaint the added two. Since you both had your acclaim files frozen, this isn't as burning of an affair as it would be if his book was open.
You should aloof address a agenda that your bedmate died, and accommodate his name, Social Security cardinal and date of birth. You should accommodate an official archetype of the afterlife certificate. Ask for a acceptance to be beatific to you.
To ability Teresa Murray, email moneymatters@plaind.com or alarm or argument 216-316-7064. She cannot acknowledge to all queries or comments.


Kelly Bryant transferred to Missouri after graduating from Clemson. With the Tigers now ineligible for the postseason, what could he do next?
The news of Missouri's football program being hit with a postseason ban for the 2019 season raised the question about the future of former Clemson quarterback Kelly Bryant, who enrolled with the school as a graduate transfer with one final season of eligibility.
The postseason ban doesn't impact his original decision to transfer: Bryant was looking for the opportunity to start on the Power Five level after losing his job to Trevor Lawrence, who as a true freshman led Clemson to the national championship. Missouri was an attractive location due to senior Drew Lock's departure and the Tigers' lack of proven in-house options to promote as his successor.
In that sense, Missouri and its productive offense could still give Bryant the chance to prove himself at quarterback in the nation's deepest conference. While not viewed as up to Lock's standard as a passer, Bryant would take over a scheme that ranked in the SEC's top three in passing yards in each of the top three seasons.
As the starter at Clemson in 2017, Bryant completed 65.8 percent of his attempts with 13 touchdowns and 8 interceptions in leading the Tigers to the College Football Playoff. In a handful of starts this past season before being replaced by Lawrence and announcing his intent to transfer, Bryant threw for 461 yards and a pair of touchdowns. In comparison, Lock led the Bowl Subdivision with 44 passing touchdowns in 2017 and accounted for 79 total scores across his final two seasons as the starter.
Bryant might give Missouri's offense a different look as a runner, however. Bryant ran for 665 yards and 11 touchdowns as the starter in 2017, showing the sort of athleticism that could provide a contrast to Lock's more standard, pocket-passing style.
Multiple reports indicate Bryant at this point plans to stay with Missouri. Though he could revisit that decision.
NCAA rules state that Bryant could still land at another program without penalty and participate in the 2019 season.
According to Bylaw 14.7.2 (c), the NCAA "may waive the one-year residence requirement" for student-athletes at an institution placed on probation from "participating in postseason competition during all of the remaining seasons of the student-athlete's eligibility." Bryant has just one final season of eligibility.
In addition, the Division I Committee on Infractions recommended that Missouri waive the restrictions on all seniors who wish to transfer.
That leaves the decision up to Bryant, whose decision would also be impacted by the fact that it is currently too late to transfer into any FBS program in time to participate in spring drills — meaning he wouldn't join another program until the summer, at which point it may be too late to grasp yet another offensive system and rise to the top of the depth chart before the start of the 2019 season.