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The Symposium will take place on the following dates:
For CLE registration, click here
For non-CLE and student registrations, click here
Zoom link will be provided to CLE registrations prior to February 4th
FLORIDA BAR CLE CREDITS AVAILABLE:
1.0 Ethics
Download the Symposium Program or View it Online
Symposium 2022 Videos
The 2022 Symposium was recorded live on Zoom on February 4 – 5, 2022. The videos are posted below for reference.
Panel I: Crypto, Blockchain, and the Future of the Regulatory Landscape
Panel II: Big Tech and Antitrust
Panel III: Tech and Venture Capital Investment
Fireside Chat: The Pros and Cons of the Technology Sector on Local Communities
Panel IV: Role of Criminal Punishment in Tech Regulation
Symposium 2022 Program Overview
This year's symposium, entitled Beyond the Valley, will examine the legal effects of technology companies' movement away from Silicon Valley and into cities like Miami. We will discuss what role attorneys will play in shaping the regulatory landscape governing this rapidly growing industry. The Symposium will not only focus on the substantive legal and regulatory issues but will also discuss the effects, positive and negative, the new industry may have on local communities. Panelists will address a variety of themes including, how new cities like Miami can avoid the negative effects tech has had in Silicon Valley on already marginalized communities, the relationship between venture capital investment and tech start-up, the role of anti-trust regulation, and cryptocurrencies and blockchain. The Law Review is looking forward to structuring the 2021 Symposium as an interdisciplinary event, where experts from the University of Miami, the greater Miami area, and across the country will provide practical insights into a broad range of issues.
Question Submission: Audience members can submit questions for the panelists here
Opening Keynote Address (1:30 – 2:30 PM)
Leigh-Ann Buchanan is a writer, strategist, and sought-after advisor on REDI (racial equity, diversity and inclusion) strategies for tech and innovation ecosystems and organizational culture. NY Times, Inc. Magazine, NPR, NBC, South Florida Times, Miami Herald, Al Jezeera, American Psychological Association, and the ABA Law Journal have featured her thoughts and contributions.
She focuses her energy on projects that bridge systemic access and opportunity gaps, with a concentration on tech, entrepreneurship and high impact leaders. As an advisor, she has supported innovation and technology leaders at Fortune 500 companies including Google, Facebook, Delta, P&G Ventures, AirBnB, HSBC and UBS. As a founder, she has launched ventures from concept to scale, raising over $4.5 million in equity free capital to scale numerous ventures.
Leigh-Ann has been recognized as one of the Most Influential Black Business Leaders in South Florida, a McClatchy Florida Influencer, one of Miami's 50 Most Powerful and Influential Black Business Leaders, a Black Enterprise Magazine Tech Trailblazer and the American Bar Association Top 40 Young Lawyers.
She is the Founder of Nyah Project and currently serves as the President of the venture studio, aīre ventures. She co-hosts the Innovation City podcast and produces storytelling projects.
Panel I: Crypto, Blockchain, and the Future of the Regulatory Landscape (2:40 – 4:10 PM)
This panel will analyze the current regulation of crypto and blockchain technologies and discuss whether they are adequate in their present state or require adjustment. Panelists will discuss what role both lawyers and the tech industry will play in shaping this new landscape. The panel will bring together a prominent SEC commissioner, an attorney and author of "Digital Assets and Blockchain Technology: US Law and Regulation" a first of its kind text book for law and graduate students, the CEO of Luminous Group a blockchain technology, growth advisory, and risk management solutions company, and a University of Arkansas School of Law professor who has published three scholarly articles on cryptocurrency regulation and acted as faculty advisory for the Arkansas Law Review Symposium on crypto.
Caroline Bradley, Professor of Law and Associate Dean for International and Graduate Programs at the University of Miami School of Law (Moderator)
Daniel Stabile, Partner at Shutts & Bowen LLP; Vice Chairperson of the Miami-Dade County Cryptocurrency Task Force; Adjunct Professor at the University of Miami School of Law
Carol Van Cleef, Chair of the Bradley Arant Boult Cummings LLP's Blockchain and Digital Assets Practice; CEO of the Luminous Group
Carol Goforth, University Professor and the Clayton N. Little Professor of Law at the University of Arkansas School of Law
Hester Peirce, U.S. Securities and Exchange Commission Commissioner
Panel II: Big Tech and Antitrust (4:20 – 5:50 PM)
This panel will focus on the ongoing confrontation between big tech companies and antitrust regulators and what relationship they should have going forward as big tech companies move into smaller tech hubs such as Miami. The panel will feature individuals situated on both sides of the discussion, those who favor increased regulation as well as those who support a free-market approach. Panelists will include the vice-president of the American Antitrust Institute, a director of the Classical Liberal Institute and guest lecturer at the University of Chicago, as well as a professor of economics at the University of Utah, an empirical labor economist by training, and his research investigates the existence and implications of employer power in labor markets, with applications to antitrust.
Hannibal Travis, Professor at the Florida International University School of Law (Moderator)
Marshall Steinbaum, Assistant Professor of Economics at the University of Utah; Senior Fellow in Higher Education Finance at Jain Family Institute
Laura Alexander, Vice President of Policy at the American Antitrust Institute
Sanjukta Paul, Assistant Professor of Law, Romano Stancroff Research Scholar at Wayne State University Law School
Wilfredo Fernandez
Panel III: Tech and Venture Capital Investment (6:00 – 7:30 PM)
This panel will look at the relationship between tech start-ups and VC investors, as well as the role lawyers play in supporting and shaping this industry. Panelists will discuss the role lawyers play currently and what they can do in the future to build a better environment not only for the incoming tech companies but for the new communities these companies are built in. Panelists will include a Greenberg Traurig co-managing partner who led the Miami office to become a leader in representing angel investors and technology companies, a partner at Morris Nichols focusing on corporate law and VC and guest lecturer on the topics at the University of Pennsylvania School of Law on venture capital, as well as Duke Law's director of law and entrepreneurship who has advised tech companies and other start-ups over the last thirteen years.
William Bratton, de la Cruz/Mentschikoff Chair in Law and Economics and Senior Lecturer at the University of Miami School of Law; Professor of Law Emeritus at the University of Pennsylvania Carey Law School (Moderator)
Jaret Davis, Co-Managing Shareholder of the Miami office of Greenberg Traurig, P.A. and member of the firm's global Executive Committee
Eric S. Klinger-Wilensky, Partner with Morris, Nichols, Arsht & Tunnell LLP; Lecturer in Law at the University of Pennsylvania Law School
Erika J.S. Buell, Clinical Professor of Law and Director of the Program in Law & Entrepreneurship at Duke University School of Law
Closing Keynote Address (9:30 – 10:30 AM)
Felice Gorordo is CEO of eMerge Americas, LLC, a venture-backed platform focused on transforming Miami into the tech hub of the Americas and organizing the largest tech event connecting the Americas held annually at the Miami Beach Convention Center. In 2019, eMerge's sixth annual conference attracted more than 16,000 attendees and 3,900 unique companies from over 40 countries. Felice is also a venture partner at the private equity fund I Squared Capital, and advisor at REEF Technology.
Felice is a successful entrepreneur, investor, and advisor. Most recently, he was President & CEO of Clearpath, Inc., (acquired by L1BRE, LLC), a venture-backed tech company focused on revolutionizing the confusing, costly and paper-based immigration filing process – just as TurboTax transformed tax filing. Following the acquisition of Clearpath, Felice served as CEO of L1BRE, a venture-backed mobility tech company with operations in the U.S. and Mexico, and Global Director of StartUp Health, a digital health venture fund and accelerator investing in "health moonshots."
Prior to that, Felice was appointed by President Barack Obama as one of fifteen 2011-2012 White House Fellows and served in the White House Office of Public Engagement and Intergovernmental Affairs working on Latino, immigration and business outreach. In this capacity, he organized the first-ever "White House Conference on Connecting the Americas," which included the participation of then-Secretary of State Hillary Clinton, several cabinet secretaries and high-level business and government leaders from throughout the Americas in lead up to the 2012 Summit of the Americas.
Felice also co-founded and leads Roots of Hope, Inc., a non-profit focused on youth empowerment in Cuba through technology and entrepreneurship. Felice worked on the re-establishment of U.S.-Cuba diplomatic ties, and served as an advisor to Grammy-winner Juanes to help organize the historic Peace without Borders concert in Cuba, which resulted in 1.2 million in attendance in 2009.
In the Administration of President George W. Bush, Felice served in various positions at the U.S. Departments of Commerce, State, and Homeland Security, including as trade specialist managing a cross-sector portfolio valued at $1.5 billion in export content.
Previously, Felice served as senior fellow at the Georgetown University Beeck Center on Social Impact and Innovation, chair-level Member of the Finance Committee for the 59th Presidential Inaugural Committee, and advisor to President Joe Biden's Cancer Initiative.
PODER Magazine has named Felice one of the "100 Poderosos" Most Influential Hispanics in the U.S. and he has appeared as a guest commentator on CNN, CNN en Español, BBC, NPR, CNBC, NBC Nightly News, Univision, and Telemundo. His views and articles have also been published in Forbes, The Economist, The Washington Post, The Wall Street Journal, The New York Times, The Miami Herald and el Nuevo Herald.
Felice serves as a Lifetime Member of the Council on Foreign Relations, and serves on the Latino Victory Fund National Committee, and the boards of the American Business Immigration Coalition and the Immigration Partnership and Coalition Fund.
Felice is a graduate of Georgetown University and started his career working in President Bush's White House Office of Political Affairs.
Fireside Chat: The Pros and Cons of the Technology Sector on Local Communities (10:45 – 11:45 AM)
This fireside chat will look at the pitfalls previous tech cities, like Silicon Valley, fell into and how new tech cities, like Miami, can avoid them. The roundtable participants will discuss issues such as housing inflation, gentrification, and homelessness caused by the success of a growing tech industry and what regulations, or changes can be implemented to avoid them in Miami
Marcia Weldon, Director of the Transactional Skills Program, Faculty Coordinator of the Business Compliance & Sustainability Concentration, and Lecturer in Law at the University of Miami School of Law (Moderator)
Leigh-Ann Buchanan, Founding Executive Director of aīre ventures; Founder of the Nyah Project
Felice Gorordo, CEO of eMerge Americas, LLC
Panel IV: Role of Criminal Punishment in Tech Regulation (12:00 – 1:30 PM)
This panel will look at the role criminal punishment plays in preventing harmful actions from tech companies and what new approaches could be taken to have a more effective role in regulation. For larger tech companies like Facebook, Amazon, Apple, or Google criminal punishments, while relevant, are a mere drop in the bucket when compared to the annual revenue they generate. Comparatively, start-ups are less likely to be able to afford any such penalty. Thus, a key question to discuss is whether the current regulatory framework disproportionately stifles new entrants into the market, creating a barrier to entry, which effectively allows larger companies to protect a growing monopoly. Panelists will include an adjunct professor at the University of Miami School of Law teaching white collar crime, as well as a federal prosecutor.
Tamara Lave, Professor of Law at the University of Miami School of Law (Moderator)
Sajjad Matin, Assistant United State Attorney
Kendall Coffey, Partner, Coffey Burlington


Make copies of this letter and send it to all known and suspected financial institutions, including banks, brokerages, private companies, and related institutions that are known to have the Decedent's accounts or might have them. If you are dealing with multi-branched or multi-state organizations, consider sending duplicate requests to any known offices Decedent may have dealt with (e.g., Merrill Lynch might have an account in Boston; but you know he also lived in Sarasota, Florida for part of the year; you should probably send a letter to the Sarasota office too, just to be prudent and exercise your proper duties as a fiduciary). This seemingly straight forward task can be very productive for estates in uncovering unsuspected assets such as old retirement accounts, stocks in private companies not otherwise known, and the like.
In addition, as the above letter suggests, you should make a list of financial people the Decedent dealt with. Since these kinds of people move around a fair amount in their career, finding out who they are, and hopefully conversing with them, may lead you to some hidden asset treasure at the companies these people left or are currently employed by. Open-ended discussions with them often bring unexpectedly positive results. In addition, speaking with some old pals of the Decedent can also give you a sense of how he or she dealt with his or her financial affairs (e.g., orderly, disorderly, tight, loose, speculative, and so on and so on). These conversations may also lead you into areas, hobbies, or other avenues where Decedent may have parked some money or investments that, over time, may have brought results unexpected by Decedent, but now productive for the Estate. Be sure to maintain a written log of all phone conversations as well as keeping track of written correspondence. A well-written log can be consulted as you go along to inspire more thoughts about old contacts as you receive new information.
Make multiple copies of this letter. Contact family members, especially heirs, for any suggestions on what avenues to pursue. Brothers and sisters are of particular help because they may know more about what the Decedent did with his or her assets in their early days and, if the Decedent's spouse is also deceased, may know more about the assets he or she may have had or would have been likely to influence Decedent to have and to hold (e.g., stock in his or her brother's company).
Be sure to include the self-addressed stamped envelope with your letter as well as following up by phone. Once again, follow-ups here can be very productive for an estate. Include a copy of the "Letters of Appointment" or "Letters of Administration," if applicable, with each set of correspondence you send out. The more official your documents look, the more productive you will be in your efforts.
As a practical matter, the larger the estimated Estate, the more worthwhile the above activities that are of an investigative as opposed to administrative nature. Also, the more the Decedent was inclined to be speculative or entrepreneurial, the more apt they would have been to have assets in unexpected places.


Comp.compilers: Re: Lemon, fallback, and vacuous tokens.
Re: Lemon, fallback, and vacuous tokens.
Error recovery has been studied and numerous papers written on it.
that allows the parse to continue as the recovery method.
pioneered and been considered succcessful.


"I'm going to go to my room" I tell tom not knowing what to do.
I stand up and start walking to my room, tom starts walking behind me. I thought he would stop at his room but he follows me to mine.
"I know" he sighs "I guess we are more alike than I thought"
We stare at one an other not knowing what to say.
I smile a little happy we are connecting.
He closes the door behind him, I walk up and open the door to see if he locked me in and he didn't which makes me happy.


text:
Q: php - Add items to a cart from remote address its a shopping cart site .
 i am saving items in cart by user ip address .
by this
$_SERVER['REMOTE_ADDR'];

If 2 users using the website on same ip address then cart mix the item .so tell me what is the alternate method.
session or temp values something .
  and other login cart item shift to the user account.

A: $_SESSION is what you're looking for. The customer IP is not a good idea. I'd recommend using sessions. You can set the "cart id" in the user session OR have the session id in your cart data structure.
This way you'll know which user is the "owner" of which cart.

redpajama_set_name: RedPajamaStackExchange


text:
Home » Life » Our Languages Abroad
19 May Our Languages Abroad
Posted at 13:34h in Life, Life Experts	by	Carrie Frais 0 Comments
When we move abroad, our languages move with us. Many parents wonder about their children becoming multilingual, or engaging in new ways of being multilingual. I did, too. My experience of multilingualism is both academic, from extensive research and teaching, and personal, as a parent of three trilingual children in a mixed family who has lived in six different countries and two continents. The headings in this article summarise some of the concerns that fellow globe-trotting parents have expressed to me over the years, about our languages abroad.
My children are too old to learn new languages
Parents may hesitate to accept a posting abroad on the grounds that their children have grown past the best before date, as it were, to learn languages. Despite widespread belief that age plays a decisive role in language learning ability, there is no evidence to support this claim. Our habits, linguistic ones included, may well be old, but our ability to break them is not, if we so wish. We can all find the motivation to learn whatever we need to learn in order to do, and enjoy, what matters to us.
One very appealing reason for children to wish to learn a new language is that their friends in a new country use it. Socialising with peers, wanting to understand them and to gain access to their circles is all the motivation that's needed. One added incentive is, of course, that the children will soon notice that they are outperforming their parents' proficiency in this new language, since mum and dad are less likely to become as immersed in undertakings mediated by local languages.
My children are too young to learn new languages
In multilingual and monolingual families alike, parents wish to spare their little ones an additional language 'burden', as many express it. Parents raising children multilingually from birth may baulk at the prospect of nurturing their toddlers in a country where none of the home languages are used. They report, for example, that the children's languages are already developing unevenly: words are late or missing altogether in one or more languages, or the children mix their languages, all of which they take to be signs of linguistic confusion and/or delay. They worry that additional languages will only compound whatever damage might already be there.
The facts are first, that multilingualism has no correlation whatsoever with language delay, which is a clinical condition affecting monolinguals and multilinguals alike. The languages of a multilingual develop differently because they are there to serve different purposes: one to talk to mum, say, another to dad, yet another to friends or to discuss outings, hobbies, homework. If multilinguals were able to use all of their languages in exactly the same way, they wouldn't need more than one language: they would be monolinguals instead. Depending on the children's use of their languages, and exposure to them, certain words will naturally appear sooner, or later, or not at all, in one language or another. And second, the facts are that language mixes are typical multilingual behaviour, not tokens of language confusion. A new language won't aggravate a 'problem', because there is no problem: multilingualism is not about what languages can do to people, it's about what people can do with their languages.
The new language will find its place alongside the ones that are already there, and wil develop in the same multilingual way. Parents raising their children monolingually, in turn, wonder whether it is wise to expose children who haven't yet acquired any language to a different one. They would prefer to 'give' one language to the children first, or wait until that language is 'in place', as they put it, before introducing a new one. Questions such as these relate to the belief that without at least one 'good' language from the outset, the children's overall cognitive development may suffer. The answer is that languages don't need to be acquired one after the other: simultaneous multilinguals are evidence of that. The answer is also that all the languages of a multilingual are good languages, because cognitive, social and emotional development draws on the whole of our linguistic repertoire, whether monolingual or multilingual.
My children will have to attend a local school, but they don't know the local language
Before the move, if time allows, tuition in the children's new language may be arranged in the home country. In the case of impending moves, contacting the new school beforehand for advice and/or language tuition upon arrival might be a good idea. Once in the new country, however, the issue of being taught in a language that one is learning at the same time soon vanishes. The children's new language will take off on its own, given their daily exposure to it, the new routines that involve it and, not least, the new friends that use it.
In the first few months, the children are likely to have trouble understanding what is going on in the classroom, at all times. But then again, who doesn't? We all remember this kind of trouble in school, and not necessarily because the school language was new to us. Schoolchildren, whether born in the country of schooling or newcomers to it, are learning to use a language in a new way, for school purposes. This means that they are also learning the language, by learning how to do new things through it.
Languages are tools, that we use to do other things with. We learn to use tools by using them, and the more we use them, the better we get at using them. The way to make a language ours is to use it, be it to attempt to express ourselves or to say that we can't do this yet, to share ignorance and knowledge through it, to be told off and praised in it, or to ask questions, and ask, and ask again, in it and about it. Which is, incidentally, exactly how babies learn their native language(s).
My children are struggling with the school language, so we will have to start using it at home
Parents who consider making this decision themselves feel at least somewhat comfortable using the school language. This may not be true of parents who are urged, by their children's school and/or teacher, to switch to the school language at home. In the latter case, one might want to ask about which mechanisms the school and/or teacher have in place to teach that language to the parents, up to which standards, and within which urgent time frame. The first thing to bear in mind is that using a language in one setting does not promote its development in other settings. Switching to the school language at home won't solve school problems with that language. Home and school are distinct environments involving distinct uses of language, which do not transfer across each other. Switching to the school language may also defeat the parents' goal of raising their children multilingually, or bar the children's access to grandparents, little cousins, and other big and small relatives and friends back home.
Using the school language at home, however, is different from switching to it. Parents can implement linguistic division of labour in the home, using the school language to assist with homework, for example. Apportioning linguistic territory in this way is typical of multilingual families, and in fact no different from learning, as we grow up, that different uses of language associate with different purposes and with different people (school, home, peers, elders), whether we're monolingual or multilingual.
My children have started mixing their languages
Mixed language, defined as the use of features attributed to one language (words, grammar, intonation) in another, is widely believed to be exclusive to multilinguals, although monolinguals use it, too. The difference is that monolingual mixes go by rather more benevolent labels like 'borrowings': words like bungalow, robot, typhoon, for example, or captain, mutton, prince, aren't indigenous to English but are nevertheless said to be English words. The observation that multilinguals mix their languages, often implying a deviation from some 'proper' linguistic expression, in fact describes typical multilingualism. Rather than evidence of deficient language command, language mixes provide evidence of proficient multilingualism.
Among very young multilinguals, reasons behind mixes are best sought in the children's motor immaturity. Beginner vocal tracts, or hands, need a long time and much hard work to acquire the sophisticated gymnastics required of fluent speech, or sign. When faced with, say, a difficult word in one language, young children naturally choose an alternative word that fulfils the same purpose but that they can manage more easily. So do all of us when learning a foreign language: children's first language(s) are as foreign to them. Very young children have no idea that they are learning what adults call 'languages', or 'different' languages – or that adults worry about such things. All they want is to get things done for them, and they do this their way.
Among older multilinguals, mixes relate to different languages having different personalities, just like people. Words or expressions in one language, rather than another, may strike us as more appropriate to what we want to say. One typical scenario is this: the children have just come home from school, and want to report their school day. The language of choice in which to do this is the school language, since the school day took place in it. The children are expected to use the home language instead but, to them, the home language lacks school-bound vocabulary and grammar. So one solution is to fill those gaps in the home language with bits and pieces of the school language.
My children refuse to use our home language(s)
One reason for this may be developmental. Very early in life, we gain awareness of ourselves as individuals, and of the world around us as independent from us. The overwhelming consequences of this realisation are perfectly expressed in the popular label 'The Terrible Threes'. From our parents' perspective, one of the terrible things that happens to us is our discovery that our peers are infinitely more fun to engage with than those oldies at home whom we'd so far regarded as the only available source of well-being. From around age 3, in other words, our children will do as we did before them, and start favouring peer company and adopting peer models. This naturally includes linguistic models, and this is why school languages are not only a powerful magnet, but a largely irresistible one.
Refusal to use the home language(s) may, alternatively, reflect the children's perception that those languages are there just to serve boring purposes like urging them to behave, wash their hands or go do their homework, or asking and answering perfunctory questions like "Can we eat now?" or "How was your day?" Choosing not to use a language may also be evidence of the Law of Least Effort, rather than a matter of refusal: if the children can get away with using a single language, because they know that their parents understand it and/or respond to them in it, why bother using several?
One way of retaining the appeal of home languages is to use them with our children in what engages them: films, music, theatre, games, books, outings. New treats experienced in the home languages may do the trick, or telling the children what our day was like, what they think of it and why, or that we'd love to cook together their favourite home recipe for dinner today.
Spoken and/or written interaction with relatives and friends back home may also help, including online, on email, on social networks. In short, anything that guides the children into (re)discovering a niche for their home languages, and that strengthens their perhaps dwindling command of them, is a good thing. Anything but language 'teaching': children learn best when they're not being taught, and the best language lessons don't target the languages themselves.
My children use only the school language among themselves now
Peer appeal among siblings is the likely reason for this development. Parents may feel despondent that 'their' languages are losing ground to a new language at home, but multilingualism isn't a zero-sum game where a perceived gain somewhere must match some loss elsewhere.
Parents may well be the initiators of language policies at home, but our children's uses of the languages around them develop a life of their own that we can neither predict nor aspire to control. In addition, if we parents have our languages, why can't our children have theirs, too, since they go on using our languages with us? Sibling bonding, including linguistic bonding, is best managed by the children themselves. It's then up to us parents to choose to treat this new language as an intruder or a welcome guest to the family.
The children may revert to using their home language(s) among themselves, for similar reasons of peer appeal, once the family returns home and schooling resumes or starts in a home language. Or they may retain their new language as their common language – time will tell. Home language policies can't be set in stone, because language uses evolve with us and for us. This is true of families who stay in place, and all the more so of families on the move.
Madalena Cruz-Ferreira is a multilingual scholar, educator and parent. She is Portuguese, married to a Swede, and based in Singapore. Her book Three is a Crowd? reports the trilingual development of their three children. She holds a PhD in Linguistics from the University of Manchester, UK, and her blog Being Multilingual deals with child and adult multilingualism in the home, in school, and in clinic, which has been her research topic for the past 40 years. She is on Twitter @MadalenaCruFer.
Moving Abroad with Kids, Multilingualism
Blog MumAbroad Life
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I'm going to be visiting Bangkok for one week, and would like to have a contingency plan in case the hotel WiFi is unusable. What are my options for a data only SIM card for one week?
Prepaid SIM card with data-only plan in Slovenia?
I will be stationed in Germany but will be traveling to Switzerland, Italy and Greece. Any ideas on a good plan that will allow me to use my iPhone for calls and data while traveling in Europe?
What are the options for a prepaid SIM card in Australia?
I will be visiting NZ this coming January and I would like to know if I will be able to get a SIM card that will work in my Blackberry Q10 while I am there?
Which carrier has extensive cellular coverage in Sicily?


We're often surprised by our credit card statements, and not just because of how much we're charging. The balance amount always seems to be a number that we wouldn't have guessed, and some purchases that we expect to see aren't on the statement, and some that we don't expect to see are listed. This is because of timing and the credit card billing cycle.
When we receive our credit card statement, the bank or Credit Card Company has taken a snap-shot of our account. The amounts are frozen in time on the statement date when the snap shot was taken, but we're actively using the card (account). This is important because the timing of payments and purchases change the balance throughout the billing cycle. The statement is one of those pieces of information that is somewhat obsolete the moment its' received. We have all had a purchase show up on our statement sooner than we expected, or experienced a delay in billing. This can make managing our finances difficult from time to time.
As an example, if I receive my statement dated June 1st, and my credit card balance is $1,000.00, this is a snap-shot in time of the account. It shows the balance on that date (June 1st), a list of transactions made during the period up until that date, the payment that I made last month, any interest charged, a minimum payment due, and the date that my payment is due. The statement shows how the account looked on June 1st from the credit card company's perspective. If I made a purchase the week before June 1st from a company that doesn't bill until they ship the merchandise, and they ship it June 2nd, then it would not appear on the statement or affect the account balance on the statement that I received.
For the example, let's say that the Payment Due Date on this account is June 15th. This is when my payment on the account needs to be received, but it's important to note that from June 1st (the day of the snap-shot) until June 15th, any charges that I make are affecting the account balance, and I wouldn't see these transactions unless I looked at the account online. Let's say that the day I decide to mail the payment (yes, I still mail some of them) is June 9th.
It was $1,000.00 on my statement, I'm mailing a $500.00 payment (but they haven't received it yet), and I've made some purchases ($225.00). The account balance will be $1,225.00 if all of the charges have hit the account.
So, the $500.00 payment that I'm sending (thinking I'm paying half the balance) will be applied to the new $1,225.00 balance when the company receives my payment. But wait, my billing cycle ends on the last day of the month. The snap-shot was taken on the 1st and the next one won't be taken until the first day in July. It's only the second week in June. There will be more changes to the account before the next snap-shot.
Time goes by and it's now June 20th and again I need gas for the car, and some toner for my home printer. I charge another $198.00, and the month (my billing cycle) isn't over yet. It ends on July 1st. In the next ten days (before July 1st) I end up charging another $80.00, which brings my total charges for the month (the billing cycle) to $503.00. How do things look now?
In July, I get a new statement (snap-shot in time) dated July 1st.
Wow, I thought that I paid half of my credit card balance. I've forgotten that the snap-shot balance is always in the past, and that I'm adding to it as I charge. My payment in this case doesn't really put a dent in the balance. In fact I'm losing ground. The new balance is higher than last month's.
When I receive my statement, I'm making a payment based on the balance shown, but setting a specific payment amount based on what I really charge on the card would be much better. Using the scenario above, if I budget a $650.00 payment each month for this credit card, and I hold my charging to $500.00 each month, then I will eventually eliminate the balance.
There are two things that will make this plan work. First, the maximum that I charge for the month must be the budgeted amount (in this case $500.00). Second, I have to maintain the $650.00 payment each month. Then regardless of the billing cycle, my payment will keep pace with my monthly charging, and at the end of seven months when the balance is eliminated, I can lower the monthly payment to $500.00 (the budgeted charging amount). This will free up $150.00 to apply to another area of my finances.
We can eliminate some financial surprises by understanding the cycles of banking and credit cards, and including them in our financial plans. There will always be months when things don't quite add up, but we'll know what's going on and how to get things back in line with our financial plans.
This entry was posted in Debt, Personal Finance and tagged Credit Cards, debt, Finance, interest. Bookmark the permalink.


text:
DETAILS: 87 minutes, featurette
AUDIO: English Mono
Subtitles: English, French, Spanish
STUDIO: MGM/ American International/ 20th Century Fox Home Video
I AM LEGEND is a classic novel from writer Richard Matheson. It has been adapted into film three times. The third film adaptation, I AM LEGEND starring Will Smith, hits theaters on December 14th. Richard Matheson has felt no one has done his book justice for film. Charlton Heston fought albino vampires in the 1970s OMEGA MAN (the second film adaptation). Arnold Schwarzenegger almost starred in a film version called AMERICAN LEGEND to be directed by Ridley Scott (BLADE RUNNER, BLACK RAIN, ALIEN, KINGDOM OF HEAVEN) before Warner Brothers nixed the film before it was to begin production.
I have read I AM LEGEND. It is a fantastic novel. I imagined it would have been perfect for actor Rutger Hauer. The first film adaptation (THE LAST MAN ON EARTH) was the result of a script by Richarad Matheson. It was rewritten (as in stndard Hollywood practice) and Richard disowned the film. he had his csreen credit changed to Logan Swanson becaause he had four kids to support (and couldn't have his name completely taken out of the credits which would result in losing future royalities). THE LAST MAN ON EARTH comes off too much as a typical Roger Cormanesque vampire film which retains some of the elements of the book only to be executed badly. Vicent Price as Dr. Robert Morgan can't even save the film from it's shortcomings. When he meets another survivor, the relationship between him and Ruth is shortchanged to get viewers to the climax. One of my favorite parts of the novel is the last man's relationship with the woman he finds.
The print used for the transfer does have some slight damage, but image detail isn't that bad.
SPECIAL FEATURES: The only extras is a short interview featurette called RICHARD MATHESON STORYTELLER; THE LAST MAN ON EARTH which has Richard talk about writing the novel and the screenplay for this film.
FINAL ANALYSIS: THE LAST MAN ON EARTH is good at times, but it doesn't do the novel justice. I am eagerly awaiting the upcoming I AM LEGEND.
this DVD review is (c)12-7-2007 David Blackwell and cannot be reprinted without permission. send all comments to lord_pragmagtic@hotmail.com
look for additional content at http://enterline-media.tripod.com and http://enterlinemedia.livejournal.com


Ignacio A. Donoso, Esq., is nationally recognized as a leading business immigration attorney, with particular expertise in visas for companies expanding operations to the U.S., start-ups and entrepreneurs.
Mr. Donoso founded Donoso & Associates in 2013, after previously working from 2010 to 2013 as Partner of FosterQuan, LLP (now Foster Global, LLP), one of the top immigration law firms in the U.S.
Mr. Donoso's career spans 18 years as a member of the State Bar of California, of which he has dedicated the last 13 years exclusively to immigration law, including business immigration, family-based immigration and removal defense. In early career, Mr. Donoso advised Fortune 100 companies on international transactions and mergers. Since 2003, Mr. Donoso has represented Fortunate 500 companies and their employees, skilled workers, professors and researchers, professionals, athletes, artists and thousands of investors in their U.S. visa planning.
Mr. Donoso has been selected to serve in many AILA National Committees including the Texas Service Center Liaison Committee, the National Annual Conference Planning Committee, and since 2013, Mr. Donoso has served on the National EB-5 Committee.
Mr. Donoso is a frequent speaker at national and international conferences on U.S. immigration law, and has frequently published on specific issues related to immigration matters. He has been interviewed on national news media including Fox Business and the New York Times, as well as local television and radio in relation on immigration issues.
Adjunct Professor, International & Comparative Law, University of Barcelona, 2002, and Visiting Professor, Universidad de los Andes 1998.
America Jewish Committee (Houston), Board of Directors, 2006 to 2009.
Frequently Interviewed on national news (e.g., Fox Business; New York Times) and local television on immigration issues.
David M. Morris is Of Counsel to I.A. Donoso & Associates, LLC. David is a nationally known business immigration attorney with over 20 years of immigration law experience. He is renowned for his representation of clients in the areas of investment, life sciences, and information technology. He is regularly invited to speak on immigration law issues at national conferences, is interviewed regularly by national media including MSNBC, and his articles have been published in leading professional journals including The Labor Lawyer and the American Bar Association's Labor and Employment Journal. David has served for many years as Chair of the National EB-5 Committee of the American Immigration Lawyers Association.
Julie Sheehy is a Legal Assistant with I.A. Donoso & Associates, LLC. She received her Bachelor of Science degree in Legal Studies from St. John's University in New York. Julie has more than seven years of business immigration experience.
Maria Cabrera is a legal assistant with I.A. Donoso & Associates, LLC. She received her B.A. in Political Science from the University of Maryland, Baltimore County. Maria is fluent in English, and Spanish and has basic proficiency in French.
Benjamin Hu is partner with I.A. Donoso & Associates, LLC. Benjamin Hu has experience in employment-based immigration, family-based immigration, and EB-5 investor immigration. Before joining I.A. Donoso & Associates, LLC, he was Of Counsel attorney to a law firm specializing in U.S. immigration law. He has been practicing law since 2009, and is licensed in Maryland and the District of Columbia. He has lived in the United Kingdom, mainland China, and the U.S. He is fluent in English and Mandarin Chinese, and is proficient in German and French.
Dennis Tristani is partner with I.A. Donoso & Associates, LLC. Dennis has experience in employment-based immigration, family-based immigration, removal defense and EB-5 investor immigration. Before joining I.A. Donoso & Associates, LLC, he was an associate attorney at WMR Immigration Law Group in Washington, D.C. He received his B.A. from Yale University in 2007, and his J.D. from the American University Washington College of Law in 2012. He has been practicing law since 2012, and is licensed to practice in Maryland. Dennis is fluent in Spanish and speaks Portuguese.
Rongrong Pan is a Senior Attorney with I.A. Donoso & Associates, LLC. Rongrong Pan has experience in employment-based immigration, family-based immigration, and EB-5 investor immigration. Before joining I.A. Donoso & Associates, LLC, he was an attorney with Visa Law Group PLLC in Washington, DC. He received his B.A. from Peking University, China in 1999, and his J.D. from the University of Maryland Francis King Carey School of Law in 2011. He has been practicing law since 2012, and is licensed in Maryland and the District of Columbia. He is fluent in English and Mandarin Chinese.
Jessica Chi is an Associate Attorney with I.A. Donoso & Associates, LLC. Jessica received her B.A. in Political Science and BA in Sociology from the University of California – Los Angeles (UCLA) and received her J.D. from the Washington and Lee University School of Law, where she was a Senior Articles Editor for the Journal of Energy, Climate, and the Environment and a Student Attorney at the Immigrant Rights Clinic. Jessica is licensed to practice law in the state of Virginia. Jessica is fluent in English and Mandarin and understands Taiwanese.
Amara Tek is a Legal Assistant with I.A. Donoso & Associates, LLC. Formerly, Amara worked as a Legal Assistant with Visa Law Group, an immigration law firm based in Washington, DC. Prior to working in the immigration field, Amara worked with survivors of domestic violence and sexual assault in the Asian/Pacific Islander communities. In this capacity, she provided direct services, training and technical assistance, and community outreach. She has a B.A. in Communication and a minor in Asian American Studies from the University of Maryland, College Park. She has basic proficiency in Cambodian.
Amy Lui is a Legal Assistant with I.A. Donoso & Associates, LLC with extensive experience in employment-based immigration, family-based immigration and EB-5 visas. Prior joining the firm, she was a Senior Paralegal with Visa Law Group in Washington, DC. She received her LL.M from the University of Pittsburgh and is licensed to practice law in Taiwan. She is fluent in English and Chinese, and has basic proficiency in Cantonese.
Sara Lindstrom is a Legal Assistant with I.A. Donoso & Associates. She has fifteen years of experience in business and family-based immigration law, including twelve years as a Senior Paralegal with Visa Law Group, and three years with Greenberg Traurig. Prior to that, Sara started her career at George Mason University, where she served as an International Student Advisor and Assistant Director of the Office of International Programs and Services, assisting in the management of the university's F-1 and J-1 visa programs for students and scholars, and organizing and presenting various programs and events. Sara has a Bachelor's degree in Communication, and a Master's degree in English (nonfiction writing) from George Mason University. She is fluent in Swedish, and has basic proficiency in Spanish and French.
Raquel Katz-Cohen is a Legal Assistant with I.A. Donoso & Associates. Before joining I.A. Donoso & Associates, Raquel worked fifteen years as Senior Legal Assistant and Legal Associate at the Inter-American Investment Corporation, an international organization in the IDB Group, where she gained extensive knowledge in loan and investment projects for financing private enterprises. Raquel received her Paralegal Certificate from Graduate School USA. Raquel has lived in Uruguay, Israel and the U.S. and is fluent in Spanish, English and Hebrew and proficient in Portuguese and French.
Kui Yang is a Legal Assistant with I.A. Donoso & Associates with extensive experience in EB-5 visas. Ms. Yang completed my Master of Laws (LL.M.) from Temple University in Philadelphia in May, 2012. Before moving to the U.S., Ms. Yang completed a Master of Laws Degree from Capital University of Economics and Business, (Beijing) in January, 2002. Ms. Yang previously worked as a judge in the People's Republic of China for 9 years. Ms. Yang is fluent in English and Mandarin Chinese.


text:
MERLIN's Colin Morgan Delivers New Magic in Season 3
By Jenna Busch, Newsarama Contributor December 17, 2010 10:28am ET
MERLIN Colin Morgan Delivers New Magic
Merlin is back! Season 3 premieres on the SyFy channel on January 7th at 10/9c and we got a chance to chat with star Colin Morgan. When we last left our magical friend, he found out that his father was a dragon lord, and the king's ward and fellow magic user Morgana (Katie McGrath) disappeared. Guess who the big bad is this season? Morgan talked about where we left him in Season 2, gives us some secrets from Season 3 and explains a pretty darn funny YouTube video that's been making it's way around the net. Note: For the purpose of clarity, I've replaced the UK 'series' with the US 'season.' Oh, and in case you were wondering, that little tidbit on Morgan's IMDB page, saying he's allergic to tomatoes and had to wear protective cream during scenes in the stocks? Not true.
Newsarama: Is it weird to be talking about Season 3 when it's already aired in the UK?
Colin Morgan: Yeah, it's really interesting looking back on the whole series. It's been another huge step for all the characters. So much has changed and I know it's just starting over in the states. It is weird revisiting it, but it's also been a learning experience to look back on it and moments that you've missed. It's like talking about good memories.
Nrama: In Season 2 you fought trolls and warlocks. You found out you were a dragon lord. So what moments can we look forward to in Season 3?
Morgan: One of the big thing that changed for Merlin at the end of Season 2 was that he met a huge, important figure from his past, his father, who he lost. But he gained from him the power of the dragon lord, which was a big step, I think, for Merlin, both emotionally and in terms of his powers and his ability. So that's where we leave Merlin on his personal quest. But we start off the third season a year in the future so a lot has happened in that year. What exactly has happened, we're not sure. We only know that there has been a search for Morgana by Uther (Anthony Head) and when she does eventually turn up, she obviously throws a huge wrench in the works for Merlin. He considered her gone, but she's back and ready to wreak havoc, which she does in abundance in the second part of the two parter opener of the third season. And that's a very big, dramatic war and Lord of the Rings-style things happen in the opener, which is very quickly contrasted by some of the more comic episodes throughout the series, one of which, Richard Wilson who plays Gaius gets possessed by a goblin. [laughs] We have other romantic confusion going on when Arthur (Bradley James) meets a princess who is actually possessed by a fairy. Also a lot more dramatic things which are elements from Arthurian legends, including the Crystal Cave and elements involving a certain kind of table [laughs] which appears much later. So lots of big things happening and comic moments with the charm the show normally has, as well as big character changes.
Nrama: According to Twitter, everyone is desperate for Merlin to hook up. Are we going to see any of that this season?
Morgan: [laughs] Well, I am too! Merlin had his first love in the second season with Freya, and she's very, very sadly lost, and I think it was a big, big blow to Merlin. There is some emotional things for Merlin coming up. But love interest-wise, it's pretty quiet. [laughs] He doesn't get a lot of love interests, or else no one is really attracted to Merlin. He's a lonely guy! [laughs]
Nrama: No! That's sad! But you have the relationship with Arthur...I'm only joking a little there.
Morgan: [laughs]
Nrama: What's the relationship with Arthur like this season?
Morgan: It's great. They're still obviously in the master/servant relationship, so that's always going to be there. That's the structure of Camelot. It has been for years and that's the way it will be, but these guys have been through so much together. There's a friendship there as well. You're seeing them at a stage that you would anyone who has known each other for three years and been through what they've been through, except these guys have much more at stake. These guys are very much aware of what their friendship pivots on, which is that he must see Arthur fulfill his destiny, as Merlin wants to fulfill his destiny. So I'd describe it as a friendship with huge things at stake. Only Merlin is the only one who is aware of it. Again, they're forging ahead. And of course there is that normal charm there that you've got with the Arthur and Merlin scenes, the servant and master scenes, but it's more about them moving towards King and advisor, I would say, which is more like the legends of old.
Nrama: So, there is a particular video of you on YouTube, which is making the rounds called, 'You're the Voice.' I love it! Do you want to explain that?
Morgan: [laughs] Well, that was actually...that video appeared...we randomly did it one day. I don't know if anyone knows how the song came about. So I will tell you. [laughs] There is a film called Hot Rod, don't know if you've seen it, but it features that song. I was singing it on set one day and I was going, who sings that song? And everyone kind of knew what it was, but didn't really know what it was about or the words. So Bradley kind of researched it and proceeded to sing it quite a lot that day, and we thought it be quite funny to mime it on camera and put music to it, which someone did...so it seems it's taken off. [laughs] I've received countless videos from people with their own versions of it, including a whole primary school class who did a music video of it. [laughs] So it's really taken off! It's crazy!
Nrama: Is there any particular research you've done from an Athurian book you've loved that you'd like to see in the fourth season?
Morgan: One of the stories that I really loved from the research that I've done is the Crystal Caves, which actually featured in the third season. That was my favorite thing, this cave filled with crystals which show reflections of the future, which Merlin was able to use. We've only seen it affect him. In the fourth season, I'd like to see him perfect it and use it as a tool. So if that's something that can be revisited, since it was one of my favorite things that we've done from my research, that would be great.
What moments are you excited to see?
INHUMANS' Final Judgment Delivered In EWING & DEL MUNDO's JUDGMENT DAY
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IDW PUBLISHING December 2017 Solicitations


text:
\section{Introduction}

The purpose of the paper is to construct a noncompact complete Riemannian manifold of nonnegative Ricci curvature
with quadratically asymptotically nonnegative sectional curvature and infinite topological type;
this gives a negative answer to the problem proposed by Jiping Sha and Zhongmin Shen \cite{SS},
which asks {\it if a complete Riemannian manifold with nonnegative Ricci curvature
and quadratically asymptotically nonnegative sectional curvature is of finite topological type}.

In order to give a more precise statement, let's first fix some notation. Let $M^n$ be an $n$-dimensional
complete noncompact manifold, $p_0\in M$ a fixed point. By $B(p_0, t)$ (resp. $S(p_0, t)$)
denote the ball (resp. sphere) with the center at $p_0$ and the radius $t$. Set
$$
{\text{diam}}(p_0; t)=\sup_i{\text{diam}}(\Sigma_i, M\setminus B(p_0, {\frac 12}t)),
$$
where $\Sigma_i$ is a connected component of $S(p_0, t)$, ${\text{diam}}(\Sigma_i, M\setminus B(p_0, {\frac 12}t))$
denotes the interior diameter of $\Sigma_i$ in $M\setminus B(p_0, {\frac 12}t)$ when $M\setminus B(p_0, {\frac 12}t)$
is considered as a metric space with the induced metric and $\Sigma_i$ as its subset.
We also define
$$
K_{p_{0}} (t) = \inf_{M^{n} \setminus B(p_{0},t)} K,
$$
where $K$ denotes the sectional curvature of $M$, and the infimum is taken over all the 2-planes at the points
in $M\setminus B(p_{0},t)$. We say that $M$ is of {\it quadratically asymptotically nonnegative curvature} if,
for all $t\geq 0$,
$$
K_{p_{0}} (t) \geq -{\frac{K_{0}}{1+t^2}},
$$
for some positive constant $K_0$.
We also say that $M$ is of finite topological type if it is homeomorphic to
the interior of a compact manifold with boundary; otherwise, infinite topological type.
Then, the main aim of this paper is to show the following

\begin{theorem} There exists a $6$-dimensional complete Riemnannian manifold $M$
with a base point $p_0$, which is of positive Ricci curvature, quadratically asymptotically nonnegative curvature,
and infinite topological type. Furthermore, it satisfies
\begin{align*}
&\lim_{t\to\infty}{\frac{{\text{vol}}(B_{t}(p_0))}{t^6}}=0~~{\text{and}}~~{\text{diam}}(p_{0};t)=O(t).
\end{align*}
\end{theorem}

Earlier examples of positive (or nonnegative) Ricci curvature and infinite
topological type have been constructed by Sha and Yang \cite{ShY},
which have sectional curvature bounded below but diameter growth of order
$O(t^{\frac 2 3})$ (also see \cite{SW}); moreover, they have infinite second Betti number.
The examples of Sha and Yang \cite{ShY, ShY1} show that in the higher dimensional case ($\ge 4$)
one cannot expect similar results of Cheeger-Gromoll \cite{CG}
in the case of positive (or nonnegative) Ricci curvature.
In other words, in order to get certain finiteness result,
some additional assumptions (e.g. volume growth or diameter growth) must be required;
in this aspect, one can refer to
\cite{AG, SW, SS, LS, dx, sso, JY} and therein. As for the 3-dim case, complete noncompact Riemannian
manifolds with nonnegative Ricci curvature were completely classified recently by G. Liu \cite{L} based on
the technique earlier developed by Schoen-Yau \cite{SY}.

After the examples of Sha and Yang, Perelman \cite{P} gave a more general construction
of 4-dimensional compact examples with positive Ricci curvature, big volume and arbitrary
large Betti numbers. Menguy applied Perelman's construction to obtain noncompact examples
of infinite topological type \cite{M1,M2}. However, all these noncompact examples are
not of quadratically asymptotically nonnegative curvature though they have positive
Ricci curvature and infinite topological type. In particular, Menguy's examples \cite{M1} have
sectional curvature bounded below by $-({\frac{g(t)} {t}})^2$; here, $t$ is the distance function
to a fixed point and $g(t)$ is a function that is going to infinity as $t$ goes to infinity.

It is easy to get higher dimensional examples with the same properties by taking
a metric product of the above 6-dimensional example with the standard spheres.
Here, we should also point out that the examples with only quadratically
asymptotically nonnegative curvature and infinite topological type has been constructed
by Abresch \cite{A}; but these examples are not of nonnegative Ricci curvature in general.

In the final of this section, we give a brief description for the idea of the construction,
which mainly benefits from the works of Perelman \cite{P} and Menguy \cite{M1, M2, M3}.
We first begin with a doubly warped product metric
$$
ds^{2}=dt^{2}+u^{2}(t)(dx^{2}+f^{2}(t,x)d\sigma^{2})+g^{2}(t)d\theta^{2},
$$
where $d\sigma^{2}, d\theta^{2}$ are the standard metric of the 2-dim sphere $S^{2}$.
For the first part, $dt^{2}+u^{2}(t)(dx^{2}+f^{2}(t,x)d\sigma^{2})$, of the metric,
we assume that it is the spherical metric for $t_{i} < t < t_{i}+2r_{i}$, here $t_i+2r_i<t_{i+1}$
(for the choice of $t_i$ and $r_i$, cf. \S 2), and $u(t)=O(t)$.
Thus, the metric on this part is similar to the compact example of Perelman \cite{P},
which is a double suspension over a small 2-sphere so that one gets a singular 4-sphere
with the set of its singular points being a circle. Then, it allows us to remove the geodesic ball
$B_{\frac 45 r_{i}}(o_{i})$($o_{i}=(t_{i}+r_{i},0)$) and glue in a $\mathbb{CP}^{2}$
via a neck, provided with $f(t,x)$ almost like $R_{0}\sin x$ ($R_{0} \ll 1$)
and smoothed near $x=0$. We can continue this surgery when $i$ goes to infinity.
Then, we get a manifold with infinite second Betti number, which is of course infinite topological type.

Actually, this is exactly the constructing process of Menguy's Example with Euclidean
volume growth \cite{M1} (i.e. the order of growth is the dimension of manifolds).
However, the radius $r_{i}$ in Menguy's examples must be of order $o(t)$
(more precisely, $r_{i}=\frac{t_i}{h(t_i)}$, here $h$ is a certain function going to infinity
as $t\to\infty$), this makes the sectional curvature to
be bounded from below by $-({\frac{h(t)}{t}})^2$ for sufficiently large $t$, which certainly is not
quadratically asymptotically nonnegative. But the factor $h(t)$ here
is essential for $-\frac{u_{tt}}{u} \ge 0$ so that Ricci curvature can be
controlled to be positive. In our construction, we set $r_{i}=rt_i=O(t)$ to get
quadratically asymptotically nonnegative sectional curvature (here $r$ is a certain positive constant, see \S 2.1).
To ensure that Ricci curvature is positive, we take the second warped product part $\times_{g(t)}S^2$
and assume the order $\gamma$ of $g(t)$ ($=O(t^{\gamma})$) being sufficiently small,
so that we can control the curvature through some new directions.
Moreover, in order to be able to do surgery by gluing $\mathbb{CP}^{2}$ via a neck on
$(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})$,
we set $g(t)$ to be constant on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})$.
We here remark that since the first warped product on $[t_i, t_i+2r_i]$ is part of a certain sphere
(i.e. $u(t)={\frac{1}{\sqrt{K_i}}}\sin(\sqrt{K_i}(t-t_i+\psi_i))$, see \S 2.1), our choice for $r_i$
will not destroy the positiveness of Ricci curvature on this part;
so, we can assume that $g$ is constant on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})$.
Clearly, by means of Sha-Shen's finiteness result \cite{SS}, the manifold constructed here
is not of Euclidean volume growth.

\section{Construction of the manifolds}

We begin with a doubly warped product manifold with boundary,
$$
Q = [t_1,+\infty) \times_{u} S^3 \times_{g} S^2,
$$
for some $t_1>0$ and certain functions $u(t)$, $g(t)$ in $t\ge t_1$, all of which will be fixed later.
Note that $S^3$ here is not necessarily a round sphere.

First, taking a sequence of intervals $[t_i, t_i +2 r_i]$ ($1 \leq i < +\infty$) to be determined later,
we will let $u(t)$ be spherical on $[t_i, t_i +2 r_i]$ and $g(t)$ be constant
$g_i$ on $[t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}]$.
In the following, we'll see that our surgeries actually arise on
$(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})\times_{u} S^3 \times_{g_i} S^2$, $1 \leq i < +\infty$.

In fact, we have surgeries only on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})\times_{u(t)} S^3$.
More precisely, we remove a geodesic ball $B_{\frac 45 r_{i}}(o_{i})$
with the center at $o_{i}=(t_{i}+r_{i},0)$ (for the precise meaning of $o_i$, see \S 2.1)
on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}) \times_{u(t)} S^3$,
and glue a $\mathbb{CP}^{2} \setminus B(p_0)$ (called a {\it core}, topologically $\mathbb{CP}^2$ deleting a ball)
via $S^3 \times [0,1]$ (called a {\it neck}) along $\partial B_{\frac 45 r_{i}}(o_{i})$ with appropriate metrics depending on $i$
and the boundary properties on $\mathbb{CP}^{2} \setminus B(p_0)$ and $S^3 \times [0,1]$ (for details, see \S 2.2).
By $P_i$ ($1 \leq i < +\infty$) denote these glued parts (both the core and the neck).
Similarly, we also glue a core via a neck along the boundary $t=t_1$ to smooth the manifold near the origin.
We denote this glued part by $P_0$.

So, our manifold is actually
$$
M = (Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2))
\cup_{\textrm{Id}} \coprod\limits_{i=0}^{+\infty} (P_i \times_{g_i} S^2)
$$
where $g_0 = g_1$ and 'Id' means gluing along the boundary through the identity map.

Here, we need to point out that actually one should also do the same surgery
on $Q$ at $o'_{i}=(t_{i}+r_{i},\pi)$, i.e. remove a geodesic ball $B_{\frac 45 r_{i}}(o'_{i})$
(isometric to $B_{\frac 45 r_{i}}(o_{i})$) with the center at $o'_{i}=(t_{i}+r_{i},\pi)$
and glue with the $P_i$ along the boundary, since the Ricci curvature is also not necessarily positive
on $B_{\frac 45 r_{i}}(o'_{i})$. In fact, our construction of $Q$ in $x$ is symmetric at $x={\frac \pi 2}$. More precisely,
$f(t,x)$ will first be constructed for $x\in [0,\frac\pi2]$ and then symmetrically extended to $[\frac\pi2,\pi]$;
moreover $B_{\frac 45 r_{i}}(o_{i})$ and $B_{\frac 45 r_{i}}(o'_{i})$ are disjoint (for details, see \S 2.1.3).
Based on this, for the sake of simplicity, in the following we will only do the surgery on $B_{\frac 45 r_{i}}(o_{i})$.

$M$ has infinite second Betti number and hence be of infinite topological type.
In the following, we will give much more details of the construction.

\subsection{Construction of $Q$}
We equip $Q=[t_1,+\infty) \times_{u(t)} S^3 \times_{g(t)} S^2$ with the following metric
$$
ds^{2}=dt^{2}+u^{2}(t)[dx^{2}+f^{2}(t,x)d\sigma^{2}]+g^{2}(t)d\theta^{2},
$$
where both $d\sigma^{2}$ and $d\theta^{2}$ are the standard metric of the sphere $S^{2}$.
Let $T, X, \Sigma_{1}, \Sigma_{2}, \Theta_{1}, \Theta_{2}$ be an orthonormal basis of the tangent space
corresponding to the directions $dt, dx, d\sigma, d\theta$ respectively.

Give some constants $c, \gamma, \alpha$ and $t_{1}$ satisfying
$$
0 < c <{\frac 13}, ~~0 < \gamma <{\frac 14}, ~~\alpha > 1, ~~t_{1} > 1.
$$
Set
$$
K=K(c)=\frac{1-c^{2}}{c^{2}},
$$
and define $\psi=\psi(c)$ by
$$
\sin(\sqrt{K}\psi)=\sqrt{1-c^{2}}.
$$
Take $r>0$ satisfying
$$
r \le r(c) = \frac{\pi}{4\sqrt{K}}-\frac{\psi}{2}.
$$
Note that $c$ first is fixed while $r$, $\gamma, \alpha$ and $t_{1}$ will be determined later. We define
$$
t_{i}=t_{1} \alpha^{i},
$$
$$
r_{i}=r t_{i},
$$
$$
\psi_{i}=\psi t_{i},
$$
$$
K_{i}=\frac{K}{t_{i}^{2}},
$$
and
$$
\Delta=\sqrt{K_{i}}(2r_{i}+\psi)=\sqrt{K}(2r+\psi).
$$

In the following, we'll give the constructions of $u, g$ and $f$ respectively.

\subsubsection{Construction of $u(t)$}
For $t_{i} < t < t_{i}+2r_{i}$, we set
$$
u(t)=\frac{1}{\sqrt{K_{i}}} \sin(\sqrt{K_{i}}(t-t_{i}+\psi_{i})),
$$

For $t_{i}+2r_{i} < t < t_{i+1}$, we set
$$
u(t)=t_{i}w(\frac{t}{t_{i}}),
$$
where $w$ is a $C^{2}$ function on $[1+2r,\alpha]$, which is independent of $i$.
Actually, we can define $w$ as follows.  First, set
$$
w(t)=\min\{\frac{\sin\Delta}{\sqrt{K}}+(t-1-2r)\cos\Delta+
\frac{c+1}{2\log\frac{\alpha}{1+2r}}(t\log\frac{t}{1+2r}-t+1+2r), ct \}
$$
and then smoothen $w(t)$ to be a $C^{2}$ function.

Since, at $t=1+2r$,
$$
\frac{\sin\Delta}{\sqrt{K}} < c(1+2r),
$$
and, at $t=\alpha$.
$$
\frac{\sin\Delta}{\sqrt{K}}+(\alpha-1-2r)\cos\Delta+
\frac{c+1}{2\log\frac{\alpha}{1+2r}}(\alpha\log\frac{\alpha}{1+2r}-\alpha+1+2r) > c\alpha,
$$
provided with $\cos\Delta \ge \frac{c+1}{\log\frac{\alpha}{1+2r}}$, which is possible for $\alpha \ge \alpha_{0}(c,r)$.
(Here, $\alpha_0$ is some constant depending on $c, r$; similarly for the following $\alpha_1, \alpha_2$.)
Thus, $u(t)$ is $C^{1}$ at the endpoints $t_{i}+2r_{i}=(1+2r)t_{i}$ and $t_{i+1}=\alpha t_{i}$.

On the other hand, if $w(t)=ct$,
$$
\begin{cases}
w_{t} \equiv c,\\
w_{tt} \equiv 0;
\end{cases}
$$
if $w(t)=\frac{\sin\Delta}{\sqrt{K}}+(t-1-2r)\cos\Delta+
\frac{c+1}{2\log\frac{\alpha}{1+2r}}(t\log\frac{t}{1+2r}-t+1+2r)$,
$$
\begin{cases}
\cos\Delta \le w_{t} \le \frac{1+3c}{2} < 1,\\
-3\frac{w_{tt}}{w}+\frac{\gamma(1-2\gamma)}{t^{2}} > 0,
\end{cases}
$$
when $\cos(\Delta) \ge \frac{c+1}{\log\frac{\alpha}{1+2r}}$ and $\alpha \ge \alpha_{1}(c,r,\gamma)$.

\vskip .2cm

\noindent
{\bf Conclusion:} When $\alpha \ge \alpha_{1}(c,r,\gamma)$, we have $u(t)$ satisfying, for $t > t_{1}$,
$$
\begin{cases}
\cos\Delta \le u_{t} \le  c, \quad -\frac{u_{tt}}{u}=\frac{K}{t_{i}^{2}}, & t_{i} < t < t_{i}+2r_{i},\\
\cos\Delta \le u_{t} \le \frac{1+3c}{2}, \quad -3\frac{u_{tt}}{u}+\frac{\gamma(1-2\gamma)}{t^{2}} > 0, & t_{i}+2r_{i} < t < t_{i+1}.
\end{cases}
$$

\subsubsection{Construction of $g(t)$}
Now, we construct the function $g(t)$, which is used to control the curvature component
$-\frac{u_{tt}}{u}$ for $t_{i}+2r_{i} < t < t_{i+1}$. More precisely, we let $\gamma<\frac 14$
and $-\frac{g_{tt}}{g} \ge \frac{\gamma(1-2\gamma)}{2t^{2}}$ on these intervals.
On the other hand, in order to make the surgery simple (i.e. we only need to consider surgeries on
$[t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}] \times_{u(t)} S^3$), we set $g(t)$ to be constants
on $t_{i}+\frac{r_{i}}{6} < t < t_{i}+\frac{11r_{i}}{6}$. Moreover, in order to make
the Ricci curvature $Ric(\Theta_{k},\Theta_{k})$ positive, we need the component
$K(\Theta_{1},\Theta_{2})=\frac{1}{g^{2}}-\frac{g_{t}^{2}}{g^{2}}$.
This is why the dimension of our examples can not be smaller than 6.

For $t_{1} < t < t_{1}+\frac{r_{1}}{6}$, we set
$$
g(t) \equiv g_1 = g(t_{1}+\frac{r_{1}}{6}) = (t_{1}+\frac{r_{1}}{6})^{\gamma} = (1+\frac{r}{6})^{\gamma}t_{1}^{\gamma}.
$$

For $t_{i}+\frac{r_{i}}{6} < t < t_{i+1}+\frac{r_{i+1}}{6}=\alpha(t_{i}+\frac{r_{i}}{6})$, we set
$$
\frac{g_{t}}{g}=\begin{cases}
0, & t_{i}+\frac{r_{i}}{6} < t < t_{i}+\frac{11r_{i}}{6},\\
\frac{6\gamma\beta}{(1+2r)rt_{i}^{2}}[t-(t_{i}+\frac{11r_{i}}{6})], & t_{i}+\frac{11r_{i}}{6} < t < t_{i}+2r_{i},\\
\frac{\gamma\beta}{t}, & t_{i}+2r_{i} < t < t_{i+1},\\
\frac{6\gamma\beta}{\alpha^{2}rt_{i}^{2}}[(t_{i+1}+\frac{r_{i+1}}{6})-t], & t_{i+1} < t < t_{i+1}+\frac{r_{i+1}}{6},
\end{cases}
$$
where $\beta=\frac{\log\alpha}{\log\alpha-\log(1+2r)+\frac{r(1+r)}{6(1+2r)}}=\beta(r,\alpha)$ ensures that
$$
g(t_{i}+\frac{r_{i}}{6})=(t_{i}+\frac{r_{i}}{6})^{\gamma}, \quad \forall i \ge 1.
$$
Thus, for $t_{i}+\frac{r_{i}}{6} < t < t_{i+1}+\frac{r_{i+1}}{6}=\alpha(t_{i}+\frac{r_{i}}{6})$,
$$
g(t) \le g(t_{i+1}+\frac{r_{i+1}}{6})=\alpha^{\gamma}(t_{i}+\frac{r_{i}}{6})^{\gamma} \le \alpha^{\gamma}t^{\gamma}
$$
and
$$
g(t) \ge g(t_{i}+\frac{r_{i}}{6})=(t_{i}+\frac{r_{i}}{6})^{\gamma} \ge \alpha^{-\gamma}t^{\gamma}.
$$
Note that for $\alpha \ge \alpha_{2}(r)$, we can have $\frac 12 \le \beta \le 2$. Then
$$
|\frac{g_{t}}{g}| \le  \frac{2(1+\frac{r}{6})\gamma}{t}
$$
and
$$
\begin{cases}
|\frac{g_{tt}}{g}| \le \frac{12\gamma(1+\frac{r\gamma}{3})}{rt_{i}^{2}}, & t_{i} < t < t_{i}+2r_{i},\\
-\frac{g_{tt}}{g} \ge \frac{\gamma(1-2\gamma)}{2t^{2}}, & t_{i}+2r_{i} < t < t_{i+1}.
\end{cases}
$$

\vskip .2cm

\noindent
{\bf Conclusion:} When $\alpha \ge \alpha_{2}(r)$, we have a $C^{1}$ function $g(t)$ satisfying,  for $t > t_{1}$,
$$
\alpha^{-\gamma}t^{\gamma} \le g(t) \le \alpha^{\gamma}t^{\gamma},
$$
$$
|\frac{g_{t}}{g}| \le  \frac{2(1+\frac{r}{6})\gamma}{t},
$$
and
$$
\begin{cases}
|\frac{g_{tt}}{g}| \le \frac{12\gamma(1+\frac{r\gamma}{3})}{rt_{i}^{2}}, & t_{i} < t < t_{i}+2r_{i},\\
-\frac{g_{tt}}{g} \ge \frac{\gamma(1-2\gamma)}{2t^{2}}, & t_{i}+2r_{i} < t < t_{i+1}.
\end{cases}
$$

\subsubsection{Construction of $f(t,x)$}
The function $f(t,x)$ here is actually $f_{\epsilon(t)}(x)$, which depends on the parameter
$\epsilon<1$. Then, we will set $f_{\epsilon}(x)$ to be almost $R_{0}\sin x$ ($R_{0}$ is sufficiently
small and determined when gluing in $\mathbb{CP}^{2}$'s) and smoothed near $x=0$.
The original idea here comes from \cite{P}, also see \cite{M1} and \cite{M2}.

Let $\phi \in C^{2}(\mathbb{R})$, and satisfy
$$
\phi(x)=\begin{cases}
1, & x \le 0,\\
0, & x \ge 1.
\end{cases}
$$
Set
$$
\phi_\epsilon(x)=\phi\big(\frac{x-\epsilon}{\epsilon^{\frac 14}-\epsilon}\big)=\begin{cases}
1, & x \le \epsilon,\\
0, & x \ge \epsilon^{\frac 14}.
\end{cases}
$$
Let $b=b(\epsilon), l=l(\epsilon)$ and $\delta=\delta(\epsilon)$ to be determined later. Then, set
$$
f_{\epsilon}(x)=\begin{cases}
\frac{\sin(lx)}{l}, & 0 \le x \le b,\\
\frac{\sin(lb)}{l}e^{(1-\epsilon)\log\frac xb}, & b \le x \le \epsilon,\\
R_{0}\sin(x+\delta\phi_{\epsilon}(x)), & \epsilon \le x \le \frac \pi2
\end{cases}
$$
and $f_{\epsilon}(x)=f_{\epsilon}(\pi-x)$ for $\frac \pi2 \le x \le \pi$.

We here remark that $f_\epsilon$ on $[{\frac \pi 2}, \pi]$ is obtained by symmetric extension at $\frac \pi 2$; so,
all the things done around $o_{i}=(t_{i}+r_{i},0)$ in the following
(including the surgery of gluing in $\mathbb{CP}^{2}$'s) can and must also be done
around $o'_{i}=(t_{i}+r_{i},\pi)$ (we have to remove $B_{\frac 45 r_{i}}(o'_{i})$
since the Ricci curvatures on these geodesic balls are not necessarily positive).

Thus, we just consider the case of $0 \le x \le \frac \pi2$ from now on
only by requiring the geodesic balls $B_{\frac 45 r_{i}}(o_{i})$ and $B_{\frac 45 r_{i}}(o'_{i})$ disjoint.
In order to make these two geodesic balls disjoint, we only need to assume
$\frac{\frac 45 r_{i}}{u(t_i)} < \frac{\pi}{2}$, which is valid since we have set
$r \le r(c) = \frac{\pi}{4\sqrt{K}}-\frac{\psi}{2} \le \frac{\pi}{4\sqrt{K}} = \frac{\pi}{4} \frac{c}{\sqrt{1-c^2}}$ and $0<c<\frac 13$.

For the function $f_\epsilon$ in $x$, we have the following two properties,
the proofs of which can be found in \cite{M2}.

\begin{proposition} (\cite{M2}, Lemma 1.22)
$\forall \eta > 0$, there exists $\epsilon_{0} > 0$ such that $\forall \epsilon < \epsilon_{0}, \forall x \in [0,\pi]$,
one has
$$
-\frac{(f_{\epsilon})_{xx}}{f_{\epsilon}} \ge 1-\eta
$$
and
$$
\frac{1-(f_{\epsilon})_{x}^{2}}{f_{\epsilon}^{2}} \ge 1-\eta.
$$
\end{proposition}

\begin{proposition} (\cite{M2}, Lemma 1.26)
There exists $\epsilon_{0} > 0$ such that $\forall \epsilon < \epsilon_{0}$, $\forall x \in [0,\frac \pi2]$,
one has
$$
A(x)=\tan(x)|\frac{(f_{\epsilon})_{x}}{f_{\epsilon}}-\cot(x)| < 2\epsilon.
$$
\end{proposition}

Now, let $\epsilon=\epsilon(t) < \epsilon_{0}$ be a smooth function satisfying
$$
\epsilon(t)=\begin{cases}
\epsilon_{i} & t_{i-1}+\frac 32 r_{i-1} < t < t_{i}+\frac 12 r_{i}\\
\epsilon_{i+1} & t_{i}+\frac 32 r_{i} < t < t_{i+1}+\frac 12 r_{i+1}.
\end{cases}
$$
where $\{\epsilon_i\}$ is a decreasing sequence of positive constants to be determined.

Set $f(t,x)=f_{\epsilon(t)}(x)$. Clearly, $f(t,x)=R_{0}\sin x$ for $\epsilon^{\frac 14} \le x \le \frac \pi2$;
also $f(t,x)=f_{\epsilon_{i}}(x)$ for $t_{i-1}+\frac 32 r_{i-1} < t < t_{i}+\frac 12 r_{i}$
and $f(t,x)=f_{\epsilon_{i+1}}(x)$ for $t_{i}+\frac 32 r_{i} < t < t_{i+1}+\frac 12 r_{i+1}$.
This shows that $f_t \neq 0$ is possible only on
$\{(t,x)~|~t_{i}+\frac 12 r_{i} \le t \le t_{i}+\frac 32 r_{i}, 0 \le x \le \epsilon^{\frac 14}\}$.

Thus, if we can show
$\{(t,x)~|~t_{i}+\frac 12 r_{i} \le t \le t_{i}+\frac 32 r_{i},
0 \le x \le \epsilon^{\frac 14}\} \subset B_{\frac 45 r_{i}}(o_{i})$, then $f_{t} \equiv 0$
outside $B_{\frac 45 r_{i}}(o_{i})$.
To prove this, we only need to show
$$
d((t_{i}+r_{i}\pm\frac 12 r_{i},\epsilon^{\frac 14}),(t_{i}+r_{i},0)) < \frac 45 r_{i}.
$$
In fact,
\begin{align*}
    & d((t_{i}+r_{i}\pm\frac 12 r_{i},\epsilon^{\frac 14}),(t_{i}+r_{i},0))\\
\le & d((t_{i}+r_{i}\pm\frac 12 r_{i},\epsilon^{\frac 14}),(t_{i}+r_{i}\pm\frac 12 r_{i},0))
+d((t_{i}+r_{i}\pm\frac 12 r_{i},0),(t_{i}+r_{i},0))\\
\le & \int_{0}^{\epsilon^{\frac 14}} u(t_{i}+\frac 32 r_{i})dx+\frac 12 r_{i}\\
\le & \epsilon^{\frac 14} \frac{t_{i}}{\sqrt{K}}+\frac 12 r_{i}
\end{align*}
Thus, in order of the above aim, we need to choose only $\epsilon < (\frac{r\sqrt{K}}{5})^{4}$.

\vskip .2cm

\noindent
{\bf Conclusion:} For $\epsilon < \epsilon_{0}(c,r)$, $f$ as a function in $x$ always
satisfies the proposition 2.1 and 2.2 above, and $f_{t} \equiv 0$ outside $B_{\frac 45 r_{i}}(o_{i})$.

\vskip .2cm
\noindent {\bf Remark:} {\it We here remark that, at some discrete points, $u(t)$, $g(t)$
and $f_\epsilon (x)$ are only $C^{1}$. However, if the manifold constructed in this manner
has positive Ricci curvature on the complement of those $C^{1}$ parts, the manifold can then
be smoothen to be a $C^{2}$ manifold of positive Ricci curvature; for this, one can refer to \cite{P}, also \cite{M1}.}

\subsection{Construction of $P_i$ ($1 \leq i < +\infty$)}
As mentioned before, each $P_i$ topologically consists of a core and a neck by glueing them together
along the corresponding boundary (component) $S^3$. In this subsection, for sake of convenience,
we introduce a "standard" $P_i'$ (topologically $P_i$) with an appropriate metric according to Perelman so that
we can do surgery on the above doubly warped product; for the details of the proofs, see \cite{P}.
Here, we'd like to remark that by "standard" we do not mean that the metrics have
an explicit or canonical expression, instead it just means that the geometry of the manifolds in question
together with the boundaries has some uniform characteristics (or properties). When we do the surgery, $P_i$ is obtained
by some appropriate rescaling of a certain standard $P_i'$.

The core topologically is $\mathbb{CP}^{2} \setminus B(p_0)$ ($\mathbb{CP}^{2}$ deleting a ball),
but equipped with a metric of positive Ricci curvature
and strictly convex boundary which is isometric to a round sphere. Precisely, this is stated as follows.
\vskip .2cm
\noindent {\bf Core:} {\it The metric can be expressed as
$$
ds^2 = dt^2 + (\sin t \cos t)^2 dx^2 + (\frac {1}{100} \cosh (\frac {t}{100}))^2 dy^2 + (\frac {1}{100} \cosh (\frac {t}{100}))^2 dz^2,
$$
where $0 \leq t \leq t_0$ is the distance from $\mathbb{CP}^{1}$ and $X$,$Y$,$Z$
is the standard invariant framing of $S^3$, satisfying $[X,Y]=2Z$, $[Y,Z]=2X$, $[Z,X]=2Y$.
Moreover, $0<t_{0}<\frac {1}{10}$ satisfies $\frac{\sin(2t_{0})}{2}=\frac{1}{100}\cosh\frac{t_{0}}{100}$,
which makes the boundary a round sphere of radius $\frac{1}{100}\cosh\frac{t_{0}}{100}$
with the normal curvatures bigger than or equal to $\frac{1}{100}\tanh\frac{t_{0}}{100}$.}
\vskip .2cm

The neck is $S^3 \times [0,1]$, equipped with a metric of positive Ricci curvature
such that the boundary component $S^{3} \times \{0\}$ is concave and also isometric to a round sphere,
while the boundary component $S^{3} \times \{1\}$ is convex and looks like a rugby. More precisely, one has the following
\vskip .2cm
\noindent {\bf Neck:} {\it Let $(S^3,g=dt^{2}+B^{2}(t)d\sigma^{2})$
be a rotationally symmetric metric satisfying

{\rm(\romannumeral1)} sectional curvature $> 1$;

{\rm(\romannumeral2)} $0 \le t \le \pi R$, $\max_t\{B(t)\}=r$,
there exists $\rho < \frac{\sinh\frac{t_{0}}{100}}{10^4}$ ($0<t_{0}<\frac {1}{10}$ s.t.
$\frac{\sin(2t_{0})}{2}=\frac{1}{100}\cosh\frac{t_{0}}{100}$) such that $0<r<\rho<R$ and $r^2<\rho^3$.

Then, there exists a metric on $S^3 \times [0,1]$ such that

1) $Ric>0$;

2) the boundary component $S^{3} \times \{0\}$ is concave, with normal curvatures
equal to $-\lambda$, and is isometric to the round sphere $S^{3}(\rho\lambda^{-1})$, for some $\lambda>0$;

3) the boundary component $S^{3} \times \{1\}$ is strictly convex, with all its normal curvatures
bigger than 1, and is isometric to $(S^{3},g)$.}
\vskip .2cm

In order to make the glued manifold have positive Ricci curvature, we need the following criterion, also see \cite{P}.

\vskip .2cm
\noindent {\bf Gluing Criterion:} {\it Let $M_{1}$, $M_{2}$ be two compact smooth manifolds
of positive Ricci curvature, with isometric boundaries $\partial M_{1} \backsimeq \partial M_{2} = X$.
Suppose that the normal curvatures of $\partial M_{1}$ are bigger than the negative of
the normal curvatures of $\partial M_{2}$. Then, the glued manifold $M_{1}\cup_{X} M_{2}$
along the boundary $X$ can be smoothed near $X$ to produce a manifold of positive Ricci curvature.}
\vskip .2cm

According to the above glueing criterion, it is easy to see that, after rescaling the core by $(\frac{1}{100}\cosh\frac{t_{0}}{100})^{-2}\cdot(\frac\rho\lambda)^2$,
we can glue the scaled core (the boundary of which is the round 3-sphere of radius $\rho\lambda^{-1}$ and convex
with the normal curvatures bigger than or equal to $(\frac{\sinh\frac{t_{0}}{100}}{10^4})\cdot\frac\lambda\rho$,
i.e. strictly bigger than $\lambda$) and the neck along the round sphere boundaries to get a "standard" $P_i'$
so that it has positive Ricci curvature and its boundary is isometric to
$(S^3, g)$ and strictly convex with the normal curvatures bigger than $1$. Note that here we only ask $(S^3, g)$
to have the characteristics (or properties) in the definition of the neck.

\subsection{Glue $P_i \times_{g_i} S^2$ with $Q \setminus \coprod\limits_{i=1}^{+\infty}
(B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2)$ ($1 \leq i < +\infty$)}

In \S 2.1, we have constructed $Q=[t_1,+\infty) \times_{u(t)} S^3 \times_{g(t)} S^2$ with the metric
$$
ds^{2}=dt^{2}+u^{2}(t)[dx^{2}+f^{2}(t,x)d\sigma^{2}]+g^{2}(t)d\theta^{2}.
$$
Moreover, we also mentioned that we only need to do surgeries
on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}) \times_{u(t)} S^3$,
since $g(t)\equiv g_i$ (a constant) on $[t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}]$.

In order to do surgery, we first need to remove a geodesic ball $B_{\frac 45 r_{i}}(o_{i})$
from $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}) \times_{u(t)} S^3$.
Note that $u(t)=\frac{1}{\sqrt{K_{i}}} \sin(\sqrt{K_{i}}(t-t_{i}+\psi_{i}))$ on $[t_{i}, t_{i}+2r_i]$.

Then, we need to check that the boundary $\partial B_{\frac 45 r_{i}}(o_{i})$
with the induced metric, after some suitable rescaling, owns the metric properties of the boundary
of a certain standard $P_i'$, i.e. the assumptions (i) and (ii) of the metric $g$ in the definition
of the neck in \S 2.2, so that we can glue the $P_i$ (obtained from such a $P_i'$ by a suitable scaling)
on $((t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}) \times_{u(t)} S^3) \setminus B_{\frac 45 r_{i}}(o_{i})$
along the isometric boundaries $\partial B_{\frac 45 r_{i}}(o_{i})$ and $\partial P_i$.

Furthermore, according to the gluing criterion in \S 2.2, in order to guarantee the resulted manifold
on $(t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6})$
to admit a metric with positive Ricci curvature, we need to check that the Ricci curvature of
each piece of manifold is positive and the normal curvatures of $\partial P_i$ are bigger
than the negatives of the normal curvatures of $\partial B_{\frac 45 r_{i}}(o_{i})$
in $((t_{i}+\frac{r_{i}}{6}, t_{i}+\frac{11r_{i}}{6}) \times_{u(t)} S^3) \setminus B_{\frac 45 r_{i}}(o_{i})$.

Obviously, when we take the product of the above resulted manifold with a two-sphere (i.e. $\times_{g_i} S^2$, $g_i$ is constant),
it still has positive Ricci curvature. Thus, if both
$Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2)$ and
$P_i \times_{g_i} S^2$ have positive Ricci curvature (actually, the later is already known of positive Ricci curvature in \S 2.2),
they can be glued together to get a manifold with positive Ricci curvature. The positiveness of Ricci curvature of
$Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2)$ will be verified in \S 3.

In the remaining part of this subsection, we shall check the following two facts.
The first one is that the normal curvatures of $\partial P_i$ are bigger than
the negatives of the normal curvatures of $\partial B_{\frac 45 r_{i}}(o_{i})$.
The second one is that the boundary $\partial B_{\frac 45 r_{i}}(o_{i})$ with the induced metric,
after some suitable rescaling, owns the characteristics of the metric $g$ in the definition of the neck in \S 2.2.

\subsubsection{Normal curvatures}

In order to prove the first fact, it is sufficient to show that for $\epsilon$ sufficiently small,
any $x \in \partial B_{\frac 45 r_{i}}(o_{i})$, and any orthogonal unit vectors $X,Y\in S_x \partial B_{\frac 45 r_{i}}(o_{i})$,
$$
K_{int}(X,Y) > \max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|^2,
$$
where $N$ is the unit outward normal vector of $\partial B_{\frac 45 r_{i}}(o_{i})$,
$\uppercase\expandafter{\romannumeral2}_{N}$ is the corresponding second fundamental form,
$K_{int}$ is the intrinsic curvature, and
$S\partial B_{\frac 45 r_{i}}(o_{i}) \subset T\partial B_{\frac 45 r_{i}}(o_{i})$ is the unit tangent bundle.

In fact, after rescaling $\partial B_{\frac 45 r_{i}}(o_{i})$ by
$\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|)^2$,
we obtain a rotationally symmetric metric $(S^{3},g)$ with curvature bigger than 1 and
the absolute value of the corresponding normal curvatures less than or equal to 1. In the next subsection,
we will check that this rotationally symmetric metric owns the characteristics of the metric $g$ of the neck in \S 2.2,
so that it is the boundary of a certain standard $P_i'$.
In particular, the normal curvatures of $\partial P_i$ in $P_i$ (obtained from the $P'_i$ by
$(\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|)^{-2}$-rescaling)
are bigger than the negative of the normal curvatures of $\partial B_{\frac 45 r_{i}}(o_{i})$.

The proof of the above inquality is similar to that in \S1.6 of \cite{M1}.
Denote by $N$ the unit outward normal vector to $\partial B_{\frac 45 r_{i}}(o_{i})$,
$$
N=T\cos\xi+X\sin\xi,
$$
and
$$
Y=X\cos\xi-T\sin\xi.
$$
Thus
$$
\uppercase\expandafter{\romannumeral2}_{N}(Y,Y)=\sqrt{K_{i}}\cot(\frac 45 \sqrt{K_{i}}r_{i}),
$$
since the $T\wedge X$ plane is isometric to the corresponding one of $S^{2}(\frac{1}{\sqrt{K_{i}}})$.
On the other hand, for $j=1, 2$,
$$
\uppercase\expandafter{\romannumeral2}_{N}(\Sigma_{j},\Sigma_{j})=\frac{u_{t}}{u}\cos\xi+\frac{f_{x}}{fu}\sin\xi.
$$

Note that if $f(t,x)=\sin x$,
$$
\frac{u_{t}}{u}\cos\xi+\frac{f_{x}}{fu}\sin\xi=\sqrt{K_{i}}\cot(\frac 45 \sqrt{K_{i}}r_{i}).
$$
Thus,
\begin{align*}
|\uppercase\expandafter{\romannumeral2}_{N}(\Sigma_{j},\Sigma_{j})-\sqrt{K_{i}}\cot(\frac 45 \sqrt{K_{i}}r_{i})| & \le A(x)\frac{\cot x |\sin\xi|}{u}\\
                    & = A(x)|\sqrt{K_{i}}\cot(\frac 45 \sqrt{K_{i}r_{i}})-\frac{u_{t}}{u}\cos\xi|\\
                    & \le A(x)(\sqrt{K_{i}}\cot(\frac 45 \sqrt{K_{i}r_{i}})+|\frac{u_{t}}{u}|)\\
                    & \le 2\epsilon[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+\frac{1+3c}{2ct_{i}}]\\
                    & = 2[\sqrt{K}\cot(\frac 45 \sqrt{K}r)+\frac{1+3c}{2c}]\frac{\epsilon}{t_{i}}\\
                    & = D(c,r)\frac{\epsilon}{t_{i}} \le D(c,r)\frac{\epsilon_i}{t_i}.
\end{align*}

Then, one has
$$
\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}\leq\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|\leq\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+D(c,r)\frac{\epsilon_i}{t_i}.
$$

For the intrinsic curvatures, by Gauss equation, we have
\begin{align*}
K_{int}(\Sigma_{1},\Sigma_{2}) & = K(\Sigma_{1},\Sigma_{2})+(\uppercase\expandafter{\romannumeral2}_{N}(\Sigma_{j},\Sigma_{j}))^{2}\\
                             & \ge \frac{1-f_{x}^{2}}{f^{2}}\frac{1}{u^{2}}-\frac{u_{t}^{2}}{u^{2}}+[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}-D(c,r)\frac{\epsilon}{t_{i}}]^{2}\\
                             & \ge \frac{1-\eta-(\frac{1+3c}{2})^{2}}{u^{2}}+[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}-D(c,r)\frac{\epsilon}{t_{i}}]^{2}\\
                             & \ge \frac{1-\eta-(\frac{1+3c}{2})^{2}}{c^{2}t_{i}^{2}}+[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}-D(c,r)\frac{\epsilon}{t_{i}}]^{2}\\
                             & > [\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+D(c,r)\frac{\epsilon_i}{t_i}]^{2},
\end{align*}
assuming $\epsilon$ sufficiently small. Similarly,
\begin{align*}
K_{int}(Y,\Sigma_{j}) & = K(X,\Sigma_{j})\cos^{2}\xi+K(T,\Sigma_{j})\sin^{2}\xi+(\uppercase\expandafter{\romannumeral2}_{N}(Y,Y)
(\uppercase\expandafter{\romannumeral2}_{N}(\Sigma_{j},\Sigma_{j}))\\
                             & \ge (-\frac{f_{xx}}{f}\frac{1}{u^{2}}-\frac{u_{t}^{2}}{u^{2}})\cos^{2}\xi+
                             (-\frac{u_{tt}}{u})\sin^{2}\xi\\
                             & \quad +\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}-D(c,r)\frac{\epsilon}{t_{i}}]\\
                             & \ge \frac{1-\eta-(\frac{1+3c}{2})^{2}}{c^{2}t_{i}^{2}}\cos^{2}\xi+
                             \frac{K}{t_{i}^{2}}\sin^{2}\xi\\
                             & \quad +\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}[\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}-D(c,r)\frac{\epsilon}{t_{i}}]\\
                             & > [\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+D(c,r)\frac{\epsilon_i}{t_i}]^{2},
\end{align*}
assuming $\epsilon$ sufficiently small.

\subsubsection{The boundary $\partial B_{\frac 45 r_{i}}(o_{i})$ with the induced metric}

Next, we prove the second fact, i.e. $\partial B_{\frac 45 r_{i}}(o_{i})$ with the induced metric,
after rescaling by $\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|)^2$,
owns the characteristics of the metric $g$ of the neck in \S 2.2.

Even though we could not write out the precise expression of the metric, we can check it in the limit case (thanks to the construction of the neck by Perelman \cite{P}, it only requires some rough characterization of the geometric shape).

Indeed, for $i$ sufficiently large, as $\epsilon$ goes to 0, the metric of $B_{\frac 45 r_{i}}(o_{i})$ converges to
$$
dt^{2}+(\frac{1}{\sqrt{K_{i}}} \sin(\sqrt{K_{i}}(t-t_{i}+\psi_{i})))^{2}[dx^{2}+(R_{0}\sin x)^{2}d\sigma^{2}].
$$
Thus the metric of $\partial B_{\frac 45 r_{i}}(o_{i})$ converges to
$$
(\frac{t_{i}\sin(\frac 45 \sqrt{K}r)}{\sqrt{K}})^{2}[dy^{2}+(R_{0}\sin y)^{2}d\sigma^{2}].
$$
Rescaled by $(\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}})^2$ (note that $\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|)^2$ converges to $(\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}})^2$ when $\epsilon$ goes to 0) and letting $t=y\cos(\frac 45 \sqrt{K}r)$, the metric becomes
$$
dt^{2}+(R_{0}\cos(\frac 45 \sqrt{K}r)\sin\frac{t}{\cos(\frac 45 \sqrt{K}r)})^{2}d\sigma^{2}.
$$
Choosing $R_{0}$ sufficiently small, the metric is exactly what we want.

\subsection{Construction of $P_0$}

Note that after gluing $P_i$ ($1 \leq i < +\infty$) with $Q$, the manifold has a boundary,
i.e. the hypersurface $t=t_{1}$. In this subsection, we will glue in a core via a neck along the boundary $t=t_1$
to smooth the manifold near the origin, based on the gluing criterion.

For $t_{1} < t < t_{1}+\frac{r_{1}}{6}$, we already have
$$
\begin{cases}
u(t) = \frac{1}{\sqrt{K_{1}}} \sin(\sqrt{K_{1}}(t-t_{1}+\psi_{1})),\\
g(t) \equiv (t_{1}+\frac{r_{1}}{6})^{\gamma}.
\end{cases}
$$
Then, for the hypersurface $t=t_{1}$, the intrinsic metric is
$$
ds^{2}=(ct_{1})^{2}[dx^{2}+f_{\epsilon_{1}}^{2}(x)d\sigma^{2}].
$$
Rescaled by $t_{1}^{-2}$ and letting $t=cx$, when $\epsilon_{1}$ goes to 0, it converges to
$$
dt^{2}+(cR_{0}\sin\frac{t}{c})^{2}d\sigma^{2}.
$$
Similarly, choosing $R_{0}$ sufficiently small, the metric owns the properties of the metric $g$ of the neck in \S 2.2.

On the other hand,
$$
\uppercase\expandafter{\romannumeral2}_{-T}(X,X)=
\uppercase\expandafter{\romannumeral2}_{-T}(\Sigma_{j},\Sigma_{j})=-\frac{1}{t_{1}},
$$
$$
K_{int}(\Sigma_{1},\Sigma_{2})=K(\Sigma_{1},\Sigma_{2})+\frac{1}{t_{1}^{2}}=
\frac{1-f_{x}^{2}}{f^{2}}\frac{1}{u^{2}} \ge \frac{1-\eta}{c^{2}t_{1}^{2}},
$$
$$
K_{int}(X,\Sigma_{j})=K(X,\Sigma_{j})+\frac{1}{t_{1}^{2}}=
-\frac{f_{xx}}{f}\frac{1}{u^{2}} \ge \frac{1-\eta}{c^{2}t_{1}^{2}}.
$$
Setting $1-\eta > c^2$, we can continue the manifold for $t \le t_{1}$ by gluing a core via a neck (i.e. $P_0$) with a metric product $\times_{g_0}S^{2}$ ($g_0=(t_{1}+\frac{r_{1}}{6})^{\gamma}$).

\section{Positiveness of Ricci curvature and quadratically asymptotic non-negativeness of curvature}

\subsection{Positive Ricci curvature}

In this subsection, we will show that the constructed manifold above has positive Ricci curvature
by taking some appropriate constants, i.e. $\eta, r, \gamma$ sufficiently small and $\alpha, t_1$ sufficiently big.

Recall that our manifold is constructed as
$$
M = (Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2))
\cup_{\textrm{Id}} \coprod\limits_{i=0}^{+\infty} (P_i \times_{g_i} S^2).
$$
As mentioned before, by our constrction and the gluing criterion, we only need to check that
$Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2)$ have positive Ricci curvature.

Since $f_{t} \equiv 0$ outside $B_{\frac 45 r_{i}}(o_{i})$, the nonzero Ricci curvatures are as follows.
\begin{align*}
Ric(T,T) & = -3\frac{u_{tt}}{u}-2\frac{g_{tt}}{g}\\
         & \ge \begin{cases}
         -3\frac{u_{tt}}{u}+\frac{\gamma(1-2\gamma)}{t^{2}} > 0, & t_{i}+2r_{i} < t < t_{i+1},\\
         3\frac{K}{t_{i}^{2}}-\frac{24\gamma(1+\frac{r\gamma}{3})}{rt_{i}^{2}} > 0, & t_{i} < t < t_{i}+2r_{i},
         \end{cases}
\end{align*}
provided with $\gamma < \gamma_{0}(c,r)$;

\begin{align*}
Ric(X,X) & =-2\frac{f_{xx}}{f}\frac{1}{u^{2}}-2(\frac{u_{t}}{u})^{2}-2\frac{u_{t}}{u}\frac{g_{t}}{g}-\frac{u_{tt}}{u}\\
         & \ge 2\frac{1-\eta}{u^{2}}-2(\frac{u_{t}}{u})^{2}--2\frac{u_{t}}{u}\frac{2(1+\frac r6)\gamma}{t}-\frac 13 \frac{\gamma(1-2\gamma)}{t^{2}}\\
         & = \frac{2}{u^{2}}[1-\eta-u_{t}^{2}-\frac{2(1+\frac r6)\gamma u_{t}u}{t}]-\frac 13 \frac{\gamma(1-2\gamma)}{t^{2}}\\
         & \ge \frac{2}{(\frac{1+3c}{2})^{2}t^{2}}\{1-\eta-(\frac{1+3c}{2})^{2}[1+2(1+\frac r6)\gamma]\}-\frac 13 \frac{\gamma(1-2\gamma)}{t^{2}}\\
         & = \frac{2}{(\frac{1+3c}{2})^{2}t^{2}}\{1-\eta-(\frac{1+3c}{2})^{2}[1+2(1+\frac r6)\gamma+\frac 16 \gamma(1-2\gamma)]\}\\
         & > 0,
\end{align*}
provided with $1-\eta > (\frac{1+3c}{2})^{2}[1+2(1+\frac r6)\gamma+\frac 16 \gamma(1-2\gamma)]$,
which is possible when $\eta < \eta_{0}(c)$ and $\gamma < \gamma_{0}(c,r,\eta)$;

\begin{align*}
Ric(\Sigma_{j},\Sigma_{j}) & =(\frac{1-f_{x}^{2}}{f^{2}}-\frac{f_{xx}}{f})\frac{1}{u^{2}}-2(\frac{u_{t}}{u})^{2}-
2\frac{u_{t}}{u}\frac{g_{t}}{g}-\frac{u_{tt}}{u},\\
                           & > 0
\end{align*}
which is totally similar to $Ric(X,X)$;

\begin{align*}
Ric(\Theta_{k},\Theta_{k}) & =\frac{1}{g^{2}}-(\frac{g_{t}}{g})^{2}-\frac{g_{tt}}{g}-3\frac{u_{t}}{u}\frac{g_{t}}{g}\\
                           & \ge \frac{1}{\alpha^{\gamma}t^{\gamma}}-\frac{4(1+\frac r6)^{2}\gamma^{2}}{t^{2}}-\frac{12\gamma(1+2r)(1+\frac{r\gamma}{3})}{rt^{2}}-
                           \frac{3(1+3c)(1+\frac r6)\gamma}{t^{2}\cos\Delta}\\
                           & > 0,
\end{align*}
provided with $t_{1} > t_{1}(c,r,\gamma)$.

Moreover, all the off-diagonal terms of the Ricci tensor vanish. Thus the Ricci curvatures are positive.

\subsection{Quadratically asymptotic non-negativeness of sectional curvature}
In this subsection, we show that our example has quadratically asymptotically nonnegative sectional curvature.

For $P_i \times_{g_i} S^2$, we only have to check that $P_i$ has quadratically asymptotically nonnegative sectional curvature. After rescaling the standard cores and necks by $\max_{Z \in S_x \partial B_{\frac 45 r_{i}}(o_{i})} |\uppercase\expandafter{\romannumeral2}_{N}(Z,Z)|)^2$ ($\leq(\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+D(c,r)\frac{\epsilon_i}{t_i})^2$),
$$
K_{p_{0}} (t) \ge -(K_{0})^{2}(\frac{\sqrt{K}\cot(\frac 45 \sqrt{K}r)}{t_{i}}+
D(c,r)\frac{\epsilon}{t_{i}})^{2} \ge -\frac{K_{0}^{2}(c,r)}{t_{i}^{2}}.
$$

For $Q \setminus \coprod\limits_{i=1}^{+\infty} (B_{\frac 45 r_{i}}(o_{i}) \times_{g_i} S^2)$,
$$
K(T,X)=K(T,\Sigma_{j})=-\frac{u_{tt}}{u} > -\frac 13 \frac{\gamma(1-2\gamma)}{t^{2}},
$$
$$
K(T,\Theta_{k})=-\frac{g_{tt}}{g} \ge -\frac{12\gamma(1+\frac{r\gamma}{3})(1+2r)}{rt^{2}},
$$
$$
K(X,\Sigma_{j})=-\frac{f_{xx}}{f}\frac{1}{u^{2}}-\frac{u_{t}^{2}}{u^{2}} \ge \frac{1-\eta-(\frac{1+3c}{2})^{2}}{\cos^{2}\Delta}\frac{1}{t^{2}} > 0,
$$
$$
K(X,\Theta_{k})=K(\Sigma_{j},\Theta_{k})=-\frac{u_{t}}{u}\frac{g_{t}}{g} \ge -\frac{(1+3c)(1+\frac r6)\gamma}{\cos\Delta}\frac{1}{t^{2}},
$$
$$
K(\Sigma_{1},\Sigma_{2})=\frac{1-f_{x}^{2}}{f^{2}}\frac{1}{u^{2}}-\frac{u_{t}^{2}}{u^{2}} \ge \frac{1-\eta-(\frac{1+3c}{2})^{2}}{\cos^{2}\Delta}\frac{1}{t^{2}} > 0,
$$
$$
K(\Theta_{1},\Theta_{2})=\frac{1}{g^{2}}-\frac{g_{t}^{2}}{g^{2}} \ge \frac{1}{t^{2\gamma}}-\frac{4(1+\frac r6)^{2}\gamma^{2}}{t^{2}} > 0.
$$
And other terms of curvature tensors are zero.

Thus, we have
$$
K_{p_{0}} (t) \ge -\frac{K_{0}^{2}(c,r,\gamma)}{t^{2}}.
$$

\vskip .2cm
\noindent {\bf Remark:} {\it
For the diameter growth, we have
$$
0 < \pi\cos\Delta = \liminf_{t \to \infty} \frac{{\text{diam}}(p_0; t)}{t}
< \limsup_{t \to \infty} \frac{{\text{diam}}(p_0; t)}{t} = \pi c.
$$
For the volume growth, we have
$$
0 < V_{1} = \liminf_{t \to \infty} \frac{{\text{Vol}}(B(p_0,t))}{t^{4+2\gamma}}
< \limsup_{t \to \infty} \frac{{\text{Vol}}(B(p_0,t))}{t^{4+2\gamma}} = V_{2} < +\infty.
$$}

redpajama_set_name: RedPajamaArXiv


text:
import Ember from 'ember';
import { moduleForComponent, test } from 'ember-qunit';
import {
  destroy,
  initAttrs,
  renderingTests,
  setOnComponent,
  setPropertiesOnComponent,
  setPropertiesOnController,
  setupComponent
} from '../../helpers/unit/component';

const { run, typeOf } = Ember;

let component;

moduleForComponent('error-field', 'Unit | Component | error field', {
  needs: ['helper:capitalize-string'],
  unit: true,

  beforeEach: function() {
    component = setupComponent(this);
  },
});

test('Rendering', function(assert) {

  assert.expect(2);

  renderingTests(assert, this, component);
});

test('Properties', function(assert) {
  const property = 'bananas';

  assert.expect(8);

  assert.ok(!!component.get('className'),
    'The component should have a default class name');

  assert.notOk(component.get('property'),
    'The component should not have a default property');

  assert.ok(typeOf(component.get('formController')) === 'instance',
    'The component should have a formController binding');

  assert.notOk(component.get('visible'),
    'Visible should be false by default');

  setPropertiesOnComponent(component, { property });

  assert.equal(component.get('property'), property,
    'The component should have a value for property');

  assert.equal(component.get('label'), property,
    'The label should default to the value of property');

  assert.ok(component.get('text').indexOf(property) > -1,
    'The error text should contain the new property name');

  this.render();

  const element = this.$();

  assert.ok(element.hasClass(component.get('className')),
    'The class name should be bound to the element');

});

test('Error binding', function(assert) {
  const property = 'apples';

  assert.expect(3);

  assert.notOk(component.get('bindingForErrors'),
    'The component should not have created a binding for validation errors');

  this.render();

  setPropertiesOnController(component, {
    'validations.apples': {
      presence: true
    }
  });

  setPropertiesOnComponent(component, { property });

  initAttrs(component);

  assert.ok(!!component.get('bindingForErrors'),
    'The component should have created a binding for validation errors');

  destroy(component);

  assert.notOk(!!component.get('bindingForErrors'),
    'The component should remove the binding for validation errors when the component is being destroyed');

});

test('The DOM', function(assert) {
  const property = 'apples';
  const error = 'should be present';

  assert.expect(1);

  setPropertiesOnController(component, {
    'validations.apples': {
      presence: true
    },
    'errors.apples': Ember.A([
      error
    ]),
  });

  setPropertiesOnComponent(component, {
    property,
    showError: true,
  });

  initAttrs(component);

  this.render();

  const layout = component.$().html().trim().toLowerCase();

  assert.equal(layout, `${property} ${error}`,
    'The component should show a user-friendly error message');

});

redpajama_set_name: RedPajamaGithub