Sandia National Laboratories was established as part of the Los Alamos Laboratory designated "Z Division" in 1945. As everyone should know, Los Alamos is the birthplace of the atomic age. In 1949, Western Electric took over administration of the Laboratories and in 1950 the Metal Trades Council established a bargaining unit at the Laboratories.
The Council is made up of 12 affiliated local unions, which include: IBB Local 4; Chemical Workers Local 387; IBEW Local 1988; LIUNA Local 16; IAM Local 794; IUOE Local 953; OPEIU Local 251; UA Local 412; SMART Local 49; IBT Local 492, Insulators Local 76, and IUPAT Local 823.


Gerald Badurek, Zdeněk Hradil, Alexander Lvovsky, Gabriel Molina-Teriza, Helmut Rauch, Jaroslav Řeháček et al.
This book is a comprehensive survey of most of the theoretical and experimental achievements in the field of quantum estimation of states and operations. Albeit still quite young, this field has already been recognized as a necessary tool for research in quantum optics and quantum information, beyond being a fascinating subject on its own as it touches upon the very conceptual foundations of quantum mechanics. The books consists of twelve extensive lectures that are essentially self-contained and modular allowing to combine various chapters as a basis for advanced courses and seminars on theoretical or experimental aspects. The last two chapters, for instance, form a self-consistent exposition on quantum discrimination problems. The book will benefit graduate students and newcomers to the field as high-level but accessible textbook, lecturers in search for advanced course material and researcher wishing to consult a modern and authoritative source of reference.


text:
Q: Difference between Embed, Embed Advanced or Dynamic Map I'm using Google Maps Javascript API to add maps to a site. Now when Google are making changes to their prices I am unsure what my limit is. This site lists the prices but how do I know if I am using the Embed, Embed Advanced or Dynamic Map?

A: I can share information obtained from the Google Maps Platform technical support. Billing SKUs can be found in their Pricing Docs
Now, referring to your question,

*

*Embed SKU refers to a basic Maps Embed API request that uses map or a map with a marker (not including the Directions mode, Street View mode, or Search mode). These requests are free of charge.

*Embed Advanced SKU refers to an advanced Maps Embed API request that uses the Directions mode, Street View mode, or Search mode. These requests are also free of charge. User interactions with the map, such as panning, zooming, or switching map layers, do not generate additional map loads. Embed Advanced were charged before, but they changed this with their recent update. For it currently says in their pricing docs that,

"All Maps Embed API requests are available at no charge with unlimited usage."

*

*Dynamic Map SKU refers to a web page or application that displays a map using the Maps JavaScript API. A map is created with the google.maps.Map() class. Any user interactions with the map, such as panning, zooming, or switching map layers, do not generate additional map loads. The creation of a Street View panorama is no longer charged as a map load, it is charged as a "Dynamic Street View".

I hope my answer clarifies your doubts.

redpajama_set_name: RedPajamaStackExchange


MANCHESTER UNITED striker Romelu Lukaku should have been banned by the Football Association for his kick on Brighton defender Gaetan Bong, says Ian Wright.
Images appeared to show Lukaku kicking out twice at Bong just before Ashley Young's strike deflected in off Lewis Dunk to secure Manchester United a 1-0 win.
The 24-year-old was left sweating on FA action as retrospective punishment for violent conduct would incur a three-match ban, but he will face no action.
The Belgium international would have been ruled out for the crucial title showdown with Manchester City next month, as well as games at Watford and Arsenal.
And former Arsenal striker Wright was surprised by the FA's decision.
"If they look back at it, how can they not see it is not right what he is done? Should he be done? Yes," Wright told BBC Radio 5 Live.
"He has kicked out at him. He has had two digs at him. Whoever is sitting in here making that decision - why haven't they pulled him up on it?"
Lukaku, who has scored 12 goals since joining United in the summer, is expected to keep his place up front for tonight's trip to Watford.


text:
WTF with Marc Maron Podcast
Summary: Marc Maron welcomes comedians, actors, directors, writers, authors, musicians and folks from all walks of life to his home for amazingly revealing conversations. Marc's probing, comprehensive interview style allows guests to express themselves in ways listeners have never heard.
Artist: Marc Maron
Copyright: 2012.
Episode 1031 - Brent Butt | File Type: audio/mpeg | Duration: 01:25:19
Canadian comedian Brent Butt talks with Marc about Saskatchewan, his hit show Corner Gas, and why elusive success in America never really bothered him.
Episode 1030 - Stephen Colbert | File Type: audio/mpeg | Duration: 01:29:03
Stephen Colbert and Marc talk about grief, sadness, anxiety, The Colbert Report, the White House Correspondents' Dinner, and The Late Show in the age of Trump.
Episode 1029 - Steve Sweeney | File Type: audio/mpeg | Duration: 01:20:58
Boston comic Steve Sweeney talks with Marc about growing up in Charlestown, doing Shakespeare, working in jails, hanging with Rodney Dangerfield, and making Sweeney Killing Sweeney.
Episode 1028 - Eve Ensler | File Type: audio/mpeg | Duration: 01:27:48
Writer and activist Eve Ensler and Marc have a profound and emotional conversation about her courageous new book The Apology.
Episode 1027 - Jamie Denbo | File Type: audio/mpeg | Duration: 01:21:08
Jamie Denbo, creator of the new show American Princess, talks with Marc about Renaissance fairs, Disney, the original UCB Theater and more.
Episode 1026 - Mavis Staples | File Type: audio/mpeg | Duration: 01:28:30
Mavis Staples talks with Marc about Gospel music, the Staple Singers, the segregated South, MLK, Bob Dylan.
Episode 1025 - Perry Farrell | File Type: audio/mpeg | Duration: 01:15:22
Perry Farrell of Jane's Addiction talks with Marc about surfing, organized crime, teenage angst, being a parent dealing with teenage angst, and more.
Episode 1024 - John Turturro | File Type: audio/mpeg | Duration: 01:23:33
John Turturro talks with Marc about family roots, studying history, Spike Lee, the Coen Brothers, and bridging the political divide.
Episode 1023 - David Letterman | File Type: audio/mpeg | Duration: 01:32:49
David Letterman joins Marc to talk about his early days in comedy, Robin Williams, Johnny Carson, Jay Leno, late night shows, and the hard work of becoming a better person.
Episode 1022 - Timothy Olyphant | File Type: audio/mpeg | Duration: 01:27:31
Actor Timothy Olyphant talks with Marc about fine arts, open mics, Go, Deadwood, Justified, and a perpetual fear of success.
Episode 1021 - Duff McKagan | File Type: audio/mpeg | Duration: 01:44:27
Duff McKagan talks with Marc about Guns N' Roses, Axl, Slash, traveling the world, learning from history, Joe Strummer, sobriety and more.
Episode 1020 - Lisa Kudrow | File Type: audio/mpeg | Duration: 01:27:06
Lisa Kudrow talks with Marc about genealogy, headaches, Jon Lovitz, Conan, The Comeback, Booksmart, her friends the Friends.
Episode 1019 - Kyle Mooney | File Type: audio/mpeg | Duration: 01:18:50
SNL's Kyle Mooney talks with Marc about high school hip hop, Brigsby Bear, internet trolls, cue cards, and digital videos.
Episode 1018 - Anjelica Huston | File Type: audio/mpeg | Duration: 01:28:30
Anjelica Huston talks with Marc about her dad, Ireland, Jack Nicholson, the Addams Family, Wes Anderson, John Wick and more.
Episode 1017 - Anna Konkle & Maya Erskine | File Type: audio/mpeg | Duration: 01:21:51
PEN15's Anna Konkle & Maya Erskine explain what it's like to play the 13-year-old versions of themselves and why they're putting themselves through the trauma of middle school again.

redpajama_set_name: RedPajamaCommonCrawl


text:
Court jurisdiction
Call: +91 99 48 66 66 22
M.H. Uma Maheshwari and Others Vs. United India Insurance Co. Ltd. and Another
by Publication | Jun 14, 2020 | Case Laws
SUPREME COURT OF INDIA
FULL BENCH
(Before : N.V. Ramana, R. Subhash Reddy and Surya Kant, JJ. )
Civil Appeal No. 2558 of 2020 [Arising out of S.L.P.(C) No.19221 of 2018]
Decided on : 12-06-2020
M.H. UMA MAHESHWARI AND OTHERS — Appellant
UNITED INDIA INSURANCE CO. LTD. AND ANOTHER — Respondent
A. Motor Vehicles Act, 1988 – Section 166 – Rash and Negligent Driving – Compensation – Accident occurred due to negligent driving of the driver of the vehicle and proceeded to quantify the compensation – Appeal allowed. (Para 5)
B. Motor Vehicles Act, 1988 – Section 166 – Reduction in compensation – Appeal against – High Court, curiously while maintaining the multiplier of 13 as per the judgment of this Court in the case of Sarla Verma, (2009) 6 SCC 121, reduced the compensation only on the ground that the deceased was aged 50 years 3 months on the date of the accident, as such the compensation is to be calculated on account of loss of dependency by granting future prospects at 15% but not 30% – So far as the application of multiplier of 13 by the Tribunal is concerned, the High Court has not interfered with the same – When the age of the deceased was considered in the group of 40 to 50 years – High Court has committed error in granting only 15% towards future prospects instead of 30% – As per the judgments of this Court primarily the age group is to be considered – Considering the age group as 40 to 50 years, when the multiplier of 13 is maintained by the High Court, there is no reason or justification for reducing the compensation by granting 15% towards future prospects – Appeal allowed. (Para 8)
Counsel for Appearing Parties
Mr. Shekhar G. Devasa Advocate, Mr. Manish Tiwari Advocate, Mr. Luv Kumar Advocate, For M/S. Devasa & Co. Advocate for the Appellant;
Ms. Neerja Sachdeva Advocate, Mr. Yash Pal Dhingra Advocate for the Respondent.
Cases Referred
Sarla Verma & Ors. vs. Delhi Transport Corporation & Anr., (2009) 6 SCC 121
National Insurance Company Limited vs. Pranay Sethi & Ors., (2017) 16 SCC 680
R. Subhash Reddy, J. – Leave granted.
This civil appeal is filed by the claimants in a claim petition filed under Section 166 of the Motor Vehicles Act, 1988 (for short, 'the Act') in MVC No.1639 of 2012 before the Motor Accident Claims Tribunal-VI and III Addl. Sr. Civil Judge, Mangalore, D.K. (for short, 'the Tribunal'), aggrieved by the judgment dated 20.07.2017 passed in Misc. First Appeal No.4903 of 2016 by the High Court of Karnataka at Bengaluru.
Necessary facts in brief are as under :
The deceased S.T. Devaraju was the husband of first appellant and father of appellant nos. 2 and 3. On 16.07.2012 when he was travelling in the car, viz., Tata Indigo Manza bearing registration no. KA19-MC-5879 to Raichur, the said car met with an accident. The deceased Devaraju suffered severe injuries and subsequently died. The deceased Devaraju was working as Commissioner of Raichur City Municipal Corporation during the relevant time.
The appellants herein, alleging that accident occurred due to rash and negligent driving of the driver of the vehicle, filed claim petition under Section 166 of the Act claiming compensation of Rs.2,00,00,000/- with interest at the rate of 12% p.a. It was the case of the appellants that the deceased was drawing monthly salary of Rs.55,000/- and he was the KGS Cadre officer selected through Public Service Commission. Further pleading that due to untimely death of the deceased, the appellants lost dependency and the deceased was having bright future, the above said claim was made. The claim was opposed by the respondents by filing the written statement. The appellants have led oral and documentary evidence before the Tribunal. The first appellant was examined as PW-1 and on their behalf the other two witnesses were examined as PW-2 and PW-3 and documents Ex.P1 to P24 were marked. On behalf of the respondents, no oral evidence was adduced and only a copy of the Insurance Policy was marked as exhibit, with consent.
The Tribunal, by considering the oral and documentary evidence on record, has recorded a finding that the accident occurred due to negligent driving of the driver of the vehicle and proceeded to quantify the compensation. Having regard to the evidence on record, the Tribunal, by recording a finding that the deceased was earning Rs.50,463/- p.m. by way of salary, by applying the principles laid down in the case of Sarla Verma & Ors. vs. Delhi Transport Corporation & Anr., (2009) 6 SCC 121 applied the multiplier of 13 and by giving 30% towards future prospects, arrived at a compensation of Rs.1,02,33,912. Out of the said sum, by deducting 1/3rd towards the personal expenditure and 10% towards income tax, the Tribunal has held that the appellant-claimants were entitled to a compensation of Rs.61,40,347.20 towards loss of dependency. By further awarding an amount of Rs.1,00,000/- towards loss of consortium to the first appellant and Rs.3,00,000/- for all the appellants towards loss of love and affection and Rs.20,000/- towards funeral expenses against the claim of Rs.2,00,000/-, the Tribunal has awarded the total compensation of Rs.65,60,347.20.
Aggrieved by the award of the Tribunal, the first respondent United India Insurance Co. Ltd. has preferred Misc. First Appeal No.4903 of 2016 before the High Court of Karnataka at Bengaluru. The award of the Tribunal was mainly assailed before the High Court on three grounds, namely, that as the deceased was over 50 years of age, the Tribunal committed error in computing the future prospects at 30%; secondly it was the case of the first respondent that as the first appellant was claiming family pension, deduction should have been made while computing the loss of dependency; and thirdly by awarding the compensation of Rs.1,00,000/- to the first appellant towards loss of consortium, the Tribunal again granted compensation of Rs.3,00,000/- to all the appellants under the head 'loss of love and affection'. The High Court, on the ground that the deceased was aged 50 years 3 months on the date of accident, has come to the conclusion that the appellants are entitled to compensation on account of loss of dependency by computing future prospects of the deceased at 15% and not 30%. Further it was held that by awarding an amount of Rs.1,00,000/- towards loss of consortium to the first appellant, the Tribunal has committed error by awarding Rs.1,00,000/- to the first appellant towards the head 'loss of love and affection'. With the aforesaid findings, the High Court has recalculated the compensation payable to the appellants at Rs.57,78,480/-, i.e., Rs.54,33,480/- towards loss of dependency; Rs.1,00,000/- towards loss of consortium; Rs.2,00,000/- towards of love and affection to the children; Rs.25,000/- towards funeral expenses and Rs.20,000/- towards transportation of dead body.
We have heard Sri Shekhar Devasa, learned counsel appearing for the appellants and Ms. Neerja Sachdeva, learned counsel appearing for the 1st respondent-Insurance Company and perused the material on record. Though notice is served on respondent no.2, he remains unrepresented.
The Tribunal, by recording a finding that the deceased was in the age group of 40 to 50 years, applied the multiplier of 13 while calculating the compensation. The High Court, curiously while maintaining the multiplier of 13 as per the judgment of this Court in the case of Sarla Verma,(supra) has reduced the compensation only on the ground that the deceased was aged 50 years 3 months on the date of the accident, as such the compensation is to be calculated on account of loss of dependency by granting future prospects at 15% but not 30%. So far as the application of multiplier of 13 by the Tribunal is concerned, the High Court has not interfered with the same. When the age of the deceased was considered in the group of 40 to 50 years, we are of the view that the High Court has committed error in granting only 15% towards future prospects instead of 30%. As per the judgments of this Court primarily the age group is to be considered. Considering the age group as 40 to 50 years, when the multiplier of 13 is maintained by the High Court, there is no reason or justification for reducing the compensation by granting 15% towards future prospects. Though the learned counsel appearing for respondent no.1-Insurance Company has submitted that the compensation towards future prospects was awarded as per the Constitution Bench judgment of this Court in the case of National Insurance Company Limited vs. Pranay Sethi & Ors., (2017) 16 SCC 680 but at the same time it is to be noticed that in the very same judgment in paragraph 59.3 while considering the grant of future prospects, this Court has specifically said that the addition should be 30% if the age of the deceased was in the age group of 40 to 50 years. For application of multiplier, the High Court has also accepted the age group of the deceased between 40 and 50 years. In that view of the matter, there is no reason for reducing the compensation by granting future prospects at 15% only. In absence of any challenge to the findings recorded by the High Court confirming the application of multiplier of 13, we are of the view that the High Court has committed error in reducing the compensation on account of loss of dependency. For loss of love and affection, when the compensation of Rs.1,00,000/- on account of loss of consortium was awarded to the first appellant, she was not entitled for another Rs.1,00,000/- towards the same but, at the same time though the appellants have claimed Rs.2,00,000/- towards transportation of dead body and funeral expenses, only an amount of Rs.20,000/- and Rs.25,000/- was awarded towards the respective heads. Taking into account the facts and circumstances of the case, we are of the view that even such grant of Rs.1,00,000/- ought not have been reduced by the High Court.
For the aforesaid reasons, we are of the view that the compensation awarded by the Tribunal is just and reasonable and the same was interfered with by the High Court without any valid grounds, as such, we allow this appeal and set aside the judgment dated 20.07.2016 passed in Misc. First Appeal No. 4903 of 2016 (MV-D) by the High Court of Karnataka at Bengaluru and restore the award dated 29.09.2015 passed in MVC No. 1639 of 2012 by the Motor Accident Claims Tribunal-VI and III Addl. Sr. Civil Judge, Mangalore. No order as to costs.
Tweets by La_Mintage


These days, almost everything is done electronically, including resumes and cover letters. However, some people like to stand out from the rest by including a handwritten cover letter with their resume. They feel it adds a personal touch that is usually missing from today's job seeker and show their attention to the small details.
Make sure you have neat, clear penmanship. After all, your letter and resume are being held up in comparison to others that have been neatly type. Don't use cursive writing, and be sure that your penmanship is easy to read, as neat as possible and isn't rife with a lot of flourishes or scribble. Write in blue or black ink; avoid red or other fancy colors.
Keep it short, simple and to the point. Don't drone on and on about yourself and your personal experiences in the letter. Just because your letter is handwritten doesn't mean it shouldn't be professional. Start out by introducing yourself, give a paragraph or two (short ones) about your professional abilities, provide contact information and close the letter with a thank you and your signature. You want to sell yourself in just a few paragraphs, so be sure to emphasize the things you do best and make a point to showcase why this company needs you.
If possible, write your cover letter on stationery. It adds a sophisticated touch and gives the personal result you were trying to achieve. Stay away from notebook or computer paper, as that can make your cover letter look sloppy, juvenile and unprofessional. Make sure the stationery you choose has a simple design and is easy to write --- and read. Stationery gives your letter a more polished look and has the appearance of a note from a friend or colleague, which is the feeling you are trying to convey with your handwritten cover letter.
When you are submitting your resume, make sure your letter is paper clipped or placed on top. If you must mail your resume, be sure to take care when folding your letter, or use a 9x12 envelope to avoid folding. After all, you don't want your ink to run or bleed on the paper. Don't staple your letter, because again, this tends to look a bit messy and unprofessional. If you can't paper clip it, then simply place the letter on top of your resume and submit it in a plain folder, with your name clearly written in the tab.
Moultry, Lynda. "How to Write a Handwritten Cover Letter for a Resume." , https://careertrend.com/how-4794646-write-handwritten-cover-letter-resume.html. 05 July 2017.


CAREER NOTES: Aleksandra Duliba continues to improve her national record, most recently with a 2:23:06 performance at the 2015 Dubai Marathon. Her personal best time of 2:21:29, however, was earned with her sixth- place performance in Boston last year. Also in 2014, she finished second in Toronto in 2:24:43. Duliba made her marathon debut with a national record win at the 2013 Los Angeles Marathon (2:26:08), and then finished fourth at the Chicago Marathon in 2:23:44. In Los Angeles, the women's elite field was given an 18-minute, 35-second lead over the elite men in a gender challenge. Duliba beat the entire field by 52 seconds to collect a $50,000 bonus for staying ahead of the handicap. The former Belarus national marathon record of 2:26:23 was held by Madina Biktagirova, and coincidentally was set in Los Angeles in 1992. Also in 2013, Duliba won the Dushanbe Half Marathon in 1:15:07 and set a 33:08.80 personal best in the 10,000m. In 2012 she won the bronze medal in the 10,000m at the Belarus National Championships and won the Szamotuly Half Marathon, the Swarzedz 10K, and the Dzialoszyn 10K.
PERSONAL NOTES: Duliba was very athletic as a youth and entered the Sport School of Olympic Reserve to specialize in cross-country skiing. She soon discovered she was very talented running on the track and made the transition. She currently trains in Kiev, Ukraine, with her fiancé Vitaliy Shafar, who is also running the Boston Marathon this year. The pair is coached by Igor Osmak and is affiliated with the Spartnik group and agent Andre Baranov. Duliba's hobbies include reading, listening to classical music, and being surrounded by nature and beautiful architecture.


Sabrblade wrote: Which one? There are four possible questions.
2. What planet do the bots come from?
3. What are the good guys called?
4. What are the bad guys called?
And those were apparently too easy because the spammers got passed them.
Though, it may not be a very good idea to discuss these questions publicly, as anyone could see this and figure out a way to bypass them.
PM me if anyone wants to discuss the questions further.


text:
Q: why the last Value Row values are repeated for the HTML Table I have a static Html Table which contains one button "Add" and when  clicking on that button it creates a new row which contains some text boxes
and it append to this table .But my Problem is say i have 3 rows and when i click "Add" we are able to create the new ROW  but the Previous row values
are copied in to new row that is being generated  .Here is the Function That iam using
    $('#btn1').click(function(){
        tr= $("#Tbl tr:last").html();
        num= $('#num').val()*1;
        first= $("#Tbl tr:last").attr('id').substr(4)*1;
        last=first+num;
        for(i=first+1;i<=last;i++) {
            var newtr= tr.replace(first,i);
            $('#Tbl').append('
'+newtr+'
');
        }
    });

How do i make the  value Selected as  "ABC" in Select box  when the new Row is being added.Iam using this statement But did not workout .How can i make the value as selected
$('#Tbl tr:last :input').find('input[name="code"]').val('ABC');

A: As @Patrick points out, you may be copying in values from the last row.  So why not clear them out after you create the new row?  Do this after the for loop:
$('#Tbl tr:last :input').val('');

This should select the input fields in the last tr of the #Tbl table and remove values from them.  This will specifically select the last row, but now I see that you're adding a variable number of rows, so lets handle that properly.
Since you're appending a bunch of rows, why not clean them up as you add them?
for(i=first+1;i<=last;i++) {
    var newtr= tr.replace(first,i);
    $('#Tbl').append('
'+newtr+'
').find(':input').val('');
}

That should handle as many rows as you can add.

A: I'm not a real jQuery hero, but why not use simply this?
$('#Tbl tr:last').after('
...
')

A: it looks to me that when you do
tr= $("#Tbl tr:last").html();

your grabbing all of the html from inside that previous row (including the values of any controls inside that row) and pasting it into the new row

A: When you type into a text box, it sets the "value" attribute in the HTML. So later, when you copy the innerHTML of the previous row, it takes those "value" attributes and duplicates them in your new row.
Instead of parsing the HTML of the previous row, you might want want to keep your data within your script and use that to generate new rows.
var nextId = 0;
var DEFAULT_ROW_HTML = "
<input type=\"text\" />	...
";
...
$("#Tbl").append(DEFAULT_ROW_HTML.replace("@ID@", nextId));

Know what I mean, jelly bean?

A: Try this:
$('#btn1').click(function(){
    $('#Tbl').each(function(){
        var $table = $(this);
        // Number of td's in the last table row
        var n = $('tr:last td', this).length;
        var tds = '
';
        for(var i = 0; i < n; i++){
            tds += ' 	';
        }
        tds += '
';
        if($('tbody', this).length > 0){
            $('tbody', this).append(tds);
        }else {
            $(this).append(tds);
        }
    });
});


text:
Home Rock THE DARKNESS – The Fonda Theater – 3/29/18
THE DARKNESS – The Fonda Theater – 3/29/18
The Darkness @ Fonda Theater in Los Angeles, CA on 3/29/2018
Review by JoJo-Anthony
Photography by Joe Schaeffer
One thing that can be said about British rockers The Darkness, is that they are definitely a band on a mission. That mission is simple…these lads want to bring back fun and chaos to Rock N' Roll. Their performance at The Fonda Theatre captured both.
I first became aware of the Darkness in 2003 when their video for "I Believe in a Thing Called Love" gained heavy rotation on American television. At first I was a bit perplexed with the band's seventies glam image, the high wailing falsetto vocals and their overtly campy posturing. This led me to question if the band was merely a parody of an age gone by or were they a hundred percent serious? Their debut album "Permission to Land" became an international hit after constant touring and they followed up their success with "One Way Ticket to Hell" in 2005. It was then that I became a devoted fan. Unfortunately the band split up the following year due to personal and health issues. Thankfully, the original members reunited in 2011 and have released three albums since then.
Marching on to the stage at the Fonda Theatre in Los Angeles, The Darkness delivered a blistering ninety minute set which featured material predominantly from their debut album and their current release "Pinewood Smile".
(https://www.amazon.com/Pinewood-Smile-Deluxe-Explicit-Darkness/dp/B073XTXWGX)
The frenzied crowd were definitely there to see the band and have a good time. Opening with "Open Fire" from their fourth album, Justin Hawkins took complete control of the stage. Hawkins is a natural and experienced front man who commands attention and respect from the audience and immediately establishes a rapport with the crowd and treats everyone as though they were merely sitting in at a band rehearsal. Along with his brother Dan on guitar, bassist Frankie Poullain and drummer Rufus Taylor, the four members perform at a high level without the aid of backing musicians or tapes. These guys are a real Rock N' Roll band.
I must admit as a lifelong Queen fan, I am a bit biased towards this band at the addition of young Taylor playing drums (son of the legendary Roger Taylor) He maintains an amazing energy and stamina throughout the performance.
Highlights included "One Way Ticket," "Barbarian," "Solid Gold," "Black Shuck," and my current favorite, "All the Pretty Girls." The band closed the show with three selections from their debut album, "Stuck in a Rut," "Get Your Hands off My Woman" and "Growing on Me". Encores included "Japanese Prisoner of Love" and the much anticipated "I Believe in a Thing Called Love". The song has definitely become a staple and the band's anthem. It was a bit amusing to hear the audience take control of the vocals while Hawkins merely strummed the chords lightly on his guitar while most of the crowd sang in a high falsetto range that could have shattered glass.
The Darkness are definitely one of the constant touring bands out there keeping the hard rock genre alive. They crossover enough to attract fans ranging from teens to baby boomers. A popular thread amongst other contemporary bands like the Struts, Vintage Trouble and Greta Van Fleet.
Judging from the response of the LA crowd, those fans have an undying affection for the band's debut album which was released over fifteen years ago. Baffling?…No, that's just the power and influence of Rock N' Roll. If it's good, there's no expiration date!
Love Is Only a Feeling
Southern Trains
Black Shuck
One Way Ticket
Givin' Up
All the Pretty Girls
Buccaneers of Hispaniola
Every Inch of You
Stuck in a Rut
Get Your Hands Off My Woman
Growing on Me
Japanese Prisoner of Love
The complete album of photos can be found here:
https://www.flickr.com/photos/jschaef00/albums/72157694328603094
The Fonda Theater
Joe Schaeffer
JoJo Anthony
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text:
\section{introduction}
\noindent
Let $X: M^n\to \mathbf{R}^{n+p}$ be an isometric immersion from  an $n$-dimensional  Riemannian
manifold $M^n$ into a Euclidean space $\mathbf{R}^{n+p}$.  One considers a smooth one-parameter
family of immersions:
$$
F(\cdot, t):M^n\to  \mathbf{R}^{n+p}
$$
satisfying $F(\cdot, 0)=X(\cdot)$ and
\begin{equation}
\bigl(\dfrac{\partial F(p,t)}{\partial t}\bigl)^N=H(p,t), \quad (p,t)\in M\times [0,T),
\end{equation}
where $H(p,t)$ denotes the mean curvature vector  of submanifold $M_t=F(M^n,t)$ at point $F(p,t)$.
The equation (1.1) is called the mean curvature flow equation.
A submanifold $X: M^n\to \mathbf{R}^{n+p}$ is said to be a self-shrinker in $\mathbf{R}^{n+p}$ if it satisfies
\begin{equation}
H=-X^N,
\end{equation}
where $X^N$ denotes the orthogonal projection into the normal bundle of $M^n$ (cf. Ecker-Huisken \cite{EH}).
Self-shrinkers play an important role in the study of the mean curvature flow since they are not only
solutions of the mean curvature flow equation, but they also describe all possible blow ups at a given
singularity of a mean curvature flow. Huisken \cite{H1} proved that the sphere of radius $\sqrt n$ is the only closed
embedded self-shrinker hypersurfaces with non-zero mean curvature. For classifications of complete non-compact
embedded self-shrinker hypersurfaces,
Huisken \cite{H2} and Colding and Minicozzi \cite{CM1}  proved that an $n$-dimensional  complete
embedded self-shrinker hypersurface with non-negative mean curvature and polynomial volume growth in  $\mathbf{R}^{n+1}$ is  a Riemannian product $S^k\times \mathbf{R}^{n-k}$,  $0\leq k < n$. Smoczyk \cite{S1} has obtained several results for complete self-shrinkers with higher co-dimensions.\par
\noindent
For study of the rigidity problem for self-shrinkers,   Le and Sesum \cite{LS} and Cao and Li \cite{CL}
have classified   $n$-dimensional  complete embedded self-shrinkers in $\mathbf{R}^{n+p}$ with  polynomial volume growth if the squared norm $|A|^2$ of the second fundamental form satisfies  $|A|^2\leq 1$. For a further study,
see  Colding and Minicozzi \cite{CM2, CM3}, Ding and Wang \cite{DW}, Ding and Xin \cite{DX1, DX2}, Wang \cite{W}  and so on.
\vskip2mm
\noindent
In \cite{CM1}, Colding and Minicozzi introduced a differential operator   $\mathfrak L$ and used it to
study self-shrinkers. The differential operator $\mathfrak L$ is defined by
\begin{equation}
\mathfrak L f=\Delta f -\langle X,\nabla f\rangle
\end{equation}
for a smooth function $f$, where $\Delta$ and $\nabla$ denote the Laplacian and the gradient
operator on the self-shrinker, respectively and $\langle \cdot, \cdot \rangle$ denotes the standard
inner product of $\mathbf R^{n+p}$.  We should notice that the differential operator $\mathfrak L$
plays a very important role in studying of $n$-dimensional  complete embedded self-shrinkers in
$\mathbf{R}^{n+p}$ with  polynomial volume growth  in order to guarantee integration by part holds
as in \cite{CM1}.
\vskip2mm
\noindent
The purpose of  this paper is to  study eigenvalues of the closed eigenvalue problem for
 the differential operator $\mathfrak L$ on compact
self-shrinkers  in $\mathbf{R}^{n+p}$ in sections 3 and 4 and  eigenvalues of the Dirichlet
eigenvalue problem of  the differential operator $\mathfrak L$ on a bounded domain with a piecewise smooth boundary  in complete
self-shrinkers  in $\mathbf{R}^{n+p}$ in section 5.  I shall adapt the idea of Cheng and Yang in \cite{CY1} for studying eigenvalues of the Dirichlet
eigenvalue problem of  the Laplacian $\Delta $  to  the differential operator $\mathfrak L$ by constructing appropriated trial functions
for the differential operator $\mathfrak L$.
Since the differential operator $\mathfrak L$ is self-adjoint
with respect to  measure $e^{-\frac{|X|^2}{2}}dv$, where $dv$ is the volume element of $M^n$
and $|X|^2=\langle X, X\rangle$,
we know that the closed eigenvalue problem:
\begin{equation}
\mathfrak L u=- \lambda u \quad \text{on} \ M^n
\end{equation}
for the differential operator $\mathfrak L$ on compact
self-shrinkers  in $\mathbf{R}^{n+p}$ has  a real and  discrete spectrum:
\begin{equation*}
0=\lambda_0<\lambda _{1} \leq  \lambda _{2} \leq \cdots \leq \lambda _{k} \leq \cdots
 \longrightarrow \infty,
\end{equation*}
where each eigenvalue is repeated according to  its multiplicity.
We shall prove the following:

\vskip 2mm
\noindent
\begin{theorem} Let $M^n$ be an $n$-dimensional compact
self-shrinker  in $\mathbf{R}^{n+p}$. Then,
eigenvalues of the closed eigenvalue problem ${\rm (1.4)}$
satisfy
\begin{equation}
\sum_{i=0}^k (\lambda_{k+1}-\lambda_{i})^2 \leq \frac{4}{n}\sum_{i=0}^k
(\lambda_{k+1}-\lambda_{i})(\lambda_{i}+\frac{2n-\min_{M^n}{|X|^2}}{4}).
\end{equation}
\end{theorem}

\begin{remark}
The sphere $S^n(\sqrt n)$ of radius $\sqrt n$ is a compact self-shrinker in $\mathbf{R}^{n+p}$.
For  $S^n(\sqrt n)$ and  for any $k$, the inequality ${\rm (1.5)}$ for eigenvalues
of the closed eigenvalue problem ${\rm (1.4)}$ becomes  equality.
Hence our results in theorem {\rm 1.1} are sharp.
\end{remark}

\noindent
Furthermore, from the recursion formula of Cheng and Yang \cite{CY}, we can obtain an upper bound for eigenvalue
$\lambda_{k}$:
\begin{theorem} Let $M^n$ be an $n$-dimensional compact
self-shrinker  in $\mathbf{R}^{n+p}$. Then,
eigenvalues of the closed eigenvalue problem ${\rm (1.4)}$
satisfy, for any $k\geq 1$,
\begin{equation*}
\lambda_{k}+\frac{2n-\min_{M^n}{|X|^2}}{4}
\leq (1+\frac {a(min\{n,k-1\})}{n})(\frac{2n-\min_{M^n}{|X|^2}}{4})k^{2/n},
\end{equation*}
where the bound of $a(m)$ can be formulated as:
$$
\left \{ \aligned
a(0)&\leq 4, \\
 a(1)&\leq 2.64,\\
 a(m)&\leq 2.2-4\log(1+\frac1{50}(m-3)),\qquad \mbox{for}\quad
 m\geq 2.
\endaligned \right .
$$
In particular, for $n\ge 41$ and $k\geq 41$, we have
 \begin{equation*}
\lambda_{k}+\frac{2n-\min_{M^n}{|X|^2}}{4}
\leq (\frac{2n-\min_{M^n}{|X|^2}}{4})k^{2/n}.
\end{equation*}
\end{theorem}
\vskip 1mm
\noindent
Results for eigenvalues of the Dirichlet eigenvalue problem of the differential operator $\mathfrak L$
are given in section 5.

\vskip 2mm
\noindent
{\bf Acknowledgements.}  We would like to express our gratitude to
the referee for valuable  suggestions and comments.

\section{Preliminaries}
\noindent
Suppose $X:M^{n}\longrightarrow \mathbf{R}^{n+p}$ is an isometric
immersion from Riemannian manifold $M^n$ into the (n+p)-dimensional
Euclidean space $\mathbf {R}^{n+p}$.  Let $\{E_A\}_{A=1}^{n+p}$ be the standard
basis of $\mathbf{R}^{n+p}$.  The position vector can be written by
$X=(x_1, x_2, \cdots, x_{n+p})$.
We choose a local orthonormal frame field
$\{e_1, e_2, \cdots,e_n, e_{n+1}, \cdots, e_{n+p} \}$ and the dual
coframe field
$\{\omega_1, \omega_2, \cdots,\omega_{n}, \omega_{n+1}, \cdots, \omega_{n+p}\}$
along $M^n$ of  $\mathbf{R}^{n+p}$ such that  $\{e_1, e_2, \cdots,e_n \}$
is a local orthonormal basis on $M^n$. Thus,  we have
$$
\omega_{\alpha}=0, \quad n+1\leq\alpha\leq n+p
$$
on $M^n$.
From the  Cartan's lemma, we have
$$
\omega_{i\alpha}=\sum_{j=1}^nh^{\alpha}_{ij}\omega_j, \quad  h^{\alpha}_{ij}=h^{\alpha}_{ji}.
$$
The  second fundamental form ${\bf h}$ of $M^n$ and the mean curvature vector $H$
are defined, respectively,  by
$$
{\bf h}=\sum_{\alpha=n+1}^{n+p}\sum_{i,j=1}^n h^{\alpha}_{ij}\omega_i\otimes \omega_je_{\alpha}
\quad H=\sum_{\alpha=n+1}^{n+p}\sum_{i=1}^n h^{\alpha}_{ii}e_{\alpha}.
$$
One considers the mean curvature flow for a submanifold  $X: M^n\to \mathbf{R}^{n+p}$. Namely,
we consider a  one-parameter family of immersions:
\begin{equation*}
F(\cdot, t):M^n\to  \mathbf{R}^{n+p}
\end{equation*}
satisfying $F(\cdot, 0)=X(\cdot)$ and
\begin{equation}
\bigl(\dfrac{\partial F(p,t)}{\partial t}\bigl)^N=H(p,t), \quad (p,t)\in M\times [0,T),
\end{equation}
where $H(p,t)$ denotes the mean curvature vector  of submanifold $M_t=F(M^n,t)$ at point $F(p,t)$.
An important class of solutions to the mean curvature flow equation (2.1) are self-similar shrinkers,
which profiles, self-shrinkers, satisfy
\begin{equation*}
H=-X^N,
\end{equation*}
which is a system of quasi-linear elliptic partial differential equations of the second order.
Here $X^N$ denotes the orthogonal projection of $X$  into the normal bundle of $M^n$.

In \cite{CM1}, Colding and Minicozzi introduced a differential operator   $\mathfrak L$ and used it to
study self-shrinkers. The differential operator $\mathfrak L$ is defined by
\begin{equation}
\mathfrak L f=\Delta f -\langle X,\nabla f\rangle
\end{equation}
for a smooth function $f$, where $\Delta$ and $\nabla$ denote the Laplacian and the gradient
operator on the self-shrinker, respectively.
For a compact self-shrinker $M^n$ without boundary,  we have
\begin{equation*}
\begin{aligned}
&\int_{M^n}f\mathfrak L u \ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{M^n}f(\Delta u -\langle X,\nabla u\rangle)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{M^n}f\text{div}( e^{-\frac{|X|^2}{2}}\nabla u)dv\\
&=\int_{M^n}u \mathfrak L f \ e^{-\frac{|X|^2}{2}}dv,\\
\end{aligned}
\end{equation*}
that is,
\begin{equation}
\begin{aligned}
&\int_{M^n}f\mathfrak L u \ e^{-\frac{|X|^2}{2}}dv
=\int_{M^n}u \mathfrak L f \ e^{-\frac{|X|^2}{2}}dv,\\
\end{aligned}
\end{equation}
for any smooth functions $u, f$.
Hence, the differential operator $\mathfrak L$ is self-adjoint with respect to the  measure $e^{-\frac{|X|^2}{2}}dv$.
Therefore, we know that the closed eigenvalue problem:
\begin{equation}
\mathfrak L u=- \lambda u \quad \text{on} \ M^n
\end{equation}
has  a real and  discrete spectrum:
\begin{equation*}
0=\lambda_0<\lambda _{1} \leq  \lambda _{2} \leq \cdots \leq \lambda _{k} \leq \cdots
 \longrightarrow \infty.
\end{equation*}
Furthermore,  we have
\begin{equation}
\mathfrak L x_A=-x_A.
\end{equation}
In fact,
\begin{equation*}
\begin{aligned}
&\mathfrak L x_A=\Delta \langle X, E_A\rangle-\langle X,\nabla x_A\rangle\\
& =\langle \Delta X, E_A\rangle-\langle X,E_A^T\rangle\\
& =\langle H, E_A\rangle-\langle X,E_A^T\rangle\\
& =-\langle X^N, E_A\rangle-\langle X,E_A^T\rangle=-x_A.
\end{aligned}
\end{equation*}
Denote the induced metric by $g$ and define $\nabla u\cdot\nabla v=g(\nabla u,\nabla v)$ for
functions $u, v$.
We get,  from (2.5),
\begin{equation}
\mathfrak L |X|^2=\sum_{A=1}^{n+p}\bigl(2x_A\mathfrak L x_A+2\nabla x_A\cdot\nabla x_A\bigl)=2(n-|X|^2).
\end{equation}
Here we have used
\begin{equation*}
\sum_{A=1}^{n+p}\nabla x_A\cdot\nabla x_A=n.
\end{equation*}
\begin{proposition} For an $n$-dimensional compact  self-shrinker $M^n$  without boundary in $\mathbf{R}^{n+p}$,
we have
\begin{equation*}
 \min_{M^n}|X|^2\leq n=\dfrac{\int_{M^n}|X|^2 \ e^{-\frac{|X|^2}{2}}dv}
{\int_{M^n} e^{-\frac{|X|^2}{2}}dv}\leq \max_{M^n}|X^N|^2.
\end{equation*}
\end{proposition}
\begin{proof}
Since $\mathfrak L$ is self-adjoint with respect to the  measure $e^{-\frac{|X|^2}{2}}dv$, from (2.6), we have
$$
n\int_{M^n} e^{-\frac{|X|^2}{2}}dv=\int_{M^n}|X|^2 \ e^{-\frac{|X|^2}{2}}dv
\geq \min_{M^n}|X|^2\int_{M^n} e^{-\frac{|X|^2}{2}}dv.
$$
Furthermore, since
\begin{equation}
\Delta |X|^2=2(n+\langle X, H\rangle)=2(n-|X^N|^2),
\end{equation}
we have
$$
n\leq \max_{M^n}|X^N|^2.
$$
It completes the proof of this proposition.
\end{proof}

\section{Universal estimates for eigenvalues}
\noindent
In this section, we give proof of the theorem 1.1.
In order to prove our theorem 1.1, we need to construct trial functions.
Thank to $\mathfrak L X=-X$. We can use  coordinate functions of the position vector $X$
of the  self-shrinker  $M^n$ to construct trial functions.

\vskip2mm
\noindent
{\it Proof of Theorem {\rm1.1}}.
For an $n$-dimensional compact  self-shrinker $M^n$  in $\mathbf{R}^{n+p}$,
the closed eigenvalue problem:
\begin{equation}
\mathfrak L u=- \lambda u \quad \text{on} \ M^n
\end{equation}
for the differential operator $\mathfrak L$ has a discrete spectrum.
For any integer $j\geq 0$, let $u_j$ be an eigenfunction corresponding to the eigenvalue $\lambda_j$
such that
\begin{equation}
\begin{cases}
\begin{aligned}
&\mathfrak L u_j=- \lambda_j u_j \quad \text{on} \ M^n\\
&\int_{M^n}u_iu_j \ e^{-\frac{|X|^2}{2}}dv=\delta_{ij}, \ \text{for any} \ i, j.
\end{aligned}
\end{cases}
\end{equation}
From the Rayleigh-Ritz inequality, we have
\begin{equation}
\lambda_{k+1} \leq \frac{\displaystyle{-\int_{M^n}}\varphi\mathfrak L \varphi  \ e^{-\frac{|X|^2}{2}}dv}
{\displaystyle{\int_{M^n}}\varphi^2 \ e^{-\frac{|X|^2}{2}}dv},
\end{equation}
for any function $\varphi$ satisfies $\int_{M^n}\varphi u_j \ e^{-\frac{|X|^2}{2}}dv$, $0\leq j\leq k$.
Since $X: M^n\to \mathbf {R}^{n+p}$ is a self-shrinker in $\mathbf {R}^{n+p}$, we have
\begin{equation}
H=-X^N.
\end{equation}
Letting $x_A$, $A=1, 2, \cdots, n+p $,  denote components of the position vector $X$,
we define, for $0\leq i\leq k$,
\begin{equation}
\varphi_i^A:=x_A u_i-\sum_{j=0}^k a_{ij}^Au_j, \quad a_{ij}^A=\int_{M^n}x_Au_iu_j \ e^{-\frac{|X|^2}{2}}dv.
\end{equation}
By a simple calculation, we obtain
\begin{equation}
\displaystyle{\int_{M^n} u_j\varphi_i^Ae^{-\frac{|X|^2}{2}}dv}=0,\hspace{5mm}i,j=0, 1,\cdots, k.
\end{equation}
From the Rayleigh-Ritz inequality, we have
\begin{equation}
\lambda_{k+1} \leq \frac{\displaystyle{-\int_{M^n}}\varphi_i^A\mathfrak L \varphi_i^A  \ e^{-\frac{|X|^2}{2}}dv}
{\displaystyle{\int_{M^n}}(\varphi_i^A)^2 \ e^{-\frac{|X|^2}{2}}dv}.
\end{equation}
Since
\begin{equation}
\begin{aligned}
&\mathfrak L \varphi_i^A=\Delta  \varphi_i^A-\langle X, \nabla  \varphi_i^A\rangle\\
&=\Delta (x_A u_i-\sum_{j=0}^k a_{ij}^Au_j)-\langle X, \nabla  (x_A u_i-\sum_{j=0}^k a_{ij}^Au_j)\rangle\\
&=x_A\Delta u_i + u_i\Delta x_A +2\nabla x_A\cdot \nabla u_i -\langle X, x_A\nabla  u_i+u_i\nabla x_A\rangle\\
&-\sum_{j=0}^k a_{ij}^A\Delta u_j +\langle X, \sum_{j=0}^k a_{ij}^A\nabla u_j)\\
&=-\lambda_ix_Au_i+u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i+\sum_{j=0}^k a_{ij}^A\lambda_ju_j,
\end{aligned}
\end{equation}
we have, from (3.7) and (3.8),
\begin{equation}
\begin{split}
&(\lambda_{k+1}-\lambda_{i})||\varphi_i^A||^2 \\
&\leq -\int_{M^n}\varphi_i^A(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i) \ e^{-\frac{|X|^2}{2}}dv:=W_i^A,
\end{split}
\end{equation}
where
$$
||\varphi_i^A||^2 =\int_{M^n}(\varphi_i^A)^2 \ e^{-\frac{|X|^2}{2}}dv.
$$
$$
$$
On the other hand, defining
\begin{equation*}
b_{ij}^A=-\int_{M^n}(u_j\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_j)u_i \ e^{-\frac{|X|^2}{2}}dv
\end{equation*}
we obtain
\begin{equation}
b_{ij}^A=(\lambda_i-\lambda_j)a_{ij}^A.
\end{equation}
In fact,
\begin{equation*}
\begin{aligned}
\lambda_ia_{ij}^A&=\int_{M^n}\lambda_iu_iu_jx_A e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}u_jx_A\mathfrak Lu_i \ e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}u_i\mathfrak L(u_jx_A) \ e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}u_i(x_A\mathfrak Lu_j+u_j\mathfrak L x_A+2\nabla x_A\cdot \nabla u_j) \ e^{-\frac{|X|^2}{2}}dv\\
&=\lambda_ja_{ij}^A+b_{ij}^A,
\end{aligned}
\end{equation*}
that is,
\begin{equation*}
b_{ij}^A=(\lambda_i-\lambda_j)a_{ij}^A.
\end{equation*}
Hence, we have
\begin{equation}
b_{ij}^A=-b_{ji}^A.
\end{equation}
From (3.6), (3.9)  and the Cauchy-Schwarz  inequality, we infer
\begin{equation}
\begin{aligned}
W_i^A&=-\int_{M^n}\varphi_i^A\bigl(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}\varphi_i^A\bigl(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
&\leq \|\varphi_i^A\|\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|.
\end{aligned}
\end{equation}
Hence, we have, from (3.9) and (3.12),
\begin{equation*}
\begin{split}
&(\lambda_{k+1}-\lambda_{i})(W_i^A)^2\\
&=(\lambda_{k+1}-\lambda_{i})\|\varphi_i^A\|^2\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|^2\\
&\leq W_i^A\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|^2.\\
\end{split}
\end{equation*}
Therefore, we obtain
\begin{equation}
(\lambda_{k+1}-\lambda_{i})^2W_i^A
\leq (\lambda_{k+1}-\lambda_{i})\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|^2.
\end{equation}
Summing  on $i$ from $0$ to $k$ for (3.13),  we have
\begin{equation}
\sum_{i=0}^k(\lambda_{k+1}-\lambda_{i})^2W_i^A
\leq \sum_{i=0}^k(\lambda_{k+1}-\lambda_{i})\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|^2.
\end{equation}
By the definition of $b_{ij}^A$ and (3.10), we have
\begin{equation}
\begin{aligned}
&\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i
-\sum_{j=0}^kb_{ij}^Au_j\|^2\\
&=\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\|^2\\
&-2\sum_{j=0}^kb_{ij}^A\int_{M^n}(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i)u_j\ e^{-\frac{|X|^2}{2}}dv
+\sum_{j=0}^k(b_{ij}^A)^2\\
&=\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\|^2-\sum_{j=0}^k(b_{ij}^A)^2\\
&=\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\|^2-\sum_{j=0}^k (\lambda_i-\lambda_j)^2(a_{ij}^A)^2.\\
\end{aligned}
\end{equation}
Furthermore, according to  the definitions of $W_i^A$ and $\varphi_i^A$, we have from (3.10)
\begin{equation}
\begin{aligned}
W_i^A&=-\int_{M^n}\varphi_i^A\bigl(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}(x_A u_i-\sum_{j=0}^k a_{ij}^Au_j)\bigl(u_i\mathfrak Lx_A
+2\nabla x_A\cdot \nabla u_i\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
&=-\int_{M^n}(x_A u_i^2\mathfrak Lx_A
+2x_Au_i\nabla x_A\cdot \nabla u_i\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
&+\sum_{j=0}^k a_{ij}^A\int_{M^n}u_j( u_i\mathfrak Lx_A
+2\nabla x_A\cdot \nabla u_i\bigl) \ e^{-\frac{|X|^2}{2}}dv\\
\end{aligned}
\end{equation}
\begin{equation*}
\begin{aligned}
&=-\int_{M^n}\bigl(x_A\mathfrak Lx_A-\frac12\mathfrak L(x_A)^2\bigl)u_i^2 \ e^{-\frac{|X|^2}{2}}dv
+\sum_{j=0}^k a_{ij}^Ab_{ij}^A\\
&=\int_{M^n}\nabla x_A\cdot \nabla x_A  u_i^2\ e^{-\frac{|X|^2}{2}}dv
+\sum_{j=0}^k (\lambda_i-\lambda_j)(a_{ij}^A)^2.\\
\end{aligned}
\end{equation*}
Since
\begin{equation}
\begin{aligned}
&2\sum_{i,j=0}^k (\lambda_{k+1}-\lambda_{i})^2 (\lambda_i-\lambda_j)(a_{ij}^A)^2\\
&=\sum_{i,j=0}^k (\lambda_{k+1}-\lambda_{i})^2 (\lambda_i-\lambda_j)(a_{ij}^A)^2-\sum_{i,j=0}^k (\lambda_{k+1}-\lambda_{j})^2 (\lambda_i-\lambda_j)(a_{ij}^A)^2\\
\end{aligned}
\end{equation}
\begin{equation*}
\begin{aligned}
&=-\sum_{i,j=0}^k (\lambda_{k+1}-\lambda_{i}+\lambda_{k+1}-\lambda_j) (\lambda_i-\lambda_j)^2(a_{ij}^A)^2\\
&=-2\sum_{i,j=0}^k (\lambda_{k+1}-\lambda_{i}) (\lambda_i-\lambda_j)^2(a_{ij}^A)^2,
\end{aligned}
\end{equation*}
from (3.14), (3.15), (3.16) and (3.17), we obtain, for any $A$, $A=1, 2, \cdots, n+p$,
\begin{equation}
\begin{aligned}
&\sum_{i=0}^k (\lambda_{k+1}-\lambda_{i})^2\int_{M^n}\nabla x_A\cdot \nabla x_A  u_i^2\ e^{-\frac{|X|^2}{2}}dv\\
&\leq \sum_{i=0}^k (\lambda_{k+1}-\lambda_{i})\|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\|^2.
\end{aligned}
\end{equation}
On the other hand, since
\begin{equation*}
\mathfrak L x_A=-x_A, \quad \sum_{A=1}^{n+p}(\nabla x_A\cdot \nabla u_i)^2=\nabla u_i\cdot\nabla u_i,
\end{equation*}
we infer, from  (2.6),
\begin{equation}
\begin{aligned}
&\sum_{A=1}^{n+p} \|u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\|^2\\
&=\sum_{A=1}^{n+p}\int_{M^n}\bigl(u_i\mathfrak Lx_A+2\nabla x_A\cdot \nabla u_i\bigl)^2\ e^{-\frac{|X|^2}{2}}dv\\
&=\sum_{A=1}^{n+p}\int_{M^n}\bigl(u_i^2(x_A)^2-4u_ix_A\nabla x_A\cdot \nabla u_i
+4(\nabla x_A\cdot \nabla u_i)^2\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\sum_{A=1}^{n+p}\int_{M^n}\bigl(u_i^2(x_A)^2-\nabla (x_A)^2\cdot \nabla u_i^2\bigl)\ e^{-\frac{|X|^2}{2}}dv
+4\int_{M^n}\nabla u_i\cdot\nabla u_i\ e^{-\frac{|X|^2}{2}}dv\\
\end{aligned}
\end{equation}
\begin{equation*}
\begin{aligned}
&=\int_{M^n}(\mathfrak L|X|^2+|X|^2) u_i^2\ e^{-\frac{|X|^2}{2}}dv+4\lambda_i\\
&=\int_{M^n}(2n-|X|^2) u_i^2\ e^{-\frac{|X|^2}{2}}dv+4\lambda_i\\
&\leq(2n-\min_{M^n}{|X|^2})+4\lambda_i.\\
\end{aligned}
\end{equation*}
Furthermore, because of
\begin{equation}
\sum_{A=1}^{n+p}\nabla x_A\cdot \nabla x_A =n,
\end{equation}
taking summation on $A$ from $1$ to $n+p$ for (3.18) and using  (3.19) and (3.20), we get
\begin{equation*}
\begin{aligned}
&\sum_{i=0}^k (\lambda_{k+1}-\lambda_{i})^2\leq \frac 4n\sum_{i=0}^k (\lambda_{k+1}-\lambda_{i})
(\lambda_i+\frac {2n-\min_{M^n}{|X|^2}}4).
\end{aligned}
\end{equation*}
It finished the proof of the theorem 1.1.
\begin{flushright}
$\square$
\end{flushright}

\section{Upper bounds for eigenvalues}

\noindent
The following recursion formula of Cheng and Yang \cite{CY} plays a very important role in order to
prove the theorem 1.2.

\vskip 2mm
\noindent
{\bf A recursion formula of Cheng and Yang}.  Let  $\mu_1 \leq  \mu_2 \leq  \dots,
\leq \mu_{k+1}$ be any positive  real numbers satisfying
\begin{equation*}
  \sum_{i=1}^k(\mu_{k+1}-\mu_i)^2 \le
 \frac 4n\sum_{i=1}^k\mu_i(\mu_{k+1} -\mu_i).
\end{equation*}
 Define
 \begin{equation*}
 \Lambda_k=\frac 1k\sum_{i=1}^k\mu_i,\qquad T_k=\frac 1k
\sum_{i=1}^k\mu_i^2, \ \ \
F_k=\left (1+\frac 2n \right )\Lambda_k^2-T_k.
\end{equation*}
Then, we have
\begin{equation}
F_{k+1}\leq C(n,k) \left ( \frac {k+1}k \right )^{\frac 4n}F_k,
\end{equation}
where
$$
C(n,k) =1-\frac 1{3n}
  \left (\frac k{k+1}\right )^{\frac
  4n}\frac {\left(1+\frac 2n\right )\left (1+
  \frac 4n\right )}{(k+1)^3}<1.
$$

\vskip 2mm
\noindent
{\it Proof of Theorem {\rm 1.2}.}
From the proposition 2.1, we know
$$
\mu_{i+1}=\lambda_{i}+\dfrac{2n-\min_{M^n} |X|^2}{4}>0,
$$
for any $i=0, 1, 2, \cdots$.  Then, we obtain from (1.5)
\begin{equation}
\sum_{i=1}^k (\mu_{k+1}-\mu_{i})^2 \leq \frac{4}{n}\sum_{i=1}^k
(\mu_{k+1}-\mu_{i})\mu_i.
\end{equation}
Thus, we know that $\mu_i$'s satisfy the condition of the above recursion formula of Cheng and Yang \cite{CY}.
Furthermore,  since
\begin{equation*}
\mathfrak L x_A=-x_A  \ \text{\rm and} \   \int_{M^n}x_A\ e^{-\frac{|X|^2}{2}}dv=0,  \quad \text{\rm for} \  A=1, 2, \cdots, n+p,
\end{equation*}
$\lambda=1$ is an eigenvalue of $\mathfrak L$ with multiplicity at least $n+p$.
Thus,
$$
\lambda_{1}\leq \lambda_{2}\leq \cdots\leq \lambda_{n+1}\leq 1.
$$
Hence,  we have
\begin{equation}
\sum_{j=1}^{n}(\mu_{j+1}-\mu_1)=\sum_{j=1}^{n}\lambda_{j}\leq n\leq 2n-\min_{M^n} |X|^2=4\mu_1
\end{equation}
because of $\min_{M^n} |X|^2\leq n$ according to the proposition 2.1.
Hence, we can prove  the theorem 1.2  as in Cheng and Yang \cite{CY} almost word by word.
For the convenience of readers, we shall give a self contained proof.
First of all, according to the  above recursion formula of Cheng and Yang, we have
\begin{equation*}
F_{k} \le  C(n,k-1) \left (\frac k{k-1} \right )^ {\frac
4n}F_{k-1} \le  k^{\frac 4n} F_1=\frac 2n k^{\frac 4n}\mu_1^2.
\end{equation*}
Furthermore, we infer,  from (4.2)
$$
\left [\mu_{k+1} -\left (1+\frac 2n\right)
 \Lambda_k \right ]^2
\le \left (1+\frac 4n\right )F_k-\frac 2n \left (1+\frac 2n\right
)\Lambda_k^2.
$$
Hence, we have
$$
\frac {\frac 2n} {\left (1+\frac 4n\right)}\mu_{k+1}^2+ \frac
{1+\frac2n}{1+\frac 4n } \left (\mu_{k+1}- \left (1+\frac
4n\right )\Lambda_k \right )^2 \le \left (1+\frac 4n\right )F_k.
$$
Thus, we derive
\begin{equation}
\mu_{k+1} \le \left (1+\frac 4n\right ) \sqrt{\frac n2 F_k}
 \le \left (1+\frac 4n\right ) k^{\frac 2n}\mu_1.
\end{equation}
Define
\begin{align*}
&a_1(n)=\frac{n(1+\frac4n)\left(1+\frac 8{n+1}+\frac 8{(n+1)^2}
\right )^{\frac 12}}{(n+1)^{\frac 2n}}-n,\\
&a_2(k,n)=\frac n{k^{\frac 2n}}\left (1+\frac
{4(n+k+4)}{n^2+5n-4(k-1)}\right)-n,\\
&a_2(k)=\max \{ a(n,k),k \le n\le 400\}, \\
 &a_3(k)=\dfrac{4}{1-\frac k{400}}-2\log k,\\
 &a(k) =\max \{a_1(k), a_2(k+1)), a_3(k+1) \}.
\end{align*}

\vskip2mm
\noindent
{\it The case 1}.  For   $k \ge n+1$, we have
\begin{equation}
\aligned \mu_{k+1}\le &\frac {\left (1+\frac 4n\right)
\left(1+\frac 8{n+1}+\frac 8{(n+1)^2}
\right )^{\frac 12}}{(n+1)^{\frac 2n}}k^{\frac 2n}\mu_1\\
&=\left (1+\frac {a_1(n)}n\right)k^{\frac 2n}\mu_1,
\endaligned
\end{equation}
where $a_1(n) \le 2.31$. In fact,
since
$\mu_{k+1}$ satisfies (4.2),  we have, from (4.1),
\begin{equation}
\mu_{k+1}^2 \le  \frac n2 \left (1+\frac 4n\right)^2F_k \le
\frac n2 \left (1+\frac 4n\right)^2 \left(\frac k{n+1}\right
)^{\frac 4n}F_{n+1}.
\end{equation}
On the other hand,
\begin{equation}
\aligned F_{n+1}=&\frac 2n\Lambda_{n+1}^2-\sum_{i=1}^{n+1}\frac
{(\mu_i-\Lambda_{n+1})^2}{n+1}\\
\le&\frac 2n\Lambda_{n+1}^2-\frac {(\mu_1-\Lambda_{n+1})^2+
\frac 1n(\mu_1-\Lambda_{n+1})^2}{n+1}\\
=&\frac 2n\left (\Lambda_{n+1}^2-\frac
{(\mu_1-\Lambda_{n+1})^2}2\right).
\endaligned
\end{equation}
It is obvious that $\Lambda_{n+1}^2-\dfrac
{(\mu_1-\Lambda_{n+1})^2}2$ is an increasing function of
$\Lambda_{n+1}$.  From (4.3), we have
\begin{equation}
\mu_{n+1}+\dots +\mu_2 \le (n+4)\mu_1.
\end{equation}
Thus,  we derive
\begin{equation}
\Lambda_{n+1} \le (1+\frac4{n+1})\mu_1.
\end{equation}
Hence, we have
\begin{equation}
\frac n2F_{n+1}\leq \left(1+\frac 8{n+1}+\frac 8{(n+1)^2}
\right)\mu_1^2.
\end{equation}
From (4.6) and (4.10), we complete the proof of  (4.5).
\vskip2mm
\noindent
{\it The case 2}.   For $k \ge 55$ and $n\ge 54$, we have
\begin{equation}
\aligned \mu_{k+1}\le k^{\frac 2n}\mu_1.
\endaligned
\end{equation}

\noindent
If $ k\ge n+1$, from the case 1, we have
$$
\mu_{k+1} \le\frac 1{(n+1)^{\frac 2n}} \left (1+ \frac
4n\right )^2 k^{\frac 2n} \mu_1.
$$
Since
\begin{equation}
\begin{aligned} &(n+1)^{\frac 2n}=\exp \left (\frac 2n\log (n+1)\right)\\
&\ge 1+\frac 2n\log (n+1)+\frac 2{n^2}(\log (n+1))^2\\
&\ge \left (1+\frac 1n\log (n+1)\right )^2,
\end{aligned}
\end{equation}
we have
\begin{equation}
\mu_{k+1} \le \left (\frac{ 1+\frac 4n}{1+\frac 1n \log
(n+1)}\right )^2 k^{\frac 2n}\mu_1.
\end{equation}
Then, when $n\ge 54$, $\log (n+1) \geq  4 $,  we have
$$
\mu_{k+1} \le k^{\frac 2n}\mu_1.
$$
On the other hand, if $k\le n$, then $ \Lambda_k \le
\Lambda_{n+1}$.
Since
\begin{align*}
\frac n2F_k&
=\Lambda_k^2-\frac{n}2\frac{\sum_{i=1}^k(\mu_i-\Lambda_k)^2}k\\
  &\leq\Lambda_k^2-\frac{n}2\frac{(\mu_1-\Lambda_k)^2
  +\dfrac{\left\{\sum_{i=2}^k(\mu_i-\Lambda_k)\right\}^2}{k-1}}k\\
&\leq\Lambda_{k}^2-\frac{(\mu_1-\Lambda_{k})^2}2\\
&\leq
\Lambda_{n+1}^2-\frac{(\mu_1-\Lambda_{n+1})^2}2\leq
(1+\frac4n)^2\mu_1^2,
\end{align*}
 we have
$$
\aligned \mu_{k+1} \le &\left (1+\frac 4n \right ) \sqrt{\frac
n2 F_k} \le \frac 1{k^{\frac 2n}}\left ( 1+\frac 4n\right )^2
      k^{\frac 2n}\mu_1      \le \left (\frac {1+\frac 4 n}{1+\frac {\log k}n}
      \right )^2
      k^{\frac 2n} \mu_1.
      \endaligned
      $$
 Here we used $ k^{\frac 2n} \geq(1+\frac {\log k}n)^2$.
By the same assertion as above, when $k\geq 55$, we also have
$$
\mu_{k+1} \leq k^{\frac 2n}\mu_1.
$$
\vskip2mm
\noindent
 {\it The case 3}.  For $k\leq 54$ and $k\leq n$, we have
 \begin{equation*}
\begin{aligned}
 \mu_{k+1}\leq (1+\frac{\max\{a_2(k),a_3(k)\}}n)k^{\frac2n}\mu_1.
\end{aligned}
\end{equation*}
Because of $k\le n$ and $k \le 54$,
from (4.3), we derive,
\begin{equation}
\mu_{k+1} \le \frac{1}{n-k+1}\{(n+5)\mu_1 -k\Lambda_k\}.
\end{equation}
Since the formula (4.2) is a quadratic inequality for $\mu_{k+1}$, we have
 \begin{equation}
  \mu_{k+1} \le \left (1+\frac 4n \right )\Lambda_k.
\end{equation}
Since the right hand side of (4.14) is a decreasing function of
$\Lambda_k$ and the right hand side of (4.15) is an increasing
function of $\Lambda_k$, for $\dfrac{1}{n-k+1}\{(n+5)\mu_1
-k\Lambda_k\} =\left (1+\frac 4n \right )\Lambda_k$,  we infer
\begin{equation}
\begin{aligned} \mu_{k+1}\le&\frac 1{k^{\frac 2n}}\left (1+\frac
{4(n+k+4)}{n^2+5n-4(k-1)}\right)
k^{\frac 2n}\mu_1\\
&=\left (1+\frac { a_2(k,n)}n\right )k^{\frac 2n}\mu_1.
\end{aligned}
\end{equation}
From the definition of  $a_2(k)=\max \{ a(n,k),k \le n\le 400\}$, when $n\leq 400$, we obtain
\begin{equation}
\mu_{k+1} \le \left (1+\frac { a_2(k)}n\right )k^{\frac2n}\mu_1.
\end{equation}
When $n > 400$ holds, from (4.4), we have
$$
\mu_{k+1}\le \left ( 1+\frac 4{n-k}\right )\mu_1.
$$
Since $n>400$ and $k\leq 54$, we know $\frac2n\log k<\frac1{50}$.
Hence, we have
\begin{align*}
k^{-\frac2n}=e^{-\frac2n\log k}
&=1-\frac2n\log k+\frac12(\frac2n\log k)^2-\cdots\\
&\leq 1-\frac2n\log k+\frac12(\frac2n\log k)^2.
\end{align*}
Therefore, we obtain
\begin{align*}
&(1+\frac{4}{n-k})k^{-\frac2n}\\
&\leq(1+\frac{4}{n-k})\left(1-\frac2n\log k+\frac12(\frac2n\log k)^2\right)\\
&\leq 1+\frac {\left (4/(1-\frac k{400})-2\log k \right)}n.
\end{align*}
Hence, we infer
\begin{equation}
\begin{aligned}
 \mu_{k+1}\le& \left ( 1+\frac 4{n-k}\right )k^{-\frac2n}k^{\frac2n}\mu_1\\
\le& \left (1+\frac {\left (4/(1-\frac k{400})-2\log k \right)}
n\right )k^{\frac 2n}\mu_1\\
=&\left (1+ \frac {a_3(k)}n\right )k^{\frac 2n}\mu_1.
\end{aligned}
\end{equation}
By Table 1 of the values of $a_1(k)$, $a_2(k+1)$ and $a_3(k+1)$
which are calculated by using Mathematica and are listed up in the
next page,  we have
 $ a_1(1) \le a_2(2) \leq a_3(2) =a(1)\le 2.64$ and, for $ k\geq 2$,
$$
a_3(k+1) \le a_2 (k+1) \le a_1(k).
$$
Hence, $a(k)=a_1(k)$ for $k\geq 2$.
Further, for $k\geq 41$, we know $a(k)<0$.
Hence, for $k\geq 2$, we derive
$$
\mu_{k+1}\leq (1+\frac{a(\min\{n,k-1\})}n)k^{\frac2n}\mu_1
$$
and for $n\geq 41$ and $k\geq 41$,  we have
$$
\mu_{k+1}\leq k^{\frac2n}\mu_1.
$$
When $k=1$, $a(0)=4$ from (4.4). It is easy to check that, when $k\ge 3$,
 by a simple calculation,
$$
a(k) \le 2.2-4\log (1+\frac{k-3}{50}).
$$
This completes the proof of the theorem 1.2.
\begin{flushright}
$\square$
\end{flushright}

\vskip 3mm
{\begin{center}{Table 1: The values of $a_1(k)$, $a_2(k+1)$ and $a_3(k+1)$}
\end{center}}
\begin{table}[htbp]
\begin{center}
\begin{tabular}{|c|rrrrrrrrrrrr} \hline
$k$                &1&2&3&4&5&6&7&8&9&10&\\ \hline
$a_1(k)\leq $  &2.31&2.27&2.2&2.12&2.03&1.94&1.86&1.77&1.69&1.61\\ \hline
$a_2(k+1)\leq $   &2.62&2.05&2.00&1.96&1.90&1.84&1.77&1.70&1.63&1.56\\ \hline
$a_3(k+1)\leq $  &2.64&1.84&1.27&0.84&0.48&0.18&-0.07&-0.30&-0.50&-0.68\\
 \hline
 \\
$k$                &11&12&13&14&15&16&17&18&19&20&\\ \hline
$a_1(k)\leq$  &1.53&1.46&1.39&1.32&1.25&1.18&1.12&1.06&1.00&0.94\\ \hline
$a_2(k+1)\leq $   &1.49&1.42&1.35&1.29&1.22&1.16&1.10&1.04&0.98&0.92\\ \hline
$a_3(k+1) \leq $  &-0.84&-0.99&-1.13&-1.26&-1.37&-1.48&-1.59&-1.68&-1.78&-1.86\\ \hline
 \\
$k$                &21&22&23&24&25&26&27&28&29&30&\\ \hline
$a_1(k)\leq $  &0.89&0.83&0.78&0.72&0.67&0.62&0.58&0.53&0.48&0.44\\ \hline
$a_2(k+1)\leq $   &0.87&0.82&0.76&0.71&0.66&0.61&0.57&0.52&0.47&0.43&\\ \hline
$a_3(k+1)\leq $  &-1.94&-2.02&-2.10&-2.17&-2.23&-2.30&-2.36&-2.42&-2.47&-2.53\\ \hline
 \\
 $k$                &31&32&33&34&35&36&37&38&39&40&41&\\ \hline
$a_1(k)\leq $  &0.39&0.35&0.31&0.27&0.23&0.19&0.15&0.11&0.07&0.03&-0.00\\ \hline
$a_2(k+1)\leq $  &0.38&0.34&0.30&0.26&0.22&0.18&0.14&0.10&0.07&0.03&-0.01\\ \hline
$a_3(k+1)\leq $  &-2.58&-2.63&-2.68&-2.72&-2.77&-2.81&-2.85&-2.89&-2.93&-2.97&-3.00\\ \hline
\end{tabular}
\end{center}
\end{table}

\vskip 2mm
\noindent

\section{The Dirichlet eigenvalue problem}

\noindent
For a bounded domain $\Omega$  with a piecewise smooth boundary $\partial \Omega$ in  an $n$-dimensional  complete
self-shrinker  in $\mathbf{R}^{n+p}$, we consider
the following Dirichlet eigenvalue problem of  the differential operator $\mathfrak L$:
\begin{equation}
\begin{cases}
\mathfrak Lu=-\lambda u  &  \text{in $\Omega$}, \\
u=0 & \text{on $\partial \Omega$}.
\end{cases}
\end{equation}
This eigenvalue problem has a real and  discrete spectrum:
\begin{equation*}
0<\lambda _{1} < \lambda _{2} \leq \cdots \leq \lambda _{k} \leq \cdots
 \longrightarrow \infty,
\end{equation*}
where each eigenvalue is repeated according to  its multiplicity.  We have  following
estimates for eigenvalues of the  Dirichlet eigenvalue problem $(5.1)$.

\begin{theorem} Let $\Omega$ be a bounded domain with a piecewise smooth boundary $\partial \Omega$
in an $n$-dimensional complete
self-shrinker  $M^n$   in $\mathbf{R}^{n+p}$. Then,
eigenvalues of the  Dirichlet eigenvalue problem $(5.1)$ satisfy
\begin{equation*}
\sum_{i=1}^k (\lambda_{k+1}-\lambda_{i})^2 \leq \frac{4}{n}\sum_{i=1}^k
(\lambda_{k+1}-\lambda_{i})(\lambda_{i}+\frac{2n-\inf_{\Omega}{|X|^2}}{4}).
\end{equation*}
\end{theorem}
\noindent
\begin{proof}. By making use of the same proof as in the proof of the theorem 1.1,
we can prove the theorem 5.1  if one notices to count the number of eigenvalues from $1$.
\end{proof}

\noindent
From the recursion formula of Cheng and Yang \cite{CY}, we can give an upper bound for eigenvalue $\lambda_{k+1}$:
\begin{theorem} Let $\Omega$ be a bounded domain with a piecewise smooth boundary $\partial \Omega$
in an $n$-dimensional complete
self-shrinker  $M^n$   in $\mathbf{R}^{n+p}$. Then,
eigenvalues of the  Dirichlet eigenvalue problem $(5.1)$ satisfy, for any $k\geq 1$,
\begin{equation*}
\lambda_{k+1}+\frac{2n-\inf_{\Omega}{|X|^2}}{4}
\leq (1+\frac {a(min\{n,k-1\})}{n})(\lambda_1+\frac{2n-\inf_{\Omega}{|X|^2}}{4})k^{2/n},
\end{equation*}
where the bound of $a(m)$ can be formulated as:
$$
\left \{ \aligned
a(0)&\leq 4, \\
 a(1)&\leq 2.64,\\
 a(m)&\leq 2.2-4\log(1+\frac1{50}(m-3)),\qquad \mbox{for}\quad
 m\geq 2.
\endaligned \right .
$$
In particular, for $n\ge 41$ and $k\geq 41$, we have
 \begin{equation*}
\lambda_{k+1}+\frac{2n-\inf_{\Omega}{|X|^2}}{4}
\leq (\lambda_1+\frac{2n-\inf_{\Omega}{|X|^2}}{4})k^{2/n}.
\end{equation*}
\end{theorem}

\begin{remark}
For the Euclidean space $\mathbf{R}^{n}$, the differential operator $\mathfrak L$ is
called Ornstein-Uhlenbeck operator in stochastic analysis. Since the Euclidean
space $\mathbf{R}^{n}$ is a complete self-shrinker in $\mathbf{R}^{n+1}$, our
theorems also give estimates for eigenvalues of the Dirichlet
eigenvalue problem of the Ornstein-Uhlenbeck operator.
\end{remark}

\noindent
For the  Dirichlet eigenvalue problem $(5.1)$, components $x_A$'s of the position vector $X$  are not eigenfunctions
corresponding to the eigenvalue $1$ because they do not satisfy the boundary condition.
In order to prove the theorem 5.2, we need to obtain the following  estimates for lower order eigenvalues.

\begin{proposition} Let $\Omega$ be a bounded domain with a piecewise smooth boundary $\partial \Omega$
 in an $n$-dimensional complete
self-shrinker  $M^n$   in $\mathbf{R}^{n+p}$. Then,
eigenvalues of the  Dirichlet eigenvalue problem $(5.1)$ satisfy
\begin{equation*}
\sum_{j=1}^n (\lambda_{j+1}-\lambda_{1}) \leq (2n-\inf_{\Omega}{|X|^2})+4\lambda_1.
\end{equation*}
\end{proposition}
\noindent

\begin{proof}
Let  $u_j$ be an  eigenfunction corresponding
to the eigenvalue $\lambda_j$ such that
\begin{equation}
\begin{cases}
\begin{aligned}
&\mathfrak L u_j=- \lambda_j u_j \quad \text{in} \ {\Omega}\\
&u_j=0,  \ \ \text{on $\partial \Omega$}\\
&\int_{\Omega}u_iu_j \ e^{-\frac{|X|^2}{2}}dv=\delta_{ij}, \ \text{for any} \ i, j=1, 2, \cdots.
\end{aligned}
\end{cases}
\end{equation}
We consider  an  $(n+p)\times (n+p)$-matrix $B=(b_{AB})$ defined
by
$$
b_{AB}=\int_{\Omega} x_{A} u_1u_{B+1}e^{-\frac{|X|^2}{2}}dv.
$$
From the orthogonalization of Gram and Schmidt, there exist an upper triangle matrix $R=(R_{AB})$ and  an orthogonal matrix $Q=(q_{AB})$ such that
$R=QB$. Thus,
\begin{equation}
R_{AB}=\sum_{C=1}^{n+p}q_{AC}b_{CB}
=\int_{\Omega} \sum_{C=1}^{n+p}q_{AC}x_C u_1u_{B+1}=0, \ \text{for} \ 1\leq B <A\leq n+p.
\end{equation}
Defining $y_A=\sum_{C=1}^{n+p}q_{AC}x_C$, we have
\begin{equation}
\int_{\Omega}y_A u_1u_{B+1}=\int_{\Omega}\sum_{C=1}^{n+p}q_{AC}x_C u_1u_{B+1}=0, \ \text{for} \ 1\leq B <A\leq n+p.
\end{equation}
Therefore,  the  functions $\varphi_{A}$ defined by
$$
\varphi_{A}=(y_A-a_A)u_1, \  a_A=\int_{\Omega} y_Au_1^2 \ e^{-\frac{|X|^2}{2}}dv,
\  \text{for}   \ \ 1\leq A \leq n+p
$$
satisfy
$$
\int_{\Omega}\varphi_{A}u_{B}=0,\qquad \mbox{for}\  1\leq B \leq A \leq n+p.
$$
Therefore, $\varphi_A$ is a trial function.
From the Rayleigh-Ritz inequality, we have, for $1\leq A\leq n+p$,
\begin{equation}
\lambda_{A+1} \leq \frac{\displaystyle{-\int_{\Omega}}\varphi_A\mathfrak L \varphi_A  \ e^{-\frac{|X|^2}{2}}dv}
{\displaystyle{\int_{\Omega}}(\varphi_A)^2 \ e^{-\frac{|X|^2}{2}}dv}.
\end{equation}
From the definition of $\varphi_{A}$, we derive
\begin{equation*}
\begin{aligned}
&\mathfrak L \varphi_A=\Delta  \varphi_A-\langle X, \nabla  \varphi_A\rangle\\
&=\Delta \{(y_A-a_A) u_1\}-\langle X, \nabla   \{(y_A-a_A) u_1\}\rangle\\
&=y_A\mathfrak Lu_1 + u_1\mathfrak Ly_A +2\nabla y_A\cdot \nabla u_1 -a_A\mathfrak Lu_1\\
&=-\lambda_1y_Au_1-u_1y_A+2\nabla y_A\cdot \nabla u_1+a_A\lambda_1u_1.
\end{aligned}
\end{equation*}
Thus, (5.5) can be written as
\begin{equation}
(\lambda_{A+1}-\lambda_1)\|\varphi_{A}\|^2
\leq \int_{\Omega} \bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv.
\end{equation}
From the Cauchy-Schwarz inequality, we obtain
$$
\left(\int_{\Omega} \bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv\right)^2\leq
\|\varphi_{A}\|^2 \|y_Au_1-2\nabla y_A\cdot\nabla u_1\|^2.
$$
Multiplying the above inequality by $(\lambda_{A+1}-\lambda_1)$, we infer, from  (5.6),
\begin{equation}
\begin{aligned}
&(\lambda_{A+1}-\lambda_1)\left(\int_{\Omega}\bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv\right)^2\\
 &\leq (\lambda_{A+1}-\lambda_1)\|\varphi_{A}\|^2 \|y_A u_1-2\nabla y_A\cdot\nabla u_1\|^2\\
&\leq \Big(\int_{\Omega} \bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv\Big)
 \|y_Au_1-2\nabla y_A\cdot\nabla u_1\|^2\\
\end{aligned}
\end{equation}
Hence, we derive
\begin{equation}
\begin{aligned}
&(\lambda_{A+1}-\lambda_1)\int_{\Omega}
\bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv
\leq   \|y_A u_1-2\nabla y_A\cdot\nabla u_1\|^2\\
\end{aligned}
\end{equation}
Since
$$
\sum_{A=1}^{n+p} y_A^2= \sum_{A=1}^{n+p} x_A^2=|X|^2,
$$
we infer
\begin{equation}
\begin{aligned}
&\sum_{A=1}^{n+p} \|y_A u_1-2\nabla y_A\cdot\nabla u_1\|^2\\
&=\sum_{A=1}^{n+p}\int_{\Omega}\bigl(y_A ^2u_1^2
-4y_Au_1\nabla y_A\cdot\nabla u_1+4(\nabla y_A\cdot\nabla u_1)^2\bigl)
 \ e^{-\frac{|X|^2}{2}}dv\\
 &=\int_{\Omega} \bigl(|X|^2u_1^2
-\nabla |X |^2\cdot\nabla u_1^2+4\nabla u_1\cdot\nabla u_1\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega} \bigl(|X|^2u_1^2+
\mathfrak L |X |^2 u_1^2+4\nabla u_1\cdot\nabla u_1\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega} (2n-|X|^2)u_1^2\ e^{-\frac{|X|^2}{2}}dv+4\lambda_1\leq (2n-\inf_{\Omega} |X|^2)+4\lambda_1.
\end{aligned}
\end{equation}
\vskip2mm
\noindent
On the other hand, from the definition of $\varphi_A$, we have
\begin{equation}
\begin{aligned}
&\int_{\Omega} \bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega}\bigl(y_A^2 u_1^2-a_Ay_Au_1^2
+2a_Au_1\nabla y_A\cdot\nabla u_1-2y_Au_1\nabla y_A\cdot\nabla u_1\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega} \bigl(y_A^2 u_1^2-a_Ay_Au_1^2
-a_A\mathfrak L y_Au_1^2+\frac{1}{2}\mathfrak L y_A^2 u_1^2\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega} \bigl(y_A^2 u_1^2+\frac{1}{2}\mathfrak L y_A^2 u_1^2\bigl)\ e^{-\frac{|X|^2}{2}}dv\\
&=\int_{\Omega} \nabla y_A\cdot\nabla y_A u_1^2\ e^{-\frac{|X|^2}{2}}dv.\\
\end{aligned}
\end{equation}
For any point $p$,  we choose
a new coordinate system $\bar X=(\bar x_1,\cdots,\bar x_{n+p})$ of $\mathbf  R^{n+p}$
given by  $X-X(p)=\bar X O$ such that
 $(\frac {\partial }{\partial \bar x_1})_p,\cdots,
 (\frac {\partial }{\partial \bar x_n})_p$ span $T_pM^n$
 and at $p$, $g\Big(\frac{\partial}{\partial \bar x_i},\frac{\partial}{\partial \bar x_j}\Big)
 =\delta_{ij}$,
 where  $O=(o_{AB})\in O(n+p)$ is an $(n+p)\times (n+p)$ orthogonal matrix.
 \begin{equation*}
 \begin{aligned}
 &\nabla y_A\cdot\nabla y_A=g(\nabla y_A, \nabla y_A)
 = \sum_{B,C=1}^{n+p}q_{AB}q_{AC}g(\nabla x_B, \nabla x_C)\\
&=\sum_{B,C=1}^{n+p}q_{AB}q_{AC}
g(\sum_{D=1}^{n+p}o_{DB}\nabla \bar x_{D}, \sum_{E=1}^{n+p}o_{EC}\nabla \bar x_E)\\
&=\sum_{B,C,D,E=1}^{n+p}q_{AB}o_{DB}q_{AC}o_{EC}
g(\nabla \bar x_{D},\nabla \bar x_E)\\
&=\sum_{j=1}^n\bigl(\sum_{B=1}^{n+p}q_{AB}o_{jB}\bigl)^2\leq 1
\end{aligned}
\end{equation*}
since $OQ$ is an   orthogonal matrix  if $Q$ and $O$ are orthogonal matrices,
that is, we have
 \begin{equation}
 \begin{aligned}
 \nabla y_A\cdot\nabla y_A\leq 1.
\end{aligned}
\end{equation}
Thus, we obtain, from (5.10) and (5.11),
\begin{equation}
\begin{aligned}
&\sum_{A=1}^{n+p}(\lambda_{A+1}-\lambda_1)
\int_{\Omega} \bigl(y_A u_1-2\nabla y_A\cdot\nabla u_1\bigl)\varphi_A \ e^{-\frac{|X|^2}{2}}dv\\
&=\sum_{A=1}^{n+p}(\lambda_{A+1}-\lambda_1)\int_{\Omega}  \nabla y_A\cdot\nabla y_A u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&=\sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1)\int_{\Omega}  \nabla y_j\cdot\nabla y_j u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&+\sum_{A=n+1}^{n+p}(\lambda_{A+1}-\lambda_1)\int_{\Omega}  \nabla y_A\cdot\nabla y_Au_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&\geq \sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1)\int_{\Omega}\nabla y_j\cdot\nabla y_j  u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&+\sum_{A=n+1}^{n+p}(\lambda_{n+1}-\lambda_1)\int_{\Omega} \nabla y_A\cdot\nabla y_Au_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&=\sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1)\int_{\Omega} \nabla y_j\cdot\nabla y_j  u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&+(\lambda_{n+1}-\lambda_1)\int_{\Omega} (n-\sum_{j=1}^{n}\nabla y_j\cdot\nabla y_j ) u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
\end{aligned}
\end{equation}
\begin{equation*}
\begin{aligned}
&=\sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1)\int_{\Omega} \nabla y_j\cdot\nabla y_j  u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&+(\lambda_{n+1}-\lambda_1)\int_{\Omega}\sum_{j=1}^{n}(1-\nabla y_j\cdot\nabla y_j ) u_1^2\ e^{-\frac{|X|^2}{2}}dv\\
&\geq \sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1).
\end{aligned}
\end{equation*}
According to (5.8), (5.9) and (5.12), we obtain
$$
\sum_{j=1}^{n}(\lambda_{j+1}-\lambda_1)\leq (2n-\inf_{\Omega} |X|^2)+4\lambda_1.
$$
This completes the proof of the proposition 5.1.
\end{proof}
\vskip 2mm

\noindent
{\it Proof of Theorem {\rm 5.2}}. By making use of the proposition 5.1 and the same proof as in the proof of the theorem 1.2,
we can prove the theorem 5.2  if one notices to count the number of eigenvalues from $1$.

\begin{flushright}
$\square$
\end{flushright}

redpajama_set_name: RedPajamaArXiv


Our comprehensive range of office space services can help your business overcome any real estate challenge throughout the entire property lifecycle; including investment acquisition or disposal; office rental, leasing and tenant representation; valuation; fit-out and workplace consultancy; property management and facilities management.
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text:
package utils

import (
	"encoding/binary"
	"math/bits"

	"github.com/apache/arrow/go/v9/arrow/bitutil"
)

// BitmapWriter is an interface for bitmap writers so that we can use multiple
// implementations or swap if necessary.
type BitmapWriter interface {
	// Set sets the current bit that will be written
	Set()
	// Clear clears the current bit that will be written
	Clear()
	// Next advances to the next bit for the writer
	Next()
	// Finish flushes the current byte out to the bitmap slice
	Finish()
	// AppendWord takes nbits from word which should be an LSB bitmap and appends them to the bitmap.
	AppendWord(word uint64, nbits int64)
	// AppendBools appends the bit representation of the bools slice, returning the number
	// of bools that were able to fit in the remaining length of the bitmapwriter.
	AppendBools(in []bool) int
	// Pos is the current position that will be written next
	Pos() int
	// Reset allows reusing the bitmapwriter by resetting Pos to start with length as
	// the number of bits that the writer can write.
	Reset(start, length int)
}

type bitmapWriter struct {
	*bitutil.BitmapWriter
}

func NewBitmapWriter(bitmap []byte, start, length int) BitmapWriter {
	return &bitmapWriter{bitutil.NewBitmapWriter(bitmap, start, length)}
}

func (b *bitmapWriter) AppendWord(uint64, int64) {
	panic("unimplemented")
}

type firstTimeBitmapWriter struct {
	buf    []byte
	pos    int64
	length int64

	curByte      uint8
	bitMask      uint8
	byteOffset   int64
	endianBuffer [8]byte
}

// NewFirstTimeBitmapWriter creates a bitmap writer that might clobber any bit values
// following the bits written to the bitmap, as such it is faster than the bitmapwriter
// that is created with NewBitmapWriter
func NewFirstTimeBitmapWriter(buf []byte, start, length int64) BitmapWriter {
	ret := &firstTimeBitmapWriter{
		buf:        buf,
		byteOffset: start / 8,
		bitMask:    bitutil.BitMask[start%8],
		length:     length,
	}
	if length > 0 {
		ret.curByte = ret.buf[int(ret.byteOffset)] & bitutil.PrecedingBitmask[start%8]
	}
	return ret
}

func (bw *firstTimeBitmapWriter) Reset(start, length int) {
	bw.pos = 0
	bw.byteOffset = int64(start / 8)
	bw.bitMask = bitutil.BitMask[start%8]
	bw.length = int64(length)
	if length > 0 {
		bw.curByte = bw.buf[int(bw.byteOffset)] & bitutil.PrecedingBitmask[start%8]
	}
}

func (bw *firstTimeBitmapWriter) Pos() int { return int(bw.pos) }
func (bw *firstTimeBitmapWriter) AppendWord(word uint64, nbits int64) {
	if nbits == 0 {
		return
	}

	// location that the first byte needs to be written to for appending
	appslice := bw.buf[int(bw.byteOffset):]

	// update everything but curByte
	bw.pos += nbits
	bitOffset := bits.TrailingZeros32(uint32(bw.bitMask))
	bw.bitMask = bitutil.BitMask[(int64(bitOffset)+nbits)%8]
	bw.byteOffset += (int64(bitOffset) + nbits) / 8

	if bitOffset != 0 {
		// we're in the middle of the byte. Update the byte and shift bits appropriately
		// so we can just copy the bytes.
		carry := 8 - bitOffset
		// Carry over bits from word to curByte. We assume any extra bits in word are unset
		// so no additional accounting is needed for when nbits < carry
		bw.curByte |= uint8((word & uint64(bitutil.PrecedingBitmask[carry])) << bitOffset)
		// check everything was transferred to curByte
		if nbits < int64(carry) {
			return
		}
		appslice[0] = bw.curByte
		appslice = appslice[1:]
		// move the carry bits off of word
		word = word >> carry
		nbits -= int64(carry)
	}
	bytesForWord := bitutil.BytesForBits(nbits)
	binary.LittleEndian.PutUint64(bw.endianBuffer[:], word)
	copy(appslice, bw.endianBuffer[:bytesForWord])

	// at this point, the previous curByte has been written, the new curByte
	// is either the last relevant byte in word or cleared if the new position
	// is byte aligned (ie. a fresh byte)
	if bw.bitMask == 0x1 {
		bw.curByte = 0
	} else {
		bw.curByte = appslice[bytesForWord-1]
	}
}

func (bw *firstTimeBitmapWriter) Set() {
	bw.curByte |= bw.bitMask
}

func (bw *firstTimeBitmapWriter) Clear() {}

func (bw *firstTimeBitmapWriter) Next() {
	bw.bitMask = uint8(bw.bitMask << 1)
	bw.pos++
	if bw.bitMask == 0 {
		// byte finished, advance to the next one
		bw.bitMask = 0x1
		bw.buf[int(bw.byteOffset)] = bw.curByte
		bw.byteOffset++
		bw.curByte = 0
	}
}

func (b *firstTimeBitmapWriter) AppendBools(in []bool) int {
	panic("Append Bools not yet implemented for firstTimeBitmapWriter")
}

func (bw *firstTimeBitmapWriter) Finish() {
	// store curByte into the bitmap
	if bw.length > 0 && bw.bitMask != 0x01 || bw.pos < bw.length {
		bw.buf[int(bw.byteOffset)] = bw.curByte
	}
}

func (bw *firstTimeBitmapWriter) Position() int64 { return bw.pos }

redpajama_set_name: RedPajamaGithub


text:
package main

import (
	"flag"
	"fmt"
	"os"
)

type Config struct {
	IndexDomain    string
	RegistryDomain string
}

type command struct {
	name     string
	argsdesc string
	desc     string
	nargs    int
	fn       func(*Config, []string)
}

var commands = [...]command{
	command{"info", "<repos_name>", "lookup repos meta-data", 1, CmdInfo},
	command{"layerinfo", "<repos_name> <layer_id>", "lookup layer meta-data", 2, CmdLayerInfo},
	command{"curlme", "<repos_name> <layer_id>", "print a curl command for fetching the layer", 2, CmdCurlme},
}

func OptParse() {
	flag.Usage = func() {
		fmt.Fprintf(os.Stderr, "Usage: %s [options] command\n", os.Args[0])
		fmt.Fprintln(os.Stderr, "")
		fmt.Fprintln(os.Stderr, "options:")
		flag.PrintDefaults()
		fmt.Fprintln(os.Stderr, "")
		fmt.Fprintln(os.Stderr, "commands:")
		for _, c := range commands {
			fmt.Fprintf(os.Stderr, "  %s %s: %s\n", c.name, c.argsdesc, c.desc)
		}
	}

	config := &Config{}
	flag.StringVar(&config.IndexDomain, "i", "https://index.docker.io", "override index endpoint")
	flag.StringVar(&config.RegistryDomain, "r", "", "override registry endpoint")
	flag.BoolVar(&Quiet, "q", false, "disable debug logs")
	flag.Parse()

	for _, c := range commands {
		if flag.Arg(0) == c.name {
			args := flag.Args()[1:]
			if len(args) != c.nargs {
				s := ""
				if c.nargs > 1 {
					s = "s"
				}
				fmt.Fprintf(os.Stderr, "%s takes %d argument%s: %s\n", c.name, c.nargs, s, c.argsdesc)
				os.Exit(2)
			}
			c.fn(config, args)
			return
		}
	}
	flag.Usage()
	os.Exit(2)
}


text:
Pien Huang
Wanyu Zhang / NPR
Pien Huang is a global health and development reporter on the Science desk. She was NPR's first Reflect America Fellow, working with shows, desks and podcasts to bring more diverse voices to air and online.
She's a former producer for WBUR/NPR's On Point and was a 2018 Environmental Reporting Fellow with The GroundTruth Project at WCAI in Cape Cod, covering the human impact on climate change. As a freelance audio and digital reporter, Huang's stories on the environment, arts and culture have been featured on NPR, the BBC and PRI's The World.
Huang's experiences span categories and continents. She was executive producer of Data Made to Matter, a podcast from the MIT Sloan School of Management, and was also an adjunct instructor in podcasting and audio journalism at Northeastern University. She worked as a project manager for public artist Ralph Helmick to help plan and execute The Founder's Memorial in Abu Dhabi and with Stoltze Design to tell visual stories through graphic design. Huang has traveled with scientists looking for signs of environmental change in Cameroon's frogs, in Panama's plants and in the ocean water off the ice edge of Antarctica. She has a degree in environmental science and public policy from Harvard.
Coronavirus FAQ: Do Airplane Passengers Not Know There's A Pandemic Going On?
By Pien Huang • Jan 15, 2021
Each week, we answer "frequently asked questions" about life during the coronavirus crisis. If you have a question you'd like us to consider for a future post, email us at goatsandsoda@npr.org with the subject line: "Weekly Coronavirus Questions."
Trump Administration Urges States To Open COVID-19 Vaccination To Everyone Over 65
By Selena Simmons-Duffin & Pien Huang • Jan 12, 2021
Updated 2:20 p.m. ET
The Trump administration is making several big changes to its COVID-19 vaccine distribution strategy, officials announced Tuesday, in a bid to jump-start the rollout and get more Americans vaccinated quickly.
The first change is to call on states to expand immediately the pool of people eligible to receive vaccines to those 65 and older, and those with underlying health conditions that make them more susceptible to COVID-19.
Here's How The U.S. Can Jump-Start Its Sluggish COVID-19 Vaccine Rollout
By Selena Simmons-Duffin & Pien Huang • Jan 8, 2021
This time last year, the world was heading into a pandemic that would upend everything and cost 1.9 million lives — and counting. The promise of the new year is that vaccines are finally here and offer a way out.
Black Doctors Use Social Media To Share Accurate Information About COVID-19 Vaccine
By Pien Huang • Jan 1, 2021
More Than 1 Million People Have Received COVID-19 Vaccinations
By Pien Huang • Dec 25, 2020
Older People, Some Essential Workers Should Get Vaccines Next, CDC Panel Says
People who are ages 75 and older or frontline essential workers should be next in line to get a COVID-19 vaccine, a federal advisory committee to the Centers for Disease Control and Prevention determined Sunday.
Those groups follow frontline health care workers and nursing home residents, who have already begun receiving the limited supplies of vaccines available.
Revisiting With People Who Got Stuck In Months-Long Wuhan Lockdown
CDC Advisers Agree With FDA: COVID-19 Vaccine Is OK For Public Use
By Pien Huang & Selena Simmons-Duffin & Carmel Wroth • Dec 12, 2020
An important federal advisory committee at the Centers for Disease Control and Prevention has added its vote of support for the Pfizer-BioNTech COVID-19 vaccine.
In an emergency meeting Saturday, the CDC's Advisory Committee on Immunization Practices voted to recommend the first COVID-19 vaccine for use for people 16 or older in the U.S, expressing hope that the vaccine would help curb the spread of the disease that has killed more than 295,000 people in the U.S.
When Might You Expect Your 1st COVID-19 Shot?
In the U.S., front-line health care workers are likely first in line to get immunized with a COVID-19 vaccine, once the FDA says yes. But what about the rest of us?


text:
Q: How to Import/Edit Data on GUI without SQL I am new to Python and would like to implement a simple Employee Management System (as shown on the attached photo) that do the following functionality
• A GUI for entering, viewing, and exporting employee data. (I have done this task)
• The ability to export data into an Excel or csv file. (I have done this task)
• The ability to import data from an Excel or csv file into the GUI.
• The ability to edit employee data on screen and save the updated result to a file
I need to implement the last two tasks (i.e., import data from an CSV file and display on the GUI Window rather than on the Python Console) as my current code load the CSV file content into the Python console .
Moreover, I would like to edit the employee data on the GUI screen and save the updated result to the csv file.
Here is my code
from csv import *
from tkinter import *
from tkinter import filedialog
import tkinter.messagebox

root = Tk()
root.title("Employee Management System")
root.geometry("700x350")
root.maxsize(width=700, height=350)
root.minsize(width=700, height=350)
root.configure(background="dark gray")

#  Define Variables
main_lst=[]

def OpenFile():
filepath=filedialog.askopenfilename()
#   print(filepath)
# OR
file=open(filepath,'r')
print(file.read())
file.close

 def Add():
 lst=[txtFullname.get(),txtAddress.get(),txtAge.get(),txtPhoneNumber.get(),txtGender.get()]
 main_lst.append(lst)
 messagebox.showinfo("Information","The data has been added successfully")

  def Save():
  with open("data_entry.csv","w") as file:
  Writer=writer(file)
  Writer.writerow(["txtFullname","txtAddress","txtAge","txtPhoneNumber","txtGender"])
  Writer.writerows(main_lst)
  messagebox.showinfo("Information","Saved succesfully")

 def Clear():
 txtFullname.delete(0,END)
 txtAddress.delete(0,END)
 txtAge.delete(0,END)
 txtPhoneNumber.delete(0,END)
 txtGender.delete(0,END)

def Exit():
wayOut = tkinter.messagebox.askyesno("Employee Management System", "Do you want to exit the
system")
if wayOut > 0:
    root.destroy()
    return

# Label Widget
labelFN = Label( root,text="Full Name", font=('arial', 12, 'bold'), bd=10, fg="white",
bg="dark blue").grid(row=1, column=0)
labelAdd = Label(root, text="Home Address", font=('arial', 12, 'bold'), bd=10, fg="white",
bg="dark blue").grid(row=2, column=0)
labelAge = Label( root,text="Age", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark
blue").grid(row=3, column=0)
labelPhone_Num = Label( root, text="Phone Number", font=('arial', 12, 'bold'), bd=10,
fg="white", bg="dark blue").grid(row=4, column=0)
labelGender = Label( text="Gender", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark
blue").grid(row=5, column=0)

# Entry Widget
txtFullname = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtFullname.grid(row=1, column=1)

txtAddress = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtAddress.grid(row=2, column=1)

txtAge = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtAge.grid(row=3, column=1)

txtPhoneNumber = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtPhoneNumber.grid(row=4, column=1)

txtGender = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtGender.grid(row=5, column=1)

# Buttons

ButtonLoad = Button(text='LoadFile', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),
width=8, fg="black",bg="dark gray", command=OpenFile).grid(row=6, column=0)

ButtonAdd = Button( text='Add Record', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),
width=8, fg="black", bg="dark gray", command=Add).grid(row=6, column=1)

ButtonSave = Button(text='Save', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8,
fg="black", bg="dark gray", command=Save).grid(row=6, column=2)

ButtonClear = Button(text='Clear', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8,
fg="black", bg="dark gray", command=Clear).grid(row=6, column=3)

ButtonExit = Button(text='Exit', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),  width=8,
fg="black",bg="dark gray", command=Exit).grid(row=6, column=4)
'''
ButtonImport = Button(text='Import', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),
width=8, fg="black",bg="dark gray", command=Import).grid(row=6, column=5)
'''

root.mainloop()

Employee Management System-Screenshot of the current output

A: In order to import data from a csv file you can use the reader method from the csv module and simply load that information into your main_lst list through your OpenFile function.  I am not sure about editing though since your UI can only show one entry at a time, it makes it difficult to know which entry you are trying to edit.
from csv import *
from tkinter import *
from tkinter import filedialog
from tkinter import messagebox

root = Tk()
root.title("Employee Management System")
root.geometry("700x350")
root.maxsize(width=700, height=350)
root.minsize(width=700, height=350)
root.configure(background="dark gray")

#  Define Variables
main_lst=[]

def OpenFile():
    filepath=filedialog.askopenfilename()
    with open(filepath, "rt") as csvfile:
        rows = reader(csvfile)
        for row in rows:
            main_lst.append(row)
        lst = [txtFullname, txtAddress, txtAge, txtPhoneNumber, txtGender]
        for i,x in enumerate(lst):
            x.insert(0, row[i])

def Add():
    lst=[txtFullname.get(),txtAddress.get(),txtAge.get(),txtPhoneNumber.get(),txtGender.get()]
    main_lst.append(lst)
    messagebox.showinfo("Information","The data has been added successfully")

def Save():
    with open("data_entry.csv","w") as file:
        Writer=writer(file)
        Writer.writerow(["txtFullname","txtAddress","txtAge","txtPhoneNumber","txtGender"])
        Writer.writerows(main_lst)
        messagebox.showinfo("Information","Saved succesfully")

def Clear():
    txtFullname.delete(0,END)
    txtAddress.delete(0,END)
    txtAge.delete(0,END)
    txtPhoneNumber.delete(0,END)
    txtGender.delete(0,END)

def Exit():
    wayOut = messagebox.askyesno("Employee Management System", "Do you want to exit the system")
    if wayOut > 0:
        root.destroy()
    return

# Label Widget
labelFN = Label( root,text="Full Name", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark blue").grid(row=1, column=0)
labelAdd = Label(root, text="Home Address", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark blue").grid(row=2, column=0)
labelAge = Label( root,text="Age", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark blue").grid(row=3, column=0)
labelPhone_Num = Label( root, text="Phone Number", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark blue").grid(row=4, column=0)
labelGender = Label( text="Gender", font=('arial', 12, 'bold'), bd=10, fg="white", bg="dark blue").grid(row=5, column=0)

# Entry Widget
txtFullname = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtFullname.grid(row=1, column=1)

txtAddress = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtAddress.grid(row=2, column=1)

txtAge = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtAge.grid(row=3, column=1)

txtPhoneNumber = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtPhoneNumber.grid(row=4, column=1)

txtGender = Entry(root, font=('arial', 12, 'bold'), bd=4, width=22, justify='left')
txtGender.grid(row=5, column=1)

# Buttons

ButtonLoad = Button(text='LoadFile', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8, fg="black",bg="dark gray", command=OpenFile).grid(row=6, column=0)

ButtonAdd = Button( text='Add Record', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8, fg="black", bg="dark gray", command=Add).grid(row=6, column=1)

ButtonSave = Button(text='Save', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8, fg="black", bg="dark gray", command=Save).grid(row=6, column=2)

ButtonClear = Button(text='Clear', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'), width=8, fg="black", bg="dark gray", command=Clear).grid(row=6, column=3)

ButtonExit = Button(text='Exit', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),  width=8, fg="black",bg="dark gray", command=Exit).grid(row=6, column=4)
'''
ButtonImport = Button(text='Import', padx=10, pady=8, bd=4, font=('arial', 12, 'bold'),
width=8, fg="black",bg="dark gray", command=Import).grid(row=6, column=5)
'''

root.mainloop()

your csv file should look like this:
example.csv
txtFullname,txtAddress,txtAge,txtPhoneNumber,txtGender
Bob,Main St.,35,18004438768,Male
Alice,1st St.,62,922-333-1253,Female


text:
Q: Test stream with wrapped array I'm trying to test that the function emitArray emits a Response.Success an its value is ['test'].
If I emit a List<String> everything works as expected, but once I wrap the result list in a Response<List<String>> the test fails.
The result is emitted, but it fails when comparing with the expected result.
I'm wondering if it's related to the implementation of == in Response.Success, I'm using the default implementation that the IDE provides.
This is not the real code I have, it's just a simple example that is easier to understand to try to identify the issue.
This is my class to test:
class ListResponse {
  final _array = BehaviorSubject<Response<List<String>>>();

  Stream<Response<List<String>>> get array => _array.stream;

  Future<void> emitArray() async {
    _array.add(Response.success(['test']));
  }

  void dispose() {
    _array.close();
  }
}

This is my test:
void main() {
  ListResponse underTest;
  setUp(() {
    underTest = ListResponse();
  });

  test('It should emit array', () {
    final array = Response.success(['test']);

    expect(
      underTest.array,
      emitsInOrder([
        array,
        emitsDone,
      ]),
    );

    underTest.emitArray();

    underTest.dispose();
  });
}

This is the error it throws:
Expected: should do the following in order:
          • emit an event that SuccessResponse<List<String>>:<SuccessResponse{value: [test]}>
          • be done
  Actual: <Instance of 'BehaviorSubject<Response<List<String>>>'>
   Which: emitted • SuccessResponse{value: [test]}
                  x Stream closed.
            which didn't emit an event that SuccessResponse<List<String>>:<SuccessResponse{value: [test]}>

This is the Response class
class Response<T> {
  Response._();

  factory Response.success(T value) = SuccessResponse<T>;
  factory Response.error(Exception error) = ErrorResponse<T>;
}

class ErrorResponse<T> extends Response<T> {
  ErrorResponse(this.error): super._();

  final Exception error;

  @override
  bool operator ==(Object other) =>
      identical(this, other) ||
          other is ErrorResponse &&
              runtimeType == other.runtimeType &&
              error == other.error;

  @override
  int get hashCode => error.hashCode;

  @override
  String toString() {
    return 'ErrorResponse{error: $error}';
  }
}

class SuccessResponse<T> extends Response<T> {
  SuccessResponse(this.value): super._();

  final T value;

  @override
  bool operator ==(Object other) =>
      identical(this, other) ||
          other is SuccessResponse &&
              runtimeType == other.runtimeType &&
              value == other.value;

  @override
  int get hashCode => value.hashCode;

  @override
  String toString() {
    return 'SuccessResponse{value: $value}';
  }
}

A:
I'm wondering if it's related to the implementation of == in Response.Success

Exactly. This particular test is failing because you can't compare Lists with ==:
abstract class List<E> implements EfficientLengthIterable<E> {
  ...
  /**
  * Whether this list is equal to [other].
  *
  * Lists are, by default, only equal to themselves.
  * Even if [other] is also a list, the equality comparison
  * does not compare the elements of the two lists.
  */
 bool operator ==(Object other);
}

As a workaround you can change the implementation to compare objects' string representations instead:
  @override
  bool operator ==(Object other) =>
      identical(this, other) ||
      other is SuccessResponse &&
          runtimeType == other.runtimeType &&
          value.toString() == other.value.toString();

Interestingly, passing unwrapped List<String>s objects passes test. That happens because StreamMatcher uses equals() from matcher package to match events, and equals() can match lists and maps. It first tries to match objects with ==, then checks whether they are Iterable/Set/Map (and deep matches them recursively), and then reports mismatch error.


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