text:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Drawing;
using FlatRedBall.Performance.Measurement;

namespace FlatRedBallProfiler.Managers
{
    #region ViewMode

    public enum ViewMode
    {
        Expaned,
        Collapsed
    }

    #endregion

    public class TreeViewManager
    {
        static TreeViewManager mSelf;
        TreeView mTreeView;
        ViewMode mViewMode;

        public static TreeViewManager Self
        {
            get
            {
                if (mSelf == null)
                {
                    mSelf = new TreeViewManager();
                }
                return mSelf;
            }
        }

        public ViewMode ViewMode
        {
            get
            {
                return mViewMode;
            }
            set
            {
                mViewMode = value;
                RefreshUI();
            }
        }

        Section SectionBasedOnViewMode
        {
            get
            {
                if (this.ViewMode == Managers.ViewMode.Expaned)
                {
                    return ProjectManager.Self.Section;
                }
                else
                {
                    return ProjectManager.Self.MergedSection;
                }
            }
        }

        public void Initialize(TreeView treeView)
        {
            mTreeView = treeView;
        }

        public void RefreshUI()
        {
            mTreeView.Nodes.Clear();

            TreeNode newNode = CreateTreeNode(SectionBasedOnViewMode);

            AppendTimeInMilliseconds(newNode);

            mTreeView.Nodes.Add(newNode);

            AddTreeNodesFor(SectionBasedOnViewMode, newNode.Nodes);

        }

        private void AddTreeNodesFor(FlatRedBall.Performance.Measurement.Section section, TreeNodeCollection treeNodeCollection)
        {
            TreeNode mostExpensive = null;
            float mostExpensiveValue = -1; // so it is always smaller than even the smallest starting value:
            double timeOfSubSections = 0;

            List<TreeNode> list = new List<TreeNode>();
            // This may be a revisited node:
            foreach (TreeNode node in treeNodeCollection)
            {
                list.Add(node);
            }

            foreach (var subSection in section.Children)
            {
                TreeNode treeNode = null;

                treeNode = CreateTreeNode(subSection);

                if (subSection.Time > mostExpensiveValue)
                {
                    mostExpensive = treeNode;
                    mostExpensiveValue = subSection.Time;
                }

                timeOfSubSections += subSection.Time;

                list.Add(treeNode);

                // adding tree nodes will just add times to existing
                // tree nodes if they already exist
                AddTreeNodesFor(subSection, treeNode.Nodes);

            }

            if (mostExpensive != null)
            {
                AppendTimeInMilliseconds(mostExpensive);
            }

            if(section.Time != 0)
            {
                double ratioAccountedFor = timeOfSubSections / section.Time;

                if (ratioAccountedFor < .98)
                {
                    TreeNode node = new TreeNode();
                    node.Text = "???? " + (100 - ratioAccountedFor*100).ToString("0.00") + "%";
                    list.Insert(0, node);
                }
            }
            treeNodeCollection.AddRange(list.ToArray());

        }

        private static void AppendTimeInMilliseconds(TreeNode mostExpensive)
        {
            mostExpensive.BackColor = Color.Orange;
            float time = ((Section)mostExpensive.Tag).Time;
            time *= 1000;
            mostExpensive.Text += " (" + time.ToString("0.00") + " ms)";
        }

        private static TreeNode CreateTreeNode(FlatRedBall.Performance.Measurement.Section subSection)
        {
            TreeNode treeNode = new TreeNode();
            treeNode.Tag = subSection;

            SetTreeNodeText(subSection, treeNode);
            return treeNode;
        }

        private static void SetTreeNodeText(FlatRedBall.Performance.Measurement.Section subSection, TreeNode treeNode)
        {
            string percentage = "100%";

            if (subSection.Parent != null)
            {
                float ratio = subSection.Time / subSection.Parent.Time;

                percentage = (ratio * 100).ToString("0.00") + "%";

            }
            string text = subSection.Name + " " + percentage;

            treeNode.Text = text;
        }
    }
}

redpajama_set_name: RedPajamaGithub


Build Html Form tutorial teaches basic to advance HTML coding. 11 chapter ebook. Get everything you need to know to build a website.
Basic to advanced build Html Form sign-up to receive your tutorial. Teaches Html coding and more importantly how to use it correctly.
Tutorial contains practice code to teach methods and techniques for building web pages from scratch. Sign up below to get your FREE 11 chapter ebook.
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How to create common tags - Control the attributes and properties of the HTML page.
tables - Allow you to divide your virtual real estate into columns and rows.
forms - Collect information from your visitors such as surveys, polls, personal info, or just plain comments.
lists - numbered or unnumbered lists like this one.
links - allow you to navigate through your website or anywhere else on the web.
images - the correct way to code your images for size, position and justification.
frames - allow you to have complete control over your web page. This page is an example.
comments - add comments to your code to help you stay organized or send a note, or describe your code for future understanding.
Meta tags - are used by the search engines to help describe (and rank) your web pages.
escape codes - html tags often come in pairs, an open and a close tag. Find out which ones need them and which ones don't.
Advice on design layout, website navigation, clean code, style, do's and don'ts, web browsers and search engine submissions.
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Make cool forms like the one above with Graphical opt-in boxes.


I turned in the cover art for UY Saga 8. Ink and watercolor on approximately 11×17, 4-ply, cold press Strathmore Bristol board.
This book will contain UY Books 29 (Two Hundred Jizo), 30 (Thieves & Spies), and 31 (The Hell Screen). Major featured characters are Kitsune, and Inspector Ishida. I rough-pencilled this idea with a 2H 3mm mechanical pencil, then determined the vanishing points for the building perspectives.
I reinforced the pencils using a B 5mm mechanical pencil, adding details to the building. This will be a night scene so I had Inspector Ishida carrying a lantern in the rough, but thought it was too distracting and eliminated it.
The borders were inked with a 3.5 technical pen and characters inked with a Koh-i-Noor Art Pen (now discontinued), using black Platinum Carbon Ink.
The background was inked with the same pen. If this was to be black and white art, I would have used a thinner pen nib to reinforce the perspective, but I figured I could do that with the color.
Pencil lines erased using Sakura SumoGrip eraser.
I cut out a mask to cover the borders, foreground and figures.
The background was watercolored with mainly browns and blues. I wanted a very limited neutral palette, using spots of brightness for effect.
Then the foreground was colored.
The mask covering the figures were removed.
More work on the central figures.
The final mask in the foreground was removed and bright scarlet was used to indicate a pool of fresh blood.
The mask protecting the borders was removed, and the cover is finished.
As with the art for all the UY Saga books, there will be a banner at the top covering 2.5 inches of the art (including Nezumi). However, the entire art will be used for the hardcovers.


Fred Diaz is no stranger to the North American auto market. Automaker Mitsubishi announced earlier this week that the company has hired Diaz as their President and CEO of their North America business. Diaz will start at the company on April 1 and current President of Mitsubishi Motors North America, Ryujro Kobashi, will head to the company's Japanese headquarters. He will be placed in charge of overseas sales.
Fred Diaz, 52, came to the automaker nearly eight months ago after nearly 30 years as an executive at multiple automotive companies. His prior position was as General Manager of Performance Optimization for the global marketing and sales division of Mitsubishi Motors. He mostly recently spent 4 years as the a senior level executive at Nissan and prior to that, he was the CEO of Ram and Chrysler de Mexico and Latin America. His knowledge of the automotive industry is perhaps bested by very few.
Diaz has in depth knowledge of the North and South American auto market and is expected to help the automaker expand their dealership network. He will be tasked with completing the company's three-year plan for growth that it recently announced. The goal is to increase their by 30%, from 100,000 units to 130,000 units. They are currently the fastest growing non-luxury brand in the United States. He will also be expected to help launch the company's electric and hybrid arm, something the company has been lacking in. Diaz has a bachelor's degree from Texas Lutheran University and a MBA from Central Michigan University.


Essentially England welcomes advertising enquiries from marketing agencies and companies for advertising that is relevant to and complements the content of our site.
Advertising on Essentially England is should adhere to the UK Code of non-broadcast Advertising, Sales Promotion and Direct Marketing (CAP Code), the Electronic Commerce (EC Directive) Regulations 2002, and teh 2006 ICC Consolidated Code of Advertising and Marketing Communications Practice (Chapter D).
Before contacting us, please check our general advertising policy for our advertising procedure and for types of adverts we are unable to accept on the site.
For content relevant to and complementing the content of the site, we currently offer the following types of advertising: banner ads, text ads and sponsored pages. Details of each option are listed below.
Banner ads: 468 x 60 banner on a page of your choice. Banner links to specified location on your site, opening a new page. Page containing banner may be shared with Google Adsense ads or other third party advertisements.
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Feature page / sponsored page: Editorial, links and creatives / images as supplied by you.
Custom-made advertising: Please contact us to discuss your requirements if you'd like something I haven't listed.


text:
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Viewing posts categorised under: Maps
Top Question in Each U.S. State
Jason C.
Maps, Moving News
American have questions! And each state has a "most popular question". Some of the questions are quite hilarious. Others are intriguing or downright scary. In this blog post, we're going to examine what some of these questions are. The questions come from Google Trends and were reported by Estately.
The Funny Questions
Let's just jump right in. In this section, we'll cover some of the more funny questions, starting with this: "Is funner a word?". Now, I know what you're thinking. It's not necessarily a "funny" question; however, it definitely related to the subject. This question comes from the residents of Minnesota.
"Is Ted Cruz the Zodiac Killer?" This is arguably a pretty funny question and topic (if you have a dark sense of humor). The Zodiac Killer is a serial killer who hasn't been caught. Internet sleuths noticed some similarities between what's known as the Zodiac Killer and Ted Cruz (former presidential candidate and current Texas Senator). Last year, Ted Cruz thought it would be funny to join in on the fun. He posted a tweet and wrote "Happy Halloween" with a photo of the Zodiac Killer's symbol writing. This question comes from Idaho. See the tweet below.
Happy Halloween pic.twitter.com/jIgTaIMzep
— Ted Cruz (@tedcruz) October 31, 2018
"What is Wyoming?". Alright, this question is being included here for one big reason: The question is coming from residents of Wyoming! Answer: the state you live in.
The Strange Questions
In this section, we're going to go over some of the more off-ball questions people have. And remember: These questions are the most popular question in that state. Keep that in mind.
"Do penguins have knees?". Sure, it's a valid question! But why would the residents of Iowa be asking it? There are no penguins in Iowa! At least there aren't in the wild. Perhaps you can find some in a zoo. If you're wondering, the answer is yes, penguins do have knees.
"How to be emo?". This great question comes from the residents of New Mexico. And this question is a good sign that business owners should be investing in Hot Topic and Spencer's locations around the state. More information regarding this question doesn't list the age of people asking the question; however, I think it's safe to assume that this question is mostly being asked by younger teens.
The Scary Questions
In this section, we'll talk about the more popular, scary questions. These aren't questions like "are ghosts real"; instead, they are questions where, if you knew who was asking them, you might want to avoid them.
And here's the first question: "Am I a psycho?". This question comes from residents of Missouri. We hope that the answer to this is "no".
"How to make meth?". This question comes from the residents of Kansas. Gosh, it's quite surprising that this is the most popular question in any state. If we were to guess, this question would be most popular in New Mexico, not Kansas. And it isn't because meth is most popular in the state. Instead, it's because that's where the fictitious setting of the popular show Breaking Bad takes place.
"How to get away with murder?". Yes, if you know someone who's asking this question, you might want to steer clear of them. This question is the most popular question in Delaware.
"How to make a baby?". We included this question in the scary section because if this question is the most popular one in any state, it is scary. This question comes from the residents of Kentucky. We're hoping that this internet search educated them on this very important question.
The Smart Questions
The questions in this section are more "normal" and would seem to be questions the average person would ask. Let's see what some of these are.
"What is quinoa?". Good question! Quinoa is an extremely delicious and healthy. This question comes from the residents of Utah.
"How to smoke salmon?". This is another great question and very fitting since it comes from the residents of Alaska. Salmon is abundant in the state and the people here are asking the right questions. At least, the most popular question in the state seems to be that.
"How to open a jar?". This is another functional question and one that might be asked by more of the elderly residents of the state. Here's a tip: if a jar is difficult to open, bang the sides of the lid with a kitchen utensil. It will help loosen it and make it easier to open. This question comes from the residents of North Carolina.
"Who won the Civil War?". It's important for residents of any state to ask questions like these. Wars such as the Civil War shouldn't be forgotten. History repeats itself; however, if people remember the past, perhaps the bad parts (civil war) won't have to be repeated. This question comes from the residents of Arkansas.
The last question we'll cover is one that has many answers, depending on who you ask: "What is the meaning of life?". This is a question that has been asked by residents in all cities, states, and countries for as long as humans have been alive. This great question comes from the residents of Hawaii.
We have the full map of questions for your reading pleasure. Check out the screenshot below to read the rest!
Featured photo by Gordon John on Pixabay
Map of questions is a screenshot from Estately
Maps of Poverty Cycles in Major US Cities
Dallas, Maps, Seattle
Does it matter where you grow up? More specifically, are the same economic opportunities available to all Americans, no matter where they were raised? There's new evidence that's able to answer these questions. The Opportunity Atlas essentially puts the roots of poverty on the map.
As an introduction, they state, "The Opportunity Atlas answers this question using anonymous data following 20 million Americans from childhood to their mid-30s. Now you can trace the roots of today's affluence and poverty back to the neighborhoods where people grew up. See where and for whom opportunity has been missing, and develop local solutions to help more children rise out of poverty."
They then prompt you to explore cities around the U.S. In this article, we're going to take a closer look at 5 major cities: Seattle, Dallas, Houston, Austin, and Chicago. Let's get started!
Take a look at the map above. The main colors here are red, white, blue and shades in between. The colors indicate household income over the course of one year. The redder the color, the less money the family makes. The bluer the color, the more the family makes. And, white is essentially the median income in the area. This goes for all maps you'll look at in this article.
A quick glance at the map shows that the worse parts, income-wise, is right in the heart of Seattle. This doesn't make that there are no high-paying jobs there. Instead, as mentioned before, it's the median income for all families living within a certain area.
There's quite a bit more red in Dallas than there is in Seattle! The good thing, however, is that the cost of living is generally lower in Dallas than it is in Seattle. Venture out to University Park and Forney, among other cities, and you'll realize areas of higher family median income.
The colors of Houston look very much like those of Dallas. I'd venture to say, however, that it appears Houston has more blue areas. The inner city to the north and south are quite red. West of the city looks very blue, especially around Sugarland and Bellaire. To the west, cities like Deer Park have a higher household income.
Austin, like you'll see next with Chicago, looks like night and day when it comes to areas around the city. To the west, it's mostly blue. To the east, it's mostly red. Think about the kids growing up to the west and to the east. Do they stay in those areas when they grow up, and make the same or a similar amount of money as their parents did?
Chicago, to the immediate west and south, is very red. Maywood is also red. Speaking of Maywood, the area has a very high crime rate. You might be wondering why the east side of Chicago is all white. It's not because the area is full of average or median income households. Instead, that's where Lake Michigan is!
Featured photo by Useche70 on Pixabay
Map photos are screenshots by RPS Relocation
Best U.S. Cities for New Year's Eve Celebrations
Joe Youngblood
Get ready for 2019! In celebration of the new year, we decided to write about the best cities for New Year's Eve celebrations. WalletHub compiled multiple data points in numeous cities to determine which ones made the cut.
Methodology of Choosing Cities
Adam McCann of WalletHub writes they used "three key dimensions: 1) Entertainment & Food, 2) Costs and 3) Safety & Accessibility. We evaluated those dimensions using 28 relevant metrics. Each metric was graded on a 100-point scale, with a score of 100 representing the most favorable conditions for New Year's Eve partiers."
Among the 28 metrics, some are:
New Year's Eve Events per Capita
Duration of 2018 Fireworks Show
Availability of Affordable Fine Dining
Average Alcoholic Beverage Price
Neighborhood Security
The 100 Best Cities
Source: WalletHub
Using the 28 metrics, WalletHub was able to rank the top 100 cities. Let's talk about the top five in order. They are: New York, Los Angeles, Atlanta, San Diego, and Las Vegas.
New York has the world-famous Times Square celebrations; however, the celebrations of the other top cities are lesser known. Near Los Angeles, the EVE party at Universal Studios Hollywood is popular. In Atlanta, the popular Peach Drop emulates the ball drop in Times Square. Big Night San Diego is a popular New Year's Eve party event. And in Las Vegas, 40,000 people watch multiple simultaneous fireworks shows from the downtown area.
Numerous cities in Texas, and Seattle, also ranked highly on the list. In Texas, some of the best cities are: San Antonio (#16), Dallas (#17), Houston (#22), Austin (#27), El Paso (#35), and Fort Worth (#46). Seattle clocks in fairly high at #15.
See the full list of the best U.S. cities for New Year's Eve celebrations below. Have you attended a New Year's Eve celebration in one of the cities below? We'd love to hear about your experience.
Featured photo by Jonathan Petersson on Pexels
Most Expensive Cities for Commuting in the USA
Dallas, Data Visualization, Maps, Moving News, Seattle
What's your daily commute like? If you live in or near a major city, traffic is probably a fact of life. Back in August, we shared how much time people of major cities waste over the course of their lives by commuting. The data was put together by EducatedDriver.org. Now, they're back with more information about your commute; however, this time, they share how much money you spend on gas and vehicle maintenance.
Recall that the average American spends about 408 days commuting. That's over an entire year of their life driving! In addition to spending time, people also spend money as they drive. Alex Lauderdale writes, "We did the math for nearly 100 major US cities, and found the average American will spend $108,727 on gas and vehicle maintenance costs associated with commuting while driving 173,203 miles to and from work in their lifetime. That's enough miles to drive around the world 7 times!"
Lifetime Cost of Commuting Map
A quick glance at the map above shows this: red, yellow, and green circles. The circles represent the lifetime cost of commuting in a certain city. The red circles indicate the most expensive cities, the yellow circles indicate more of the average-costing cities, and the green circles indicate cities that are least expensive for commuting.
The lifetime cost of commuting is most in Atlanta. The average round-trip there is 26 miles and the average cost of commuting there is a whopping $182,886. Let's take a look at some other areas.
Texas has six cities on the map: Dallas, Houston, San Antonio, Austin, El Paso, and McAllen. Texas is the only state that has two large red circles which cover Dallas and Houston. Check out a photo of Texas below.
California is the state with the most cities on the map. The ten circles cover the following cities: Sacramento, Stockton, San Francisco, San Jose, Fresno, Bakersfield, Oxnard, Los Angeles, Riverside, and San Diego. We often hear about how bad the traffic can be in Los Angles; however, the city has a yellow circle. As a matter of fact, California has no red circles at all. See the photo of the state below.
Top 30 Expensive Cities for Commuting
Where are the most expensive cities for commuting? You have to click each circle to determine exactly how expensive the city is for commuting. To make better sense the map, we turned a large part of the data into a list.
Below, you'll find the top 30 most expensive cities for commuting. As stated earlier, Atlanta is the most expensive, followed closely by Dallas, Houston, and Phoenix. Check it out as the list goes from most expensive to least:
Atlanta, GA – $182,886
Dallas, TX – $174,314
Houston, TX – $174,314
Phoenix, AZ – $162,883
Nashville, TN – $157,168
Detroit, MI – $148,595
Birmingham, AL – $144,309
Chicago, IL – $142,880
St. Louis, MO – $142,880
Charlotte, NC – $138,594
Minneapolis, MN – $135,736
Indianapolis, IL – $131,450
Orlando, FL – $130,021
Knoxville, TX – $130,021
Washington, DC – $130,021
Jacksonville, FL – $130,021
Riverside, CA – $130,021
Seattle, WA – $128,592
Columbia, SC – $128,592
Kansas City, MO – $127,163
Memphis, TN – $127,163
San Antonio, TX – $125,734
Los Angeles, CA – $125,734
Jackson, MS – $124,306
Richmond, VA – $124,306
Cincinnati, OH – $124,306
Austin, TX – $122,877
Baltimore, MD – $122,877
Columbus, OH – $122,877
Miami, FL – $122,877
The Methodology of the Map
How did EducatedDriver.org come up with the data used to compile the map? Alex Lauderdale writes, "For the purposes of this study, we assumed the average person starts full-time work at 18 (some people start earlier, others a bit later). We also know the average retirement age is 63 in the United States. That works out to a total of 45 years working a full-time job."
This tells us that the data isn't precise; however, it's essentially as estimate based upon common knowledge such as the average retirement age and total years the average person works.
Alex continues, "From there, we operated based on the assumption most people work about 250 days per year, accounting for 2 weeks yearly vacation and time off. That adds up to a whopping 11,250 days of working/commuting over a career."
Again, we're seeing best estimates based upon data averages. Not everyone has exactly 2 weeks of vacation each year. Some people have more vacation than that and not everyone uses all of their vacation days.
Finally, Alex pulled data about the average round trip distances as well as how much it costs per mile to drive a vehicle. He writes, "we used data from the US Census Bureau on average daily round trip commute distances as well as data from AAA on the total cost per mile of operating a vehicle (60.8 cents per mile for the average sedan when gas, insurance, and maintenance costs are considered)."
How does your city commute compare with the rest of the country? Do you live near a red, yellow, or green circle? We're interested in your experience and want to know if it lines up with the data! Use the comments area below to share your story.
The Best Cities in the USA for Sports in 2018 Ranked
The World Series, Super Bowl, and Stanley Cup are among the pinnacles of sports. These championship events give way to heightened emotion, excitement, and, ultimately, victory or defeat plunging cities into mourning, jubilation, and sometimes anarchy-esque chaos. The larger cities across the country normally have a healthy amount of sports teams and fans that eagerly pursue winning one of these crowns. Others, not so much. Which U.S. cities are the best for sports? In an attempt to answer this, Wallet Hub put together 2018's best cities for sports.
Adam McCann of WalletHub explains the methodology in picking the best cities. In a nutshell, he writes, "each sports category was assigned a weight corresponding with the total percentage of adults in the U.S. who claim to follow that particular sport, according to The Global Sports Media Consumption Report".
The sports we are going to highlight are football, basketball, baseball, hockey, and soccer.
Mapping the Best Cities for Sports
Wallet Hub ran the data to create a visualization of what their findings look like on a map. As you can see, circles — some larger than others — dot the landscape. The larger circles indicate best sports cities that overall rank lower on the list. The smaller circles, like Boston, Chicago, and Los Angeles, indicate cities that are the best of the best when it comes to sports cities.
At a quick glance, it looks like the best cities listed are more or less in tune with the general population size. For example, circles are found up and down the east coast stretching all the way to the Midwest. After that, there are cities few and far between until arriving on the west coast. On the east coast, California has the most of these best cities for sports.
The Best Cities for Football
The Philadelphia Eagles won 2018's Super Bowl and the city ranks 8th on the list of best cities for football. The Green Bay Packers, a legendary team that is owned by the fans in a city of just over 100,000 people, have the top spot. The #2 slot goes to the Steel City where the Pittsburgh Steelers and their fan are always revved up for game day. Dallas takes the #3 spot even though the professional NFL team that has their name is actually about 20.5 miles away in Arlington, TX. The Cowboys and their fans call themselves "America's Team" which may have been more true in the 90's than it is today where "Most of Texas' team" is likely more fitting. Still the city of Dallas' mood often fits with how well the Cowboys played the week before and much of the area pays close attention to every decision the team makes. The New England Patriots out of Boston, MA sit just outside of the top 3 spots at #4. The team has competed in the Super Bowl eight times in the last twenty years, out of those appearances they have won five NFL championships.
Here are the top 30 best cities for football:
The Best Cities for Basketball
The #1 Los Angeles Lakers were to the 2000s what the Chicago Bulls were to the 1990s. Coach Phil Jackson and players Shaquille O'Neal and Kobe Bryant propelled the Lakers to three straight championship wins. The Lakers won five championships from 2000-2010. #2 Boston Celtics saw legendary basketball champion Larry Bird play for the team; however, the Celtics were good even before he joined. The Celtics won the NBA Finals an amazing 8 times in a row from 1959-1966. The #3 Golden State Warriors (Oakland, CA) are the hottest team in basketball right now. They won 2018's NBA Finals. The Warriors also won in 2015 and 2017. Steve Curry is to the Warriors what Michael Jordan was to the Chicago Bulls.
Here are the top 30 best cities for basketball:
The Best Cities for Baseball
The Boston Red Sox won 2018's World Series and ranks at number 8 in the best cities for baseball. The Red Sox have competed in the championship game twice in the past 10 years. They won both times. Their rivals are the New York Yankees. #1 New York is home to the Yankees and Mets. #2 Los Angeles is home to the Angels and they've only had one World Series win (2002). The team became more popular after the 1994 movie "Angeles in the Outfield". The #3 St. Louis Cardinals have won numerous championship games; however, the lost in 2003 to the current champs (Boston Red Sox).
Here are the top 30 best cities for baseball:
Buies Creek, NC
The Best Cities for Hockey
The Washington Capitals won 2018's Stanley Cup and the city ranks at number 10 in the best cities for hockey. The Chicago Blackhawks are notable because they won the championship game three times in the last ten years. The #1 Boston Bruins have played in the Stanley Cup 19 times and won 6 of them. The #2 Detroit Red Wings, however, won even more championship games. The Red Wings have played in the Stanley Cup 24 times and won 11 of them. Finally, the #3 Pittsburgh Penguins have played in the finals only 6 times; however, the Penguins won 5 out of 6 of those games. That gives the Penguins a better win percentage over the Bruins and Red Wings.
Here are the top 30 best cities for hockey:
Durham, NH
The Best Cities for Soccer
Toronto FC is 2018's MLS Cup winner. While a North America Soccer team, Toronto is not applicable to the best cities map because it only covers the United States. The top spot for soccer is held by Los Angeles whose team, LA Galaxy, has won the MLS Cup three times in the last ten years. #1 LA Galaxy is popular in part due to having international soccer star David Beckham as a former player. He helps the Galaxy win two straight MLS Cups (2011 and 2012). The #2 Orlando City Lions only became a team 5 years ago, in 2013. The demographic and local support for soccer lead to the creation of the team. Finally, the #3 Seattle Sounders is also a relatively new team. The Sounders were formed 11 years ago, in 2007, for the same reasons as the creation of the Orlando Lions. Soccer is becoming more popular in the U.S.!
Here are the top 30 best cities for soccer:
Dallas, Seattle, Las Vegas, Kansas City, and Chicago
If you'be been following our blog, you know that these cities are great places to live for various reasons. Where do they stand as the best cities for sports? Check them out below.
Dallas—Football: 3; Basketball: 13; Baseball: 177; Hockey: 19; Soccer: 16
Seattle—Football: 12; Basketball: 83; Baseball: 24; Hockey: 450; Soccer: 3
Las Vegas—Football: 197; Basketball: 198; Baseball: 89; Hockey: 18; Soccer: 43
Kansas City—Football: 450; Basketball: 450; Baseball: 450; Hockey: 450; Soccer: 45
Chicago—Football: 27; Basketball: 8; Baseball: 5; Hockey: 5; Soccer: 10
Featured photo is a screenshot of Wallet Hub sports map by RPS Relocation
All sports photos by CC0 Creative Commons on Pixabay
Map of best sports cities is embedded from WalletHub
How the USA Uses Its Land
Data Visualization, Maps
The United States has one of largest land masses of any country in the world. Our country has vast mountains, plains, deserts. Puerto Rico—a territory of the United States—is home to the only tropical rain forest in North America. What is all of this land used for? Bloomberg pulled data from a number of sources to determine to answer this question.
Sure, a percentage of land is allotted for residential or business use. With the amount of traffic in some areas, it can seem as if our country is overpopulated; however, as Bloomberg learned from the data, that's not the case.
Six Land Types
Bloomberg authors Dave Merrill and Lauren Leatherby write, "Using surveys, satellite images and categorizations from various government agencies, the U.S. Department of Agriculture divides the U.S. into six major types of land. The data can't be pinpointed to a city block—each square on the map represents 250,000 acres of land. But piecing the data together state-by-state can give a general sense of how U.S. land is used".
These six land types are:
Pasture/range
Cropland
Special Use
The map clearly shows urban areas are a minority compared to the rest, just by looking at the colored squares. Miscellaneous and special use areas have fewer squares than forest, cropland, or pasture and range zones. Bloomberg was able to collect the acreage for each land type.
Acreages of the Land Types
Imagine if you were able to view each land type together on the map. That's what you get in the photo above. Here, it's proven that urban areas are indeed the smallest and pasture and ranges are the largest. How many acreages does each land type contain? Here's the breakdown:
Urban: 69.4M acres
Miscellaneous: 68.9M acres
Special Use: 168.6M acres
Cropland: 391.5M acres
Forest: 538.6M acres
Pasture/range: 654M acres
"Four in five Americans live, work and play" in urban areas, write Merrill and Leatherby. Given this fact, most Americans would be able to fit in a relatively small area of the country, in theory. Urban areas are growing quickly, however. Merrill and Leatherby continue, "The U.S. is becoming more urban—at an average rate of about 1 million additional acres a year. That's the equivalent of adding new urban area the size of Los Angeles, Houston and Phoenix combined. U.S. urban areas have more than quadrupled since 1945".
Food Production and Land
Around one-fifth of the land in our country is agricultural land and most of it is located in the Midwest. However, only a small percentage of agricultural land is currently used to feed Americans. Some of the land is used for ethanol production or livestock feed. Other land is used to cultivate crops with the goal of exporting them to other countries. Here's a breakdown of what our cropland is used for:
Livestock feed: 127.4M acres
Food we eat: 77.3M acres
Other grain & feed exports: 62.8M acres
Idle/fallow: 52M acres
Ethanol, biodiesel: 38.1M acres
Wheat exports: 21.5M acres
Cotton/non-food: 13.6M acres
Here's a photo of this data visualized:
Let's zoom in on the food we eat. In the photo below, you can see wheat, non-citrus fruits, vegetables, rice, sugarcane, and a few other types of foods. Wheat, soybeans, peanuts, and oilseeds appear to be about half of what we eat. Barley, maple syrup, potatoes, and citrus are among the least of what we consume.
Featured photo by CC0 Creative Commens on Pixabay
Bloomberg data map photos are screenshots by RPS Relocation
The Green Energy Map of the USA
Energy helps makes the world turn. Maybe that why every state in our country produces it; however, not all energy is equal when it comes to being green and environmentally friendly. Reddit user Dr_Engineerd shared a map of the United States. Specifically, the map details the percentage of electricity produced by non-fossil fuel sources. The data comes from U.S. Energy Information Administration. Let's take a closer look at the map!
The Non-fossil Fuel Energy Map
Each state on the map is depicted by a certain color. The redder states are those which have a low percentage of non-fossil fuel energy production. The bluer states, on the other hand, of those which have a higher percentage of non-fossil fuel energy production.
Most of the states in the Midwest, with the exception of Illinois and South Dakota, lean towards the red end of the color spectrum. This means the electricity of those states is produced by a higher amount of fossil fuel energy. The northern states on both coasts trend bluer. The electricity there is produced by more non-fossil fuel energy.
Highest and Lowest States by Percentage
A cursory glance at the map will give you an idea where each state stands. After all, the color of the state indicates the type of fuel sources it uses (fossil fuels versus non-fossil fuels). To get a better idea of fuel sources, it's best to get the actual percentage of electricity produced by non-fossil fuel sources. Below are the states which have the most, and least, percentages of electricity produced by non-fossil fuel sources.
Top 5 non-fossil fuel states:
Vermont (99.6%)
Washington (85.55%)
Idaho (81.8%)
New Hampshire (77%)
Oregon (75.1%)
Bottom 5 non-fossil fuel states:
Delaware (1.6%)
West Virginia (4.6%)
Rhode Island (4.8%)
Indiana (6.2%)
Kentucky (6.9%)
Comments and Thoughts
Reddit user Dr_Engineerd explains the types of non-fossil fuel energy sources. Solar, wind, biomass, geothermal, and hydroelectric are common types of green energy. Nuclear energy also falls into this category in part because it's emission-free.
Dr_Engineerd also shared what the map would look like without including nuclear. See it below.
Another Reddit user, jswynn5, has an interesting take on the energy product of Texas. Jswynn5 writes, "Side note, I used to live in Texas, they are the largest wind energy producers in the US. Unfortunately it doesn't come close to how much fossil fuel energy they produce. Just interesting facts".
Featured photo by Pixabay on Pexels
Non-fossil fuel energy map and Reddit comments are screenshots by RPS Relocation
The Green Energy Map of Canada
How green is Canada's energy? When it comes to energy, let's just say it depends on the province. Like the individual states in the United States, the provinces in Canada are diverse next to each other. Reddit user PaulsEggo posted an image visualizing how green each province is.
The Province Map
The provinces vary in color from red to light orange to light green to green. The red provinces are those which have a low percentage of electricity produced by renewables and uranium. The green, on the other side of the spectrum, are provinces which have a high percentage of electricity produced by renewables and uranium. Here are exact percentages for each color:
Red: 0-20%
Light Orange: 21-40%
Light Green: 41-94%
Green: 95+%
Territorial Statistics Chart
Reddit user PaulsEggo created the map visualization; in addition, he created the following provincial and territorial statistics chart.
It's important to note that uranium and nuclear energy are considered clean energy. The chart is based on the data PaulsEggo collected from the National Energy Board (NEB). NEB writes, "Canada is rich in energy supply. From the hydroelectric reservoirs of British Columbia, Manitoba, Ontario, and Quebec to the hydrocarbon resources of western Canada and offshore East Coast. From the uranium deposits in northern Saskatchewan to the abundant wind and solar potential across the country. This wealth of resources ranks Canada as the fifth largest natural gas producer, sixth largest crude oil producer, second largest hydro producer and sixth for share of renewable electricity generation."
Alberta, Saskatchewan, and Geography
Alberta and Saskatchewan are among the provinces colored in red in the map. One Reddit user, Tamer_, commented "They're still burning 87% of the coal and 72% of the natural gas used for power in Canada".
Another Reddit user, kylethsmith, commented how geography plays a role in having clean energy. They write, "The map is purely a result of geography. The vast majority of the green energy reflected in this map is hydro power, which is abundant in places like BC and Quebec. Ontario has some great hydro electric assets, too, along with the population density to make building a large nuclear plant viable. Saskatchewan is flat. There aren't a lot of waterfalls on the prairies to use for hydroelectricity. It is also sparsely populated, making the investment of tens of billions of dollars in a nuclear plant uneconomical".
Featured photo by Daniel Joseph Petty on Pexels
Canada visualization map, statistics chart, and Reddit comments are screenshots by RPS Relocation
Dallas Map Startup Shows Critical Data in One Place
Dallas, Data Visualization, Maps
Knowledge is power! And in our Information Age, data is the new form of knowledge. This data is being used to make humanity better. Technological advances are moving at the speed of Moore's Law which sees computer power doubling every two years.
Data can be used to improve the lives of people at the federal, state, and local levels. One of the ways to determine where data will encourage positive action is to actually see the data. This can be accomplished by creating a data visualization on a map.
For Dallas, Texas, this such data visualization has become a reality. A substantial number of data sets have been collected with the goal of making the city better. This visualization is called TheMap. D Magazine gives us some background of how the map came to be.
Improving Dallas with a Map
Robert Mundinger, a longtime resident of Dallas, noticed problems in the areas he lived in. Matt Goodman of D Magazine wanted to learn more about TheMap from Mundinger.
Matt writes, TheMap is "an open-source, interactive web portal that allows users to access an array of data sets that can help people better understand all sorts of things about Dallas. The data sets can be used to generate maps. It's all exportable. "A kid's educational outcome is dependent on education and crime and health and access to food, access to internet, libraries," Mundinger says. "All these variables affect each other so much that there needs to be one place where people can look at them together and analyze them."
What Mundinger wants to achieve in simple: Highlight at-risk areas of the city. In addition, and while he was at it, Mundinger went ahead and included more data set that might be of use.
There are neighborhoods in Dallas which have a high concentration of poverty. This is included in the map visualization. There are intersections that have more fatal pedstrian crashes than others. This is also included in TheMap. There's even data about where there are working females over the age of 16. The information contained within TheMap is comprehensive!
Let's take a closer look at the data sets available for viewing.
Data Sets in TheMap
Police stations, fire stations, neighborhood associations, poverty areas—these are some of the data sets that are included in TheMap. Take a look at the column on the left to see them all. TheMap defaults to Popular Maps whe you visit the website.
You can otherwise click on Education, Civic, Housing, or other major types of data sets. Check out the row near the top to see more.
Once you click a data set, the information will be uploaded to the map. Let's say you click on police departments. Pinpoints of every police department in Dallas will show up on the map. You can click them individually to get the name, address, and more information about the particular police department.
Bus Stops in Dallas
One of my favorite visualizations in TheMap is seeing where DART bus stops are. DART—which stands for Dallas Area Rapid Transit—seemingly has bus stops all over the city. A quick glance shows what can be estimated to over 100 bus stops.
This data set is surely important to Mundinger. After all, he's concerned with increasing the access of public services to all residents of Dallas, including those below the poverty line. With TheMap, Dallas city officials can combine areas of poverty with bus stop locations. If there's an area which is lacking of public transportation, DART could add a new bus stop there.
Combining Datasets in TheMap
One of the most powerful and helpful features of TheMap is the ability to combine and layer multiple datasets. To do this, you'll need to create an account which only takes a minute.
Let's say I want to see the locations of libraries and recreation centers. People considering a move to Dallas may enjoy both of these placs and want to live near them. See the screenshot of TheMap to see how it looks when you combine data sets.
Notice that recreation centers and libraries are shown on the left side. In the screenshot, they're both toggled as being "Visible" in TheMap. You can toggle either of them on or off to show their exact locations. Or, you can click on each pin to gain further information.
Final Thoughts on TheMap
Robert Mundinger has done a great, free public service to the city of Dallas. City officials should be using TheMap before they make decisions that will affect the residents of the city. For example, a new police station can be built near an area of high crime.
Dallas isn't the only city that can benefit from using TheMap. Mundinger, or people inspired by him, can take this idea and apply it to virtually any city. It would all be in the name of improving the lives of residents over an extended period of time.
Featured photo by Aksonsat Uanthoeng on Pexels
TheMap photos are screenshots by RPS Relocation
Dallas City Bond Projects Map
Dallas, Maps
The size of Texas is massive and the entire southeastern part rests on the Gulf of Mexico. Cities like Corpus Christi, Galveston, and Houston have felt the wrath of hurricanes. Given the central location of Dallas, one might assume flooding isn't a major issue; however, it has been at times. To combat this, Dallas Reports detailed how the city is working on a number of flood prevention projects.
Projects in the Works
The projects are made possible from the passage of Dallas's 2017 bond package. Specifically, these projects are covered under Proposition D. Jesus Jimenez of Dallas News writes "the city of Dallas has 84 projects in the works to improve flood protection, storm drainage, and erosion control."
It's important to remember that flood protection is only one aspect of minimizing potential damage to structures. Storm drainage and erosion control are equally important. For example, as an area takes on a large amount of water, there needs to be a mechanism in place for that water to be routed somewhere. Erosion control mitigates the instances of land, coastal, and construction disappearance.
Jesus Jimenez continues, "Sarah Standifer, assistant director of stormwater operations for the city of Dallas, said that while the last bond package sought to address citywide improvements, Proposition D of this bond package focuses on more neighborhood-specific improvements." This tells me, without delving into the results of the last bond package, that citywide improvements were through to be sufficient. That's why Dallas is now moving on to neighborhood-specific improvements. But which neighborhoods will be chosen? Let's take a look at that next.
Dallas' Flood Prevention Map
Google Maps has created a snapshot of areas where flood prevention projects are taking place. This map allows residents of Dallas access to track the progress of the ongoing projects.
Click on one of the blue pins to get a description of the work being done. For example, I clicked on Turnpike Distribution IV IH30-Commerce. The description says "Design and construction of connection of the improved channel in the Turnpike Distribution Center". For another example, I clicked Teakwood Elmridge Drainage Relief. The description says "Design and construction – Replace existing system at the Teakwood and Elmridge area with a 100-year system."
How Project Locations are Chosen
Do neighborhoods in and around Dallas get an equal amount of flood prevention aid? That's not how it's outlined in Proposition D. Jesus Jimenez writes "The process to identify projects funded through Proposition D included analyzing calls to 311, flood studies, community calls and emails, system inspection, and drainage master plans."
This is the most logical way to determine which neighborhoods should get flood prevention aid. After all, places that have no history of flooding or aren't vulnerable shouldn't get chosen over places that have a history of flooding. Departments of the city keep records of areas that have been impacted by flooding. They use these records to determine which neighborhoods will be getting the flood prevention aid.
The decision rested with more than only what areas have flooded in the past. Jesus Jimenez continues, "Projects then went through a two-step evaluation involving technical criteria with a focus on public safety, and balancing criteria with a focus on supporting economic development and enhancing quality the of life."
Economic development areas are just as important to keep safe as it is to keep neighborhoods safe. If business areas are wiped out, people may not be able to afford their homes; therefore, there has to be a balance between sending aid to neighborhoods and business areas.
Featured image by Pixabay on Pexels
Dallas city bonds projects map is a screenshot by RPS Relocation
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redpajama_set_name: RedPajamaCommonCrawl


text:
// go-rdb
package core

const TupleCapacity = 100
const RowCapacity = 5000

func StreamToRelation(s Stream, n *Node) (*Relation, error) {
	return GetRelation(s, make([][]Value, 0, RowCapacity), n)
}

func GetRelation(s Stream, buf [][]Value, n *Node) (*Relation, error) {
	err := s.Init(n)
	if err != nil {
		return nil, err
	}
	result := &Relation{
		Attrs: NewSchema(),
		Data:  buf,
	}
	lastTuple := NewTuple()
	for s.HasNext() {
		tuple, err := s.Next()
		if err != nil {
			return nil, err
		}
		if tuple == nil {
			continue
		}
		lastTuple = tuple
		result.Data = append(result.Data, tuple.Data)
	}
	result.Attrs = lastTuple.Schema
	s.Close()
	return result, nil
}

/*
func makeValues(t *Tuple) []Value {
	m := make([]Value, 0, t.Len())
	for _, col := range t.Attrs {
		m = append(m, t.Get(col.Name))
	}
	return m
}
*/


text:
Jabung is een bestuurslaag in het regentschap Magetan van de provincie Oost-Java, Indonesië. Jabung telt 2826 inwoners (volkstelling 2010).

Plaats in Oost-Java

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
\begin{abstract}
We evaluate numerically certain multiple integrals representing nearest and next-nearest neighbor correlation functions of the spin-1/2 $XXZ$ Heisenberg infinite chain at finite temperatures.
\end{abstract}
\pacs{05.30.-d, 75.10.Pq}
\maketitle

\section{Introduction}
\label{intsec}
The determination of correlation functions is an important task both from a theoretical and from an experimental point of view. In scattering experiments correlation functions are measured, and detailed insight into the microscopic structure of the sample can be obtained. Scattering experiments may thus well decide about the adequateness of a given (simplified) theoretical model, once its correlations functions are calculated. On the other hand, taken for granted that a theoretical model adequately describes a certain sample, the knowledge of its correlation functions is useful to estimate the quality of the experimental setup. The principal problems on the theoretical side are threefold: To deal with many-body interactions between the particles, to account for finite temperatures and to carry out the thermodynamic limit of a macroscopic sample size. With regard to these intricate difficulties, techniques which allow for the exact calculation of correlation functions of at least some typical interacting quantum systems in the thermodynamic limit are highly welcome.

Progress into this direction has recently been achieved by one of the authors in collaboration with A. Kl\"umper and A. Seel (GKS) \cite{gks1,gks2,gks3}. In these works, the one-dimensional  spin-1/2 $XXZ$ Heisenberg chain with periodic boundary conditions in a magnetic field,
\begin{eqnarray}
H=J \sum_{j=1}^L\left[S_j^xS_{j+1}^x + S_j^yS_{j+1}^y+\Delta \left(S^z_jS^z_{j+1}-\frac14\right)\right]-h\sum_{j=1}^L S^z_j\label{defh}\;,
\end{eqnarray}
was considered. Formulae for the correlation functions were derived by combining the quantum-transfer-matrix (QTM) approach \cite{suz85,suz87,kl92rsos,kl93} to the thermodynamics of integrable systems with certain results for matrix elements of the $XXZ$-chain \cite{kor82,sla89} which proved to be useful before \cite{kit00,kit02} at $T=0$. In the QTM approach, the free energy of the system is expressed in terms of an auxiliary function defined in the complex plane \cite{kl93}. This auxiliary function is determined uniquely as a solution of a non-linear integral equation. GKS found that essentially the same auxiliary function can be used to calculate $m$-point correlation functions for arbitrary temperatures $T$, again in the thermodynamic limit. As a result, all the correlation functions of spatial range $m$ are given in terms of $m$-fold integrals over combinations of the auxiliary function taken at different integration variables.

In this work, the integral equations are solved numerically for all temperatures at $h=0$ and the multiple integrals are calculated for $m=2,3$, where we restrict ourselves to $\Delta\geq 1$ for $m=3$. In particular, we focus on $\langle S_j^x S_{j+m-1}^x\rangle_T$ and on the emptiness formation probability $P(m)$. In order to estimate the quality of our data we compare them with other exact results: first of all the nearest-neighbor correlators can be obtained independently of the multiple integral representation by taking the derivative of the free energy with respect to $\Delta$. Furthermore, very recently \cite{tsu05}  a high temperature expansion of the emptiness formation probability $P(3)$ at $\Delta=1$ has been performed up to the order $(J/T)^{42}$, starting from the multiple integral representation. At $T=0$, closed expressions are available \cite{tak77,boo01,boo03,boo05,kat03,kat04} as well. There are, moreover, the $XX$-limit ($\Delta\to0$) and the Ising limit ($J\to 0,\;\Delta\to\infty,\; J\Delta$ fixed) where the correlation functions are known over the whole range of temperatures \cite{mcc68,shi01,bax82}. All these independent results provide useful tests for the accuracy of our numerics. The main error we observe is due to the discretization of the integrals, an error which, in principle, can be made arbitrarily small by increasing the numerical accuracy.

Besides the expected crossover behaviour from low to high temperatures, we find a surprising feature in $\langle S_j^x S_{j+m-1}^x\rangle_T$ for $\Delta>1$: the antiferromagnetic correlation is maximal at an intermediate temperature $T_0$, and not, as naively expected, at $T=0$. We explain this behaviour from a competition between quantum and thermodynamical fluctuations.

This paper is organized as follows: in the next section we give some details about the numerical procedure. Results are presented separately in the third section. The article ends with a summary and an outlook.
\section{Numerical procedure}
\label{num}
Let $A_{1 \ldots m}$ be an operator which acts on sites $1,\ldots,m$ of the spin chain. In order to calculate $\langle A_{1 \ldots m}\rangle_T$, the operator $A_{1 \ldots m}$ is expanded in terms of the elementary matrices $e_\alpha^\beta$,
\begin{eqnarray}
e_1^1=\left(\begin{array}{cc}1 & 0\\0&0\end{array}\right)\,,\;
e_1^2=\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)\,,\;
e_2^1=\left(\begin{array}{cc}0 & 0\\1&0\end{array}\right)\,,\;
e_2^2=\left(\begin{array}{cc}0 & 0\\0&1\end{array}\right)\,,\nonumber
\end{eqnarray}
such that
\begin{eqnarray}
A_{1 \ldots m}=A_{\alpha_1\ldots\alpha_m}^{\beta_1\ldots\beta_m} e^{\;\;\;\alpha_1}_{1\,\beta_1}\,e^{\;\;\;\alpha_2}_{2\,\beta_2}\,\ldots\,\,e^{\;\;\;\alpha_m}_{m\,\beta_m}\nonumber.
\end{eqnarray}
Then it suffices to calculate the expectation value $\langle e_{1\;\beta_1}^{\;\;\;\alpha_1}\ldots e_{m\;\beta_m}^{\;\;\;\;\alpha_m}\rangle_T$ which defines the density matrix of the chain segment consisting of sites $1,\ldots,m$.

In \cite{gks3} the following multiple integral representation for the density matrix was has been obtained \footnote{Eq. \refeq{gen} was formulated as a conjecture in \cite{gks3} and a proof was only sketched. By now a complete proof is available for the general case \cite{has05}.}:
\begin{eqnarray}
\langle e_{1\;\beta_1}^{\;\;\;\alpha_1}\ldots e_{m\;\beta_m}^{\;\;\;\;\alpha_m}\rangle_T&=&
\left[\prod_{j=1}^{|\alpha^+|}\int_{\mathcal C}\frac{\mbox d \omega_j}{2\pi\text{i} (1+\mathfrak a(\omega_j))}\right]\left[\prod_{j=|\alpha^+|+1}^{m}\int_{\mathcal C}\frac{\mbox d \omega_j}{2\pi\text{i} (1+{\overline{{\mathfrak{a}}}}(\omega_j))}\right]\nonumber\\
& &\times\frac{(-1)^m\det_m\left[\frac{1}{(b-1)!}\frac{\partial^{b-1}}{\partial \xi^{b-1}}G\left(\omega_a,\xi\right)|_{\xi=0}\right]_{ab}}{\prod_{1\leq j<k\leq m} \sinh\left(\omega_k-\omega_j+\eta\right)}\nonumber\\
& &\times\left[\prod_{j=1}^{|\alpha^+|}\sinh^{\alpha_j^+-1}(\omega_{|\alpha^+|-j+1}-\eta)\sinh^{m-\alpha_j^+}\omega_{|\alpha^+|-j+1}\right]\nonumber\\
& &\times\left[\prod_{j=1}^{m-|\alpha^+|}\sinh^{\beta_j^--1}(\omega_{|\alpha^+|+j}+\eta)\sinh^{m-\beta_j^-}\omega_{|\alpha^+|+j}\right]\label{gen}\; .
\end{eqnarray}
Here $(\alpha_n)^m_{n=1}$ and $(\beta_n)^m_{n=1}$ denote sequences of up- and down-spins, where $1\equiv \uparrow$ and $2\equiv \downarrow$. The position of the $j$th up-spin (down-spin) in the sequence $(\alpha_n)^m_{n=1}$ ($(\beta_n)^m_{n=1}$) is $\alpha_j^+$ ($\beta_j^-$) and the number of up- (down-) spins in the corresponding sequences is defined as $|\alpha^+|$ ($|\beta^-|$). The contour $\mathcal C$ in \refeq{gen} is given in \cite{gks1}; it depends on the value of $\Delta=:\cosh \eta$. The parameter $\eta$ may be purely real and positive (massive case, $\Delta>1$), or imaginary with $\eta=:\text{i} \gamma$, $\pi/2\geq\gamma\geq 0$ (massless case, $0\leq \Delta\leq 1$). The auxiliary functions $\mathfrak a, \bar \mathfrak a, G$ which occur in the integrand in \refeq{gen} are solutions of the following integral equations:
\begin{eqnarray}
\ln \mathfrak a(\lambda)&=&-\beta h- \frac{J\beta \sinh\eta}{2}\, \hat\kappa_{\frac{\eta}{2}}(\lambda+\eta/2)-\int_\mathcal C\frac{\mbox d \omega}{2\pi\text{i}}\hat\kappa_\eta(\lambda-\omega)\ln(1+\mathfrak a(\omega))\label{lna}\\
G(\lambda,\xi)&=&  \hat\kappa_{\frac{\eta}{2}}(\lambda-\xi-\eta/2)+\int_\mathcal C\frac{\mbox d \omega}{2\pi\text{i}} \,  \hat\kappa_\eta(\lambda-\omega)\frac{G(\omega,\xi)}{1+\mathfrak a(\omega)} \label{g}\\
\hat \kappa_\eta(\lambda)&=& \frac{\sinh( 2\eta)}{ \sinh(\lambda+\eta)\sinh(\lambda-\eta)}\nonumber\\
{\overline{{\mathfrak{a}}}}&:=&1/\mathfrak a\nonumber\,,
\end{eqnarray}
with $\beta:=1/T$. Note that the $T$- and $h$-dependence of the correlation functions enters through the auxiliary functions, whereas the $m$-dependence is encoded in the $m$-fold integral. Thus the evaluation of \refeq{gen} requires two steps: solving equations \refeq{lna}, \refeq{g} and then calculating the multiple integrals. In this work, the following correlation functions will be calculated numerically at $h=0$:
\begin{eqnarray}
 P(2) &:=&\langle e_{j\;1}^{\;\;\;1}e_{j+1\;1}^{\;\;\;\;\;\;\;1}\rangle_T=\frac14+\langle S^z_jS^z_{j+1}\rangle_T\nonumber\\
\langle S^x_jS^x_{j+1}\rangle_T &=&\frac12\langle e_{j\;1}^{\;\;\;2}e_{j+1\;2}^{\;\;\;\;\;\;\;1}\rangle_T\nonumber\\
 P(3) &:=&\langle e_{j\;1}^{\;\;\;1}e_{j+1\;1}^{\;\;\;\;\;\;\;1}e_{j+2\;1}^{\;\;\;\;\;\;\;1}\rangle_T=\frac14+\frac12\langle S^z_jS^z_{j+1}\rangle_T+\langle S^z_jS^z_{j+2}\rangle_T\nonumber\\
\langle S^x_jS^x_{j+2}\rangle_T &=&\frac12\langle e_{j\;1}^{\;\;\;2}\left(e_{j+1\;1}^{\;\;\;\;\;\;\;1}+e_{j+1\;2}^{\;\;\;\;\;\;\;2}\right)e_{j+2\;2}^{\;\;\;\;\;\;\;1}\rangle_T\nonumber.
\end{eqnarray}
The function $P(m)=\langle e_{1\;1}^{\;\;\;1}\ldots e_{m\;1}^{\;\;\;\;1}\rangle_T$ is referred to as the emptiness formation probability.

For the numerical treatment, it is convenient to reformulate equations \refeq{lna}, \refeq{g} such that the integrations are done along straight lines parallel to the real or imaginary axes. This procedure is due to Kl\"umper (cf. the book \cite{book} and references therein). We review the main steps here in order to introduce a notation which will prove to be convenient for our purposes.

The integration contour $\mathcal C$ is chosen according to fig. \ref{fig1}a) (\ref{fig1}b))  in the massive (massless) case.
\begin{figure}
\begin{center}
\vspace{-1cm}
 \includegraphics[scale=0.8]{contour}
\caption{Integration contours a) in the massive b) in the massless case.}
\label{fig1}
\end{center}
\end{figure}
In the massive (massless) regime, Re $\mathcal C_\pm=\pm\eta/2\mp \epsilon$ (Im $\mathcal C_\pm=\pm\gamma/2\mp \epsilon$), where $\epsilon$ is a small positive quantity; the limit $\epsilon\to 0$ is included implicitly. According to eqs.\ \refeq{lna}, \refeq{g} the functions $\ln \mathfrak a$, $G$ are given in the part of the complex plane enclosed by $\mathcal C$, once they have been calculated on $\mathcal C$. However, for the calculation of correlators, \refeq{gen}, these functions are only needed {\em on} $\mathcal C$. Since in the massive case, all functions are periodic with period $\text{i} \pi$, integrals along $\mathcal C_{1,2}$ in opposite directions cancel each other. We thus have to calculate the auxiliary functions along two straight lines $\mathcal C_\pm$ parallel to the imaginary (real) axis in the massive (massless) case.

Let us first focus on the calculation of the auxiliary functions in the massless case, the massive case is treated analogously afterwards. Subsequently, we comment on the calculation of the multiple integrals with the help of the auxiliary functions.
\subsection{Massless case}
We first concentrate on $0<\gamma\leq \pi/2$ and deal with the isotropic case $\gamma=0$ at the end of this section.
It is convenient to define
\begin{subequations}
\begin{eqnarray}
\mathfrak b(x)&:=&\mathfrak a(x+\text{i} \gamma/2),\; {\overline{{\mathfrak{b}}}}(x):=  {\overline{{\mathfrak{a}}}}(x-\text{i}\gamma/2)\label{lnb}\\
\mathfrak B&:=& 1+\mathfrak b, \;  {\overline{{\mathfrak{B}}}}:= 1+{\overline{{\mathfrak{b}}}} \label{lnbb}\; .
\end{eqnarray}
\end{subequations}
Note that ${\overline{{\mathfrak{b}}}}=\mathfrak b^*|_{h\to-h}$. We now perform a ''particle-hole-transformation'' in \refeq{lna} by substituting
\begin{eqnarray}
\begin{array}{ll}
\ln(1+\mathfrak a(x+\text{i} \gamma/2))=\ln \mathfrak B(x)& \mbox{on  } \mathcal C_+\\
\ln(1+\mathfrak a(x-\text{i} \gamma/2))=\ln {\overline{{\mathfrak{B}}}}(x)+\ln \mathfrak b(x-\text{i} \gamma)& \mbox{on  } \mathcal C_- \; .
\end{array}
\end{eqnarray}
Taking account of the simple pole of $\ln \mathfrak b(x)$ at $x=-\text{i} \gamma/2$, we arrive at
\begin{eqnarray}
\ln \mathfrak b(x)&=&-\beta h+\frac{J \beta \sinh(\text{i} \gamma)}{2}\,  \hat \kappa_{\text{i} \gamma/2} (x)+\frac{1}{2\pi\text{i}}\,\left[\hat \kappa_{\text{i} \gamma}*\ln \mathfrak B-\hat \kappa^{(+)}_{\text{i} \gamma}*\ln {\overline{{\mathfrak{B}}}}-\hat \kappa_{\text{i} \gamma}*\ln \mathfrak b\right](x)\label{lnb1}
\end{eqnarray}
with the definitions
\begin{eqnarray}
\hat \kappa_{\text{i} \gamma}^{(+)}(x)&:=&\hat \kappa_{\text{i} \gamma}(x+\text{i} \gamma-\text{i} \epsilon)\nonumber\\
\left[f*g\right](x)&:=&\int_{-\infty}^{\infty} f(x-y)g(y)\,\mbox d y\label{conv1}\; .
\end{eqnarray}
The integrations in \refeq{lnb1} are now convolutions along the real axis, such that one can express $\ln \mathfrak b$ in terms of $\ln \mathfrak B$, $\ln {\overline{{\mathfrak{B}}}}$ by solving an algebraic equation in Fourier space and transforming back to direct space. The result is
\begin{subequations}
\label{subb}
\begin{eqnarray}
\ln \mathfrak b(x)&=& -\frac{\pi}{2(\pi-\gamma)}\,\beta h- \frac{J\beta\sin\gamma}{2} \,d(x) + \left[\kappa *\ln \mathfrak B-\kappa_+*\ln {\overline{{\mathfrak{B}}}}\,\right](x)\label{lnb2}\\
\ln {\overline{{\mathfrak{b}}}}&=& \ln \mathfrak b^*|_{h\to -h}\label{lnbb2}
\end{eqnarray}
\end{subequations}
with
\begin{eqnarray}
d(x)&=&\frac{\pi}{\gamma\,\cosh\frac{\pi}{\gamma} x}\label{d1}\\
\kappa(x)&=& \int_{-\infty}^{\infty} \frac{\sinh\left(\frac\pi2-\gamma\right)k\,\text{e}^{\text{i} k x}}{2\cosh\frac\g2k \, \sinh\left(\frac\pi2-\frac\g2\right)k}\, \frac{\mbox d k}{2\pi}\nonumber\\
\kappa_+ (x)&=& \kappa(x+ \text{i} \gamma- \text{i} \epsilon)\nonumber
\end{eqnarray}
Note that the shift by $\epsilon$ in \refeq{lnb2} is necessary to ensure integrability. In a similar fashion, eq.\ \refeq{g} is treated. Therefore we introduce the functions
\begin{eqnarray}
G_\mathfrak b(x):=G(x+\text{i} \gamma/2,\xi)\,,\; G_{\overline{{\mathfrak{b}}}}:=-G(x-\text{i} \gamma/2,\xi)\label{defg},
\end{eqnarray}
where we suppressed the dependence on $\xi$ on the left-hand sides. It is now convenient to substitute
\begin{eqnarray}
\frac{G}{1+\mathfrak a}&=&\left\{\begin{array}{ll}
\displaystyle-\frac{G_{\overline{{\mathfrak{b}}}}}{1+{\overline{{\mathfrak{b}}}}^{-1}}& \mbox{on   } \mathcal C_-\\[0.4cm]
\displaystyle-\frac{G_\mathfrak b}{1+\mathfrak b^{-1}}+G_\mathfrak b& \mbox{on   } \mathcal C_+
\end{array}\right.\label{tran1}\\
\frac{G}{1+{\overline{{\mathfrak{a}}}}}&=&\left\{\begin{array}{ll}
\displaystyle\frac{G_{\overline{{\mathfrak{b}}}}}{1+{\overline{{\mathfrak{b}}}}^{-1}}-G_{\overline{{\mathfrak{b}}}}& \mbox{on   } \mathcal C_-\\[0.4cm]
\displaystyle\frac{G_\mathfrak b}{1+\mathfrak b^{-1}}& \mbox{on   } \mathcal C_+\,.
\end{array}\right.\label{tran2}
\end{eqnarray}
Then
\begin{subequations}
  \label{geq}
  \begin{eqnarray}
  G_\mathfrak b(x)&=& \text{i}\, d(x-\xi)+\left[\kappa *\frac{G_\mathfrak b}{1+\mathfrak b^{-1}}-\kappa_+*\frac{G_{\overline{{\mathfrak{b}}}}}{1+{\overline{{\mathfrak{b}}}}^{-1}}\right](x)\label{gb}\\
  G_{\overline{{\mathfrak{b}}}}&=& G_\mathfrak b^*|_{h\to -h}\label{gbb} .
  \end{eqnarray}
\end{subequations}
Using the substitution rules \refeq{tran1}, \refeq{tran2}, the general density matrix element \refeq{gen} is expressed in terms of $G_\mathfrak b/(1+\mathfrak b^{-1})$, $G_{\overline{{\mathfrak{b}}}}/(1+{\overline{{\mathfrak{b}}}}^{-1})$, $G_\mathfrak b$, $G_{\overline{{\mathfrak{b}}}}$. This expression is rather lengthy: For an $m$-point-function, one obtains a sum of $3^m$-many terms, each one an $m$-fold integral. Here we would like to spare the reader the general expression, but rather make the following comments:
\begin{itemize}
\item Eqs.\ \refeq{subb}, \refeq{gb} are solved numerically by iteration; the convolutions are done by applying the Fast-Fourier-Transform algorithm. Errors then arise from the truncation and the discretization of the integrals.
\item  In \refeq{gen} derivatives of $G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$ with respect to $\xi$ enter. These can be calculated straightforwardly by taking the $n$-th derivative of \refeq{geq} with respect to $\xi$ which results in linear integral equations for $\partial^n_\xi G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$.
\item The formulation in terms of the functions $G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$ is particularly suited for low temperatures. At $T=h=0$, $1/(1+\mathfrak b^{-1})=1/(1+{\overline{{\mathfrak{b}}}}^{-1})=0$ such that $G_{\mathfrak b}=G_{\overline{{\mathfrak{b}}}}$ is given by the inhomogeneity in \refeq{gb}. Furthermore, only one of the $3^m$-many terms in \refeq{gen} remains (namely the one containing only combinations of $G_\mathfrak b, \,G_{\overline{{\mathfrak{b}}}}$). Then in this term all functions are known. This is the well-known result of Jimbo et al. \cite{jm96} which has first been obtained for $\Delta>1$ in \cite{jim92} (for a proof in the critical case and an extension to $h\neq 0$ see also \cite{kit00}).
\end{itemize}
The functions for the isotropic case $\gamma=0$ are obtained by rescaling $x\to \gamma x$, $G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}\to \gamma G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$. Then in \refeq{gen} algebraic functions occur instead of hyperbolic ones, and the integration kernel $\kappa$ takes the form
\begin{eqnarray}
\kappa(x)&=& \int_{-\infty}^{\infty} \frac{\text{e}^{\text{i} k x}}{1+\text{e}^{|k|}} \, \frac{\mbox d k}{2\pi}\nonumber.
\end{eqnarray}
\subsection{Massive case}
Let us go through the changes which have to be done in the massive case with respect to the massless case. Analogously to \refeq{lnb}, \refeq{lnbb}, we define the functions
\begin{eqnarray}
 \mathfrak b(x)&=& \mathfrak a(\text{i} x+\eta/2),\; {\overline{{\mathfrak{b}}}}(x)= {\overline{{\mathfrak{a}}}}(\text{i} x-\eta/2)\nonumber\\
\mathfrak B&:=&1+\mathfrak b, \; {\overline{{\mathfrak{B}}}}:= 1+{\overline{{\mathfrak{b}}}}\nonumber\; .
\end{eqnarray}
We use the same symbols $\mathfrak b,{\overline{{\mathfrak{b}}}}$ etc. as in the massless case. Note that here $-\pi/2\leq x \leq \pi/2$ and the $\mathfrak b$-functions are periodic on the real axis with period $\pi$. Accounting for the pole of $\ln \mathfrak b(x)$ at $x=\text{i} \eta/2$ we find the following equations
\begin{eqnarray}
\ln \mathfrak b(x)&=& -\beta h-\frac{J \beta\sinh\eta}{2} \, \hat \kappa_{\eta/2}(x) +\frac{1}{2\pi}\left[\hat \kappa_\eta *\ln \mathfrak B-\hat \kappa_\eta *\ln \mathfrak b-\hat \kappa^{(-)}_\eta *\ln {\overline{{\mathfrak{B}}}}\,\right](x)\label{lnbm}\\
\ln {\overline{{\mathfrak{b}}}}(x)&=&\left.\ln \mathfrak b^*\right|_{h\to-h}\nonumber,
\end{eqnarray}
where the kernel and the convolutions are now defined as
\begin{eqnarray}
\hat\kappa_\eta(x)&=&\frac{\sinh(2\eta)}{\sin(x+\text{i} \eta)\, \sin(x-\text{i} \eta)}\nonumber\\
\left[f*g\right](x)&=&\int_{-\pi/2}^{\pi/2} f(x-y)g(y)\, \mbox d y\label{conv2}.
\end{eqnarray}
Because of periodicity, eq.\ \refeq{lnbm} is manipulated further in Fourier space after performing a discrete Fourier transformation. Upon transforming back one obtains
\begin{eqnarray}
\ln \mathfrak b(x)&=& -\frac{\beta h}{2}-\frac{J\beta\sinh\eta}{2}\,d(x) +\left[\kappa *\ln \mathfrak B-\kappa_-*\ln{\overline{{\mathfrak{B}}}}\,\right](x)\label{lnbm2}\\
d(x)&=&\frac{2K}{\pi}\text{dn} \left(\frac{2Kx}{\pi},\text{i}\frac{\eta}{\pi}\right)\label{d2}\\
\kappa(x)&:=& \sum_{n=-\infty}^\infty \frac{\text{e}^{\text{i} 2n x}}{\pi\left(1+\text{e}^{2  \eta|n|}\right)}\,\nonumber\\
\kappa_- (x)&:=& \kappa(x- \text{i} \eta+ \text{i} \epsilon).
\end{eqnarray}
The driving term \refeq{d2} is written in terms of the Jacobi elliptic function dn$(x,\tau)$ \cite{ww}, with Fourier series expansion
\begin{eqnarray}
\text{dn}(x,\tau)&=& \frac{\pi}{2K(\tau)}\sum_{n=-\infty}^{\infty} \, \frac{\text{e}^{\text{i} \pi n x/K(\tau)}}{\cos n\pi \tau},\label{defdn}
\end{eqnarray}
in the strip $\left|\Im (x)\right|<\Im(\tau\,K(\tau))$ of the complex plane. In \refeq{defdn} the constant $K(\tau)$ is defined through $K(\tau):=\frac{\pi}{2} \vartheta_3^2(0,\tau)$
and $\vartheta_3$ is one of Jacobi's Theta-functions \cite{ww}. In very close analogy one derives equations for $G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$, defined by
\begin{eqnarray}
G_\mathfrak b(x)&:=& G(\text{i} x+\eta/2,\xi)\,,\;G_{\overline{{\mathfrak{b}}}}(x):= -G(\text{i} x-\eta/2,\xi) \nonumber\; ,
\end{eqnarray}
where, as in the massless case, the $\xi$-dependence is not noted explicitly in $G_{\mathfrak b,{\overline{{\mathfrak{b}}}}}$.
The symbolical substitutions \refeq{tran1}, \refeq{tran2} still hold, with $\mathcal C_{\pm}$ defined in fig.\ \ref{fig1}a), so that
\begin{eqnarray}
G_\mathfrak b(x)&=&-d(x-\xi)+\left[\kappa *\frac{G_\mathfrak b}{1+\mathfrak b^{-1}}-\kappa_-*\frac{G_{\overline{{\mathfrak{b}}}}}{1+{\overline{{\mathfrak{b}}}}^{-1}}\right](x)\label{gm}\\
G_{\overline{{\mathfrak{b}}}}&=& G_\mathfrak b^*|_{h\to -h}\label{gmbb} .
\end{eqnarray}
\subsection{Calculating the integrals}
Let us consider eq.\ \refeq{gen}, where the substitutions implied by \refeq{tran1}, \refeq{tran2} have been performed. One observes that due to the product of hyperbolic functions in the denominator in eq.\ \refeq{gen}, multiple poles may occur. We explain in the following how to treat these poles numerically and show that contributions due to these poles cancel each other if there is more than one pole. For definiteness we treat the massless case, however, the arguments are directly transferable to the massive case.

The product $\prod_{1\leq a<b\leq m} \sinh(\omega_a-\omega_b-\text{i} \gamma)$ with
\begin{eqnarray}
\omega_\alpha=\left\{\begin{array}{ll}
x_\alpha+\text{i} \gamma/2 & \mbox{on   } \mathcal C_+\\
x_\alpha-\text{i} \gamma/2 & \mbox{on   } \mathcal C_-
\end{array}\right.
\end{eqnarray}
acquires a zero if there are $j,k$ such that $\omega_j=x_j+\text{i} \gamma/2-\text{i} \epsilon$, $\omega_k=x_k-\text{i} \gamma/2+\text{i} \epsilon$. This leads to a factor
\begin{eqnarray}
\frac{1}{\sinh(x_j-x_k-\text{i} \epsilon)}&=& \text{P}\left[\frac{1}{\sinh(x_j-x_k)}\right]+\text{i} \pi \delta(x_j-x_k)\label{pol1}\; .
\end{eqnarray}
When taking the principal value numerically, one has to account for the regular part at $x_j=x_k$, defined by $f_{reg}(x_j,x_j)=0$:
\begin{eqnarray}
\text{P} \int_{-\infty}^{\infty} \frac{f(x_j,x_k)}{\sinh(x_j-x_k)}\, \mbox d x_k&=&\int_{-\infty}^{x_j-\epsilon}+\int_{x_j+\epsilon}^\infty \frac{f(x_j,x_k)}{\sinh(x_j-x_k)}\,\mbox d x_k -2\epsilon \partial_{x_k} f_{reg}(x_j,x_k)|_{x_k=x_j}\, \label{pol2}.
\end{eqnarray}
The last terms in eqs.\ \refeq{pol1}, \refeq{pol2} are additional terms caused by the pole.

Now consider the case of multiple poles. Set $\omega_j=x_j+\text{i} \gamma/2$ in the product $\prod_{1\leq a<b\leq m} \sinh(\omega_a-\omega_b-\text{i} \gamma)$ and all other $\omega_k=x_k-\text{i} \gamma/2$. Then if $j=m-1$, there is one simple pole. If $j<m-1$, the product $\sinh(x_j-x_{m-1}-\text{i} \epsilon)\sinh(x_j-x_m-\text{i} \epsilon)$ occurs in the denominator. It is not difficult to show that the additional contributions due to these two poles either vanish directly  or cancel each other. One should be aware of the fact that the $\delta$-function in \refeq{pol1} yields no contribution if $f\equiv f_{reg}$. This is the case for $T=0$, where all functions are real and the determinant in \refeq{gen} vanishes if any two arguments of the auxiliary functions are equal.

We now address the question of possible simplifications of the multiple integral representation \refeq{gen}. Obviously, in the case $m=2$ the double integration has to be equal to one single integration, since the nearest-neighbor correlators are obtained by taking the derivative of the free energy per lattice site $f$ with respect to the anisotropy $\Delta$, and the free energy is given by a single integral according to \cite{kl93}. Especially
\begin{subequations}
\begin{eqnarray}
\langle S_j^zS_{j+1}^z\rangle_T&=& \frac14+\frac{\partial_\Delta f }{J} \label{nn1}\\
\langle S_j^xS_{j+1}^x\rangle_T&=& \frac{u}{2J}  -\frac{\Delta}{2} \langle S_j^zS_{j+1}^z-1/4\rangle_T\label{nn2}\\
\langle S_j^zS_{j+1}^z\rangle_T&=&\langle S_j^xS_{j+1}^x\rangle_T=\frac{1}{12}+\frac{u}{3 J},\qquad \Delta=1\label{nn3},
\end{eqnarray}
\end{subequations}
where $u=\partial_\beta(\beta f)$ is the inner energy per lattice site and $f$ is given by
\begin{eqnarray}
f= e_0-\frac{T}{2\pi}\left[d*\ln\mathfrak B{\overline{{\mathfrak{B}}}}\right](0). \nonumber
\end{eqnarray}
Here $d(x)$ is given by eq.\ \refeq{d1} (eq.\ \refeq{d2}) in the massless (massive) case; the convolution for both cases is defined in eq.\ \refeq{conv1} (eq.\ \refeq{conv2}). The ground state energy is given by
\begin{eqnarray}
e_0&=&\left\{
\begin{array}{ll}
\displaystyle -\frac{J\,\sin\gamma}{2}\,\int_{-\infty}^{\infty} \frac{\sinh\left(\frac\pi2-\frac\g2\right) k}{2\cosh\frac{\gamma k}{2} \, \sinh \frac{\pi k}{ 2}}\, \mbox d k& \mbox{massless}\\[0.4cm]
\displaystyle -\frac{J \sinh\eta}{2} \sum_{n=-\infty}^\infty \frac{\text{e}^{-\eta |n|}}{\cosh\eta n}& \mbox{massive.}
\end{array}\right.\nonumber
\end{eqnarray}
Eqs.\ \refeq{nn1}, \refeq{nn2} also allow us to extract the leading order in $T$ at low temperatures. We consider here the case of vanishing magnetic field, $h=0$. From \cite{klpei}, using dilogarithms, one finds for $T\to 0$ in the massless case
\begin{eqnarray}
f=e_0-T^2\frac{\gamma}{3J\sin \gamma}\nonumber.
\end{eqnarray}
Thus for $T\to 0$
\begin{subequations}
\begin{eqnarray}
\langle S^z_jS^z_{j+1}\rangle_T&=& \langle S^z_jS^z_{j+1}\rangle_{T=0} +(T/J)^2 \,\frac{\left(1-\gamma\,\cot\gamma\right)}{3 \sin^2 \gamma}\label{aatt}\\
\langle S^x_jS^x_{j+1}\rangle_T&=& \langle S^x_jS^x_{j+1}\rangle_{T=0} +(T/J)^2\,\frac{\left(\gamma-\cot\gamma+\gamma\,\cot^2\gamma\right)}{6 \sin\gamma}\label{xxtt}\\
\langle S^z_jS^z_{j+1}\rangle_T&=&\langle S^x_jS^x_{j+1}\rangle_T=\langle S^z_jS^z_{j+1}\rangle_{T=0}+\frac{T^2}{9 J^2},\; \Delta=1\label{aattiso}.
\end{eqnarray}
\end{subequations}
The correlator $\langle S^z_jS^z_{j+1}\rangle_{T=0}$ as a function of $\Delta$ is plotted in fig.\ 3 of ref.\ \cite{joh73}. This result has been extended to all temperatures in the range $-1\leq \Delta\leq 1$ in \cite{fab98}, fig.\ 2. Note from \refeq{aatt}-\refeq{aattiso} that the sign of the $T^2$-coefficient is positive, so that the antiferromagnetic correlations first decrease with increasing temperature.

In the massive case, for finite $\eta$, the leading temperature-dependent order may be obtained by a saddle-point integration, very similar to \cite{joh72}. Then
\begin{eqnarray}
f&=&-e_0-A^{1/2} T^{3/2} \text{e}^{-B/T}\nonumber\\
A&=&\frac{k'}{2J  k^2 K\sinh\eta}\nonumber\\
B&=& J\frac{k'\,K}{\pi} \sinh\eta\nonumber
\end{eqnarray}
with the moduli $k^{1/2}:=\vartheta_2(0,\text{i}\eta/\pi)/\vartheta_3(0,\text{i}\eta/\pi)$ and $\left(k'\right)^{1/2}:=\vartheta_4(0,\text{i}\eta/\pi)/\vartheta_3(0,\text{i}\eta/\pi)$. It is now not difficult to show that
\begin{eqnarray}
\langle S^z_jS^z_{j+1}\rangle_T&=& \langle S^z_jS^z_{j+1}\rangle_{T=0} +C T^{1/2}\text{e}^{-B/T} \nonumber\\
\langle S^x_jS^x_{j+1}\rangle_T&=& \langle S^x_jS^x_{j+1}\rangle_{T=0} +D T^{1/2} \text{e}^{-B/T}\nonumber\\
C&=& A^{1/2} \frac{k'K}{\pi}\left(\coth\eta+\sum_{n=1}^\infty\frac{1}{\cosh^2\eta(n-1/2)}\right)>0\nonumber\\
D&=& \frac{1}{2J}\left(A^{1/2} B-J\, C \cosh\eta\right)<0\nonumber\; .
\end{eqnarray}
At first sight, the last inequality is surprising: The antiferromagnetic correlations of $\langle S^x_jS^x_{j+1}\rangle_T$ increase with increasing temperature. Since $\lim_{T\to\infty}\langle S^x_jS^x_{j+1}\rangle_T=0$, we expect a minimum at finite temperature. This will be confirmed in the next section. Intuitively, this result can be understood as follows: At $T=0$, the Ising-like anisotropy leads to a Ne\'el-like ground state, which enhances the alignment of the spins along the $z$-direction. An initial increase of the temperature reduces $\left|\langle S^z_jS^z_{j+1}\rangle_T\right|$, so that quantum fluctuations of the spins are favored, and $\left|\langle S^x_jS^x_{j+1}\rangle_T\right|$ increases. At high enough temperatures the spins become uncorrelated, so that both correlation functions decay to zero.

As far as the low-temperature behavior of correlation functions in the massive regime is concerned, we would like to note that using the saddle-point integration technique, one finds that
\begin{eqnarray}
\langle e_{1\;\beta_1}^{\;\;\;\alpha_1}\ldots e_{m\;\beta_m}^{\;\;\;\alpha_m}\rangle_T&=&\langle e_{1\;\beta_1}^{\;\;\;\alpha_1}\ldots e_{m\;\beta_m}^{\;\;\;\alpha_m}\rangle_{T=0}+C_m T^{1/2}\text{e}^{-B/T}+\mathcal O\left(T^{3/2}\text{e}^{-B/T}, T\text{e}^{-2B/T}\right)\nonumber,
\end{eqnarray}
for arbitrary $m$, with some coefficient $C_m$.

Up to now, we were not able to derive eqs.\ \refeq{nn1}, \refeq{nn2} from \refeq{gen}. Still, the latter equations provide a useful test for the accuracy of our numerical integration scheme for $m=2$, as explained in more detail in the following section. For $m>2$, simplifications of \refeq{gen} have not been performed yet for any finite temperatures except at $\Delta=0$ and $ \Delta\to\infty$, see below. However, for $T=0$, where the integrand is known explicitly, the multiple integrals have been simplified considerably in the isotropic regime \cite{boo01,boo03,boo05}. In the anisotropic regime, the multiple integrals have been reduced to single integrals for $m\leq 4$ \cite{kat03,kat04}. These works allow us to estimate the numerical accuracy at $T=0$. In the other extreme, at $T=\infty$, the spins are uncorrelated so that in this limiting case we also have explicit results which can be checked. Note, however, that it is rather non-trivial to reproduce these high-temperature results from the general formula \refeq{gen}. This was done only very recently by Tsuboi and Shiroishi in \cite{tsu05}, where a high-temperature expansion of $P(m)$ has been performed to high order in $1/T$ for the isotropic case. This is one further result which will serve as a reference point for estimating the accuracy of our numerics for $m=3$.

The cases $\Delta=0$ (free fermions) and $\Delta\to\infty$ such that $J\cdot \Delta=:c$ finite (Ising limit) allow for substantial simplifications. Details will be explained in a separate publication \cite{gs05}, here we only give the results. For $\Delta=0$, that is $\gamma=\pi/2$, the convolutions in \refeq{lnb2}, \refeq{gb} disappear, so that all functions are known explicitly for all temperatures. Then
\begin{eqnarray}
P(m)=\det_m\left[\int_{-\pi}^\pi \frac{\mbox d p}{2\pi} \frac{\text{e}^{\text{i}(a-b)p}}{1+\exp\left(\beta J\cos p-\beta h\right)}\right]_{ab},\nonumber
\end{eqnarray}
in agreement with \cite{shi01}.
In the other extreme, the Ising limit, the driving term and integration kernels become constant ($d(x)\to 1$, $\kappa(x)\to1/(2\pi)$) such that the auxiliary functions do not depend on $x$ and the integrations are trivial. One finds \cite{bax82,suz02,gs05}
\begin{eqnarray}
P(m)&=&\frac{1+\langle\sigma^z\rangle_T}{2}\left[\frac{1+\langle\sigma^z\rangle_T}{2}+\frac{1-\langle\sigma^z\rangle_T}{2}\frac{\lambda_2}{\lambda_1}\right]^{m-1}\label{is}\\
\lambda_{1,2}&=&\text{e}^{-\beta c/4} \left(\cosh\frac{\beta h}{2}\pm \sqrt{\sinh^2\frac{\beta h}{2}+\text{e}^{\beta c}}\right)\nonumber\\
\langle\sigma^z\rangle_T&=&\frac{\sinh\frac{\beta h}{2}}{\sqrt{\sinh^2\frac{\beta h}{2}+\text{e}^{\beta c}}}\nonumber .
\end{eqnarray}

\section{Results}
We first present results of the next-neighbor correlation functions, comparing data from the double integration with those from the single integration. Results for next-nearest neighbors will be given subsequently. Correlation functions for longer distances can be evaluated in the same way, however, the numerical cost increases exponentially with the number of integrations. That is why for the time being in this feasibility study we do not go beyond $m=3$.

\subsection{Nearest neighbors}
Figs.\ \ref{fig2}, \ref{fig3} show $P(2)$, $\langle S^x_jS^x_{j+1}\rangle_T$, as calculated from the double integral, for $P(2)$ together with the relative error with respect to the data obtained from the derivative of the free energy according to \refeq{nn1}, \refeq{nn2}. The single integration results can be considered as exact; the numerical error is less than 8 digits for all temperatures. We checked that the relative error for $\langle S^x_jS^x_{j+1}\rangle_T$ vanishes at low temperatures as for $P(2)$; at high temperatures, the absolute error is about $10^{-4}$.
\begin{figure}
\begin{center}
\vspace{-2cm}
 \includegraphics[angle=-90, scale=0.5]{./FIG/aasymdintdiva}
\vspace{-1cm}
\includegraphics[angle=-90, scale=0.5]{./FIG/erroraa}
\caption{a) Emptiness formation probability $P(2)$ for nearest neighbors; the upper curve corresponds to the $XX$ limit $\eta=\text{i} \pi/2$. The inset shows the $T^2$-coefficient at low temperatures in the massless case. Straight lines indicate the exact values according to eqs.\ \refeq{aatt}, \refeq{aattiso}. b) Relative error of $P(2)$. The curve labeled by ``$\eta =5$ Ising'' shows the relative difference between the $\eta=5$-values of a) and the Ising limit eq.\ \refeq{is}.}
\label{fig2}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
 \includegraphics[angle=-90, scale=0.5]{./FIG/xxdiva}
\caption{Correlation function $\langle S^x_jS^x_{j+1}\rangle$ for nearest neighbors; the lower curve corresponds to $\eta=\text{i} 0.8\pi/2$. The inset shows the $T^2$-coefficient at low temperatures in the massless case. Straight lines indicate the exact values according to eq.\ \refeq{xxtt}.}
\label{fig3}
\end{center}
\end{figure}

In both cases, the error increases with increasing temperature, reaching maximal values at $T\approx J$ near the isotropic point. Generally, the error is considerably larger at high than at low temperatures. This is due to two reasons:
\begin{itemize}
\item At finite temperature, the auxiliary functions are given numerically from the solutions of the integral equations, whereas at $T=0$, they are known explicitly. The numerical iteration scheme necessarily induces errors due to the truncation and discretization of the integrals.
\item The additional contribution from poles in eq.\ \refeq{pol1} increases with increasing temperature and causes an additional error in the integrations.
\end{itemize}
We have verified that the error is reduced by decreasing the distance between two successive integration points. We conclude that the errors are mainly due to the truncation and discretization of the integrals and can be made arbitrarily small at the expense of increasing computation time.

Considering the low-temperature behavior of $\langle S^x_jS^x_{j+1}\rangle_T$, one finds an initial decrease from $T=0$ with a minimum at $T_0$, this is illustrated in fig.\ \ref{figmass}.  We commented on this phenomenon in the previous section. From fig.\ \ref{figmass} one observes that in the Ising limit, the maximum of $\langle S^x_jS^x_{j+1}\rangle_T/\langle S^x_jS^x_{j+1}\rangle_{T=0}$ is located at $T_0 /(\Delta J)\approx 0.168$.

\subsection{Next-Nearest neighbors}
We calculated next-nearest neighbor correlation functions for the isotropic and the massive case. In the massless case, the auxiliary functions in \refeq{gen} are multiplied by a kernel which, depending on the integration variables, can increase exponentially, so that the auxiliary functions have to be determined extremely accurately over the whole integration range. This problem is still controllable for nearest neighbors, but becomes severe for $m>2$. In the isotropic case, however, one deals with algebraic functions, which are much better to handle numerically. At anisotropy parameter $\Delta>1$ the integration range is finite, so that those problems do not occur.

In fig.\ \ref{fig4} we show the emptiness formation probability $P(3)$. In the isotropic case, we compare our data with the high-temperature expansion of \cite{tsu05} and with Quantum Monte Carlo (QMC) data which were obtained in \cite{tsu05}. The relative error of our data shown in fig.\ \ref{fig4} is larger than in fig.\ \ref{fig2}, which is due to the fact that the data in \refeq{fig4} were calculated by using only half of the number of integration points compared to the $m=2$ case. We checked at single temperature values that the error is of the same order as in the $m=2$ case when the calculations are done with the same numerical accuracy. However, the computational costs would be unreasonably high. At lowest temperatures, a comparison with \cite{boo01,kat03,kat04} yields relative errors of the order $2\cdot10^{-4}$.
\begin{figure}
\begin{center}
\vspace{-2cm}
 \includegraphics[angle=-90, scale=0.5]{./FIG/aaadiva}
\vspace{-1cm}
 \includegraphics[angle=-90, scale=0.5]{./FIG/erroraaa}
\caption{a) Emptiness formation probability $P(3)$ for next-nearest neighbors. Shown are results from the 3-fold integral, from QMC \cite{tsu05} and high-temperature expansion (HTE) combined with Pad\'e approximation \cite{tsu05}. b) Relative error of the data from the 3-fold integral with respect to the QMC and the HTE-data.}
\label{fig4}
\end{center}
\end{figure}

As a second example, $\langle S^x_{j}S^x_{j+2}\rangle_T$ is depicted in fig.\ \ref{fig5}. At lowest temperatures we again compare our data with the explicit results of \cite{kat03,kat04} which yields a relative error of $7\cdot 10^{-3}$. At highest temperatures, $\langle S^x_{j}S^x_{j+2}\rangle_T=0$. The absolute error here is of the order $10^{-3}$. As in the $m=2$-case we observe an increase of the antiferromagnetic correlation with increasing $T$, reaching a maximum at some finite temperature $T_0$ in the massive regime. Fig.\ \ref{figmass} illustrates this behaviour, where $\langle S^x_jS^x_{j-m+1}\rangle_T/\langle S^x_jS^x_{j-m+1}\rangle_{T=0}$ is depicted for $m=2,3$ and $\Delta>1$.
\begin{figure}
\begin{center}
 \includegraphics[angle=-90, scale=0.5]{./FIG/sx1sxdiva}
\caption{The correlation function $\langle S^x_jS^x_{j+2}\rangle_T$ for $\eta=0,1$. Note that at low temperatures for $\Delta>1$, the correlation increases, before it starts to decrease to zero.}
\label{fig5}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
 \includegraphics[angle=-90, scale=0.5]{./FIG/maxmass}
\caption{The correlation function $\langle S^x_jS^x_{j+m-1}\rangle_T$ divided by its zero-temperature value $\langle S^x_jS^x_{j+m-1}\rangle_{T=0}$ for $m=2,3$ in the massive regime $\Delta>1$. The dotted horizontal line gives the $T=0$-value 1. The temperature is normalized with respect to $J\Delta$.}
\label{figmass}
\end{center}
\end{figure}
\section{Summary and Outlook}
We demonstrated that an accurate numerical evaluation of the formula \refeq{gen} is possible by considering the special cases $m=2$ with $\Delta \geq 0$ and $m=3$ with $\Delta\geq 1$. The numerical error was estimated precisely by comparing with alternative approaches. We have found that this error is only due to the discretization and, for $\eta =\text{i} \gamma$, the truncation of the integrals. For $m=3$ in the massless case we encountered problems hard to deal with numerically. Also the computational costs of an extension of the numerics to $m>3$ become very high if one wants to maintain the accuracy. Thus, the next step should be to address the question of in how far the multiple integrals can be simplified. Here we would like to mention two approaches.

In \cite{tsu05} an high-temperature expansion (HTE) has been applied to calculate $P(m)$. The excellent agreement of these data with those obtained by numerically evaluating the multiple integrals directly (cf. fig.\ \ref{fig4}), even at temperatures below the crossover region highlights the applicability of the HTE method. On the other hand, in the works \cite{boo01,boo04,kat03,kat04} the multiple integrals have been reduced to single integrals at $T=0$. The case $m=2$ shows that simplifications are possible also at finite $T$ (compare also \cite{gks2}), where the integrand in \refeq{gen} is given implicitly by integral equations. Further progress may be expected from extending the recent results \cite{boo04,boo04b} for the inhomogeneous generalizations of the correlation functions of the $XXZ$ chain to finite $T$. This issue is the subject of current research.

\section*{Acknowledgments}
We are very grateful to M. Shiroishi and Z. Tsuboi for sending us their QMC- and HTE-data. We would also like to acknowledge very helpful and stimulating discussions with A. Kl\"umper and A. Seel.
\newpage

redpajama_set_name: RedPajamaArXiv


text:
FALL TV PREVIEW – NETWORK LINE-UPS
The 2011 Fall TV Preview – Primetime Line-Up… Or: How to pass your time during the Mad Men hiatus. Perhaps the networks learned a valuable lesson on over-hyped shows that only lasted one season, because I am noticing a quieter summer marketing blitz this session.
For instance, it seems that the failure of NBC's The Event to capture that Lost feel (something ABC tried unsuccessfully with Flash Forward) or The Cape to capture any coherent feel at all has the networks fearfully sneaking in a Whitney scene to the audience. It's as if they're the smoke monster and wondering, "Will they kill and devour this just like Law and Order: Los Angeles?" Are they hoping that folks will actually know who Whitney Cummings is before they have them deciding between her and Grey's Anatomy on Thursdays? Good luck with that.
Here's an interesting fact: Each of the major networks including ABC, NBC, CBS, and Fox (sorry CW, you lost your only viable option with me when you ended Smallville) all cancelled seven shows. A coincidence I'm sure.
CBS seems the most pleased with their incumbent line-up as they are only brining in six new shows this fall with Person of Interest worthy of the most attention. Jim Caviezel stars as an ex-covert operative given a second chance by a magnanimous billionaire who possesses the resources and technology to right wrongs before they occur. Caviezel carries himself with a quiet yet severe dignity that is enticing, but even with pedigree producer J.J. Abrams, the Thursday night slot is a tough nut to crack.
ABC, like NBC, has unseated their prior year's tent poles with V and Detroit 1-8-7 being the most notable pink slips. Unlike CBS, however, ABC is bringing in thirteen new shows (four at mid-season), including a Charlie's Angels reboot starring three beautiful women who look just different enough to bring in the widest possible audience. Whatever, at least you know exactly what you are getting with that show unlike Pan Am, which in beautiful HD looks like Catch Me if You Can but is definitely trying more for the pseudo-sleazy-sexy-charming allure of Mad Men. Creator Jack Orman had previously written for Jag and was a co-creator of the failed Dr. Vegas; neither bodes well until you consider Matthew Weiner used to write for Andy Richter Controls the Universe. If Orman can make 60's sheik work as Weiner did with 'Mad Men', it should do well with a broader network audience.
Fox also seems to like what they have as they are only premiering seven new shows (two at midseason), which does not include the latest two-night two and a half hour Simon Cowell juggernaut each week. Dinosaurs are the other big story at Fox as Terra Nova is set for Monday nights. Executive produced by Steven Spielberg (who has had some success with dino vehicles in the past) and with a budget that essentially promises lots and lots of dinosaur action, sci-fi fans and budding paleontologists alike should be satiated.
Interesting fact number two: Each major network cancelled or ended a police procedural style drama. Now we are only left with about seventy-five cop shows.
Back to NBC, who along with its reluctant promotion of Whitney (and the aforementioned excision of last year's most promising) is also premiering Prime Suspect starring Maria Bello, based on the fantastic British series starring Helen Mirren. If producer/writer/director Peter Berg can just stick to the script, and not try to model AMC's The Killing, then they should be fine. Otherwise they'd better hope to siphon some of the Pan Am crowd with their own 1960's homage The Playboy Club, with Eddie Cibrian as Don Drap…er, Nick Dalton.
Bottom line, it depends on what you are looking for. The networks are bowing shows of all shapes and sizes, from the foreboding Grimm on NBC to comedy New Girl starring indie darling Zooey Deschanel on Fox. ABC is even delving back into the surreal with Once Upon a Time from Lost producers Adam Horowitz and Edward Kitsis. There seems to be something for everybody… if you believe the hype.