id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
7700 | https://zapple.tech/blog/types-of-automation-testing/importance-of-automated-sanity-testing/ | Importance of automated sanity testing - Step-by-Step Guide
Request a quote
Pricing
Portfolio
ServicesBY PLATFORMSWeb Test Automation Services Mobile test automation Desktop test automation Server-side test automation BY TESTING TYPESFunctional testing Regression testing Integration testing Performance testing Security testing BY ACTIVITIESTest automation strategy Audit and Assessment Custom framework development Automated scripts development Automated testing maintenance Tools integration BY APPROACHAQA outsourcing AQA outstaffing QA Outsourcing Automated QA managed service HIRE HIRE Hire Test Automation Engineers Hire QA Engineers Dedicated QA Team
IndustriesE-commerce Financial Healthcare Travel and Accommodation
ResourcesBlog Trainings Ebooks Virtual summit Meetups
About us
Get in touch
Search
Pricing
Portfolio
ServicesBY PLATFORMSWeb Test Automation Services Mobile test automation Desktop test automation Server-side test automation BY TESTING TYPESFunctional testing Regression testing Integration testing Performance testing Security testing BY ACTIVITIESTest automation strategy Audit and Assessment Custom framework development Automated scripts development Automated testing maintenance Tools integration BY APPROACHAQA outsourcing AQA outstaffing QA Outsourcing Automated QA managed service HIRE HIRE Hire Test Automation Engineers Hire QA Engineers Dedicated QA Team
IndustriesE-commerce Financial Healthcare Travel and Accommodation
ResourcesBlog Trainings Ebooks Virtual summit Meetups
About us
Get in touch
Importance Of Automated Sanity Testing
Sergey Almyashev
AQA ProcessImportance of Test AutomationTypes of Automation Testing
23/02/2022
According to the GMInsights research, the value of the QA services market in 2020 exceeded $40 billion and is projected to grow by 7% annually, reaching $60 billion by 2027. This rapid growth highlights the increasing demand and critical importance of testing in the creation of modern complex systems and single products. In today’s competitive market, if a company releases the first software version containing performance or interface errors, customers are likely to respond negatively, which could severely impact the brand’s reputation and drive users to competitors. This makes it essential to ensure rigorous quality control right from the early stages of development. The cost of a mistake can be prohibitive, especially for startups, where even minor errors can lead to significant financial and reputational damage.
Given these stakes, automated sanity testing becomes a vital tool in the QA process. ZappleTech experts have thoroughly studied the QA industry and compiled detailed information on its importance and role in the viability of IT products. Their findings emphasize how automated sanity testing can effectively identify errors at all stages of application development, ensuring that products meet the high standards expected by users. From the article, you will gain deeper insights into the process of sanity testing automation and how it contributes to maintaining the reliability and performance of software projects.
Table of Contents
Automated Sanity Testing: Goals, Objectives, Blind Spots, Frequency, and Problems
Definition and Purpose of Sanity Testing
Sanity vs. smoke comparison: different approach and philosophy
Goals and Objectives of Sanity Testing
Frequency of automated sanity testing of an IT product
When to Perform Sanity Testing?
Blind spots and regression testing issues:
Sanity vs. smoke comparison: different approach and philosophy
An example of using Sanity during testing
Automated sanity tests: advantages, tools, profit
Tools to optimize the sanity check process
Let’s summarize the software development life cycle
Automated Sanity Testing: Goals, Objectives, Blind Spots, Frequency, and Problems
Sanity testing is one of the most crucial aspects of quality assurance (QA), playing a vital role in identifying problematic areas in software that either fail to function properly or do not perform required actions. This type of testing is essential in ensuring that the basic functionality and core functionalities of an application are working as expected after minor changes or bug fixes. However, relying on spontaneous or ad-hoc approaches to sanity testing is a fundamentally flawed practice that can lead to missed issues and inadequate coverage.
The goals and objectives of sanity testing include quickly verifying that recent code changes have not adversely affected the critical functionalities of the software. By focusing on specific areas of concern, sanity testing ensures that any new defects are caught early in the development cycle, preventing them from escalating into more significant problems later on.
Despite its importance, sanity testing has its blind spots. Since it is typically a targeted and narrow scope test, it may overlook some issues that a broader testing approach might catch. Therefore, it’s essential to complement sanity testing with more comprehensive testing methods to ensure full coverage.
The frequency of sanity testing depends on the pace of development and the frequency of code changes. Ideally, it should be performed regularly, especially after every significant update or bug fix. This helps maintain a high level of software quality and reduces the risk of introducing new defects.
However, manual sanity testing can be time-consuming and prone to human error. To address this, many teams are turning to automated sanity testing. Automated sanity testing allows for quicker execution and more consistent results, ensuring that critical areas are tested thoroughly and efficiently every time changes are made.
Sanity testing is needed for:
Checking the finished build and each component.
Identification of non-working functions.
Comparison of results using the “white box” method.
Surface testing of each item.
Obtaining information about the progress of an IT product.
Defining major and minor bugs.
Automated sanity testing helps you quickly verify that recent code changes haven’t broken existing functionality, ensuring continuous stability throughout the development process. This keeps your project on track and minimizes disruptions.
Mykhailo Poliarush
CEO, ZappleTech Inc.
Automated sanity testing helps you quickly verify that recent code changes haven’t broken existing functionality, ensuring continuous stability throughout the development process. This keeps your project on track and minimizes disruptions.Mykhailo PoliarushCEO, ZappleTech Inc. A bit like a kind of regression testing, isn’t it? In fact, a health check is its component, although it makes sense to consider them separately, not combined in a complex.
Sanity testing tasks:
Reducing non-working functions.
Cursory check of the product at every stage.
Indication of development progress and its effectiveness.
Additional collection of information about unresolved bugs.
Spontaneous checking between sprints or during them.
Study of the main problems, analysis, and prevention.
Performance testing is provided by the “white box” method. Although limited in scope, it improves the final quality, reduces data scatter in the reports, and facilitates faster error recovery.
Definition and Purpose of Sanity Testing
Sanity testing is a type of software testing that involves verifying the functionality of a software application or system after minor changes or bug fixes. The primary purpose of sanity testing is to ensure that the changes made to the software have not introduced any new bugs or issues that could affect its overall functionality. This type of testing is a crucial step in the software development life cycle, as it helps to identify and fix defects early on, reducing the risk of downstream problems and ensuring that the software meets the required quality standards. By focusing on specific areas of concern, sanity testing provides a quick and efficient way to validate that the software remains stable and reliable after updates.
Sanity vs. smoke comparison: different approach and philosophy
Many people compare these two types of testing, which is a common mistake. While sanity testing reveals superficial problems only, smoke testing is used to scan a wide area at all stages of development. Sanity testing focuses on verifying specific functionalities after changes are made, ensuring that a known result is achieved. In contrast, smoke testing determines whether the basic functions of a software application are fully operational or if they are interrupted at some stage.
In addition to traditional sanity testing, automated sanity testing plays a crucial role in modern software development. This approach allows for faster and more consistent checks, ensuring that key functionalities work as expected after each code change. By incorporating automated sanity testing into the process, developers can quickly identify any issues, further distinguishing it from the broader scope of smoke testing.
Let’s take a look at the main differences:
This table demonstrates the difference between these types of checks. They cannot be compared because they are complementary and focus the attention of QA specialists on different areas of the product. Both tests are vital and cannot be mutually exclusive.
Goals and Objectives of Sanity Testing
The primary goals and objectives of sanity testing are to:
Verify that the changes made to the software have not introduced any new bugs or issues.
Ensure that the software functions as expected after minor changes or bug fixes.
Identify and fix defects early on, reducing the risk of downstream problems.
Build confidence in the software, allowing developers to move forward with further testing and deployment.
Reduce the overall cost of software development by identifying and fixing defects early on.
By achieving these goals, sanity testing helps maintain the integrity and quality of the software throughout the development process.
Frequency of automated sanity testing of an IT product
Automated sanity testing plays a crucial role in maintaining the stability of an IT product during its development cycle. The frequency of automated tests, including automated sanity testing, can significantly impact the overall efficiency and quality of the project. The more often it is conducted, the better it ensures that the core functionalities of the software remain intact after minor changes or bug fixes.
It’s advisable to start preparing for automated sanity testing from the project’s inception, but its true effectiveness becomes apparent after the first builds are compiled. If builds are compiled daily, the frequency of these tests should match to ensure that any issues are caught early. However, it’s essential to balance this with productivity, as each testing cycle can take around 20 minutes.
Although automated sanity testing is essential, it’s often performed as a secondary process. This is because other testing types, such as regression or functional testing, are prioritized for their deeper analysis of the product. Automated sanity testing, however, addresses the need for quick validation of key features, ensuring that the product remains stable even as development progresses. We’ll delve into how this process works and its benefits in detail later.
By automating sanity tests, you can rapidly assess core features without manual intervention, saving valuable time and resources. This allows your team to focus on more complex testing tasks, increasing overall efficiency.
Sergey Almyashev
COO, ZappleTech Inc.
When to Perform Sanity Testing?
Sanity testing should be performed after minor changes or bug fixes have been made to the software. This can include:
After a new build has been released.
After bug fixes have been implemented.
After minor changes have been made to the software.
Before further testing and deployment.
By conducting sanity testing at these critical junctures, developers can ensure that the software remains stable and functional, paving the way for more comprehensive testing and successful deployment.
Blind spots and regression testing issues:
So, testing starts only after compiling a working build, which is the main problem of this solution. Unlike the others, performance checking is about looking for deviations between the actual and expected results. The intermediate state of an IT product is a blind spot that neutralizes attempts to investigate the functionality for errors. This blind spot often complicates the identification of issues that could have been caught earlier in the development cycle, making it crucial to incorporate automated sanity testing to continuously monitor and verify core functionalities even during these intermediate stages.
The second problem with performance testing is superficiality. You cannot set more advanced parameters for the scenario, only the expectation/reality comparison. Actually, it is not a critical disadvantage because the test is a component of regression testing performed to get additional information about the product during sprints and spontaneous checks. However, integrating automated sanity testing within this process can help address this issue by providing consistent, automated checks that ensure the system’s critical functions remain stable, even with the introduction of new code.
Sanity vs. smoke comparison: different approach and philosophy
Many people compare these two types of testing, which is a common mistake. While sanity testing reveals superficial problems only, a smoke test is used to scan a wide area at all stages of development. Sanity testing focuses on verifying specific functionalities after changes are made, ensuring that a known result is achieved. In contrast, smoke tests determine whether the basic functions of a software application are fully operational or if they are interrupted at some stage.
In addition to traditional sanity testing, automated sanity testing plays a crucial role in modern software development. This approach allows for faster and more consistent checks, ensuring that key functionalities work as expected after each code change. By incorporating automated sanity testing into the process, developers can quickly identify any issues, further distinguishing it from the broader scope of smoke testing.
Let’s take a look at the main differences:
This table demonstrates the difference between these types of checks. They cannot be compared because they are complementary and focus the attention of QA specialists on different areas of the product. Both tests are vital and cannot be mutually exclusive.
An example of using Sanity during testing
Imagine a marketplace project where multiple sellers operate, each paying a commission to the platform’s owner for every transaction. For instance, the platform might take a 10% commission on sales, plus an additional 1% for processing fees when payments are made using a bank card. One critical aspect to check is whether these commissions are calculated automatically and correctly transferred to the owner’s account.
Let’s break this down with a specific example. We start with a transaction amount of $350 and use two virtual cards: one for the seller and one for the intermediary. After processing the transaction, we check the balances. If the seller ends up with $311.5 and the intermediary with $38.5, we consider the test successful, and the results are accurate. However, if the expected amounts don’t match, this indicates a potential bug in the system, and further testing is required. The issue is then flagged and assigned to developers to be fixed.
While this type of testing is essential, repeating it manually every day can quickly become tedious for a specialist, leading to delays or even neglect of the task. This is where automated sanity testing comes into play. By automating this process and incorporating it into regular sprints, you can ensure that these checks are performed consistently without the risk of human error or oversight.
Incorporating automated sanity testing into your testing strategy not only saves time but also enhances the reliability of your marketplace platform, ensuring that financial transactions are always processed accurately and efficiently.
Automated sanity tests: advantages, tools, profit
Automated sanity testing plays a crucial role in ensuring the stability and reliability of software products. By employing automated tests to automate the processes that are typically performed manually and spontaneously, you can significantly enhance the efficiency of testing and reduce the likelihood of human error. This type of testing focuses on verifying that new changes or updates to the codebase do not introduce any unforeseen issues or bugs. It’s particularly beneficial after minor code modifications, where a quick validation is necessary to ensure that everything still functions as expected.
One of the primary advantages of automated sanity testing is the speed at which tests can be executed. Automated tools can run these tests much faster than a human could, allowing for more frequent testing and quicker feedback. This is especially important in agile development environments, where code changes are frequent, and the need for rapid testing is paramount.
Moreover, automated sanity testing tools can be integrated into your continuous integration and continuous deployment (CI/CD) pipelines, further streamlining the development process. By catching issues early in the development cycle, you can avoid costly delays and ensure that your product meets the highest quality standards before it reaches the end users.
In addition to improving testing efficiency, automated sanity testing also contributes to overall cost savings. By reducing the need for manual testing, you can allocate resources more effectively and focus on more complex testing scenarios that require human insight. This not only saves time but also increases the overall profitability of your project by ensuring that your software is reliable and ready for release.
In conclusion, incorporating automated sanity testing into your QA strategy is essential for maintaining the stability and quality of your software products. The advantages it offers, from faster testing cycles to cost savings, make it an indispensable tool in modern software development. By automating sanity tests, you can rapidly assess core features without manual intervention, saving valuable time and resources. This allows your team to focus on more complex testing tasks, increasing overall efficiency.
Sergey Almyashev COO, ZappleTech Inc.
Sanity testing automation has many advantages over traditional manual error checking:
It speeds up checking and comparing the expected result with the actual one.
Testing is parallelized, so it is possible to run multiple scripts for different functions simultaneously.
There is no need for strict control over checking each element.
More time to concentrate on more important and urgent tasks.
Decreased reliance on regression testing between sprints due to continuous troubleshooting.
The final quality of testing and its effectiveness are improved in terms of stability and the number of fixes before release.
To optimize the development and QA service, you need to run automated sanity testing during sprints, starting from the early stages of project development.
Do the implementation team and the customer benefit from sanity testing automation? What do you think?
Automated sanity tests provide immediate feedback after code changes, enabling faster detection of critical issues. This helps in addressing problems early, reducing the risk of larger defects down the line.
Mikhail Bodnarchuk
CDO, ZappleTech Inc.
Tools to optimize the sanity check process
To optimize the sanity check process, it’s crucial to consider tools that are specifically designed to enhance efficiency and accuracy through automated tests. Given the limited testing area involved in sanity checks, almost any basic automation tool can be effective, provided it works seamlessly with the interface, handles input/output data efficiently, and accurately analyzes the results. When selecting a tool, ensure that it aligns with your technical stack, product type, and specific features.
In our projects, we emphasize the use of classic software and automation environments that have consistently demonstrated flexibility, versatility, and reliability. This approach not only streamlines the process but also integrates automated sanity testing into the workflow, ensuring that the software maintains stability and meets quality standards before moving forward in the development cycle.
Best tools for sanity testing automation:
Selenium is one of the most popular products for automating interface and functionality testing. Works with a wide range of technologies and supports scripting.
Cypress is a widespread test automation tool. It supports algorithm-based customized scenarios and has a recording/playback feature, which makes it versatile for testing web solutions.
Playwright is a good cross-platform alternative to other web product testing apps. It supports extended syntax, all popular browsers, and many programming languages.
Let’s summarize the software development life cycle
--------------------------------------------------- Many people believe that automated tests in sanity testing automation are merely a subset of regression testing and, as such, aren’t necessary as a separate process. However, we strongly disagree. Our team considers automated sanity testing to be a crucial step in the QA process, offering an extra layer of validation that ensures your product’s core functionality is intact after any changes. This proactive approach allows us to catch and address issues early, reducing the likelihood of more significant problems arising later in the development cycle.
Automated sanity testing serves as a quick, efficient check that saves time and resources, ultimately contributing to a smoother and more reliable product launch. By integrating this process into your QA strategy, we can help you deliver a higher-quality product that meets customer expectations and performs flawlessly under real-world conditions.
Automated sanity tests provide immediate feedback after code changes, enabling faster detection of critical issues. This helps in addressing problems early, reducing the risk of larger defects down the line. Mikhail Bodnarchuk CDO, ZappleTech Inc.
Need to test an IT product? Order this service from our manager, and we will provide you with quality services that will speed up your product’s release, ensuring a quicker path to market success and a more substantial return on investment. Whether you’re working on a small update or a major release, our automated sanity testing service will help you achieve your goals with confidence and precision.
4.5/5 - (31 votes)
automated testingbest practicesguideimportance of testingsanity testing
-
OUR CASES
View All Cases
Table of Contents
Automated Sanity Testing: Goals, Objectives, Blind Spots, Frequency, and Problems
Definition and Purpose of Sanity Testing
Sanity vs. smoke comparison: different approach and philosophy
Goals and Objectives of Sanity Testing
Frequency of automated sanity testing of an IT product
When to Perform Sanity Testing?
Blind spots and regression testing issues:
Sanity vs. smoke comparison: different approach and philosophy
An example of using Sanity during testing
Automated sanity tests: advantages, tools, profit
Tools to optimize the sanity check process
Let’s summarize the software development life cycle
Continue Reading
Previous post Automation Testing for E-Learning Platforms
Next post What Is Unit Testing?
Related Posts
-
AQA Virtual Summit/MeetupsAutomated testingImportance of Test Automation
Szilard Szell – How Testing is Evolving in DevOps and how shall we evolve ourselves
03/05/2024
-
AQA ProcessImportance of Test AutomationTest Automation Frameworks
How to Get Started Automating Mobile App Testing?
12/04/2024
Reinventing the way you create automated tests, say hi to the very powerful AQA services available on the market, made by Zapple.Tech.
© 2024 Zapple.Tech. All rights reserved.
SOLUTIONS
Web test automationMobile test automationAPI & Serverside test automationPerformance testingTool and solutions developmentTestOps and DevOps
INDUSTRIES
E-commerceFinancialHealthcareTravel and Accommodation
MAIN
PricingServicesPortfolioContact us
RESOURCES
TrainingsEbooksBlogVirtual summitMeetups
HIRE
Hire test automation engineersHire QA engineersDedicated QA team
Close
Search
Hit enter to search or ESC to close
Search for: |
7701 | https://en.wiktionary.org/wiki/tercentennial | tercentennial - Wiktionary, the free dictionary
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigation
Main Page
Community portal
Requested entries
Recent changes
Random entry
Help
Glossary
Contact us
Special pages
Feedback
If you have time, leave us a note.
Search
Search
[x] Appearance
Appearance
move to sidebar hide
Text
Small Standard Large
This page always uses small font size
Width
Standard Wide
The content is as wide as possible for your browser window.
Color (beta)
Automatic Light Dark
This page is always in light mode.
Donations
Preferences
Create account
Log in
[x] Personal tools
Donations
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
[x] Toggle the table of contents
Contents
move to sidebar hide
Beginning
1 EnglishToggle English subsection
1.1 Etymology
1.2 Noun
1.2.1 Related terms
tercentennial
[x] 5 languages
Eesti
Malagasy
Русский
தமிழ்
Tiếng Việt
Entry
Discussion
Citations
[x] English
Read
Edit
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Create a book
Download as PDF
Printable version
In other projects
From Wiktionary, the free dictionary
English
[edit]
Etymology
[edit]
See tricentennial.
Noun
[edit]
tercentennial (pluraltercentennials)
The three-hundredth anniversary of an event; tricentennial. 2006 was the tercentennial of Benajmin Franklin's birth.
Related terms
[edit]
tercentenary
Retrieved from "
Categories:
English lemmas
English nouns
English countable nouns
English terms with usage examples
English terms suffixed with -ennial
Hidden categories:
Pages with entries
Pages with 1 entry
This page was last edited on 28 September 2024, at 04:32.
Definitions and other text are available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy.
Privacy policy
About Wiktionary
Disclaimers
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Search
Search
[x] Toggle the table of contents
tercentennial
5 languagesAdd topic |
7702 | https://economics.stackexchange.com/questions/13910/why-are-elasticities-defined-as-logarithmic-derivatives | Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Economics
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Why are elasticities defined as logarithmic derivatives?
Ask Question
Asked
Modified 8 years, 11 months ago
Viewed 26k times
6
$\begingroup$
In my economics class, we often compute the elasticity of $Y$ with respect to $X$, $$\eta = \frac{\partial \log Y}{\partial \log X}.$$ You can compute this from the slope of a line fit to a log-log plot.
Why is it more natural to consider this quantity than the much simpler quantity $$\eta' = \frac{\partial Y}{\partial X}$$ which is just as easy to measure? There seems to be some implicit assumption that $\eta$ is perhaps 'more applicable', maybe that it's less sensitive to the value of $X$ than $\eta'$ is, so it's more useful for extrapolation. For example, I've seen a demand curve extrapolated by assuming that $\eta$ is constant, yielding a power law. But why not assume $\eta'$ is constant, yielding a line? Neither seems particularly more natural to me.
Why is $\eta$ a more useful quantity than $\eta'$ in economics?
microeconomics
elasticity
Share
Improve this question
asked Oct 25, 2016 at 1:32
knzhouknzhou
21511 gold badge22 silver badges77 bronze badges
$\endgroup$
Add a comment |
4 Answers 4
Reset to default
13
$\begingroup$
If I understand your question, first the elasticity haven't units. The problem with $\partial Y/\partial X$ is that if you change the measure units the result is different. Is less problematic to express it in percentages: $$ El_X(Y)=\frac{\Delta Y/Y}{\Delta X/X}$$ if $\Delta Y \to 0$ when $\Delta X \to 0$ then you obtain $$El_X(Y)=\frac{dY}{dX}\frac{X}{Y}$$ if $Y=Y(X,W,...,Z)$ you change $d$ for $\partial$.
Second, like you know $d \log(x)=\frac{dx}{x}$ thus $$ El_X(Y)=\frac{d Y}{d X}\frac{X}{Y}=\frac{d\log(Y)}{d\log(X)}.$$
Share
Improve this answer
edited Oct 25, 2016 at 12:31
answered Oct 25, 2016 at 3:20
Luis SalazarLuis Salazar
17666 bronze badges
$\endgroup$
Add a comment |
3
$\begingroup$
First, elasticity measures the responsiveness of quantity demanded or quantity supplied when a change in price occurs.
These measurements are made in percentage change form.
From my perspective, the main reason you are computing elasticity using $log$ is because doing this puts your data in percentage terms. Given that elasticity is a ratio of percentage change in quantity demanded/supplied to the percentage change in price, this would be the most plausible explanation.
You will see using $log$ to transform data into percentage terms in Econometrics courses.
Update: I don't think it is more "natural" to phrase things as percentages. But percentages can give more information. They give relative information. For example, Suppose you have \$100 and I have \$10. A third person gives each of us \$1, now you have \$101 and I have \$11. In absolute terms, we both received a dollar, but in percentage terms, your cash grew by 1% and my cash grew by 10%.
I hope this helps!
Share
Improve this answer
edited Oct 25, 2016 at 17:08
RegressForward
3,4871313 silver badges3131 bronze badges
answered Oct 25, 2016 at 3:05
user10862user10862
3111 bronze badge
$\endgroup$
3
$\begingroup$ But then my question just becomes, why is it more natural to phrase things as percentages? $\endgroup$
knzhou
– knzhou
2016-10-25 03:06:16 +00:00
Commented Oct 25, 2016 at 3:06
$\begingroup$ pct.(%) change in Demand (Q) w.r.t. (with relation to) pct.(%) change in Price - is Point Elasticity of Demand... NB it icandiff in each point of Demand curve, aka with each step of differentiation $\endgroup$
JeeyCi
– JeeyCi
2024-10-07 06:10:17 +00:00
Commented Oct 7, 2024 at 6:10
$\begingroup$ pct. change helps to avoid the need of normalization of P & Q before modelling $\endgroup$
JeeyCi
– JeeyCi
2024-10-07 06:15:14 +00:00
Commented Oct 7, 2024 at 6:15
Add a comment |
1
$\begingroup$
One extremely function throughout economics is the Cobb-Douglas functional form (eg. $y = bx_1^a x_2^b$). This won't just show up in utility functions, but also production functions or growth functions (like in the Solow Swan model).
Taking the log-log form here has a few implications. First, it drops $a$ and $b$ from exponents to linear coefficients, which gives you a form $lny = b+alnx_1+blnx_2$. This is important, because you can run a linear regression on this functional form, but not the original one. You can see an example of that in this paper.
It also has applications in empirical microeconomics; you can test to see if a firm has increasing, decreasing or constant returns to scale if you find $a+b > or < 1$. This can have important implications; it can answer some questions like "should we break up these big companies?"
Second it makes the math nicer in some cases. This can actually be useful in some cases; in Maximum likelihood estimation having a more mathematically friendly functional form can save you hours of computation time finding maxima.
Share
Improve this answer
edited Oct 25, 2016 at 3:11
answered Oct 25, 2016 at 2:39
Matt RMatt R
12577 bronze badges
$\endgroup$
2
$\begingroup$ But then why is Cobb-Douglas realistic? There are many other reasonable functions. $\endgroup$
knzhou
– knzhou
2016-10-25 02:42:38 +00:00
Commented Oct 25, 2016 at 2:42
$\begingroup$ There are of course alternatives, but this functional form is mathematically convenient, easily extendable, and empirically supported. So it's generally a reasonable starting point in many applications. $\endgroup$
Matt R
– Matt R
2016-10-25 02:56:37 +00:00
Commented Oct 25, 2016 at 2:56
Add a comment |
1
$\begingroup$
First, the interpretation of $\eta=\frac{\partial\log y}{\partial\log x}$ and $\eta'=\frac{\partial y}{\partial x}$ are different. $\eta$ is the ratio of percent change and $\eta'$ is the ratio of absolute change. But you already know that. The real question is not why we define elasticity as a ratio of percent changes rather than absolute changes in economics, because that's how we use the word elasticity in everyday life: suppose rubber band A is 10 inches long and can be stretched by 1 inch when force F is applied, and rubber band B is 1 inch long and can be stretched by 0.5 inch when the same force F is applied. We would say rubber band B is more elastic than A, because we don't care about absolute change, but relative change when defining elasticity.
So I think your question is why is elasticity $\eta$ more applicable/useful than $\eta'$? My answer is that $\eta$ is not more useful/applicable/natural. Both $\eta$ and $\eta'$ are useful in different applications.
Take the price elasticity example. At unit price \$1, consumer A would buy 10 apples, and consumer B would buy 5 apples. At unit price \$2, consumer A would buy 6 apples, and consumer B would buy 1 apple. That is, when the price of apples increases by \$1, both consumers will reduce their purchase by 4 apples. For both consumers, $\eta'=\frac{\Delta\text{apples}}{\Delta\text{price}}=4$. This is a useful quantity if what we wanted to know is what happens to the sales of apples if price increased by \$1. But if we wanted to know which consumer would respond more dramatically to the price change, $\eta'$ is not a good measure. Because it seems consumer B is more sensitive to the price changes: she would cut her apple consumption by 80%, compared with only 40% reduction of consumer A. So $\eta$ is a better measure of this sensitivity or elasticity with respect to price changes.
In your demand curve extrapolation example, assuming constant elasticity $\eta$ is probably closer to truth than assuming constant $\eta'$. If you assume constant $\eta'$, the demand curve is a straight line. This effectively means price change from \$1 to \$2 will induce same change in quantity demanded as the price change from \$100 to \$101. But this is not supported by either evidence or common sense. Human brain does not seem to work this way. In this sense, relative changes do seem to be relevant in more economic applications than the absolute changes.
Share
Improve this answer
edited Oct 26, 2016 at 0:05
answered Oct 25, 2016 at 19:38
PaulPaul
48833 silver badges99 bronze badges
$\endgroup$
Add a comment |
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
microeconomics
elasticity
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
Estimating Price Elasticity from sales
Why does marginal cost (derivative of total cost) differ from variable cost at each level?
Why does quantity supplied increase with price in economics?
Mathematical analysis of an economic model - stuck in calculations
2 Question on uncertainity
Is there any formal definition of a relationship between resource availability and demand?
0 Why are constant price elasticities reported?
0 Is there a standard term for the elasticity of an isoquant?
3 Different elasticities of substitution
Hot Network Questions
What is this chess h4 sac known as?
Does the curvature engine's wake really last forever?
How long would it take for me to get all the items in Bongo Cat?
Checking model assumptions at cluster level vs global level?
Is it ok to place components "inside" the PCB
How to rsync a large file by comparing earlier versions on the sending end?
Find non-trivial improvement after submitting
Proof of every Highly Abundant Number greater than 3 is Even
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
Origin of Australian slang exclamation "struth" meaning greatly surprised
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
Analog story - nuclear bombs used to neutralize global warming
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Are there any alternatives to electricity that work/behave in a similar way?
What is the meaning of 率 in this report?
How many stars is possible to obtain in your savefile?
Is there a specific term to describe someone who is religious but does not necessarily believe everything that their religion teaches, and uses logic?
Clinical-tone story about Earth making people violent
My dissertation is wrong, but I already defended. How to remedy?
How can the problem of a warlock with two spell slots be solved?
How to home-make rubber feet stoppers for table legs?
How to locate a leak in an irrigation system?
Bypassing C64's PETSCII to screen code mapping
Riffle a list of binary functions into list of arguments to produce a result
more hot questions
Question feed |
7703 | https://www.shmoop.com/study-guides/yellow-wallpaper/the-narrator.html | The Narrator in The Yellow Wallpaper Character Analysis | Shmoop
We have changed our privacy policy. In addition, we use cookies on our website for various purposes. By continuing on our website, you consent to our use of cookies. You can learn about our practices by reading our privacy policy.
See PlansLogin
More on The Yellow Wallpaper
Topics
Character Roles (Protagonist, Antagonist...)
Tools of Characterization
Intro
See All
### Summary
See All
Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8
Part 9
Part 10
Part 11
Part 12
Themes
See All
Freedom and Confinement
Madness
Gender
Literature and Writing
Society and Class
Quotes
See All
Freedom and Confinement
Madness
Gender
Literature and Writing
Society and Class
Characters
See All
The Narrator
The Narrator Timeline
The Narrator Quotes
John
The Woman in the Wallpaper
Mary
Jennie
Analysis
See All
Symbolism, Imagery, Allegory
Setting
Narrator Point of View
Genre
Tone
Writing Style
What’s Up With the Title?
What’s Up With the Ending?
Plot Analysis
Three Act Plot Analysis
Trivia
Steaminess Rating
Allusions
Questions
See All
### Quizzes
See All
### Best of the Web
See All
### Teaching
See All
### Lit Glossary
See All
### Table of Contents
See All
The Narrator
Back
More
Character Analysis
Introducing… Jane Doe
The first line of “The Yellow Wallpaper” does double duty, introducing both the setting of the story—a home for someone else’s ancestors—and the story’s narrator:
It is very seldom that mere ordinary people like John and myself secure ancestral homes for the summer. (3)
We know that the narrator is 1) a woman (because she is married to a man, and this is 1890), 2) probably middle class (“mere ordinary people”), and 3) has a husband named John. All of this is actually pretty aggressively anonymous: the narrator has no name and is married to a guy who might as well have no name, since “John” doesn’t really give us any clues about who he is or where he might be from.
The fact that these three traits—her gender, her class, and her marriage—all make it into the first sentence (while her name doesn’t) suggests that these general characteristics may be more important to the unfolding of the plot than her actual identity or personal history (about which we learn very little).
Why are these traits so important? Because they provide the context central conflicts that drive the story. The narrator is a woman of sensitive temperament, and she is also a writer. She has been ill, and her illness has placed her in a weak position in relation to domineering John.
As her husband and as her physician (a situation we think the American Medical Association would find problematic), John makes all of the narrator’s decisions for her, which really irritates her:
If a physician of high standing, and one’s own husband, assures friends and relatives that there is really nothing the matter with one but temporary nervous depression—a slight hysterical tendency—what is one to do? (3)
So, “one’s own husband” is badmouthing the narrator to her friends and refusing to take her seriously. That might have us crawling up (or maybe into?) the walls, too.
And what about class? Well, our Jane Doe apparently is of a social position that means she doesn’t have to work. She may be middle class enough that staying at an ancestral home is a new thing for her, but she’s still definitely on the upper end of the social spectrum. Maybe this lack of labor would be lucky for her if she was allowed to do anything else, but her illness has restricted her activities pretty much entirely.
Bored Out of Her Gourd
John has prescribed absolute rest—he won’t even let her look after her baby. And while his intentions may be good, he’s driving the narrator nuts from boredom. How do we know this? She tells us so within the first page:
I did write for a while in spite of [John telling me not too]; but it does exhaust me a good deal—having to be so sly about it, or else meet with heavy opposition. (3-4)
The writing that she actively enjoys has been forbidden, so she has to do it in secret, leaving her tired and wrung out.
There’s a binary between John (professional man) and our narrator (unemployed woman); there’s also one between our narrator (lady of leisure) and her sister-in-law, Jennie (who takes care of the housework and the baby). Jennie may not have much power in the household, but she does have one thing that the narrator envies: an occupation.
There comes John's sister. Such a dear girl as she is, and so careful of me! I must not let her find me writing. She is a perfect and enthusiastic housekeeper, and hopes for no better profession. I verily believe she thinks it is the writing which made me sick! (8)
Jennie’s housekeeping gives her an authority over the narrator (who is, after all, doing zilch for anyone) that makes the narrator feel self-conscious and suspicious of Jennie's motives. As the story goes on, the narrator grows to resent John and Jennie more and more. With nothing to do but brood over how she has been wronged—and how incredibly ugly her wallpaper is—the narrator’s paranoia has a field day.
Something else we can’t help but notice in these previous passages: neither John nor Jennie thinks much of the activity of writing. They identify it as another symptom of the narrator’s nervousness, her emotional nature. But the narrator rebels against them without much fanfare right from the beginning, simply by stealing time to write “The Yellow Wallpaper.”
The narrator’s mental health hinges not only on whether she has work to do, but what kind of work it is. She wants to write and isn’t allowed, something that “does exhaust her a good deal” (3). The subtle undermining of her confidence as a writer doesn’t exactly help to repair the damaged relationships she shares with her husband and her sister-in-law, sending her further into a frenzy of paranoia that leads to her mounting obsession with the design of the paper on her bedroom wall.
So – Just What Is Wrong With Her?
Well, “The Yellow Wallpaper” remains as ambiguous and unclear about the narrator’s illness as it does about her identity, so it’s tough to say what, exactly, is wrong with her. Still, we know mental illness is going to be an issue right from the first page because, once again, the narrator lets us know explicitly:
_You see [John, the narrator’s husband] does not believe I am sick! And what can one do? […]
So I take phosphates or phosphites—whichever it is, and tonics, and journeys, and air, and exercise, and am absolutely forbidden to "work" until I am well again.
Personally, I disagree with [John and her physician brother’s] ideas.
Personally, I believe that congenial work, with excitement and change, would do me good.
But what is one to do? (3)_
We can get a lot about the character from this passage: first, she’s pretty alienated from her own treatment (“phosphates or phosphites—whichever it is”? In our experience, anyone who really believes in the medicine she's taking isn’t going to forget what it’s called). Also, the antagonism between the narrator and pretty much everyone around her becomes apparent when she’s all, “Personally, I disagree!”
Second, the narrator feels ill, but her ailment doesn’t manifest physically. This is definitely an ongoing tension (as though their marriage needed more baggage) between the narrator and her physician husband, who “scoffs openly at any talk of things not to be felt and seen and put down in figures” (3).
Whenever the narrator tells John that she feels worse and worse, all he replies is that her body is getting better and better (though, why it was weak in the first place, we don’t know. Perhaps her baby had a difficult birth?). He refuses to acknowledge that her mind could be sick even though her body is healthy.
We know from the outset that the narrator resents her husband’s treatment and continues to feel unsettled even as she recovers physically. So, she’s not starting the story in a great place mentally; the yellow wallpaper isn’t so ugly that it can drive totally sane people out of their minds. But—our narrator isn't totally sane.
The story does a great job of suggesting the claustrophobic conditions that make her condition worse: note that the bed in her terrible attic room is nailed down, underlining how trapped the narrator is. She’s not allowed to do anything that might “upset” her—in other words, that might give her release for her emotions: she’s not allowed to write, she’s not allowed to work, and she’s not allowed to travel. (Is anyone else reminded of Nurse Ratched’s treatments in One Flew Over the Cuckoo’s Nest? And look how that turned out.)
The only thing that the narrator has left to do is to speculate about the ugly, irregular wallpaper of the attic room. As she loses stability, the wallpaper’s importance to her as the one puzzle she has to occupy herself with becomes greater and greater.
In fact, as she grows more certain that she gets the wallpaper as no one else does, the people she knows become correspondingly less understandable. Without any outlet beyond the wallpaper, the narrator’s anger builds, leading her to increasingly paranoid speculations about John and Jennie:
The fact is I am getting a little afraid of John. He seems very queer sometimes, and even Jennie has an inexplicable look.(13)
As she slowly aligns herself with the wallpaper and distances herself from John and Jennie, the narrator starts to recognize another woman creeping behind the pattern of the paper – and it’s then that everything really hits the fan. (For more on the creeping woman and the narrator’s final breakdown, see the Woman in the Wallpaper Character Analysis.)
A Society Lady
So, let’s talk for a second about this building obsession that drives our narrator to “get to work” (17) peeling away at the walls imprisoning a woman she sees behind the yellow wallpaper’s maddening pattern.
There is a moment, right at the end when the narrator slips into the persona of this creeping paper woman:
_I don't like to look out of the windows even—there are so many of those creeping women, and they creep so fast.
I wonder if they all come out of that wallpaper as I did?
But I am securely fastened now by my well-hidden rope—you don't get me out in the road there!
I suppose I shall have to get back behind the pattern when it comes night, and that is hard!
It is so pleasant to be out in this great room and creep around as I please! (18)_
Do you notice anything weird about this passage (besides the obvious)? The narrator now believes that she is a woman recently freed from behind the bars of an ugly wallpaper pattern. But she’s still talking (despite the exclamation points) like a well-bred society lady. Where’s the cursing? How is she still making any kind of sense?
In a lot of ways, “The Yellow Wallpaper” reads like a real Gothic novel – oppressive husband, unfair social restrictions, a crazy lady—but the narrator, even in the depths of her breakdown, still writes like Edith Wharton.
Compare this to, say, that famous madwoman in the attic, Bertha in Charlotte Bronte’s Jane Eyre, whose insanity makes her unable to speak. The narrator of this story, by contrast, is not only still totally verbally skilled, but remember—she’s writing about her own breakdown. It’s like the narrator has been so indoctrinated by social codes of the day that, even in the midst of a complete mental collapse, she still remembers to keep her journal and to be polite, both to herself and to poor, fainting John.
There’s a logical progression to this story that doesn’t surrender to the total mental breakdown of its heroine. This tips us off that “The Yellow Wallpaper,” again, may be less interested in the particulars of its protagonist’s mental state and more interested in protesting larger social issues like unjust treatment of the mentally ill and of women.
A Narrator We Can Believe In?
It may not seem like an obvious choice to end our character analysis by asking whether the person in question is, in fact, a character, but in the case of the nameless narrator of “The Yellow Wallpaper" it’s worth thinking about. Because the narrator isn’t just a character, she’s also a plot device: for all of her apparent passivity, she really dominates the story, carefully sculpting not only what we know about her, but also what we think of her.
Lots of novels (check out Shmoop on Moby Dick or The Great Gatsby for examples) use the limited perspective of one marginal character to narrate (and make sense of) the (usually more exciting) people around him. It’s a two-for-one deal. Not only do we, the readers, see the crazy shenanigans of a charismatic central character, like Moby Dick’s Captain Ahab or The Great Gatsby’s Jay Gatsby, but we also get one peripheral character (Ishmael in Moby Dick, Nick in Gatsby) telling us what he thinks of said shenanigans.
“The Yellow Wallpaper,” though, offers pretty much no outside perspective on the actions of the story’s heroine. It’s like if Bob Costas, instead of sitting at NBC commenting on, say, gymnast Shawn Johnson’s performance at a competition, suddenly decided to put on a red leotard and jump on the balance beam himself.
And after balancing his heart out, he wouldn’t stop there . Then he’d move to the judge’s bench and decide his own scores. Like imaginary Gymnast Bob, the narrator of “The Yellow Wallpaper” isn’t just the story’s main actor; she also provides all the background information and analysis made available to the reader.
Consolidating the narrator and the central character into one person is a really high-stakes way to tell a story. Why? Well, to return to Gymnast Bob, would you trust the opinion of someone who evaluated his own performance, without anyone to support his claims? The rewards are just as great as the risks, though: we tend to believe first person narrators (because, after all, we have nothing to go on but what they give us), and they have a lot of power over the reader’s interpretation of events.
Let’s take a look at some quotes from the narrator about her husband John to see the subtle way she shapes the reader’s feelings about the people in her life.
John laughs at me, of course, but one expects that in a marriage.(3)
I sometimes fancy that in my condition if I had less opposition and more society and stimulus—but John says the very worst thing I can do is to think about my condition, and I confess it always makes me feel bad.(4)
_Dear John! He loves me very dearly, and hates to have me sick. I tried to have a real earnest reasonable talk with him the other day, and tell him how I wish he would let me go and make a visit to Cousin Henry and Julia.
But he said I wasn’t able to go, nor able to stand it after I got there; and I did not make out a very good case for myself, for I was crying before I had finished._(10)
We know, from these passages, that 1) the narrator and John have frequent disagreements over her condition, 2) John doesn’t take her seriously, and 3) their arguments make the narrator cry.
The narrator’s powerlessness, emotional distress, and pleas to John that she needs to get away—all of these make us sympathize with her, as does her effort to convince herself (and the reader) that she still loves the man in spite of his treatment of her. (“Dear John”? Come on.)
John comes across as a big ol’ jerk for not believing his wife when she tells him she’s still feeling bad. And that’s the genius of the narration of “The Yellow Wallpaper”: it seems natural, even inevitable, that we believe John’s a horrible person.
But who’s telling us this information? His discontented, desperately unhappy wife. And we’re not saying you shouldn’t believe her—John seems like an utter tool—but we do want to point out that it is also strategic, in a story challenging contemporary mental health treatment, to make the ailing narrator a total victim and the husband/doctor kind of a monster.
The Narrator Timeline
Back
More
Cite This Page
Tired of ads?
Join today and never see them again.
Get Started
×Close
Cite This Source
Close
Site Map
Help
About Us
Jobs
Partners
Affiliates
Colleges
Terms of Use
Privacy
© 2025 Shmoop University. All rights reserved. We speak student®
Instagram
Facebook
Twitter
Linkedin
Logging out…
Logging out...
You've been inactive for a while, logging you out in a few seconds...
I'm Still Here!
W hy's T his F unny?
CLOSE
Study Guides
Literature
Poetry
Movies
Bible
Mythology
Historical Texts
Music
Lit Movements
History
Math
Biology
Biography
Students
Study Guides
Shakespeare
Quotes
Videos
Grammar
Teachers
Teachers Home
Test Prep
Courses
Subscription
Teaching Guides
Teaching Blog
Parents
Parents Home
Homeschool
College Resources
Districts
Districts Home
Heartbeat
Test Prep
Intervention
Courses
Pro Services
Schedule Demo
Case Studies
Shmoop
About Us
Careers
Awards
News
Advertising
Contact
FAQs
BlogSupport |
7704 | https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/2%3A_Logic/2.8%3A_Multiple_Quantiers | 2.8: Multiple Quantifiers - Mathematics LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
2: Logic
MTH 220 Discrete Math
{ }
{ "2.1:_Propositions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.2:_Conjunctions_and_Disjunctions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.3:_Implications" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.4:_Biconditional_Statements" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.5:_Logical_Equivalences" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.6_Arguments_and_Rules_of_Inference" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.7:_Quantiers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.8:_Multiple_Quantiers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1:_Introduction_to_Discrete_Mathematics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2:_Logic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3:_Proof_Techniques" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4:_Sets" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5:_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6:_Relations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7:_Combinatorics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8:_Big_O" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Appendices : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Wed, 05 Feb 2020 20:15:48 GMT
2.8: Multiple Quantifiers
24120
24120
Judy P Dean
{ }
Anonymous
Anonymous
2
false
false
[ "article:topic", "authorname:hkwong", "license:ccbyncsa", "showtoc:yes" ]
[ "article:topic", "authorname:hkwong", "license:ccbyncsa", "showtoc:yes" ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Expand/collapse global hierarchy
1. Home
2. Campus Bookshelves
3. Monroe Community College
4. MTH 220 Discrete Math
5. 2: Logic
6. 2.8: Multiple Quantifiers
Expand/collapse global location
2.8: Multiple Quantifiers
Last updated Feb 5, 2020
Save as PDF
2.7: Quantifiers
3: Proof Techniques
Page ID 24120
Harris Kwong
State University of New York at Fredonia via OpenSUNY
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Multiple Quantifiers
1. Negation with Multiple Quantifiers
Summary and Review
Exercises \PageIndex{}
Multiple Quantifiers
Multiple quantifiers can be used. With more than one quantifier, the order makes a difference.
Example 2.8.1
When multiple quantifiers are present, the order in which they appear is important. Determine whether these two statements are true or false.
∀x∈Z∃y∈R∗(xy<1)
∃y∈R∗∀x∈Z(xy<1)
Here, R∗ denotes the set of all nonzero real numbers.
Answer
1. To prove that the statement is true, we need to show that no matter what integer x we start with, we can always find a nonzero real number y such that xy<1. For x≤0, we can pick y=1, which makes xy=x≤0<1. For x>0, let y=1 x+1, then xy=x x+1<1. This concludes the proof that the first statement is true.
2. Let y=1. Can we find an integer x such that xy≥1? Definitely! For example, we can set x=2. This counterexample shows that the second statement is false. NOTE: the statement is false, but this is not a valid explanation. Do you see why? What do you need to show an existence statement is false?
hands-on Exercise 2.8.1
True or false: ∃y∈R∀x∈Z(xy<1)?
Example 2.8.2
Many theorems in mathematics can be expressed as quantified statements. Consider
“If x is rational and y is irrational, then x+y is irrational.”
This is same as saying
“Whenever x is rational and y is irrational, then x+y is irrational.”
The keyword “whenever” suggests that we should use a universal quantifier. (2.8.1)∀x,y(x is rational∧y is irrational⇒x+y is irrational). It can also be written as (2.8.2)∀x∈Q∀y∉Q(x+y is irrational). Although this form looks complicated and seems difficult to understand (primarily because it is quite symbolic, hence appears to be abstract and incomprehensible to many students), it provides an easy form for negation. See the discussion below.
The fact that an implication can be expressed as a universally quantified statement sounds familiar.
Negation with Multiple Quantifiers
We shall learn several basic proof techniques in Chapter 3. Some of them require negating a logical statement. Since many mathematical results are stated as quantified statements, it is necessary for us to learn how to negate a quantification. The rule is rather simple. Interchange ∀ and ∃, and negate the statement that is being quantified. In other words,
(2.8.3)∀xp(x)―≡∃x p(x)―,and∃xp(x)―≡∀x p(x)―.
If we have ∀x∈Z, we only change it to ∃x∈Z when we take negation. It should not be negated as ∃x∉Z. The reason is: we are only negating the quantification, not the membership of x. In symbols, we write
(2.8.4)∀x∈Zp(x)―≡∃x∈Zp(x)―.
The negation of “∃x∈Zp(x)” is obtained in a similar manner.
Example 2.8.11
We find (2.8.5)∀x∈Z∃y∈R∗(xy<1)―≡∃x∈Z∀y∈R∗(xy≥1), and (2.8.6)∃y∈R∗∀x∈Z(xy<1)―≡∀y∈R∗∃x∈Z(xy≥1). Remember that we do not change the membership of x and y.
Example 2.8.12
The statement
“All real numbers x satisfy x 2≥0”
can be written as, symbolically, ∀x∈R(x 2≥0). Its negation is ∃x∈R(x 2<0). In words, it says “There exists a real number x that satisfies x 2<0.”
Summary and Review
Symbolically, here's how to negate statements with quantifiers:∀xp(x)―≡∃x p(x)―,and∃xp(x)―≡∀x p(x)―.
In general, “∀x∃yp(x,y)” is NOT the same as “∃y∀xp(x,y)”, so order makes a difference.
Exercises \PageIndex{}
Exercise 2.8.1
Determine whether these statements are true or false:
There is an integer m such that both m/2 is an integer and, for every integer k, m/(2k) is not an integer.
For every integer n, there exists an integer m such that m>n 2.
There exists a real number x such that for every real number y, xy=0.
Answer
(a) false (b) true (c) true
Exercise 2.8.2
Negate the following statements:
For all real numbers x, there exists an integer y such that p(x,y) implies q(x,y).
There exists a rational number x such that for all integers y, either p(x,y) or r(x,y) is true.
For all integers x, there exists an integer y such that if p(x,y) is true, then there exists an integer z so that q(x,y,z) is true.
Exercise 2.8.3
Find the negation (in simplest form) of each symbolic statement.
∀x<0∧x∈R∀y,z∈R(yxz)
∀x∈Z[p(x)∨q(x)]
∀x,y∈R[p(x,y)⇒q(x,y)]
Answer
(a) ∃x<0∧x∈R∃y,z∈R(y<z∧xy≤xz)
(b) ∃x∈Z[p(x)―∧q(x)―]
(c) ∃x,y∈R[p(x,y)∧q(x,y)―]
Exercise 2.8.4
For this statement, (i) represent it in symbolic form, (ii) find the symbolic negation (in simplest form), and (iii) express the negation in words.
For all real numbers x and y there exists an integer z such that 2z=x+y.
Exercise 2.8.5
For each statement, (i) represent it in symbolic form, (ii) find the symbolic negation (in simplest form), and (iii) express the negation in words.
For all real numbers x and y, x+y=y+x.
For every positive real number x there exists a real number y such that y 2=x.
There exists a real number y such that, for every integer x, 2x 2+1>x 2y.
Answer
(a)
∀x,y∈R(x+y=y+x)
∃x,y∈R(x+y≠y+x)
There exist real numbers x and y such that x+y≠y+x.
(b)
∀x∈R+∃y∈R(y 2=x)
∃x∈R+∀y∈R(y 2≠x)
There exists a positive real number x such that for all real numbers y, y 2≠x.
(c)
∃y∈R∀x∈Z(2x 2+1>x 2y)
∀y∈R∃x∈Z(2x 2+1≤x 2y)
For every real number y, there exists an integer x such that 2x 2+1≤x 2y.
Some students may not be familiar with matrices. A matrix is rectangular array of numbers. Matrices are important tools in mathematics. The product of two matrices of appropriate sizes is defined in a rather unusual way. It is the peculiar way that two matrices are multiplied that makes matrices so useful in mathematics. The square of a matrix is of course the product of the matrix with itself. It is well-defined only when the matrix is a square matrix. As it turns out, the order of multiplication of two matrices is important. In other words, given any two matrices A and B, it is not always true that AB=BA.↩
This page titled 2.8: Multiple Quantifiers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .
Back to top
2.7: Quantifiers
3: Proof Techniques
Was this article helpful?
Yes
No
Recommended articles
2.1: Propositions
2.2: Conjunctions and Disjunctions
2.3: Implications
2.4: Biconditional Statements
2.5: Logical Equivalences
Article typeSection or PageAuthorHarris KwongLicenseCC BY-NC-SAShow Page TOCyes
Tags
This page has no tags.
© Copyright 2025 Mathematics LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status× |
7705 | https://www.youtube.com/watch?v=4X0SGGrXDiI | 2. First-Order Equations
MIT OpenCourseWare
5940000 subscribers
1915 likes
Description
199307 views
Posted: 15 Jan 2015
MIT 2.087 Engineering Mathematics: Linear Algebra and ODEs, Fall 2014
View the complete course:
Instructor: Gilbert Strang
In this lecture, Prof. Strang explains first-order equations.
License: Creative Commons BY-NC-SA
More information at
More courses at
81 comments
Transcript:
The following
content is provided under a Creative
Commons license. Your support will help MIT
OpenCourseWare continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Well,
OK, Professor Frey invited me to give the
two lectures this week on first order equations, like
that one, first order dy dt. And the lectures next week will
be on second order equation. So we're looking for, you could
say, formulas for the solution. We'll get as far as we
can with formulas, then numerical methods. Graphical methods take over
in more complicated problems. This is a model problem. It's linear. I chose it to have
constant coefficient a, and let me check the units. Always good to see the
units in a problem. So let me think of this
y, as the money in a bank, or bank balance, so y as in
dollars, and t, time, in years. So we're looking at the ups
and downs of bank balance y. The rate of change, so the
units then are dollars per year. So every term in the equation
has to have the right units. So y is in dollars,
so the interest rate a is percent per
year, say 6% a year. So a could be 6%-- that's
dimensionless-- per year, or half a percent per
month if we change. So if we change
units, the constant a would change from 6 to a half. But let's stay with 6. And then q of t represents
deposits and withdrawals, so that's in dollars
per year again. Has to be. So that's continuous. We think of the deposits
and the interest as being computed continuously
as time goes forward. So if that's a constant-- and
I'll take that case first, q equal 1-- that would
mean that we're putting in, depositing $1 per year,
continuously through the year. So that's the model that comes
from a differential equation. A difference equation would
give us finite time steps. So I'm looking for the solution. And with constant
coefficients, linear, we're going to get a formula
for the solution. I could actually deal
with variable interest rate for this one
first order equation, but the formula becomes messy. But you can still do it. After that point, for a
second order equations like oscillation,
or for a system of several equations
coupled together, constant coefficients is
where you can get formulas. So let's go with that case. So how to solve that equation? Let me take first of all, a
constant, constant source. So I think of q as
the source term. To get one nice formula, let me
take this example, ay plus 1, let's say. How do you find y
of t to solve that? And you start with some
initial condition y of 0. That's the opening deposit
that you make at time 0. How to solve that equation? Well, we're looking
for a solution. And solutions to linear
equations have two parts. So the same will happen
in linear algebra. One part is a solution
to that equation, so we're just looking
for one, any one, and we'll call it a
particular solution. And the associated null
equation, dy dt equal ay. So this is an equation
with q equals 0. That's why it's called null. And it's also
called homogeneous. So more textbooks use that
long word homogeneous, but I use the word null
because it's shorter and because it's the same
word in linear algebra. So let me call yn
the null solution, the general null solution. And y, I'm looking here for
a particular solution yp, and I'm going to-- here's
the key for linear equations. Let me take that off and
focus on those two equations. How does solving the null
equation, which is easy to do, help me? Why can I, as I plan
to do, add in yn to yp? I just add the two equations. Can I just add
those two equations? I get the derivative of yp
plus yn on the left side. And I have a times yp plus yn. And that is a critical moment
there when we use linearity. I had a yp a yn, and I
could put them together. If it was y squared, yp
squared and yn squared would not be the same
as yp plus yn squared. It's the linearity that
comes, and then I add the 1. So what do I see from this? I see that yp plus yn
also solves my equation. So the whole family of
solutions is 1 yp plus any yn. And why do I say any yn? Because when I find
one, I find more. The solutions to
this equation are yn could be e to the at,
because the derivative of e to the at does bring
down a factor a. But you see, I've left space
for any multiple of e to the at. This is where that long
word homogeneous comes from. It's homogeneous means I can
multiply by any constant, and I still solve the equation. And of course, the
key again is linear. So now I have-- well, you could
say I've done half the job. I've found yn, the general yn. And now I just have
to find one yp, one solution to the equation. And with this source
term, a constant, there's a nice way to
find that solution. Look for a constant solution. So certain right
hand sides, and those are the like the
special functions for the special source terms
for differential equation, certain right hand
sides-- and I'm just going to go down a
list of them today. The next one on the
list-- can I tell you what the next one
on the list will be? y prime equal ay. I use prime for--
well, I'll write dy dt, but often I'll write y prime. dy dt equal ay plus
an exponential. That'll be number two. So I'm just preparing
the way for number two. Well, actually number
one, this example is the same as that exponential
example with exponent s equal 0, right? If s is 0, then I
have a constant. So this is a special
case of that one. This is the most
important source term in the whole subject. But here we go
with a constant 1. So we've got yn. And what's yp? I just looked to see. Can I think of one? And with these
special functions, you can often find a
solution of the same form as the source term. And in this case,
that means a constant. So if yp is a constant,
this will be 0. So I just want to
pick the constant that makes this thing 0. And of course, their right hand
side is 0 when yp is minus 1 over a. So I've got it. We've solved that
equation, except we didn't match the
initial condition yet. Let me if you take
that final step. So the general y is any
multiple, any null solution, plus any one particular
solution, that one. And we want to match it
to y of 0 at t equals 0. So I want to take that solution. I want to find that
constant, here. That's the only remaining
step is find that constant. You've done it in the homework. So at t equals 0,
y of 0 is-- at t equals 0, this is C. This
is the minus 1 over a. So I learn what the C has to be. And that's the final step. C is bring the 1 over
a onto that side, so C will be y of 0 e to the
at minus 1 over a e to at. And here we had
a minus 1 over a. Well, it'll be plus
1 over a e to the at. So now I've just put in the C,
y of 0 plus 1 over a. y of 0 plus 1 over a has gone
in for C. And now I have to subtract this 1 over a. Here, I see a 1 over a,
so I can do it neatly. 11 minutes, 43 seconds Got a solution. We can check it, of course. At t equals 0, this
disappears, and this is y of 0. And it has the form. It's a multiple of e to the
at and a particular solution. So that's a good one. Notice that to get the
initial condition right, I couldn't take C to be y of
0 to get the initial condition right. To get the initial
condition right, I had to get that, this
minus 1 over a in there. Good for that one? Let me move to the
next one, exponentials. So again, we know that the
null equation with no source has this solution e to the at. And I'm going to suppose
that the a in e to the at in the null solution
is different from the s in the source function,
which will come up in the particular solution. So you're going to
see either the st in the particular
solution and an e to the at in the null solution. And in the case when s equals
a, that's called resonance, the two exponents are the
same, and the formula changes a little. Let's leave that case for later. How do I solve this? I'm looking for a
particular solution because I know the
null solutions. How am I going to get
a particular solution of this equation? Fundamental observation,
the key point is it's going to be a
multiple of e to the st. If an exponential goes in, then
that will be an exponential. Its derivative will
be an exponential. I'll have e to the
st's everywhere. And I can get the number right. So I'm looking for y try. So I'll put try, knowing it's
going to work, as some number times e to the st.
So this would be like the exponential response. Response, do you know
that word response? So response is the solution. The input is q, and
the response is Y. And here, the input
is e to the st, and the response is a multiple
of e to the st. So plug it in. The timed derivative
will be Y. Taking the derivative will bring down
a 1. e to the st equals aY. A aY e to the st plus 1 e to
the st. Just what we hoped. The beauty of
exponentials is that when you take their derivatives,
you just have more exponential. That's the key thing. That's why exponential is
the most important function in this course, absolutely
the most important function. So it happened here. I can cancel e to the
st, because every term has one of them. So I'm seeing that--
what am I getting for Y? Getting a very important
number for Y. So I bring aY onto
this side with sY. On this side I just have a 1. Maybe it's worth putting
on its own board. Y is, so Ys aY comes with a
minus, and the 1, 1 over-- so Y was multiplied
by s minus a. That's the right quantity to
get a particular solution. And that 1 over s
minus a, you see why I wanted s to
be different from a. I If s equaled a in that
case, in that possibility of resonance when the two
exponents are the same, we would have 1 over 0, and we'd
have to look somewhere else. The name for that-- this has to
have a name because it shows up all the time. The exponential
response function, you could call it that. Most people would call
it the transfer function. So any constant coefficient
linear equation's going to have a transfer
function, easy to find. Everything easy, that's
what I'm emphasizing, here. Everything's straightforward. That transfer function
tells you what multiplies the exponential. So the source was here. And the response is here, the
response factor, you could say, the transfer function. Multiply by 1 over s minus a. So if s is close to a,
if the input is almost at the same exponent as the
natural, as the null solution, then we're going to
get a big response. So that's good. For a constant coefficient
problem second order, other problems we can find
that response function. It's the key function. It's the function
if we have, or if we were to look at Laplace
transforms, that would be the key. When you take
Laplace transforms, the transfer function shows up. Then when you take inverse
Laplace transforms, you have to find what function
has that Laplace transform. So did we get the-- we
got the final answer then. Let me put it here. y is e
to the st times this factor. So I divide by s minus a. A nice solution. Let me also anticipate
something more. An important case for e to
the st is e to the i omega t. e to the st, we think about
as exponential growth, exponential decay. But that's for positive
s and negative s. And all important in
applications is oscillation. So coming, let me say,
coming is either late today or early Wednesday will
be s equal i omega, so where the source term
is e to the i omega t. And alternating, so this
is electrical engineers would meet it constantly from
alternating voltage source, alternating current source,
AC, with frequency omega, 60 cycles per
second, for example. Why don't I just
deal with this now? Because it involves
complex numbers. And we've got to take a little
step back and prepare for that. But when we do it,
we'll get not only e to the i omega t,
which I brought out, but also, it's real part. You remember the great
formula with complex numbers, Euler's formula,
that e to the i omega t is a combination of cosine
omega t, the real part, and then the imaginary
part is sine omega t. So this is looking
like a complex problem. But it actually solves two
real problems, cosine and sine. Cosine and sine will be on our
short list of great functions that we can deal with. But to deal with them neatly,
we need a little thought about complex numbers. So OK if I leave
e to the i omega t for the end of the list, here? So I'm ready for another
one, another source term. And I'm going to pick
the step function. So the next example is going to
be dy dt equals ay plus a step. Well, suppose I
put H of t there. Suppose I put H of t. And I ask you for the
solution to that guy. So that step function,
its graph is here. It's 0 for negative time,
and it's 1 for positive time. So we've already solved
that problem, right? Where did I solve this equation? This equation is
already on that board. Because why? Because H of t is
for t positive. That's the only
place we're looking. This whole problem, we're
not looking at negative t. We're only looking
at t from 0 forward. And what is H of
t from 0 forward? It's 1. It's a constant. So that problem, as it stands,
is identical to that problem. Same thing, we have a 1. No need to solve that again. The real example is when
this function jumps up at some later time T. Now I have
the function is H of t minus T. Do you see that, why the step
function that jumps at time T has that formula? Because for little
t before that time, in here, this is--
what's the deal? If little t is
smaller than big T, then t minus T is
negative, right? If t is in here, then
t minus capital T is going to be a
negative number. And H of a negative number is 0. But for t greater than capital
T, this is a positive number. And H of a positive number is 1. Do you see how if you
want to shift a graph, if you want the graph
to shift, if you want to move the starting time,
then algebraically, the way you do it is to change t to
t minus the starting time. And that's what I want to do. So physically, what's
happening with this equation? So it starts with
y of 0 as before. Let's think of a bank balance
and then other things, too. If it's a bank balance, we put
in a certain amount, y of 0. We hope. And that grew. And then starting
at time, capital T, this switch turns on. Actually, physically,
step function is really often describing a
switch that's turned on, now. This source term act
begins to act at that time. And it acts at 1. So at time capital T we start
putting money into our account. Or taking it out, of course. If this with a minus sign,
I'd be putting money in. Sorry, I would start with
some money in, y of 0. I would start with money in. Yeah, actually, tell
me what's the solution to this equation that
starts from y of 0? What's the solution up until
the switch is turned on? What's the solution before
this switch happens, this solution while
this is still 0? So let's put that part
of the answer down. This is for t smaller
than T. What's the answer? 26 minutes, 17 seconds This is all common sense. It's coming fast, so I'm
asking these questions. And when I asked that question,
it's a sort of indication that you can really
see the answer. You don't need to go back
to the textbook for that. What have we got here? Yeah? AUDIENCE: Is it the null
solution [INAUDIBLE]? PROFESSOR: It'll be this guy. Yeah, the particular
solution will be 0. Right, the particular solution
is 0 before this is on. I'm sorry, the
null solution is 0, and the particular solution,
well, the particular solution is a guy that starts right. I don't know. Those names were not important. And then the question
is-- so it's just our initial deposit growing. Now, all I ask, what
about after time T? What about after time T? For t after time T, and
hopefully, equal time T, what do you think
y of t will be? Again, we want to
separate in our minds the stuff that's starting
from the initial condition from the stuff that's piling
up because of the source. So one part will be that guy. I haven't given the
complete answer. But this is continuing to grow. And because it's linear, we're
always using this neat fact that our equation is linear. We can watch things
separately, and then just add them together. So I plan to add
this part, which comes from initial
condition to a part that-- maybe we can guess it--
that's coming from the source. And how do we have any
chance to guess it? Only because that particular
source, once it's turned on, jumps to a constant
1, and we've solved the equation for a constant 1. Let me go back here. I think our answer
to this question-- so this is like just first
practice with a step function, to get the hang of
a step function. So I'm seeing this same y of
0 e to the at in every case, because that's what happens
to the initial deposit. I'll say grow, assuming
the bank's paying a positive interest rate. And now, where did
this term comes from? What did that term represent? AUDIENCE: The money
that [INAUDIBLE]. PROFESSOR: The money that, yeah? AUDIENCE: They had
each of [INAUDIBLE]. PROFESSOR: The money
that came in and grew. It came in, and then
it grew by itself, grew separately from
that these guys. So the initial condition
is growing along. And the money we put
in starts growing. Now, the point is what? That over here, it's going
to look just like that. So I'm going to have a 1 over a. And I'm going to have
something like that. But can you just guess
what's going to go in there? When I write it down,
it'll make sense. So this term is representing
what we have at time little t, later on, from the
deposits we made, not the initial one, but
the source, the continuing deposits. And let me write it. It's going to be a 1 over a
e to the a something minus 1. It's going to look
just like that guy. When I say that guy, let
me point to it again-- e to the at minus 1. But it's not quite
e to the at minus 1. What is it? AUDIENCE: t minus [INAUDIBLE]. PROFESSOR: t minus
capital T, because it didn't start until that time. So I'm going to leave that as,
like, reasonable, sensible. Think about a step function
that's turned on a capital time T. Then it grows from that time. Of course, mentally, I
never write down a formula like that without
checking at t equal to T, because that's the one important
point, at t equal capital T. What is this at
t equal capital T? It's 0. At t equal capital T, this is
e to the 0, which is 1 minus 1 altogether 0. And is that the right answer? At t equal capital T is 0,
should I have nothing here? Yes? No? Give me a head shake. Should I have nothing
at t equal capital T? I've got nothing. e to the 0 minus
1, that's nothing? Yes, yes that's the right thing. Because at capital T, the
source has just turned on, hasn't had time to
build up anything, just that was the
instant it turned on. So that's a step function. A step function is a
little bit of a stretch from an ordinary
function, but not as much of a stretch as its derivative. In a way, this is like the
highlight for today, coming up, to deal with not only a step
function, but a delta function. 32 minutes, 55 seconds I guess every author
and every teacher has to think am I going
to let this delta function into my course or into the book? And my answer is yes. You have to do it. You should do it. Delta functions are--
they're not true functions. As we'll see, no
true function can do what a delta function does. But it's such an
intuitive, fantastic model of things happening over
a very, very short time. We just make that
short time into 0. So we're saying with
the delta function, we're going to say that
something can happen in 0 time. Something can happen in 0 time. It's a model of, you know,
when a bat hits a ball. There's a very short time. Or a golf club hits a golf ball. There's a very
short time interval when they're in contact. We're modeling that by 0
time, but still, the ball gets an impulse. Normally, for 0 time, if you're
doing things continuously, what you do over 0
time is no importance. But we're not doing things
continuously, at all. So here we go. You've seen this guy, I think. But if you haven't,
here's the time to see it. So the delta function
is the derivative of-- so I've written three
important functions up here. Let me start with
a continuous one. That function, the ramp
is 0, and then the ramp suddenly ramps up to t. Take its derivative. So the derivative, the
slope of the ramp function is certainly 0 there. And here, the slope is 1. So the slope jumped from 0 to 1. The slope of the ramp
function is the step function. Derivative of ramp equals step. Why don't I write
those words down? Derivative of ramp equals step. So there is already
the step function. In pure calculus,
the step function has already got a
little question mark. Because at that point, the
derivative in a calculus course doesn't exist,
strictly doesn't exist, because we get a
different answer 0 on the left side from the
answer, 1 on the right side. We just go with that. I'm not going to
worry about what is its value at that point. It's 0 up for t negative,
and it's 1 for t positive. And often, I'll take it
1 for t equals 0, also. Usually, I will. That's the small problem. Now, the bigger problem is the
derivative of the-- so this is now the derivative
of the step function. So what's the derivative
of this step function? Well, the derivative along
there is certainly 0. The derivative along
here is certainly 0. But the derivative,
when that jumped, the derivative, the
slope was infinite. That line is vertical. Its slope is infinite. So at that one point, you have
an affinity, here, delta of 0. You could say delta
of 0 is infinite. But you haven't
said much, there. Infinite is too vague. Actually, I wouldn't
know if you gave me infinite or 2 times infinite. I couldn't tell the difference. So I'll put it in quotes,
because it sort of gives us comfort. But it doesn't mean much. What does mean much? Somehow that's important. Can I tell you how to
work with delta functions, how to think about
delta functions? It's the right way to
think about delta function. So here's some comment
on delta function. Giving the values of the
function, 0, and infinity, and 0, is not the best. What you can do with
a delta function is you can integrate it. You can define the
function by integrals. Integrals of things are nice. Do you think in your mind
when you take derivatives, as we did going left to right,
we were taking derivatives. The function was getting crazy. When we go right to left, take
integrals, those are smoothing. Integrals make
functions smoother. They cancel noise. They smooth the function out. So what we can do is to take the
integral of the delta function. We could take it from
any negative number to any positive number. And what answer would we get? What would be the right,
well, the one thing people know about the
delta function is-- and actually, it's the
key thing-- the integral of the delta function. Again, I'm integrating
the delta function from some negative number
up to some positive number. And it doesn't matter where n
is, because the function is 0 there and there. But what's the answer here? Put me out of my misery. Just tell me the number
I'm looking for, here, the integral of
the delta function. Or maybe you haven't met it. AUDIENCE: [INAUDIBLE]. PROFESSOR: It's? It's the one good
number you could guess. It's 1. Now, why is it 1? Because if the delta function
is the derivative of the step function, this should be the
step function evaluated between N and P. This should be
the step function, , here, minus the step function, there And what is the step function? You have to keep it straight. Am I talking about
the delta function? No, right now, I've
integrated it to get H of t. So this is H of P at
the positive side, minus H of N. That's
what integration's about. And what do I get? 1, because H of P, the
step function here, H is 1. And here, it's 0, so I get 1. Good, that's the
thing that everybody remembers about
the delta function. And now I can make sense out of
2 delta function, 2 delta of t. That could be my source. So if 2 delta of
t was my source, what's the graph
of 2 delta of t? Again, it's 0 infinite 0. You really can't tell from
the infinity what's up, but what would be the
integral of 2 delta of t, the integral of 2 delta
of t or some other? Well, let me put in the 2, here? What's the integral of 2
delta of t, would be 2H of t. Keep going. What do I get here? AUDIENCE: 2. PROFESSOR: It would
be 2 of these guys, 2 of these, 2 of these, 2. All right? So we made sense out of the
strength of the impulse, how hard the bat hit the ball. But of course, we
need units in there. We have to have units. And here, the value
for that unit was 2. Now, I'm going to-- because
this is really worth doing with delta functions. I didn't ask at the start
have you seen them before. But they are worth seeing. And they just take
a little practice. But then in the
end, delta functions are way easier to work with than
some complicated function that attempts to model this. We could model that by some
Gaussian curve or something. All the integrations would
become impossible right away. We could model it by
a step function up and a step function down. Then the integrations
would be possible. But still, we have
this finite width. I could let that
width go to 0 and let the height go to infinity. And what would happen? I'd get the delta function. So that's one way to create a
delta function, if you like. If you're OK with
step functions, then one way to create delta
is to take a big step up, step down, and then let the
size of the step grow and the width of
the steps shrink. Keep the area 1, because
area is integral. So I keep this,
that little width, times this big
height equal to 1. And in the end, I get delta. Now again, my point is
that delta functions, that you really understand them. What you can legitimately do
with them is integrate them. But now in later problems,
we might have not a 1 or a 2, but a function in here, like
cosine t, or e to the t, or q of t. Can I practice with those? Can I put in a function f of t? I didn't leave enough
space to write f of t, so I'm going to put it in
here. f of t delta of t dt. And I'm going to go
for the answer, there. My question is what
does that equal? You see what the question is? I got my delta function,
which I only just met. And I'm multiplying it by
some ordinary function. f of t gives us no problems. Think of cosine t. Think of e to the t. What do you think is the
right answer for that? What do you think
is the right answer? And this tells you
what the delta function is when you see this. What do I need to know about
f of t to get an answer, here? Do I need to know what f
is at t equals minus 1? You could see from
the way my voice asked that question that
the answer is no. Why do I not care
what f is at minus 1? Yeah? AUDIENCE: Because you're
multiplying by [INAUDIBLE]. PROFESSOR: Because
I'm multiplying by somebody that's 0. And similarly, at f equal minus
1/2, or at f equal plus 1/3, all those f's make
no difference, because they're
all multiplying 0. What does make a difference? What's the key information
about f that does come into the answer? f at? At just at that one point, f at? AUDIENCE: [INAUDIBLE] PROFESSOR: 0, f at
0 is the action. The impulse is happening. The bat's hitting the ball. So we're modeling
rocket launching, here. We're launching in 0 seconds
instead of a finite time. So in other words,
well, I don't know how to put this answer down
other than just to write it. I guess I'm hoping
you're with me in seeing that
what it should be. Can I just write it? All that matters
is what f is at t equals 0, because that's
where all the action is. And that f of 0,
if f of 0 was the 2 that I had there a little while
ago, then the answer will be 2. If f of 0 is a 1, if the
answer is f of 0 times 1-- and I won't write times 1. That's ridiculous. Now we can integrate
delta functions, not just a single integral of
delta, but integral of a function, a nice
function times delta. And we get f of 0. So can I just, while we're on
the subject of delta functions, ask you a few examples? What is the integral of
e to the t delta of t dt? AUDIENCE: It's 1. PROFESSOR: Yeah, say it again? AUDIENCE: It's 1. PROFESSOR: It's 1. It's 1, right. Because e to the t, at the only
point we care about, t equal 0 is 1. And what if I change
that to sine t? Suppose I integrate
sine t times delta of t? What do I get now? I get? AUDIENCE: 0. PROFESSOR: 0, right. And actually, that's
totally reasonable. This is a function, which is
yeah, it's an odd function. Anyway, sine, if I switch t to
negative t, it goes negative. 0 is the right answer. Let me ask you this one. What about delta of t squared? Because if we're up for a delta
function, we might square it. Now we've got a
high-powered function, because squaring this
crazy function delta of t gives us something truly crazy. And what answer would
you expect for that? AUDIENCE: 1. PROFESSOR: Would you expect 1? So this is like? I'm just getting
intuition working, here, for delta functions. What do you think? I'm looking at the energy
when I square something. OK, so we had a guess of 1. Is there another guess? Yeah? AUDIENCE: A third? PROFESSOR: Sorry? AUDIENCE: 1/3. PROFESSOR: 1/3, that's
our second guess. I'm open for other
guesses before I-- OK, we have a rule here for f of t. And now what is the f of t that
I'm asking about in this case? It's delta of t, right? If f of t is delta of t,
then that would match this. And therefore, the
answer should match. Do you see what I'm
shooting for, yeah? AUDIENCE: It'd be infinity? PROFESSOR: It'd be infinity. It would be infinity. That's delta of t squared
is that's an infinite energy function. You never meet it, actually. I apologize, so so
write it down there. I could erase it right
away because you basically never see it. It's infinite energy. Well, I think you'd see it. I mean, we're really going back
to the days of Norbert Wiener. When I came to the
math department, Norbert Wiener was still
here, still alive, still walking the hallway by touching
the wall and counting offices. And hard to talk to, because
he always had a lot to say. And you got kind of
allowed to listen. So anyway, Wiener
was among the first to really use delta
functions, successfully use delta functions. Anyway, this is the big one. This is the big one. Now, so what's all that about? I guess I was trying
to prepare by talking about this function
prepare for the equation when that's the source. So dy equal ay plus
a delta function. Let me bring that delta
function in at time T. So how do you interpret
that equation? So like part of this
morning's lecture is to get a first
handle on an impulse. So let me write that
word impulse, here. Where am I going to write it? So delta is an impulse. That's our ordinary
English word for something that happens fast. And y of t is the
impulse response. 52 minutes, 53 seconds And this is the most important. Well, I said e to the st
was the most important. How can I have two most
important examples? Well, they're a tie, let's say. e to the st is the most
important ordinary function. It's the key to
the whole course. Delta of t, the impulse,
is the important one because if I can solve
it for a delta function, I can solve it for anything. Let's see if we can solve
it for a delta function, a delta function, an impulse
that starts at time T. Again, I'm just going to start
writing down the solution and ask for your help
what to write next. So what do you expect as a
first term in the solution? So I'm starting
again from y of 0. Let's see if we can
solve it by common sense. So how do I start
the solution to this? Everybody sees what
this equation is saying. I have an initial deposit of
y of 0 that starts growing. And then at time capital
T I make a deposit. At that moment, at that
instant, I make a deposit of 1. That's an instant deposit of 1. Which is, of course,
what I do in reality. I take $1 to the bank. They've got it now. At time T, I give them that $1,
and it starts earning interest. So what about y of t? What do you think? What's the first term
coming from y of 0? So the term coming
from y of 0 will be y of 0 to start
with, e to at. That takes care of the y of 0. Now, I need something. It's like this, plus
I need something that accounts for what
this deposit brings. So up until time
T, what do I put? So this is for t smaller
than T and t bigger than T. So what goes there? For t smaller than
T, what's the benefit from the delta function? 0, didn't happen yet. For t bigger than T,
what's the benefit from the delta function? AUDIENCE: [INAUDIBLE]. PROFESSOR: For t bigger
than T, well, that's right. OK, but now I've
made that deposit at time capital T.
Whatever's going there is whatever I'm
getting from that deposit. At time capital
T, I gave them $1, and they start paying
interest on it. What's going to go there? So if I gave them $1 at that
initial time, so that $1 would have been part of y of 0. What did I get at a later time? e to the at. Now I'm waiting. I'm giving them the
dollar at time capital T, and it starts growing. So what do I have
at a later time, for t later than capital T? What has that $1 grown into? e to the a times the--
right, it's critical. It's the elapsed time. It's the time since the deposit. Is that right? So what do I put here? AUDIENCE: t minus capital T? PROFESSOR: t minus
capital T, good. Apologies to bug you about
this, but the only way to learn this stuff from a
lecture is to be part of it. So I constantly ask you, instead
of just writing down a formula. I think that looks good. So suddenly, what does this
amount to at t equal capital T? Maybe I should allow
t equal capital T. At t equal capital T,
what do I have here? AUDIENCE: 1. PROFESSOR: 1. That's my $1. At t equal capital
T, we've got $1. And later it's grown. So we have now solved. We have found the
impulse response. We have found the
impulse response when the impulse happened at
capital T. That was good going. Now, I've given you
my list of examples with the pause on
the sine and cosine. I pause on the sine and
cosine because one way to think about sine and cosine
is to get into complex numbers. And that's really for next time. But apart from that, we've
done all the examples, so are we ready? Oh yeah, I'm going to try for
the big thing, the big formula. So this is the key
result of section 1.4, the solution to this equation. So I'm going back to
the original equation. And just see if we can write
down a formula for the answer. So let me write
the equation again. dy dt is ay plus some source. I think we can write down
a formula that looks right. And we could then actually
plug it in and see, yeah, it is right. So what's going to
go into this formula? We got enough examples, so now
let's go for the whole thing. So y of t, first of
all, comes whatever depends on the
initial condition. So how much do we have
at a later time when our initial deposit was y of 0? So that's the one we've
seen in every example. Every one of these things
has this term growing out of y of 0. So let me put that in again. So the part that grows out of
y of 0 is y of 0 e to the at. That's OK. So that's what the initial. So our money is coming from two
sources, this initial deposit, which was easy, and this
continuous, over time deposit, q of t. And I have to ask
you about that. That's going to be like
the particular solution, the particular solution that
comes from the source term. This is the solution it comes
from the initial condition. So what do you think
this thing looks like? I just think once we see
it, we can say, yeah, that makes sense. So now I'm saying what? If we've deposited q of
t in varying amounts, maybe a constant for a while,
maybe a ramp for awhile, maybe whatever, a step, how am
I going to think about this? So at every time t
equal to s, so I'm using little t for
the time I've reached. Right? Here's t starting at 0. Now, let me use s for
a time part way along. So part way along, I input. I deposit q of s. I deposit it at time s. And then what does it do? That money is in the
bank with everybody else. It grows along with
everything else. So what's the growth factor? What's the growth factor? This is the amount I
deposited at time s. And how much has
it grown at time t? This is the key question,
and you can answer it. It went in a time s. I'm looking at time t. What's the factor? AUDIENCE: Is it e
to the a t minus s. PROFESSOR: e to the a t minus s. So that's the contribution
to our balance at time t from our input at time s. But now, I've been
inputting all the way along. s is running all the
way from here to here. So finish my formula. Put me out of my misery. Or it's not misery, actually. Its success at this moment. What do I do now? I? AUDIENCE: Integrate. PROFESSOR: I integrate, exactly. I integrate. I integrate. So all these deposits went in. They grew that amount
in the remaining time. And I integrate from 0
up to the current time t. So you see that formula? Have a look at it. This is a general formula, and
every one of those examples could be found
from that formula. If q of s was 1, that was
our very first example. We could do that integration. If q of s was e to the--
anyway, we could do every one. I just want you to see that
that formula makes sense. Again, this is what grew out
of the initial condition. This is what grew out of
the deposit at time s. And the whole point of
calculus, the whole point of learning [? 1801 ?],
the integral equation part, the integrals part,
is integrals just add up. This term just adds up
all the later deposits, times the growth factor
in the remaining time. And as I say, if I took q of s
equal 1-- the examples I gave are really the examples where
you can do the integral. If q of s is e to the i omega
s, I can do that integral. Actually, it's not hard to do
because e to the at doesn't depend on s. I can bring an e to the
at out in this case. That formula is just
worth thinking about. It's worth understanding. I didn't, like, derive it. And the book does, of course. There's something called
an integrating factor. You can get at this
formula systematically. I'd rather get at it
and understand it. I'm more interested
in understanding what the meaning of that
formula is than the algebra. Algebra is just a
goal to understand, and that's what I
shot for directly. And as I say, that
the book also, early section of the book,
uses practice in calculus. Substitute that in
to the equation. Figure out what is dy dt. And check that it works. It's worth actually
looking at that end of what you need to know from
calculus It's is. You should be able
to plug that in for y and see that solves
the equation. Right, now I have enough
time to do cosine omega t. But I don't have enough time
to do it the complex way. So let me do as a final
example, the equation. Let me just think. I don't know if I have
enough space here. I'm now going to do dy dt--
can I call that y prime to save a little space-- equal
ay plus cosine of t. I'll take omega to be 1. Now, how could we
solve that one? I'm going to solve it
without complex numbers, just to see how easy
or hard that is. And you'll see,
actually, it's easy. But complex numbers
will tell us more. So it's easy, but
not totally easy. So what did I do in
the earlier example if the right hand side
was a 1, a constant? I look for the solution
to be a constant. If the right hand side
was an exponential, I look for the solution
to be an exponential. Now, my right hand side, my
source term, is a cosine. So what form of the solution
am I going to look for? I naturally think,
OK, look for a cosine. We could try y equals
some number M cosine t. Now, you have to see what
goes wrong and how to fix it. So if I plug that in,
looking for M the same way I look for capital Y
earlier, I plug this in, and I get aM cosine t cosine t. But what do I get for y prime? Sine t. And I can't match. I can make it work. I can't make a sine there
magic a cosine here. So what's the solution? How do I fix it? I better allow my
solution to include some sine plus N sine t. So that's the problem with
doing it, keeping things real. I'll push this
through, no problem. But cosine by itself won't work. I need to have sines there,
because derivatives bring out sines. So I have a combination
of cosine and sine. I have a combination
of cosine and sine. So the complex method
will work in one shot because e to the i omega t is a
combination of cosine and sine. Or another way to say it
is when I see cosine here, that's got two exponentials. That's got e to the it
and e to the-- anyway. Let's go for the real one. So I'm going to plug
that into there. So I'll get sines
and cosines, right? When I plug this
into there, I'll have some sines
and some cosines, and I'll just match
the two separately. So I'm going to
get two equations. First of all, let me say
what's the cosine equation? And then what's
the sine equation? So when I match cosine
terms, what do I have? What cosine terms do I
get out of y prime, here? The derivative. Well, the derivative
of cosine is a sine. That that's not a cosine term. The derivative of
sine is cosine. I think I get, if I
just match cosines, I think I get an N cosine. N cosine t equal ay. How many cosines do I
have from that term? ay has an M cosine t. I think I have an aM,
and here I've got 1. That was a natural
step, but new to us. I'm matching the cosines. I have on the left side, with
this form of the solution, the derivative will
have an N cosine t. So I had N cosines, aM
cosines, and 1 cosine. Now, what if I match signs? What happens there? We're pushing more
than an hour, so hang on for another five
minutes, and we're there. Now, what happens if
I match sines, sine t? How do I get sine t in y prime? So take the derivative of
that, and what do you have? AUDIENCE: Minus [INAUDIBLE]. PROFESSOR: Minus M sine t. That tells me how many
sine t's are in there. And on the right
hand side, a times y, how many sine t's
do I have from that? AUDIENCE: You have N t's. PROFESSOR: N, good thinking. And what about from this term? None, no sine there. So I have two equations by
matching the cosines and sines. Once you see it, you
could do it again. And we can solve
those equations, two ordinary, very simple
equations for M and N. Let's see if I make space. Why don't I do it here,
so you can see it. So how do I solve
those two equations? Well, this equation gives me--
easy-- gives me M as minus aN. So I'll just put that in for N. So I have N equals aM. But M is minus aN. I think I've got minus
a squared N plus that 1. 1 hour, 14 minutes, 6 seconds All I did was solve the
equation, just by common sense. You could say by linear
algebra, but linear algebra's got a little more
to it than this. So now I know M, and now I know
N. So now I know the answer. y is M, so M is minus aN. Oh, well, I have to figure
out what N is, here. What is N? This is giving me N, but
I better figure it out. What is N from that
first equation? And then I'll plug in. And then I'm quit. AUDIENCE: [INAUDIBLE]. PROFESSOR: 1 over, yeah. AUDIENCE: 1 plus a squared. PROFESSOR: 1 plus
a squared, good. Because that term goes
over there, and we have 1 plus a squared. So now y is M cosine t. So M is minus aN. So minus aN is 1 over 1
plus a squared cosine t. Is that right? That was the cosines. And we had N sine t. But N is just 1-- I think
I just add the sine t. Have I got it? I think so. Here is the N sine t,
and here is the M cos t. It was just algebra. Typical of these
problems, there's a little thinking and
then some algebra. The thinking led us to this. The thinking led us to the fact
we needed cosines in there, as well as cosines. But then once we did
it, then the thinking said, OK, separately match the
cosine terms and the sine term. And then do the algebra. Now, I just want to
do this with complex. So y prime equals
ay plus e to the it. 1 hour, 16 minutes, 44 seconds To get an idea, you see the two. And then I have
to talk about it. You see, I'm only going
to go part way with this and then save it for Wednesday. But if I see this, what
solution do I assume? This is like an e to the st. I
assume y is some Y e to the it. See, I don't have cosines
and sines anymore. I have e to the it. And if I take the
derivative of e to the it, I'm still in the
e to the it world. So I do this. I plug it in. Uh-huh, let me leave
that for Wednesday. We have to have some
excitement for Wednesday. So we'll get a complex
answer, and then we'll take the real part to
solve that problem. So we've got two steps,
one way or the other way. Here, we had two steps because
we had to let sines sneak in. Here, we have two steps
because I could solve it, and you could solve
that right away. But then you have to
take the real part. I'll leave that. Is there questions? Do you want me to recap
quickly what we've done. AUDIENCE: Yes. PROFESSOR: I try to
leave on the board enough to make a recap possible. Everything was
about that equation. We have only
solved-- I shouldn't say only-- we have solved the
constant coefficient, model constant coefficient,
first order equation. Wednesday comes
nonlinear equation. This one today was
strictly linear. So what did we do? We solved this equation,
first of all, for q equal 1; secondly, for q equal
e to the st; thirdly, for q equal a step; fourthly
for q equal-- where is it? Where is that delta of t? Maybe it's here. Ah, it got erased. So the fourth guy was y prime
equal ay plus delta of t, or delta of t minus
capital T. So those were our four examples. And then what did we finally do? So if we're recapping,
compressing, we're compressing
everything into two minutes. We solved those four
examples, and then we solved the general problem. And when we solved
the general problem, that gave us this integral,
which my whole goal was that you should understand that
this should seem right to you. This is adding up
the value at time t from all the inputs
at different times s. So to add them up, we
integrate from 0 to t. And finally, we
returned to the question of cos t, all
important question. But awkward question,
because we needed to let sine t in there too. |
7706 | https://learn.microsoft.com/en-us/answers/questions/5049770/find-the-angle-between-two-headings-(in-degrees) | Find the angle between two headings (in degrees) - Microsoft Q&A
Skip to main content
Microsoft Ignite
November 17–21, 2025
Session catalog is live—personalize your event experience across AI, cloud, and more.
Register now
Dismiss alert
This browser is no longer supported.
Upgrade to Microsoft Edge to take advantage of the latest features, security updates, and technical support.
Download Microsoft EdgeMore info about Internet Explorer and Microsoft Edge
Learn
Suggestions will filter as you type
Sign in
Profile
Settings
Sign out
Learn
Documentation
All product documentation
Azure documentation
Dynamics 365 documentation
Microsoft Copilot documentation
Microsoft 365 documentation
Power Platform documentation
Code samples
Troubleshooting documentation
Free to join. Request to attend.
Microsoft AI Tour
Take your business to the AI frontier.
Training
All training
Azure training
Dynamics 365 training
Microsoft Copilot training
Microsoft 365 training
Microsoft Power Platform training
Credentials
Career paths
AI-generated Plans
Free to join. Request to attend.
Microsoft AI Tour
Take your business to the AI frontier.
Q&A
Ask a question
Azure questions
Windows questions
Microsoft 365 questions
Microsoft Outlook questions
Microsoft Teams questions
Popular tags
All questions
Free to join. Request to attend.
Microsoft AI Tour
Take your business to the AI frontier.
Topics
Artificial intelligence
Learning hub to build AI skills
Compliance
Compliance resources you need to get started with your business
DevOps
DevOps practices, Git version control and Agile methods
Learn for Organizations
Curated offerings from Microsoft to boost your team’s technical skills
Platform engineering
Tools from Microsoft and others to build personalized developer experiences
Security
Guidance to help you tackle security challenges
Assessments
Interactive guidance with custom recommendations
Student hub
Self-paced and interactive training for students
Educator center
Resources for educators to bring technical innovation in their classroom
Free to join. Request to attend.
Microsoft AI Tour
Take your business to the AI frontier.
Suggestions will filter as you type
Sign in
Profile
Settings
Sign out
Q&A
Questions
Tags
Help
More
Questions
Tags
Help
Ask a question
Add
Add to Collections Add to plan
Add to Collections
Share via
Facebookx.comLinkedInEmail
Find the angle between two headings (in degrees)
Anonymous
Aug 19, 2020, 11:00 PM
I have a worksheet several columns. One of the columns is for wind direction and another column is for runway heading. I am trying to find a formula to calculate the difference between these two degrees. The calculation should return the crosswind component in degrees. I don't need to know the speed of the crosswind though. Just the degrees. For example, If the runway heading is 190 and the wind is coming from 160, the crosswind component will be 30 degrees. If the runway heading is 190 and the wind is coming from 030, the crosswind component will be 160 degrees.
Expand table
| Runway Heading | Wind Direction |
--- |
| 240 | 270 |
| 240 | 230 |
| 240 | 290 |
| 010 | 320 |
Microsoft 365 and Office | Excel | For home | Windows
Microsoft 365 and Office | Excel | For home | Windows
A family of Microsoft spreadsheet software with tools for analyzing, charting, and communicating data.
Sign in to follow Follow
Locked Question.This question was migrated from the Microsoft Support Community. You can vote on whether it's helpful, but you can't add comments or replies or follow the question.
0 comments No comments Report a concern
I have the same question I have the same question 0{count} votes
3 answers
Sort by: Most helpful
Most helpfulNewestOldest
Anonymous Aug 19, 2020, 11:38 PM Hi chrisodum27,
Thank you for reaching out to us. I am Thuy, an Independent Advisor and Microsoft user like you.
For this task, you can apply the formula as below to get expect output
=ABS($A2-$B2)
assuming that $A2 stores 240 and $B2 stores 270. Then you can drag the formula for the rest of the column for Angle column.
Hope this helps and please let me know if you need more assistance.
Regards Please sign in to rate this answer. Yes No 0 comments No comments Report a concern
2. Anonymous Aug 20, 2020, 2:56 AM I tried that and it worked for most scenarios but not all. The problem started when I had a runway heading of 010 and wind out of 280, for example. This formula would return a difference of 270 degrees. Although this is correct, it doesn't really portray a crosswind component. I did find something a formula that seems to be working though.
=MIN(MOD(Y5-Z5,360),360-MOD(Y5-Z5,360))
I still have some data to apply it to and make sure it works for everything. I do appreciate you trying to help out though. Please sign in to rate this answer. Yes No 4 people found this answer helpful. 0 comments No comments Report a concern
3. Anonymous Aug 20, 2020, 3:04 AM Hi chrisodum27,
I'm glad to hear that you finally got a promising solution for your task.
Let me know if you need more assistance.
Regards Please sign in to rate this answer. Yes No 0 comments No comments Report a concern
Additional resources
English (United States)
Your Privacy Choices
Theme
Light
Dark
High contrast
AI Disclaimer
Previous Versions
Blog
Contribute
Privacy
Terms of Use
Code of Conduct
Trademarks
© Microsoft 2025 |
7707 | https://physics.stackexchange.com/questions/281098/zeeman-paschen-back-effect | quantum mechanics - Zeeman - Paschen-Back effect - Physics Stack Exchange
Join Physics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Physics helpchat
Physics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Zeeman - Paschen-Back effect
Ask Question
Asked 9 years ago
Modified9 years ago
Viewed 8k times
This question shows research effort; it is useful and clear
4
Save this question.
Show activity on this post.
I've recently came to know that there is two effects of an atom in external magnetic field (assume the magnetic field is constant in time and direction).
One of them is the Zeeman effect the second is Paschen-Back effect. The effects differ only by how strong the magnetic field is.
So my question is how to distinguish between them? On what basis we identify a strong or weak magnetic field? Mathematically , How do I know if I have some B 0 B 0 is considered Weak or Strong field?
Thanks.
quantum-mechanics
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Improve this question
Follow
Follow this question to receive notifications
edited Sep 19, 2016 at 18:17
JonshJonsh
asked Sep 19, 2016 at 11:55
JonshJonsh
45 1 1 silver badge 4 4 bronze badges
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
4
Save this answer.
Show activity on this post.
We say strong or weak magnetic field in the sense of it's effect in the coupling of the orbital and spin angular momenta.
If the applied magnetic field strength is low, we mean that it leaves the orbital and spin angular momentum in the coupled state (i.e., the L−S L−S coupling dominates than the interaction of the magnetic field with the magnetic dipole), we state that the field strength is weak and in such cases, the splitting of spectral lines is explained using Zeeman effect.
But, when the applied magnetic field is so intense, such that it breaks down the coupling between the spin and orbital angular momentum, the splitting of spectral lines is described using Paschen-Back effect.
Update:
Zeeman effect:
Here the spin-orbit interaction dominates over the effect of the external magnetic field, L⃗L→ and S⃗S→ are not separately conserved, only the total angular momentum J⃗=L⃗+S⃗J→=L→+S→ is.
The Hamiltonian of the system is :
H=H 0+V M H=H 0+V M
where H 0 H 0 is the unperturbed Hamiltonian and V M V M is the perturbed hamiltonain due to the presence of a magnetic fielf B⃗B→.
V M=−μ⃗⋅B⃗V M=−μ→⋅B→
which represents the interaction energy of the magnetic dipole. Here we can approximate μ⃗μ→ is solely due to the spin and orbital angular moment:
μ⃗≈−μ B g J⃗h¯=−μ B(g l L⃗+g s S⃗)h¯μ→≈−μ B g J→h¯=−μ B(g l L→+g s S→)h¯
In the case of L−S L−S coupling, one can sum over all electrons in the atom:
g J⃗=⟨(g l L⃗+g s S⃗)⟩g J→=⟨(g l L→+g s S→)⟩
In the case of Zeeman effect, the perturbation V M V M is very less. In such weak magnetic fields, the spin and orbital angular momentum vectors precesses around the fixed (since it is conserved) total angular momentum vector J⃗J→. In such case, we can represent the time-averaged orbital and spin angular momenta as a projection onto the direction of J⃗J→:
S⃗a v g=S⃗⋅J⃗J 2 L⃗a v g=L⃗⋅J⃗J 2 S→a v g=S→⋅J→J 2 L→a v g=L→⋅J→J 2
Hence
⟨V M⟩=μ B h¯J⃗(g l L⃗⋅J⃗J 2+g s L⃗⋅J⃗J 2)⟨V M⟩=μ B h¯J→(g l L→⋅J→J 2+g s L→⋅J→J 2)
After making some manipulations, we finally gt
V M=μ B B e x t m j g j V M=μ B B e x t m j g j
where g j g j is the Lande g g-factor given by
g j=[1+(g s−1)j(j+1)−l(l+1)+s(s+1)2 j(j+1)]g j=[1+(g s−1)j(j+1)−l(l+1)+s(s+1)2 j(j+1)]
m j m j is the z z- component (assuming the external magnetic field points in the z z-direction) of total angular momentum, μ B μ B is the Bohr magneton and B e x t B e x t is the strength of the external magnetic field.
The Zeeman first order correction to the energy is
Δ E z=μ B g j B e x t m j Δ E z=μ B g j B e x t m j
From this, the magnetic field strength can be written as:
B e x t=Δ E z μ B g j m j B e x t=Δ E z μ B g j m j
i.e., if we know the energy involved in the transition giving rise to Zeeman effect, we can find out the external magnetic field required for that transition (even though it is unusual to write the equation as above).
Paschen_Back effect:
Here the splitting of atomic energy levels take place in a strong magnetic field such that the L−S L−S coupling is broken. Hence the effect is a strong field limit to the Zeeman effect. When the magnetic-field perturbation significantly exceeds the spin-orbit interaction, one can safely assume [H 0,S]=0[H 0,S]=0. In such a case,
Δ E z=E z−E 0=B z μ B(m l+g s m s)Δ E z=E z−E 0=B z μ B(m l+g s m s)
where E 0 E 0 corresponds to the unperturbed part of the energy eigen value of the Hamiltonian (or the ground state if you prefer). This gives
B z=Δ E z μ B(m l+g s m s)B z=Δ E z μ B(m l+g s m s)
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
edited Sep 20, 2016 at 0:56
answered Sep 19, 2016 at 12:46
UKHUKH
5,031 1 1 gold badge 20 20 silver badges 36 36 bronze badges
8
1 Thanks for your answer. But still, how mathematically you can define Strong and Weak magnetic field? I thought maybe somehow to look at the μ b(L⃗+2 S⃗)⋅B μ b(L→+2 S→)⋅B and e 2 m c[B×r]2 e 2 m c[B×r]2 terms and to define it by them...Jonsh –Jonsh 2016-09-19 12:51:41 +00:00 Commented Sep 19, 2016 at 12:51
Please see: hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html and hyperphysics.phy-astr.gsu.edu/hbase/quantum/paschen.html#c1UKH –UKH 2016-09-19 14:26:50 +00:00 Commented Sep 19, 2016 at 14:26
1 I already saw this. As you may noticed, it is not answering my question.Jonsh –Jonsh 2016-09-19 15:38:00 +00:00 Commented Sep 19, 2016 at 15:38
You asked on what basis we say whether a field is strong or not so that when to use Zeeman effect and Paschen-Back effect. The answer is if the field is strong enough so that the the orbital interaction with the magnetic field dominates the L-S interaction, then we say the field is strong. Otherwise the field is weak UKH –UKH 2016-09-19 15:55:40 +00:00 Commented Sep 19, 2016 at 15:55
1 If you have something else in your mind, please specify that in your question UKH –UKH 2016-09-19 15:56:32 +00:00 Commented Sep 19, 2016 at 15:56
|Show 3 more comments
Your Answer
Thanks for contributing an answer to Physics Stack Exchange!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
quantum-mechanics
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
14Why was the Stark effect discovered much later than the Zeeman effect?
1Zeeman effect in a magnetic field with multiple components
4What's the most general approach to Zeeman effect?
1Zeeman Effect and Lande g g-factor
1Zeeman effect vs Stern Gerlach effect
1L⃗L→, S⃗S→ and J⃗J→ of the atom or electron when considering the Zeeman effect
3Why does Weak-field Zeeman effect break all degeneracy in the Hydrogen atom?
Hot Network Questions
alignment in a table with custom separator
An odd question
Origin of Australian slang exclamation "struth" meaning greatly surprised
Why do universities push for high impact journal publications?
Exchange a file in a zip file quickly
Is it ok to place components "inside" the PCB
Should I let a player go because of their inability to handle setbacks?
Another way to draw RegionDifference of a cylinder and Cuboid
Does the curvature engine's wake really last forever?
Matthew 24:5 Many will come in my name!
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Why include unadjusted estimates in a study when reporting adjusted estimates?
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
ICC in Hague not prosecuting an individual brought before them in a questionable manner?
How can I show that this sequence is aperiodic and is not even eventually-periodic.
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Why are LDS temple garments secret?
в ответе meaning in context
How to start explorer with C: drive selected and shown in folder list?
For every second-order formula, is there a first-order formula equivalent to it by reification?
Can a cleric gain the intended benefit from the Extra Spell feat?
How do you emphasize the verb "to be" with do/does?
Xubuntu 24.04 - Libreoffice
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Physics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7708 | https://alistairsavage.ca/symfunc/notes/Savage-SymmetricFunctions.pdf | Symmetric Functions Alistair Savage Department of Mathematics and Statistics University of Ottawa This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License Contents Preface 4 1 Preliminaries 5 1.1 The symmetric group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 1.2 Partitions and compositions . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 1.3 Graded rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.4 Formal power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 2 The ring of symmetric functions 20 2.1 Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.2 Symmetric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 2.3 Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.4 Elementary symmetric functions . . . . . . . . . . . . . . . . . . . . . . . . .
29 2.5 Complete homogeneous symmetric functions . . . . . . . . . . . . . . . . . .
33 2.6 Power sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 3 Schur functions 42 3.1 Alternating polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 3.2 Schur polynomials and Schur functions . . . . . . . . . . . . . . . . . . . . .
46 3.3 The Hall inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 3.4 Skew Schur functions and semistandard tableaux . . . . . . . . . . . . . . .
56 3.5 Transition matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 3.6 The Littlewood-Richardson rule . . . . . . . . . . . . . . . . . . . . . . . . .
71 3.7 The Murnaghan–Nakayama rule . . . . . . . . . . . . . . . . . . . . . . . . .
78 4 Adjoint operators and Hopf algebras 81 4.1 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 4.2 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85 4.3 Hopf algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 4.4 Hopf algebra structure on the ring of symmetric functions . . . . . . . . . .
95 5 The boson-fermion correspondence 102 5.1 The Heisenberg algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
102 5.2 Bosonic Fock space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
106 5.3 Fermionic Fock space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108 2 Contents 3 5.4 Bosonization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
114 5.5 Fermionization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 6 Applications to representation theory 126 6.1 Characters of finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . .
126 6.2 Induction and restriction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
130 6.3 Characters of symmetric groups . . . . . . . . . . . . . . . . . . . . . . . . .
133 6.4 Specht modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142 6.5 Representations of general linear groups . . . . . . . . . . . . . . . . . . . .
146 Index of notation 149 Preface These are notes for the topics course Symmetric Functions (MAT 4995/5327) at the Uni-versity of Ottawa, which took place in Winter 2022. We first study the basics of symmetric functions, including the important topic of Schur functions. We essentially cover the ma-terial found in the first chapter of [Mac15], but we often give more detail and assume less background from the reader. We then cover some more advanced topics, including the Hopf algebra structure on the ring of symmetric functions, the boson-fermion correspondence, and applications to the representation theory of the symmetric groups and general linear groups.
We assume that the reader has a basic knowledge of the theory of rings and modules as covered, for example, in Rings and Modules (MAT 3143), or its French equivalent Anneaux et Modules (MAT 3543), at the University of Ottawa.
I would like to thank the students of the course for making it an enjoyable experience. I also thank Darij Grinberg for very useful comments and corrections.
Alistair Savage Course website: 4 Chapter 1 Preliminaries In this first chapter we cover some background that will be used throughout the course. In particular, we will discuss the symmetric group, partitions, compositions, graded rings, and formal power series. Our discussion will be brief. Throughout these notes we let N := {0, 1, 2, . . . } denote the set of natural numbers, including zero.
1.1 The symmetric group We begin recalling a few important facts about the symmetric group. For n ∈N, let Sn denote the symmetric group on n letters. Thus Sn is the group of permutations of {1, 2, . . . , n}.
By convention S0 is the trivial group. The group Sn has order n!.
If π ∈Sn, we can write it in one-line notation as π = π(1) π(2) · · · π(n).
Alternatively, we may write it in cycle notation, which describes the effect of repeatedly applying the permutation to the elements of {1, 2, . . . , n}.
Example 1.1.1. Consider the element π ∈S6 given by π(1) = 6, π(2) = 1, π(3) = 4, π(4) = 3, π(5) = 5, π(6) = 2.
Then in one-line notation we have π = 6 1 4 3 5 2 and in cycle notation we have π = (1 6 2)(3 4)(5).
We sometimes omit cycles of length one in cycle notation, writing π = (1 6 2)(3 4).
5 6 Preliminaries The inverse of an element can be given easily in cycle notation by reversing the order of the cycles. For example, π−1 = (2 6 1)(4 3).
△ Note that we have a natural inclusion {1, 2, . . . , n} ⊆{1, 2, . . . , n + 1}. Thus, we can view Sn as the groups of Sn+1 that fixes n + 1. In this way we have a chain of groups S0 ⊆S1 ⊆S2 ⊆S3 ⊆· · · .
Define S∞:= [ n∈N Sn.
Thus, S∞is the set of all bijections π: {1, 2, 3, . . . } →{1, 2, 3, . . . } such that there exists n ∈N with π(m) = m for m ≥n.
Exercises.
1.1.1. Give an example of a bijection π: {1, 2, 3, . . . } →{1, 2, 3, . . . } such that π / ∈S∞.
1.1.2. Suppose π ∈Sk and that, for i ∈N, π has mi cycles of length i. Prove that the number of permutations σ ∈Sk that commute with π (that is, such that σπ = πσ) is Y i≥1 imimi!.
Hint: Write π in cycle notation. What are the cycles of σπσ−1?
1.2 Partitions and compositions In the theory of symmetric functions, many mathematical objects are labelled by partitions and compositions. In this section we define these concepts and prove some basic results concerning them that will be used in the sequel.
Definition 1.2.1 (Partition). A partition λ of a nonnegative integer k ∈N is a weakly decreasing sequence (λj)∞ j=1 of nonnegative integers such that P∞ j=1 λj = k. If λ is a partition of k, then we often write λ ⊢k and |λ| = k, and we sometimes call k the size of λ. We let Par(k) denote the set of partitions of k and let Par := S k∈N Par(k) denote the set of all partitions. We denote the unique partition (0, 0, . . . ) of 0 by ∅.
A partition has finitely many nonzero terms, and we typically omit zero terms when writing λ. For example (5, 2, 2, 1), (5, 2, 2, 1, 0), (5, 2, 2, 1, 0, 0), and (5, 2, 2, 1, 0, 0, . . . ) Partitions and compositions 7 are all the same partition of 10, which we will usually write as (5, 2, 2, 1). The nonzero terms in a partition λ are called the parts of λ. The length of λ, denoted ℓ(λ), is the number of parts of λ. For n ∈N, we let Parn(k) := {λ ∈Par(k) : ℓ(λ) ≤n} and Parn := [ k∈N Parn(k).
We will often view elements λ of Parn as weakly decreasing sequences (λ1, . . . , λn) of length n, since we have λj = 0 for j > n by definition. It follows immediately from the definitions that Parn(k) ⊆Parm(k) and Parn ⊆Parm for n ≤m and that Par(k) = [ n∈N Parn(k) and Par = [ n∈N Parn.
For any partition λ and i ∈N, the multiplicity of i in λ, denoted mi(λ), is the number of parts of λ equal to i. In other words, mi(λ) = |{j ∈N : λj = i}|, where |X| denotes the cardinality of a finite set X. We will sometimes write a partition by writing each of its distinct parts once, with an exponent indicating the multiplicity of that part. For example we might write (8, 8, 7, 5, 5, 5, 5, 3, 2, 2, 1, 1, 1) as (82, 71, 54, 31, 22, 13).
We will often think about partitions geometrically as follows. The Young diagram, also known as the Ferrers diagram, of a partition λ is a stack of boxes in which each row is left-justified and the j-th row has λj boxes. We will use the English convention, in which the first row is the top row. Thus, the Young diagram of the partition (5, 4, 2, 2, 2, 1, 1) is (1.1) .
Some references use the French convention, where the first row is the bottom row. (There is also a Russian convention!) We will often identify a partition with its Young diagram.
The conjugate of a partition λ is the partition λ′ whose Young diagram is the transpose of the Young diagram of λ, that is, the Young diagram obtained by reflection in the main diagonal. For example if λ = (5, 4, 2, 2, 2, 1, 1), whose Young diagram is given in (1.1), then the conjugate partition λ′ = (7, 5, 2, 2, 1) has Young diagram .
8 Preliminaries It follows that λ′ j is the number of boxes in the j-th column of λ, or equivalently (1.2) λ′ j = |{i : λi ≥j}|.
In particular, we have λ′ 1 = ℓ(λ) and λ1 = ℓ(λ′). We also have λ′′ = λ.
For k ∈N, we define the dominance partial ordering (sometimes called the natural partial ordering) ≤on Par(k) as follows: for λ, µ ∈Par, (1.3) λ ≤µ ⇐ ⇒λ1 + · · · + λi ≤µ1 + · · · + µi for all i ≥1.
We write λ < µ if λ ≤µ and λ ̸= µ.
Example 1.2.2. If k = 6, we have (16) < (2, 14) and (2, 2, 2) < (4, 2).
However, (3, 13) and (23) are not comparable.
△ Definition 1.2.3 (Weak composition). A weak composition α of k ∈N is a sequence (αj)∞ j=1 of nonnegative integers such that P∞ j=1 αj = k. (By contrast, a composition of k ∈N is a finite sequence (αj)n j=1 of positive integers such that Pn j=1 αj = k.) We let WComp(k) denote the set of weak compositions of k and let WComp = S k∈N WComp(k) denote the set of all weak compositions. We denote the unique composition (0, 0, . . . ) of 0 by ∅. For α ∈WComp, we define |α| = ∞ X j=1 αj.
For k, n ∈N, we define WCompn(k) := {α ∈WComp(k) : αj = 0 for j > n} and WCompn := [ k∈N WCompn(k).
As for partitions, we will often view elements α of WCompn as sequences (α1, . . . , αn) of length n. We have WCompm(k) ⊆WCompn(k) and WCompm ⊆WCompn for m ≤n and WComp(k) = [ n∈N WCompn(k) and WComp = [ n∈N WCompn.
Note that every partition is a weak composition but not the other way around since we require partitions to be weakly decreasing sequences. For example (2, 0, 1, 1) is a weak composition of 5, but is not a partition of 5.
The symmetric group Sn acts on the set WCompn(k) as follows. For α = (α1, . . . , αn) and π ∈Sn, define (1.4) απ := (απ(1), απ(2), . . . , απ(n)).
Partitions and compositions 9 This defines a right action of Sn on WCompn(k); see Exercise 1.2.1. For α ∈WCompn(k), we define (1.5) αSn := {απ : π ∈Sn}.
(We use the notation αSn instead of Snα since we have a right action of Sn.) For every α ∈WCompn(k), there exists a permutation π ∈Sn such that απ ∈Parn(k).
(This permutation π is unique if and only if α has no repeated terms.) It follows that [ α∈Parn(k) αSn = {απ : α ∈Parn(k), π ∈Sn} = WCompn(k).
Similarly, the set S∞acts on the set WComp(k).
Precisely, for α = (α1, α2, . . . ) ∈ WComp(k) and π ∈S∞, we define απ := (απ(1), απ(2), . . . ) and αS∞= {απ : π ∈S∞}.
For every α ∈WComp(k), there exists a (not unique!) permutation π ∈S∞such that απ ∈Par(k) and we have a disjoint union (1.6) G λ∈Par(k) λS∞= {λπ : λ ∈Par(k), π ∈S∞} = WComp(k).
The dominance partial ordering (1.3) can be naturally extended to a partial order on weak compositions. (Note that we will still reserve the term dominance partial ordering to refer to partitions only.) Namely, for α, β ∈WComp, we define (1.7) α ≤β ⇐ ⇒α1 + · · · + αi ≤β1 + · · · + βi for all i ∈N.
Lemma 1.2.4. Suppose α ∈WComp. Then α ∈Par ⇐ ⇒απ ≤α for all π ∈S∞.
Proof. Suppose that α ∈Par. Then α1 ≥α2 ≥· · · . Suppose β = απ for some π ∈S∞. For i ∈N, α1 + · · · + αi is the sum of the i greatest elements of α, which are the same as the i greatest elements of β. Thus β1 + · · · + βi ≤α1 + · · · + αi.
Hence απ = β ≤α.
Conversely, suppose απ ≤α for all π ∈S∞. Then, for i ≥1, we have (α1, α2, . . . , αi−1, αi+1, αi, αi+2, αi+3, . . . ) ≤α.
It follows that, for i ≥1, α1 + · · · + αi−1 + αi+1 ≤α1 + · · · + αi−1 + αi and so αi+1 ≤αi. Hence α ∈Par.
10 Preliminaries Exercises.
1.2.1. If G is a group, let Gop denote the opposite group. Thus, Gop = {gop : g ∈G}, gophop = (hg)op, g, h ∈G.
A right action of a group G on a set X is a group homomorphism Gop →SX, where SX denotes the group of permutations of X. By contrast, a left action of G on X is a group homomorphism G →SX.
(a) Show that (1.4) defines a right action of Sn on WCompn(k). In other words, show that απ1π2 = (απ1)π2 for all α ∈WCompn(k) and π1, π2 ∈Sn.
(b) Show that, for n ≥3, (1.4) does not define a left action of Sn on WCompn(k).
(c) How could you modify (1.4) to give a natural left action of Sn on WCompn(k)?
1.2.2. Prove that, for k ≤5, the dominance partial ordering on Par(k) is a total ordering.
1.2.3. A partition λ is self-conjugate if λ′ = λ. Prove that, for k ∈N, the number of self-conjugate partitions of k is equal to the number of partitions of k into distinct, odd parts.
Hint: Think about Young diagrams.
1.2.4. For k ∈N, define the lexicographic ordering <lex on Par(k) as follows: For λ, µ ∈ Par(k), we write λ <lex µ if and only if there exists some i ∈N such that λj = µj for all j < i, and λi < µi.
This is a total order on Par(k). Prove that, for λ, µ ∈Par(k), λ < µ = ⇒λ 0. This is called the trivial gradation on R.
(b) The polynomial ring Z[x] is graded by degree. More precisely, we have a gradation Z[x] = L k∈N Z[x]k, where Z[x]k consists of the homogeneous polynomials of degree k, together with the zero polynomial.
(c) The polynomial ring Z[x, y] is also graded by total degree.
△ If R is a graded ring, then by the definition of direct sum, every element r ∈R can be written uniquely as a sum r = X i∈N ri, ri ∈Ri, where only finitely many of the ri are nonzero. The ri are called the homogeneous components of r.
Definition 1.3.4 (Graded subring). Suppose R is a graded ring. A graded subring of R is a subring S of R such that (1.9) S = M k∈N Sk, Sk = S ∩Rk.
Equivalently, S is a graded subring of R if it is a subring of R and the homogeneous compo-nents of all elements of S are again elements of S. If S is a graded subring of R, then it is itself a graded ring with gradation (1.9).
Example 1.3.5. Recall that, if R is a ring and X ⊆R, then the subring of R generated by X is the smallest subring of R containing X. Consider the case R = Z[x, y], graded by total degree. Then the subring generated by x + y, which consists of all polynomials in x + y with integer coefficients, is a graded subring of R, while the subring generated by x + y2 is not a graded subring of R. See Exercise 1.3.1.
△ Definition 1.3.6. Suppose R is a graded ring. A graded ideal of R is an ideal I of R such that I = M k∈N Ik, Ik = I ∩Rk.
Equivalently, I is a graded ideal of R if it is an ideal of R and the homogeneous components of all elements of I are again elements of I.
12 Preliminaries Example 1.3.7. Consider the ring Z[x, y], graded by total degree. Then the ideal I = xZ[x, y] generated by x is a graded ideal of Z[x, y]. Precisely, we have I = M k∈N Ik, Ik = I ∩Z[x, y]k = xZ[x, y]k−1.
However, the ideal (x + 1)Z[x, y] is not a graded ideal. See Exercise 1.3.2.
△ Definition 1.3.8 (Homomorphism of graded rings). Suppose R and S are graded rings. A homomorphism of graded rings from R to S is a ring homomorphism σ: R →S such that σ(Rk) ⊆Sk for all k ∈N.
If, in addition, σ is an isomorphism, then we say it is an isomorphism of graded rings. A homomorphism of graded rings σ: R →R is an endomorphism of graded rings; it is an automorphism of graded rings if it is also an isomorphism.
If σ: R →S is a homomorphism of graded rings, then we have its homogeneous compo-nents σk : Rk →Sk, k ∈N, obtained from the restriction of σ to Rk. These are homomorphisms of additive groups. We often write σ = P k∈N σk since σ(r) = X k∈N σk(rk), r ∈R.
(Here r = P k∈N rk is the decomposition of r into its homogeneous components.) This sum is always well defined since only finitely many terms are nonzero.
Lemma 1.3.9. If σ: R →S is a homomorphism of graded rings, then its kernel ker σ is a graded ideal of R.
Proof. We know from ring theory that ker σ is an ideal of R. It remains to show that the homogeneous components of ker σ are again elements of ker σ. Suppose r = P k∈N rk ∈ker σ.
Then we have 0 = σ(r) = X k∈N σ(rk).
Since σ is a homomorphism of graded rings, we have σ(rk) ∈Sk for all k ∈N. Since the sum S = L k∈N Sk is direct, this implies that σ(rk) = 0 for all k ∈N. In other words rk ∈ker σ for all k ∈N, as desired.
Example 1.3.10. Suppose R is a commutative ring and consider the polynomial ring R[x], graded by degree. For any a ∈R, we have the evaluation homomorphism R[x] →R, f(x) 7→f(a).
Since x has degree 1 and a has degree zero if a ̸= 0, the evaluation homomorphism is graded if and only if a = 0.
△ Graded rings 13 Proposition 1.3.11. If R is a graded ring and X is a set of graded ring automorphisms of R, then RX := {r ∈R : σ(r) = r for all σ ∈X} is a graded subring of R.
Proof. We leave it as an exercise (Exercise 1.3.3) to verify that RX is a subring of R. It remains to show that the homogenous components of elements of RX are again elements of RX. Suppose r ∈RX. Then, for σ ∈X, X k∈N rk = r = σ(r) = X k∈N σ(rk).
Since σ(rk) ∈Rk, it follows from the uniqueness of the decomposition of an element of R as a sum of homogeneous components that σ(rk) = rk. Hence rk ∈RX for all k ∈N, as desired.
Many of the important results on rings can be generalized to the setting of graded rings.
For example, the quotient of a graded ring by a graded ideal is naturally a graded ring (see Exercise 1.3.7) and one has graded versions of the isomorphism theorems for rings.
We conclude this section with a brief discussion of associative algebras, which are more general than rings. We fix a commutative ring k, which is often called the ground ring. A unital associative k-algebra R is a ring that is also a k-module in such a way that the two additions (the ring addition and the module addition) are the same operation and the scalar multiplication satisfies a(xy) = (ax)y = x(ay) for all a ∈k, x, y ∈R.
In these notes, the term k-algebra (or simply algebra, if the ground ring k is clear from the context) will always mean unital associative k-algebra. Since a Z-module is the same as an abelian group, a Z-algebra is simply a ring. This is the sense in which associative algebras generalize rings.
Examples 1.3.12.
(a) The ring of polynomials k[x] with coefficients in k is a k-algebra.
(b) For n ≥1, the ring of matrices Matn(k) with entries in k is a k-algebra.
(c) If G is a finite group, the group ring kG is a k-algebra.
△ A graded k-algebra is a k-algebra R with a direct sum decomposition (1.8) such that each Rk is a k-submodule of R, and such that RkRl ⊆Rk+l for all k, l ∈N. All of the results of this section on graded rings have natural analogues for k-algebras.
Examples 1.3.13. We can repeat Examples 1.3.3 with k-algebras in place of rings.
(a) Any k-algebra R can be given a gradation by letting R0 = R and Ri = 0 for i > 0.
This is called the trivial gradation on R.
(b) The k-algebras k[x] and k[x, y] are graded by degree.
△ 14 Preliminaries Exercises.
1.3.1. Consider the ring Z[x, y] graded by total degree.
(a) Show that the subring generated by x + y is a graded subring of R.
(b) Show that the subring generated by x + y2 is not a graded subring of R.
1.3.2. Prove that (x + 1)Z[x, y] is not a graded ideal of Z[x, y], where the gradation is given by total degree.
1.3.3. Suppose R is a ring and X is a set of ring automorphisms. Show that RX := {r ∈R : σ(r) = r for all σ ∈X} is a subring of R.
1.3.4. Prove that the multiplicative unit in a graded ring is homogeneous of degree zero.
1.3.5. Suppose u is a homogeneous invertible element in a graded ring R. Show that u has degree zero.
1.3.6. Suppose u is a homogeneous invertible element in a graded ring R. Show that the map ρu : R →R, ρu(r) = uru−1, r ∈R, is a graded ring automorphism of R.
1.3.7. Suppose that I is a graded ideal of a graded ring R. Show that L k∈N Rk/Ik has a natural induced structure of a graded ring.
1.4 Formal power series In this section we briefly introduce the concept of formal power series. We will use this formalism in the sequel to concisely encode relations between various sequences of symmetric functions. Throughout this section R denotes a commutative ring.
The ring of formal power series with coefficients in R is RJtK := ( ∞ X n=0 antn : an ∈R for all n ∈N ) .
More precisely, elements of RJtK are functions f : N →R, where we write such a function as P∞ n=0 f(n)tn. The unit element of R is 1. (Here we are writing 1 to denote the formal Formal power series 15 power series P antn, where a0 = 1, the multiplicative identity of R, and an = 0 for r > 0.) Addition and multiplication are given by (1.10) ∞ X n=0 antn + ∞ X n=0 bntn = ∞ X n=0 (an + bn)tn, ∞ X n=0 antn ! ∞ X n=0 bntn !
= ∞ X n=0 n X i=0 aibn−i !
tn.
Note that, for the multiplication of two formal power series, the coefficient of each tn involves a finite sum and hence is well defined.
The use of the word formal in our terminology comes from the fact that we do not worry ourselves with issues of convergence. The formal power series P∞ n=0 antn is sometimes called the generating function of the sequence (an)∞ n=0. Note that we have a natural inclusion of rings R[t] ⊆RJtK.
We denote by R× the group of units of R: R× = {a ∈R : ab = 1 for some b ∈R}.
The following proposition describes the units of RJtK.
Proposition 1.4.1. We have RJtK× = ( ∞ X n=0 antn ∈RJtK : a0 ∈R× ) .
Proof. Let a(t) = P∞ n=0 antn ∈RJtK. If a(t) ∈RJtK×, then there exists P∞ n=0 bntn ∈RJtK such that 1 = ∞ X n=0 antn ∈RJtK ! ∞ X n=0 bntn ∈RJtK !
= a0b0 + ∞ X n=1 n X i=0 aibn−i !
tn.
It follows that a0b0 = 1, and hence a0 ∈R×.
Conversely, suppose that a0 ∈R×. Then we define a sequence (bn)∞ n=0 recursively by b0 = a−1 0 , bn = −a−1 0 n X i=1 aibn−i, n ≥1.
It is then straightforward to verify that ∞ X n=0 antn ! ∞ X n=0 bntn !
= 1.
Example 1.4.2. Suppose a ∈R. Then, by Proposition 1.4.1, 1 −at ∈RJtK×. We leave it as an exercise (Exercise 1.4.3) to show that △ (1.11) (1 −at)−1 = ∞ X n=0 antn.
16 Preliminaries For n ∈N, let p(n) denote the number of partitions of n. No simple explicit formula for p(n) is known. However, the sequence (p(n))∞ n=0 satisfies some recurrence relations that can be used to efficiently compute p(n). As the following result shows, these recurrence relations can be encoded as a product expression for the generating function of the sequence.
Proposition 1.4.3. If p(n; k) denotes the number of partitions λ ⊢n with largest part λ1 ≤k, then (1.12) ∞ X n=0 p(n; k)tn = k Y j=1 1 1 −tj .
Proof. We prove the result by induction on k. When k = 0, both sides of (1.12) are equal to 1. Now, for k ≥1, we have p(n; k) = p(n; k −1) + p(n −k; k −1) + p(n −2k; k −1) + · · · , where p(n −lk; k −1) counts the number of partitions λ ⊢n with mk(λ) = l. Thus p(n; k)tn = p(n; k −1)tn + tkp(n −k; k −1)tn−k + t2kp(n −2k; k −1)tn−2k + · · · .
Summing over n gives ∞ X n=0 p(n; k)tn = 1 + tk + t2k + · · · ∞ X n=0 p(n; k −1)tn = k Y j=1 1 1 −tj , where we used the induction hypothesis in the last step.
Taking the limit at k →∞then gives a generating function for the sequence (p(n))∞ n=0.
Corollary 1.4.4. We have (1.13) ∞ X n=0 p(n)tn = ∞ Y j=1 1 1 −tj .
Proof. First note that the right-hand side of (1.13) a well-defined formal power series, even though it involves an infinite product. This is because, to compute the coefficient of tk in Q∞ j=1 1 1−tj , only the factors 1 1−tj for j ≤k contribute. Then the result follows from (1.12) and the fact that p(n; k) = p(n) for k ≥n.
Our discussion of formal power series can be generalized to any number of indeterminates, even infinitely many. Since the case of countably many indeterminates x1, x2, . . . will be important for us, let us discuss it here. It is useful to use multi-index notation for monomials.
For α = (α1, α2, . . . ) ∈WComp, we define the monomial (1.14) xα := ∞ Y i=1 xαi i .
Formal power series 17 Note that since all but finitely many of the exponents αi are equal to zero, this is actually a finite product. The degree of this monomial is |α|. For example x(5,3,0,4) = x5 1x3 2x4 4 has degree |(5, 3, 0, 4)| = 12.
Define (1.15) RJx1, x2, . . .K := ( X α∈WComp aαxα : aα ∈R for all α ∈WComp ) .
This is a ring with multiplication given by (1.16) X α∈WComp aαxα ! X β∈WComp bβxβ !
= X α,β∈WComp aαbβxα+β, where α + β = (α1 + β1, α2 + β2, . . . ). The infinite sum on the right-hand side of (1.16) is well defined since, for any γ ∈WComp, there exists only finitely many pairs (α, β) such that α + β = γ, and so the coefficient of xγ is finite.
For k ∈N, define (1.17) RJx1, x2, . . .Kk := X α∈WComp(k) aαxα : aα ∈R for all α ∈WComp(k) .
This is the R-submodule of RJx1, x2, . . .K consisting of zero and formal power series of degree k. Note that (1.18) RJx1, x2, . . .K ̸= M k∈N RJx1, x2, . . .Kk, since the right-hand side consists of formal power series of bounded total degree. For example, ∞ X n=0 xn 1 ∈RJx1, x2, . . .K but ∞ X n=0 xn 1 / ∈ M k∈N RJx1, x2, . . .Kk.
Exercises.
Throughout these exercises, R denotes a commutative ring.
1.4.1. The ring of formal Laurent series with coefficients in R is R( (t) ) = ( ∞ X n=N antn : N ∈Z, an ∈R for all n ∈N ) .
Note that the exponents n appearing in any given element of R( (t) ) are bounded below (by some N ∈Z), but we allow arbitrarily low bounds.
18 Preliminaries (a) Show that R( (t) ) is a ring with the usual addition and multiplication (that is, as in (1.10), but with lower bound N in the sums).
(b) Show that ( ∞ X n=−∞ antn : an ∈R for all n ∈Z ) is not a ring under the usual addition and multiplication.
1.4.2. Is the ring RJtK graded by degree?
1.4.3. Prove (1.11).
1.4.4. Recall that the Fibonacci sequence (fn)∞ n=0 is defined recursively by f0 = f1 = 1, fn = fn−1 + fn−2 for n ≥2.
(a) Use the recursion relation to find a generating function for the Fibonacci sequence.
(b) Show that, for n ∈N, fn = an −bn √ 5 , where a = 1 + √ 5 2 , b = 1 − √ 5 2 .
The next four exercises aim to justify some manipulations we will do with power series in the sequel. For the remainder of the exercises we work over a field k and fix a formal power series A(t) = P n≥0 antn ∈kJtK.
1.4.5. If a0 = 1, show that A(t)−1 = ∞ X n=0 (1 −A(t))n.
Hint: Write A(t) = 1 −(1 −A(t)) and compare to (1.11).
1.4.6. Recall the Taylor series for the natural logarithm: log(1 + x) = P n≥1(−1)n−1 xn n . If a0 = 1 (so that the constant term of A(t) is 1), we define log A(t) := ∞ X n=1 (−1)n−1(A(t) −1)n n .
Prove that this is indeed a well-defined formal power series.
1.4.7. Define the formal derivative of A(t) to be A′(t) = d dtA(t) := X n≥1 nantn−1.
If a0 = 1, prove that d dt log A(t) = A′(t)/A(t).
Formal power series 19 1.4.8.
(a) Suppose a0 = 0. Prove that exp A(t) := X n≥0 1 n!A(t)n is a well-defined formal power series and that its constant term is 1. Furthermore, show that exp X n≥1 antn = Y n≥1 exp (antn) .
(b) Show that, if a0 = 0, then log exp A(t) = A(t).
(c) Show that if a0 = 1, then exp log A(t) = A(t).
Chapter 2 The ring of symmetric functions In this chapter we introduce the main object of study in the course: the ring of symmetric functions. We will give two descriptions of this ring, one as a certain type of limit and another more direct definition. Both definitions will be useful in what follows. After defining the ring of symmetric functions, we will introduce several important bases and generating sets for this ring.
2.1 Symmetric polynomials Let Poln := Z[x1, . . . , xn] denote the ring of polynomials in n indeterminates with integer coefficients. This is a unique factorization domain. It is also a graded ring with gradation Poln = M k≥0 Polk n, where Polk n consists of homogeneous polynomials of degree k, together with the zero poly-nomial. (Note that we use superscripts to indicate the grading, instead of subscripts as in Section 1.3.) It is useful to use multi-index notation for monomials. For α ∈WCompn, we define the monomial xα := xα1 1 xα2 2 · · · xαn n .
The degree of this monomial is |α| := n X i=1 αi.
The monomials xα, α ∈WCompn, form a Z-basis for Poln. Similarly, the monomials xα, α ∈WCompn(k), form a basis for Polk n.
For π ∈Sn, the map (2.1) f 7→πf := f(xπ(1), xπ(2), . . . , xπ(n)), f ∈Poln, 20 Symmetric polynomials 21 is a ring automorphism of Poln. This defines a left action of Sn on Poln; see Exercise 2.1.1.
We say that a polynomial f ∈Poln is symmetric if it is invariant under the action of Sn. In other words, f is symmetric if πf = f for all π ∈Sn.
The ring of symmetric polynomials is the subring of Poln consisting of symmetric polynomi-als: Symn := PolSn n := {f ∈Poln such that πf = f for all π ∈Sn}.
By Proposition 1.3.11, Symn is a graded ring: (2.2) Symn := M k∈N Symk n, Symk n := (Polk n)Sn.
Let λ ∈Parn and recall the definition of λSn from (1.5). The corresponding monomial symmetric polynomial is (2.3) mλ(x1, . . . , xn) := X α∈λSn xα.
Remark 2.1.1. Note that (2.3) is not equal to P π∈Sn xαπ in general since there may be π1, π2 ∈Sn with απ1 = απ2, and we only want to count these terms once. For example m(3,2,2)(x1, x2, x3, x4) = x3 1x2 2x2 3 + x3 1x2 2x2 4 + x3 1x2 3x2 4 + x2 1x3 2x2 3 + x2 1x3 2x2 4 + x3 2x2 3x2 4 + x2 1x2 2x3 3 + x2 1x3 3x2 4 + x2 2x3 3x2 4 + x2 1x2 2x3 4 + x2 1x2 3x3 4 + x2 2x2 3x3 4 but X π∈S4 x(3,2,2)π = 2m(3,2,2)(x1, x2, x3, x4).
△ We have mλ(x1, . . . , xn) ∈Symk n for λ ∈Parn(k).
Define mλ(x1, . . . , xn) = 0 for ℓ(λ) > n.
Proposition 2.1.2. The monomial symmetric polynomials (2.4) mλ(x1, . . . , xn), λ ∈Parn(k), form a Z-basis of Symk n.
Proof. We must show that the elements (2.4) are linearly independent and span Symk n. To see that they are linearly independent, note that, for any monomial xα, there is only one partition λ that is a permutation of α, and xα is a term in mλ(x1, . . . , xn). It follows that, 22 The ring of symmetric functions for partitions λ ̸= µ, the polynomials mλ(x1, . . . , xn) and mµ(x1, . . . , xn) have no terms in common. It follows that X λ∈Parn(k) aλmλ(x1, . . . , xn) = 0 ⇐ ⇒(aλ = 0 for all λ).
Hence the elements (2.4) are linearly independent.
Next we prove that the elements (2.4) span Symk n. Suppose f ∈Symk n. We prove that f lies in the span of (2.4) by induction on the number of nonzero terms in f. The base case f = 0 is clear, so we suppose f ̸= 0. Choose a term axα, a ∈Z \ {0}, of f. Since f is symmetric, the terms axβ also appear in f for all permutations β of α. Let λ be the unique permutation of α that is a partition. Then f −amλ ∈Symk n has fewer terms than f. By the induction hypothesis, f −amλ can be written as a Z-linear combination of the elements (2.4); hence the same is true for f.
Corollary 2.1.3.
(a) The monomial symmetric polynomials mλ(x1, . . . , xn), λ ∈Parn, form a Z-basis of Symn.
(b) If n ≥k, then monomial symmetric polynomials mλ(x1, . . . , xn), λ ∈Par(k), form a Z-basis of Symk n. In particular, Symk n is a free Z-module of rank p(k), the number of partitions of k.
Proof. Part (a) follows from Proposition 2.1.2 and (2.2). Part (b) follows from Proposi-tion 2.1.2 and the fact that, when n ≥k, we have Parn(k) = Par(k) since ℓ(λ) ≤n for all partitions λ of k.
Exercises.
2.1.1. For a group G, recall the definition of oppose group Gop from Exercise 1.2.1. A right action of a group G on a ring R is a group homomorphism Gop →Aut R, where Aut R denotes the group of ring automorphisms of R. By contrast, a left action of G on R is a group homomorphism G →Aut R.
(a) Show that (2.1) defines a left action of Sn on Poln. In other words, show that π1(π2f) = (π1π2)f for all f ∈Poln and π1, π2 ∈Sn.
Symmetric functions 23 (b) Show that, for n ≥3, (2.1) does not define a right action of Sn on Poln.
(c) How could you modify (2.1) to give a natural right action of Sn on Poln?
(Compare this exercise to Exercise 1.2.1.) 2.1.2. Show that, for α ∈WCompn and π ∈Sn, we have (2.5) π (xα) = xαπ−1 .
The next three problems use the following notation. For a subset X ⊆Sn, define PolX n = {f ∈Poln : πf = f for all π ∈X}.
For a subset P ⊆Poln, define G(P) = {π ∈Sn : πf = f for all f ∈P}.
2.1.3. Prove that G(P) is a subgroup of Sn for all subsets P ⊆Poln.
2.1.4. Is it true that PolG(P) n = P for all subrings P ⊆Poln, n ∈N?
Prove or give a counterexample.
2.1.5. It is true that G(PolH n ) = H for all subgroups H ⊆Sn, n ∈N? Prove or give a counterexample.
2.2 Symmetric functions It turns out that, in many ways, the symmetric polynomials behaves better in the limit where we have an infinite number of indeterminates x1, x2, . . . . This limit, which we must be careful to define precisely, leads to the concept of a symmetric function.
For m ≥n ≥0, consider the ring homomorphism Polm →Poln, f(x1, . . . , xm) 7→f(x1, . . . , xn, 0, . . . , 0).
Restriction to Symm gives a ring homomorphism (2.6) ρm,n : Symm →Symn.
Note that, for l ≥m ≥n ≥0, we have ρm,n ◦ρl,m = ρl,n.
Proposition 2.2.1. For m ≥n ≥0 and λ ∈Par, we have ρm,n (mλ(x1, . . . , xm)) = mλ(x1, . . . , xn).
In particular, ρm,n (mλ(x1, . . . , xm)) = 0 if ℓ(λ) > n.
24 The ring of symmetric functions Proof. Let A be the (possibly empty) set of permutations α = (α1, . . . , αm) of λ = (λ1, . . . , λm) such that αi = 0 for all n < i ≤m. Let B be the set of permutations α = (α1, . . . , αm) of λ = (λ1, . . . , λm) such that αi ̸= 0 for some n < i ≤m. Then mλ(x1, . . . , xm) = X α∈A xα + X α∈B xα, and so ρm,n (mλ(x1, . . . , xm)) = X α∈A ρm,n(xα) + X α∈B ρm,n(xα) = X α∈A xα = mλ(x1, . . . , xn).
Note that, if ℓ(λ) > n, then A = ∅, and so ρm,n (mλ(x1, . . . , xm)) = 0.
Corollary 2.2.2. For all m ≥n ≥0, the ring homomorphism ρm,n is surjective.
Proof. This follows from Propositions 2.1.2 and 2.2.1.
Restricting ρm,n to Symk n, we obtain its homogeneous components (see Section 1.3) ρk m,n : Symk m ↠Symk n, k ≥0, m ≥n.
Lemma 2.2.3. The map ρk m,n is bijective if m ≥n ≥k.
Proof. The proof of this lemma is left as Exercise 2.2.1.
We now take the inverse limit (2.7) Symk = lim ← − n Symk n of the Z-modules Symk n with respect to the homomorphisms ρk m,n. By definition, (2.8) Symk := ( f = (fn)∞ n=0 ∈ ∞ Y n=0 Symk n : ρk m,n(fm) = fn for all m ≥n ) .
For k, n ≥0, consider the projection ρk n : Symk →Symk n, ρk n(f) = fn.
Proposition 2.2.4. The projection ρk n is an isomorphism for all n ≥k.
Proof. By Lemma 2.2.3, any element f ∈Symk is uniquely determined by fk since we must have fn = ρk k,n(fk) for n ≤k and fn = (ρk n,k)−1(fk) for n > k.
Symmetric functions 25 By Proposition 2.2.1, for any λ ∈Par(k), the sequence of monomial symmetric poly-nomials (mλ(x1, . . . , xn))∞ n=0 is an element of Symk, which we call the monomial symmetric function corresponding to λ and denote by mλ. We think of it as a formal sum (2.9) mλ = X α∈λS∞ xα ∈Symk.
Note that, in contrast with (2.3), the above sum is infinite as long as λ ̸= ∅. This is because we are considering permutations of the infinite sequence (λ1, λ2, . . . ) rather than the finite sequence (λ1, . . . , λn). For example, m(2,1) = X i̸=j x2 i xj = x2 1x2 + x1x2 2 + x1x2 3 + x2 1x3 + x2x2 3 + x2 2x3 + · · · .
However, when we apply ρk n for some n ∈N, only finitely many terms survive. More precisely, for λ ∈Par(k), ρk n(mλ) = mλ(x1, . . . , xn) for all n ≥0.
Now define (2.10) Sym := ∞ M k=0 Symk.
Thus, Sym is the free Z-module generated by the monomial symmetric functions mλ for all partitions λ. We have surjective Z-module homomorphisms (2.11) ρn := ∞ M k=0 ρk n : Sym →Symn, n ≥0.
If we define, for n, k ≥0, (2.12) Sym≤k := k M l=0 Syml and Sym≤k n := k M l=0 Syml n, then the restriction of ρn to Sym≤k yields a homomorphism of Z-modules ρ≤k n : Sym≤k →Sym≤k n which is an isomorphism when k ≤n.
Proposition 2.2.5.
(a) The monomial symmetric functions mλ, λ ∈Par(k), form a basis of Symk. In particular, Symk is a free Z-module of rank p(k), the number of partitions of k.
(b) The monomial symmetric functions mλ, λ ∈Par, form a basis of Sym.
Proof. Part (a) follows from Proposition 2.2.4 and Corollary 2.1.3(a). Then part (b) follows from (2.10).
26 The ring of symmetric functions So far we have only described Sym as a Z-module. However, as the next result shows, it is naturally a ring.
Proposition 2.2.6. There is a unique structure of a graded ring on Sym such that the ρn are ring homomorphisms.
Proof. We first prove uniqueness. Suppose we have defined a multiplication on Sym such that the ρn are ring homomorphisms. Let f, g ∈Sym. Then there exists some k ≥0 such that f, g ∈Sym≤k and we must have ρ2k(fg) := ρ2k(f)ρ2k(g) ∈Sym≤2k.
Now, since ρ2k is an isomorphism when restricted to Sym≤2k, we must have (2.13) fg = ρ≤2k 2k −1 (ρ2k(f)ρ2k(g)) .
This completes the proof of uniqueness.
Existence is then seen by taking (2.13) as the definition of the multiplication in Sym. We leave it as exercise Exercise 2.2.2 to verify that the structure we have defined satisfies the axioms of a ring.
The graded ring Sym is called the ring of symmetric functions. The term ‘function’ is used here in a formal sense, to distinguish this ring from the ring of symmetric polynomials.
We do not view elements of Sym as actual functions.
It was important that we first took the inverse limit in (2.7) of the graded pieces Symk n and then took the direct sum in (2.10). This construction ensures that elements of Sym are infinite sums of monomials with bounded degree. What we are doing is taking the inverse limit of the Symn in the category of graded rings. (General inverse limits in a category are defined in terms of a universal property.) If we had instead taken the inverse limit of the Symn directly (the inverse limit in the category of rings) our ring would include elements such as the infinite product Q∞ i=1(1 + xi).
To illustrate this difference further, let us discuss another approach to the definition of the ring of symmetric functions. Recall the definitions of ZJx1, x2, . . .K and ZJx1, x2, . . .Kk from (1.15) and (1.17). The group S∞acts on each ZJx1, x2, . . .Kk in the usual way: πf(x1, x2, . . . ) := f(xπ(1), xπ(2), . . . ), f(x1, x2, . . . ) ∈ZJx1, x2, . . .Kk, π ∈S∞.
Then Symk = (ZJx1, x2, . . .Kk)S∞, Sym = M k∈N (ZJx1, x2, . . .Kk)S∞.
Note that (2.14) Sym ̸= ZJx1, x2, . . .KS∞.
(Compare to (1.18).) For example, ∞ Y i=1 (1 + xi) ∈ZJx1, x2, . . .KS∞ but ∞ Y i=1 (1 + xi) / ∈Sym.
Tableaux 27 Warning 2.2.7. The definition of the ring of symmetric functions given in [Egg19, Def. 1.19] seems to be incorrect. The ring defined there is ZJx1, x2, . . .KS∞. See Warning 2.4.8.
Remark 2.2.8. We have worked in this section over the coefficient ring Z. We could have instead worked over any other commutative ring. In particular, we will denote by SymQ the ring of symmetric functions over Q, which will be important for us later on. We can also obtain SymQ by an extension of scalars: SymQ ∼ = Q ⊗Z Sym. (Students not familiar with tensor products can find the basics in [Sav, Ch. 3].) △ Exercises.
2.2.1. Prove Lemma 2.2.3. Furthermore, prove that ρk m,n is not bijective if m > n and k > n.
2.2.2. Show that the multiplication on Sym defined in Proposition 2.2.6 satisfies the defining axioms for a ring.
2.2.3. For n ∈N, let Rn = Z[x]/xnZ[x] be the quotient of Z[x] by the ideal generated by xn. Then, for all m ≥n ≥0, we have a projection φm,n : Rm →Rn, φm,n(f + xmZ[x]) = f + xnZ[x], f ∈Z[x].
It is clear that φm,n ◦φl,m = φl,n for all l ≥m ≥n ≥0. Identify the inverse limit lim ← − n Rn := ( f = (fn)∞ n=0 ∈ ∞ Y n=0 Rn : φm,n(fm) = fn for all m ≥n ) with a ring we studied in Chapter 1.
2.2.4. Suppose k, l ∈N with k > l. Write m(k)m(l) as a linear combination of mλ, λ ∈Par.
2.2.5. Write m(4)m(3)m(2)m(1) as a linear combination of mλ, λ ∈Par.
2.3 Tableaux Before we continue our discussion of symmetric functions, we introduce a combinatorial tool that will prove useful in our discussions.
For a partition λ, a tableau (plural tableaux) of shape λ is a filling of a Young diagram of λ with positive integers. (These are also called Young tableaux.) For example (2.15) T1 = 5 1 3 2 1 1 2 3 , T2 = 1 3 4 5 1 2 8 6 , and T3 = 1 1 1 1 1 1 1 1 28 The ring of symmetric functions are tableaux of shape (4, 3, 1). We let shape(T) denote the shape of a tableau T. The weight wt(T) of a tableau T (also called its content) is the composition µ = (µ1, . . . , µr) where µi is the number of entries of T equal to i. For example, for the tableaux in (2.15), we have wt(T1) = (3, 2, 2, 0, 1), wt(T2) = (2, 1, 1, 1, 1, 1, 0, 1), wt(T3) = (8).
Note that | wt(T)| = | shape(T)|.
We say that a tableau is row strict if its entries are strictly increasing along each row.
For λ ∈Par, we let RowStrict(λ) denote the set of row-strict tableaux of shape λ, and we let RowStrict(λ, n) denote the set of row-strict tableaux of shape λ with entries chosen from the set {1, 2, . . . , n}. For example, for the tableaux of (2.15) and λ = (4, 3, 1), we have T1 / ∈RowStrict(λ), T2 ∈RowStrict(λ, 8), T2 ∈RowStrict(λ), T2 / ∈RowStrict(λ, 5), T3 / ∈RowStrict(λ).
We say that a tableau is row nondecreasing if its entries are weakly increasing along each row. For λ ∈Par, let RowWeak(λ) denote the set of row-nondecreasing tableaux of shape λ. We let RowWeak(λ, n) denote the set of such tableaux with entries chosen from the set {1, 2, . . . , n}. For example, for the tableaux of (2.15) and λ = (4, 3, 1), we have T1 / ∈RowWeak(λ), T2 ∈RowWeak(λ, 8), T3 ∈RowWeak(λ).
We say that a tableau is semistandard if it is row nondecreasing and its entries are strictly increasing down each column. For example (2.16) T4 = 1 1 2 4 7 2 3 3 6 and T5 = 1 3 4 5 7 2 6 9 8 are semistandard tableaux. For λ ∈Par, we let SST(λ) denote the set of semistandard tableau of shape λ and we let SST(λ, n) denote the set of such tableaux with entries chosen from the set {1, 2, . . . , n}.
We say that a tableau is standard if it is row strict, its entries are strictly increasing down each column, and it contains each integer 1, . . . , n exactly once, where n is the number of boxes in T. Thus, a standard tableau has weight wt(T) = (1n). For example, for the tableaux in (2.16), T5 is standard, but T4 is not. For λ ∈Par, we let ST(λ) denote the set of standard tableaux of shape λ.
If T is a tableau, we define the monomial (2.17) xT := Y j∈T xj = xwt(T) where the first product is over the entries in T. For example, for the tableaux of (2.15), we have xT1 = x3 1x2 2x2 3x5, xT2 = x2 1x2x3x4x5x6x8, and xT3 = x8 1.
We will use this convention to give explicit descriptions of important symmetric functions as sums of monomials xT over certain tableaux.
Elementary symmetric functions 29 Exercises.
2.3.1. Find all elements of SST((2, 2, 1), 3). Which are standard?
2.3.2. Find all elements of SST((2, 1), 3). Which are standard?
2.4 Elementary symmetric functions In this section we introduce our first important generating set for the ring of symmetric functions: the elementary symmetric functions. The fact that these give an algebraically independent set of generators for Sym is the fundamental theorem of symmetric functions (Theorem 2.4.6).
For r ∈N, the r-th elementary symmetric function er is the sum of all products of r distinct indeterminates xi. Thus, (2.18) e0 = 1, er := X 1≤i1<i2<···<ir xi1xi2 · · · xir = m(1r), r ≥1.
We adopt the convention that er = 0 for r < 0.
For r, n ∈N, the r-th elementary symmetric polynomial in n indeterminates is (2.19) er(x1, . . . , xn) = ρn(er) = X 1≤i1<i2<··· n.
Convention 2.4.1. From now on, whenever we define a symmetric function f ∈Sym, we adopt the convention that f(x1, . . . , xn) = ρn(f).
Lemma 2.4.2. The generating function for the elementary symmetric polynomials in n indeterminates is n X r=0 er(x1, . . . , xn)tr = n Y i=1 (1 + xit).
Proof. We prove the result by induction on n. When n = 0, the result is clear when we recall the convention that the product Qn i=1(1 + xit) = 1 when n = 0. Now suppose the result holds for some n ∈N. Then we have n+1 Y i=1 (1 + xit) = n X r=0 er(x1, . . . , xn)tr !
(1 + xn+1t) 30 The ring of symmetric functions = 1 + n+1 X r=1 (er(x1, . . . , xn) + er−1(x1, . . . , xn)xn+1) tr = n+1 X r=0 er(x1, . . . , xn+1)tr, where we have use the fact that, for r ≥1, er−1(x1, . . . , xn)xn+1 + er(x1, . . . , xn) = X 1≤i1<i2<···<ir=n+1 xi1xi2 · · · xir + X 1≤i1<i2<···<ir≤n xi1xi2 · · · xir = er(x1, . . . , xn+1).
It follows from Lemma 2.4.2 that the generating function for the elementary symmetric functions is (2.21) E(t) := ∞ X r=0 ertr = ∞ Y i=1 (1 + xit).
Remark 2.4.3. Let us take a moment to discuss how to interpret the infinite product occur-ring on the right-hand side of (2.21). We want to interpret it as an element of SymJtK. So it should be a formal power series with coefficients in Sym. Elements of Sym are sequences f = (fn)∞ n=0 such that ρm,n(fm) = fn for all m ≥n ≥0. For each r ∈N, the coefficient of tr in Q∞ i=1(1+xit) is the symmetric function f = (fn)∞ n=0 where fn is the coefficient of tr in the finite product Qn i=1(1 + xit). By Lemma 2.4.2, we have fn = er(x1, . . . , xn), and so f = er.
This justifies the equality (2.21).
△ For a partition λ = (λ1, λ2, . . . ), define (2.22) eλ := eλ1eλ2 · · · .
Note that this is a finite product since λ has finitely many nonzero terms. Our next goal is to show that the eλ, λ ∈Par, form a basis of Sym. To do this, we work out how to express the eλ in terms of the monomial symmetric functions, which we know form a basis of Sym by Proposition 2.2.5. Recall the definition of the conjugate partition (1.2) and the dominance ordering (1.3).
Proposition 2.4.4. Let λ ∈Par(k) and let λ′ be the conjugate partition. Then (2.23) eλ = mλ′ + X µ<λ′ Mλ,µ(e, m)mµ.
where, for µ ∈Par(k), µ < λ′, Mλ,µ(e, m) is the number of row-strict Young tableaux of shape λ and weight µ.
Proof. Let λ = (λ1, . . . , λk). We have eλ = eλ1eλ2 · · · eλk, eλj = X ij,1<ij,2<···<ij,λj xij,1xij,2 · · · xij,λj , 1 ≤j ≤k.
Elementary symmetric functions 31 Fix a particular choice of the indices ij,1 < ij,2 < · · · < ij,λj, 1 ≤j ≤k, and consider the corresponding term (2.24) xα = k Y j=1 xij,1xij,2 · · · xij,λj in the product eλ1eλ2 · · · eλk. For each j ∈{1, . . . , k}, write the indices ij,1 < · · · < ij,λj in the j-th row of the Young diagram of λ: (2.25) i1,1 i1,2 · · · · · · · · · i1,λ1 i2,1 i2,2 · · · · · · · · · i2,λ2 .
.
.
.
.
.
ij,1 · · · ij,λj .
.
.
.
.
.
iλ′ 2,2 .
.
.
iλ′ 1,1 Since the entries are strictly increasing along each row, all entries ≤r must occur in the leftmost r columns of the diagram. Hence the total number of entries ≤r is less than or equal to the total number of boxes in the first r columns. Since αr is the number of entries equal to r, this implies that α1 + · · · + αr ≤λ′ 1 + · · · + λ′ r.
Thus we have α ≤λ′.
Now, since α ≤λ′ for all terms (2.24) and eλ is S∞invariant, it follows from Lemma 1.2.4 that only mµ for µ ≤λ′ appear when we write eλ in the basis of monomial symmetric functions. Thus eλ = X µ≤λ′ Mλ,µ(e, m)mµ, where Mλ,µ(e, m) is the number of row-strict Young tableaux of shape λ and weight µ. If µ = λ′, then the only such filling is the one in which each entry in (2.25) is equal to its column number, that is, ij,l = l for all j, l. Thus Mλ,λ′(e, m) = 1.
The proof of Proposition 2.4.4 gives us an explicit combinatorial description of the ele-mentary symmetric functions in terms of Young tableaux.
32 The ring of symmetric functions Proposition 2.4.5. For any partition λ and n ≥1, we have (2.26) eλ = X T∈RowStrict(λ) xT and eλ(x1, . . . , xn) = X T∈RowStrict(λ,n) xT.
Proof. The first equation follows from the proof of Proposition 2.4.4. Since eλ(x1, . . . , xn) = ρn(eλ), the second equation then follows from the fact that ρn(xT) = 0 if T has any entries greater than n.
Recall that elements of a commutative ring R are algebraically independent over a subring S ⊆R if the elements do not satisfy any polynomial equation with coefficients in S.
Theorem 2.4.6 (Fundamental theorem of symmetric functions). The elementary symmetric functions eλ, λ ∈Par, form a basis of Sym. Furthermore, Sym = Z[e1, e2, . . . ] and the er, r > 0, are algebraically independent over Z.
Proof. For the first statement of the theorem, it suffices to prove that, for k ∈N, the eλ, λ ∈Par(k), form a basis of Symk. Fix a total order on the basis mλ, λ ∈Par(k), of Symk that refines the dominance ordering on Symk. (For instance, we can take the lexicographic ordering; see Exercise 1.2.4.) Then Proposition 2.4.4 implies that matrix expressing the eλ′, λ ∈Par(k), in terms of this ordered basis is triangular with ones on the diagonal, and hence invertible. It follows every monomial symmetric function can be written as a Z-linear combination of elementary symmetric functions. Thus the elementary symmetric functions eλ, λ ∈Par, form a basis of Sym.
This implies that every element of Sym is uniquely expressible as a polynomial in the er, proving the second statement of the theorem.
Corollary 2.4.7 (Fundamental theorem of symmetric polynomials). For n ∈N, the eλ(x1, . . . , xn), λ ∈Parn, form a basis of Symn. Furthermore, Symn = Z[e1(x1, . . . , xn), . . . , en(x1, . . . , xn)] and the elementary symmetric polynomials e1(x1, . . . , xn), . . . , en(x1, . . . , xn) are algebraically independent over Z.
Proof. It follows from Theorem 2.4.6 and (2.20) that Symn is generated by the elementary symmetric polynomials e1(x1, . . . , xn), . . . , en(x1, . . . , xn). By Proposition 2.1.2, the rank of Symk n as a Z-module is the cardinality of Parn(k). Thus, the eλ(x1, . . . , xn), λ ∈Parn(k), must be linearly independent. Hence e1(x1, . . . , xn), . . . , en(x1, . . . , xn) are algebraically in-dependent over Z.
Theorem 2.4.6 implies that the ring Sym of symmetric functions is isomorphic, as a ring, to a polynomial ring in infinitely many indeterminates. You might be tempted to think that this implies Sym is not so interesting! However, it is the relationship between various elements of Sym (for example, different bases) that makes the theory of symmetric functions so rich. For example, in the next section will introduce another set of generators for Sym, and then study their relation to the elementary symmetric functions.
Complete homogeneous symmetric functions 33 Warning 2.4.8. It is crucial for Theorem 2.4.6 that we defined symmetric functions as we did, instead of defining them to be elements of ZJx1, x2, . . .KS∞. The elementary symmetric functions e1, e2, . . . do not generate ZJx1, x2, . . .KS∞as a ring; see Exercise 2.4.1. Thus, the statement at the top of [Egg19, p. 38] is incorrect (since the definition of symmetric functions in that reference is incorrect); see Warning 2.2.7.
Exercises.
2.4.1. Prove that the elementary symmetric functions e1, e2, . . . do not generate ZJx1, x2, . . .KSn as a ring. Hint: Find an element of ZJx1, x2, . . .KSn that cannot be written as a polynomial in the e1, e2, . . . .
2.4.2. Write e(2,2) and e(2,1,1) as linear combinations of monomial symmetric functions.
2.4.3. Prove that, for all 0 ≤k ≤n, we have ek(1, . . . , 1 | {z } n entries ) = n k .
Here the notation ek(1, . . . , 1) means evaluating the elementary symmetric polynomial ek(x1, . . . , xn) at (x1, . . . , xn) = (1, . . . , 1).
2.4.4. Find and prove a formula for Mλ,(n)(e, m), where λ ⊢n.
2.4.5. Find and prove a formula for Mλ,(1n)(e, m), where λ ⊢n.
2.4.6. Find and prove a formula for Mλ,(n−1,1)(e, m), where λ ⊢n.
2.5 Complete homogeneous symmetric functions For r ∈N, the r-th complete homogeneous symmetric function hr is the sum of all monomials of total degree r in the indeterminates x1, x2, . . . , so that (2.27) h0 = 1, hr := X 1≤i1≤i2≤···≤ir xi1xi2 · · · xir = X λ∈Par(r) mλ, r ≥1.
Note that h1 = e1.
We adopt the convention that hr = 0 for r < 0.
34 The ring of symmetric functions As usual, for n ∈N, we then define the r-th complete homogeneous symmetric polynomial in n indeterminates to be hr(x1, . . . , xn) := ρn(hr) = X 1≤i1≤i2≤···≤ir≤n xi1xi2 · · · xir.
Proposition 2.5.1. The generating function for the complete homogeneous symmetric func-tions is (2.28) H(t) := X r≥0 hrtr = ∞ Y i=1 1 1 −xit.
Proof. The proof of this result is left as Exercise 2.5.1.
It follows from (2.21) and (2.28) that (2.29) H(t)E(−t) = 1.
Comparing coefficients of powers of t, this implies that (2.30) n X r=0 (−1)rerhn−r = 0 for all n ≥1.
(We see here the power of the approach of generating functions!) By Theorem 2.4.6, we may define a homomorphism of graded rings (2.31) ω: Sym →Sym, ω(er) = hr, r ∈N.
Recall that an endomorphism is an involution if its square is the identity map (equivalently, it is an automorphism and equal to its own inverse.) Proposition 2.5.2. We have (2.32) ω(hr) = er, for all r ∈N.
In particular, the map ω is an involution.
Proof. Once we prove the first statement, we have ω2(en) = en for all n ∈N, and hence the second follows statement follows from Theorem 2.4.6.
We prove that ω(hn) = en for all n ∈N by induction on n. Since ω is a ring homomor-phism, we have ω(h0) = ω(1) = 1 = e0, and so the result holds for n = 0. Now suppose that n ≥1 and ω(hr) = er for r < n. By (2.30), we have hn = − n X r=1 (−1)rerhn−r and en = (−1)n+1 n−1 X r=0 (−1)rerhn−r.
Complete homogeneous symmetric functions 35 Thus ω(hn) = − n X r=1 (−1)rω(er)ω(hn−r) = − n X r=1 (−1)rhren−r = − n−1 X r=0 (−1)n−rhn−rer = (−1)n+1 n−1 X r=0 (−1)rerhn−r = en, where, in the third equality, we replaced r by n −r.
Theorem 2.5.3. We have Sym = Z[h1, h2, . . . ] and the complete homogeneous symmetric functions hr, r > 0, are algebraically independent over Z.
Proof. By Proposition 2.5.2, ω is an automorphism. Thus the result follows from Theo-rem 2.4.6.
Remark 2.5.4. If we work with symmetric polynomials in n indeterminates, we still have a mapping ω: Symn →Symn, ω(er) = hr, 1 ≤r ≤n.
Applying ρn to (2.30) shows that n X r=0 (−1)rer(x1, . . . , xn)hn−r(x1, . . . , xn) = 0.
Thus, one can repeat the argument in the proofs of Proposition 2.5.2 and Theorem 2.5.3 to see that Symn = Z[h1(x1, . . . , xn), . . . , hn(x1, . . . , xn)], with h1, . . . , hn algebraically independent. However, even though er(x1, . . . , xn) = 0 for r > n, the complete homogeneous symmetric polynomials hn+1(x1, . . . , xn), hn+2(x1, . . . , xn), . . .
are nonzero. For example, h3(x1, x2) = x3 1 + x2 1x2 + x1x2 2 + x2 3 = 2h1(x1, x2)h2(x1, x2) −h1(x1, x2)3.
In particular, we do not have ω(er) = hr for r > n. This is example of how symmetric functions are often simpler than symmetric polynomials.
△ For λ = (λ1, λ2, . . . ) ∈Par, we define (2.33) hλ = hλ1hλ2 · · · .
By Theorem 2.5.3, the hλ, λ ∈Par, form a Z-basis of Sym. Define the forgotten symmetric functions (2.34) fλ = ω(mλ), λ ∈Par.
36 The ring of symmetric functions The name “forgotten” comes from the fact that these symmetric functions have no simple direct description and thus tend not to appear much in applications. We now have four bases of Sym: {eλ : λ ∈Par}, {hλ : λ ∈Par}, {mλ : λ ∈Par}, {fλ : λ ∈Par}.
The automorphism ω interchanges the first two bases and the last two bases.
It follows from Proposition 2.2.5 that we can write each hλ in terms of the monomial symmetric functions. The next result gives an explicit formula.
Proposition 2.5.5. For λ ∈Par(k), we have (2.35) hλ = X µ⊢|λ| Mλ,µ(h, m)mµ, where, for µ ⊢|λ|, Mλ,µ(h, m) is the number of row-nondecreasing Young tableaux of shape λ and weight µ.
Proof. The proof of this proposition is left as Exercise 2.5.2.
We can now give an explicit combinatorial description of the complete homogeneous symmetric functions in terms of Young tableaux. (Compare to Proposition 2.4.5.) Proposition 2.5.6. For any partition λ and n ≥1, we have (2.36) hλ = X T∈RowWeak(λ) xT and hλ(x1, . . . , xn) = X T∈RowWeak(λ,n) xT.
Proof. The proof of this proposition is analogous to that of Proposition 2.4.5. In particular, to prove the first equality, one alters the proof of Proposition 2.4.4 to work with the hλ instead of the eλ. The first equality also follows from Proposition 2.5.5. Then applying ρn gives the second equality.
Exercises.
2.5.1. Prove (2.28) by following the method of the proof of Lemma 2.4.2 and (2.21).
2.5.2. Prove Proposition 2.5.5.
2.5.3. Write h(2,2) and h(2,1,1) as linear combinations of monomial symmetric functions.
2.5.4. Prove that, for all 0 ≤k ≤n, we have hk(1, . . . , 1 | {z } n entries ) = n + k −1 k .
Here the notation hk(1, . . . , 1) means evaluating the elementary symmetric polynomial hk(x1, . . . , xn) at (x1, . . . , xn) = (1, . . . , 1).
Power sums 37 2.5.5. Find and prove a formula for Mλ,(n)(h, m), where λ ⊢n.
2.5.6. Find and prove a formula for Mλ,(1n)(h, m), where λ ⊢n.
2.5.7. Find and prove a formula for Mλ,(n−1,1)(h, m), where λ ⊢n.
2.5.8. Prove that, for all n ≥1, en = det(h1−j+k)1≤j,k≤n, where (h1−j+k)1≤j,k≤n is the matrix with (j, k) entry equal to h1−j+k.
2.5.9. Prove that, for all n ≥1, hn = det(e1−j+k)1≤j,k≤n, where (e1−j+k)1≤j,k≤n is the matrix with (j, k) entry equal to e1−j+k.
2.6 Power sums In this section we introduce another important set of symmetric functions. As we will see, these new symmetric functions are most well-behaved when we work over the rational num-bers instead of the integers. We will make use in this section of the results of Exercises 1.4.5 to 1.4.8 on formal power series.
For r ≥1, the r-th power sum is (2.37) pr := ∞ X i=1 xr i = m(r).
As usual, for n ∈N, we define pr(x1, . . . , xn) := ρn(pr) = xr 1 + xr 2 + · · · + xr n.
It turns out that many formulas are more natural when we consider the scaled power sums pr r .
Proposition 2.6.1. The generating function for the sequence of scaled powers sums in n indeterminates is ∞ X r=1 pr(x1, . . . , xn) r tr = log n Y j=1 1 1 −xjt !
.
Proof. Recall that log 1 1−x = −log(1 −x) = P∞ r=1 1 rxr. Thus we have ∞ X r=1 pr(x1, . . . , xn) r tr = ∞ X r=1 n X j=1 1 r(xjt)r = n X j=1 log 1 1 −xjt = log n Y j=1 1 1 −xjt !
.
38 The ring of symmetric functions Now, taking the limit as n →∞, we obtain the generating function for the power sums pn: (2.38) P(t) := ∞ X r=1 pr r tr = log ∞ Y j=1 1 1 −xjt !
(2.28) = log H(t).
(See Remark 2.4.3 for a discussion on how we interpret this type of infinite product of generating functions.) Remark 2.6.2. Note that, in [Mac15, p. 23], P(t) denotes the generating function P∞ r=1 prtr−1, which is the formal derivative of our generating function (2.38).
△ Note that (2.39) P ′(t) = d dt log H(t) = H′(t)/H(t).
Also, by (2.29), E(t) = H(−t)−1, and so E′(t) = H(−t)−2H′(−t) Thus (2.40) E′(t)/E(t) = H′(−t)/H(−t) = P ′(−t).
Proposition 2.6.3. For n ≥1, we have nhn = n X r=1 prhn−r, (2.41) nen = n X r=1 (−1)r−1pren−r.
(2.42) (The equations (2.41) and (2.42) are known as Newton’s identities.) Proof. From (2.39), we have H′(t) = P ′(t)H(t). Thus ∞ X n=1 nhntn−1 = ∞ X n=1 pntn−1 ! ∞ X m=0 hmtm !
= ∞ X n=1 n X r=1 prhn−rtn−1.
Equating coefficients yields (2.41). The proof of (2.42) is analogous.
For a partition λ = (λ1, λ2, . . . ) ∈Par, we define (2.43) pλ = pλ1pλ2 · · · .
Recall, from Remark 2.2.8, that SymQ denotes the symmetric functions with coefficients in Q.
Power sums 39 Theorem 2.6.4. We have SymQ = Q[p1, p2, . . . ] and the power sums p1, p2, . . . are algebraically independent over Q. Equivalently, the pλ, λ ∈Par, form a basis for SymQ.
Proof. The identity (2.41) allows us to solve for the power sums recursively in terms of the complete homogeneous symmetric functions and vice versa: (2.44) pn = nhn − n−1 X r=1 prhn−r and hn = 1 n n X r=1 prhn−r.
It follows that SymQ = Q[h1, h2, . . . ] = Q[p1, p2, . . . ], where the first equality is Theorem 2.5.3.
Furthermore, if the pr satisfied a polynomial equation over Q then we can replace them by their expressions in terms of the hr to see that the hr satisfy a nontrivial polynomial equation over Q. (See Exercise 2.6.1.) Then, clearing denominators, we would have that the hr are algebraically dependent over Z. But this contradicts Theorem 2.5.3. Hence the pr are algebraically independent over Q.
Remark 2.6.5. It is crucial in Theorem 2.6.4 that we work over Q instead of Z. For example, h2 = 1 2 p2 1 + p2 does not have integral coefficients when expressed in terms of the pλ. For this reason, when we really want to work over the integers, the powers sums can be problematic.
△ Proposition 2.6.6. For n ∈N and λ ∈Par, we have (2.45) ω(pn) = (−1)n−1pn and ω(pλ) = (−1)|λ|−ℓ(λ)pλ, where ω is the automorphism of Sym defined in (2.31).
Proof. We prove the first equality by induction on n. Since ω(p1) = ω(e1) = h1 = p1, the result holds for n = 1. Then, for n ≥2, we have, using (2.44), ω(pn) (2.44) = nω(hn) − n−1 X r=1 ω(pr)ω(hn−r) = nen − n−1 X r=1 (−1)r−1pren−r (2.42) = (−1)n−1pn, completing the proof of the induction step. Then, for λ ∈Par, we have ω(pλ) = ω(pλ1) · · · ω(pλℓ(λ)) = (−1)|λ|−ℓ(λ)pλ.
It follows from Proposition 2.2.5 that we can write each pλ in terms of the monomial symmetric functions. The next result gives an explicit formula.
40 The ring of symmetric functions Proposition 2.6.7. For k ≥1 and λ ∈Par(k), we have (2.46) pλ = X µ⊢k Mλ,µ(p, m)mµ, where Mλ,µ(p, m) is the number of Young tableaux of shape λ and weight µ in which the entries in each row are constant.
Proof. The proof of this result is left as Exercise 2.6.2.
We conclude this section by deducing explicit formulas for the hn and en as linear com-binations of the pλ.
Recall that, for λ ∈Par, mi(λ) denotes the multiplicity of i in λ.
Define (2.47) zλ := Y i≥1 imi(λ)mi(λ)!
(See Exercise 1.1.2 for a combinatorial interpretation of zλ.) Proposition 2.6.8. We have (2.48) H(t) = X λ∈Par pλ zλ t|λ| and E(t) = X λ∈Par (−1)|λ|−ℓ(λ)pλ zλ t|λ|.
Proof. By (2.38), we have H(t) = exp P(t) = exp X r≥1 pr r tr = ∞ Y r=1 exp pr r tr = ∞ Y r=1 ∞ X mr=0 (prtr)mr mr!rmr = X λ∈Par pλ zλ t|λ|.
We now obtain the second equation in (2.48) by applying ω and using (2.45).
Corollary 2.6.9. For n ∈N, we have (2.49) hn = X λ∈Par(n) pλ zλ and en = (−1)n X λ∈Par(n) (−1)ℓ(λ)pλ zλ .
Proof. This follows from equating coefficients in (2.48).
Exercises.
2.6.1. Prove that if f is a nonzero polynomial in the pr, then, after using (2.44) to replace all pr by their expressions in terms of the hr, we obtain a nonzero polynomial. Hint: Use the fact that pr is equal to rhr plus a polynomial in h1, . . . , hr−1.
2.6.2. Prove Proposition 2.6.7.
Power sums 41 2.6.3. Write p(2,2,1) and p(2,2,2) as linear combinations of monomial symmetric functions.
2.6.4. Find and prove a formula for Mλ,(n)(p, m), where λ ⊢n.
2.6.5. Find and prove a formula for Mλ,(1n)(p, m), where λ ⊢n.
2.6.6. Find and prove a formula for Mλ,(n−1,1)(p, m), where λ ⊢n.
2.6.7. Using Newton’s identity (2.42), prove that pn = e1 1 0 0 · · · 0 2e2 e1 1 0 · · · 0 .
.
.
.
.
.
...
...
...
.
.
.
.
.
.
.
.
.
...
1 0 .
.
.
.
.
.
e1 1 nen en−1 en−2 · · · · · · e1 and n!en = p1 1 0 0 · · · 0 p2 p1 2 0 · · · 0 .
.
.
.
.
.
...
...
...
.
.
.
.
.
.
.
.
.
...
...
0 pn−1 pn−2 · · · · · · p1 n −1 pn pn−1 · · · · · · p2 p1 .
Chapter 3 Schur functions In this chapter we introduce Schur functions, which form what is probably the most impor-tant basis of the ring of symmetric functions. In most applications of symmetric functions to other areas of mathematics, such as geometry and representation theory, the Schur functions play a crucial role. We begin with a discussion of alternating polynomials, which we then use these to define Schur functions. Next we define a natural bilinear form on the space of symmetric functions and examine the behaviour of the Schur functions with respect to this bi-linear form. We then give a description of Schur functions in terms of semistandard tableaux similar to the tableaux descriptions of the other bases of the ring of symmetric functions we studied in Chapter 2. We also give a complete description of the transition matrices between all the bases we have seen so far. Finally, we prove the important Littlewood–Richardson rule, which gives a method of computing the product of Schur functions.
3.1 Alternating polynomials An inversion for π ∈Sn is a pair (i, j) with 1 ≤i < j ≤n such that π(i) > π(j). Let inv(π) denote the number of inversions of π. The sign of π is defined to be sgn(π) := (−1)inv(π).
Alternatively, if we write π as a product of m transpositions (recall that a transposition is a cycle of length two) then sgn(π) = (−1)m. (Note that m is not unique, but its parity is.) We have sgn(πσ) = sgn(π) sgn(σ) for all π, σ ∈Sn.
Example 3.1.1. If π = (1 3 5) ∈S5, then the set of inversions is {(1, 2), (1, 5), (2, 5), (3, 4), (3, 5), (4, 5)}.
Thus inv(π) = 6 and sgn(π) = (−1)6 = 1.
We can write π = (1 3)(3 5) = (2 4)(1 3)(3 5)(2 4), and both expressions involve an even number of permutations.
△ 42 Alternating polynomials 43 We say a polynomial f ∈Poln is alternating, or skew-symmetric, if πf = sgn(π)f for all π ∈Sn.
Let Altn := {f ∈Poln : πf = sgn(π)f for all π ∈Sn} denote the Z-module of all alternating polynomials in n indeterminates. Note that Altn is not a subring of Poln. In fact, if f, g ∈Altn, then π(fg) = (πf)(πg) = sgn(π)f sgn(π)g = fg for all π ∈Sn, and so fg ∈Symn. On the other hand, if f ∈Symn and g ∈Altn, then π(fg) = (πf)(πg) = sgn(π)fg for all π ∈Sn, and so fg ∈Altn. It follows that Altn is a Symn-module.
We can construct alternating polynomials by antisymmetrizing monomials as follows. If α ∈WCompn, define (3.1) aα = aα(x1, . . . , xn) := X π∈Sn sgn(π)xαπ.
Then aα is alternating (Exercise 3.1.4).
Example 3.1.2. If n = 3 and α = (5, 3, 2), then aα = X π∈S3 xαπ = x5 1x3 2x2 3 −x5 1x2 2x3 3 −x3 1x5 2x2 3 −x2 1x3 2x5 3 + x2 1x5 2x3 3 + x3 1x2 2x5 3.
△ Lemma 3.1.3. For α ∈WCompn, we have aα = det(x αj i )1≤i,j≤n = det xα1 1 xα2 1 · · · xαn 1 xα1 2 xα2 2 · · · xαn 2 .
.
.
.
.
.
.
.
.
.
.
.
xα1 n xα2 n · · · xαn n .
Proof. Recall that the determinant of a matrix B = (bi,j)1≤i,j≤n can be written as (3.2) det B = X π∈Sn sgn(π) n Y i=1 bi,π(i) !
.
Thus det(x αj i )1≤i,j≤n = X π∈Sn sgn(π) n Y i=1 x απ(i) i !
= X π∈Sn sgn(π)xαπ = aα.
44 Schur functions Remark 3.1.4. Note that if two parts of α are equal, say αi = αj for some 1 ≤i < j ≤n, then, for π = (i j), we have sgn(π) = −1, and so aα = −aαπ = −aα.
Hence aα = 0. (This can also be seen using Lemma 3.1.3 since, if αi = αj, then two columns of the matrix there are equal; hence the determinant is zero.) It follows that there are no repeated terms in the sum (3.1) unless aα = 0. (Compare to Remark 2.1.1.) It also follows that, for any α ∈WCompn with aα ̸= 0, there exists a unique π ∈Sn such that απ ∈Parn.
Then aα = sgn(π)aαπ, and so we can restrict our attention to aλ where λ1 > λ2 > · · · > λn.
△ Let (3.3) δn = (n −1, n −2, . . . , 1, 0) ∈Parn.
For α, β ∈WCompn, define α + β = (α1 + β1, α2 + β2, . . . , αn + βn) ∈WCompn.
Lemma 3.1.5. We have a bijection of sets Parn →{λ ∈Parn : λ1 > λ2 > · · · > λn}, λ 7→λ + δn.
Proof. The proof of this lemma is left as Exercise 3.1.5.
Proposition 3.1.6. The elements (3.4) aλ+δn(x1, . . . , xn), λ ∈Parn, form a Z-basis of Altn.
Proof. We must show that the elements (3.4) are linearly independent and span Altn. Our proof will be similar to the proof of Proposition 2.1.2. First, it follows from Remark 3.1.4 that, for λ, µ ∈Parn with λ ̸= µ, the polynomials aλ+δn and aµ+δn have no terms in common.
Hence these elements are linearly independent.
Next we prove that the elements (3.4) span Altn. Suppose f ∈Altn. We prove that f lies in the span of (3.4) by induction on the number of nonzero terms in f. The base case f = 0 is clear, so we suppose f ̸= 0. Choose a nonzero term cxα, c ∈Z \ {0}, of f. Since f is alternating, the terms sgn(π)cxαπ also appear in f for all π ∈Sn. By Remark 3.1.4 and Lemma 3.1.5, we have απ = λ + δn for some unique λ ∈Parn, π ∈Sn.
Then cαλ+δn(x1, . . . , xn) contains the term sgn(π)cxα, and so f −sgn(π)caλ+δn(x1,...,xn) ∈Altn has fewer terms than f. By the induction hypothesis, f −caλ+δn(x1, . . . , xn) can be written as a Z-linear combination of the elements (3.4); hence the same is true for f.
Alternating polynomials 45 Lemma 3.1.7. Every element of Altn is divisible by Q 1≤i<j≤n(xi −xj).
Proof. Suppose f ∈Altn. For 1 ≤i < j ≤n, interchanging xi and xj multiplies f by −1. It follows that substituting xi for xj in f gives zero: f(x1, . . . , xj−1, xi, xj+1, . . . , xn) = 0.
Thus, by the factor theorem for polynomials, f is divisible by xi −xj. Now, since xi −xj is degree one, it is an irreducible polynomial. Therefore, the xi −xj, 1 ≤i < j ≤n, are all coprime. Hence, since Poln is a unique factorization domain, f is divisible by their product Q 1≤i<j≤n(xi −xj).
Corollary 3.1.8. We have (3.5) aδn(x1, . . . , xn) = det xn−1 1 xn−2 1 · · · x1 1 xn−1 2 xn−2 2 · · · x2 1 .
.
.
.
.
.
.
.
.
.
.
.
xn−1 n xn−2 n · · · xn 1 = Y 1≤i<j≤n (xi −xj).
(The polynomial (3.5) is called a Vandermonde determinant.) Proof. The first equality is Lemma 3.1.3. By Lemma 3.1.7, we have aδn(x1, . . . , xn) = c Y 1≤i n.
Proof. The proof of this result is left as Exercise 3.2.1.
Schur polynomials and Schur functions 47 Remark 3.2.3. It is stated in [Mac15, p. 41] that aα(x1, . . . , xn, 0) = aα(x1, . . . , xn) for α ∈ WCompn. However, this is false. See Exercise 3.2.2.
△ It follows from Lemma 3.2.2 that, for ℓ(λ) ≤n, ρn+1,n (sλ(x1, . . . , xn+1)) = aλ+δn(x1, . . . , xn)en(x1, . . . , xn) aδn(x1, . . . , xn)en(x1, . . . , xn) = sλ(x1, . . . , xn) in the notation of (2.6). Therefore, if we define (3.8) sλ(x1, . . . , xn) = 0 for ℓ(λ) > n, we have ρn+1,n (sλ(x1, . . . , xn+1)) = sλ(x1, . . . , xn) for all λ ∈Par.
Thus, the sequence sλ(x1, . . . , xn) ∞ n=0 defines an element of Sym|λ|; see (2.8). For λ ∈Par, the element (3.9) sλ := sλ(x1, . . . , xn) ∞ n=0 is called the Schur function corresponding to the partition λ.
Theorem 3.2.4.
(a) The Schur polynomials sλ(x1, . . . , xn), λ ∈Parn(k), form a Z-basis of Symk n.
(b) The Schur polynomials sλ(x1, . . . , xn), λ ∈Parn, form a Z-basis of Symn.
(c) The Schur functions sλ, λ ∈Par(k), form a Z-basis of Symk.
(d) The Schur functions sλ, λ ∈Par, form a Z-basis of Sym.
Proof. By definition, the alternating function aλ+δn(x1, . . . , xn) is the image under the map (3.6) of the Schur polynomial sλ(x1, . . . , xn). Then part (b) follows from Propositions 3.1.6 and 3.1.9. Since sλ is homogeneous of degree |λ|, part (a) follows immediately. Finally, (c) and (d) follow from Proposition 2.2.4 since Park n = Parn when n ≥k.
It follows from Theorems 2.4.6 and 2.5.3 that the Schur functions can be expressed as polynomials in the elementary and complete homogeneous symmetric functions.
The following theorem gives these expressions explicitly.
Theorem 3.2.5 (Jacobi–Trudi identities). For λ ∈Par, we have (3.10) sλ = det(hλi−i+j)1≤i,j≤n = det hλ1 hλ1+1 · · · hλ1+n−1 hλ2−1 hλ2 · · · hλ2+n−2 .
.
.
.
.
.
...
.
.
.
hλn−n+1 hλn−n+2 · · · hλn for n ≥ℓ(λ) and (3.11) sλ = det(eλ′ i−i+j)1≤i,j≤m = det eλ′ 1 eλ′ 1+1 · · · eλ′ 1+m−1 eλ′ 2−1 eλ′ 2 · · · eλ′ 2+m−2 .
.
.
.
.
.
...
.
.
.
eλ′ m−m+1 eλ′ m−m+2 · · · eλ′ m for m ≥ℓ(λ′).
(Recall that er = hr = 0 for r < 0.) 48 Schur functions We split the proof of Theorem 3.2.5 into pieces. We begin with a technical lemma. Then we prove that the determinants appearing in (3.10) and (3.11) are equal. Finally, we prove (3.10).
Lemma 3.2.6. Suppose λ is partition, and let n ≥ℓ(λ), m ≥ℓ(λ′). Then the n+m numbers λi + n −i, 1 ≤i ≤n, n −1 + j −λ′ j, 1 ≤j ≤m, are a permutation of {0, 1, 2, . . . , n + m −1}.
Proof. The Young diagram of λ (coloured green in the diagram below) is contained in the Young diagram of the partition (mn), which is a n × m rectangle.
n | {z } m Number the segments of the boundary (coloured blue in the diagram above) between λ and its complement in (mn) with the numbers 0, 1, . . . , n + m −1, starting at the bottom. The vertical segments are then numbered λi + n −i, 1 ≤i ≤n, and (considering the transposed diagram) the horizontal segments are numbered (n + m −1) −(λ′ j + m −j) = n −1 + j −λ′ j, 1 ≤j ≤m.
Next we prove that the determinants appearing in (3.10) and (3.11) are equal. In fact, for future use, we prove a slightly more general result.
Lemma 3.2.7. For λ, µ ∈Par with ℓ(λ), ℓ(µ) ≤n and ℓ(λ′), ℓ(µ′) ≤m, we have (3.12) det(hλi−µj−i+j)1≤i,j≤n = det(eλ′ i−µ′ j−i+j)1≤i,j≤m.
Proof. Let N be a positive integer and consider the (N + 1) × (N + 1) matrices H = (hi−j)0≤i,j≤N, E = (−1)i−jei−j 0≤i,j≤N .
Both H and E are lower triangular with 1’s on the diagonal. Thus det(H) = det(E) = 1.
Furthermore, (2.30) implies that H and E are mutually inverse.
Let λ be a partition of length ℓ(λ) ≤n such that ℓ(λ′) ≤m, with n + m = N + 1. Let Hλ be the n × n submatrix of H formed by entries whose row index is λi + n −i, 1 ≤i ≤n, and column index is µj + n −j, 1 ≤j ≤n. The, by Lemma 3.2.6, the m × m submatrix of E formed by deleting these rows and column is the submatrix of E formed by entries whose row index is n −1 + i −λ′ i, 1 ≤i ≤m, and column index is n −1 + j −µ′ j, 1 ≤j ≤m. Then, by a generalization of Cramer’s rule, we have Schur polynomials and Schur functions 49 (3.13) det hλi−µj−i+j 1≤i,j≤n = det(Hλ) = (−1)|λ|+|µ| det (−1)λ′ i−µ′ j−i+jeλ′ i−µ′ j−i+j 1≤i,j≤m .
(See [GAE02] for a proof of this generalization of Cramer’s rule. The matrices A, X, and Y in [GAE02] are our H, E and the identity matrix, respectively. Then det(A) = 1 in [GAE02, (2)], and we expand the determinant of AY appearing there along the columns coming from the identity matrix, then take the transpose.) Using the fact that multiplying a row or column of a matrix by −1 multiplies the determinant by −1, we see that the minus signs on the right-hand side of (3.13) cancel, giving (3.12).
Note that taking µ = ∅in (3.12) shows that the determinants appearing in (3.10) and (3.11) are equal. We can now give the proof of the Jacobi–Trudi identities.
Proof of Theorem 3.2.5. In light of (3.12), it suffices to prove (3.10). Assume n ≥ℓ(λ). It suffices to work with n indeterminates x1, . . . , xn. (See Exercise 3.2.3.) So we will write hr for hr(x1, . . . , xn) and er for er(x1, . . . , xn).
For 1 ≤k ≤n, let e(k) r = er(x1, . . . , xk−1, xk+1, . . . , xn) and define the matrix M = (−1)n−ie(j) n−i 1≤i,j≤n .
We first prove that, for α ∈WCompn, we have (3.14) Aα = HαM where Aα = (xαi j )1≤i,j≤n, Hα = (hαi−n+j)1≤i,j≤n.
Let E(k)(t) = n−1 X r=0 e(k) r tr = Y 1≤i≤n i̸=k (1 + xit).
Then, using (2.28), we have H(t)E(k)(−t) = 1 1 −xkt.
Comparing coefficients of tαi gives n X j=1 hαi−n+j · (−1)n−je(k) n−j = xαi k , which implies (3.14).
Now, taking determinants of both sides of (3.14) and using Lemma 3.1.3 gives aα = det(Aα) = det(Hα) det(M) for all α ∈WCompn.
50 Schur functions Note that Hδn is triangular with 1’s on the diagonal, and hence det(Hδn) = 1. Therefore det(M) = aδn, and so (3.15) aα = aδn det(Hα).
Taking α = λ + δn and noting that Hλ+δn = (hλi−i+j)1≤i,j≤n then gives (3.10).
Corollary 3.2.8. For all n ∈N, we have (3.16) s(n) = hn and s(1n) = en.
Proof. The first equality follows from taking λ = (n) in (3.10) and the second equality follows from taking λ = (1n) in (3.11).
Corollary 3.2.9. We have (3.17) ω(sλ) = sλ′ for all λ ∈Par, where ω is the involution of Sym defined in (2.31).
Proof. Since the determinant of a matrix is a polynomial in its entries, we have ω(sλ) = det (ω(hλi−i+j))1≤i,j≤n = det (eλi−i+j)1≤i,j≤n = sλ′, where we have used the fact that λ′′ = λ.
Exercises.
3.2.1. Prove Lemma 3.2.2.
3.2.2. Show that, for all n ≥1, there exists α ∈WCompn such that aα(x1, . . . , xn, 0) ̸= aα(x1, . . . , xn).
Hint: Find an example where the left-hand side is zero, but the right-hand side is not.
3.2.3. Note that the Jacobi–Trudi identity (3.10) holds for any n ≥ℓ(λ). If m ≥ℓ(λ), explain the relationship between the matrix appearing in (3.10) when n = m and n = m+1.
Why do these values of n give the same formula for sλ?
Now do the same analysis for the identity (3.11). Note that this analysis justifies working in n variables in the proof of Theorem 3.2.5 since, when n ≥|λ|, we can apply Proposition 2.2.4.
3.2.4. Write the Schur function s(2,1) as a linear combination of the complete homogeneous symmetric functions, the elementary symmetric functions, and the monomial symmetric functions.
3.2.5. Write the Schur function s(3,1) as a linear combination of the complete homogeneous symmetric functions, the elementary symmetric functions, and the monomial symmetric functions.
The Hall inner product 51 3.3 The Hall inner product In this section we introduce a natural bilinear form on the space of symmetric functions and explore how the various bases of Sym that we have found behave with respect to this bilinear form. We will follow the presentation of [Mac15, §I.4] but include more detail than is found there.
Definition 3.3.1 (Hall inner product). We define a Z-valued bilinear form ⟨, ⟩on Sym, called the Hall inner product, by requiring that the bases (hλ)λ∈Par and (mλ)λ∈Par are dual to each other: (3.18) ⟨hλ, mµ⟩= δλµ := ( 1 if λ = µ, 0 if λ ̸= µ.
The inner product of two arbitrary elements can then by computed by writing them in these bases and using Z-bilinearity. We also use the notation ⟨, ⟩to denote the corresponding Q-valued bilinear form on SymQ.
Remark 3.3.2. We will see later a motivation for the definition of the Hall inner product coming from the representation theory of the symmetric group; see Remark 6.3.5.
△ Let x = (x1, x2, . . . ) and y = (y1, y2, . . . ) be two sequences of independent indeterminates.
We will use the notation f(x) and f(y) to denote symmetric functions in the xi and yi, respectively.
For the proofs that follow, it will be useful to introduce the following notation.
For α ∈Zn, define (3.19) hα(x) := hα1(x)hα2(x) · · · .
Since hr(x) = 0 for r < 0, we see that hα(x) = 0 unless α ∈WComp. Furthermore, for α ∈WComp, hα(x) is equal to hλ(x), where λ is the unique partition that is a permutation of α.
Proposition 3.3.3. We have (3.20) ∞ Y i,j=1 (1 −xiyj)−1 = X λ∈Par hλ(x)mλ(y) = X λ∈Par mλ(x)hλ(y).
Proof. Taking t = yj in (2.28), and then taking the product over j ≥1 gives (3.21) ∞ Y i,j=1 (1 −xiyj)−1 = ∞ Y j=1 H(yj) = ∞ Y j=1 ∞ X r=0 hr(x)yr j !
= X α∈WComp hα(x)yα.
Now, by (1.6), we have (3.22) WComp = G λ∈Par λS∞.
52 Schur functions Thus X α∈WComp hα(x)yα = X λ∈Par X α∈λS∞ hα(x)yα = X λ∈Par hλ(x) X α∈λS∞ yα = X λ∈Par hλ(x)mλ(y).
This completes the proof of the first equality in (3.20).
The second equality follows by interchanging the xi and yi.
Note that the expressions for Q∞ i,j=1(1−xiyj)−1 in (3.20) involve the dual bases (hλ)λ∈Par and (µλ)λ∈Par. The next result makes this observation more precise.
Proposition 3.3.4. For each k ≥0, let (uλ)λ∈Par(k) and (vλ)λ∈Par(k) be Q-bases of Symk Q.
Then the following conditions are equivalent: (a) ⟨uλ, vµ⟩= δλµ for all λ, µ; (b) P λ∈Par uλ(x)vλ(y) = Q∞ i,j=1(1 −xiyj)−1.
Proof. Since the (hλ)λ∈Par(k) and (mλ)λ∈Par(k) form Q-bases of Symk, we have, for λ ∈Par(k), (3.23) uλ = X ρ∈Par(k) cλρhρ and vµ = X σ∈Par(k) dµσmσ for some cλρ, dµσ ∈Q.
Then ⟨uλ, vµ⟩= X ρ∈Par(k) cλρdµρ.
Thus (3.24) (a) ⇐ ⇒ X ρ∈Par(k) cλρdµρ = δλµ for all λ, µ ∈Par(k), k ∈N.
On the other hand, by (3.20), condition (b) is equivalent to X λ∈Par(k) uλ(x)vλ(y) = X ρ∈Par(k) hρ(x)mρ(y).
Using (3.23), we have X λ∈Par(k) uλ(x)vλ(y) = X λ,ρ,σ∈Par(k) cλρdλσhρ(x)mσ(y).
Thus (3.25) (b) ⇐ ⇒ X ρ∈Par(k) hρ(x)mρ(y) = X λ,ρ,σ∈Par(k) cλρdλσhρ(x)mσ(y) ⇐ ⇒ X λ∈Par(k) cλρdλσ = δρσ for all ρ, σ ∈Par(k), k ∈N.
If we define the matrices Ck = (cλµ)λ,µ∈Par(k) and Dk = (dλµ)λ,µ∈Par(k) .
The Hall inner product 53 Then, letting M ⊺denote the transpose of a matrix M, we have (a) (3.24) ⇐ ⇒CkD ⊺ k = I ⇐ ⇒DkC ⊺ k = I ⇐ ⇒C ⊺ kDk = I (3.25) ⇐ ⇒(b), where, in the second bi-implication, we took transposes of both sides.
Proposition 3.3.4 tells us that, to find other dual bases, we can look for other expressions for the infinite product Q∞ i,j=1(1 −xiyj)−1. This is our next goal.
Proposition 3.3.5. We have (3.26) ∞ Y i,j=1 (1 −xiyj)−1 = X λ∈Par z−1 λ pλ(x)pλ(y).
Proof. For r ∈N, the r-th power sum in the variables xiyj, i, j ≥1, is pr(xy) := ∞ X i,j=1 (xiyj)r = ∞ X i=1 xr i ! ∞ X i=1 yr j !
= pr(x)pr(y).
It follows that, for λ ∈Par, we have pλ(xy) = pλ1(xy)pλ2(xy) · · · = pλ1(x)pλ1(y)pλ2(x)pλ2(y) · · · = pλ(x)pλ(y).
Now, by (2.28) and (2.48), we have (3.27) ∞ Y i=1 (1 −xit)−1 = X λ∈Par z−1 λ pλ(x)t|λ|.
Using this equality, but with the set of variables xiyj, i, j ≥1, instead of the set of variables xi, i ≥1, gives ∞ Y i,j=1 (1 −xiyjt)−1 = X λ∈Par z−1 λ pλ(xy)t|λ| = X λ∈Par z−1 λ pλ(x)pλ(y)t|λ|.
Taking t = 1 gives (3.26).
Proposition 3.3.6 (Cauchy’s formula). We have (3.28) ∞ Y i,j=1 (1 −xiyj)−1 = X λ∈Par sλ(x)sλ(y).
Proof. As usual, it suffices to prove the result over finite sets of variables x = (x1, . . . , xn) and y = (y1, . . . , yn). Recall the definition of the alternating symmetric functions aα from Section 3.1 and the partition δn = (n −1, n −2, . . . , 1, 0). Taking xk = yk = 0 for k > n in (3.21), then multiplying both sides by aδn(x)aδn(y), we have aδn(x)aδn(y) n Y i,j=1 (1 −xiyj)−1 54 Schur functions = aδn(x) X α∈WCompn π∈Sn hα(x) sgn(π)yα+(δn)π = aδn(x) X β∈WCompn π∈Sn sgn(π)hβ−(δn)π(x)yβ, where, in the second equality, we changed to a sum over β = α + (δn)π. Now, by (3.15), we have aβ(x) = aδn(x) det (hβi−n+j(x))1≤i,j≤n = aδn(x) det hβi−δn j (x) 1≤i,j≤n (3.2) = aδn(x) X π∈Sn sgn(π) n Y i=1 hβi−δn π(i)(x) = aδn(x) X π∈Sn sgn(π)hβ−(δn)π(x).
Combining this with the above, we have aδn(x)aδn(y) n Y i,j=1 (1 −xiyj)−1 = X β∈WCompn aβ(x)yβ (3.22) = X µ∈Par X π∈Sn aµπ(x)yµπ = X µ∈Par X π∈Sn sgn(π)aµ(x)yµπ = X µ∈Par aµ(x)aµ(y) = X λ∈Par aλ+δn(x)aλ+δn(y), where, in the last equality, we used the fact that aµ = 0 unless µ = λ + δn for some λ ∈Par. Finally, we divide both sides by aδn(x)aδn(y) and use the definition (3.7) of the Schur polynomials.
We can now draw several conclusions about the bilinear form.
Theorem 3.3.7.
(a) We have (3.29) ⟨pλ, pµ⟩= δλµzλ for all λ, µ ∈Sym.
In particular, the pλ, λ ∈Par, form an orthogonal basis of SymQ.
(b) We have (3.30) ⟨sλ, sµ⟩= δλµ, for all λ, µ ∈Sym.
In particular, the Schur functions form an orthonormal basis of Sym.
(c) The bilinear form ⟨, ⟩is symmetric and positive definite.
(d) The involution ω of (2.31) is an isometry. In other words, (3.31) ⟨ω(u), ω(v)⟩= ⟨u, v⟩ for all u, v ∈SymQ.
(e) We have (3.32) ⟨eλ, fµ⟩= δλµ, for all λ, µ ∈Par, where fµ = ω(mµ), as in (2.34). In other words, the bases (eλ)λ∈Par and (fλ)λ∈Par are dual.
The Hall inner product 55 Proof.
(a) It follows from Proposition 3.3.4 and (3.26) that ⟨z−1 λ pλ, pµ⟩= δλµ for all λ, µ ∈Par.
Then we multiply both sides by zλ and use bilinearity of the form.
(b) This follows immediately from Proposition 3.3.4 and (3.28).
(c) The fact that the form is symmetric follows from the symmetry of (3.30). Further-more, since ⟨, ⟩admits an orthonormal basis, it is positive definite.
(d) Using (3.17), we have, for all λ, µ ∈Par, ⟨ω(sλ), ω(sµ)⟩= ⟨sλ′, sµ′⟩= δλ′µ′ = δλµ = ⟨sλ, sµ⟩.
Since the sλ, λ ∈Par, form a basis of Sym, (3.31) follows.
(e) For all λ, µ ∈Par, we have ⟨eλ, fµ⟩= ⟨ω(hλ), ω(mµ)⟩= ⟨hλ, mµ⟩= δλµ.
We can now deduce an important characterizations of the Schur functions.
Proposition 3.3.8. The Schur functions sλ, λ ∈Par(k), are the unique orthonormal basis of Symk up to sign and reordering.
Proof. Suppose (uλ)λ∈Par(k) is another orthonormal basis for Symk. Then, for all λ ∈Par(k), we have uλ = X µ∈Par(k) cλµsµ for some cλµ ∈Z.
So C = (cλµ)λ,µ∈Par(k) is an integer matrix. (We view this as a matrix by choosing some ordering of Par(k).) Furthermore C is orthogonal, since it is the change of basis matrix between two orthonormal bases. However, the only orthogonal integer matrices are signed permutation matrices (that is, matrices that have exactly one entry ±1 in each row and column, with all other entries being zero); see Exercise 3.3.1. Thus the uλ are a permutation of the sλ up to sign.
Exercises.
3.3.1. Prove that the only orthogonal integer matrices are signed permutation matrices (that is, matrices that have exactly one entry ±1 in each row an column, with all other entries being zero).
3.3.2. Recall the forgotten symmetric functions fλ from (2.34). Express P λ∈Par eλ(x)fλ(y) as an infinite product.
3.3.3. Compute ⟨en, hn⟩for n ∈N.
56 Schur functions 3.3.4. Compute ⟨hn, pn⟩for n ∈N.
3.3.5. Compute ⟨en, pn⟩for n ∈N.
3.3.6. Give a combinatorial formula for ⟨hλ, hµ⟩, λ, µ ∈Par.
More precisely, show that ⟨hλ, hµ⟩is equal to the number of Young tableaux satisfying certain conditions. Hint: Use Proposition 2.5.5.
3.4 Skew Schur functions and semistandard tableaux In Propositions 2.4.5 and 2.5.6 we gave explicit expressions for the elementary and complete homogeneous symmetric functions in terms of tableaux.
These expressions were closely related to the expressions of these symmetric functions as linear combinations of monomial symmetric functions given in Propositions 2.4.4 and 2.5.5. In this section, our goal is to develop such expressions for Schur functions. In turns out that this is most easily done if we slightly generalize our definition of Schur functions, to consider skew Schur functions. We follow the presentation of [Mac15, §I.5] but add some additional details.
Since the Schur functions form an orthonormal basis for Sym, every symmetric function is uniquely determined by its scalar product with the sλ. In particular, we have (3.33) f = X λ∈Par ⟨f, sλ⟩sλ for all f ∈Sym.
Definition 3.4.1 (Skew Schur functions). For λ, µ ∈Par, define the skew Schur function (3.34) sλ/µ := X ν∈Par ⟨sλ, sµsν⟩sν ∈Sym.
Remark 3.4.2. The equation (3.34) implies that sλ/µ is obtained by acting on sλ with the operator adjoint to multiplication by sµ. More precisely, for f ∈Sym and ν, λ ∈Par, we have (3.35) ⟨sν/λ, f⟩ (3.34) = X ρ∈Par ⟨sν, sλsρ⟩⟨sρ, f⟩= sν, sλ X ρ∈Par ⟨sρ, f⟩sρ + = ⟨sν, sλf⟩.
We will discuss these adjoint operators in more detail in Section 4.2.
△ Define integers cλ µν for λ, µ, ν ∈Par by (3.36) sµsν = X λ∈Par cλ µνsλ.
Since sλ is homogeneous of degree |λ|, we have (3.37) cλ µν = 0 if |λ| ̸= |µ| + |ν|.
The integers cλ µν are called Littlewood–Richardson coefficients, and we will explore them further in Section 3.6.
Skew Schur functions and semistandard tableaux 57 Lemma 3.4.3. Suppose λ, µ ∈Par. Recall that ∅= (0, 0, 0, . . . ) denotes the zero partition.
(a) We have sλ/∅= sλ.
(b) We have (3.38) sλ/µ = X ν∈Par cλ µνsν.
(c) The skew Schur function sλ/µ is homogeneous of degree |λ| −|µ|. In particular, it is zero if |λ| < |µ|.
Proof.
(a) Since s∅= 1, we have sλ/∅ (3.34) = X ν∈Par ⟨sλ, sν⟩sν (3.33) = sλ.
(b) We have sλ/µ (3.34) = X ν∈Par ⟨sλ, sµsν⟩sν (3.36) = X ν,σ∈Par cσ µν⟨sλ, sσ⟩sν (3.30) = X ν∈Par cλ µνsν.
(c) This follows from (3.37) and (3.38).
By Theorems 2.4.6 and 2.5.3, the skew Schur functions can be written as polynomials in the hr or the er. The following result, which is a generalization of the Jacobi–Trudi identities (Theorem 3.2.5), gives these expression explicitly.
Proposition 3.4.4. For λ, µ ∈Par, we have (3.39) sλ/µ = det hλi−µj−i+j 1≤i,j≤n for n ≥ℓ(λ), ℓ(µ) and (3.40) sλ/µ = det eλ′ i−µ′ j−i+j 1≤i,j≤m for m ≥ℓ(λ′), ℓ(µ′).
Proof. By (3.12), the determinants appearing in (3.39) and (3.40) are equal, and so it suffices to prove (3.39).
Let x = (x1, x2, . . . ) and y = (y1, y2, . . . ) be two sets of variables. Then X λ∈Par sλ/µ(x)sλ(y) (3.38) = X λ,ν∈Par cλ µνsν(x)sλ(y) (3.36) = X ν∈Par sν(x)sµ(y)sν(y).
Now, by (3.20) and (3.28), we have X ν∈Par sν(x)sν(y) = ∞ Y i,j=1 (1 −xiyj)−1 = X ν∈Par hν(x)mν(y).
58 Schur functions Thus we have (3.41) X λ∈Par sλ/µ(x)sλ(y) = X ν∈Par hν(x)mν(y)sµ(y).
Now work over n indeterminates y = (y1, . . . , yn), so that sums above are only over partitions of length ≤n (recall (3.8)). Then, using the definition (3.7) of the Schur polynomials and multiplying both sides of (3.41) by aδn, we have X λ∈Parn sλ/µ(x)aλ+δn(y) = X ν∈Parn hν(x)mν(y)aµ+δn(y) (3.1) = X ν∈Parn hν(x)mν(y) X π∈Sn sgn(π)y(µ+δn)π (2.3) = X α∈WCompn hα(x)yα X π∈Sn sgn(π)y(µ+δn)π = X α∈WCompn X π∈Sn hα(x) sgn(π)yα+(µ+δn)π, where we used the definition (3.19) of hα for α ∈Zn. Equating coefficients of yλ+δn, we have sλ/µ(x) = X π∈Sn sgn(π)hλ+δn−(µ+δn)π(x) = X π∈Sn sgn(π) n Y i=1 h(λ+δn)i−(µ+δn)π(i) (3.2) = det hλi−µj−i+j 1≤i,j≤n .
Corollary 3.4.5. For all λ, µ ∈Par, we have (3.42) ω(sλ/µ) = sλ′/µ′.
Proof. Choose n ≥ℓ(λ), ℓ(µ). Then ω(sλ/µ) (3.39) = det ω(hλi−µj−i+j) 1≤i,j≤n = det eλi−µj−i+j 1≤i,j≤n (3.40) = sλ′/µ′.
Define a relation ⊇on Par as follows: for λ, µ ∈Par, we define (3.43) λ ⊇µ ⇐ ⇒λi ≥µi for all i.
Equivalently, λ ⊇µ if and only if the Young diagram of λ contains the Young diagram of µ. If λ ⊇µ, then we identify λ/µ with a skew Young diagram (sometimes just called a skew diagram), which is the set theoretic difference of the Young diagrams of λ and µ. For example, if λ = (6, 5, 3, 1) and µ = (4, 3, 2) then we can nest their Young diagrams as follows: Skew Schur functions and semistandard tableaux 59 Then the skew diagram λ/µ is drawn as follows: (3.44) A path in a skew diagram θ is a sequence x0, x1, . . . , xm of boxes in θ such that xi−1 and xi have a common side, for 1 ≤i ≤m. A subset φ of θ is connected if any two squares in φ can be joined by a path in φ. The connected components of θ are the maximal connected subsets of θ, which are themselves skew diagrams. For example, the skew diagram (3.44) has three connected components.
The conjugate of a skew diagram λ/µ is the skew diagram λ′/µ′. We say that a skew diagram is a horizontal strip if it has at most one square in each column. Thus λ/µ is a horizontal strip if and only if λ′ i −µ′ i ≤1 for all i ≥1. For example, the skew diagram (3.44) is not a horizontal strip, but the following skew diagram is: We say that λ/µ is a vertical strip if it has at most one square in each row. Equivalently, λ/µ is a vertical strip if and only if λ′/µ′ is a horizontal strip.
Proposition 3.4.6. Suppose λ, µ ∈Par.
(a) We have (3.45) sλ/µ = 0 unless λ ⊇µ.
(b) If θ1, θ2, . . . , θr are the components of the skew diagram λ/µ, then sλ/µ = Qr i=1 sθi.
Proof. Fix n ≥ℓ(λ), ℓ(µ).
(a) Suppose λ ⊉µ. Then λr < µr for some r, and so λi ≤λr < µr ≤µj for 1 ≤j ≤r ≤i ≤n.
Therefore, for 1 ≤j ≤r ≤i ≤n, we have λi −µj −i + j < 0 and so hλi−µj−i+j = 0.
Therefore, in the matrix appearing in (3.39), the first r entries of the last n −r + 1 rows are zero. This implies that these rows are linearly dependent, and thus the determinant is zero.
(b) Note that as we move down the rows of the skew diagram λ/µ, we start a new connected component whenever µr ≥λr+1 for some 1 ≤r < n. In this case, we have λi ≤λr+1 ≤µr ≤µj for 1 ≤j ≤r < i ≤n, and so λi −µj −i + j < 0 for 1 ≤j ≤r < i ≤n.
60 Schur functions Therefore, the matrix (hλi−µj−i+j)1≤i,j≤n can be written in block form A C 0 B , where A is of size r × r and B is of size (n −r) × (n −r). Thus sλ/µ = det(hλi−µj−i+j)1≤i,j≤n = det(A) det(B) = det(hλi−µj−i+j)1≤i,j≤r det(hλi−µj−i+j)r+1≤i,j≤n = sθ1sθ2, where θ1 = (λ1, . . . , λr)/(µ1, . . . , µr) and θ2 = (λr+1, . . . , λn)/(µr+1, . . . , µn).
The result then follows by induction.
Example 3.4.7. Suppose λ/µ is a horizontal strip and let ν1, . . . , νr be the lengths of the components of the strip. Then, by Proposition 3.4.6, we have sλ/µ = s(ν1)s(ν2) · · · s(νr) (3.16) = hν = hν1hν2 · · · hνr.
Similarly, if λ/µ is a vertical strip, then sλ/µ = eν = eν1eν2 · · · eνr, where νi are the lengths of the components of the strip.
△ In finitely many variables, we have an even stricter condition for the skew Schur functions to be nonzero.
Lemma 3.4.8. We have sλ/µ(x1, . . . , xn) = 0 unless 0 ≤λ′ i −µ′ i ≤n for all i ≥1.
Proof. We already know that sλ/µ = 0 if λ′ r < µ′ r for some r ≥1 by Proposition 3.4.6(a).
Now suppose that λ′ r −µ′ r > n for some r ≥1. Then we have λ′ i −µ′ j ≥λ′ r −µ′ r > n for 1 ≤i ≤r ≤j ≤n.
Therefore λ′ i −µ′ j −i + j > n for 1 ≤i ≤r ≤j ≤n.
Since ek(x1, . . . , xn) = 0 for k > n, this implies that the matrix eλ′ i−µ′ j−i+j 1≤i,j≤n has an r × (n + 1 −r) block of zeros in the top-right corner. So it has block form A 0(r−1)×1 0(r−1)×(n−r) B 01×1 01×(n−r) C D E , where 0p×q denotes a p × q zero matrix, A has size (r −1) × (r −1), and E has size (n −r) × (n −r). It follows that sλ/µ(x1, . . . , xn) = det eλ′ i−µ′ j−i+j 1≤i,j≤n = det(A) det(01×1) det(E) = 0.
Skew Schur functions and semistandard tableaux 61 Let x = (x1, x2, . . . ), y = (y1, y2, . . . ), z = (z1, z2, . . . ) be three sets of independent variables and let sλ(x, y) denote the Schur function corresponding to λ in the set of variables (x1, y1, x2, y2, . . . ).
Proposition 3.4.9. For λ, µ ∈Par with λ ⊇µ, we have (3.46) sλ/µ(x, y) = X ν:λ⊇ν⊇µ sλ/ν(x)sν/µ(y).
Proof. We have X λ,ν∈Par sλ/ν(x)sλ(z)sν(y) (3.38) = X λ,ν,µ∈Par cλ νµsµ(x)sλ(z)sν(y) (3.36) = X ν,µ∈Par sµ(x)sν(z)sµ(z)sν(y) (3.28) = ∞ Y i,k=1 (1 −xizk)−1 ! ∞ Y j,k=1 (1 −yjzk)−1 !
= ∞ Y k=1 ∞ Y i=1 (1 −xizk)−1 ∞ Y j=1 (1 −yjzk)−1 !
(3.28) = X λ∈Par sλ(x, y)sλ(z).
Comparing coefficients of sλ(z) then gives (3.47) sλ(x, y) = X ν∈Par sλ/ν(x)sν(y) (3.45) = X ν⊆λ sλ/ν(x)sν(y).
This is the special case of (3.46) where µ = ∅. Then X µ⊆λ sλ/µ(x, y)sµ(z) = sλ(x, y, z) = X ν⊆λ sλ/ν(x)sν(y, z) = X µ⊆ν⊆λ sλ/ν(x)sν/µ(y)sµ(z).
In the first equality above, we have used (3.47) after replacing the variables x with (x1, y1, x2, y2, . . . ) and the variables y with z. The second and third equalities are is similar.
Finally, comparing coefficients of sµ(z) yields (3.46).
Corollary 3.4.10. Suppose x(1), . . . , x(n) are n sets of variables and λ, µ ∈Par. Then (3.48) sλ/µ x(1), . . . , x(n) = X (ν) n Y i=1 sνi/νi−1 x(i) , where the sum is over all sequences (ν) = (ν0, ν1, . . . , νn) of partitions such that µ = ν0 ⊆ν1 ⊆· · · ⊆νn = λ.
62 Schur functions Proof. This follows from Proposition 3.4.9 by induction on n.
Recall the definition of semistandard tableaux from Section 2.3. We define semistandard skew tableaux in an analogous way. Precisely, for λ, µ ∈Par with λ ⊇µ, a semistandard skew tableau of shape λ/µ is a filling of the boxes of the skew diagram λ/µ with positive integers such that the entries are weakly increasing along rows and strictly increasing down columns. We let SST(λ/µ) denote the set of semistandard skew tableaux of shape λ/µ and, for n ≥1, we let SST(λ/µ, n) denote the set of such skew tableaux with entries chosen from the set {1, 2, . . . , n}. The weight wt(T) of a skew tableau is defined as for usual tableaux and we again use the notation xT := xwt(T).
Now note that a sequence (ν) = (ν0, ν1, . . . , νn) of partitions with µ = ν0 ⊆ν1 ⊆· · · ⊆νn = λ such that each νi/νi−1 is a horizontal strip is equivalent to a semistandard skew tableau of shape λ/µ with entries chosen from {1, 2, . . . , n}. Namely, the skew tableau νi corresponds to the set of boxes with entries ≤i. For example, the semistandard skew tableau 1 1 3 4 1 2 4 2 3 5 2 3 1 2 3 corresponds to the sequence , , , , , , where we have indicated in gray the boxes that have been added at each step.
Theorem 3.4.11 (Tableau description of skew Schur functions). For λ, µ ∈Par and n ≥1, we have (3.49) sλ/µ = X T∈SST(λ/µ) xT and sλ/µ(x1, . . . , xn) = X T∈SST(λ/µ,n) xT.
Proof. It suffices to prove the second equality, since first then follows by taking the limit n →∞. For 1 ≤i ≤n, let x(i) consist of the single variable xi. Then Corollary 3.4.10 gives sλ/µ(x1, . . . , xn) = X (ν) n Y i=1 sνi/νi−1(xi).
Skew Schur functions and semistandard tableaux 63 Now, for a single variable xi, it follows from Lemma 3.4.8 that sνi/νi−1(xi) is zero unless νi/νi−1 is a horizontal strip, in which case sνi/νi−1(xi) = x|νi|−|νi−1| i . Thus, each sequence (ν) appearing in the sum above corresponds to a tableau, and we have sλ/µ(x1, . . . , xn) = X (ν) n Y i=1 x|νi|−|νi−1| i = X T∈SST(λ/µ,n) xT.
Corollary 3.4.12. For λ ∈Par and n ≥1, we have (3.50) sλ = X T∈SST(λ) xT and sλ(x1, . . . , xn) = X T∈SST(λ,n) xT.
Proof. This follows from taking µ = ∅in (3.49).
Example 3.4.13. The set SST((2, 1, 1), 4) consists of the tableaux 1 1 2 3 , 1 2 2 3 , 1 3 2 3 , 1 4 2 3 , 1 1 2 4 , 1 2 2 4 , 1 3 2 4 , 1 4 2 4 , 1 1 3 4 , 1 2 3 4 , 1 3 3 4 , 1 4 3 4 , 2 2 3 4 , 2 3 3 4 , 2 4 3 4 .
Thus s(2,1,1)(x1, x2, x3, x4) = 3x1x2x3x4 + x2 1x2x3 + x1x2 2x3 + x1x2x2 3 + x2 1x2x4 + x1x2 2x4 + x1x2x2 4 + x2 1x3x4 + x1x2 3x4 + x1x3x2 4 + x2 2x3x4 + x2x2 3x4 + x2x3x2 4.
△ Definition 3.4.14 (Kostka numbers). For λ, µ, ν ∈Par with λ ⊇µ, the Kostka number Kλ/µ,ν is the number of semistandard skew tableaux of shape λ/µ and weight ν. We write Kλ,µ for Kλ/∅,µ.
The Kostka numbers are the coefficients giving the skew Schur functions in terms of the monomial symmetric functions.
Proposition 3.4.15. For λ, µ ∈Par with λ ⊇µ, we have (3.51) sλ/µ = X ν∈Par Kλ/µ,νmν.
64 Schur functions Proof. This follows from Theorem 3.4.11.
Example 3.4.16. From Example 3.4.13, we see that K(2,1,1),(14) = 3, K(2,1,1),(2,1,1) = 1, K(2,2,1),(2,2) = 0, K(2,2,1),(3,1) = 0, K(2,2,1),(4) = 0.
Thus s(2,1,1)(x1, x2, x3) = 3m(14) + m2,1,1.
△ Exercises.
3.4.1. Write the Schur function s(2,2) as a linear combination of the monomial symmetric functions.
3.4.2. Write the Schur function s(3,2) as a linear combination of the monomial symmetric functions.
3.4.3. Write the skew Schur function s(3,2)/(1) as a linear combination of monomial symmetric functions.
3.4.4. Write the skew Schur function s(3,3,1)/(2,1) as a linear combination of monomial sym-metric functions.
3.4.5. Suppose λ, µ, ν ∈Par with λ ⊇µ. Show that ⟨sλ/µ, hν⟩= Kλ/µ,ν and ⟨sλ/µ, eν⟩= Kλ′/µ′,ν.
3.5 Transition matrices In Propositions 2.4.4, 2.5.5, 2.6.7 and 3.4.15 we have given explicit formulas expressing bases for symmetric functions in terms of the monomial symmetric functions. Our goal in this section to complete this picture and give formulas for converting from any basis to any other. We follow the presentation of [Mac15, §6].
We will work with matrices whose rows and columns are indexed by partitions of a positive integer n. In doing so, we will order the partitions of n in reverse lexicographic order, so that (n) comes first and (1n) comes last. Note from Exercise 1.2.4 that (3.52) λ > µ = ⇒λ >lex µ, but the converse implication does not hold.
We say that a matrix M = (Mλ,µ)λ,µ∈Par(n) is dominant upper triangular if Mλ,µ ̸= 0 = ⇒λ ≥µ.
Transition matrices 65 By (3.52), a dominant upper triangular matrix is upper triangular in the usual sense. We say the matrix is dominant upper unitriangular if it is dominant upper triangular and Mλ,λ = 1 for all λ ∈Par(n). Similarly, it is dominant lower triangular if Mλ,µ ̸= 0 = ⇒λ ≤µ and dominant lower unitriangular if we also have Mλ,λ = 1 for all λ ∈Par(n).
Let UTn (respectively, UT ⊺ n) denote the set of dominant upper (respectively, lower) uni-triangular matrices with integer entries, indexed by the partitions of n.
Lemma 3.5.1. The sets UTn and UT ⊺ n are groups with respect to matrix multiplication.
Proof. Suppose M, N ∈UTn. Then (MN)λ,µ = X ν∈Par(n) Mλ,νNν,µ = X ν∈Par(n) λ≥ν≥µ Mλ,νNν,µ is zero unless these exists ν ∈Par(n) such that λ ≥ν ≥µ. Such a ν exists if and only if λ ≥µ. Furthermore, we have (MN)λ,λ = Mλ,λNλ,λ = 1. Hence MN ∈UTn.
Since the determinant of a triangular matrix is the product of its diagonal entries, we have det(M) = 1, and so M is invertible. Since M is dominant upper unitriangular, it can be transformed to the identity matrix using only row operations that add a multiple of rows indexed by a partition λ to rows indexed by partitions µ < λ. Since this same series of row operations transforms the identity matrix to M −1, we see that M −1 ∈UTn. This completes the proof that UTn is a group.
It remains to prove that UT ⊺ n is a group. Note first that A ∈UTn ⇐ ⇒A ⊺∈UT ⊺ n.
Suppose A, B ∈UT ⊺ n. Then B ⊺A ⊺∈UTn = ⇒AB = B ⊺A ⊺⊺∈UT ⊺ n.
Furthermore A is invertible since A⊺∈UTn is. Finally, (A−1) ⊺= (A ⊺)−1 ∈UTn = ⇒A−1 ∈UT ⊺ n.
Hence UT ⊺ n is also a group.
Let J = (Jλ,µ)λ,µ∈Par(n) be the transposition matrix with entries Jλ,µ = ( 1 if λ′ = µ, 0 otherwise.
Strictly speaking, we have a transposition matrix J for each n. But we will use the same notation J for all of these; the n will be clear from the context. Note that (3.53) J2 = I and J ⊺= J, where I denotes the identity matrix.
66 Schur functions Lemma 3.5.2. Suppose λ, µ ∈Par(n). Then λ ≥µ ⇐ ⇒µ′ ≥λ′.
Proof. Since λ′′ = λ, it suffices to prove one implication. Suppose that µ′ ̸≥λ′. We wish to show that λ ̸≥µ.
Let i ≥1 be the smallest index such that (3.54) λ′ 1 + · · · + λ′ i > µ′ 1 + · · · + µ′ i.
Thus λ′ 1 + · · · + λ′ j ≤µ′ 1 + · · · + µ′ j, 1 ≤j ≤i −1, and so λ′ i > µ′ i.
Let l = λ′ i and m = µ′ i. Since |λ| = |µ|, it follows from (3.54) that (3.55) λ′ i+1 + λ′ i+2 + · · · < µ′ i+1 + µ′ i+2 + · · · .
Since λ′ i+1 + λ′ i+2 + · · · is the number of boxes in the diagram of λ lying to the right of the i-th column, we have λ′ i+1 + λ′ i+2 + · · · = l X j=1 (λj −i).
Similarly, µ′ i+1 + µ′ i+2 + · · · = m X j=1 (µj −i).
Therefore it follows from (3.55) that (3.56) m X j=1 (µj −i) > l X j=1 (λj −i) ≥ m X j=1 (λj −i), where the second inequality holds since l > m and λj ≥i for 1 ≤j ≤l. It follows from (3.56) that µ1 + · · · + µm > λ1 + · · · + λm.
Hence λ ̸≥µ, as desired.
Corollary 3.5.3. A matrix M is dominant upper triangular (respectively, dominant upper unitriangular) if and only if JMJ is dominant lower triangular (respectively, dominant lower unitriangular).
Proof. Let N = JMJ. Then Nλ,µ = Mλ′,µ′. Then the result follows from Lemma 3.5.2.
If (uλ)λ∈Par(n) and (vλ)λ∈Par(n) are two Q-bases of Symn Q, we let M(u, v) = (Mλ,µ(u, v))λ,µ∈Par(n) denote the matrix of coefficients in the equations (3.57) uλ = X µ∈Par(n) Mλ,µ(u, v)vµ, λ ∈Par(n).
We call M(u, v) the transition matrix from the basis (uλ)λ∈Par(n) to the basis (vλ)λ∈Par(n). It is an invertible matrix of rational numbers. For an invertible matrix M, we define M −⊺:= M ⊺−1 = M −1⊺.
Transition matrices 67 Proposition 3.5.4. Suppose (uλ)λ∈Par(n), (vλ)λ∈Par(n), and (wλ)λ∈Par(n) are all Q-bases of Symn Q. Let (u′ λ)λ∈Par(n) and (v′ λ)λ∈Par(n) be the bases dual to (uλ)λ∈Par(n) and (vλ)λ∈Par(n), respectively, with respect to the Hall inner product. Then the following equalities holds.
M(u, v)M(v, w) = M(u, w), (3.58) M(v, u) = M(u, v)−1, (3.59) M(u′, v′) = M(v, u) ⊺= M(u, v)−⊺, (3.60) M(ω(u), ω(v)) = M(u, v).
(3.61) Proof. The proof of this proposition is left as Exercise 3.5.1.
Let (3.62) K = M(s, m).
Note that this notation matches (3.51) and recall from Definition 3.4.14 that Kλ,µ, called a Kostka number, is the number of semistandard tableaux of shape λ and weight µ.
Theorem 3.5.5. The transition matrices between the bases (eλ)λ∈Par(n), (hλ)λ∈Par(n), (mλ)λ∈Par(n), (fλ)λ∈Par(n), and (sλ)λ∈Par(n) are given in the following table, where the entry in row u and column v is M(u, v): e h m f s e I K⊺JK−⊺ K⊺JK K⊺K K⊺J h K⊺JK−⊺ I K⊺K K⊺JK K⊺ m K−1JK−⊺ K−1K−⊺ I K−1JK K−1 f K−1K−⊺ K−1JK−⊺ K−1JK I K−1J s JK−⊺ K−⊺ K JK I All of the matrices in this table have integer entries.
Proof. By (3.18) and Theorem 3.3.7, we have s′ = s, e′ = f, f ′ = e, h′ = m, m′ = h.
Thus M(s, h) (3.60) = M(s′, h′)−⊺= M(s, m)−⊺(3.62) = K−⊺.
By (3.17), we have M(ω(s), s) = J.
Thus M(s, e) (3.61) = M(ω(s), ω(e)) (2.31) = M(ω(s), h) (3.58) = M(ω(s), s)M(s, h) = JK−⊺.
Finally, M(s, f) (3.60) = M(s′, f ′)−⊺= M(s, e)−⊺= JK−⊺−⊺= J−⊺K (3.53) = JK.
68 Schur functions We can now use (3.58) and (3.59) to compute all the other entries of the given table, going via the Schur functions. For example M(m, e) (3.58) = M(m, s)M(s, e) (3.59) = M(s, m)−1M(s, e) = K−1JK−⊺.
The fact that all the matrices have integer entries follows from the fact that all of the bases are bases of Sym over Z.
We can now deduce various triangularity properties of our various bases.
Proposition 3.5.6.
(a) The matrix K is dominant upper unitriangular.
(b) The matrices M(s, h) and M(h, s) are dominant lower unitriangular.
(c) The matrices M(s, m) and M(m, s) are dominant upper unitriangular.
Proof.
(a) Suppose T is a semistandard tableau of shape λ and weight µ. Then, for each r ≥1, there are µ1 + · · · + µr entries ≤r in T. These must all be located in the top r rows of T since the entries of T strictly increase down columns. Thus µ1 + · · · + µr ≤λ1 + · · · + λr for all r ≥1.
Hence µ ≤λ. Thus K is dominant upper triangular. The same argument also shows that Kλ,λ = 1 for all λ, and so K is dominant upper unitriangular.
(b) Since K is dominant upper unitriangular, K⊺is dominant lower unitriangular. Thus K−⊺is also dominant lower unitriangular by Lemma 3.5.1.
Hence M(s, h) = K−⊺and M(h, s) = K⊺are both dominant lower unitriangular.
(c) This follows immediately from the fact that M(s, m) = K, M(m, s) = K−1, and K is dominant upper unitriangular.
We can give additional combinatorial interpretations of the transition matrices M(e, m) and M(h, m). (Compare to Propositions 2.4.4 and 2.5.5.) If M is a matrix, its i-th row sum is the sum of the entries in its i-th row. Similarly, its i-th column sum is the sum of the entries in its i-th column.
Proposition 3.5.7. Suppose λ, µ ∈Par(n).
(a) M(e, m)λ,µ = P ν∈Par(n) Kν,λKν′,µ is the number of matrices of zeros and ones whose i-th row sum is λi and whose i-th column sum is µi.
(b) M(h, m)λ,µ = P ν∈Par(n) Kν,λKν,µ is the number of matrices of nonnegative integers whose i-th row sum is λi and show i-th column sum is µi.
Proof.
(a) First note that M(e, m)λ,µ = X ν,σ∈Par(n) (K ⊺)λ,νJν,σKσ,µ = X ν,σ∈Par(n) Kν,λδν′σKσ,µ = X ν∈Par(n) Kν,λKν′,µ.
Transition matrices 69 Now, for µ ∈Par(n), consider the coefficient of the monomial xµ in eλ = eλ1eλ2 · · · . Each monomial in eλi is of the form Q j x aij j , where aij ∈{0, 1}, and P j aij = λi. Thus we have Y i,j x aij j = Y j x µj j , and hence P i aij = µj. Thus the matrix (aij) has i-th row sum equal to λi and j-th column sum equal to µj.
(b) The proof is analogous. The only difference is that we replace eλ by hλ, and thus the exponents aij can now be arbitrary nonnegative integers.
Theorem 3.5.5 is missing the basis of power sum symmetric functions. We now consider the transition matrices involving the Q-basis (pλ)λ∈Par. Let (3.63) L = M(p, m), so that pλ = X µ∈Par(n) Lλ,µmµ, λ ∈Par(n).
By Proposition 2.6.7, Lλ,µ is the number of Young tableaux of shape λ and weight µ in which the entries in each row are constant. It follows that, if Lλ,µ ̸= 0, then each µi is a sum of parts of λ. If this is the case, we say that λ is a refinement of µ. For example (5, 5, 4, 3, 3, 2, 1, 1, 1, 1) is a refinement of (9, 7, 6, 2, 1, 1), where the blue parts of the first partition have been added to obtain the blue part of the second partition, and similarly for the other colours.
The notion of refinement can be explained in an equivalent way as follows. For partitions λ and µ, we let λ ∪µ denote the partition obtained by taking all the parts of λ and µ and arranging them in weakly decreasing order. Thus, for example, (6, 4, 3, 3, 1) ∪(7, 5, 4, 3, 2, 1, 1) = (7, 6, 5, 4, 4, 3, 3, 3, 2, 1, 1, 1).
Then λ is a refinement of µ if λ is of the form S i≥1 λ(i), where each λ(i) is a partition of µi.
We write λ ≤R µ if λ is a refinement of µ. It is straightforward to verify that ≤R is a partial order on Par(n) for all n ≥0.
Lemma 3.5.8. For λ, µ ⊢n, we have λ ≤R µ = ⇒λ ≤µ.
Proof. Suppose λ, µ ⊢n and λ ≤R µ.
Thus we have a function f : {1, 2, . . . , ℓ(λ)} → {1, 2, . . . , ℓ(µ)} such that µj = X i:f(i)=j λi, 1 ≤j ≤ℓ(µ).
Then, for 1 ≤k ≤ℓ(λ), we have λ1 + · · · + λk ≤ X i∈f({1,2,...,k}) µi ≤µ1 + · · · + µk, where the last inequality holds since f({1, 2, . . . , k}) contains at most k elements and µ1 + . . . + µk is the sum of the k largest parts of µ.
70 Schur functions Corollary 3.5.9. The matrix L is dominant lower triangular.
Proof. As discussed above, Proposition 2.6.7 implies that Lλ,µ = 0 unless λ ≤R µ. Thus, by Lemma 3.5.8, Lλ,µ = 0 unless λ ≤µ. Hence L is dominant lower triangular.
Theorem 3.5.10. The transition matrices involving the power sum symmetric functions are given by M(p, e) = εzL−⊺, (3.64) M(p, f) = εL, (3.65) M(p, h) = zL−⊺, (3.66) M(p, s) = LK−1, (3.67) where ε and z are diagonal matrices given by ελ,µ = δλµ(−1)|λ|−ℓ(λ), zλ,µ = δλµzλ.
Proof. By (3.18) and Theorem 3.3.7, the bases dual to (hλ)λ∈Par(n) and (pλ)λ∈Par(n) are (mλ)λ∈Par(n) and (z−1 λ pλ)λ∈Par(n), respectively. Note also that M(zu, v) = zM(u, v) and M(εu, v) = εM(u, v) for any bases (uλ)λ∈Par(n) and (uλ)λ∈Par(n) of Symn Q. (See Exercise 3.5.2.) Thus M(p, h) (3.60) = M(z−1p, m)−⊺= (z−1L)−⊺= z ⊺L−⊺= zL−⊺, M(p, e) (3.61) = M(ω(p), ω(e)) (2.45) = (2.31) M(εp, h) = εM(p, h) = εzL−⊺, M(p, f) (3.61) = M(ω(p), ω(f)) (2.45) = (2.34) M(εp, m) = εM(p, m) = εL, M(p, s) (3.58) = M(p, m)M(m, s) (3.59) = M(p, m)M(s, m)−1 = LK−1.
We will see later (in Proposition 6.3.8) that M(p, s) gives the character table of the symmetric group.
Exercises.
3.5.1. Prove Proposition 3.5.4.
3.5.2. Suppose that n ≥0 and that (uλ)λ∈Par(n) and (vλ)λ∈Par(n) are two Q-bases of Symn Q.
Furthermore, suppose that D = (Dλ,µ)λ,µ∈Par(n) is a diagonal matrix. Show that M(Du, v) = DM(u, v).
The Littlewood-Richardson rule 71 3.5.3. By Proposition 3.5.6(a) and Exercise 1.2.4, we have Kλ,µ ̸= 0 = ⇒λ ≥lex µ.
Show that the converse of this implication is false in general. That is, find λ, µ ∈Par, |λ| = |µ|, such that λ >lex µ but Kλ,µ = 0.
3.5.4. In this exercise, we consider partitions of n = 3.
(a) Write the partitions of n in reverse lexicographic order.
(b) Write down the matrices J, K, L, ε, and z.
(c) Compute M(h, f), M(s, e), M(p, f), and M(p, h).
(d) Write p(1n) as a linear combination of the forgotten symmetric functions.
(e) Write pn as a linear combination of the complete homogeneous symmetric functions.
3.5.5. Repeat Exercise 3.5.4 for n = 4.
3.6 The Littlewood-Richardson rule Our goal in this section is to prove the Littlewood–Richardson rule, which tells us how to write the product of two Schur functions as a linear combination of Schur functions. There are many different proofs of this important result. We will follow the relatively short proof found in [Ste02]. Other proofs can be found in [Egg19, Ch. 10] and [Mac15, §I.9].
Before proving the full Littlewood–Richardson rule, we begin with two important spe-cial cases, known as Pieri’s formulas, which are much easier to prove. These formulas are corollaries of the following result.
Proposition 3.6.1. For λ, µ ∈Par, we have hµsλ = X ν⊇λ Kν/λ,µsν and eµsλ = X ν⊇λ Kν′/λ′,µsν.
Proof. We have hµsλ = X ν∈Par ⟨sν, hµsλ⟩sν (3.35) = X ν⊇λ ⟨sν/λ, hµ⟩sν (3.51) = X ν⊇λ ρ∈Par Kν/λ,ρ⟨mρ, hµ⟩sν (3.18) = X ν⊇λ Kν/λ,µsν.
This proves the first equality. Applying ω then gives the second equality.
Corollary 3.6.2 (Pieri’s formula). For λ ∈Par and r ∈N, we have (3.68) hrsλ = X ν sν, where the sum is over all partitions ν ⊇λ such that ν/λ is a horizontal strip with r boxes.
72 Schur functions Proof. Consider Proposition 3.6.1 with µ = (r). Since Kν/λ,(r) is the number of semistandard skew tableaux of shape ν/λ whose entries are all 1s, we have Kν/λ,(r) = ( 1 if ν/λ is a horizontal strip, 0 otherwise.
The result follows.
Corollary 3.6.3 (Dual Pieri’s formula). For λ ∈Par and r ∈N, we have (3.69) ersλ = X ν sν, where the sum is over all partitions ν ⊇λ such that ν/λ is a vertical strip with r boxes.
Proof. We apply the involution ω to (3.68).
Example 3.6.4. The ways of adding a horizontal strip with 2 boxes to (3, 2, 2) are , , , , .
Thus we have h2s(3,2,2) = s(5,2,2) + s(4,3,2) + s(4,2,2,1) + s(3,2,2,2).
On the other hand, the ways of adding a vertical strip with 2 boxes to (3, 2, 2) are , , , , .
Thus we have e2s(3,2,2) = s(4,3,2) + s(4,2,2,1) + s(3,3,3) + s(3,3,2,1) + s(3,2,2,1,1).
△ We now turn our attention to the full Littlewood–Richardson rule. A key element in the proof will be a certain set of involutions on tableaux, which we now describe.
Suppose T is a semistandard tableau and fix a positive integer j. We say that an entry j in T is free if there is no entry j + 1 in the same column. Similarly, we say that an entry j + 1 in T is free if there is no entry j in the same column. In any given row, the free entries must occur in consecutive columns. (See Exercise 3.6.1.) We define βj(T) to be the tableau obtained from T by interchanging the number of free j’s and (j + 1)’s in each row (always keeping j’s to the left of (j +1)’s, so that the tableau remains semistandard). The operation βj is called a Bender–Knuth involution. Since the non-free j’s and (j + 1)’s occur in pairs, we see that (3.70) wt(βj(T)) = wt(T)σj, where σj is the transposition of j and j + 1.
The Littlewood-Richardson rule 73 Example 3.6.5. Fix j = 3. In the following tableau, of all the entries equal to 3 or 4, the free entries are coloured green and the non-free entries are coloured red: T = 1 1 1 1 1 2 2 2 2 2 2 3 4 6 2 2 2 2 2 3 3 3 3 3 4 4 5 7 3 3 3 4 4 4 5 5 6 6 4 4 5 5 6 7 6 Applying the Bender–Knuth involution, we obtain β3(T) = 1 1 1 1 1 2 2 2 2 2 2 3 3 6 2 2 2 2 2 3 3 4 4 4 4 4 5 7 3 3 3 3 4 4 5 5 6 6 4 4 5 5 6 7 6 .
We have wt(T) = (5, 11, 9, 8, 5, 5, 2) and wt(βj(T)) = (5, 11, 8, 9, 5, 5, 2) = wt(T)σ3.
△ For a skew tableau T and positive integer j, let T≥j denote the subtableau of T consisting of columns j, j + 1, . . . . We define Tj analogously. For example, if T = 1 2 1 2 4 2 2 3 1 2 3 then T≥3 = 1 2 1 2 4 2 2 3 , T>3 = 1 2 1 2 4 2 3 , and T<3 = 1 2 3 .
Proposition 3.6.6. For all λ ∈Parn and all µ, ν ∈Par such that ν ≤µ, we have (3.71) aλ+δn(x1, . . . , xn)sµ/ν(x1, . . . , xn) = X T aλ+wt(T)+δn(x1, . . . , xn), where the sum is over all T ∈SST(µ/ν, n) such that λ + wt(T≥j) ∈Parn for all j ≥1.
74 Schur functions Proof. Fix λ ∈Parn and µ, ν ∈Par such that ν ≤µ. Since sµ/ν is a symmetric function, we have sµ/ν(x1, . . . , xn) = π−1sµ/ν(x1, . . . , xn) (2.5) = X T∈SST(µ/ν,n) xwt(T)π for all π ∈Sn.
Thus (3.72) aλ+δnsµ/ν(x1, . . . , xn) = X π∈Sn X T∈SST(µ/ν,n) sgn(π)x(λ+δn+wt(T))π = X T∈SST(µ/ν,n) aλ+wt(T)+δn.
We say that T ∈SST(µ/ν, n) is a gardyloo if λ + wt(T≥j) / ∈Parn for some j ≥1. Therefore, T is a gardyloo if and only if (3.73) λk + wt(T≥j)k < λk+1 + wt(T≥j)k+1 for some k, j. (Since wt(T) is the weight of T, wt(T)k is the number of entries of T equal to k.) Our goal is to show that all gardyloo terms cancel in (3.72).
Suppose T is a gardyloo. Let j be the largest integer such λ + wt(T≥j) / ∈Parn. Then choose the smallest k such that (3.73) holds. Thus we have (3.74) λk + wt(T>j)k ≥λk+1 + wt(T>j)k+1 and λk + wt(T≥j)k < λk+1 + wt(T≥j)k+1.
It follows that wt(T≥j)k+1 −wt(T>j)k+1 > wt(T≥j)k −wt(T>j)k.
Now, wt(T≥j)i −wt(T>j)i is the number of i’s in the j-th column of T. Since the entries in T strictly decrease down columns, this number is either zero or one. So we must have wt(T≥j)k+1 −wt(T>j)k+1 = 1 and wt(T≥j)k −wt(T>j)k = 0.
It then follows from (3.74) that the j-th column of T contains a k + 1 but no k, and that (3.75) λk + wt(T≥j)k + 1 = λk+1 + wt(T≥j)k+1.
Let T ∗denote the tableau obtained from T by applying the Bender–Knuth involution βk to the subtableau T<j of T, leaving the remainder of T unchanged. This involves changing some subset of the entries of T<j from k to k + 1 and vice-versa. Since the j-th column of T contains a k + 1 but no k, the tableau T ∗is semistandard. Furthermore, since T≥j = (T ∗)≥j, we have (T ∗)∗= T. Hence T 7→T ∗is an involution on the set of gardyloos.
Now we compare the contributions of T and T ∗to the sum (3.72). It follows from (3.70) that wt(T wt(T≥j)i. Hence wt(T≥j) / ∈Par, and so T is not a Littlewood–Richardson tableau.
Conversely, suppose T is not a Littlewood–Richardson tableau. Let j be the smallest integer such that wt(T≥j) / ∈Par. Then there exists some i such that wt(T≥j)i+1 > wt(T≥j)i.
Then the entries i and i + 1 in T≥j are arranged as in (3.77), with all i entries preceding the gray box in word(T) lying in T≥j. (Any entry i in a column < j and in a row above the grey box would have to have an entry i + 1 immediately below it, contradicting our choice of j.) Hence word(T) is not a lattice word.
It follows from the Littlewood–Richardson rule that the product of two Schur functions is a nonnegative integer linear combination of Schur functions. Later we will see how this fact also follows from a representation theoretic interpretation of the Littlewood–Richardson rule; see Remark 6.3.7. These applications to representation theory will also help explain the importance of the Littlewood–Richardson rule.
Exercises.
3.6.1. Prove that for a semistandard tableau T and a fixed positive integer j, the corre-sponding free entries in any row must lie in consecutive columns.
3.6.2. Consider the tableau T = 1 1 1 1 2 2 2 3 3 3 4 2 2 3 3 3 4 4 5 3 4 4 5 5 6 5 8 .
Compute β1(T), β2(T), β4(T) and β9(T).
78 Schur functions 3.6.3. Show that, for α ∈WCompn and σ ∈Sn, we have σaα = aασ−1.
3.6.4. Deduce the dual Pieri formula (3.69) directly from the Littlewood–Richardson rule, as in Example 3.6.10.
3.6.5. Prove that cλ µν = cλ νµ for all λ, µ, ν ∈Par.
3.7 The Murnaghan–Nakayama rule Our goal in this section is to prove the Murnaghan–Nakayama rule, which tells us how to write the product of a power sum and a Schur function as a linear combination of Schur functions.
This result will be important in our proof of the boson-fermion correspondence. When we discuss the connection between symmetric functions and characters of the symmetric group, we will see that this rule also gives us a combinatorial method to compute the irreducible characters of the symmetric group.
A skew Young diagram is a border strip if it is connected and contains no 2 × 2 block of squares. Thus the successive rows (or columns) of a border strip overlap by exactly one square. For example, (3.78) , , , , and are border strips, but , , , and are not, since the first two contain 2 × 2 boxes and the last two are not connected. The height ht(θ) of a border strip θ is one less than the number of rows it contains. For example, the heights of the border strips in (3.78) are 4, 0, 3, 2, and 4, respectively.
Proposition 3.7.1 (Murnaghan–Nakayama rule). For all λ ∈Par and r ≥1, we have (3.79) prsλ = X µ (−1)ht(µ/λ)sµ, where the sum is over all partitions µ ⊇λ such that µ/λ is a border strip with r boxes.
Proof. Fix λ ∈Par and r ≥1. It suffices to consider finitely many indeterminates x1, x2, . . . , xn, where n ≥|λ| + r. Let χi ∈WCompn be the composition with 1 in the i-th position and 0’s elsewhere. Then we have (3.80) praλ+δn = X π∈Sn n X i=1 sgn(π)x(λ+δn)π+rχi = X π∈Sn sgn(π) n X i=1 x(λ+δn+rχπ−1(i))π = X π∈Sn sgn(π) n X i=1 x(λ+δn+rχi)π = n X i=1 aλ+δn+rχi.
The Murnaghan–Nakayama rule 79 Now, we would like to rearrange the terms of λ + δn + rχi in descending order. Since λ ∈Parn, the terms of λ + δn are strictly decreasing. (See Lemma 3.1.5.) Thus we have λ1 + n −1 > λ2 + n −2 > · · · > λn.
The composition λ + δn + rχi only differs from λ + δn in the i-th position. Thus, in order to arrange the terms of λ + δn + rχi in descending order, we only need to move its i-th term λi + n −i + r into the correct position. If λ + δn + rχi has two equal terms, then the alternating polynomial aλ+δn+rχi in (3.80) is zero. Otherwise, there exists some j ≤i such that λj−1 + n −j + 1 > λi + n −i + r > λj + n −j.
That is, we must move the term λi + n −i + r to the j-th position in order to arrange the terms in descending order. (We adopt the convention here that λ0 = ∞when j = 1, so that the first inequality above is automatically satisfied.) Then we have aλ+δn+rχi = (−1)i−jaµ+δn, where µ = (λ1, . . . , λj−1, λi + j −i + r, λj + 1, . . . , λi−1 + 1, λi+1, . . . , λn) ∈Parn.
Note that µk = λk−1 + 1 for j < k ≤i.
So the Young diagrams look something like where λ consists of the white boxes and µ consists of the white and gray boxes. In particular, we see that µ/λ is a border strip with r boxes. In the above example, j = 4 (the first row containing a gray box) and i = 7 (the last row containing a gray box). The values of i for which λ + δn + rχi has two equal terms correspond precisely to those i where one cannot add to λ a border strip whose bottom row is the i-th row.
It follows from the above discussion that praλ+δn = X µ (−1)ht(µ/λ)aµ+δn, where the sum is over all partitions µ ⊆λ such that µ/λ is a border strip with r boxes.
Then the result follows from (3.7) after dividing both sides by aδn.
80 Schur functions Exercises.
3.7.1. Write p1s(5,3,2,2) as a linear combination of Schur functions.
3.7.2. Write p3s(4,4,2,1) as a linear combination of Schur functions.
3.7.3. Use the Murnaghan—Nakayama rule to write the power sum pr, r ≥1, as a linear combination of Schur functions.
Chapter 4 Adjoint operators and Hopf algebras So far we have considered Sym only as a ring. In this chapter we turn our attention to additional structure on Sym.
We begin with a brief discussion of tensor products over commutative rings, since this concept is necessary for the additional structure. We then explore adjoint operators with respect the Hall inner product, which are a crucial ingredient of the Heisenberg algebra that we will discuss in the next chapter. Finally, we define Hopf algebras and define a Hopf algebra structure on Sym. In addition to being important in its own right, this structure will play an important role in our discussion of the Heisenberg algebra and applications to representation theory.
4.1 Tensor products The material in this and following chapters will refer to tensor products of modules over a commutative ring k. (In our case, k will usually be either Z or Q.) In this section we give a brief treatment of such tensor products. We give their definition and some of their most important properties. We omit proofs, which can be found, for example, in [Sav, Ch. 3].
Fix a commutative ring k. Let U and V be k-modules. Let F be the free k-module with basis U × V . Let G be the submodule of F generated by the elements (u + u′, v) −(u, v) −(u′, v), (u, v + v′) −(u, v) −(u, v′), (cu, v) −c(u, v), (u, cv) −c(u, v), for u, u′ ∈U, v, v ∈V , c ∈k. The tensor product of U and V over k is U ⊗k V := F/G.
If the ground ring k is clear from the context, we sometimes write U ⊗V for U ⊗k V . We set u ⊗v := (u, v) + G, u ∈U, v ∈V.
Therefore, every element of U ⊗k V can be written in the form t = n X i=1 ui ⊗vi, n ∈N, u1, . . . , un ∈U, v1, . . . , vn ∈V.
Keep in mind that such a representation is not unique in general.
81 82 Adjoint operators and Hopf algebras Warning 4.1.1. It is not true that every element of U ⊗V is of the form u ⊗v for some u ∈U, v ∈V . An element of the form u ⊗v is called a simple tensor. Every element can be written as a finite sum of simple tensors. Therefore, when defining a k-linear map whose domain is U ⊗V , we will often define it on simple tensors. There is then at most one way to extend it to all of U ⊗V .
It follows from the definition that u ⊗v is linear in each entry: u ⊗(v + v′) = u ⊗v + u ⊗v′, (u + u′) ⊗v = u ⊗v + u′ ⊗v, (4.1) c(u ⊗v) = cu ⊗v = u ⊗cv.
(4.2) Thus, (4.3) h: U × V →U ⊗R V, h(m, n) = m ⊗n, is an k-bilinear map. The tensor product is uniquely characterized by the following universal property.
Proposition 4.1.2 (Universal property of the tensor product). For every k-module W and bilinear map φ: U × V →W, there exists a unique k-module homomorphism ˜ φ: T →W such that ˜ φ ◦h = φ; that is, the diagram U × V U ⊗k V W h φ ˜ φ commutes.
Proof. For a proof, see [Sav, Prop. 3.2.2].
Defining a map on a tensor product can be subtle, since elements of a tensor product are equivalence classes. See Exercise 4.1.2. Typically the best way to define a map whose domain is a tensor product U ⊗k V is to use Proposition 4.1.2. More precisely, you define the map on U × V , show that the map is bilinear, and then use Proposition 4.1.2 to conclude that the map factors through U ⊗k V .
Example 4.1.3. Fix m, n ∈Z>0. Let’s prove that we have an isomorphism of Z-modules (Z/mZ) ⊗Z (Z/nZ) ∼ = Z/ gcd(m, n)Z.
For x ∈Z, let [x]k denote its image in Z/kZ. Let d = gcd(m, n) and consider the map φ: (Z/mZ) × (Z/nZ) →Z/dZ, φ([a]m, [b]n) = [ab]d.
We leave it as Exercise 4.1.1 to prove that this map is well defined (that is, independent of the choice of representatives a and b). Since it is clearly Z-bilinear, it induces a map ˜ φ: (Z/mZ) ⊗Z (Z/nZ) →Z/ gcd(m, n)Z, ˜ φ([a]m ⊗[b]n) = [ab]d.
Tensor products 83 Since ˜ φ([a]m ⊗n) = [a]d, for all a ∈Z, the map ˜ φ is surjective. It remains to show that it is injective. Suppose ˜ φ k X i=1 [ai]m ⊗[bi]n !
= k X i=1 [aibi]d = 0.
Then Pk i=1 aibi ∈dZ and so Pk i=1 aibi = dz for some z ∈Z. Choose x, y ∈Z such that d = xm + yn. (Here we use the fact mZ + nZ = gcd(m, n)Z.) Then we have k X i=1 ([ai]m ⊗[bi]n) = k X i=1 ([ai]m ⊗bin) = k X i=1 (bi[ai]m ⊗n) = k X i=1 ([aibi]m ⊗n) = [dz]m ⊗n = d[z]m ⊗n = xm[z]m ⊗n + yn[z]m ⊗n = m ⊗n + [yz]m ⊗nn = d + [yz]m ⊗n = d.
Thus ˜ φ is injective.
△ The following two propositions summarize the most important properties of the tensor product.
Proposition 4.1.4 (Properties of the tensor product I). Suppose U, V , and W are k-modules. In the following statements, all isomorphisms are isomorphisms of k-modules and all tensor products are over k.
(a) We have an isomorphism U ⊗V ∼ = V ⊗U, u ⊗v 7→v ⊗u.
(b) We have an isomorphism (U ⊗V ) ⊗W ∼ = U ⊗(V ⊗W), (u ⊗v) ⊗w 7→u ⊗(v ⊗w).
(c) We have an isomorphism U ⊗(V ⊕W) ∼ = (U ⊗V ) ⊕(U ⊗W), u ⊗(v + w) 7→u ⊗v + u ⊗w, u ∈U, v ∈V, w ∈W.
84 Adjoint operators and Hopf algebras (d) Viewing k as a module over itself via multiplication, we have an isomorphism (4.4) k ⊗U ∼ = U, c ⊗u 7→cu.
Proof. Proofs of these statements can be found in [Sav, §3.3].
In light of Proposition 4.1.4(b), we will often omit the parentheses in multiple tensor products. For example, we will write U ⊗V ⊗W, and elements of this triple tensor product will be written as finite linear combinations of simple tensors u ⊗v ⊗w, for u ∈U, v ∈V , w ∈W. Similarly, we will often identify k ⊗U, U ⊗k, and U.
Proposition 4.1.5 (Properties of the tensor product II). Suppose U and V are k-modules.
In the following statements, all isomorphisms are isomorphisms of k-modules and all tensor products are over k.
(a) If U and V are generated, as k-modules, by the sets X and Y respectively, then U ⊗V is generated by {u ⊗v : u ∈X, v ∈Y }.
(b) If U and V are free modules with bases X and Y respectively, then U ⊗V is a free module with basis {u ⊗v : u ∈X, v ∈Y }.
(c) If k is a field and U and V are finite-dimensional vector spaces over k, then U ⊗V is a finite-dimensional vector space over k, and dim(U ⊗V ) = (dim U)(dim V ).
(d) We have an isomorphism km ⊗kn ∼ = kmn.
Proof. Proofs of these statements can be found in [Sav, §3.3].
Example 4.1.6. Our main examples will be the Z-algebra Sym and the Q-algebra SymQ. It follows from Proposition 4.1.5(b) that uλ ⊗vµ, λ, µ ∈Par, is a basis of Sym ⊗Z Sym, where (uλ)λ∈Par and (vλ)λ∈Par are any bases of Sym. For example eλ ⊗hµ, λ, µ ∈Par, is a basis of Sym ⊗Sym. If we instead work over Q, then we may also use the Q-basis (pλ)λ∈Par consisting of power sum symmetric functions.
△ We can also take tensor products of k-module homomorphisms. Suppose f1 : U1 →V1 and f2 : U2 →V2 are k-module homomorphisms. Then we have a k-module homomorphism (4.5) f1 ⊗f2 : U1 ⊗U2 →V1 ⊗V2, (f1 ⊗f2)(u1 ⊗u2) := f1(u1) ⊗f2(u2).
Adjoint operators 85 (We then extend to finite sums of simple tensors by linearity.) One can check that the map f1 ⊗f2 is a well-defined k-module homomorphism; see [Sav, §3.4].
If A and B are k-algebras, then their tensor product A ⊗B is a naturally a k-algebra with multiplication given by (a1 ⊗b1)(a2 ⊗b2) = a1b1 ⊗a2b2, a1, a2 ∈A, b1, b2 ∈B.
For example, Sym ⊗Sym is a Z-algebra. We can define an inner product on this algebra by (4.6) ⟨u1 ⊗v1, u2 ⊗v2⟩:= ⟨u1, u2⟩⟨v1, v2⟩, where the angled brackets on the right-hand side denote the Hall inner product.
Exercises.
4.1.1. Prove that the map φ of Example 4.1.3 is well defined.
4.1.2. Suppose k is a field and U and V are nonzero finite-dimensional vector spaces. Is the map φ: U ⊗V →U, φ(u ⊗v) = u, well defined?
4.1.3. Let n ∈Z, n ≥1. Prove that Q ⊗Z (Z/nZ) is the zero Z-module.
4.1.4. Suppose V and W are finite-dimensional vector spaces over a field k.
Let V ∗= Homk(V, k) be the dual space of V . Show that the map ψ: W ⊗k V ∗→Homk(V, W), ψ(w ⊗f)(v) = f(v)w, w ∈W, f ∈V ∗, v ∈V, is an isomorphism of vector spaces.
4.2 Adjoint operators In this section we discuss the adjoint operators, with respect to the Hall inner product, of multiplication by symmetric functions. These adjoint operators will form “half” of the Heisenberg algebra that we will introduce in Section 5.1.
Every ring acts on itself by multiplication. In particular we have an injective ring homo-morphism (4.7) Sym →End(Sym), f 7→(v 7→fv), sending f ∈Sym to the operation of multiplication by f. (See Exercise 4.2.1.) 86 Adjoint operators and Hopf algebras Recall the Hall inner product ⟨, ⟩from Definition 3.3.1. We deduced several important properties of this inner product in Theorem 3.3.7. In particular, ⟨, ⟩is symmetric and positive definite. This allows us to define the adjoint of any linear operator on Sym. In particular, we have the adjoint f ⊥of multiplication by any symmetric function f, as described in the following proposition.
Proposition 4.2.1. For each f ∈Sym there exists a unique linear operator f ⊥∈End(Sym) satisfying (4.8) ⟨f ⊥u, v⟩= ⟨u, fv⟩ for all u, v ∈Sym.
The map Sym →End(Sym), f 7→f ⊥, is an injective ring homomorphism.
Proof. Let f ∈Sym. We first prove existence of a linear operator f ⊥satisfying (4.8). Define f ⊥∈End(Sym) by (4.9) f ⊥u := X λ∈Par ⟨u, fsλ⟩sλ, u ∈Sym.
The fact that f ⊥is a linear map follows easily from the fact that ⟨, ⟩is bilinear. (Exer-cise 4.2.2.) For all v ∈Sym we have ⟨f ⊥u, v⟩= X λ∈Par ⟨u, fsλ⟩⟨sλ, v⟩= u, f X λ∈Par ⟨sλ, v⟩sλ + = ⟨u, fv⟩.
Thus f ⊥satisfies (4.8).
Next we show uniqueness. Suppose that T ∈End(Sym) satisfies ⟨Tu, v⟩= ⟨u, fv⟩ for all u, v ∈Sym.
Then, for all u, v ∈Sym, we have T −f ⊥ u, v = ⟨Tu, v⟩−⟨f ⊥u, v⟩= ⟨u, fv⟩−⟨u, fv⟩= 0.
Thus (T −f ⊥)(u) = 0 for all u, and hence T = f ⊥.
Next we show that the map f 7→f ⊥is injective. Suppose that f ∈Sym and f ⊥= 0.
Then 0 = ⟨f ⊥f, 1⟩= ⟨f, f⟩.
Since ⟨, ⟩is positive definite, this implies that f = 0.
It remains to show that the map f 7→f ⊥is a ring homomorphism. For all u, v, f, g ∈Sym, we have ⟨(f + g)⊥u, v⟩= ⟨u, (f + g)v⟩= ⟨u, fv⟩+ ⟨u, gv⟩= ⟨f ⊥u, v⟩+ ⟨g⊥u, v⟩= ⟨(f ⊥+ g⊥)u, v⟩.
Adjoint operators 87 It then follows from the uniqueness proved above that (f + g)⊥= f ⊥+ g⊥. Similarly, ⟨(fg)⊥u, v⟩= ⟨u, fgv⟩= ⟨u, gfv⟩= ⟨g⊥u, fv⟩= ⟨f ⊥g⊥u, v⟩, and so (fg)⊥= f ⊥g⊥. Finally, ⟨1⊥u, v⟩= ⟨u, 1v⟩= ⟨1u, v⟩, and so 1⊥= 1.
Lemma 4.2.2. For f ∈Symn, we have f ⊥(Symm) ⊆Symm−n.
(We say that f ⊥is an operator of degree −n.) Proof. We leave the proof of this result as Exercise 4.2.3.
We can reinterpret the skew Schur functions in terms of adjoint operators.
Lemma 4.2.3. For λ, µ ∈Par, we have (4.10) sλ/µ = s⊥ µ sλ.
Proof. We have sλ/µ (3.34) = X ν∈Par ⟨sλ, sµsν⟩sν = X ν∈Par ⟨s⊥ µ sλ, sν⟩sν = s⊥ µ sλ.
Let x = (x1, x2, . . . ) and y = (y1, y2, . . . ) be two sets of indeterminates.
Lemma 4.2.4. For all f ∈Sym, we have (4.11) f(x, y) = X µ∈Par (s⊥ µ f)(x)sµ(y).
Proof. First note that sλ(x, y) (3.46) = X µ∈Par sλ/µ(x)sµ(y) = X µ∈Par (s⊥ µ sλ)(x)sµ(y).
Thus, (4.11) holds for f = sλ, λ ∈Par. Since both sides of (4.11) are linear in f, and the Schur functions form a basis of Sym, it follows that (4.11) holds for all f ∈Sym.
Lemma 4.2.5. For λ ∈Par and n ≥1, we have h⊥ λ mµ = ( mν if µ = λ ∪ν, 0 if µ is not of the form λ ∪ν for some ν ∈Par.
88 Adjoint operators and Hopf algebras Proof. Since the basis (hλ)λ∈Par is dual to the basis (mλ)λ∈Par by (3.18), the coefficient of mν when we write h⊥ λ mµ in the basis of monomial symmetric functions is ⟨h⊥ λ mµ, hν⟩. Then the result follows from the computation ⟨h⊥ λ mµ, hν⟩= ⟨mµ, hλhν⟩= ⟨mµ, hλ∪ν⟩ (3.18) = δµ,λ∪ν.
We now turn our attention to the operators p⊥ n . Since the power sum symmetric functions are only a basis over Q, we work over Q for the remainder of the section.
Recall from Theorems 2.4.6, 2.5.3 and 2.6.4 that SymQ = Q[e1, e2, . . . ] = Q[h1, h2, . . . ] = Q[p1, p2, . . . ].
Thus, for n ≥1, we have linear operators (4.12) ∂ ∂en , ∂ ∂hn , ∂ ∂pn ∈End(SymQ).
The operator ∂ ∂en acts on an element f ∈SymQ by acting on its expression as a polynomial in the elementary symmetric functions, and similarly for ∂ ∂hn and ∂ ∂pn .
Example 4.2.6. Using the transition matrix M(e, h) = K⊺JK−⊺from Theorem 3.5.5, we can compute e3 = h(3) −2h(2,1) + h(1,1,1) = h3 −2h2h1 + h3 1.
Thus ∂e3 ∂h1 = ∂ ∂h1 h3 −2h2h1 + h3 1 = −2h2 + 3h2 1.
△ Proposition 4.2.7. For n ≥1, we have (4.13) p⊥ n = n ∂ ∂pn .
Proof. Since the pλ, λ ∈Par, form a basis of SymQ, it suffices to prove that (4.14) ⟨p⊥ n pλ, pµ⟩= n ∂pλ ∂pn , pµ for all λ, µ ∈Par.
We have ⟨p⊥ n pλ, pµ⟩= ⟨pλ, pnpµ⟩= ⟨pλ, pµ∪(n)⟩ (3.29) = ( zλ if λ = µ ∪(n), 0 otherwise.
On the other hand ∂pλ ∂pn = ( mn(λ)pµ if λ = µ ∪(n) for some µ, 0 if mn(λ) = 0.
Thus, the equality (4.14) holds when λ has no part equal to n.
On the other hand, if λ = µ ∪(n), we have ∂pλ ∂pn , pµ = mn(λ)⟨pµ, pµ⟩ (3.29) = mn(λ)zµ.
Adjoint operators 89 Since zλz−1 µ = Y i≥1 imi(λ)mi(λ)!
! Y i≥1 imi(µ)mi(µ)!
!−1 = nmn(λ), we have ⟨p⊥ n pλ, pµ⟩= zλ = nmn(λ)zµ = n ∂pλ ∂pn , pµ , as desired.
If k is a commutative ring and A is a k-algebra, a k-derivation (or simply a derivation, if the ground ring k is clear from the context) of A is a k-linear map D: A →A satisfying the Leibniz rule: D(ab) = D(a)b + aD(b), a, b ∈A.
A derivation is uniquely determined by its action on a set of generators, since one can inductively use the Leibniz rule to compute its value on any product of the generators.
Corollary 4.2.8. The operator p⊥ n is a Q-derivation of SymQ.
Proof. This follows from (4.13) and the fact that the partial derivative ∂ ∂pn satisfies the Leibniz rule.
Proposition 4.2.9. For n ≥1, we have (4.15) p⊥ n = ∞ X r=0 hr ∂ ∂hn+r = (−1)n−1 ∞ X r=0 er ∂ ∂en+r .
Proof. Since p⊥ n is a derivation by Corollary 4.2.8, it suffices to check the given equalities on a set of generators. For the first equality, we check on the generators hm, m ≥1. First note that, for all λ ∈Par(n), we have ⟨hn, pλ⟩ (2.49) = X µ∈Par(n) z−1 µ ⟨pµ, pλ⟩ (3.29) = 1.
Thus, for all m ≥n ≥1, λ ∈Par(m −n), we have ⟨p⊥ n hm, pλ⟩= ⟨hm, pnpλ⟩= 1 = ⟨hm−n, pλ⟩.
Since p⊥ n hm ∈Symm−n by Lemma 4.2.2, and the pλ, λ ∈Par(m−n), form a basis of Symm−n, it follows that p⊥ n hm = hm−n = ∞ X r=0 hr ∂hm ∂hn+r .
This proves the first equality in (4.15).
Next, for λ ∈Par(m −n), we have ⟨p⊥ n em, pλ⟩= ⟨em, pnpλ⟩ (3.31) = ⟨ω(em), ω(pnpλ)⟩ 90 Adjoint operators and Hopf algebras (2.31) = (2.45)(−1)n−1⟨hm, pnω(pλ)⟩ = (−1)n−1⟨p⊥ n hm, ω(pλ)⟩ = (−1)n−1⟨hm−n, ω(pλ)⟩ (3.31) = (−1)n−1⟨ω(hm−n), pλ⟩ = (−1)n−1⟨em−n, pλ⟩.
Since p⊥ n em ∈Symm−n by Lemma 4.2.2, and the pλ, λ ∈Par(m−n), form a basis of Symm−n, it follows that p⊥ n em = (−1)n−1em−n = (−1)n−1 ∞ X r=0 er ∂em ∂en+r .
This proves the second equality in (4.15).
Remark 4.2.10. Note that even though the sums in (4.15) are infinite, all but finitely many terms act as zero on any given element of Sym. Thus, these infinite sums are well-defined operators on Sym.
△ The involution ω of (2.31) is compatible with adjoints in the following sense.
Proposition 4.2.11. For all f, g ∈Sym, (4.16) ω(f ⊥g) = ω(f)⊥ω(g).
Proof. For all f, g, h ∈Sym, we have ω(f ⊥g), h = f ⊥g, ω(h) = ⟨g, fω(h)⟩ (3.31) = ⟨ω(g), ω(f)h⟩= ω(f)⊥ω(g), h .
Since this holds for all h ∈Sym, the result follows.
Exercises.
4.2.1. Show that (4.7) is an injective ring homomorphism.
4.2.2. Show that f ⊥, as defined by (4.9), is a linear map Sym →Sym.
4.2.3. Prove Lemma 4.2.2.
4.2.4. Suppose that k is a commutative ring, A is a k-algebra, and D: A →A is a k-derivation. Prove that D(1A) = 0.
4.2.5. Show that, for f, g ∈Sym, we have (f ⊗g)⊥= f ⊥⊗g⊥. Here (f ⊗g)⊥denotes the operator Sym ⊗Sym →Sym ⊗Sym adjoint, with respect to the inner product (4.6), to multiplication by f ⊗g.
Hopf algebras 91 4.2.6. Prove that {f ⊥: f ∈SymQ} = SpanQ ∂k ∂pi1∂pi2 · · · ∂pik : k ∈N, i1, . . . , ik ≥1 .
Here we view both sides as subspaces of EndQ(Sym).
4.3 Hopf algebras In Section 4.4 will define the structure of a Hopf algebra on the ring of symmetric functions.
In preparation for this, we give in this section the definition of a Hopf algebra. For a more detailed treatment, see, for example, [DNR01].
Throughout this section we fix a commutative ring k. (In most of our applications, we will take k = Z or k = Q.) All tensor products are over k. If U and V are two k-modules, we let flip: U ⊗V ∼ = − →V ⊗U, flip(u ⊗v) = v ⊗u, be the isomorphism of Proposition 4.1.4(a).
Definition 4.3.1 (Algebra). A (unital associative) algebra over k is an k-module A together with k-linear maps ∇: A ⊗A →A and η: k →A, such that the following two diagrams commute: (4.17) A ⊗A ⊗A A ⊗A A ⊗A A ∇⊗id id⊗∇ ∇ ∇ A A ⊗A A ⊗A A η⊗id id⊗η id ∇ ∇ The map ∇is called the multiplication (or product) of A and η is the unit of A. We often refer to A itself as being an algebra, leaving the maps ∇and ε implied. We say that A is commutative if ∇◦flip = ∇.
Note that, in (4.17), when we write η ⊗id: A →A ⊗A, we really mean the composite A ∼ = − − → (4.4) k ⊗A η⊗A − − →A ⊗A, and similarly for id ⊗η: A →A ⊗A. When comparing Definition 4.3.1 to the definition of k-algebra given in Section 1.3, the unit η corresponds to the map η: k →A, η(c) = c1A, where 1A is the multiplicative identity of A. It is then straightforward to see that, if A1 and A2 are algebras, a k-linear map φ: A1 →A2 is a morphism of algebras (also called a 92 Adjoint operators and Hopf algebras homomorphism of algebras) in the usual sense if and only if the following diagrams commute: A1 ⊗A1 A1 A2 ⊗A2 A2 ∇1 φ⊗φ φ ∇2 k A1 A2 η1 η2 φ Here ∇i and ηi denote the multiplication and unit of Ai, respectively, for i = 1, 2.
Reversing all the arrows in the diagrams (4.17) gives the following definition of a coalge-bra.
Definition 4.3.2 (Coalgebra). A (counital coassociative) coalgebra over k is a k-module C together with k-linear maps ∆: C →C ⊗C and ε: C →k, such that the following two diagrams commute: (4.18) C C ⊗C C ⊗C C ⊗C ⊗C ∆ ∆ id⊗∆ ∆⊗id C C ⊗C C ⊗C C ∆ ∆ id id⊗ε ε⊗id The map ∆is called the comultiplication (or coproduct) of C and ε is the counit of C. We often refer to C itself as a coalgebra, leaving the maps ∆and ε implied. We say that C is cocommutative if flip ◦∆= ∆.
Note that, in (4.18), when we write ε ⊗id: C ⊗C →C, we really mean the composite C ⊗C ε⊗id − − →k ⊗C ∼ = − − → (4.4) C, and similarly for id ⊗ε: C ⊗C →C.
It is useful to define a consistent notation for writing out the comultiplication explicitly.
If C is a coalgebra and c ∈C, we write (4.19) ∆(c) = X (c) c(1) ⊗c(2).
This is known as Sweedler notation.
Definition 4.3.3 (Coalgebra morphism). Suppose (C1, ∆1, ε1) and (C2, ∆2, ε2) are coalge-bras. A k-linear map φ: C1 →C2 is a morphism of coalgebras if the following diagrams commute: C1 ⊗C1 C1 C2 ⊗C2 C2 φ⊗φ ∆1 φ ∆2 k C1 C2 ε1 φ ε2 Hopf algebras 93 Definition 4.3.4 (Bialgebra). A bialgebra over k is a tuple (B, ∇, η, ∆, ε) such that (B, ∇, η) is an algebra, (B, ∆, ε) is a coalgebra, and the following diagrams commute (4.20) B ⊗B B B ⊗B B ⊗B ⊗B ⊗B B ⊗B ⊗B ⊗B ∆⊗∆ ∇ ∆ id⊗flip ⊗id ∇⊗∇ (4.21) B ⊗B B k ∇ ε⊗ε ε k B B ⊗B η η⊗η ∆ B k k ε η id We often refer to B itself as a bialgebra, leaving the maps ∇, η, ∆, and ε implied.
A morphism of bialgebras is a map that is both a morphism of algebras and a morphism of coalgebras.
Remark 4.3.5. Commutativity of the diagrams (4.20) and (4.21) corresponds to the state-ment that ∇and η are morphisms of coalgebras or, equivalently, that ∆and ε are morphisms of algebras.
△ Definition 4.3.6 (Hopf algebra). A Hopf algebra H over k is a bialgebra over k, together with a k-linear map S : H →H, called the antipode such that the following diagram com-mutes: (4.22) H ⊗H H ⊗H H K H H ⊗H H ⊗H S⊗id ∇ ε ∆ ∆ η id⊗S ∇ If H and H′ are Hopf algebras, then a map φ: H →H′ is a morphism of Hopf algebras if φ is a morphism of bialgebras. It then follows automatically (see [DNR01, Prop. 4.2.5]) that f ◦S = S′ ◦f, where S and S′ are the antipodes of H and H′ respectively.
Remark 4.3.7. It can be shown (see [DNR01, Rem. 4.2.3]) that if a bialgebra admits an antipode, then the antipode is unique.
Therefore, the antipode is not really additional structure. Being a Hopf algebra is a property that a bialgebra may possess. Furthermore, the antipode is an antihomomorphism of algebras and coalgebras (see [DNR01, Prop. 4.2.6]).
In other words, S ◦η = η, S ◦∇= ∇◦flip ◦(S ⊗S), (4.23) ε ◦S = ε, ∆◦S = (S ⊗S) ◦flip ◦∆.
(4.24) The equations (4.23) are equivalent to the statements △ (4.25) S(1) = 1, S(ab) = S(b)S(a) for all a, b.
94 Adjoint operators and Hopf algebras Examples 4.3.8.
(a) If G is a group, then the group algebra kG is a cocommutative Hopf algebra with ∆g = g ⊗g, ε(g) = 1, S(g) = g−1 for all g ∈G.
It is commutative if and only if G is abelian.
(b) If V is a vector space over a field k, then the tensor algebra T(V ) := L∞ n=0 V ⊗n is an algebra with multiplication given by tensor product and multiplicative identity given by the identity element of V ⊗0 ∼ = k. It is also a cocommutative Hopf algebra with ∆1 = 1 ⊗1, ∆v = v ⊗1 + 1 ⊗v, ε(v) = 0, S(v) = −v, for all v ∈V.
We extended these definitions to higher tensor powers using the fact that ∆and ε must be morphisms of algebras and S is an antihomomorphism of algebras. This Hopf algebra is commutative if and only if V has dimension less than or equal to 1.
(c) If g is a Lie algebra, then its universal enveloping algebra U(g) is a cocommutative Hopf algebra with ∆a = a ⊗1 + 1 ⊗a, ε(a) = 0, S(a) = −a, for all a ∈g.
We then extend these definitions to all of U(g) as above. This Hopf algebra is commu-tative if and only if g is an abelian Lie algebra.
(d) Quantized enveloping algebras provide examples of Hopf algebras that are neither commutative nor cocommutative.
△ An element g in a coalgebra is said to be grouplike if g ̸= 0 and ∆g = g ⊗g.
For example, in the Hopf algebra kG of Example 4.3.8(a), the elements of the group G are grouplike. (In fact, this is where the term grouplike comes from.) An element a in a coalgebra is said to be primitive if ∆a = a ⊗1 + 1 ⊗a.
For example, in the Hopf algebra U(g) of Example 4.3.8(c), the elements of g are primitive.
Remark 4.3.9. The main motivation behind the definition of bialgebras and Hopf algebras is the additional structure they provide to modules. If A and B are k-algebras, U is an A-module, and V is a B-module, then U ⊗V is naturally an (A ⊗B)-module with action given by (a ⊗b)(u ⊗v) = au ⊗bv, a ∈A, b ∈B, u ∈U, v ∈V.
Thus, if U and V are both A-modules, U ⊗V is naturally an (A ⊗A)-module. However, if A is a bialgebra, then U ⊗V is also an A-module with action given by (4.26) a · (u ⊗v) := ∆(a)(u ⊗v), extended by linearity. Furthermore, if A is a Hopf algebra, then the dual V ∗= Homk(V, k) is also an A-module, with action △ (4.27) (af)(v) := f(S(a)v), a ∈A, f ∈V ∗, v ∈V.
Hopf algebra structure on the ring of symmetric functions 95 Examples 4.3.10.
(a) Suppose G is a group and U and V are kG-modules. Then U ⊗V and V ∗are kG-modules with actions given by g · (u ⊗v) = gu ⊗gv, (gf)(v) = f(g−1v), g ∈G, u ∈U, v ∈V, f ∈V ∗.
(b) Suppose g is a Lie algebra and U and V are g-modules. Then U ⊗V and V ∗are g-modules with actions given by x · (u ⊗v) = xu ⊗v + u ⊗xv, (xf)(v) = −f(xv), x ∈g, u ∈U, v ∈V, f ∈V ∗.
Here we use the fact that a g-module is equivalent to a U(g)-module.
△ Exercises.
4.3.1. Prove that ε(g) = 1 for every grouplike element g in a coalgebra.
4.3.2. Prove that the set of grouplike elements of a Hopf algebra is a group under the multiplication ∇of the bialgebra.
4.3.3. Prove that the set of primitive elements of a coalgebra over k is a k-submodule.
4.3.4. In the notation of Remark 4.3.9, prove that (4.26) defines the structure of an A-module on U ⊗k V and that (4.27) defines the structure of an A-module on V ∗.
4.3.5. Suppose (C, ∆, ε) is a counital coassociative coalgebra over a field k. Show that the dual space C∗= Homk(C, k) is a unital associative algebra with multiplication ∇: C∗⊗C∗→C∗, ∇(f ⊗g)(x) = (f ⊗g) ◦∆(x), f, g ∈C∗, x ∈C, extended by linearity, and unit η: k →C∗, η(a) = aε, a ∈k.
4.4 Hopf algebra structure on the ring of symmetric func-tions In this section we endow the ring of symmetric functions with the structure of a Hopf algebra.
This structure is interesting in its own right, and will also be important in the applications to representation theory that we will discuss in the sequel. Throughout this section we work over the ground ring Z.
We can identify Sym ⊗Sym 96 Adjoint operators and Hopf algebras with the ring of functions of bounded degree in two sets of indeterminates x = (x1, x2, . . . ) and y = (y1, y2 . . . ) that are symmetric in each set separately via the map f ⊗g 7→f(x)g(y), f, g ∈Sym.
This map sends the basis mλ ⊗mµ, λ, µ ∈Par, to the basis mλ(x)mµ(y), λ, µ ∈Par. On the other hand, for any f ∈Sym, the function f(x, y) is symmetric in (x1, y1, x2, y2, · · · ).
Therefore it also lies in Sym ⊗Sym. So we have a comultiplication map (4.28) ∆: Sym →Sym ⊗Sym, (∆f)(x, y) = f(x, y).
We also define a counit map (4.29) ε := ρ0 : Sym →Z, where ρ0 is the map of (2.11) and we recall that Sym0 = Z. Thus ε(f) is the constant term of f. Finally, we define an antipode (4.30) S : Sym →Sym, S(hn) = (−1)nen, extended to all of Sym by (4.25).
Theorem 4.4.1. The maps ∆, ε, and S endow Sym with the structure of a commutative and cocommutative Hopf algebra.
Proof. We first show that (Sym, ∆, ε) is a coalgebra. To do this, we identify Sym⊗Sym⊗Sym with the ring of functions in three sets of indeterminates x = (x1, x2, . . . ), y = (y1, y2, . . . ), and z = (z1, z2, . . . ) that are symmetric in each set of variables separately. Then, for f ∈ Sym, we have (id ⊗∆) ◦∆(f) (x, y, z) = f(x, y, z) = (∆⊗id) ◦∆(f) (x, y, z), and so the first diagram in (4.18) commutes. Next, note that (id ⊗ε) ◦∆(f) is the constant term, with respect to the y, of f(x, y). This is precisely, f(x). Thus (id ⊗ε) ◦∆(f) = f.
Similarly, (ε ⊗id) ◦∆(f) = f.
Thus the second diagram in (4.18) commutes. This completes the proof that (Sym, ∆, ε) is a coalgebra.
Next we consider (4.20). For f, g ∈Sym, we have ∆◦∇(f ⊗g) = ∆(fg) = (fg)(x, y).
On the other hand (∇⊗∇)◦(id ⊗flip ⊗id) ◦(∆⊗∆)(f ⊗g) = (∇⊗∇) ◦(id ⊗flip ⊗id)(f(x, y) ⊗g(x, y)) Hopf algebra structure on the ring of symmetric functions 97 = (fg)(x, y) = ∆◦∇(f ⊗g).
Thus, diagram (4.20) commutes.
Next we compute that, for f, g ∈Sym, ε ◦∇(f ⊗g) = ε(fg) = ε(f)ε(g), since the constant term of fg is the product of the constant terms of f and g. So the first diagram in (4.21) commutes. It is straightforward to verify that the last two diagrams in (4.21) also commute. (See Exercise 4.4.1.) Proposition 4.4.2. For all n ≥1, we have (4.31) ∆hn = n X r=0 hr ⊗hn−r, ∆en = n X r=0 er ⊗en−r, ∆pn = pn ⊗1 + 1 ⊗pn.
Proof. By definition, hn(x, y) is the sum of all monomials of total degree n in the indetermi-nates x1, y1, x2, y2, . . . . Any such monomial is a product of a monomial in the xi of degree r times a monomial of degree n −r in the yi, for some 0 ≤r ≤n. Thus (∆hn)(x, y) = hn(x, y) = n X r=0 X 1≤i1≤i2≤···≤ir 1≤j1≤j2≤···≤jn−r xi1xi2 · · · xiryj1yj2 · · · yjn−r = n X r=0 hr(x)hn−r(y).
Similarly, (∆en)(x, y) = en(x, y) = n X r=0 X 1≤i1<i2<···<ir 1≤j1<j2<··· 1, where we recall the convention that hr = 0 for r < 0.
4.4.7. Prove that, for all m, n ≥1, (4.37) e⊥ men = en−m and e⊥ mhn = hn if m = 0, hn−1 if m = 1, 0 if m > 1, where we recall the convention that er = 0 for r < 0.
Chapter 5 The boson-fermion correspondence In this section we explore the boson-fermion correspondence, which is a precise relationship between a natural representation, known as bosonic Fock space, of the infinite rank Heisen-berg algebra, and a natural representation, known as fermionic Fock space, of the infinite rank Clifford algebra. We begin by defining the Heisenberg algebra and its action on bosonic Fock space. We then define the Clifford algebra and its action on fermionic Fock space.
Finally we describe the two halves of the boson-fermion correspondence: bosonization and fermionization.
5.1 The Heisenberg algebra In this section we construct the Heisenberg algebra from the ring of symmetric functions.
We start with a rather abstract definition, as the Heisenberg double of the Hopf algebra of symmetric functions, and then give a more concrete presentation that matches the usual definition of the Heisenberg algebra.
Recall from (4.7) that we have an injective ring homomorphism (5.1) Sym →End(Sym), f 7→f +, where f +(v) = fv, sending f ∈Sym to the operation f + of multiplication by f. We let Sym+ denote the image of this map. Recall also, from Proposition 4.2.1, that we have an injective ring homomorphism (5.2) Sym →End(Sym), f 7→f −:= f ⊥, where f ⊥is the operator adjoint to f + with respect to the Hall inner product.
We let Sym−denote the image of this map. Thus Sym+ and Sym−are both subrings of End(Sym) isomorphic to Sym.
Definition 5.1.1 (Heisenberg double). We define the Heisenberg double Heis of Sym to be the subring of End(Sym) generated by Sym+ and Sym−. In other words Heis is the subring of End(Sym) generated by the operations of multiplication by elements of Sym and the adjoints of these operations with respect to the Hall inner product.
We can also work over Q and define HeisQ as above, replacing Sym by SymQ everywhere.
Then we have HeisQ := Q ⊗Z Heis.
102 The Heisenberg algebra 103 Remark 5.1.2. The Heisenberg double is a more general construction than what we have presented here. More generally, one can form a Heisenberg double from a pair of dual Hopf algebras. Since Sym is self-dual as a Hopf algebra (Remark 4.4.7), we are able to take the Heisenberg double of Sym with itself.
△ We define an involution (5.3) ω: Heis →Heis, ω(a) := ω ◦a ◦ω, a ∈Heis, where, on the right-hand side, ω is the involution of Sym from (2.31).
We leave it as Exercise 5.1.1 to show that (5.4) ω(f ±) = ω(f)± for all f ∈Sym.
Thus, under the inclusion (5.1) of Sym in Heis, the involution (5.3) extends the involution (2.31). This justifies our use of the same notation for both.
Our goal now is to give a presentation of the Heisenberg double. In other words, we want a description in terms of generators and relations.
We begin by computing a key commutation relation.
Proposition 5.1.3. For f, g ∈Sym, we have (5.5) f −g+ = X (f) f ⊥ (1)g + f − (2), where we have used Sweedler notation (4.19).
Proof. For f, g, h ∈Sym, (f −g+)h = f ⊥(gh) (4.34) = X (f) f ⊥ (1)g f ⊥ (2)h = X (f) f ⊥ (1)g + f − (2) h.
Proposition 5.1.4. We have an isomorphism of Z-modules β : Sym ⊗Z Sym ∼ = − →Heis, f ⊗g 7→f +g−.
Proof. Since the map Sym × Sym →Heis, (f, g) 7→f +g−, is bilinear, it induces the map β, which is therefore well defined. By definition, elements of Heis are linear combinations of products of f ±, f ∈Sym. Proposition 5.1.3 allows us to move elements of the form f + to the left of elements of the form f −. Thus the map β is surjective.
It remains to prove that β is injective. Suppose, towards a contradiction, that β(a) = 0 for some a ∈Sym ⊗Sym, a ̸= 0. By Proposition 4.1.5(b), we can write a = X λ,µ∈Par aλµsλ ⊗sµ, aλµ ∈Z, 104 The boson-fermion correspondence where only finitely many of the aλµ are nonzero. Choose ν minimal, with respect to the partial order ⊇, amongst the partitions such that aλν ̸= 0 for some λ. Then we have 0 = β(a)(sν) = X λ,µ∈Par aλµsλs⊥ µ sν (4.10) = X λ,µ∈Par aλµsλsν/µ (3.45) = X λ∈Par aλνsλ.
Since the Schur functions are linearly independent, this implies that aλν = 0 for all λ ∈Par, which is a contradiction. Hence β is injective.
Corollary 5.1.5. If (uλ)λ∈Par and (vλ)λ∈Par are two bases of Sym (respectively, SymQ), then u+ λ v− µ , λ, µ ∈Par, is a Z-basis of Heis (respectively, a Q-basis of HeisQ).
Proof. This follows immediately from Proposition 5.1.4.
Propositions 5.1.3 and 5.1.4 allow us to find many different presentations of Heis and HeisQ. We can choose any set of generators for Sym+, any set of generators for Sym−, and then use Proposition 5.1.3 to compute the relations satisfied by these elements. The following theorem is one example of this.
Theorem 5.1.6. As a Q-algebra, HeisQ is generated by p± n , n ≥1, subject to the relations (5.6) p+ mp+ n = p+ n p+ m, p− mp− n = p− n p− m, p− mp+ n = p+ n p− m + nδmn, m, n ≥1.
Proof. Let H denote the Q-algebra generated by p± n , n ≥1, subject to the relations (5.6).
In HeisQ, we have p+ mp+ n = p+ n p+ m, p− mp− n = p− n p− m, and p− mp+ n (5.5) = (4.31) (1⊥pn)+p− m + (p⊥ mpn)+1−(4.13) = p+ n p− m + nδmn.
Thus we have a map H →HeisQ, p± n 7→p± n . This map is surjective by Theorem 2.6.4 and injective by Corollary 5.1.5. (See Remark 5.1.7.) The infinite rank Heisenberg algebra is the Q-algebra generated by p± n , n ≥1, subject to the relations (5.6). Thus, Theorem 5.1.6 states that the Heisenberg double is isomorphic to the infinite rank Heisenberg algebra. From now on, we will identify the two, and simply refer to HeisQ as the Heisenberg algebra. The relations (5.6), which are sometimes called the canonical commutations relations, are the relations satisfied by position and momentum operators in quantum mechanics. These relations imply the Heisenberg uncertainty principle.
Remark 5.1.7. The Heisenberg Lie algebra h is the Lie algebra spanned (over Q) by p± n , n ≥1, and c, with Lie bracket given by [p+ m, p+ n ] = 0 = [p− m, p− n ], [p− m, p+ n ] = nδmnc, [c, p± n ] = 0, m, n ≥1.
Then HeisQ is the quotient of the universal enveloping algebra of h by the relation c = 1. It follows from the Poincaré–Birkhoff–Witt theorem that this quotient is isomorphic to Q[p+ 1 , p+ 2 , . . . ] ⊗Z Q[p− 1 , p− 2 , . . . ] as a Z-module.
△ The Heisenberg algebra 105 Remark 5.1.8. In order to work with the generators p± n , we need to work over Q instead of over Z (see Remark 2.6.5). However, if we instead choose generators e± n or h± n , we can give a presentation of Heis, which is an integral form of the Heisenberg algebra. See Exercises 5.1.2 to 5.1.4.
△ The definition of a Z-graded ring is obtained from Definition 1.3.1 by replacing N by Z everywhere, and the definition of a Z-graded algebra is analogous. The Heisenberg algebra is a Z-graded algebra, with gradation given by (5.7) HeisQ = M n∈Z Heisn Q, Heisn Q := {a ∈HeisQ : aSymm Q ⊆Symm+n Q for all m ∈N}, where we adopt the convention that Symm = 0 for m < 0.
Proposition 5.1.9. For all n ∈Z, we have Heisn Q = SpanQ{s+ λ s− µ : λ, µ ∈Par, |λ| −|µ| = n}.
Proof. By Corollary 5.1.5, s+ λ s− µ , λ, µ ∈Par, is a Q-basis of HeisQ. Thus, it suffices to show that s+ λ s− µ ∈Heis|λ|−|µ| Q for all λ, µ ∈Par.
Suppose f ∈Symm Q . Then, by Lemma 4.2.2, we have s⊥ µ f ∈Symm−|µ| Q . Thus s+ λ s− µ f = sλs⊥ µ f ∈Symm+|λ|−|µ| Q .
Exercises.
5.1.1. Prove that the involution ω of (5.4) satisfies ω(f ±) = ω(f)± for all f ∈Sym.
5.1.2. Prove that h− me+ n = e+ n h− m + e+ n−1h− m−1 and e− mh+ n = h+ n e− m + h+ n−1e− m−1 for all m, n ∈N.
5.1.3. Prove that h− mh+ n = min(m,n) X r=0 h+ n−rh− m−r and e− me+ n = min(m,n) X r=0 e+ n−re− m−r for all m, n ∈N.
106 The boson-fermion correspondence 5.1.4. Prove that s− λ s+ µ = X ν⊆λ ρ⊆µ cλ νρs+ µ/ρs− ν for all λ, µ ∈Par.
5.1.5. Let m, n ∈N. In HeisQ, write p− me+ n and p− mh+ n as linear combinations of elements of the form f +g−, f, g ∈Sym.
5.2 Bosonic Fock space In this section we define bosonic Fock space, which is a natural action of the Heisenberg algebra on the space of symmetric functions. Throughout this section we will work over the field Q of rational numbers. (However, see Remark 5.2.4.) Recall that if A is a Q-algebra, then a left A-module is equivalent to a Q-algebra homo-morphism (5.8) A →EndQ(V ) for some Q-module V . Here EndQ(V ) denotes the Q-algebra of Q-linear endomorphisms of V . The map (5.8) is called a representation of A.
Since HeisQ is, by definition, a subring of EndQ(SymQ), we have the inclusion map HeisQ , →EndQ(SymQ).
This is called the Fock space representation of the Heisenberg algebra HeisQ, and the HeisQ-module SymQ is called bosonic Fock space.
Remark 5.2.1. The term boson refers to a type of subatomic particle with the property that many particles can occupy the same state. If we view pn as a particle in state n, then, for example, p3 1p2 2p5 corresponds to a system with three particles in state 1, two particles in state 2, and one particle in state 5. Soon we will encounter fermionic Fock space, where there can be at most one particle in each state. See Remark 5.3.4.
△ Recall that a left ideal I of a Q-algebra A is maximal if the only left ideal of A properly containing I is equal to A itself.
Lemma 5.2.2. The set I−:= SpanQ{f +g−: f ∈SymQ, g ∈Symn Q, n ≥1} is a maximal left ideal of HeisQ.
Proof. The set I−is clearly closed under addition. For p, f, g ∈SymQ, h ∈Symn Q, n ≥1, we have p+f − g+h− (5.5) = X (f) p+ f ⊥ (1)g + f − (2)h−= X (f) pf ⊥ (1)g + f(2)h −∈I−.
Bosonic Fock space 107 Hence I−is a left ideal of Heis.
It remains to show that I−is maximal. It follows from Corollary 5.1.5 that we have a decomposition of Q-modules HeisQ = I−⊕Sym+ Q.
Thus I−̸= HeisQ. Now suppose J is a left ideal of HeisQ properly containing I−. Then it follows from the above decomposition that J contains a nonzero element f + ∈Sym+ Q.
Amongst all such elements, choose one for which f has the smallest degree. If we can show that f is a nonzero constant polynomial, then J contains the multiplicative identity 1, and so J = HeisQ, and we are done.
Suppose, towards a contradiction, that f is nonconstant. By Theorem 2.6.4, f is non-constant polynomial in the power sums pn. So there exists n ∈N such that ∂f ∂pn ̸= 0. Now note that p− n f + (5.5) = (4.31) (p⊥ n f)+ + f +p− n , and so n ∂f ∂pn + (4.13) = (p⊥ n f)+ = p− n f + −f +p− n ∈J, (We use here the fact that f +p− n ∈I−⊆J by the definition of I−, and p− n f + ∈J since J is a left ideal.) Since the degree of n ∂f ∂pn is less than the degree of f, this contradicts our choice of f.
Recall that the annihilator of an element v in an A-module V is AnnA(v) = {a ∈A : av = 0}.
The annihilator AnnA(v) is a left ideal of A. If V is a HeisQ-module, then we call a nonzero element v ∈V a highest weight vector if I−v = 0. In other words, v is a highest weight vector if I−⊆AnnHeis(v). For example, any nonzero constant is a highest weight vector in SymQ.
The following result gives an important characterization of bosonic Fock space. It is closely related to the Stone–von Neumann theorem. Recall that a nonzero A-module V is simple if its only submodules are {0} and V .
Theorem 5.2.3. Bosonic Fock space is a simple HeisQ-module. Furthermore, any HeisQ-module generated by a highest weight vector is isomorphic to bosonic Fock space.
Proof. Since I−is maximal left ideal of HeisQ by Lemma 5.2.2, the quotient HeisQ/I−is a simple HeisQ-module. Since bosonic Fock space is generated by the highest weight element 1 ∈SymQ, it thus suffices to prove that any HeisQ-module generated by a highest weight vector is isomorphic to HeisQ/I−.
Suppose V is a HeisQ-module generated by a highest weight vector v. (Since highest weight vectors are nonzero by definition, this implies that V is nonzero.) Then we have a surjective homomorphism of left HeisQ-modules φ: HeisQ →V, a 7→av.
108 The boson-fermion correspondence By the definition of a highest weight vector, we have I−⊆ker(φ). Since I−is maximal by Lemma 5.2.2, and ker(φ) ̸= HeisQ since V is nonzero, it follows that I−= ker(φ). Thus, by the first isomorphism theorem for modules, we have V ∼ = HeisQ/I−.
It follows immediately from the definition (5.7) of the Z-grading on HeisQ that bosonic Fock space is a graded module, in the sense that (5.9) Heism Q Symn Q ⊆Symm+n Q , m, n ∈Z.
Remark 5.2.4. Throughout this section we worked over the ground ring Q. It is also possible to work over Z with some minor modifications. In particular, bosonic Fock space is no longer a simple Heis module over Z, since nSym is a Heis-submodule of Sym for all n ∈Z.
△ Exercises.
5.2.1. Suppose that f = X λ∈Par cλpλ for some cλ ∈Q, λ ∈Par, only finitely many of which are nonzero. Choose λ of maximal size amongst the partitions such that cλ ̸= 0. Show that p− λ f (4.13) = ℓ(λ) Y i=1 λi ∂ ∂pλi f is a nonzero constant polynomial.
5.2.2. Give an example of a HeisQ-module that is not generated by a highest weight vector.
5.3 Fermionic Fock space In this section we define the infinite rank Clifford algebra and its natural action on fermionic Fock space. Throughout this section we work over the field Q of rational numbers.
Definition 5.3.1 (Clifford algebra). The infinite rank Clifford algebra Cl is the Q-algebra generated by ψ± i , i ∈Z, subject to the relations (5.10) ψ+ i ψ+ j + ψ+ j ψ+ i = 0, ψ− i ψ− j + ψ− j ψ− i = 0, ψ+ i ψ− j + ψ− j ψ+ i = δij, for all i, j ∈Z.
Fermionic Fock space 109 Let V be the Q-module with basis vi, i ∈Z. Then define F to be the Q-module with basis (5.11) vi0 ∧vi1 ∧vi2 ∧· · · , i0 > i1 > i2 > · · · , ik+1 = ik −1 for k ≫0.
We view these basis elements as formal symbols, which we call semi-infinite wedges. (They are also sometimes called semi-infinite monomials.) We refer to F as the semi-infinite wedge space of V . The condition ik+1 = ik −1 for k ≫0 means that there exists N ∈N such that ik+1 = ik −1 for all k ≥N. We will also consider symbols as in (5.11), but without the condition i0 > i1 > i2 > · · · , using the convention (5.12) vi0 ∧· · · ∧vil ∧vil+1 ∧vil+2 ∧vil+3 ∧· · · = −vi0 ∧· · · ∧vil ∧vil+2 ∧vil+1 ∧vil+3 ∧· · · (We allow l = −1 here, corresponding to interchanging the first two terms.) In other words, interchanging neighbouring terms introduces a minus sign. Note that this convention implies that (5.13) vi0 ∧vi1 ∧· · · = 0 if ik = il for some k ̸= l.
So a semi-infinite wedge with repeated terms is zero. Then any semi-infinite wedge vi0 ∧vi1 ∧· · · , ik+1 = ik −1 for k ≫0, is equal to zero or a semi-infinite wedge of the form (5.11), up to sign, since we can reorder its terms.
In order to simplify notation, we define (5.14) vi := vi0 ∧vi1 ∧· · · where i = (i0, i1, . . . ), ik+1 = ik −1 for k ≫0.
The condition ik+1 = ik−1 for k ≫0 implies that there exists some c ∈Z such that ik = c−k for k ≫0. For c ∈Z, define I(c) := {(i0, i1, . . . ) : i0 > i1 > · · · , ik = c −k for k ≫0}, I = [ c∈Z I(c).
Then we have a decomposition of Q-modules (5.15) F = M c∈Z F(c), F(c) := SpanQ vi, i ∈I(c) .
Nonzero elements of F(c) are said to be of charge c.
Lemma 5.3.2. For c ∈Z, we have a bijection of sets Par →I(c), λ 7→(λ1 + c, λ2 + c −1, λ3 + c −2, . . . ).
Proof. The proof of this lemma is left as Exercise 5.3.1.
110 The boson-fermion correspondence For c ∈Z and λ ∈Par, define v(c) λ := vλ1+c ∧vλ2+c−1 ∧vλ3+c−2 ∧· · · .
It follows from Lemma 5.3.2 that, for c ∈Z, (5.16) F(c) has basis v(c) λ , λ ∈Par.
Our goal in this section is to endow F with the structure of a simple Cl-module.
Proposition 5.3.3 (Fermionic Fock space). Semi-infinite wedge space F is a Cl-module under the action (5.17) ψ+ i vi = vi ∧vi, ψ− i vi = ( (−1)kvi0 ∧· · · ∧vik−1 ∧vik+1 ∧· · · if ik = i, 0 if ik ̸= i for all k ∈N, extended by linearity, for i = (i0, i1, . . . ), i0 > i1 > i2 > · · · , ik+1 = ik −1 for k ≫0.
Viewed as a Cl-module with this action, F is called fermionic Fock space.
Proof. To prove that F is a Cl-module with the given action, we must verify that the relations (5.10) are satisfied. We leave it as Exercise 5.3.2 to verify that the formulas (5.17) also hold when i = (i0, i1, . . . ), ik+1 = ik −1 for k ≫0, has no repeated terms. That is, it is not necessary to suppose that i0 > i1 > i2 > · · · .
Now, for all i, j ∈Z and all semi-infinite wedges vi, i = (i0, i1, . . . , ), we have ψ+ i ψ+ j vi = vi ∧vj ∧vi (5.12) = −vj ∧vi ∧vi = −ψ+ j ψ+ i vi.
Thus the first relation in (5.10) is satisfied. We leave it as Exercise 5.3.3 to verify that the second relation in (5.10) is satisfied.
It remains to consider the third relation in (5.10). First consider the case i ̸= j. We have ψ+ i ψ− j vi = ( (−1)kvi ∧vi0 ∧· · · ∧vik−1 ∧vik+1 ∧· · · if ik = j, 0 if ik ̸= j for all k ∈N, and ψ− j ψ+ i vi = ( (−1)k+1vi ∧vi0 ∧· · · ∧vik−1 ∧vik+1 ∧· · · if ik = j, 0 if ik ̸= j for all k ∈N.
Hence ψ+ i ψ− j vi = −ψ− j ψ+ i vi, as desired. Now consider the case i = j. We have (5.18) ψ+ i ψ− i vi = ( vi if ik = i for some k ∈N, 0 if ik ̸= i for all k ∈N, Fermionic Fock space 111 and (5.19) ψ− i ψ+ i vi = ( 0 if ik = i for some k ∈N, vi if ik ̸= i for all k ∈N.
Hence (ψ+ i ψ− i + ψ− i ψ+ i )vi = vi, as desired.
Remark 5.3.4. The operators ψ+ i are called wedging or creation operators, and the operators ψ− i are called contraction or annihilation operators. A fermion is a type of subatomic particle with the property that at most one particle can occupy each state. This is sometimes referred to as the Pauli exclusion principle. If we view vi as a particle in state i, then this property corresponds to (5.13). Compare to Remark 5.2.1.
△ Lemma 5.3.5. For all i, c ∈Z, we have ψ+ i F(c) ⊆F(c+1), ψ− i F(c) ⊆F(c−1).
Proof. We leave the proof of this lemma as Exercise 5.3.5.
Proposition 5.3.6. The elements (5.20) ψ− i1 · · · ψ− irψ+ j1 · · · ψ+ jsψ− k1 · · · ψ− kt, r, s, t ≥0, 0 ≥i1 > · · · > ir, j1 > · · · > js, k1 > · · · > kt > 0, form a Q-basis of Cl.
Proof. The defining relations (5.10) allow us to reorder the generators ψ± i up to terms with fewer generators. It follows that the elements (5.20) span Cl. It remains to prove that they are linearly independent.
Since we may reorder the ψ± i appearing in the elements (5.20), up to a sign and shorter expressions, it suffices to prove that the elements (5.21) ψ+ i1 · · · ψ+ irψ− j1 · · · ψ− js, r, s ≥0, i1 > · · · > ir, j1 > · · · > js, are linearly independent. (See Exercise 5.3.4.) Suppose, towards a contradiction, that we have a linear dependence relation amongst these elements. Let W denote the finite set of elements (5.21) appearing in this linear dependence relation with nonzero coefficient. So we can write this relation in the form X w∈W aww = 0, aw ∈Q.
Choose k ∈Z less than all the indices appearing in elements of W. Now choose an element ψ+ i1 · · · ψ+ irψ− j1 · · · ψ− js ∈W with s minimal, and define i = vj1 ∧· · · ∧vjs ∧vk ∧vk−1 ∧· · · .
112 The boson-fermion correspondence Then we leave it as Exercise 5.3.6 to show that (5.22) X w∈W awwvi ̸= 0, which is a contradiction.
Let J denote the Q-span of the elements (5.20) for which t > 0 or (ja ≤0 for some 1 ≤a ≤s).
In other words, J is the Q-span of the elements (5.20) that annihilate v(0) ∅= v0 ∧v−1 ∧v−2 ∧· · · ∈F.
Proposition 5.3.7. The set J is a left ideal of Cl, and we have an isomorphism of Cl-modules (5.23) Cl/J ∼ = − →F, a + J 7→av(0) ∅.
Proof. Using Proposition 5.3.6, we see that J = AnnCl v(0) ∅ . Thus J is the kernel of the Cl-module homomorphism (5.24) Cl →F, a 7→av(0) ∅.
Hence J = AnnCl(v(0) ∅) is a left ideal of Cl. Since every element vi, i ∈I, can be obtained from v(0) ∅ by applying wedging and contracting operators, the map (5.24) is also surjective.
Thus we have an isomorphism (5.23) by the first isomorphism theorem for modules.
We would like an analogue of Theorem 5.2.3 for fermionic Fock space. We say that a nonzero element v in a Cl-module is a vacuum vector if (5.25) ψ+ i v = 0 for all i ≤0 and ψ− i v = 0 for all i > 0.
For example, the vector v(0) ∅is a vacuum vector.
Theorem 5.3.8. Fermionic Fock space F is a simple Cl-module. Furthermore, any Cl-module generated by a vacuum vector v is isomorphic to fermionic Fock space via an iso-morphism sending v(0) ∅ to v.
Proof. Suppose U is a nonzero Cl-submodule of F. We must show that U = F. We first show that U contains a semi-infinite wedge. Let u be a nonzero element of U. Then, after multiplying by an element of Q if necessary, we have u = vi + X j∈X cjvj, where cj ∈Q \ {0}, and the sum is over a finite set X not containing i. We can choose u so that the cardinality of X is as small as possible. We aim to show that X is empty.
Fermionic Fock space 113 Suppose, towards a contradiction, that X is nonempty. Then choose k ∈X. Since i ̸= k, we can choose r ∈Z such that r appears in the sequence i but not in k, or vice versa. If r appears in i but does not appear in k, then, by (5.18), we have ψ+ r ψ− r u = vi + X j∈X′ cjvj where X′ is a subset of X not containing k. In particular X′ is proper subset of X, contra-dicting our choice of u. Similarly, if r appears in k but does not appear in i, then, by (5.19), we have ψ− r ψ+ r u = vi + X j∈X′ cjvj for a proper subset X′ of X.
We now know that U contains some semi-infinite wedge vi. Then is easy to see that, using the operators ψ± i , i ∈Z, we can obtain all other semi-infinite wedges, up to sign. Thus U contains all semi-infinite wedges, and so U = F, as desired.
Now suppose V is a Cl-module generated by a vacuum vector v, and consider the homo-morphism of Cl-modules κ: Cl →V, a 7→av, a ∈Cl.
The kernel of κ contains the left ideal J. Since κ is not the zero map and J is maximal, we must have ker(κ) = J. On the other hand, since V is generated by v, the map κ is also surjective. Thus we have an isomorphism of Cl-modules Cl/J ∼ = − →V, a + J 7→av, a ∈Cl.
Then it follows from (5.23) that we have an isomorphism of Cl-modules F ∼ = − →V, av(0) ∅7→av, a ∈Cl.
We conclude this section by defining a natural inner product on F that will correspond, under the boson-fermion correspondence, to the Hall inner product on the ring of symmetric functions. We define this inner product on F by (5.26) ⟨, ⟩: F × F →Q, ⟨vi, vj⟩= δij, i, j ∈I, extended by bilinearity.
Proposition 5.3.9. With respect to the inner product (5.26), we have (ψ± i )⊥= ψ∓ i for all i ∈Z. In other words, (5.27) ⟨ψ± i u, v⟩= ⟨u, ψ∓ i v⟩ for all i ∈Z, u, v ∈F.
Proof. We leave the proof of this result as Exercise 5.3.7.
114 The boson-fermion correspondence Exercises.
5.3.1. Prove Lemma 5.3.2.
5.3.2. Show that the formulas (5.17) also hold when i = (i0, i1, . . . ), ik+1 = ik −1 for k ≫0, has no repeated terms. That is, it is not necessary to suppose that i0 > i1 > i2 > · · · .
5.3.3. Show that the action defined in Proposition 5.3.3 satisfies the second relation in (5.10).
5.3.4. Let ⪰1 and ⪰2 be two total orders on the set Y = {ψ± i : i ∈Z}.
Then, for k ∈{1, 2}, let Yk = {z1z2 . . . zl : l ≥0, z1 ≻k z2 ≻· · · ≻k zl} ⊆Cl.
Prove that Y1 is linearly independent if and only if Y2 is linearly independent. Hint: The relations (5.10) allow you to reorder products z1z2 . . . zl up to shorter products.
5.3.5. Prove Lemma 5.3.5.
5.3.6. Prove (5.22).
5.3.7. Prove Proposition 5.3.9.
5.4 Bosonization We are now ready to present the first half of the boson-fermion correspondence, known as bosonization. This consists of defining an action of the Heisenberg algebra on fermionic Fock space, and identifying the resulting Heis-module with a sum of countably many copies of bosonic Fock space.
For r ∈Z, r ̸= 0, consider the infinite sum (5.28) Pr := X j∈Z ψ+ j+rψ− j of products of elements of Cl. Since the sum is infinite, this is not an element of Cl. However, suppose i = (i0, i1, . . . ), i0 > i1 > · · · , ik+1 = ik −1 for k ≥N.
Then (5.13) implies that Prvi = (PN k=0 vi0 ∧vi1 ∧· · · ∧vik−1 ∧vik+r ∧vik+1 ∧· · · if r < 0, PN+r+1 k=0 vi0 ∧vi1 ∧· · · ∧vik−1 ∧vik+r ∧vik+1 ∧· · · if r > 0.
Bosonization 115 In particular, Pr is well-defined as an element of EndQ(F) since all but finitely many of the terms ψ+ j+rψ− j act as zero on any given vi. It follows immediately from Lemma 5.3.5 that (5.29) PrF(c) ⊆F(c) for all c, r ∈Z, r ̸= 0.
Recall that if U and V are vector spaces equipped with inner products, then a linear map f : U →V is an isometry if ⟨f(u1), f(u2)⟩= ⟨u1, u2⟩ for all u1, u2 ∈U.
If f is also a vector space isomorphism, then we say that f is an isometric isomorphism.
Theorem 5.4.1 (Bosonization). For all c ∈Z, the map (5.30) A(c) : F(c) →SymQ, v(c) λ 7→sλ, is an isometric isomorphism. Furthermore, for all r ≥1, the diagram (5.31) F(c) F(c) SymQ SymQ A(c) P±r A(c) p± r commutes. In particular, HeisQ →EndQ F(c) , p± r 7→P±r, r ≥1, is a representation of HeisQ and, under this action, F(c) is isomorphic to bosonic Fock space.
Proof. The fact that A(c) is an isomorphism of vector spaces follows immediately from (5.16) and the fact that the Schur functions are a basis of SymQ. It is an isometric isomorphism by (3.30) and (5.26). Thus, it remains to prove that (5.31) commutes. To do this, we compare the action of Pr on v(n) λ to the Murnaghan–Nakayama rule (Proposition 3.7.1).
Fix j, r ∈Z, r ≥1, and λ ∈Par. Let i = (λ1 + c, λ2 + c −1, λ3 + c −2, . . . ) be the element of I(c) corresponding to λ under the bijection of Lemma 5.3.2, so that v(c) λ = vi.
If j does not appear in the sequence i, then ψ+ j+rψ− j vi = 0. Otherwise, ψ+ j+rψ− j vi is obtained from vi by replacing the entry j by j + r. Then, repeating the argument in the proof of Proposition 3.7.1 shows that X j∈Z ψ+ j+rψ− j vi = X µ (−1)ht(µ/λ)v(c) µ , where the sum is over all partitions µ ⊇λ such that µ/λ is a border strip with r boxes. This proves that (5.31) commutes for Pr and p+ r .
116 The boson-fermion correspondence It remains to prove that (5.31) commutes for P−r and p− r . For all λ, µ ∈Par, we have D A(c) P−rv(c) λ , sµ E = D A(c) P−rv(c) λ , A(c) v(c) µ E = D P−rv(c) λ , v(c) µ E (since A(c) is an isometry) (5.28) = X j∈Z D ψ+ j−rψ− j v(c) λ , v(c) µ E (5.27) = X j∈Z D v(c) λ , ψ+ j ψ− j−rv(c) µ E (5.28) = D v(c) λ , Prv(c) µ E = D A(c) v(c) λ , A(c) Prv(c) µ E (since A(c) is an isometry) = D A(c) v(c) λ , p+ r A(c) v(c) µ E = D p− r A(c) v(c) λ , sµ E (since p− r = p⊥ r ).
Since the sµ, µ ∈Par, span Sym, it follows that A(c) P−rv(c) λ = p− r A(c) v(c) λ for all λ ∈Par.
Thus A(c)P−r = p− r A(c) by (5.16).
It follows from Theorem 5.4.1 that F = L c∈Z F(c) is isomorphic to a direct sum of countably many copies of bosonic Fock space.
Let Q[q, q−1] denote the ring of Laurent polynomials in q with rational coefficients. Then SymQ[q,q−1] := Q[q, q−1] ⊗Q SymQ can be identified with the ring of symmetric functions with coefficients in Q[q, q−1] and we have SymQ[q,q−1] = M c∈Z qcSymQ.
We can extend the action of HeisQ on SymQ to an action on SymQ[q,q−1] by x(av) = axv, x ∈HeisQ, a ∈Q[q, q−1], v ∈SymQ.
Furthermore, we extend the Hall inner product to SymQ[q,q−1] by ⟨qcf, qdg⟩= δc,d⟨f, g⟩, c, d ∈Z, f, g ∈SymQ.
Then we have the following corollary of Theorem 5.4.1.
Corollary 5.4.2. We have an isometric isomorphism of HeisQ-modules (5.32) A: F ∼ = − →SymQ[q,q−1], v(c) λ 7→qcsλ, c ∈Z, λ ∈Par.
Fermionization 117 Exercises.
5.4.1. Consider (5.28) with r = 0. Is P0 a well-defined operator on F?
5.4.2. It follows from Theorem 5.4.1 and (5.6) that PrP−r = P−rPr −r for r ≥1. So what is wrong with the following computation?
PrP−r = X j,k∈Z ψ+ j+rψ− j ψ+ k−rψ− k (5.10) = − X j,k∈Z ψ+ j+rψ+ k−rψ− j ψ− k + X j,k∈Z δj,k−rψ+ j+rψ− k (5.10) = − X j,k∈Z ψ+ k−rψ+ j+rψ− k ψ− j + X j∈Z ψ+ j+rψ− j+r (5.10) = X j,k∈Z ψ+ k−rψ− k ψ+ j+rψ− j − X j,k∈Z δk,j+rψ+ k−rψ− j + X j∈Z ψ+ j+rψ− j+r = P−rPr − X j∈Z ψ+ j ψ− j + X j∈Z ψ+ j+rψ− j+r = P−rPr.
5.4.3. It follows from Theorem 5.4.1 and (5.6) that PrPs = PsPr for r, s ∈Z \ {0}, r ̸= s.
Prove this relation directly using (5.10) and (5.28).
5.4.4. Suppose U and V are Q-vector spaces equipped with inner products. (By definition, the inner product is bilinear, symmetric, and positive definite.) Prove that every isometry f : U →V is injective.
5.5 Fermionization In this section we present the second half of the boson-fermion correspondence, known as fermionization. This consists of defining an action of the Clifford algebra on SymQ[q,q−1] (which is a countable direct sum of copies of bosonic Fock space), and showing that the isometric isomorphism A of (5.32) is also an isomorphism of Cl-modules.
Much of our presentation follows [Mac15, p. 95, Exercise I.5.29].
Recall the generating functions E(t) and H(t) from (2.21) and (2.28), and define the formal power series (5.33) E+(t) = X n≥0 e+ n tn, E−(t) = X n≥0 e− n tn, H+(t) = X n≥0 h+ n tn, H−(t) = X n≥0 h− n tn.
These are all elements of End(Sym)JtK. They can also be viewed as maps Sym →SymJtK.
118 The boson-fermion correspondence Lemma 5.5.1. Viewing H±(t) and E±(t) as maps Sym →SymJtK, we have (5.34) H±(t) ◦ω = ω ◦E±(t), where, on the right-hand side, ω is the involution of SymJtK that applies ω (from (2.31)) to each coefficient.
Proof. For all f ∈Sym, we have H+(t) ◦ω(f) = X n≥0 hnω(f)tn (2.31) = X n≥0 ω enf tn = ω ◦E+(t)(f) and H−(t)◦ω(f) = X n≥0 h⊥ n (ω(f))tn (4.16) = X n≥0 ω ω(hn)⊥f tn (2.32) = X n≥0 ω e⊥ n f tn = ω◦E−(t)f.
Lemma 5.5.2. The formal power series E−(t) and H−(t) are ring homomorphisms Sym → Sym[t].
Proof. For any f ∈Sym, we have e− n f = e⊥ n f = 0 for all but finitely many n.
Thus, E−(t)f ∈Sym[t]. Similarly, H−(t) is a map from Sym to Sym[t]. It is clear that they are k-linear maps and that they send 1 to 1.
So it remains to check that they respect multiplication. Indeed, for f, g ∈Sym, we have E−(t)(fg) = X n≥0 e⊥ n (fg)tn (4.34) = (4.31) X n≥0 n X r=0 (e⊥ r f)(e⊥ n−rg)tn = X n≥0 e⊥ n f tn ! X m≥0 e⊥ mg tm !
= E−(t)f E−(t)g .
Hence E−(t) is a ring homomorphism.
Now, by (5.34), we have H−(t) = ω ◦E−(t) ◦ω.
Since ω is a ring homomorphism, it follows that E−(t) is also a ring homomorphism.
In what follows, we consider two indeterminates t and u. Recall from (1.11) that (1 −tu)−1 = X n≥0 tnun.
Proposition 5.5.3. We have H−(t)H+(u) = (1 −tu)−1H+(u)H−(t), (5.35) H−(t)E+(u) = (1 + tu)E+(u)H−(t), (5.36) E−(t)E+(u) = (1 −tu)−1E+(u)E−(t), (5.37) E−(t)H+(u) = (1 + tu)H+(u)E−(t).
(5.38) Fermionization 119 Proof. We first compute H−(t)H(u) = X m,n≥0 h⊥ mhn tmun (4.36) = X m,n≥0 (hn−mun−m)(tu)m = (1 −tu)−1H(u).
Since H−(t) is a ring homomorphism by Lemma 5.5.2, we have, for f ∈Sym, H−(t)H+(u)f = H−(t) (H(u)f) = H−(t)H(u) H−(t)f = (1 −tu)−1H(u)H−(t)f.
Thus (5.35) is satisfied.
Next we compute H−(t)E(u) = X m,n≥0 h⊥ men tmun (4.36) = X n≥0 (en + ten−1)un = (1 + tu)E(u).
Since H−(t) is a ring homomorphism by Lemma 5.5.2, we have, for f ∈Sym, H−(t)E+(u)f = H−(t) (E(u)f) = H−(t)E(u) H−(t)f = (1 + tu)E(u)H−(t)f.
Thus (5.36) is satisfied.
Next, we compose (5.35) on the left and right with ω to obtain ω ◦H−(t)H+(u) ◦ω = (1 −tu)−1ω ◦H+(u)H−(t) ◦ω = ⇒ω ◦H−(t) ◦ω2 ◦H+(u) ◦ω = (1 −tu)−1ω ◦H+(u) ◦ω2 ◦H−(t) ◦ω (5.34) = ⇒E−(t)E+(u) = (1 −tu)−1E+(u)E−(t).
Relation (5.38) can be similarly deduced from (5.36).
Remark 5.5.4. Equating coefficients of tmun, relations (5.35) to (5.38) correspond to Exer-cises 5.1.2 and 5.1.3.
△ Next we define (5.39) B+(t) = X n∈Z B+ n tn = H+(t)E−(−t−1), B−(t) = X n∈Z B− n tn = E+(−t−1)H−(t).
Thus (5.40) B+ n = X r≥0 (−1)rh+ n+re− r , B− n = X r≥0 (−1)re+ r h− n+r.
The operators B± n are called Bernstein operators. Note that they are not elements of Heis since the sums are infinite. However, they are elements of EndQ(SymQ) since, when they act on any element of SymQ, all but finitely many terms act as zero. In this sense, they are similar to the operators Pr from (5.28).
Note also that B±(t) are not formal power series, since we allow negative exponents.
Thus, when multiplying them by other series, we must verify that the product is well-defined.
When we write an equation involving such a product, we are implicitly asserting that these products are well defined.
If t1, t2, . . . , tn are independent variables, define B+(t1, . . . , tn) = H+(t1) · · · H+(tn)E−(−t−1 1 ) · · · E−(−t−1 n ).
When n = 1, this reduces to the B+(t) from (5.39).
120 The boson-fermion correspondence Lemma 5.5.5. For n ≥1, B+(t1) · · · B+(tn) = Y 1≤i<j≤n 1 −t−1 i tj B+(t1, . . . , tn).
(When n = 1, we have Q 1≤i<j≤n 1 −t−1 i tj = 1 by convention.) Proof. This follows from repeated use of (5.38). We leave the details as Exercise 5.5.2.
Corollary 5.5.6. For n ≥1, (5.41) B+(t1) · · · B+(tn)1 = Y 1≤i<j≤n 1 −t−1 i tj H(t1) · · · H(tn).
Proof. This follows immediately from Lemma 5.5.5 and the fact that E−(t)1 = 1.
The next result shows that the Bernstein operators are creation operators for Schur functions.1 Proposition 5.5.7. For λ ∈Parn, we have (5.42) B+ λ1B+ λ2 · · · B+ λn1 = sλ.
Proof. We have B+(t1) · · · B+(tn)1 (5.41) = Y 1≤i 0, Ψ− i 1 (5.52) = q−1B− i 1 (5.40) = q−1 X r≥0 (−1)re+ r h− i+r1 = 0.
Thus 1 is a vacuum vector.
For any c ∈Z, define Φc = Ψ+ c · · · Ψ+ 2 Ψ+ 1 if c > 0, 1 if c = 0, Ψ− c+1 · · · Ψ− −1Ψ− 0 if c < 0, and ϕc = ψ+ c · · · ψ+ 2 ψ+ 1 if c > 0, 1 if c = 0, ψ− c+1 · · · ψ− −1ψ− 0 if c < 0.
124 The boson-fermion correspondence Then, for c > 0, we have Φc1 (5.52) = qc(B+ 0 )c1 (5.40) = qc.
Similarly, for c < 0, we have Φc1 (5.52) = qc(B− 0 )−c1 (5.40) = qc.
Since we clearly have Φ01 = 1, it follows that Φc1 = qc for all c ∈Z. We also have (5.56) ϕcv(0) ∅= v(c) ∅ for all c ∈Z.
(See Exercise 5.5.5.) Now, for c ∈Z and λ = (λ1, . . . , λk) ∈Par, we have Ψ+ λ1+cΨ+ λ2+c−1 · · · Ψ+ λk+c−k+1Φc−k1 = Ψ+ λ1+cΨ+ λ2+c−1 · · · Ψ+ λk+c−k+1qc−k (5.52) = qcB+ λ1B+ λ2 · · · B+ λk−1B+ λk1 (5.42) = qcsλ.
Since qcsλ, c ∈Z, λ ∈Par, span SymQ[q,q−1], it follows that 1 generates SymQ[q,q−1] as a Cl-module.
Theorem 5.3.8 now implies that we have an isomorphism of Cl-modules A′ : F ∼ = − → SymQ[q,q−1] sending v(0) ∅to 1. Then we have A′ v(c) λ = A′ ψ+ λ1+cψ+ λ2+c−1 · · · ψ+ λk+c−k+1ϕc−kv(0) ∅ = Ψ+ λ1+cΨ+ λ2+c−1 · · · Ψ+ λk+c−k+1Φc−k1 = qcsλ.
Hence A′ = A, and so (5.55) commutes.
It follows from (2.38), (5.39) and (5.51) that Ψ+(t) = X i∈Z Ψ+ i ti = Tq exp ∞ X n=1 tn n p+ n !
exp − ∞ X n=1 t−n n p− n !
and (5.57) Ψ−(t) = X i∈Z Ψ− i t−i = q−1T −1 exp − ∞ X n=1 t−n n p+ n !
exp ∞ X n=1 tn n p− n !
.
(5.58) These expressions are known as vertex operators. They give the Clifford algebra generators ψ± i in terms of the standard Heisenberg algebra generators p± n .
Fermionization 125 Exercises.
5.5.1. Show that, for all n ∈Z, the operators B+ n and B− n are adjoint with respect to the Hall inner product. In other words, show that ⟨B+ n f, g⟩= ⟨f, B− n g⟩ for all f, g ∈SymQ.
5.5.2. Prove Lemma 5.5.5 5.5.3. With φ(tu) as in (5.43), show that f(t)φ(tu) = f(u−1)φ(tu) for all f(t) ∈Q[t].
5.5.4. Prove (5.46).
5.5.5. Prove (5.56).
5.5.6. Recall the definition of the ring of formal Laurent series from Exercise 1.4.1. Show that E−(−t−1)H+(t) is a well-defined map from Sym to Sym( (t) ), but H−(−t−1)H+(t) is not.
Chapter 6 Applications to representation theory In this final chapter, we discuss important applications of the theory of symmetric functions to representation theory. We begin by giving a brief overview of the character theory of finite groups. We then give a precise relationship between the characters of the symmetric group and the ring of symmetric functions. This connection, together with our knowledge of symmetric functions, allows us to deduce representation theoretic results. In particular, we give a combinatorial formula for the character table for the symmetric groups and explicit constructions of all their irreducible representations. We conclude with a very brief discussion of how Schur functions also give characters for polynomial representations of the general linear group.
6.1 Characters of finite groups In this section we give a brief summary of the theory of characters of finite groups. In order to keep our treatment as concise as possible, we restrict our attention to working over a ground field of characteristic zero. More details on the material found here, including proofs, can be found, for instance, in [DF04, Ch. 19, 20].
Throughout this section, G is a finite group and k is an arbitrary field of characteristic zero unless otherwise specified. The group algebra of G, denoted kG, is the k-vector space with basis G, with multiplication given by X x∈G axx ! X y∈G byy !
= X x,y axby(xy), ax, by ∈k.
A representation of G over k is a homomorphism of algebras ρ: kG →Endk(V ) where V is a finite-dimensional vector space over k. (The assumption of finite dimensionality is not necessary in general. However, we will assume through this chapter that all repre-sentation are finite dimensional.) Such a representation defines a G-module structure on V given by x · v := ρ(x)(v), x ∈G, v ∈V.
126 Characters of finite groups 127 Conversely, every G-module V defines a representation of G by ρ X x∈G axx !
(v) := X x∈G ax(x · v), ax ∈k.
Note that, by definition, a G-module is the same as a kG-module.
The degree of a representation is the dimension of the corresponding G-module V . A representation is said to be irreducible if the corresponding G-module is simple. Otherwise it is said to be reducible. Two representations are equivalent, or isomorphic, if the corresponding modules are isomorphic. Equivalently, representations ρ1 : kG →Endk(V1) and ρ2 : kG → Endk(V2) are equivalent if there exists an isomorphism of k-vector spaces f : V1 →V2 such that the diagram V1 V1 V2 V2 ρ1(x) f f ρ2(x) commutes for all x ∈G.
Proposition 6.1.1 (Maschke’s theorem). Every G-module is a direct sum of simple modules.
Proof. For a proof, see, for example, [Sav, Th. 4.6.3].
For the remainder of this section, let Z be an associative unital k-algebra. For example, we can have Z = k. Let F(G, Z) denote the ring of functions from G to Z with pointwise multiplication. An element f ∈F(G, Z) is a class function if it is constant on conjugacy classes, that is, if f y−1xy = f(x) for all x, y ∈G.
We let C(G, Z) denote the set of class functions on G with values in Z. It is straightforward to verify that this is a subring of F(G, Z) and that dimk C(G, k) is the number of conjugacy classes of G; see Exercise 6.1.3.
If ρ: G →Endk(V ) is a representation, the character of ρ is the function χρ = χV : G →k, χρ(x) = tr ρ(x), where tr ρ(x) is the trace of the matrix ρ(x) with respect to some basis of V . Every character is a class function (Exercise 6.1.1) and equivalent representations have the same character (Exercise 6.1.2). A character is said to be irreducible or reducible when the corresponding representation is irreducible or reducible, respectively. The degree of a character is the degree of the corresponding representation. Since ρ(1) is the identity linear transformation of V and the trace of an n × n identity matrix is n, we see that χρ(1) is the degree of ρ.
For class functions f, g ∈C(G, C), define (6.1) ⟨f, g⟩C G := 1 |G| X x∈G f(x)g(x), where ¯ z denotes the complex conjugate of z ∈C. We leave it as Exercise 6.1.4 to show that (6.1) is a Hermitian inner product on C(G, C). In other words, for all a, b ∈C, we have 128 Applications to representation theory • ⟨af1 + bf2, g⟩C G = a⟨f1, g⟩C G + b⟨f2, g⟩C G, • ⟨f, g⟩C G = ⟨g, f⟩C G, and • if f ̸= 0, then ⟨f, f⟩C G > 0.
We define the norm of a class function f to be ∥f∥:= q ⟨f, f⟩C G.
Proposition 6.1.2. Assume k = C.
(a) Two representations are equivalent if and only if they have the same character.
(b) The irreducible characters are an orthonormal basis for C(G, C).
(c) If χ1 and χ2 are characters corresponding to G-modules V1 and V2, respectively, then χ1 + χ2 is the character of V1 ⊕V2 and χ1χ2 is the character of V1 ⊗V2. (Recall the definition of the G-module structure on V1 ⊗V2 from Examples 4.3.10(a).) (d) If χ is any character of G, then χ(x) is a sum of roots of unity in C and χ(x−1) = χ(x) for all x ∈G.
It follows from Proposition 6.1.2(b) that the number of pairwise inequivalent irreducible characters of G is equal to the number of conjugacy class of G. Hence, by Proposition 6.1.2(a), the number of isomorphism classes of simple G-modules is also equal to the number of con-jugacy classes of G.
The character table of G is the table whose rows are labelled by irreducible representations of G and whose columns are labelled by conjugacy classes of G. The entry in the row labelled by the irreducible representation ρ and the column labelled by the conjugacy class C is χρ(x), where x ∈C. Thus the character table of G completely describes the characters of G.
Characters give a method for computing the decomposition of a G-module into a sum of simple modules as follows. Suppose we are working over k = C, and let χ1, . . . , χr be the irreducible characters of G, with corresponding modules V1, . . . , Vr. If V is a G-module, then we have χV = r X i=1 ⟨χV , χi⟩C Gχi and ⟨χV , χi⟩C G ∈N for 1 ≤i ≤n.
Then we have an isomorphism of G-modules V ∼ = r M i=1 V ⊕⟨χV ,χi⟩C G i , where V ⊕n = V ⊕· · · ⊕V | {z } n summands In other words, ⟨χV , χi⟩C G gives the multiplicity of Vi in the decomposition of V . In particular, a class function is the character of some representation if and only if it is a nonnegative integer linear combination of irreducible characters.
An integer linear combination of characters is called a virtual character. In other words, if χ1, . . . , χr are the irreducible characters of G, then a virtual character is a class function of the form f = r X i=1 aiχi, a1, . . . , ar ∈Z.
Characters of finite groups 129 Since ∥f∥= r X i=1 a2 i !1/2 , it follows that a virtual character has norm 1 if and only if it is an irreducible character up to sign. (We can then check the sign by computing f(1), which is positive if f is a character.) Note that this is not true for arbitrary class functions.
Let R(G, C) denote the Z-module of complex virtual characters. It follows from Proposi-tion 6.1.2(c) that R(G, C) is closed under multiplication, and hence is a subring of C(G, C).
Lemma 6.1.3. The irreducible complex characters are the unique orthonormal basis of R(G, C) up to sign and reordering.
Proof. The proof of this lemma is almost identical to the proof of Proposition 3.3.8.
It follows from Proposition 6.1.2(d) that (6.2) ⟨f, g⟩C G = 1 |G| X x∈G f(x)g(x−1), f, g ∈R(G, C).
Motivated by this, for any associative unital k-algebra Z, any functions f, g: G →Z, (6.3) ⟨f, g⟩G = 1 |G| X x∈G f(x)g(x−1).
Thus ⟨f, g⟩G = ⟨f, g⟩C G for f, g ∈R(G, C), but these can differ for general class functions f, g ∈C(G, C). Since we will be most interested in the subring R(G, C), we will work from now on with the inner product (6.3) since it has the advantage of being well defined over any k-algebra Z.
Now suppose G1 and G2 are both finite groups. If f1 ∈R(G1, C) and f2 ∈R(G2, C), then (6.4) f1 × f2 : G1 × G2 →C, (f1 × f2)(x1, x2) = f1(x1)f2(x2), x1 ∈G1, x2 ∈G2, is a class function on G1 × G2. Furthermore, if f1 and f2 are characters, then so is f1 × f2.
More precisely, if fi is the character of the Gi-module Vi, for i ∈{1, 2}, then f1 × f2 is the character of the (G1 × G2)-module V1 ⊗V2, with action given by (g1, g2)(v1 ⊗v2) = g1v1 ⊗g2v2, g1 ∈G1, g2 ∈G2, v1 ∈V1, v2 ∈V2, extended by linearity. Thus, we have a bilinear map R(G1, C) × R(G2, C) →R(G1 × G2, C).
The (G1 × G2)-module V1 ⊗V2 is simple if and only if V1 and V2 are both simple. It follows that we have an isomorphism of algebras (6.5) R(G1, C) ⊗R(G2, C) ∼ = − →R(G1 × G2, C), χ1 ⊗χ2 7→χ1 × χ2, for characters χ1 of G1 and χ2 of G2, extended by Z-linearity.
130 Applications to representation theory Exercises.
6.1.1. Prove that characters are class functions.
6.1.2. Prove that equivalent representations have the same character.
6.1.3. Prove that C(G, Z) is a subring of F(G, Z) and that dimk C(G, k) is the number of conjugacy classes of G.
6.1.4. Prove that (6.1) is a Hermitian inner product on the space of class functions.
6.2 Induction and restriction In order to connect the representation theory of the symmetric groups to the ring of sym-metric functions, it is necessary to consider all of the symmetric groups Sn, n ∈N, at once.
The main tools for doing this are the operations of induction and restriction, which allow us to move between representations of different symmetric groups. In this section we introduce these operations. Further details, including proofs, can be found, for instance, in [DF04, §19.3].
Throughout this section we fix a field k of characteristic zero and an associative unital k-algebra Z. (One important example will be k = Z = C.) We also fix a subgroup H of a finite group G.
If V is a G-module, then we can restrict the action to obtain an H-module, which we denote by ResG H(V ). If χ is the character of V , then the restricted character ResG H(χ) is the character of the restricted module ResG H(V ). We can extend this to a ring homomorphism ResG H : C(G, Z) →C(H, Z), ResG H(f)(x) = f(x), f ∈C(G, Z), x ∈H.
We would now like to define a natural map in the other direction: C(H, Z) →C(G, Z).
Given f ∈C(H, Z), we can extend it to a function f : G →Z (which continue to denote by the same symbol f) by declaring (6.6) f(x) = 0 if x ∈G, x / ∈H.
In general, this extended function will not be a class function. However, we can fix this by averaging over G, defining the induced class function (6.7) IndG H(f)(x) = 1 |H| X y∈G f(y−1xy), f ∈C(H, Z), x ∈G.
We leave it as Exercise 6.2.1 to show that this defines a class function on G. In fact, when f ∈R(H, C), we would like to show that IndG H(f) ∈R(G, C). This will follow from the Induction and restriction 131 fact that, if χ is a character, then so is the induced character IndG H(χ). Namely, it is the character of an induced module, which we now describe.
We begin with a more general construction. Let k be an arbitrary field, and let B be an associative unital k-algebra. If M is a right B-module and V is a left B-module, then we define M ⊗B V to be the k-vector space generated by the elements M × V , modulo the subspace W generated by the elements (m + m′, v) −(m, v) −(m′, v), (m, v + v′) −(m, v) −(m, v′), (mb, v) −(m, bv), for m, m′ ∈M, v, v′ ∈V , and b ∈B. We set m ⊗v := (m, v) + W, m ∈M, v ∈V.
This notion of tensor product has a universal property analogous to the one of Proposi-tion 4.1.2, as we now describe. Consider the k-bilinear map h: M × V →M ⊗B V, h(m, n) = m ⊗n.
If W is a k-vector space, we say that a map φ: M × V →W is B-balanced if φ(mb, v) = φ(m, bv) for all m ∈M, b ∈B, v ∈V.
Proposition 6.2.1. For every k-module W and every k-bilinear, B-balanced map φ: M × V →W, there exists a unique k-linear map ˜ φ: M ⊗B V →W such that ˜ φ ◦h = φ; that is, the diagram M × V M ⊗B V W h φ ˜ φ commutes.
Proof. For a proof, see [Sav, Prop. 3.5.3].
Now suppose that A and B are both associative unital k-algebras. An (A, B)-bimodule is a k-vector space M that is a left A-module, a right B-module, and the two actions commute: (am)b = a(mb) for all a ∈A, b ∈B, m ∈M.
Then M ⊗B V is naturally an A-module, with action given by a(m ⊗v) = (am) ⊗v, a ∈A, m ∈M, v ∈V.
In this way, the bimodule M gives us a way of turning B-modules into A-modules.
Remark 6.2.2. Note that any k-vector space V is a (k, k)-bimodule, where the right and left action are both given by scalar multiplication. Then, if A = B = k, the above notion of tensor product agrees with the one we discussed in Section 4.1. The crucial property in this case is that k is commutative, which makes left and right k-modules equivalent.
△ 132 Applications to representation theory We now consider a special case of the above construction. The group algebra kG is both a left and a right module over itself, via multiplication. These two actions commute, and so kG is a (kG, kG)-bimodule. We can restrict the action on the right, and consider kG as a (kG, kH)-bimodule. If V is an H-module, then we have the induced module IndH G(V ) := kG ⊗kH V.
Recall that the left cosets of H in G are the sets xH = {xy : y ∈H}, x ∈G.
Let x1, . . . , xm be a complete set of left coset representatives. Then kG is free right kH-module of rank m: kG = x1kH ⊕· · · ⊕xmkH.
If V is an H-module with k-basis v1, . . . , vn, the induced module IndH G(V ) has k-basis xi ⊗vj, 1 ≤i ≤m, 1 ≤j ≤n.
In particular, dimk IndG H(V ) = m dimk V.
In addition, for f ∈C(H, Z) and x ∈G, we have (6.8) IndG H(f)(x) = 1 |H| X y∈G f y−1xy = m X i=1 f x−1 i xxi , (The first equality is the definition (6.7), and the second is left as Exercise 6.2.3.) Proposition 6.2.3. If H is a subgroup of a finite group G and V is an H-module, then IndG H(χV ) = χIndG H(V ).
In other words, the induced character IndG H(χV ) is the character of the induced module IndG H(V ).
It follows from Proposition 6.2.3 that induction yields a k-linear map IndG H : R(H, C) →R(G, C).
We conclude this section with an important result in representation theory that states that this map is adjoint to restriction.
(In the language of category theory, they are adjoint functors.) Proposition 6.2.4 (Frobenius reciprocity). Suppose H is a subgroup of the finite group G, Z is a k-algebra, f ∈C(H, Z), and g ∈C(G, Z). Then IndG H(f), g G = f, ResG H(g) H .
Characters of symmetric groups 133 Proof. We have IndG H(f), g G (6.3) = 1 |G| X x∈G IndG H(f)(x)g(x−1) (6.7) = 1 |G||H| X x,y∈G f(y−1xy)g(x−1) = 1 |G||H| X y∈G X x∈yGy−1 f(y−1xy)g(x−1) (since yGy−1 = G) = 1 |G||H| X x,y∈G f(x)g(yx−1y−1) = 1 |G||H| X x,y∈G f(x)g(x−1) (since g is a class function) = 1 |H| X x∈G f(x)g(x−1) = 1 |H| X x∈H f(x)g(x−1) (since f(x) = 0 for x ∈G, x / ∈H) (6.3) = f, ResG H(g) H .
Exercises.
6.2.1. Prove that the induced class function IndG H(f), as defined in (6.7), is a class function directly from the definition.
6.2.2. Show that if H is a subgroup of an abelian group G and f ∈C(H, k), then IndG H(f) = |G| |H|f. Here, on the right-hand side, f denotes the extended function (6.6).
6.2.3. Prove the second equality in (6.8).
6.3 Characters of symmetric groups In this section we study the characters of the symmetric groups and describe a precise connection with the ring of symmetric functions. In particular, we will see that irreducible characters correspond to Schur functions. Once we have established this connection, our knowledge of symmetric functions allows us to deduce facts about the representation theory of the symmetric groups. Our approach is guided by the presentation of [Mac15, §I.7].
For m, n ∈N, we have an injective group homomorphism Sm × Sn , →Sm+n, (π, σ) 7→π × σ, 134 Applications to representation theory where Sm permutes the elements {1, . . . , m} and Sn permutes the elements {m+1, . . . , m+ n}. Precisely, for π ∈Sm and σ ∈Sn, we have (π × σ)(k) = ( π(k) if 1 ≤k ≤m, σ(k −m) + m if m < k ≤m + n.
We will identify Sm×Sn with its image under the above map and, in this way, view Sm×Sn as a subgroup of Sm+n.
Recall from Section 1.1 that every permutation π ∈Sn factors uniquely as a product of disjoint cycles. If the lengths of these cycles are ρ1, ρ2, . . . , ordered so that ρ1 ≥ρ2 ≥· · · , then ρ(π) := (ρ1, ρ2, . . . ) is a partition of n called the cycle type of π. For ρ ∈Par(n), the elements of Sn of cycle type ρ form a conjugacy class of Sn. Thus, the conjugacy classes of Sn are indexed by partitions of n. Let Cρ ⊆Sn denote the conjugacy class of elements of cycle type ρ.
By convention, S0 is the trivial group.
Let R = ∞ M n=0 Rn, Rn = R(Sn, C).
Note that, since S0 = {1} is the trivial group, we have R0 = Z, spanned by the trivial character (that is, the character of the trivial representation of the trivial group). We can endow R with the structure of a ring as follows. If f ∈Rm and g ∈Rn, then f × g, as defined in (6.4), is a virtual character of Sm × Sn. We then define (6.9) f ⋆g = IndSm+n Sm×Sn(f × g) ∈Rm+n.
In this way, we have defined a bilinear multiplication ⋆: Rm × Rn →Rm+n.
We leave it as Exercise 6.3.1 to verify that, with this multiplication, R is a commutative, associative, unital, graded ring. Note that ⋆is not the same as pointwise multiplication (denoted by juxtaposition), which gives a multiplication on each Rn.
We can also endow R with a scalar product. Recall the inner product (see (6.2)) (6.10) Rn × Rn →C, ⟨f, g⟩Sn = 1 n!
X π∈Sn f(π)g(π−1).
If f, g ∈R, we define ⟨f, g⟩= ∞ X n=0 ⟨fn, gn⟩Sn, where f = P n≥0 fn, g = P n≥0 gn with fn, gn ∈Rn.
Let SymC := C ⊗Z Sym and Symn C := C ⊗Z Symn, Characters of symmetric groups 135 We will view Sym as a subring of SymC via the inclusion Sym , →SymC, f 7→1 ⊗f.
Now define Ψ ∈C(Sn, Symn C) by Ψ: Sn →Symn C, Ψ(π) = pρ(π).
If π ∈Sm and σ ∈Sn, then ρ(π × σ) = ρ(π) ∪ρ(σ). Hence (6.11) Ψ(π × σ) = Ψ(π)Ψ(σ).
Next we define a Z-linear map (6.12) ch: R →SymC, ch(f) = ⟨f, Ψ⟩Sn = 1 n!
X π∈Sn f(π)Ψ(π), f ∈Rn, extended by linearity. (Note that Ψ(π−1) = Ψ(π) for all π ∈Sn, which is why there is no inverse in (6.12), as opposed to (6.3).) We call ch(f) the characteristic of f and call ch the characteristic map.
Proposition 6.3.1. The characteristic map (6.12) is a homomorphism of algebras.
Proof. Suppose f ∈Rm and g ∈Rn. Then ch(f ⋆g) (6.9) = (6.12) D IndSm+n Sm×Sn(f × g), Ψ E Sm+n = D f × g, ResSm+n Sm×Sn(Ψ) E Sm×Sn (Proposition 6.2.4) (6.12) = 1 m!n!
X π∈Sm σ∈Sn (f × g)(π × σ)Ψ(π × σ) (6.4) = (6.11) 1 m!n!
X π∈Sm σ∈Sn f(π)g(σ)Ψ(π)Ψ(σ) = ⟨f, Ψ⟩Sm⟨g, Ψ⟩Sn = ch(f) ch(g).
Lemma 6.3.2. For f ∈Rn, (6.13) ch(f) = X ρ∈Par(n) z−1 ρ fρpρ, where fρ is the value of f at elements of cycle type ρ, and zρ is as defined in (2.47).
136 Applications to representation theory Proof. We have ch(f) = 1 n!
X ρ∈Par(n) X π∈Cρ f(π)Ψ(π) = 1 n!
X ρ∈Par(n) |Cρ|fρpρ.
It thus suffices to prove that (6.14) n!
|Cρ| = zρ = Y i≥1 imi(ρ)mi(ρ)!.
Fix σ ∈Sn of cycle type ρ and consider the homomorphism of groups Sn →Sn, π 7→πσπ−1.
The image of this map is precisely the conjugacy class Cρ. The kernel is the set X of elements of Sn that commute with σ. Thus, we have |Cρ| = |Sn| |X| = n!
zρ , where, in the last equality, we used Exercise 1.1.2. This proves (6.14).
Let ηn be the character of the trivial one-dimensional representation of Sn. Thus ηn(π) = 1 for all π ∈Sn.
Let εn be the character of the one-dimensional sign representation of Sn. Thus, εn(π) = sgn(π) for all π ∈Sn.
Note that it follows from the definitions that η0 = ε0 = 1. We leave it as Exercise 6.3.2 to show that, if π ∈Sn has cycle type ρ, then (6.15) εn(π) = (−1)n−ℓ(ρ) for all π ∈Sn.
Now, for λ = (λ1, . . . , λk) ∈Par, let Sλ := Sλ1 × · · · × Sλk and define the following elements of Rn: ηλ := ηλ1 ⋆· · · ⋆ηλk = IndSn Sλ(ηλ1 × · · · × ηλk), (6.16) ελ := ελ1 ⋆· · · ⋆ελk = IndSn Sλ(ελ1 × · · · × ελk), (6.17) χλ := det ηλi−i+j 1≤i,j≤n , (6.18) where, in computing the determinant, we use the product ⋆. Note that χλ is an integer linear combination of characters. In other words, χλ ∈R is a virtual character. However, we will soon see (Theorem 6.3.6) that, in fact, it is an irreducible character.
Characters of symmetric groups 137 Proposition 6.3.3. For λ ∈Par, (6.19) ch(ηλ) = hλ, ch(ελ) = eλ, ch(χλ) = sλ.
Proof. First note that ch(ηn) (6.13) = X ρ∈Par(n) z−1 ρ pρ (2.49) = hn.
Thus ch(ηλ) = ch(ηλ1) · · · ch(ηλk) = hλ1 · · · hλk = hλ Similarly, ch(εn) (6.13) = (6.15) X ρ∈Par(n) (−1)n−ℓ(ρ)z−1 ρ pρ (2.49) = en, and so ch(ελ) = ch(ελ1) · · · ch(ελk) = eλ1 · · · eλk = eλ.
Finally, since ch is a ring homomorphism, we have ch χλ = det ch(ηλi−i+j) = det (hλi−i+j) (3.10) = sλ.
Theorem 6.3.4. The characteristic map (6.12) is an isometric algebra isomorphism of R onto Sym.
Proof. For f, g ∈Rn, we have ⟨ch(f), ch(g)⟩ (6.13) = X ρ,ρ′∈Par(n) z−1 ρ z−1 ρ′ fρgρ′⟨pρ, pρ′⟩ (3.29) = X ρ∈Par(n) z−1 ρ fρgρ.
On the other hand, ⟨f, g⟩Sn = 1 n!
X π∈Sn f(π)g(π−1) = 1 n!
X ρ∈Par(n) |Cρ|fρgρ (6.14) = X ρ∈Par(n) z−1 ρ fρgρ.
Thus ch is an isometry.
Now, since ch is an isometry, we have ⟨χλ, χµ⟩= ⟨sλ, sµ⟩ (3.30) = δλµ for all λ, µ ∈Par.
It follows from Lemma 6.1.3 that the χλ, λ ∈Par(n), are, up to sign, irreducible characters of Sn. Since the number of conjugacy classes in Sn is equal to the number of partitions of n, the χλ are, in fact, all of the irreducible characters of Sn, up to sign. Therefore the χλ, λ ∈Par(n), form a basis of Rn. Hence ch is an isomorphism of Rn onto Symn for all n ∈N.
It follows that ch is an isomorphism of R onto Sym.
Remark 6.3.5. Theorem 6.3.4 gives a representation theoretic motivation for Definition 3.3.1 of the Hall inner product.
More precisely, the Hall inner product is precisely the inner product on Sym that corresponds to the natural inner product on virtual characters of the symmetric group.
△ 138 Applications to representation theory Recall the definition of a standard tableau from Section 2.3.
Theorem 6.3.6.
(a) The irreducible characters of Sn are χλ, λ ∈Par(n), as defined in (6.18).
(b) The degree of χλ is Kλ,(1n), the number of standard tableaux of shape λ.
Proof. We have shown in the proof of Theorem 6.3.4 that the irreducible characters of Sn are χλ, λ ∈Par(n), up to sign. It remains to prove that, for λ ∈Par, it is χλ, and not −χλ, that is a character. To do this, it suffices to prove that χλ(1) is positive. We will show that χλ(1) = Kλ,(1n), which also proves part (b).
Note that sλ (6.19) = ch(χλ) (6.13) = X µ∈Par(n) z−1 µ χλ µpµ, where χλ µ is the value of χλ on elements of cycle type µ. Thus (6.20) ⟨sλ, pρ⟩= X µ∈Par(n) z−1 µ χλ µ⟨pµ, pρ⟩ (3.29) = χλ ρ.
Therefore, χλ(1) = χλ (1n) = ⟨sλ, pn 1⟩.
Since the Schur functions sλ, λ ∈Par(n), form a basis of Symn, it follows that hn 1 = pn 1 = X λ∈Par(n) ⟨sλ, pn 1⟩sλ = X λ∈Par(n) χλ(1)sλ.
Thus, by Theorem 3.5.5 and Definition 3.4.14, χλ(1) = M(1n),λ(h, s) = (K ⊺)(1n),λ = Kλ,(1n) is the number of standard tableaux of shape λ.
Remark 6.3.7. It follows from the Littlewood–Richardson rule (Theorem 3.6.9) that a prod-uct of two Schur functions is equal to a linear combination of Schur functions with nonnegative integer coefficients. We now have a representation theoretic explanation for this fact. For µ ∈Par(m) and ν ∈Par(n), we have sµsν (6.19) = ch(χµ) ch(χν) = ch(χµ ⋆χν).
By definition, the product χµ ⋆χν is the character of an induced representation. Thus, it is a positive integer linear combination of irreducible characters of Sm+n. It follows that sµsν is a positive integer linear combination of Schur functions.
△ Proposition 6.3.8. The transition matrix M(p, s) is the character table of Sn. Equivalently, for ρ ∈Par(n), (6.21) pρ = X λ∈Par(n) χλ ρsλ.
Thus χλ ρ is equal to the coefficient of xλ+δn in aδnpρ.
Characters of symmetric groups 139 Proof. The equation (6.21) follows immediately from (6.20). Multiplying both sides by aδn, and using (3.7), we have aδnpρ = X λ∈Par(n) χλ ρaλ+δn.
Then the final assertion in the statement of the proposition follows from the fact that the coefficient of xλ+δn in aλ+δn is 1.
Combined with the Murnaghan–Nakayama rule (Proposition 3.7.1), the above results lead to a combinatorial method of computing the character tables of the symmetric groups that we now describe. Recall the definition of a border strip from Section 3.7. We define a border-strip tableau to be a tableau such that • the entries in every row and column are weakly increasing, and • for each i, the boxes with entry i form a border strip.
For example (6.22) 1 1 1 2 2 2 3 5 1 2 2 2 3 3 3 5 2 2 3 3 3 4 4 2 3 3 4 4 4 6 3 3 4 4 4 4 4 4 is a border-strip tableau. For λ, ρ ∈Par(n), n ∈N, we let BST(λ, ρ) denote the set of border-strip tableaux of shape λ and weight ρ. The height ht(T) of a border-strip tableau T is the sum of the heights of the border strips in T. For example, the height of border-strip tableau in (6.22) is 1 + 3 + 4 + 4 + 1 + 0 = 13.
The following result is sometimes also referred to as the Murnaghan–Nakayama rule.
Proposition 6.3.9. For λ, ρ ∈Par(n), we have χλ ρ = X T∈BST(λ,ρ) (−1)ht(T) Proof. Repeated application of the Murnaghan–Nakayama rule (Proposition 3.7.1) shows that pρ = pρn · · · pρ1s∅= X µ∈Par X T∈BST(µ,ρ) (−1)ht(T)sµ.
Thus χλ ρ (6.20) = ⟨sλ, pρ⟩= X µ∈Par X T∈BST(µ,ρ) (−1)ht(T)⟨sλ, sµ⟩ (3.30) = X T∈BST(λ,ρ) (−1)ht(T).
140 Applications to representation theory Remark 6.3.10. It follows from Proposition 6.3.9 that the characters of the symmetric groups are integer valued.
In fact, we will prove something even stronger.
Namely, the irre-ducible representations of the symmetric groups can be constructed over the integers; see Remark 6.4.4. This is not true for arbitrary groups; see Exercise 6.3.5.
△ The next result shows that the involution ω of (2.31) corresponds to multiplication by the character of the sign representation.
Proposition 6.3.11. For f ∈Rn, we have (6.23) ω(ch(f)) = ch(εnf).
(Note that εnf is the pointwise product, and not the ⋆product.) Proof. It suffices to prove the result when f = χλ, λ ∈Par(n), since these span Rn. First note that, for ρ, λ ∈Par(n), we have χλ′ ρ (6.20) = ⟨sλ′, pρ⟩ (3.31) = ⟨ω(sλ′), ω(pρ)⟩ (3.17) = (2.45) (−1)n−ℓ(ρ)⟨sλ, pρ⟩ (6.15) = (6.20) εn ρχλ ρ.
Thus χλ′ = εnχλ and so ω ch(χλ) = ω(sλ) = sλ′ = ch χλ′ = ch εnχλ , as desired.
We saw in Section 4.4 that the ring of symmetric functions can be naturally endowed with the structure of a Hopf algebra. Using Theorem 6.3.4, we can deduce the corresponding structure of a Hopf algebra on R, so that the characteristic map is also an isomorphism of Hopf algebras.
Define linear maps ∆: R →R ⊗R, ∆f = n X k=0 ResSn Sk×Sn−k(f), (6.24) ε: R →C, ε(f) = ⟨f, 1⟩, (6.25) S : R →R, S(f) = (−1)nεnf.
(6.26) for f ∈Rn, then extended by linearity. Note that we view ∆f as an element of R ⊗R using the isomorphism (6.5).
Proposition 6.3.12. The maps ∆, ε, and S endow R with the structure of a Hopf algebra, and the characteristic map (6.12) is an isomorphism of Hopf algebras.
Proof. Since we already know from Theorem 6.3.4 that the characteristic map is an isometric algebra isomorphism, it suffices to show that the maps ∆, ε, and S, as defined in (6.24) to (6.26), correspond, under the characteristic map, to the comultiplication, counit, and antipode of SymC.
Characters of symmetric groups 141 As in (4.6), we define an inner product on R × R by ⟨f1 ⊗f2, g1 ⊗g2⟩= ⟨f1, g1⟩⟨f2, g2⟩.
Then, for f ∈Rn, g ∈Rk, h ∈Rn−k, we have ⟨(ch ⊗ch) ◦∆f), ch(g) ⊗ch(h)⟩ = ⟨∆f, g ⊗h⟩ (since ch is an isometry) (6.24) = D ResSn Sk×Sn−k(f), g ⊗h E = D f, IndSn Sk×Sn−k(g ⊗h) E (by Proposition 6.2.4) (6.9) = ⟨f, g ⋆h⟩ = ⟨ch(f), ch(g ⋆h)⟩ (since ch is an isometry) = ⟨ch(f), ch(g) ch(h)⟩ (since ch is a ring homomorphism) (4.33) = ⟨∆◦ch(f), ch(g) ⊗ch(h)⟩.
Thus (ch ⊗ch) ◦∆f = ∆◦ch(f).
We also have ε ◦ch(f) = ⟨ch(f), 1⟩ (since ⟨ch(f), 1⟩is the constant term of ch(f)) = ⟨f, 1⟩ (since ch is an isometry) (6.25) = ε(f) = ch ◦ε(f) (since ch is an algebra homomorphism and ε(f) ∈k).
Finally, for f ∈Rn, S ◦ch(f) = (−1)nω(ch(f)) (6.23) = (−1)n ch(εnf) (6.26) = ch ◦S(f), where, in the first equality, we used Exercise 4.4.3.
Exercises.
6.3.1. Prove that, with the multiplication ⋆of (6.9), R is a commutative, associative, unital, graded ring.
6.3.2. Prove that if π ∈Sn has cycle type ρ, then εn(π) = (−1)n−ℓ(ρ), were εn is the character of the sign representation of Sn.
142 Applications to representation theory 6.3.3. Compute the character table of S3.
6.3.4. Compute the character table of S4.
6.3.5. Let Gn be the cyclic group of order n.
(a) What are the conjugacy classes of Gn?
(b) Describe all the irreducible representations of Gn.
(c) Show that, for n > 2, the characters of Gn are not all real-valued.
6.4 Specht modules It follows from Theorem 6.3.6 that the irreducible Sn-module are enumerated by partitions of n.
Our goal in this section is to construct these modules explicitly.
We follow the procedure outlined in [Mac15, Example 15, p. 124].
A priori, we should work over the complex numbers to find all irreducible representations of a finite group. However, as we will see, the symmetric groups have the remarkable property that we can construct all of their irreducible representations over the integers. We first work over Q, and find enough irreducible representations to know that we have found them all. We then show that one can, in fact, work over Z.
It follows from our knowledge of transition matrices (Theorem 3.5.5, Lemma 3.5.2, and Proposition 3.5.6(a)) that hλ = sλ + X µ>λ Kµλsµ and eλ′ = sλ + X µ<λ Kµ′λ′sµ.
Applying the inverse of the characteristic map ch from (6.12) and using (6.19), we have ηλ = χλ + X µ>λ Kµλχµ and ελ′ = χλ + X µ<λ Kµ′λ′χµ.
These equations give the decomposition of the induced modules Hλ := IndSn Sλ(trivλ) and Eλ′ := IndSn Sλ′(signλ′) into irreducible modules. Here trivµ denotes trivial Sµ-module and signµ denotes the sign Sµ-module. It follows that the irreducible Sn-module with character χλ is the unique common irreducible component of Hλ and E′ λ. We would like to describe this module explicitly.
If A is a ring and x, y ∈A, then Ay is an A-module with action given by left multiplication, and Axy is a submodule of Ay. In particular, Axy is the image of Ax under the A-module homomorphism A →Ay, z 7→zy, z ∈A.
Thus Axy is also a quotient of Ax.
Let λ ∈Par(n), and let T be any standard tableau of shape λ. Thus the entries of T are 1, 2, . . . , n. Let R (respectively C) be the subgroup of Sn that stabilizes each row Specht modules 143 (respectively, column) of T. (We say that π ∈Sn stabilizes a subset X ⊆{1, 2, . . . , n} if π(X) = X.) Thus we have group isomorphisms R ∼ = Sλ and C ∼ = Sλ′.
Let A = QSn be the group algebra of Sn and define s, a ∈A by (6.27) s = X π∈R π and a = X π∈C sgn(π)π.
Then (QR)s = Qs is the trivial QR-module, and we leave it as Exercise 6.4.2 to show that we have an isomorphism of Sn-modules (6.28) Hλ = A ⊗QR Qs →As, x ⊗cs 7→xcs, x ∈A, c ∈Q.
Similarly, we leave it as Exercise 6.4.3 to show that we have an isomorphism of Sn-modules (6.29) Eλ′ = A ⊗QC Qa →Aa, x ⊗cs 7→xca, x ∈A, c ∈Q.
Now let e = as ∈A.
Any element of R ∩C stabilizes each row and column of T and thus must be the identity.
In other words R ∩C = {1} is the trivial subgroup of Sn. It follows that the products πσ, π ∈R, σ ∈C, are all distinct (Exercise 6.4.4). Thus e = X π∈R σ∈C sgn(σ)πσ ̸= 0.
It then follows from the above discussion that Mλ := Ae is a nonzero submodule of As ∼ = Hλ that is isomorphic to a quotient of Aa ∼ = Eλ′. Thus Mλ is a common irreducible component of Hλ and Eλ′ and so it is the irreducible Sn-module with character χλ.
Our goal is to give an explicit construction of the irreducible modules Mλ, λ ∈Par, by identifying Mλ with a space of polynomials. For a tableau T of size n and weight (1n), let mT := xd(1) 1 · · · xd(n) n ∈Q[x1, . . . , xn], where d(i) = r −1 if the entry i lies in the r-th row of T. Then define fT := Y (i,j) (xi −xj), where the product is taken over all pairs (i, j) such that j lies above, and in the same column as, i in the tableau T.
In other words, fT is the product of the Vandermonde determinants (see (3.5)) corresponding to the columns of T. It then follows as in Lemma 3.1.3 and Exercise 3.1.1 that (6.30) fT = X π∈C sgn(π)πmT = amT.
Recall from Section 2.3 that ST(λ) denotes the set of standard tableaux of shape λ.
144 Applications to representation theory Theorem 6.4.1. For λ ∈Par(n), the subspace (6.31) SpanQ{fT : T is a tableau of shape λ and weight (1n)} ⊆Poln is an Sn-submodule of Poln isomorphic to the irreducible module Mλ.
Furthermore, the polynomials fT, T ∈ST(λ), form a basis of this Sn-module.
Proof. Consider the map (6.32) A →Q[x1, . . . , xn], u 7→umT.
Since d(i) = d(j) if and only if i and j lie in the same row of T, it follows that smT = |R|mT.
Restriction of (6.32) to Ae thus yields a surjective homomorphism of A-modules θ: Mλ = Aas →AasmT = AamT (6.30) = AfT.
Since Mλ is a simple A-module, ker(θ) is an A-submodule of Mλ, and θ is not the zero map, it follows that ker(θ) = 0. Thus θ is an isomorphism of Sn-modules.
It remains to prove that the fT, T ∈ST(λ), form a basis of (6.31). We know from Theorem 6.3.6(b) that the dimension of Mλ is the number of standard tableaux of shape λ.
So it suffices to show that the fT, T ∈ST(λ), are linearly independent.
Define an ordering on monomials xα, α ∈Nn, by declaring that xα < xβ if α precedes β in the lexicographic order on Nn. That is, xα < xβ ⇐ ⇒∃i ∈{1, . . . , n} such that α1 = β1, . . . , αi−1 = βi−1, αi < βi.
Now suppose that T is a standard tableaux of shape λ and π ∈C. Let i ∈{1, 2, . . . , n} be the smallest element such that π(i) ̸= i. Then i < π−1(i), and so d(i) < d(π−1(i)), since T is standard. Thus mT = xd(1) 1 · · · xd(n) n < xd(1) 1 · · · xd(i−1) i−1 xd(π−1(i)) i · · · xd(π−1(n)) n = πmT.
It follows that (6.33) fT (6.30) = X π∈C sgn(π)πmT = mT + X α dαxα where dα ∈Q and the sum is over α such that mT < xα.
Let T1, . . . , Tr be the standard tableaux of shape λ. Since the monomials mT1, . . . , mTr are all distinct, we may choose the order of the Ti such that mT1 < mT2 < · · · < mTr.
Suppose we have a nontrivial linear dependence relation (6.34) r X i=1 cifTi = 0.
Let i be the smallest index such that ci ̸= 0. It follows from (6.33) that the coefficient of mTi in the left-hand side of (6.34) is equal to ci. But then the left-hand side cannot be equal to zero. This contradiction shows that the fT1, . . . , fTr are linearly independent, as desired.
Specht modules 145 The realization of Mλ given in Theorem 6.4.1 is called the Specht module corresponding to the partition λ. From now on, we use the notation Mλ to denote this specific realization.
Remark 6.4.2. Although we worked in Theorem 6.4.1 over the field Q, the proof remains valid over any field. In other words, if k is any field, then the fT, T ∈ST(λ), form a basis of a kSn-submodule of k[x1, . . . , xn].
△ Proposition 6.4.3. Let λ ∈Par(n). For each π ∈Sn, the matrix representing the action of π on Mλ in the basis fT, T ∈ST(λ), has integer entries.
Proof. As in the proof of Theorem 6.4.1, let T1, . . . , Tr denote the standard tableaux of shape λ. It follows from Theorem 6.4.1 that, for all tableaux of shape λ, fT is a linear combination of the fT1, . . . , fTr. Suppose fT = r X i=1 cifTi, c1, . . . , cr ∈Q.
Let m be the lowest common denominator of the ci. Then, for each i ∈{1, . . . , r}, we have ci = mi/m for some mi ∈Z, and so (6.35) mfT = r X i=1 mifTi.
If m > 1, then choose a prime number p dividing m and reduce (6.35) modulo p. Since not all the mi are divisible by p, it follows that the fTi are linearly dependent over the field Z/pZ. But this contradicts Remark 6.4.2.
It follows from the above that, if T is any tableau of shape λ, then fT is an integer linear combination of fT1, . . . , fTr. Since πfTi = fT, where T is obtained from Ti by permuting its entries by π, the proposition follows.
Remark 6.4.4. It follows from Remark 6.4.2 that the fT, T ∈ST(λ), are linearly independent over Z (since any linear dependence relation over Z is a linear dependence relation over the field Q). Thus, for λ ∈Par, SpanZ{fT : T ∈ST(λ)} is a free Z-module of rank equal to the number of standard tableaux of shape λ. By Propo-sition 6.4.3, the ring ZSn acts on this space. In this way, the Specht modules are defined over the integers. We can then reduce modulo any prime p to obtain modules for(Z/pZ)Sn.
Unlike the situation in characteristic zero, the Specht modules can be reducible in positive characteristic. For a prime number p, we say that a partition λ is p-regular if it does not have p nonzero parts of the same size. If λ is a p-regular partition, then the Specht module Mλ, defined over Z/pZ, has a unique irreducible quotient, and these irreducible quotients form a complete set of irreducible representations. However, there remain many unanswered questions about the representation theory of the symmetric group in positive characteristic and this is an active area of research. For a short discussion of the positive characteristic case, see [Wil].
△ 146 Applications to representation theory Exercises.
6.4.1. With s and a defined as in (6.27), show that s2 = |R|s and a2 = |C|a.
6.4.2. Prove that (6.28) is an isomorphism of Sn-modules.
6.4.3. Prove that (6.29) is an isomorphism of Sn-modules.
6.4.4. Suppose that H and K are subgroups of a group G such that H ∩K = {1}. Show that the products xy, x ∈H, y ∈K, are all distinct.
6.5 Representations of general linear groups In this final section, we explain how Schur functions appear as characters of simple modules for the general linear Lie algebra and general linear Lie group. Our treatment is very brief.
We refer the reader to the many books on Lie theory (for example, [Hal15]) for further details.
Fix n ∈N. Throughout this section we work over the field C of complex numbers. The general linear Lie algebra gln(C) is vector space of complex n × n matrices, together with a binary operation, called the Lie bracket, given by [X, Y ] = XY −Y X, X, Y ∈gln(C).
A representation of gln(C) is a complex vector space V , together with a linear map ρ: gln(C) →EndC(V ) such that ρ([X, Y ]) = ρ(X)ρ(Y ) −ρ(Y )ρ(X), X, Y ∈gln(C).
We define Xv := ρ(X)(v), X ∈gln(C), v ∈V.
Then V , together with this action, is called a gln(C)-module.
The Cartan subalgebra h of gln(C) is the Lie subalgebra consisting of diagonal matrices.
(Note that the Lie bracket of any two diagonal matrices is zero. We say that h is an abelian Lie algebra.) For 1 ≤i ≤n, let εi : h →C, X = (Xij) 7→Xii be the linear map that returns the (i, i)-entry of a diagonal matrix. Then define Λ := n M i=1 Zεi ⊆h∗.
Representations of general linear groups 147 If V is a gln(C)-module and λ ∈h∗, then the λ-weight space of V is Vλ := {v ∈V : Xv = λ(X)v for all X ∈h}.
We say that V is a weight module if V = M λ∈Λ Vλ.
An element v ∈V is a highest-weight vector of weight λ ∈Λ if v ∈Vλ and v is annihilated by all strictly upper triangular matrices. The module V is a highest-weight module of highest weight λ if it is generated by a highest weight vector of weight λ.
Every finite-dimensional gln(C)-module is a sum of simple modules (modules containing no nontrivial submodules). Furthermore, for each λ ∈Λ+ := ( n X i=1 λiεi : λ1 ≥λ2 ≥· · · ≥λn ) , there exists a unique (up to isomorphism) simple highest-weight module V λ of highest weight λ. The V λ, λ ∈Λ+, are a complete list of finite-dimensional simple gln(C)-modules, up to isomorphism. Thus, to completely understand the finite-dimensional representation theory of gln(C), one needs only understand the V λ, λ ∈Λ+.
If V is a weight module, we define its character to be ch(V ) = X λ∈Λ (dimC Vλ)xλ ∈Poln, where we view λ ∈Λ as an element of WCompn by identifying Pn i=1 λiεi with (λ1, . . . , λn).
Thus, the character of V records the dimensions of the weight spaces of V .
The Weyl character formula for gln(C) states that (6.36) ch V λ = P π∈Sn sgn(π)x(λ+δn)π P π∈Sn sgn(π)x(δn)π = aλ+δn aδn = sλ.
In other words, the character of the simple module with highest weight λ is precisely the Schur function sλ.
Remark 6.5.1. The Weyl character formula is a much more general result that holds for other Lie algebras. We have presented here the special case of this formula for gln(C). In fact, we have presented a form that uses the Weyl denominator formula. The partition δn is typically denoted ρ in the context of Lie theory.
△ The general linear group GLn(C) is the group of complex n × n invertible matrices. We have a surjective exponential map exp: gln(C) →GLn(C), exp(X) = ∞ X k=0 Xk k! .
148 Applications to representation theory This map allows one to move between representations of gln(C) and GLn(C). In particular, the simple polynomial representations of GLn(C) are also indexed by partitions λ ∈Parn. If ρλ : gln(C) →EndC(V λ) is the representation of gln(C) corresponding to λ ∈Parn, then it follows from (6.36) that, for a diagonal matrix H ∈gln(C), we have (6.37) tr eρλ(H) = P π∈Sn sgn(π)e(λ+δn)π(H) P π∈Sn sgn(π)e(δn)π(H) .
This expression is sometimes also referred to as the character of ρλ. (Compare to the defi-nition of character from Section 6.1.) Index of notation ∪, 69 ∂ ∂en , 88 ∂ ∂hn , 88 ∂ ∂pn , 88 λ ⊇µ, 58 ⟨, ⟩, 51, 113, 134 ⟨, ⟩C G, 127 ≤, 8, 9 ≤R, 69 ⊗, 81 ⋆, 134 ⊢, 6 A, 116 aα, 43 A(c), 115 απ, 8 Altn, 43 B±(t), 119 B± n , 119 BST(λ, ρ), 139 C, 142 C(G, Z), 127 ch, 135 χλ, 136 χρ, 127 Cl, 108 Cρ, 134 ∆, 92, 140 δn, 44 Eλ, 142 ε, 70, 92 ελ, 136 εn, 136 ε, 140 er, 29 er(x1, . . . , xn), 29 E(t), 30 E±(t), 117 η, 91 ηλ, 136 ηn, 136 F, 109 f ⊥, 86 F(c), 109 F(G, Z), 127 fλ, 35 πf, 21 fρ, 135 fT, 143 GLn(C), 147 gln(C), 146 Gop, 10 Heis, 102 Hλ, 142 hλ, 35 hr, 33 hr(x1, . . . , xn), 34 H±(t), 117 ht(T), 139 ht(λ), 78 I, 109 I(c), 109 IndG H, 130, 132 inv(π), 42 J, 65, 112 K, 67 Kλ,µ, 63 149 150 Index of notation Kλ/µ,ν, 63 kG, 126 L, 69 λ′, 7 |λ|, 6 λ/µ, 58 ℓ(λ), 7 M −⊺, 66 Mλ,µ(e, m), 30 Mλ,µ(h, m), 36 Mλ,µ(p, m), 40 mi(λ), 7 Mλ, 143 mλ, 25 mλ(x1, . . . , xn), 21 mT, 143 M(u, v), 66 N, 5 [n], 5 ∇, 91 Par, 6 Park, 7 Park(n), 7 Par(k), 6 p(n), 16 p(n; k), 16 Poln, 20 Polk n, 20 Pr, 114 pr, 37 Ψ, 135 Ψ±(t), 122 ψ± i , 108 Ψ± i , 122 P(t), 38 R, 134 R×, 15 R, 142 ResG H, 130 R(G, C), 129 ρm,n, 23 ρn, 25 ρk n, 24 Rn, 134 RowStrict, 28 RowWeak, 28 RJtK, 14 RX, 13 RJx1, x2, . . .K, 17 S, 93, 140 shape(T), 28 S∞, 6 Sλ, 136 sλ, 47 sλ/µ, 56 sλ(x1, . . . , xn), 46 Sn, 5 sgn(π), 42 SST(λ), 28 SST(λ, n), 28 ST(λ), 28 Sym, 25 Sym+, 102 Sym−, 102 Symk, 24 Symn, 21 Symk n, 21 SymQ, 27 T, 122 UTn, 65 v(c) λ , 110 vi, 109 ω, 34 WComp, 8 WComp(k), 8 WCompn, 8 WCompn(k), 8 word(T), 76 wt(T), 28 xα, 16, 20 z, 70 zλ, 40 Bibliography [DF04] D. S. Dummit and R. M. Foote.
Abstract algebra.
John Wiley & Sons, Inc., Hoboken, NJ, third edition, 2004.
[DNR01] S. Dăscălescu, C. Năstăsescu, and Ş. Raianu.
Hopf algebras: An introduction, volume 235 of Monographs and Textbooks in Pure and Applied Mathematics. Marcel Dekker, Inc., New York, 2001.
[Egg19] E. S. Egge. An introduction to symmetric functions and their combinatorics, vol-ume 91 of Student Mathematical Library. American Mathematical Society, Provi-dence, RI, 2019. doi:10.1090/stml/091.
[GAE02] Z. Gong, M. Aldeen, and L. Elsner. A note on a generalized Cramer’s rule. Linear Algebra Appl., 340:253–254, 2002. doi:10.1016/S0024-3795(01)00469-4.
[Hal15] B. Hall. Lie groups, Lie algebras, and representations, volume 222 of Graduate Texts in Mathematics.
Springer, Cham, second edition, 2015.
An elementary introduction. doi:10.1007/978-3-319-13467-3.
[Mac15] I. G. Macdonald. Symmetric functions and Hall polynomials. Oxford Classic Texts in the Physical Sciences. The Clarendon Press, Oxford University Press, New York, second edition, 2015. With contribution by A. V. Zelevinsky and a foreword by Richard Stanley, Reprint of the 2008 paperback edition [ MR1354144].
[Sav] A. Savage.
MAT 5141 – Algebra I, Lecture notes.
URL: https:// alistairsavage.ca/mat5141/notes/MAT5141-Algebra_I.pdf.
[Ste02] J. R. Stembridge. A concise proof of the Littlewood-Richardson rule. Electron. J.
Combin., 9(1):Note 5, 4, 2002. URL: 9/Abstracts/v9i1n5.html.
[Wil] M. Wildon. Representation theory of the symmetric group. URL:
ma.rhul.ac.uk/~uvah099/Maths/Sym/SymGroup2014.pdf.
151 |
7709 | https://www.desmos.com/calculator/u8wt5rnw9n | Vectors | Desmos
Loading...
Vectors
Save Copy
Log In Sign Up
To use this vector calculator simply enter the x and y value of your two vectors below. Make sure to separate the x and y value with a comma. I put an example below so you can see how it is done.
1
Next to add/subtract/dot product/find the magnitude simply press the empty white circle next to the "ADDITION" if you want to add the vectors and so on for the others.
2
To find the value of the resulting vector if you're adding or subtracting simply click the new point at the end of the dotted line and the values of your vector will appear.
3
Expression 4: "v" Subscript, 1 , Baseline equals left bracket, negative 3 , negative 1 , right bracket v 1=−3,−1
2 element list
4
Expression 5: "v" Subscript, 2 , Baseline equals left bracket, 1 , 1 , right bracket v 2=1,1
2 element list
5
ADDITION
6
SUBTRACTION
10
DOT PRODUCT
14
MAGNITUDE
17
23
24
25
34
powered by
powered by
"x"x
"y"y
"a" squared a 2
"a" Superscript, "b" , Baseline a b
7 7
8 8
9 9
divided by÷
functions
((
))
less than<
greater than>
4 4
5 5
6 6
times×
| "a" ||a|
,,
less than or equal to≤
greater than or equal to≥
1 1
2 2
3 3
negative−
A B C
StartRoot, , EndRoot
pi π
0 0
..
equals=
positive+ |
7710 | https://ltd.aruplab.com/Tests/Pub/2008460 | RBC Band 3 Protein Reduction in Hereditary Spherocytosis | ARUP Laboratories Test Directory
Skip to main content
News
Careers
Contact Us
|
Connect Sign In
News
Careers
Contact Us
Testing
Test Directory Test Updates Clinical Decision Support | ARUP Consult® Resources Featured Test Categories Search Our Test Directory
Search
Search our extensive Laboratory Test Directory to find test codes, ordering recommendations, specimen stability information, Test Fact Sheets, and more. Changes and New Tests Explore the most recent updates to our Laboratory Test Directory in one convenient location. Review important information about upcoming or current Hotlines, CPT code changes, new tests, and immediate activations.
Test Resources Find general guidance on ARUP specimen preparation and handling, specimen transportation, test turnaround times, critical alerts, billing practices, patient forms, and more. Changes and New Tests Explore the most recent updates to our Laboratory Test Directory in one convenient location. Review important information about upcoming or current Hotlines, CPT code changes, new tests, and immediate activations. CPT Code Updates Find information about our annual and biweekly updates to CPT codes.
Test Resources Find general guidance on ARUP specimen preparation and handling, specimen transportation, test turnaround times, critical alerts, billing practices, patient forms, and more.
Performed/Reported Find information about updates to the Performed schedule and the Reported turnaround time. ARUP Consult®Our clinical decision support resource, ARUP Consult, empowers clinicians with robust information, algorithms, and tools for informed patient care decisions. Choose the Right Test
Testing Algorithms
Test Fact Sheets Test Resources Find specimen instructions, test turnaround times, patient forms, test codes, and more. Specimen Handling
CPT Codes
Z-Codes
LOINC Codes
Test Turnaround Time and Contingency Plan
Critical Values and Urgent Notifications
View All Test Resources Comprehensive Diagnostic Solutions Discover the benefits of our testing for each featured test category. Explore highlighted updates, articles, video spotlights, and more.
This section highlights our specialized offerings and is separate from our Laboratory Test Directory. Autoimmune Neurologic Disease
Genetics
Hematopathology
Infectious Disease
View All
Search the Laboratory Test Directory
Clinical Decision Support | ARUP Consult®
Services
Healthcare Advisory Services Drug and Clinical Research Services Research and Innovation Tailored Solutions and Advisory Services Optimize your laboratory by partnering with ARUP Healthcare Advisory Services. Discover our solutions, including process streamlining, outreach business planning, laboratory stewardship, and more. Laboratory Outreach Solutions and Business Planning
Workflow Solutions and Space Planning
Laboratory and Pharmacy Stewardship Solutions
Laboratory Analytics and Reporting Tools | ARUP AnalyticsDx TM
View All Advisory Services For Researchers: Clinical Research Services Our team of experienced project managers is available to assist you in companion diagnostic development, clinical trial studies, academic research, and expert pathology review. Clinical and Academic Research Our Clinical Trials group supports a variety of research efforts, including academic research initiatives, and serves clients from central laboratories, clinical research organizations (CROs), in vitro device (IVD) manufacturers, and more. Discover all of our services.
Pharmaceutical Services ARUP offers companion diagnostics and central pathology review services as part of our comprehensive pharmaceutical services. Explore our capabilities and more.
View All Clinical Research Services ARUP Institute for Research and Innovation in Diagnostic and Precision Medicine™We accelerate the development and adoption of cutting-edge diagnostic and prognostic technologies advancing patient care through groundbreaking science and engineering.The Sherrie Perkins Research and Innovation Collaboration Grant The Sherrie Perkins Research and Innovation Collaboration Grant funds cutting-edge research in laboratory medicine.
Education Center
Continuing Education ARUP Expert Insights Education Events MLS Faculty and Student Resource Center Free Continuing Education Credits ARUP provides free CME, P.A.C.E.®, and Florida credits for laboratory professionals. CME Credit
P.A.C.E.® Credit
Florida Credit
View All Courses Expert-Led Spotlight Video Series Take your learning to the next level with these short lectures. Videos are typically less than 10 minutes and cover a variety of topics such as test utilization, testing technologies, new test explanations, and more. Podcast Episodes Stay up to date on trending laboratory medicine topics with our LabMind podcast.
Research Publications Explore our experts’ most recent publications in peer-reviewed sources.
Posters and Presentations Request a poster presentation from a recent conference or event. Live Webinars Attend our live webinars for insightful, interactive conversations around the latest trends and advancements in laboratory medicine. Upcoming Regional Conferences Join our experts at upcoming educational conferences and ARUP-sponsored events. Program agendas, registration information, event locations, and more can be found here.Request an ARUP Speaker Bring our experts directly to your facility with our diverse educational offerings. Choose from a variety of topics, delivered in person or via webinar. Access Free Online Resources Enhance your medical laboratory science (MLS) program with our free online resource center. Integrate these curated resources into your curriculum to enrich valuable student-faculty engagement.
About
Contact Us About Us Careers News Employee Resources Compliance Contact Us Contact our support team any time through our 24/7 phone assistance, live chat options, or directly via email.
Media Coverage and Inquiries Contact an ARUP media specialist about photos, press releases, interviews, quotes, and more. Sales and Account Inquiries Request to speak to your local account executive or to set up a new account with ARUP.
Healthcare Advisory Services Request Contact an ARUP healthcare consultant to learn more about how you can optimize your laboratory operations, generate more revenue, and improve patient care. About Us ARUP Laboratories is a national, academic, and nonprofit reference laboratory at the forefront of diagnostic medicine. With more than 40 years of experience supporting clients through unparalleled quality and service, our goal is to continually drive positive change for healthcare systems and their patients.Learn More Our Experts Explore the breadth and depth of expertise available from ARUP’s 150+ medical directors, genetic counselors, pathologists, specialty scientists, and consultants, most of whom are on-site. Explore Our Culture ARUP offers an enriching work environment with exceptional benefits, flexible hours, and an inclusive company culture that fosters pride among employees. Start Your Journey With ARUP Explore exciting career opportunities across a variety of fields and experience levels. Open positions are posted in one convenient location so you can easily find your perfect fit.
Browse All Open Positions Explore the ARUP Newsroom Stay informed about the latest from ARUP. Explore press releases, news articles, our Magnify publication, and more. Press Releases
Recent Articles
Media Resources
Magnify Magazine Employee Benefits ARUP employees can review their employee benefits, including medical coverage and open enrollment information, long-term absence arrangements, employee assistance programs, and other resources. Wellness Center Employees can access a team of dedicated wellness coaches for private or group classes, nutrition information, recovery activities, and more.
Family Health Clinic Employees and their dependents can access free primary care, occupational healthcare, immunization, mental health resources, and more at the Family Health Clinic.
Learning Management Systems Login Licensure and Accreditations ARUP is a College of American Pathologists (CAP)-, Clinical Laboratory Improvement Amendments (CLIA)-, and International Organization for Standardization (ISO) 15189-accredited laboratory. Additionally, ARUP meets all regulations for labeling non-FDA approved test methods offered by ARUP.
Medicare Coverage Information Learn about the restrictions around Medicare billing and reimbursement before you order an ARUP test. Public Health Reporting Learn about ARUP's policy regarding public health reporting and how we stay in compliance with state regulations.
Privacy and HIPAA Review our policies and practices for HIPAA and GDPR compliance.
ARUP Compliance Policy
Browse Tests A-Z
ABCDEFGHIJKLMNOPQRSTUVWXYZAllergens
Breadcrumb
ARUP Home
Lab Test Directory
RBC Band 3 Protein Reduction in Hereditary Spherocytosis
10-20-2025
Click to view changes
2008460
RBC Band 3 Protein Reduction in Hereditary Spherocytosis
RBC BAND3
Choose the Right Test
ARUP Consult® assists with test selection and interpretation
Hemolytic Anemias
Hemolytic Anemias Testing Algorithm
Go to ARUP Consult
Example Reports
Abnormal
Borderline
Normal
Interface Map
Interface Map
COMPONENT DESCRIPTION TEST TYPE INFECTIOUS UNIT OF MEASURE NUMERIC MAP LOINC
2008461
RBC Band 3 Protein Reduction in HS
Resultable
N
33048-0
For questions regarding the Interface Map, please contact interface.support@aruplab.com.
Download to Excel
Time Sensitive
Additional Technical Information
Ordering Recommendation
Recommendations when to order or not order the test. May include related or preferred tests.
Use to confirm diagnosis of hereditary spherocytosis when hemolytic anemia and spherocytes are present.
New York DOH Approval Status
Indicates whether a test has been approved by the New York State Department of Health.
This test is New York state approved.
Specimen Required
Patient PreparationInstructions patient must follow before/during specimen collection.
CollectSpecimen type to collect. May include collection media, tubes, kits, etc.
Lavender (EDTA) or green (sodium or lithium heparin).
Specimen PreparationInstructions for specimen prep before/after collection and prior to transport.
Transport 4 mL whole blood in the original container. (Min: 0.5 mL)
Storage/Transport TemperaturePreferred temperatures for storage prior to and during shipping to ARUP. See Stability for additional info.
Refrigerated.
Unacceptable ConditionsCommon conditions under which a specimen will be rejected.
Clotted or hemolyzed specimens. Specimens older than 7 days.
RemarksAdditional specimen collection, transport, or test submission information.
Specimens must be analyzed within 7 days of collection.
StabilityAcceptable times/temperatures for specimens. Times include storage and transport time to ARUP.
Ambient: 3 days; Refrigerated: 7 days; Frozen: Unacceptable
Methodology
Process(es) used to perform the test.
Qualitative Flow Cytometry
Performed
Days of the week the test is performed.
Sun-Sat
Reported
Expected turnaround time for a result, beginning when ARUP has received the specimen.
1-3 days
Reference Interval
Normal range/expected value(s) for a specific disease state. May also include abnormal ranges.
Normal
Interpretive Data
May include disease information, patient result explanation, recommendations, or details of testing.
This test can be used to confirm a suspected diagnosis of hereditary spherocytosis (HS). HS is a common inherited hemolytic anemia characterized by the presence of spherical erythrocytes (spherocytes). HS is diagnosed based on family history and clinical features, along with clinical laboratory tests, including peripheral smear examination, osmotic fragility (OF), flow cytometry, or by genetic testing (Hereditary Hemolytic Anemia Panel Sequencing, ARUP test code 2012052).
Band 3 (or solute carrier family 4 member 1, SLC4A1) is the most abundant transmembrane protein found in human red blood cells (RBC). Eosin-5-maleimide (EMA) dye binds to band 3 on intact RBC's. A reduction of fluorescence intensity will be seen in hereditary spherocytosis. This test by flow cytometry has been reported to have a sensitivity of 93 percent for a diagnosis of HS. Congenital dyserythropoietic anemia type II, Southeast Asian ovalocytosis and hereditary pyropoikilocytosis are rare disorders that may also show a positive result.
Compliance Category
Laboratory Developed Test (LDT)
Note
Additional information related to the test.
Hotline History
History of test changes published on ARUP Hotlines for the last two years
View Hotline History
Hotline History
Date of Change
Test Name Change
Methodology
Performance/Reported Schedule
Specimen Requirements
Reference Interval
Interpretive Data
Note
CPT Code
Component Change
Other Interface Change
New Test
Inactive
08/19/2024
X
N/A
CPT Codes
The American Medical Association Current Procedural Terminology (CPT) codes published in ARUP's Laboratory Test Directory are provided for informational purposes only. The codes reflect our interpretation of CPT coding requirements based upon AMA guidelines published annually.
CPT codes are provided only as guidance to assist clients with billing. ARUP strongly recommends that clients confirm CPT codes with their Medicare administrative contractor, as requirements may differ.
CPT coding is the sole responsibility of the billing party. ARUP Laboratories assumes no responsibility for billing errors due to reliance on the CPT codes published.
88184
Are you an ARUP Client? Click here for your pricing.
Components
Components of test
| Component Test Code | Component Chart Name | LOINC |
---
| 2008461 | RBC Band 3 Protein Reduction in HS | 33048-0 |
Component test codes cannot be used to order tests. The information provided here is not sufficient for interface builds; for a complete test mix, please click the sidebar link to access the Interface Map.
Aliases
Other names that describe the test. Synonyms.
Hereditary Spherocytosis, EMA, BAND 3, Osmotic Fragility
HS hemolytic anema assay
RBC Band 3 Protein Reduction in Hereditary Spherocytosis
We value your Feedback.
Feedback
I want to provide feedback regarding
Comment
First Name
Last Name
Email
Client ID
CAPTCHA
Get new captcha!
What code is in the image?
Enter the characters shown in the image.
This question is for testing whether or not you are a human visitor and to prevent automated spam submissions.
Leave this field blank
Specimen
Details
Clinical
ARUP Consult
Live Chat
About
About
Careers
News
Press Releases
Testing
Laboratory Test Directory
Choose the Right Test
Resources
Education Center
Continuing Education
Expert Videos
Podcast Episodes
Events
Contact Us
24/7 Phone Support
Sales Inquiries
By continuing to browse our website, you agree to the use of cookies in accordance with our Privacy Policy. Agree
© 2025 ARUP Laboratories. All Rights Reserved.
Live chat:Live Chat with Client Services |
7711 | https://www.onlinemathlearning.com/fraction-word-problems-g5.html | Fraction Word Problems (video lessons, examples and solutions)
OnlineMathLearning.com - Do Not Process My Personal Information
If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties.
Personal Data Processing Opt Outs
CONFIRM
×
Fraction Word Problems
In these lessons, we will learn how to solve fraction word problems using models or block diagrams (Singapore Math).
Learning Software
Calculus Courses
Discover more
Mathematics
Math
Video
Algebra Textbooks
Online Math Tutoring
Purchase math workbooks
Purchase GRE preparation materials
Science
Hire IELTS preparation tutor
Buy chemistry lab equipment
Share this page to Google Classroom
Related Pages
Fraction Word Problems: Examples
Harder Fraction Word Problems
Singapore Math Lessons
Fraction Problems Using Algebra
More Math Word Problems
Word problem involving addition and subtraction of proper unlike fractions. Understand and solve word problems on fractions using the popular model method of Singapore.
Example:
Mala has 1/2 liter of water in a bottle. She drinks 1/3 liter of water from the bottle.
a) How much water is left in the bottle?
b) If she fills another 1/6 liter of water into the bottle, how much water is there in the bottle now?
Show Video Lesson
Example of a word problem involving addition, subtraction and multiplication of fractions and whole numbers. The model method has been used to understand the problem and help derive the solution.
Example:
1/4 of the girls in a class have short hair. 2/3 of the girls in the class have medium-length hair while the rest of the girls in the class have long hair.
a) What fraction of the girls in the class have long hair?
b) If there are 36 girls in a class, how many of them have long hair?
Show Video Lesson
Word problem involving addition of mixed fractions (mixed numbers). Explained using the model method (bar method or bar diagrams).
Example:
Tim started from his home and drove one and one-half hours to meet his uncle with whom he spent three and one-quarter hours. Then he drove 1/5 of an hour to see his grandmother and spent two and one-half hours with her. Finally, he drove back home in one and three-fifths hours. Altogether, how many hours did he spend outside before returning home?
Show Video Lesson
Word problem on multiplication and subtraction of fractions. Explained using Singapore’s model method of problem solving.
Example:
Lily had 4/5 m of cloth. She used 3/4 of it to make handkerchiefs. How much cloth had she left?
Show Video Lesson
Discover more
Mathematics
Math
Video
Find math educational games
Educational assessment
Online Math Tutoring
Math Education Resources
General Science
Purchase math workbooks
Buy TOEFL test prep
Discover more
Math
Mathematics
Video
Worksheet Generators
Classroom Tools
Buy TOEFL test prep
Math Lesson Plans
Teacher Training Programs
Assessments
Purchase calculus online course
Word problem on subtraction and multiplication of fractions solved using models.
Example:
Rose had some rice. She gave 1/4 of it to her sister and cooked 1/6 of the remaining.
a) What fraction of the rice did she cook?
b) If she cooked 1/2 kg of rice, how much rice had she at first?
Show Video Lesson
Fractions word problem involving multiplication of a mixed number by a whole number using the model method.
Example:
A dictionary is 3 and one-third cm thick. Find the height of the stack made by 20 such dictionaries when placed on top of each other.
Show Video Lesson
Word problem to explain the concept of fraction as division and to show relation between fractions and decimals.
Example:
Pam packs 4 kg of onions equally into 10 bags. What is the mass of onions in eack bag?
Give your answer as a fraction in its simplest form.
Give your answer as a decimal.
Show Video Lesson
Word problem to explain addition and division of fractions using models. Also shows the relation between fractions and decimals.
Example:
3/4 kg of candies and 6/7 kg of cookies are shared equally among three children. What is the total mass of candies and cookies that each child gets? Give your answer as a decimal rounded to two decimal places.
Show Video Lesson
Discover more
Mathematics
Math
Video
Learning Software
Buy geometry tools kit
Worksheet Generators
Buy science project kits
Math Study Guides
Buy SAT practice tests
Homeschooling Resources
Word problem to show how to multiply a fraction and a whole number. Employs Singapore’s model method of problem solving.
Example:
Sam has 260 erasers. He had 3/4 as many sharpeners as erasers. He had 5 times as many markers as sharpeners. How many markers did Sam have?
Show Video Lesson
Example:
Nina is making 5 curtains and 3 cushion covers. She uses two and one-half m of fabric for each curtain and one and one-third m of fabric for each cushion cover. How much fabric does she need altogether? Each meter of fabric costs $38. How much did Nina spend on the fabric?
Show Video Lesson
Measurement word problem involving fractions.
Example:
Roy and Sandy have the same amount of paint at first. Sandy spills 2/5 of her paint. After Roy uses 1/3 of his paint to paint a wall, he still had 1/3 liter more paint than Sandy. How much paint did Roy and Sandy each had at first?
Show Video Lesson
Word problem to show the application of fractions and operations on fractions in practice. Uses the popular model method of Singapore to understand and solve the problem.
Example:
There 2442 Class A and Class B seats at a concert. 2/3 of the total number of Class A seats is equal to 1/4 of the total number of Class B seats. How many Class A seats are there?
Show Video Lesson
Try out our new and fun Fraction Concoction Game.
Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level.
×
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
Discover more
Math
Mathematics
Video
Classroom Tools
Calculus Courses
School supplies
Math Study Guides
educators
Video Lessons
Educational assessment
×
Discover more
Mathematics
Video
Math
Online learning platforms
Geometry tool sets
Learning resources subscription
College prep math
Online math courses
Programming language courses
Science experiment kits
Back to Top | Interactive Zone | Home
Copyright © 2005, 2025 - OnlineMathLearning.com.
Embedded content, if any, are copyrights of their respective owners.
×
Discover more
Math
Mathematics
Video
Classroom Tools
Buy TOEFL test prep
Homeschooling Resources
Buy chemistry lab equipment
Home
Math By Grades
Back
Pre-K
Kindergarten
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Grade 6
Grades 7 & 8
Grades 9 & 10
Grades 11 & 12
Basic Algebra
Intermediate Algebra
High School Geometry
Math By Topics
Back
Arithmetic
Algebra
Geometry
Trigonometry
Statistics
Probability
Word Problems
Pre-Calculus
Calculus
Set Theory
Matrices
Vectors
Math Curriculum
Back
NY Common Core
Illustrative Math
Common Core Standards
Singapore Math
Free Math Worksheets
Back
Worksheets by Topics
Worksheets by Grades
Printable & Online
Common Core Worksheets
Interactive Zone
Math Tests
Back
SAT Math
ACT Math
GMAT
GRE Math
GED Math
High School-Regents
California Standards
Math Fun & Games
Back
Math Trivia
Math Games
Fun Games
Mousehunt Guide
Math Video Lessons
Back
College Algebra
College Calculus
Linear Algebra
Engineering Math
Test/Exam Prep
Back
SAT Preparation
ACT Preparation
GMAT Preparation
GRE Preparation
GED Preparation
ESL, IELTS, TOEFL
GCSE/IGCSE/A Level
Back
KS3/CheckPoint 1 Science
KS3/CheckPoint 2 Science
KS3/CheckPoint 3 Science
GCSE/IGCSE Maths
GCSE/IGCSE Biology
CGSE/IGCSE Chemistry
CGSE/IGCSE Physics
A-Level Maths
Others
Back
Biology
Chemistry
Science Projects
High School Biology
High School Chemistry
High School Physics
Animal Facts
English Help
English For Kids
Programming
Calculators
Privacy Policy
Tutoring Services
What's New
Contact Us |
7712 | https://www.sausd.us/cms/lib/CA01000471/Centricity/Domain/2269/Worksheet%2051%20-%20Geometric%20Series.pdf | ©t j2Y0U1V4T YKVu5tBaY KSzoifMt7wearr5el FLMLPCy.v z MAIlDle 5rMiTgihYtusB Or4eusaevrhvYeCdy.n 1 BMga7dkep owPiBtyhK cIFnMfPiinqijtFeg SAUlAgpe2biraaa Q2n.p Worksheet by Kuta Software LLC ©m F2Y0Z1F4U 9KkuXtyaJ YSeoJf8tEwRatrYeK YLhLnCh.r M vA1lslZ hrjivgThktMsp KrfeAsieAr6vce1d7.F Math Analysis Honors - Worksheet 51 Geometric Series Given two terms in a geometric sequence find the explicit formula and the recursive formula.
1) a3 = −50 and a4 = −250 2) a2 = −9 and a5 = 1 3 3) a6 = −96 and a3 = 12 4) a5 = 1024 and a2 = −16 Evaluate each geometric series described.
5) Σ k = 1 8 3 ⋅ 5 k − 1 6) Σ k = 1 7 2 ⋅ 5 k − 1 Evaluate the related series of each sequence.
7) −4, −8, −16, −32, −64 8) 1, 5, 25, 125, 625 Evaluate each geometric series described.
9) a1 = −1, r = 4, n = 8 10) a1 = −2, r = −3, n = 9 Determine the number of terms n in each geometric series.
11) a1 = 4, r = −3, S n = −728 12) a1 = −1, r = 2, S n = −15 13) a1 = −3 5, r = 1 2, S n = −21 20 14) a1 = −4, r = 5, S n = −124 Determine if each geometric series converges or diverges.
15) 4 − 8 + 16 − 32... 16) 6 + 4 + 8 3 + 16 9 ... 17) 3 + 9 + 27 + 81... 18) −9.5 − 7.6 − 6.08 − 4.864... Evaluate each infinite geometric series described.
19) Σ n = 1 ∞ ( −1 3) n − 1 20) Σ n = 1 ∞ 3 ⋅ 3 n − 1 21) Σ k = 1 ∞ 0.2 ⋅ 0.4 k − 1 22) Σ i = 1 ∞ 16 ⋅ ( 1 4) i − 1 ©P X25071x4O HKJuNtXap SS7o9fJtnwCaSrEe7 VLCLCCi.y n WA9lzlM OrBiqg4hUtTsC zrjeSs0e8r8vkeGdF.F X 4MoaidfeQ 8wGizt0ht kIknRfji4nkiRtZeB qAJl9g8evbArUac N23.9 Worksheet by Kuta Software LLC Answers to 1) Explicit: a n = −2 ⋅ 5 n − 1 Recursive: a n = a n − 1 ⋅ 5 a1 = −2 2) Explicit: a n = 27 ⋅ ( −1 3) n − 1 Recursive: a n = a n − 1 ⋅ −1 3 a1 = 27 3) Explicit: a n = 3 ⋅ (−2) n − 1 Recursive: a n = a n − 1 ⋅ −2 a1 = 3 4) Explicit: a n = 4 ⋅ (−4) n − 1 Recursive: a n = a n − 1 ⋅ −4 a1 = 4 5) 292968 6) 39062 7) −124 8) 781 9) −21845 10) −9842 11) 6 12) 4 13) 3 14) 3 15) Diverges 16) Converges 17) Diverges 18) Converges 19) 3 4 20) No sum 21) 1 3 22) 64 3 |
7713 | https://www.quora.com/Why-do-astronauts-feel-weightlessness-inside-an-orbiting-spacecraft | Why do astronauts feel weightlessness inside an orbiting spacecraft? - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Physics
Human Space Exploration
Orbiting
Astronauts
About Gravity
Weightlessness
Spacecraft
Microgravity
Space and Travel
5
Why do astronauts feel weightlessness inside an orbiting spacecraft?
All related (82)
Sort
Recommended
Caleb Dyrud
Licensed Part 107 UAS Pilot · Author has 502 answers and 625.3K answer views
·7y
Originally Answered: Why does an astronaut in a space capsule orbiting the earth feel weightless? ·
Not because there is zero gravity like most people think. There is about 90% of Earth's gravity on the ISS. The reason they feel weightless is the same as any drop on a roller coaster. They're falling. That's how orbiting works. You are falling towards the mass, but you are moving sideways fast enough that by the time you would have hit the mass, you moved to the side of it.
Upvote ·
99 26
9 2
Sponsored by Google Ads
Google AI finds ready-to-buy shoppers this peak season.
Turn browsing into sales this peak season with an AI-powered campaign from Google Ads.
Sign Up
9 7
Related questions
More answers below
Why do astronauts feel weightlessness in a moving satellite?
Why are astronauts weightless in space?
Astronaut feels weightlessness inside the spacecraft. Why?
Why does a man inside an artificial satellite feel weightlessness?
How do astronauts describe the feeling of weightlessness?
C Stuart Hardwick
Award-Winning Scifi Author, Analog regular · Author has 13.6K answers and 204.4M answer views
·Updated 8y
Originally Answered: Why are astronauts weightless in space? ·
Technically they aren’t, it just feel that way.
Let’s say you go to sleep on night with all the window shades pulled down, and while you are sleeping, aliens show up and carefully cut your house loose, pile it on top of their flying saucer and carry it up to 100,000 feet.
In the morning, you wake up none the wise until you stumble out the door to get the paper and toddle off the edge of the saucer, right?
Now imagine if, instead of going for the paper, you go to brush your teeth. There you are in front of the mirror, paste on the brush (wondering what happened to the water) when the flying saucer
Continue Reading
Technically they aren’t, it just feel that way.
Let’s say you go to sleep on night with all the window shades pulled down, and while you are sleeping, aliens show up and carefully cut your house loose, pile it on top of their flying saucer and carry it up to 100,000 feet.
In the morning, you wake up none the wise until you stumble out the door to get the paper and toddle off the edge of the saucer, right?
Now imagine if, instead of going for the paper, you go to brush your teeth. There you are in front of the mirror, paste on the brush (wondering what happened to the water) when the flying saucer beams itself out from under you. So here you are, you, the house, the tooth brush, your drinking glass all plummeting to your doom together. Are you weightless? You are not, you just can’t feel your weight because everything around you is falling together. When the house hits the ground, suddenly you’ll feel your weight again. But in the meantime, you might say you are weightless even though you still have the same mass and Earth still has the same hold on you.
Astronauts in space are exactly like that. When you are in orbit, you are falling constantly, but you are moving away at the same rate you are falling, so you feel weightless. But you still have the same mass and the Earth still has almost the same pull, and you are enjoying only the illusion of weightlessness.
For a more detailed explanation of orbits, see Go Jump Off a Planet.
Upvote ·
99 16
Assistant
Bot
·
1y
Astronauts feel weightlessness inside an orbiting spacecraft because both the spacecraft and the astronauts are in a state of free fall towards Earth. This phenomenon can be explained by the following points:
Orbiting as Free Fall: An orbiting spacecraft is essentially in free fall towards Earth. While it is constantly falling due to gravity, it also has a horizontal velocity that keeps it moving forward. This combination of falling and forward motion creates a curved path around the Earth, resulting in an orbit.
Acceleration Due to Gravity: Even at the altitude of the International Space Statio
Continue Reading
Astronauts feel weightlessness inside an orbiting spacecraft because both the spacecraft and the astronauts are in a state of free fall towards Earth. This phenomenon can be explained by the following points:
Orbiting as Free Fall: An orbiting spacecraft is essentially in free fall towards Earth. While it is constantly falling due to gravity, it also has a horizontal velocity that keeps it moving forward. This combination of falling and forward motion creates a curved path around the Earth, resulting in an orbit.
Acceleration Due to Gravity: Even at the altitude of the International Space Station (ISS), gravity is still about 90% as strong as it is at Earth's surface. However, because the spacecraft is in free fall, both it and the astronauts inside are accelerating towards Earth at the same rate.
Relative Motion: Inside the spacecraft, there is no contact force acting on the astronauts. Since they are falling along with the spacecraft, they do not experience the normal force that we feel when standing on solid ground. This absence of a supporting force creates the sensation of weightlessness.
Microgravity Environment: The condition is often referred to as microgravity, which indicates that while gravity is present, the effects are significantly reduced due to the continuous free fall.
In summary, astronauts experience weightlessness because they are in a state of free fall along with their spacecraft, creating the sensation of floating despite the presence of gravity.
Upvote ·
9 1
Jared Maloney
BBA/Finance. HAPPY (Health and Psychophysiology) research. · Author has 2.5K answers and 9.8M answer views
·7y
Originally Answered: Why do we feel weightless in space? ·
Unlike what the other person commented, there is gravity in space. Think about it: if there was no gravity in the space around the Earth (e.g., in what we called the “space around the Earth where stuff can orbit,” then why would the Sun and Earth interact such that the Earth orbited the Sun? Short answer: if there was no gravity, the Earth wouldn’t orbit the Sun. It does, though. So, that can’t be
Continue Reading
Unlike what the other person commented, there is gravity in space. Think about it: if there was no gravity in the space around the Earth (e.g., in what we called the “space around the Earth where stuff can orbit,” then why would the Sun and Earth interact such that the Earth orbited the Sun? Short answer: if there was no gravity, the Earth wouldn’t orbit the Sun. It does, though. So, that can’t be the answer.
—
The answer is that we, and other objects, which are in free fall, do not experience gravity. The other person commented that we have mass, but no weight. That part of his/her answer was spot on.
—
What I’m about to tell you is a very famous experiment; it’s usually attributed to Einstein. And, you can test the thing if you want; this is not some funny “thought experiment” that can only be tested in your head; you can do this in real life, and I bet I won’t have much trouble convincing you that you’ve already done the experiment and just not noticed it!
So, we need a bathroom weigh scale (one of those cheap things you buy at Wal-Mart and get mad at if you step on it too much). We need an elevator; we need you, and at the end of the experiment, we’re going to need an evildoer.
So, let’s begin.
You are in an elevator on the ground floor standing on a bathroom scale. The scale reads 150 pounds, and you agree with this amount; this is how much you think you weigh, so you, and the scale, are in agreement.
The building you are in has 20 floors. Press the button for the 20th floor. The elevator starts going up…
What does the scale read? → It reads > 150 pounds. Now, you appear to weigh more!
In a frenzy you hit the emergency stop button and press the button to make the elevator go back down to the ground floor.
As the elevator goes down, what does the scale read? → It reads < 150 pounds. Now, you appear to weigh less than you think you weigh.
—
Once you get down the ground floor, you think about getting off; but, now the scale reads 150 pounds again. The elevator is standing still; you are back where you started; and, you weigh the same amount, again per the scale.
Why this is bothers you, so you decide to try this again, going for the gold. So, you press the button for the 20th floor again.
As you go up, you, indeed, notice that the scale registers you weigh more than 150 pounds.
When you get to the 20th floor, the elevator stops. Upon coming to rest, your bathroom scale settles back again on 150 pounds.
So, no matter if you are on the ground floor or the top floor — as long as the ele...
Upvote ·
9 3
9 2
Related questions
More answers below
Why does there appear to be weightlessness (zero gravity) in spacecrafts orbiting around Earth? In a geosynchronized spacecraft, wouldn't they have essentially the same weight as if they were on Earth's surface?
Can astronauts feel weightlessness whilst being in orbit of Earth (from within their spacecraft)? If so, how would they describe this feeling (e.g., “I felt lighter”)?
Why is an astronaut in an orbiting spacecraft not in zero gravity although they’re weightless?
How do astronauts navigate and control spacecraft while weightless?
What did astronauts on Apollo missions feel during weightlessness?
Curtis Quick
High School Physics Teacher/College Counselor · Author has 2.3K answers and 1.7M answer views
·7y
Astronauts orbiting the Earth in a spacecraft feel weightless because nothing is stopping their fall due to the force of gravity. On Earth we feel weight. Weight is the force of the ground holding us up against the pull of gravity. If it were not for this force, we would sink thorough the ground toward the center of the Earth.
For an astronaut, there is no force to hold them up against the pull of gravity. You might think the floor of the spacecraft would do this, but it too is falling due to the force of gravity. Since the astronaut and the spacecraft are both falling and nothing is holding th
Continue Reading
Astronauts orbiting the Earth in a spacecraft feel weightless because nothing is stopping their fall due to the force of gravity. On Earth we feel weight. Weight is the force of the ground holding us up against the pull of gravity. If it were not for this force, we would sink thorough the ground toward the center of the Earth.
For an astronaut, there is no force to hold them up against the pull of gravity. You might think the floor of the spacecraft would do this, but it too is falling due to the force of gravity. Since the astronaut and the spacecraft are both falling and nothing is holding them up against the pull of gravity, they are both in free-fall and feel weightless.
The only reason the astronaut and the spacecraft do not plunge down to crash on the surface of the Earth is because they are moving across the Earth's surface so fast that the net direction of motion is a curved path that bends down toward the Earth but never makes it there. In effect they are always falling, but also always missing the Earth, as the surface of the Earth is also bending away (due to the curvature of the Earth) at the same rate as the motion of the spacecraft.
Upvote ·
99 38
9 2
Sponsored by Hudson Financial Partners
Do I need someone to manage my wealth?
If your net worth is high wealth management might be for you, let's schedule a quick call and find out!
Contact Us
9 1
Robert Frost
Instructor and Flight Controller at NASA · Author has 9.5K answers and 230.3M answer views
·4y
Originally Answered: Why are astronauts weightless when orbiting the Earth? ·
The astronauts are in free fall, so there is no reaction force (such as we experience from the ground) to provide the sensation of weight.
Imagine an astronaut on Earth standing on a scale. The scale would indicate weight because gravity is pulling the astronaut towards the center of the Earth but the Earth isn’t willing to get out of the way. Weight is the perception of the Earth not getting out of the way. Now imagine that same astronaut in space, orbiting the Earth. He or she has the same scale, and stands upon it. Will it show weight? No. It won’t show weight because the astronaut and scale
Continue Reading
The astronauts are in free fall, so there is no reaction force (such as we experience from the ground) to provide the sensation of weight.
Imagine an astronaut on Earth standing on a scale. The scale would indicate weight because gravity is pulling the astronaut towards the center of the Earth but the Earth isn’t willing to get out of the way. Weight is the perception of the Earth not getting out of the way. Now imagine that same astronaut in space, orbiting the Earth. He or she has the same scale, and stands upon it. Will it show weight? No. It won’t show weight because the astronaut and scale are falling together and there is no contact with the Earth to allow the Earth to push back.The ISS and its crew are in free fall. They are falling towards the center of the Earth but since they are also moving quickly, sideways, they keep missing the Earth. Since there is no Earth surface to push back against the ISS or its crew, there is weightlessness. A sky-diver would experience the same thing if not for the resistance of the air below them.
Upvote ·
99 73
9 9
99 10
Robert Frost
Instructor and Flight Controller at NASA · Author has 9.5K answers and 230.3M answer views
·2y
Originally Answered: Why do astronauts feel weightless in their spaceship when they leave Earth’s gravitational pull? ·
No astronaut (or cosmonaut) has ever left “Earth’s gravitational pull”. The farthest an astronaut has been, from Earth, is 248,655 miles (400,171 km). That was achieved during Apollo 13, as the astronauts swung around the far side of the Moon. They were just 158 miles (254 km) from the surface of the Moon. The Moon orbits the Earth, because it is very much within Earth’s gravitational pull. The astronauts were very much within Earth’s gravitational pull. The astronauts on the International Space Station (ISS) experience just ten percent less gravity than you do, sitting on your chair. That 10%
Continue Reading
No astronaut (or cosmonaut) has ever left “Earth’s gravitational pull”. The farthest an astronaut has been, from Earth, is 248,655 miles (400,171 km). That was achieved during Apollo 13, as the astronauts swung around the far side of the Moon. They were just 158 miles (254 km) from the surface of the Moon. The Moon orbits the Earth, because it is very much within Earth’s gravitational pull. The astronauts were very much within Earth’s gravitational pull. The astronauts on the International Space Station (ISS) experience just ten percent less gravity than you do, sitting on your chair. That 10% reduction is caused by them being farther from the center of Earth than are you.
Astronauts feel weightless when the are in space because they are weightless at that time. Weight is simply the sensation of being pulled towards the center of the Earth but not being able to get there because the ground won’t get out of the way. Drop a ball and that ball is weightless from the moment you let it go until the moment it strikes the ground and the ground pushes back, not letting it fall further. It again gas weight. A skydiver is weightless. An astronaut in space is akin to a falling skydiver, simply higher up. The astronaut is falling downwards, like the skydiver, but with one very significant difference. The astronaut is also moving sideways at great speed. When we add that horizontal motion to the vertical motion of falling, the result is that the astronaut is traveling a curved path around the Earth - essentially falling but continually missing the Earth.
Upvote ·
99 41
9 2
9 1
Sponsored by Grammarly
92% of professionals who use Grammarly say it has saved them time
Work faster with AI, while ensuring your writing always makes the right impression.
Download
999 207
Bill Otto
Former Consulting Optical Physicist and Systems Engineer (1977–2019) · Author has 8.7K answers and 80.3M answer views
·6y
Originally Answered: Are astronauts weightless in space? ·
No. There is gravity in space, and astronauts are pulled by that gravity. What is missing is something fixed like the surface of the Earth. When say we feel heavy, we mean that we feel being pulled toward the Earth with something under our feet resisting our falling.
Gravity is not missing in space. But a fixed resistance is missing. That’s why astronauts have the illusion of being weightless. They don't feel the weight because nothing resists their falling.
Are astronauts falling toward the Earth in orbit? Yes they are. That's why they are going in an arc instead of a straight line.
If you throw
Continue Reading
No. There is gravity in space, and astronauts are pulled by that gravity. What is missing is something fixed like the surface of the Earth. When say we feel heavy, we mean that we feel being pulled toward the Earth with something under our feet resisting our falling.
Gravity is not missing in space. But a fixed resistance is missing. That’s why astronauts have the illusion of being weightless. They don't feel the weight because nothing resists their falling.
Are astronauts falling toward the Earth in orbit? Yes they are. That's why they are going in an arc instead of a straight line.
If you throw a baseball, once you let go of it is it falling? Yes, of course. It does not go in a straight line but in an arc (actually an ellipse that can be approximated with a parabola) due to gravity.
If you were falling, would you feel weightless? Yes, of course.
In orbit, astronauts are being pulled by Earth's gravity about 90% as hard as when they at sea level. But, like the baseball, they are going in an arc or an elliptical orbit. The spacecraft or space station is going in the same orbit, so it can not resist the gravity of the Earth and the astronaut does not feel his own weight even though the weight is still there.
It takes a lot of words to say all this. To make it easier to say, we have adopted a short hand way to say it. We call it “weightlessness” or “zero g”. We don't mean that literally. What we mean is that it feels the same as if they were weightless or in no gravity. (Incidentally, this is the type situation that led Einstein to pursue general relativity.)
There is an airplane used for astronaut training called the “Vomit Comet” and it was used to film sequences in Apollo 13. When it goes into a dive so that its vertical rate accelerates to match gravity, the people inside experience “zero g”. As soon as the plane levels off, it is clear that gravity is still alive and well in the aircraft. Nevertheless, the passengers may talk about being weightless or being in zero g. Again, they don’t mean it literally.
Upvote ·
99 18
9 2
9 1
Robert Frost
Instructor and Flight Controller at NASA · Author has 9.5K answers and 230.3M answer views
·6y
Originally Answered: Are astronauts in the International Space Station weightless? ·
Yes. The astronauts are in free fall, so there is no reaction force (such as we experience from the ground) to provide the sensation of weight.
Imagine an astronaut on Earth standing on a scale. The scale would indicate weight because gravity is pulling the astronaut towards the center of the Earth but the Earth isn’t willing to get out of the way. Weight is the perception of the Earth not getting out of the way. Now imagine that same astronaut in space, orbiting the Earth. He or she has the same scale, and stands upon it. Will it show weight? No. It won’t show weight because the astronaut and
Continue Reading
Yes. The astronauts are in free fall, so there is no reaction force (such as we experience from the ground) to provide the sensation of weight.
Imagine an astronaut on Earth standing on a scale. The scale would indicate weight because gravity is pulling the astronaut towards the center of the Earth but the Earth isn’t willing to get out of the way. Weight is the perception of the Earth not getting out of the way. Now imagine that same astronaut in space, orbiting the Earth. He or she has the same scale, and stands upon it. Will it show weight? No. It won’t show weight because the astronaut and scale are falling together and there is no contact with the Earth to allow the Earth to push back.The ISS and its crew are in free fall. They are falling towards the center of the Earth but since they are also moving quickly, sideways, they keep missing the Earth. Since there is no Earth surface to push back against the ISS or its crew, there is weightlessness. A sky-diver would experience the same thing if not for the resistance of the air below them.
Upvote ·
999 118
9 2
9 4
Sponsored by All Out
Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out.
Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association.
Shop Now
999 620
Richard Kent
7y
Originally Answered: Why do we feel weightlessness during orbiting the earth? ·
I assume you mean weightlessness in environments as the ISS (International Space Station). They feel the ~90% of the gravity we do, yet are weightless. This is because the ISS orbits the earth at 17,000 MPH. Any less, and their orbit would decay and they would fall back to earth. Gravity always pulls them in, however if they travel at a certain speed (~17000 mph) Earth curves faster away from them than they fall.
Thus, because ISS astro/cosmonauts are always falling, Earth’s gravity is canceled out by the orbital speed and they are weightless.
Upvote ·
9 5
Dean Carpenter
Hobbyist Physicist · Author has 7.2K answers and 6.3M answer views
·7y
Originally Answered: Why does an astronaut in a space capsule orbiting the earth feel weightless? ·
When you stand on the ground, the ground is ststionary, so when gravity tries to accelerate you, it pushes you against the ground, and you feel pressure.
If you are in an elevator and it is dropped, you start falling, and the car you are in falls at the same rate. It is accelerating at the exact same rate you are, so you move in unison. As a result you have no pressure holding you to the ground. You are in freefall, which is what people think of as weightlessness .
When you are in orbit, you are just in freefall. The only difference between this and the elevator is you also are travelling sidewa
Continue Reading
When you stand on the ground, the ground is ststionary, so when gravity tries to accelerate you, it pushes you against the ground, and you feel pressure.
If you are in an elevator and it is dropped, you start falling, and the car you are in falls at the same rate. It is accelerating at the exact same rate you are, so you move in unison. As a result you have no pressure holding you to the ground. You are in freefall, which is what people think of as weightlessness .
When you are in orbit, you are just in freefall. The only difference between this and the elevator is you also are travelling sideways fast enough that you won't hit the earth, and instead gravity pulls you around in a circle. But you and your capsule are moving at the same speed and are pulled by gravity the same amount, so you can move within it as if you were weightless.
Upvote ·
9 2
Andrew Forrest
Chief Engineer at Solstad Offshore (2005–present) · Author has 6.9K answers and 26.8M answer views
·6y
Originally Answered: Astronauts feel they are weightless inside spacecraft. Why? ·
Astronauts feel they are weightless inside spacecraft. Why?
If a spacecraft is motionless with respect to the nearest massive body, they will experience the effect of that body’s gravity and begin moving directly towards it, falling down in effect. Once they are moving freely towards the massive body (planet, star, etc), the craft will not experience any force, as the net effect is zero once the acceleration due to gravity stops and the free fall speed is constant. This is exactly what would happen inside an elevator that had it’s cables cut and safety devices removed. The elevator cage would f
Continue Reading
Astronauts feel they are weightless inside spacecraft. Why?
If a spacecraft is motionless with respect to the nearest massive body, they will experience the effect of that body’s gravity and begin moving directly towards it, falling down in effect. Once they are moving freely towards the massive body (planet, star, etc), the craft will not experience any force, as the net effect is zero once the acceleration due to gravity stops and the free fall speed is constant. This is exactly what would happen inside an elevator that had it’s cables cut and safety devices removed. The elevator cage would fall unencumbered towards the Earth, and would experience no internal forces until the abrupt stop at the bottom. If you were a passenger in the lift, you would also feel no forces upon you, it would be as if gravity was turned off, even though you are most definitely under it’s influence. This would be made apparent at the bottom of the lift shaft.
If we go back to the hypothetical spacecraft, falling towards a planet or star, the crew inside would similarly feel no effect of the body’s gravity, but they could see it approaching. If the craft has engines, then it can either point them away from the gravity well and try to overcome the energy of falling, which is very substantial, or they could conversely be used to push the craft sideways which uses a lot less energy. If there is enough thrust to reach orbital velocity (7.8 km/s for Earth), then the craft will end up travelling so fast sideways that it’s centripedal force exactly matches the gravitational force, and the vessel is now in a stable orbit. The craft and it’s crew experience no forces upon them as they are effectively in “free fall”, and all external forces are balanced out.
If this balance is changed, then both the craft and the crew will feel the acceleration forces act upon them, and a sense of gravity will return for as long as the acceleration is applied. The more force that is used to accelerate the craft, the more apparent gravity is felt by the occupants, which is felt in the opposite direction to the thrust.
Upvote ·
Loretta B. DeLoggio
insatiable curiosity and a pretty good memory · Author has 18.5K answers and 38.6M answer views
·2y
Originally Answered: Why do astronauts feel weightless in their spaceship when they leave Earth’s gravitational pull? ·
Most of what we feel when we’re moving is an illusion.
Imagine you’re on a plane, and get up to move. The minute you stand up, you’re not thrown backwards at 500 miles an hour, smashing into the back wall. Why not? Because you’re moving at the same speed as the plane surrounding you.
But you don’t feel yourself moving forward at an enormous rate? You did, when you started moving. You were pushed back into the seat as the vehicle moved forward; acceleration creates the “feeling” of weight. Once you get up to a constant speed — stop accelerating — you don’t feel the weight of forward motion any mo
Continue Reading
Most of what we feel when we’re moving is an illusion.
Imagine you’re on a plane, and get up to move. The minute you stand up, you’re not thrown backwards at 500 miles an hour, smashing into the back wall. Why not? Because you’re moving at the same speed as the plane surrounding you.
But you don’t feel yourself moving forward at an enormous rate? You did, when you started moving. You were pushed back into the seat as the vehicle moved forward; acceleration creates the “feeling” of weight. Once you get up to a constant speed — stop accelerating — you don’t feel the weight of forward motion any more. And you won’t feel any change of weight until the force changes in the opposite direction. When the plane decelerates, you feel yourself being thrown toward the front window, which is where you’d actually end up if you weren’t wearing your safety belt.
When you’re talking about space flight, you have to think about motion toward or away from the earth, as well as forward or backward relative to the earth.
If a rocket goes straight up with enough force, it will reach “escape velocity” — the speed where the acceleration upward is stronger than the pull of gravity downward. (Where it goes after that is where the engineers programmed it to go; I don’t think there’s an automatic answer. It’s like a baseball player’s hitting a ball with a bat: knowing that it’s hit hard enough to move doesn’t tell you where it’s going.)
Now, just as an airplane moves forward at a rate to get where it’s going, but you don’t feel like Evel Knievel
jumping the Grand Canyon, there’s a vertical equilibrium. If you go too slowly (more slowly than “escape velocity” — less than 25,000 miles an hour,
you will fall back down to earth. So whether you’re tossing a ball up in the air or launching a rocket, if it moves at less than 25,000 mph, it will fall back down to earth. It won’t fall down to the same place it took off from; the earth moved a foot or a thousand miles during that vertical takeoff and landing; but you’ll come back down to earth. If you move faster than escape velocity, you’ll leave the earth — forever, unless you know how to plot a course that makes the ship turn back.
But what if you’re moving upward at that speed that’s too fast to stay on the earth, but too slowly to leave it — you’re just below escape velocity? That’s when you’re at “orbital velocity.” Your speed up and down and your speed forward and backward are both at equilibrium. What you’re calling “weightless” is “motionless.” You get yourself into position, by pushing gently up, down, left, right, until you get to where you want to be — and then you stay there! It’s really no different from walking down the plane to row 14, sliding over to seat A, and settling yourself. You feel motionless, even though you’re moving the whole time.
The coolest demonstration of this was provided to me by Robert Frost several years ago. Start this video:
And at 00:20 (20 seconds) you can see the magic of equilibrium in all directions! Gravity is pulling down at the same force that the rocket is travelling up, and the rocket is moving “forward” at a constant speed — neither accelerating nor decelarating — so the yo yo doesn’t move forward or backward. Yo Yo Man, astronaut Don Pettit, is moving; but he’s been trained how to move so minutely that he doesn’t crash into things. And he’s taught himself to use his yo yo with the same kind of minute movements. He probably also has a short string, so the yo yo isn’t too likely to hit something important.
You probably think of that as weightlessness because they call it that in movies. But what it really should be called is relative motionlessness. You’re not going anywhere within a vehicle that’s going somewhere very quickly.
I hope I got the right, because Robert Frost taught me all of it, in bits and pieces. over the last seven years. Robert, if I messed up, I apologize.
Footnotes
Evel Knievel - Wikipedia
Escape velocity - Wikipedia
Upvote ·
9 1
Related questions
Why do astronauts feel weightlessness in a moving satellite?
Why are astronauts weightless in space?
Astronaut feels weightlessness inside the spacecraft. Why?
Why does a man inside an artificial satellite feel weightlessness?
How do astronauts describe the feeling of weightlessness?
Why does there appear to be weightlessness (zero gravity) in spacecrafts orbiting around Earth? In a geosynchronized spacecraft, wouldn't they have essentially the same weight as if they were on Earth's surface?
Can astronauts feel weightlessness whilst being in orbit of Earth (from within their spacecraft)? If so, how would they describe this feeling (e.g., “I felt lighter”)?
Why is an astronaut in an orbiting spacecraft not in zero gravity although they’re weightless?
How do astronauts navigate and control spacecraft while weightless?
What did astronauts on Apollo missions feel during weightlessness?
Why does a person feel weightless in a spacecraft?
Do astronauts feel weightless on the ISS?
Why do astronauts hold onto objects on board the spacecraft if they are weightless in space?
Do astronauts experience weightlessness during space transit? If so, what are they falling towards?
What happens to astronauts when their spacecraft leaves orbit? Do they experience zero gravity?
Related questions
Why do astronauts feel weightlessness in a moving satellite?
Why are astronauts weightless in space?
Astronaut feels weightlessness inside the spacecraft. Why?
Why does a man inside an artificial satellite feel weightlessness?
How do astronauts describe the feeling of weightlessness?
Why does there appear to be weightlessness (zero gravity) in spacecrafts orbiting around Earth? In a geosynchronized spacecraft, wouldn't they have essentially the same weight as if they were on Earth's surface?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025 |
7714 | https://www.pandafastener.com/es/the-way-to-keep-nuts-from-loosening.html | La forma de evitar que las nueces se aflojaran
Hogar
Productos
Tornillos Allen
Perno
Cojones
Tornillos
Ancla
Varillas roscadas y pernos
Arandelas
Sobre nosotros
Nuestra fábrica
Nuestro cliente
Control de calidad
Servicio
Muestra gratuita
Inspecciones de terceros
Personalización
Soporte post-ventas
Desarrollo de la investigación
Noticias
Noticias de la compañía
Noticias de la industria
Contáctenos
Please Choose Your Language
English
العربية
Français
Русский
Español
Português
Noticias
estamos aquí para mantenerlo actualizado con los últimos
acontecimientos en la industria del sujetador.
Hogar » Noticias » Noticias de la industria » La forma de evitar que las nueces se aflojaran
Productos principales
Pernos hexadecimales de óxido negro
0
0
Azul blanco galvanizado chapado en zinc Din7991 CSK CABEZA FLA HEX PERNOS
0
0
Tornillo de tapa de la cabeza del zinc galvanizado con galvanizado de acero
Marca de marca: PandaFastener estándar: DIN912 Finis: Material liso:...
0
0
Enviarnos un mensaje
Enviarnos un mensaje
Entregar
La forma de evitar que las nueces se aflojaran
Vistas: 100 Autor: L Tiempo de publicación: 2024-01-12 Origen: Sujetador de panda
Los pernos son una herramienta de accesorio común. Sin embargo, su uso a largo plazo puede causar muchos problemas. Estos incluyen relajación de conexión, fuerza de sujeción inadecuada y óxido de pernos. La holgura de conexión del perno tendrá un impacto en la eficiencia y la calidad del procesamiento de piezas. ¿Cómo se puede hacer el perno para apretar?
Se usan comúnmente tres técnicas antigüense: fricción, mecánica y permanente.
1.Nueces
El principio de la tuerca de mermelada de la encimera es: cuando las tuercas dobles se atascan, se crean dos superficies de fricción. Uno es entre la nuez, el sujetador y la tuerca. Durante la instalación, la precarga se aplica a la primera superficie de fricción al 80% de la cantidad de precarga que se aplica a la segunda superficie de fricción. Durante los choques y vibraciones, la fricción en la primera superficie de fricción desaparece. Sin embargo, en el mismo momento, la primera tuerca se comprime, lo que hace que la fricción aumente en la segunda superficie de fricción. Para aflojar la tuerca, debe superar las fuerzas de fricción de primera y segunda. La primera fuerza de fricción está disminuyendo mientras que la segunda fricción aumenta. Se mejorará el anti-loosening.
TANG PRINGA ALTIENTO: Los sujetadores de rosca Tang también tienen tuercas dobles que están antigacios, pero las tuercas giran en direcciones opuestas. La fricción en la primera superficie del sujetador disminuye cuando se aplican choques y vibraciones.
La primera tuerca en el diagrama (a la derecha) tiene una tendencia a aflojar y retirarse, es decir, la tuerca gira a la izquierda. La segunda tuerca gira en una dirección opuesta desde la primera (rotación izquierda en la imagen), de modo que la fuerza solía aflojar la primera tuerca se convierte directamente en la fuerza necesaria para apretar la segunda. La nuez no se aflojará de nuevo.
2. Técnica de hilo en forma de cuña de 30 grados
La parte inferior del hilo negativo tiene un bisel en forma de cuña de 30 grados. Cuando el perno y la tuerca se apretan, la punta se presiona firmemente contra la muesca en forma de cuña de la rosca negativa. Esto crea una fuerte fuerza de bloqueo.
La fuerza normal aplicada al contacto entre hilos debido al ángulo diferente de los dientes es de 60 grados, no 30 grados, como lo sería para hilos ordinarios. La presión de hilo de cuña normal de 30 grados es obviamente mucho mayor que la fuerza de fijación, por lo que la fricción resultante de esto debe aumentar.
Esquema de hilo dtflock
La siguiente figura muestra que la fuerza indicada por dos flechas en forma de PA es la misma para ambos hilos. El hilo tradicional de 60 grados tiene una presión promedio de 1.15pa, mientras que el hilo de 30 grados tiene la cuña en la parte inferior biselada en un ángulo de 30 grados. Se han cambiado tanto el ángulo como el tamaño de la presión normal del hilo.
De esta manera, entre el hilo tradicional de 60 grados y 30 grados, la relación de presión normal es 12: 7. La fricción contra la aflojamiento aumenta en consecuencia. El hilo de 30 grados puede eliminar las fuerzas de hilo que no son uniformes.
3.Nueces de bloqueo automático
Mencionamos que los hilos de cuña de 30 grados que discutimos anteriormente deben incluirse en esta categoría.
Los productos de tuercas de autobout se dividen en tuercas de alta resistencia para máquinas de construcción de carreteras, equipos de minería, maquinaria vibratoria y equipos y tuercas de nylon para aeroespacial, aviación y tanques.
Para evitar el Tornillo de aflojamiento, las cuñas se insertan entre la tuerca.
4. Adhesivo de bloqueo de listas
El adhesivo de bloqueo de hilo es una mezcla de acrilato (iniciador) (iniciador), copromotor (inhibidor de la polimerización), tinte, relleno y estabilizador. en una determinada proporción.
Para apretar la tuerca al par especificado, pase el perno a través de una captura de captura, luego aplique adhesivo de bloqueo de rosca a las roscas en la parte atractiva.
Si la profundidad del orificio del tornillo es mayor que la longitud del perno, es necesario que el adhesivo de bloqueo se aplique a las roscas del perno. El perno debe ensamblarse y apretarse al par especificado.
Para la condición del agujero ciego: deje caer el adhesivo de bloqueo en la parte inferior del agujero ciego, luego aplique el adhesivo de bloqueo a las roscas del perno, ensamblándolo y apriételo al par especificado; Si la abertura del agujero ciego está hacia abajo, simplemente aplique el adhesivo de bloqueo a las roscas del perno, y no hay necesidad de aplicar el adhesivo dentro del agujero ciego.
Después de ensamblar todas las demás partes, el pegamento de bloqueo se aplica a la parte del perno que se involucra con la tuerca. La tuerca se ensamblará y apretará a la fuerza especificada.
Tornillos ajustables): después del ensamblaje y apriete al par especificado, deje caer el adhesivo de bloqueo en el compromiso roscado para que penetre por sí mismo. Tornillos ajustables: deje caer el adhesivo de bloqueo en el compromiso roscado después del ensamblaje y apriete el par especificado.
5.Arandelas de bloqueo anti-bucle de cuña de cuña apilada
Las arandelas de bloqueo de cuña tienen serraciones radiales que están cubiertas por la superficie con las que están en contacto directo. El desplazamiento de la lavadora solo puede ser causado por cargas dinámicas cuando el sistema de bloqueo está expuesto a ellos.
La arandela de bloqueo de cuña se puede extender más en la dirección del grosor que el desplazamiento longitudinal del perno en la dirección del rosco.
El método de bloqueo de cuña aprieta los pernos utilizando la fuerza de sujeción en lugar de la fricción.
Heico-Lock tiene una historia de 120 años. Los productos incluyen cuña arandelas de bloqueo (incluido el bloqueo de anillo), cuña Tuercas de bloqueo (incluyendo 316 acero inoxidable y acero de carbono recubierto con Dacromet) y 254SMO (acero al carbono).
6. Tuerca ranurada, pasador de camarilla
Inserte la pinza Cotter en el orificio del perno y la ranura en la tuerca después de que la tuerca se haya apretado. Luego, gire el extremo de la pinza Cotter para evitar la rotación relativa entre el perno y la tuerca.
La cotterpin se monta como se muestra en la imagen a continuación.
Las tuercas ranuradas se usan con los pernos personados por el tornillo y los pasadores de chineo para detener el perno que gira en relación con la tuerca.
7. Anti-loosening de alambre en tándem
El anti-loosening de alambre de acero en tándem es colocar el alambre de acero en el orificio de la cabeza del perno, los pernos en serie, juegan el papel de la restricción mutua. Esta relajación es muy confiable, pero el desmontaje es más problemático.
Este método de antigüense a menudo se usa en aviones y cohetes.
Los cables monocatenarios generalmente se usan comúnmente para distribuir pequeños grupos de tornillos a intervalos cerrados o en lugares muy inaccesibles, como se muestra en la figura.
8. Arandela de parada
Una vez que la tuerca se aprieta, se debe doblar una lavadora de parada única o doble para descansar contra él y su conector, respectivamente, para asegurar y bloquear la tuerca. Si dos pernos requieren doble bloqueo, también se pueden usar lavadoras de doble freno para que ambas tuercas frenen entre sí.
9.Arandelas de primavera
Las lavadoras de primavera emplean el principio anti-loosening produciendo fuerza elástica continua después del aplanamiento; Esto habilita las conexiones roscadas entre Tuercas y pernos para mantener fuerzas de fricción que crean resistencia, evitando así el aflojamiento.
Al mismo tiempo, las esquinas afiladas de las aberturas de la lavadora de resorte están incrustadas en ambas superficies de perno y parte conectada respectivamente, evitando así la rotación entre perno y parte conectada.
10. Tecnología de fijación por fundición en caliente
La tecnología de fijación de fusión en caliente, sin agujeros previos a la apertura, permite que los perfiles cerrados se conecten directamente con el adhesivo de fusión en caliente, con un gran éxito en la industria automotriz.
Esta tecnología de fijación de fusión en caliente es una especie de proceso de moldeo en frío a través del centro del equipo para apretar el eje de la rotación de alta velocidad del motor para conectarse al calor fricional generado por la deformación plástica del material de la lámina, auto-tocado y atornillado.
Los pasos y procesos de proceso de conexión de fijación de fusión en caliente incluyen seis etapas: rotación (calentamiento) → Penetración → Hole de paso → roscas de ritmo → roscas de atornillado → fijación.
Tan pronto como una fuerza alcanza el noventa por ciento, su color cambia de amarillo a verde; Luego, una vez que alcanza el cien por ciento, se vuelve negro.
12. Precintoring
Las conexiones aturdidas de alta resistencia generalmente no requieren medidas adicionales antigacio, porque los pernos de alta resistencia generalmente requieren la aplicación de una gran precarga, creando una presión fuerte entre la tuerca y la conexión, con un par de fricción para detener la rotación de la tuerca, para mantener Nuez de soltarse.
Lo anterior es PandaFastener propuso 12 medidas específicas para evitar el ¡Bolt de Loosening, además, también proporcionamos una variedad de servicios y productos de sujetadores de alta calidad, si necesita contactarnos!
nuecesarandelasPerno
Tabla de contenido
Noticias relacionadas
Las diferencias entre los estándares ASME y ASTM para los sujetadores
Los 10 mejores proveedores de sujetadores en el mercado argentino
Los 10 mejores proveedores de sujetadores en el mercado brasileño
Enlaces rápidos
Sobre nosotros
Servicio y soporte
Noticias
Contáctenos
Categoría de productos
Tornillos Allen
Perno
Cojones
Tornillos
Ancla
Varillas roscadas y pernos
Arandelas
Contáctenos
Al sur de la aldea de Ximingyang, la ciudad de Linmingguan, el distrito de Yongnian, la ciudad de Handan, la provincia de Hebei, China
Teléfono: +86-181-3221-6205
WhatsApp: +86-181-3221-6205
Correo electrónico:admin@pandafastener.com
Enviarnos un mensaje
Enviarnos un mensaje
Entregar
Copyright 2025 Panda sujetador todos los derechos reservados.
Mapa del sitio | Política de privacidad | Apoyado por Leadong.com
top |
7715 | https://www.mee.gov.cn/gkml/hbb/bgth/200910/W020080416421857880537.pdf | GB 12997-200□ 代替GB 12997-1991 水质 采样方案设计技术规定 Water quality －Technical regulation on the design of sampling programmes （征求意见稿） 200□-□□-□□发布 200□-□□-□□实施 中华人民共和国国家标准 国家质量监督检验检疫总局 环 境 保 护 部 发布 I 目 次 前言...............................................................II 1 适用范围..........................................................1 第一篇 采样目标的确定 2 引言..............................................................1 3 要求..............................................................2 4 与可变性有关的特殊考虑............................................3 第二篇 采样点的选择 5 引言..............................................................3 6 一般的安全预防措施................................................3 7 采样的专门注意事项................................................4 8 天然水的各种采样情况..............................................6 9 工业用水采样......................................................9 10 工业废水........................................................10 11 污水和污水厂出水................................................10 12 暴雨污水和地面径流..............................................11 第三篇 采样频率和采样时间 13 引言............................................................11 14 采样方案的类型..................................................12 15 统计研究........................................................12 16 异常的变化......................................................14 17 采样的持续时间和混合样品........................................14 第四篇 水流的测量及其在水质方面的应用 18 引言............................................................15 19 水质控制中流量测量的必要性......................................16 20 水流测量方法....................................................17 II 前 言 为贯彻 《中华人民共和国环境保护法》 和 《中华人民共和国水污染防治法》 ， 保护环境，保障人体健康，加强对水质采样的规范和指导，制定本标准。 本标准规定了各种水体包括废水、 底部沉积物和污泥的质量控制、 质量表征、 采样技术要求、污染物鉴别采样方案的原则。 本标准的技术内容采用ISO 5667-1-2006《水质 采样 第1 部分：采样方 案设计指导》中相关内容。 自本标准实施之日起，《水质 采样方案设计技术规定》（GB 12997-91） 废止。 本标准为指导性标准。 本标准由环境保护部科技标准司组织制定。 本标准起草单位：中国环境监测总站、辽宁省环境监测中心站。 本标准自200□年□□月□□日起实施。 本标准由环境保护部解释。 1 水质 采样方案设计技术规定 1 适用范围 本标准规定了各种水体包括废水、底部沉积物和污泥的质量控制、质量表征、采样技术 要求、污染物鉴别采样方案的原则。 本标准适用于各种水体包括废水、底部沉积物和污泥的采样方案设计。 第一篇 采样目标的确定 2 引言 本篇强调在进行水、 底部沉积物和污泥采样方案设计时必须考虑的比较重要的因素。 采 样和检验的主要目的是测定其有关的物理、化学、生物和放射性参数。 在表征水体、底部沉积物和污泥的质量时，不可能检验其整体，必须采集样品，并且要 采取一切措施， 预防样品在采集和分析的间隔内发生变化。 当采集含悬浮固体或者含难混溶 的有机液体的多相样品时，还会遇到特殊的问题。 确定采样地点、采样时机、采样频率、采样持续时间、样品处理和分析的要求时主要取 决于采样目标。所以在设计采样方案之前，要首先确定采样目标。在设计采样方案时还要对 详尽程度、适宜的精密度、以及表达形式和提供结果的方式给予考虑，比如浓度或负荷、最 大值和最小值、算术平均值、中位数等。确保样品采样设计方案可以评价被统计采样错误和 化学分析错误影响的数据错误。此外，还要编制有定义参数的目录和确定相应的分析方法。 它们将对采样和输送样品时的保护进行指导。在保证获得所需资料的前提下，要注重效率。 采样目标可区分为以下三种(详见第14 章)： a.质量控制检测 需要进行短期过程的校正时由管理部门决定。 b.质量特性检测 用于表明质量， 多数情况作为研究项目的组成部分， 以达到长期质量控制目的或指出发 展趋势。 c.污染源的鉴别 2 采样方案的目标可由质量特性检测变为质量控制检测， 比如， 当硝酸盐浓度接近限值时 需要提高采样频率，这样就可由较长时期的质量表征变为短期的质量控制方案。 3 要求 要求可分为以下两类： 3.1 一般要求 在选定的测点 (例如水体的表面或里层)确定特定参数的浓度水平的数量级(或负荷)或 直观表达底部沉积物的特性。 3.2 特定要求 详细地确定整个或部分水体中所研究的物理或化学参数的浓度水平或者负荷分布及有 意义的生物种类。通常把这些参数变化的研究与时间、流量、工厂工艺、气候条件因素等结 合考虑。 还可以细分为以下更具体的采样情况： a.测定水对某种用途的使用性。 如检验井水能否用作冷却、 锅炉给水、 工艺用水或者饮用水。 b.研究排放污染物 （包括偶然泄露） 对所受纳水体的影响。 排放污染物除了增加污染负荷外， 还导致其他反应，如化学沉淀或产生气体等。 c.评价水、污水、工业废水处理厂的性能和管理。比如，评价进入废水处理厂负荷的波动和 长期的变化；测定处理过程各阶段的处理效率，提供净化后水的质量数据，控制使用净水剂 的浓度；控制那些可能损害企业构筑物或设备的物质等。 d.研究河口淡水径流和海水对河口环境的影响， 提供混合类型及因潮汐和淡水流动的变化引 起咸淡分层情况的资料。 e.测定工业生产过程中产品的损失。 这些资料对评定全厂物料衡算、 测量废水排放量都是需 要的。 f.测定锅炉水、 蒸汽冷凝水和其他回水的质量。 对这些水是否能用于预定目的可行性做出评 价。 g.调节工业冷却水系统的运行操作，使水得到最佳利用，与此同时，尽量减少锅垢，把腐蚀 降低到最低限度。 h.研究大气污染物对雨水质量的影响。 它为研究空气质量提供有价值的资料。 它还可以指出 有些问题是否会发生。如暴露的电触点是否会出问题。 i.评价地面物质输入对水质的影响。这些影响或来自天然的存在中的物质，或来自化肥、农 药，或农业化学品的污染，或两者兼而有之。 3 j.评价底部沉积物的积集和释放对水体中或底部沉积物中水生生物的影响。 k.研究导流、河流调节，不同河流间河水的相互转移对天然水道的影响。比如，在河水调节 期间各种不同质量水体的比例在不断发生变化，导致河水质量波动。 l.评价水质在配水系统中发生的变化。引起这些变化的因素很多，比如：污染、从新水源引 水、生物的生长、水垢的沉积或金属的溶解。 在某些情况下环境状况是相当稳定的。 可从简单的采样方案中获得需要的数据， 然而大 多数监测点的质量特性在不断地发生变化。因此，要想得到理想的评价需要进行连续采样。 虽然，连续采样不仅代价太高，而且在许多情况下也不现实。一些特殊情况的采样方案详见 第 4 章。 4 与可变性有关的特殊考虑 4.1 当待测水质项目的浓度出现大幅度、急剧的变化时所要求的采样方案是复杂的。这些 变化可由温度的极端变化、流态、污水厂运行状况所引起。除非有特殊需要，应避免在系统 的边界或靠近边界部位采样。 4.2 评价一个大的集水面积是很复杂的，即使浓度变化缓慢，而且变化不大显著时，也是 这样。 4.3 消除或减少由采样过程本身造成待测水质项目浓度的变化，要求在采样至分析期间把 变化降低到最低限度。 4.4 如果待测水质项目在采样和检验期间稳定，能很好地反映整个周期内平均组成的最好 指标。但是混合样对确定瞬时峰值的情况价值不大。 第二篇 采样点的选择 5 引言 本篇论述采样实践中所遇到的各种情况和它们对选择采样点所产生的影响， 鉴于安全防 护的重要性和普遍性，在各种情况下都要重视安全。本篇予以专门论述。 6 一般的安全预防措施 6.1 在水体和底部沉积物中进行采样时，会遇到各种危害人体安全和健康的情况。为了保 护人体不受伤害，要采取措施避免吸入有毒气体，防止通过口腔和皮肤吸收有毒物质。 负责设计采样方案和负责实施采样操作的人员， 必须考虑相应的安全要求。 在采样过程 中采样人员要了解应采取的必要的防护措施。 4 6.2 为了保证工作人员、仪器的安全，必须考虑气象条件。在大面积水体上采样时，要使 用救生圈和救生绳。在冰层覆盖的水体采样之前，要仔细检查薄冰层的位置和范围。当采用 水下整装呼吸装置或其他潜水器具时，则应经常检查和维护这些器具的可靠性。 6.3 采样船要坚固，在各种水域中采样时都要防止商船和捕捞船只靠近。例如：要正确使 用信号旗，以表明正在进行的工作性质。 6.4 尽可能避免从不安全的河岸等危险地点采样， 如果不能避免， 要采取相应的安全措施， 并注意不要单人行动。 如果河岸条件不是采样研究特殊要求的， 应尽量采取在桥上采样来代 替河岸边采样。 6.5 要选择任何气候条件下都能方便地进行频繁采样的地点，在某些情况下必须考虑到可 能的自然危害，如有毒的枝叶、兽类和爬行动物。危险物质应贴上标签。 6.6 安装在河岸上的仪器和其他设备，为了防止洪水淹没或破坏行为，需要采取适当的防 护措施。 6.7 为了防止一些偶然情况的出现，如：一些工业废水可能具有腐蚀性，或者含有有毒或 易燃物质，污水中也可能含有危害的气体、微生物、病害或动物，如变形虫或蠕虫。在采样 期间，必须采取一些特殊的防护措施。 6.8 当采样人员进入有毒气体环境中时，要使用气体防毒面具、呼吸、苏醒器具和其他安 全设备。此外，在进入封闭空间之前，要测量氧气的浓度和可能存在的有毒蒸汽和毒气。 6.9 在采集蒸汽和热排放物时，需特别谨慎。应使用认可的技术。 6.10 处理放射性样品要特别小心，必须采用专门的技术。 6.11 在水中或者靠近水使用电动采样设备时有触电的危险。因此，在安排工业步骤、采样 点的选定、设备的维护保养时，防止这种危险的发生。 7 采样的专门注意事项 7.1 采样方案的设计 根据不同的采样目的， 采样网络可以是单点也可扩展到整个流域。 一个干流网络应包括 潮区界以内的各采样点，较大支流的汇入口和主要污水或者工业废水的排放口。 在设计高质量的采样网络时，通常要做好主要水文站的流量测量（见第四篇）。 7.2 采样点的定位 只有固定采样点位才能对不同时间所采集的样品进行对比。 大多数河流的采样点可参照河岸地貌特点标定。 5 确定非封闭海湾以及海岸边的采样点时寻找容易识别的固定目标作参照。在船上采样， 使用仪器为采样点定位。可以使用地图或其他一些标准图表定位。 7.3 水流的特征 从充分混合的湍流中取样最为理想。 只要有可能就要把层流诱发成湍流。 但是诱发的湍 流会引起某些检测项目浓度的变化，采集测定溶解气体，易挥发物质的样品时，不能把层流 诱发成湍流。 7.4 水流的特征随时间变化 水流可从展流变成湍流， 反之亦然。 可能出现从本水系的其他部分流来的逆流水能给采 样点带来污染。 7.5 流体的组分随时间变化 流体的组分是变化的，随时可能出现不连续的“团状”物，如可溶性污染物、固体物、 挥发性物质、或者漂浮的油层膜。 7.6 从管道中采样 用适当大小的管子（如：抽取多相液体时，管的最小公称内径为 25mm）从管道中抽取 样品。液体在管中的线速度要大，足够保证液体呈湍流的特征，避免液体在管内水平方向流 动。 7.7 液体的性质 液体可能具有腐蚀性和磨蚀性，因此要考虑使用耐腐蚀和耐磨材料。对于长期采样，可 寻找一种容易替换，对样品无显著污染的配件，以代替昂贵的耐化学腐蚀的仪器设备。 7.8 采样系统内出现的温度变化 采样系统内长期或者短期内的温度变化可能引起样品性质的变化， 这种变化可能影响到 采样设备的使用。 7.9 测定悬浮固体物的采样 悬浮物可以分散在遍及液体深度的任一部位。如果可能，可借湍流条件使固-液混合均 匀；从理论上讲，线速度应足以引起湍流。采样应该在等动力下进行。如果做不到，可在流 体的整个断面上取一系列样品。应注意到，在采样期间，悬浮固体物的粒径分布在整个采样 过程中可能发生变化。 7.10 测定挥发性物质的采样 采样泵的吸入高度要小，管网系统要严密。把最先抽出的样品放掉一部分，以保证所采 集的样品具有代表性。 6 7.11 不同密度的混合水 在层流中，水因密度不同而产生分层，比如：在冷水层上面产生一个温水层，盐水上面 有淡水层。 7.12 有害液体 必须注意有毒液体、有毒烟雾的出现，以及可能发生爆炸气体的积集。 7.13 气象环境的影响 有时气象环境的变化给水质带来明显的差异。 要注意这些变化， 并在整理检测结果时予 以修正。 8 天然水的各种采样情况 8.1 降水 为化学分析而收集降水样品时，所选采样点应位于避免外界物质污染的地方。比如，应 避开烟尘、化肥、农药等污染。采样仪器最好放在草坪上。 如果样品被冻或者含有雪或雹之类， 最好用电加热器为采样漏斗加热保温。 如果现场无 法进行加热保温，则可将全套设备移到高于 0℃的低温环境解冻。 8.2 河口、沿海岸水体、海洋 8.2.1 范围和深度 要明确被测水面的边界范围和考虑与邻近水面的相互关系。 选择采样点采样部位时应考 虑到潮汐的流向以及风、海水密度、海底粗糙度、离海岸线的距离对潮汐流向的影响。要考 虑航行对水体流向和水质产生的影响。此外，还要研究局部排放对采样所产生的影响。 8.2.2 船只的使用 在适宜的气候条件下，在整个检测期间，应保证采样船能到达所要求的采样位置。 8.2.3 冰覆盖层。在冻层下面和4℃水体顶部形成0～3℃的冷水层(约5mm)。由于温度梯 度明显，生物群体也可能分层。 8.3 河流和溪流 8.3.1 混合 如果在采样点存在着明显的束流或者分层， 为了确定束流或者分层的性质和范围， 需要 进行横向和纵向系列采样。 8.3.2 选点 7 要选择能提供有代表性样品的点。 采样点应选在水质发生明显变化或者河流有重要用途 的地点，例如，采样点设在汇流口、主要排放或吸水处。在溢流堰或只产生局部影响的小排 出口通常不布点。 测点最好选在可得到流量数据的地点。水质监测设备可安装在河流水文站。 在汇合点从支流中采样时要特别小心， 要避免混入来自主河道的水。 当监测排出液对水 体产生影响时，就要在排放点的上、下游同时采样。要认真研究排放水和承纳水的混合情况 及混合对下游所带来的影响，在下游采样要延伸适当距离，以便评价排放对水体的影响。 8.3.3 潮汐河段 涨潮和落潮时采样点的布设不同。 8.4 运河 通常，对河流的研究大体上适用于运河，但要特别注意下述因素。 8.4.1 流量 水流的方向是可变的。流速可能发生明显的变化。流量的变化取决于航运的频繁情况 (船闸起闭次数)和气象条件的变化等，前者的影响大于后者。 8.4.2 分层和束流 静止状态时，运河中的分层和束流比河流明显。而当船只驶过的短时间内，水质，特别 是悬浮固体的浓度发生明显的变化。 8.5 水库和湖泊 除了入口外，还要在所有有用的泄水点和泄水深度采样。水体有热分层，其不同深度可 能存在明显的质量差异。生态研究需要更详细的采样方案，而且还需要流量和气象资料。 8.6 地下水体 8.6.1 被抽出的地下水 尽管从个别采样点所采样品不能代表整个含水层的质量， 但在评价水能否适用于某种用 途时，样品只能来自个别采样点。 8.6.2 含水层水 为了评价蓄水层的水质采样时， 只要有可能， 要在采样前把取样井或钻孔中的存水抽出。 井和钻孔中的水也会分层，为了评价分层情况需要采取附加样品，并要记下地下采样深度。 井和钻孔的套衬材料易腐蚀，采样之前要进行彻底抽吸，清除系统中积集的腐蚀产物。 在含水层的各个预定深度采集代表性样品时， 对监测井的每一层深度或者那些分散井眼的每 一层深度，应使用采样管采样。 8 8.7 河流、河口、海洋、湖泊和水库的底部沉积物 所制定的采样方案应考虑到沉积物组分纵、 横方向的变化， 必须取得有关底部沉积物的 深度和不同深度上沉积物组成的数据。 采水样时的许多重要因素，如船只的使用，也适用于底部沉积物的采样。 底层通常是不均匀的。为了提供有代表性的评价参数，应保证采集足够数量的样品。 8.8 饮水 8.8.1 供水 在供水中检测消毒剂的残余量时， 采样点应选择在全部反应完成之后和消毒剂的残余量 未发生任何损耗之前。比如，二氧化硫与过量的氯反应后监测余氯。 在供水中，为了常规生物检验需要采样，并遵从适当的预防措施。 通常，从与泵体连接的水龙头上采样。水龙头不应有附件，并能用火焰消毒。采样管的 材质可根据试验要求进行选择。如：铜管因导致水中铜离子的增加因而降低了细菌计数。为 了保证样品直接进入容器，容器应放在水龙头的下面对准龙头，但不能与之接触。 8.8.2 配水池 应在尽可能靠近配水池的水龙头下采样。 许多配水池的进水和出水使用同一个管道。 因 此，只有当配水池管处于出水时才能采样。 8.8.3 配水系统 从自来水用户所使用的水龙头上采样是最好的办法。采样前应移去龙头上的防溅湿装 置，采样时不能使用带有混合式的龙头。在干线和支线管道采样可利用消防栓和冲洗处。此 外，为细菌学检验采集样品时要特别小心。 8.8.4 饮水处理过程中所产生的污泥 大多数饮水处理厂所生成的污泥为氢氧化铝或氢氧化铁， 但也有一些处理厂产生石灰软 化泥或者生物污泥。 这些样品可在凝聚槽混凝沉淀池内的不同深度采取， 也可在浓缩池内采 取。因样品的特殊性在取出后几分钟内就会发生明显变化，因此采样后要尽量少搅动，尽快 检验。 8.9 浴场 从天然浴场采样，按照水库和湖泊采样方法进行。使用循环水系统的游泳池，应该从进 口、出口和水体中分别采样。 9 工业用水采样 9.1 上水 9 上水包括饮用水、 河水、 井水。 由于水源不同， 水质随时发生变化， 但在给定的时间内， 通常它们的组成是均质的。 这些水通过一个普通的管道系统进入工厂， 不存在特殊的采样情 况。 当同时存在非饮用工业供水系统时，要用适当的标志加以区分，以避免搞错采样点。为 了检查水是否可以饮用，要准备一些采样设备。 如果需要各水体混合物的质量数据，采样之前必须保证水体充分混合。 9.2 锅炉系统的水 9.2.1 处理厂的水 在处理厂的设计阶段， 应仔细考虑采样点的方位、 各处理阶段过滤池的进口和出口的采 样设备。当存在悬浮固体时，取样之前应将采样管彻底清洗。 当测定水中溶解气体 (如氧和二氧化碳)采样时，为了避免逸失必须使用特殊的采样技 术。 如果使用除气塔洗除二氧化碳， 那么在随后的样品处理中就要避免二氧化碳的逸失或补 充。采样管应完全浸没于水中，避免吸进气体。 9.2.2 锅炉给水和锅炉火 在蒸汽冷凝循环系统的许多采样点上采集的水样只含有痕量待测物质。 因此， 要特别小 心，避免从采样到分析过程中样品受到污染。 通常的采样系统用不锈钢制成， 采样系统要有完善的结构， 能经受住所承受的运转压力。 如果用长采样管采集高温高压锅炉给水， 为了安全， 最好在靠近采样点的地方冷却采样管中 的样品。 当用物理和化学方法除气时，通常需设两个采样点，一个在加化学药品之前，检验物理 方法除气效率，在第二个点检验总的除气效率。 所设计的锅炉采样点要保证能采到锅炉水的代表性样品。对于某些分析，如痕量金属， 它们可能部分或全部的以颗粒形式存在，在这种情况下应该使用等动力采样探头。 9.2.3 蒸汽冷凝水 在工业上控制蒸汽的质量非常重要。 通常需要从蒸汽冷凝液的回路上， 过热蒸汽或者加 压湿蒸汽中采样。所使用的采样探头，附有不锈钢冷却器。要注意防止采样和分析期间样品 受到污染。 9.2.4 冷却水 主要有三类冷却系统： a.敞开式蒸发冷却系统； 10 b.直流式（单程式）冷却系统； c.闭路循环冷却系统。 在敞开式蒸发系统中， 进水和循环水通常都要采样， 通常在进水口设一个采样点就够了。 但是就冷却系统本身而言，为了获得所需要的数据资料，则必须同时在几个点上采样。使用 生物杀虫剂处理时，则直接在冷却塔的水池中采样。从理论上讲，最好的采样系统是等动力 系统。 直流式冷却系统的采样点设在进水口和出水口处，闭路系统的采样点设在低处。 10 工业废水 10.1 采样点 工业废水的采样必须考虑废水的性质和每个采样点所处的位置。 通常， 用管道或者明沟 把工业废水排放到远而偏僻、人们很难达到的地方。但在厂区内，排放点容易接近，有时必 须采用专门采样工具通过很深的人孔采样。 为了安全起见， 最好把人孔设计成无需人进入的 采样点。 从工厂排出的废水中可能含有生活污水， 采样时应予以考虑所选采样点要避开这类 污水。 如果废水被排放到氧化塘或贮水池，那么情况就类似于湖泊采样。 10.2 废水的性质 在一些工业废水中（个别工厂不经稀释就直接排放）某些组分的浓度很难确定，需要专 门研究。例如，含石油或润滑油、高悬浮固体含量、强酸废水、易燃液体或气体的废水。 当各种不同行业的废水排入同一公共管道时， 为了采到符合要求的样品， 要进行充分的混合。 10.3 工业用水和废水处理的污泥 处理工业用水所产生的化学污泥的范围很广，有些污泥含有毒金属，或放射性物质。废 水处理厂产生的生物污泥的采样详见 （11.1.2） 。 采集这类样品时， 要采取相应的安全措施。 11 污水和污水厂出水 进入污水厂的污水，处理过程各阶段的水以及处理后的出水都需要采样。 11.1 采样点的选择 11.1.1 液体 污水的组成随时间发生明显的变化， 因此在一个过程的每个阶段选择采样点时， 尤其是 在原水中采样时需要特别认真。 污水可能蓄存在横截面很大的涵洞中， 其组成可随深度和沿 涵洞直径发生很大变化。 不同来源的污水可能混合不均， 而且在低流速下污水中的悬浮物可 能沉降。为了确认这些变化，在选择采样点之前要实施一个预采样方案，由预采样获得的资 11 料来决定常规采样点的位置。在许多情况下在不同采样点需要采 2 个或30 个常规样品，混 合这些样品得到一个综合样。 当表面有漂浮物质时，如石油或者润滑油，不能按常规办法 采样，而要从表层下面采样。 原污水常常需要过筛并将大的颗粒粉碎之后才能采样，以避 免样品中出现大的颗粒。不过，在使用自动采样器的地方，采样部位可选在预处理的上流断 面，为防止堵塞自动采样器，在采样器的入口安装滤网和小破碎机。 在处理广选择原污水的采样部位时， 必须把工厂内部的回流液考虑在内。 最好采集两个 样品，一个样品包括所有液体，代表工厂的总负荷，另一个样品不包括回流液，用于衡量外 源的负荷。如果实际上采集不到所述样品，可以用分别采样的办法计算出污水的组成。 11.1.2 污泥 需要在沉淀池、消化池、氧化塘或者干燥床采集污泥样品。 由于原污泥和消化污泥均匀性差及存在有大颗粒物，所以采样时相当困难。 用导管采样时，为了减少堵塞的可能性，采样管的内径不应小于 50mm。取样时间间隔 要短。当从池、氧化塘或者干燥床采样时，要从各种深度和位置采集大量样品。难于接近的 采样点采用专门设备。 对于以上各种情况，适宜用统计学方法确定采样频率。 12 暴雨污水和地面径流 出现暴雨污水和地面径流排放时，接纳水道的流量很大，有效稀释相当大，暴雨污水的 溢流可以控制。由于种种原因，地表径流可能被污染，甚至当水道内水流很大的情况下，溢 流对水道内的水质也构成严重威胁。 由于暴雨污水和地面径流的排放具有间歇性质， 在排放期内质量变化非常明显。 因此给 采样带来一些特殊的问题。 由于对污水管道或者不渗水表面的冲刷， 最初排放出来的污水水 质是很坏的。在这种情况下，最好使用自动采样装置。自动采样装置有许多优点，它定时采 样， 按规定的流量起动， 能够采到有代表性的样品。 在许多情况下， 希望按流量的比例采样。 通常暴雨污水中的固体物未达到浸渍化和沉降， 性质非常不均匀， 给采集有代表性的样品带 来困难， 同时也增加了阻塞设备的可能性。 在选择采样技术和采样设备时对这一点要认真考 虑。必须收集整个调查期间的有关降水量和必要的气温资料。 第三篇 采样频率和采样时间 13 引言 12 通常需要水质可能发生变化全过程的资料， 为此要不时的采样， 使所采样品足以反映水 质及其变化，但也要考虑到代价要小。相反，按主观想象确定采样频率或者仅从分析和采样 的工作量考虑，会导致盲目采样或过于频繁的采样。当非正常状态出现的时候，有必要增加 采样的频次，例如：在植物开始生长的过程中，或者在一条河的涨潮期和藻花时期等。为了 统计长时间的发展趋势，需要增加采样的频次，这样采集的样品结果才是可以利用的。而且 这些样品应该及时称量，以至于密集采样时期可以得到适当的重量。 14 采样方案的类型 采样方案有质量控制、 质量表征和污染源鉴别三种类型， 用于质量控制的检测可用于质 量表征，反之亦然。 14.1 质量控制方案 质量控制通常就是对一个或几个规定范围的环境要素的浓度进行检查。 检查结果决定是 否要即时采取措施。 所确定的采样频率要比连续测量之间出现的超过控制限的显著偏离允许 几率要大。确定采样频率的两个基本因素是： a.在预期条件下，偏离的大小和持续的时间； b.在预期条件下，出现偏离的概率。 通常，对这些因素只能给出近似的定义，但是合理的估价将能获得一个工作值，用以推 算采样频率。 14.2 质量表征方案 这些方案是针对评价一个和比较多的统计参数。 这些参数表明在某一期间内的浓度及其 变化。例如：平均值或者中值表示结果的总趋势，标准偏差表示变率。这些结果可以作为调 查研究的一部分或者表征那些不需要控制，仅有长期控制意义的水质项目。 14.3 污染源调查方案 编制这些方案是为了测定不知来源的污染排放物的特征。 通常， 对本底或污染物性质的 了解是编制方案的基础。污染出现的周期与采样频率要一致。 污染源调查采样方案不同于质量控制、 质量表征的采样方案， 它的采样频率比污染物出 现的频率要高的多。 15 统计研究 15.1 采样方案的确认 在任何采样方案中， 只有在做好认真的准备工作之后， 才能正确地确定采样时间和频率。 为了提供统计技术需要的数据，在准备工作中要提高采样频率。如果水质容易发生变化，无 13 论是随机的还是有规律的变化，所得到的值对于统计参数值，如平均值、标准偏差、最大值 等仅为真实参数的估计值，两者之间有差异。在纯随机变化情况，估计值和真值的差值可用 统计学算出。差值随样品个数的增加而降低。在采样频率确定后，数据要定期检查，以便根 据需要进行改变。 在以下 15.2 至15.5 各条中的论述是把一个统计方法应用于一个统计参数、平均值的 例子，并假定正态分布是适用的。 15.2 置信区间 实际上，n 个结果的平均值的置信区间L 限定了一个范围，位于这个范围的真实平均值 可在给定的置信水平上。 15.3 置信水平 置信水平是在计算出来的置信区间 L 范围内， 含有真实平均值的概率。 由一个样品的n 次结果计算出来的、浓度的均值为X 的置信区间，意味着该区包含真实平均值X 的机会是 100 次中有95 次。在能有效采取大量系列样品情况下，该区包含X 的频率接近95％。 15.4 置信区间的测定和样本数 对随机取样，样本数为 n，真实平均值的估计值μ 和标准偏差σ 分别用算术平均值X 和S 来代替，可按式(1)计算： 1 ) ( 1 2 − − = ∑ = n X X S n i i = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −∑ ∑ = = n i n i i i X n X n 1 2 1 2 1 1 1 …………………（1） 式中： X i —— 某一单位。 当 n 足够大时(见15.1 条)，S 的δ的数值差很小，并且由样本数n 计算得到的X 的置 信区间是X ±K／n。其中K 值由下表给出，K 值取决于所采纳的置信水平。 置信水平，％ 99 98 95 90 80 68 50 K 2.58 2.33 1.96 1.64 1.28 1.00 0.6 在选定的置信水平，为给定的置信区间 L 测定均值X，需要样品的数量为(2Kδ／L) 2。 只有当δ已知时，此式才能成立。尽管S 基于相当大的样品数，对K 值不会有什么差别，但 又采用估计值S 时，需要较多的样品数。严格的说，当样品数小于30 的时候，“K”应该被 t 值代替（t 值可以从t 分布函数百分率表得到）。 14 15.5 水质的随机变化和系统变化 随机变化通常既有正态分布又有对数正态分布， 而系统 变化可能是趋向性变化， 也可能是周期性变化或者两者变化的复合， 在同一个水体中不同的 环境要素的变化性质是不同的。 如果随机变化占优势， 尽管采样次数对质控目标可能是重要的， 但对于统计学通常是不 重要的。如果出现周期性变化，无论对整个检测周期，还是对要测的最大或最小浓度值，采 样次数都是重要的。在整个趋向期间，采样次数应以大致相同的间距分开。对于上述每种情 况所需样品的数量主要从统计学上考虑。 如果不存在系统变化， 或者与随机波动比较时又很 小，那么需要采集的样品数量要足够大，才能满足给定条件下环境要素均值的允许误差。 比如，如果应用正态分布，根据上述内容，在选定的置信水平下 n 个结果均值的置信区间 由式（2）给出： n K L σ 2 = ……………………………………………………………………………………（2） 式中：δ——频率分布的标准偏差。 如果要求的置信区间是均值的 10％， 所要求的置信水平是95％， 均值的标准偏差是20 ％，那么： n 20 96 .
1 2 10 × × = 于是： 84 .
7 = n 因此： n≈61 这表明，如测定周期为 1 个月，每天采两个样品，如果周期为1 年，每星期采1～2 个 样品。 16 异常的变化 在出现异常情况时，如制造厂的开工，河水的洪峰过程，或者各种藻类疯长时期，必须 提高采样频率。在预测长期变化趋向时，只有增加采样频率时，才可运用这些采样结果。 17 采样的持续时间和混合样品 在一个周期内，如果仅仅平均质量有意义，只要待测项目是稳定的，那么延长样品的收 集时间是有利的， 并且采样周期最好与研究的周期相同。 这个原则类似于制备混合样品的原 则。这两种方法减少了分析工作量，但却损失了对质量变化的了解。 第四篇 水流的测量及其在水质方面的应用 15 18 引言 18.1 总则 对污水和废水处理的控制及其用数学模型管理天然水体提高了流量测量的重要性。如： 不进行流量测量就不能评价污染负荷。 本篇提出了在确定采样方案时， 必须考虑的流量因素。 然而，流量的测量通常不由水质检验专家进行，所以本篇不涉及测量的具体细节。 流量测量包括三个方面： a.流向； b.流速； c.流量； d.流动结构； e.横断面积。 18.2 流向 大多数内陆水系，水流是不稳的，但流向是明显的，航道和排泄渠道水流的流向是随时 间而变化的。 掌握含水层中地下水径流的流向， 对于评价含水层被污染的程度以及选择采样 的位置都是很重要的。 在废水处理过程中， 处理池中水流动的模式影响到池中物质的混合和 悬浮物质的沉降，要考虑到水流模式以确保所采集的样品具有代表性。 在河口和沿海水体中， 经常需要测量水流方向并把它看作采样方案的主要部分。 水流的 方向和速度受潮流的影响，非常易变。而潮流又受到气象条件及其他因素的影响。 18.3 流速 流速是很重要的，可用以： a.计算流量； b.计算平均速度和迁移时间。 就水质而言， 迁移时间是指某一水团通过一定距离所需要的时 间； c.评价湍流影响及由流速导致的水体混合。 18.4 流量 流量指单位时间内流过某一点的流体的体积。有关流量的平均值和极限值的资料对废 水、污水和水处理厂的设计、运转以及为保护天然水系制定合理的质量极限是不可缺少的。 18.5 流动结构 16 流动结构可以显著影响横向和纵向的混合速率， 对于流动是在某个指定的槽或者在几个 槽，或者是否存在漩涡的情况，都应该认真评估，样品应该从一个或混合好的槽中采集。例 如：在复合槽或者漩涡中的流动结构说明样品可能是没有代表性。 18.6 横断面积 采样横截面可以是近似矩形或在边缘的一个深槽， 尺寸的宽窄和深浅也可以选择。 这些 特性可以影响混合和腐蚀，同时也可以改变成天然河流和人造水槽的时间。 19 水质控制中流量测量的必要性 19.1 处理厂的负荷 评价工厂的处理负荷需要流量数据。 流量数据可以在进入污水工程系统的排放点以及在 污水厂内部测量得到。如果污水的数量或质量随时间变化，那么要确切估量工厂负荷，需要 对排放量进行连续流量记录。 至于混合样品， 根据采样时间记录到的流量将样品按比例混合 制成。公共下水道中废水的收费与排放污染物的质量和数量成比例。 19.2 稀释效应 要控制向公共污水管道排放有毒有害的物质， 以免工作人员和污水管线及工艺过程受到 危害，与此同时要充分利用提供的稀释条件。 在考虑排放对天然水道和水质限值可能产生的影响时， 必须计算稀释能力。 在上述情况 下，以及当系统中的其他污水所产生的稀释很小时，有关排放的数据非常有价值。 19.3 污染物通量的计算 通量的计算广泛地应用于确定允许排放量和评价河流宽窄对水质的影响。 通量的计算是 模拟整个河流和河口地区质量的基础。 计算的依据是具有代表性的排放资料或者平均流量排 放资料，而动态模拟技术需要连续流量数据的流量频率的测算。 19.4 污染物质的迁移和转化的速度 如果污染物的排放浓度随时间而变化， 那么只有了解污染物从排放点迁移的速度才能正 确估计污染物的扩散和降解情况。因此，在确定一河流或河口地区的采样方案时，要力图在 沿河道流动的同一水体中采样。 当污染物偶然泄漏进入水体时， 了解污染物到达下游所需的 时间对估价污染影响是极其重要的。 19.5 与流量相关的待测物 发现某些水质待测物的浓度，如暂时硬度和氯化物，在某些情况下，通常在一定限度范 围内与流速有关。如果掌握了流速和浓度的相关性，仅仅测量流速就能评价水中待测物。但 要经常核查这种相关性是否发生变化。 17 19.6 地下水 在评价地下水源受污染的情况和转化的程度时， 要求掌握地下水流动的方向和速度的有 关资料。在评价地下污染时可以利用这些资料，从而避免了昂贵的地下水采样。 20 水流测量方法 20.1 测量可以是间断式的，如在河口用浮筒测量，在河流中使用直读式流量计；或者采用 连续式的，如大多数排放流量计。 20.2 测流向和流速可以使用： a.浮标， b.浮筒和其他漂移物， c.化学示踪剂(包括染料)； d.微生物示踪剂； e.放射性示踪剂。 20.3 流速的测量还可采用： a.直读式和自动记录式流量计； b.流速仪； c.超声波技术； d.电磁技术； e.气动技术。 20.4 测定流量时可采用： 20.4.1 流速的测量可按20.2 条或20.3 条所述，在一已知横截面积的明渠中进行。 20.4.2 直接机械方式。例如采用翻斗或标准水表。 20.4.3 在水流中的某一构筑物上，进行水位的测量，如在水道堰上测定水位。采用的方法 有: a.用规准尺进行目测； b.利用浮标、电阻变化、压力差、照相或声变方法进行自动测定。 20.4.4 下列方法适用封闭管道测量： a.通过文氏管颈部产生压力差； b.通过孔板产生压力差； e.扬水率乘以扬水时间； d.电磁、超声波及其他技术。 18 20.4.5 稀释测量。用于天然水系流量的现场测量。 |
7716 | https://abeuiuc.github.io/rot_matrices_1.html | Robot Modeling and Control
Robot Modeling
Introduction
Rotation Matrices
Translation Matrices
DH Parameters
Inverse Kinematics
Rotation Matrices
Linear AlgebraRepresenting RotationsTraining examples
Representing Rotations
In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. For example, using the convention below, the matrix.
Cos(Θ)-Sin(Θ)
Sin(Θ)Cos(Θ)
This matrix rotates points in the xy-plane counterclockwise through an angle θ about the origin of the Cartesian coordinate system. To perform the rotation using a rotation matrix R, the position of each point must be represented by a column vector v, containing the coordinates of the point. A rotated vector is obtained by using the matrix multiplication Rv.
As previously seen, in the example:
X'
Y'
Cos(Θ)-Sin(Θ)
Sin(Θ)Cos(Θ)
X
Y
This operation the same as writing the following formulas :
Basic 3D Rotations
A basic rotation (also called elemental rotation) is a rotation about one of the axes of a Coordinate system. The following three basic rotation matrices rotate vectors by an angle θ about the x, y, or z axis, in three dimensions, using the right-hand-rule which codifies their alternating signs. It's very important for you to familiarize yourself with the right-hand-rule because these matrices can be taken as clockwise or counter-clockwise and it is important to have a commune convention when working with others.
The following example shows a rotation of θ = 90 done around the Z-axis.
As expected, a vector facing the X-axis, when rotated around the Z-axis by 90 degrees, is now facing the Y-axis.
Multiple Rotations
When we are required to use multiple elementary rotations, we need to find a way to combine these rotations in a simpler format. Again using matrices for this purpose really simplifies things enormously.
For example, let's say we need to rotate a reference frame by the Y axis followed by the Z axis, as shown in the below picture.
In order to achieve this multiple rotation matrix, we simply need to figure out by how much we need to rotate each axis and simply multiple them.
Now we have a combination of elementary rotations which is in a very compact format. And all we need to do to go from reference frame A to B is to calculate B = RA
Now it's your turn to practice! Please go to the next page in order to find some practice examples.
Copyright © Tony Grift & Volga Karakus, 2018 |
7717 | https://stackoverflow.com/questions/20728460/is-it-possible-to-create-custom-operators-in-javascript | Stack Overflow
About
For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the company Visit the blog
Collectives¢ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Is it possible to create custom operators in JavaScript?
Ask Question
Asked
Modified 1 year, 1 month ago
Viewed 24k times
50
During the Math classes we learned how to define new operators. For example:
(, ), x y = x + 2y
This defines law. For any real numbers x and y, x y is x + 2y.
Example: 2 2 = 2 + 4 = 6.
Is possible to define operators like this in JavaScript? I know that a function would do the job:
function foo (x, y) { return x + 2 y; }
but I would like to have the following syntax:
var y = 2 2; // returns 6
instead of this:
var y = foo(2, 2);
Which is the closest solution to this question?
javascript
operators
Share
Improve this question
edited Dec 22, 2013 at 14:04
IonicÄ BizÄuIonicÄ BizÄu
asked Dec 22, 2013 at 10:36
IonicÄ BizÄuIonicÄ BizÄu
114k9595 gold badges311311 silver badges488488 bronze badges
5
4
Another name for this is infix function notation - the answer is no, you cannot
Eric
– Eric
2013-12-22 10:39:45 +00:00
Commented Dec 22, 2013 at 10:39
You could try to exploit valueOf with existing operators, but that's about it in JavaScript.
– elclanrs
2013-12-22 10:42:02 +00:00
Commented Dec 22, 2013 at 10:42
I would argue that it's a really bad idea to want to define an operator for which the symbol does not feature on my keyboard
Eric
– Eric
2013-12-22 10:55:24 +00:00
Commented Dec 22, 2013 at 10:55
No. You can't do that in JavaScript. However, you can do it in Haskell. First line: infixl 6 . Second line: x y = x + 2 y.
Aadit M Shah
– Aadit M Shah
2013-12-22 14:15:11 +00:00
Commented Dec 22, 2013 at 14:15
Technically you can, by writing your own lexical parser and defining your own name of script type in the block. This technique is used quite widely. Some well-known examples include Google Traceur.
Derek 朕會功夫
– Derek 朕會功夫
2014-08-22 22:32:41 +00:00
Commented Aug 22, 2014 at 22:32
Add a comment |
12 Answers 12
Reset to default
32
The short answer is no. ECMAScript (the standard JS is based on) does not support operator overloading.
~~As an aside, in ECMAScript 7, you'll be able to overload a subset of the standard operators when designing custom value types. Here is a slide deck by language creator and Mozilla CTO Brendan Eich about the subject. This won't allow arbitary operators, however, and the overloaded meaning will only be applied to value types.~~ <- haha that ended up not happening.
It is possible to use third party tools like sweet.js to add custom operators though that'd require an extra compilation step.
I've answered with a solution from outside JavaScript using esprima - this is changing JavaScript and extending it, it's not native.
Share
Improve this answer
edited Dec 20, 2017 at 8:03
answered Dec 22, 2013 at 10:39
Benjamin GruenbaumBenjamin Gruenbaum
277k9090 gold badges524524 silver badges524524 bronze badges
9 Comments
IonicÄ BizÄu
IonicÄ BizÄu
Can you add a reference? How will I define such an operator after it will be implemented?
Eric
Eric
I doubt any version of ECMAScript comes with the ability to define new symbolic operators - that's normally reserved for languages like haskell. I suspect it'll only come with operator overloading
Benjamin Gruenbaum
Benjamin Gruenbaum
@Eric what Haskell does is not magic. There is no trouble introducing backticks for infix notation in JS :) However yes, the current proposal discusses operator overloading.
Eric
Eric
@BenjaminGruenbaum: Sure, it's not magic, but defining symbolic operators is a slippery slope to perl
Eric
Eric
I see no mention on the ECMAScript page on value objects suggesting that user code can declare new value types
|
8
Given the somewhat new tagged template literals feature that was added in ES6 one can create custom DSLs to handle embedded expressions such as these including different algebraic symbols.
ex. (run in stackblitz)
function math(strings, x, y) { // NOTE: Naive approach as demonstration const operator = strings.replace(/\s/gi, ""); if (operator == "") { return x + 2 y; } else if (operator == "^") { return Math.pow(x, y); } else { return `Unknown operator '${operator}'`; } } console.log(math`${2} ${2}`)
Note that since tagged template literals don't necessarily return strings as results they can return more complex intermediate AST like structures to build up an expression that can then be further refined and then interpreted while keeping close to the domain specific language at hand. I haven't found any existing library that does this yet for Javascript but it should be an interesting and quite approachable endeavor from what it appears from what I know of tagged template literals and usage in such places as lit-html.
Share
Improve this answer
answered Apr 14, 2019 at 3:33
jpiersonjpierson
17.5k1616 gold badges109109 silver badges152152 bronze badges
Comments
5
No. You can't do that in JS.
The closest you can have IMO is to implement your own object which has a chainable interface, aka "fluent" syntax. That way you can operate as if you were speaking out in sequence.
var eq = new YourEquationObject(); // Then operate like 1 - 2 3 eq.add(1).sub(2).mul(3);
Details are up to you though. Just giving out an idea.
Share
Improve this answer
answered Dec 22, 2013 at 10:41
JosephJoseph
120k3131 gold badges186186 silver badges237237 bronze badges
Comments
5
You can add pseudo-operators via methods on Number.prototype:
Object.defineProperty(Number.prototype, 'myOp', { value: function(that) { return this + 2 that; } });
Then all of this syntax will work
alert( (2).myOp(2) ) alert( 2 .myOp(2) ) alert( 2..myOp(2) ) alert( 2.0.myOp(2) )
2.myOp(2) does not work because the period is treated as a decimal point
Share
Improve this answer
answered Dec 22, 2013 at 11:08
EricEric
98k5454 gold badges257257 silver badges389389 bronze badges
Comments
3
No. JavaScript does not support operator overloading . but you can make a class method for doing this
var mathClass = function(value){ this.value = value; } mathClass.prototype.toLaw = function(){ return 2 this.value; } var y = new mathClass(2) 2 + y.toLaw(); //2 + 2 y
Share
Improve this answer
edited Dec 22, 2013 at 10:54
answered Dec 22, 2013 at 10:43
Azadeh RadkianpourAzadeh Radkianpour
97099 silver badges1919 bronze badges
1 Comment
Eric
Eric
You can go one step further and add these to Number.prototype, as in my answer
2
Read the comments below the answer.
Apparently you can't. Here is something close :
function exec(input) { return Function( 'return ' + input.replace(/( [\d.]+)/g, '+ 2 $1') + ';' )(); } exec('2 2'); // 6
Share
Improve this answer
edited Dec 23, 2013 at 16:04
answered Dec 22, 2013 at 10:46
user1636522user1636522
5 Comments
Eric
Eric
(1 + 1) (1 + 1) will ruin your day
Eric
Eric
Paperscript does this quite throughly
IonicÄ BizÄu
IonicÄ BizÄu
@Eric Good approach, but '2 2' is a string. I would like not to be a string.
Eric
Eric
@ã¤ãªãã«ãã¶ã¦: The idea is you mark a script tag as language="myjavascript", and then have a small script to translate your code into raw javascript at runtine
Eric
Eric
To do it properly, you'll want a javascript AST library like acorn or esprima. If you want to add new symbolic operators, you'll probably have to tweak the parser a little
2
Set of compiled to JS languages support custom operators.
I would highlight ReasonML (ocaml-syntax-readable-by-js-folks) and Bucklescript (ocaml-to-js-compiler) which makes custom operators look neat:
For example an operator to concatenate strings can look like:
let (>|<) = (list, seperator) => Belt.List.reduce(list, "", (a, b) => a ++ seperator ++ b);
which can then be used like:
[Styles.button, Media.Classes.atLeastTablet] >|< " "
The downside of all this is the fact it has to be written in such compiled-to-js language, and it comes with lots of pros and cons, but usually those languages have the appeal of tons of nice syntactic sugar you don't get in js/ts
Share
Improve this answer
answered May 25, 2020 at 19:08
ambientlightambientlight
7,35233 gold badges5151 silver badges6262 bronze badges
Comments
1
The slightly longer then the short one is that Yes you can, but its a bit more involved then what you did in Math class
To extend the language with new constructs you can use a transformer like or any of the others. You need to define your syntax, write the parser to parse your statement and finally add the actual code to do the math parts. When these parts is in place you have made a new language that works just as javascript but with the added support of the operator.
Its really not that hard to add new syntax, here is facebooks example how they add => arrow function syntax
Share
Improve this answer
edited Dec 22, 2013 at 11:10
answered Dec 22, 2013 at 10:54
krskrs
4,1642121 silver badges2222 bronze badges
Comments
1
A somewhat late addition based on @iUrii answer using Number.prototype
Using the shorthand square bracket notation [], Number operations can be shown as:
Number.prototype["Ó"] = function(y){ return this + 2 y; };
Then, the operation requested can be written (fairly condensed) as:
var custom_op_result = (2)"Ó"; // Expands out to 2 + 2 2 = 6; // console.log( custom_op_result ) Returns 6
Rather long combinations can also be written using complex notation, variables, parentheses, ect... such as (with above):
Number.prototype["Ï "] = function(y){ return this / y; }; Number.prototype["Ö"] = function(y){ return y this; }; var starting_number = 2; var custom_op_result = ( starting_number"Ï " + (2)"Ö""Ó") ) 2; // Expands out to ( 2 / 2 + (2 + 2 2) 2 ) 2 = 74 // console.log( custom_op_result ) Returns 74
Or... If you want to get really crazy, and possibly confusing you can even assign the operators to variables individually like so:
var Ó = "Ó"; var Ï = "Ï "; var Ö = "Ö"; var custom_op_result = (2)ÓÏ Ö; // Expands out to 2( (2+22) /2 ) // NOTE: Third property actually works backwards to expected // Would have "expected" 32, except it is reversed as 23 // console.log( custom_op_result ) Returns 8
Here's the simple version as a clickable example:
```
Quick demo of math operators
Initial Text That Should Be Replaced
``` |
7718 | http://www.sci.brooklyn.cuny.edu/~mate/misc/hyperbolic_functions.pdf | Hyperbolic functions∗ Attila M´ at´ e Brooklyn College of the City University of New York November 30, 2020 Contents 1 Hyperbolic functions 1 2 Connection with the unit hyperbola 2 2.1 Geometric meaning of the parameter . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 3 Complex exponentiation: connection with trigonometric functions 3 3.1 Trigonometric representation of complex numbers . . . . . . . . . . . . . . . . . . . .
4 3.2 Extending the exponential function to complex numbers. . . . . . . . . . . . . . . . .
5 1 Hyperbolic functions The hyperbolic functions are defined as follows sinh x = ex −e−x 2 , cosh x = ex + e−x 2 , tanh x = sinh x cosh x = ex −e−x ex + e−x sech x = 1 cosh x = 2 ex + e−x , . . .
The following formulas are easy to establish: cosh2 x −sinh2 x = 1, sech2 x = 1 −tanh2 x, (1) sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y; the analogy with trigonometry is clear. The differentiation formulas also show a lot of similarity: (sinh x)′ = cosh x, (cosh x)′ = sinh x, (tanh x)′ = sech2 x = 1 −tanh2 x, (sech x)′ = −tanh x sech x.
Hyperbolic functions can be used instead of trigonometric substitutions to evaluate integrals with quadratic expressions under the square root. For example, to evaluate the integral Z √ x2 −1 x2 dx ∗Written for the course Mathematics 1206 at Brooklyn College of CUNY.
1 for x > 0, we can use the substitution x = cosh t with t ≥0.1 We have √ x2 −1 = sinh t and dt = sinh x dx, and so Z √ x2 −1 x2 dx = Z sinh t cosh2 t sinh t dt = Z tanh2 t dt = Z (1 −sech2 t) dt = t −tanh t + C.
Finally, we need to write the result in term of x. Denoting the inverse of the function cosh by arcosh by the requirement that t = arcosh x if x = cosh t and t ≥0,2 we can find t by solving the equation x = cosh t, i.e., x = (et + e−t)/2 for t. Multiplying the latter equation by et, we obtain a quadratic equation for et: (et)2 −2x(et) + 1 = 0.
Solving this equation for et, we obtain et = x ± p x2 −1.
The product of the two solutions is the constant term of the equation, i.e., 1; so one solution must be greater than 1 and the other must be less than 1. Clearly, the solution given by the + sign will be the one that is greater than 1; this solution will give a value of t > 0; the other solution will give t < 0. We want the former solution; that is, we have (2) t = arcosh x = ln x + p x2 −1 .
Further, tanh t = sinh t cosh t = √ x2 −1 x .
Substituting these into the result obtained for the above integral, we obtain that Z √ x2 −1 x2 dx = ln x + p x2 −1 − √ x2 −1 x + C.
2 Connection with the unit hyperbola Consider the parametric equations (3) x = cosh t, y = sinh t What curve do this equations describe? According to the first equation in (1), we have x2 −y2 = 1.
This equation represents a hyperbola, called the unit hyperbola, with center at the origin, and with a horizontal real axis. The parametric equations represent only the right branch of the hyperbola, since cosh t > 0 for all t. Every point on this right branch is represented by the parametric equations, since, if (x, y) is a point on this right branch, then we obtain this point with t = arcosh x is y ≥0 and t = −arcosh x if y < 0 (see (2)).
1Note that cosh t > 0 for all t, so this substitution works only in case x > 0 (in which case, we need to assume that x ≥1 so as not to have a negative number under the square root). The case x < 0 can be handled by the substitution −x = cosh t. The stipulation t ≥0 is used to determine t uniquely, given x; note that cosh t = cosh(−t).
2Note we have cosh t = cosh(−t); thus t is not uniquely determined by the stipulation that x = cosh t. For this reason, we required t ≥0 above, so that t be uniquely determined by x. The prefix ar in arcosh is an abbreviation of area – see the next section where the connection of hyperbolic functions with the unit hyperbola is discussed. This is in contract with the inverse trigonometric functions where the prefix arc is used, and it literally stands for an arc of the unit circle.
2 2.1 Geometric meaning of the parameter The meaning the parameter t in equations (3) is half the area that segment OP ′ sweeps with the point O being the origin or the coordinate system, and P ′ being the point with coordinates x′ = cosh t′ and y′ = sinh t′ as t′ changes between 0 and t, the area being positive if t (or, which is, the same, y) is positive, and it being negative if t (or y) is negative. 3 In fact, this is not hard to show.
To this end, assume for simplicity that t0 > 0.4 Using polar coordinates, let T be the polar angle of the point (x0, y0) determined equations (3) for the given t0. We will convert equations 3) to polar coordinates by using r = p x2 + y2, θ = arctan x y ; the latter equation will give the correct value of θ only if x > 0, which is the case for the right branch of the hyperbola: (4) r = p cosh2 t + sinh2 t, θ = arctan tanh t.
Using the area formula in polar coordinates, the area of the hyperbola segment is A = 1 2 Z T 0 r2 dθ, where r = f(θ) is the equation describing the right branch of the hyperbola in polar coordinates.
Fortunately, there is no need to determine the function f, nor there is a need to determine the upper limit T in this integral, since we can change to the variable t described in equations (4). We then have dθ = 1 1 + tanh2 t · sech2 t dt = 1 sinh2 t + cosh2 t dt.
Hence we obtain A = 1 2 Z T 0 (cosh2 t + sinh2 t) · 1 sinh2 t + cosh2 t dt = 1 2 Z T 0 dt = T 2 , as we wanted to show.
3 Complex exponentiation: connection with trigonometric functions There is a close connection between hyperbolic functions and trigonometric functions. The formulas, to be explained later, stating these connection connection, are (5) sin x = eix −e−ix 2i , cos x = eix + e−ix 2 , where i is the imaginary unit, that is, i = √−1. To make sense of these formulas, one needs to know what is to be meant by ez when z is a complex number.
The usual way of extending ez to complex numbers proceeds by the use of infinite series.5 This approach is technically efficient but has no intuitive appeal. After seeing this approach, one is still mystified by equations (5). Here we present a different approach.
3Note the analogy with trigonometric functions, where, for example, cos t represents the x-coordinate corresponding to an arc t in the unit circle. An arc t in the unit circle is, however, associated with a circular sector of area t/2, completing the analogy with hyperbolic functions.
4The argument below works without change even if t0 ≤0, but it is easier to visualize in case t0 > 0.
5By considering the Taylor series of ez, sin z, and cos z, and considering them valid also for complex values of z.
3 3.1 Trigonometric representation of complex numbers A complex number is a number of the form a + bi, where a and b are real numbers. It is customary to graph this number on the coordinate plane as the point (a, b), and calling the coordinate plane used to graph complex numbers the complex plane.6 One defines the absolute value of a complex number as its distance from the origin of the coordinate system, that is, (6) |a + bi| = p a2 + b2, and its argument as the angle between the positive x axis and the ray from the origin toward the point (a, b), angles corresponding to counter-clockwise rotation of the positive x axis to the latter ray are considered positive. The argument of a point is only determined upto an additive multiple of 2π except for the number 0 = (0, 0), which can have any argument. For the argument of a number (a, b) we have (7) arg(a + bi) = arctan b a + 2kπ or arg(a + bi) = arctan b a + (2k + 1)π for some integer (positive, negative, or zero) k; the first equation is valid if a > 0, and the second one if a < 0; unfortunately, this formula does not work in case b = 0.7 Writing ρ = |(a, b)| and θ = arg(a, b), it is clear that a = ρ cos θ and b = ρ sin θ, and so (8) a + bi = ρ(cos θ + i sin θ).
The right-hand side is called the trigonometric form of the complex number on the left.
Let z1 = ρ1(cos θ1 + i sin θ1) and z2 = ρ2(cos θ2 + i sin θ2) be two complex numbers. Then it follows from the addition formulas for sin and cos that (9) z1z2 = ρ1ρ2(cos(θ1 + θ2) + i sin(θ1 + θ2)).
This means that |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2.
These equations can also be easily verified directly, the latter by using geometric arguments, and then these equations can be used the prove the addition formulas for sin and cos.
From (9) one can easily establish de Moivre’s formula (10) (ρ(cos θ + i sin θ))n = ρn(cos nθ + i sin nθ).
6In fact, the complex number a+bi is interpreted as the ordered pair (a, b) of real numbers, so one need not wonder about what mysterious meaning √−1 may have. Then one defines operations on pairs of numbers; for example, one puts (a, b) · (c, d) = (ac −bd, ad + bc), which corresponds to the multiplication formula (a + bi)(c + di) = ac −bd + (ad + bc)i.
Addition, subtraction, and division can be defined in a similar way.
7The fact that the formula does not always work is an inconvenience in numerical calculations. To remedy this, several programming languages provide the function atan2(); this function gives a value in the range [−π, π], and atan2(a,b) returns a value for any values of a and b (for example, the C programming language distinguishes between +0 and −0, and in C this function is defined even if a or b or both are equal to ±0). See e.g.
for details.
4 3.2 Extending the exponential function to complex numbers.
We are about to extend the natural exponential function ex to complex exponents. Occasionally, it will be convenient to use exp x or exp(x) instead of ex, especially when the exponent x is replaced by a complicated expression. Our starting point will be the equation (11) exp x = lim n→∞ 1 + x n n , valid for every real x, where n is running over positive integers. The easiest way to prove this is to establish this limit for a continuous real variable t with the aid of l’Hospital’s rule: lim t→+∞ 1 + x t t = lim t→+∞exp ln 1 + x t t = lim t→+∞exp t ln 1 + x t = exp lim t→+∞t ln 1 + x t = exp lim t→+∞ ln (1 + x/t) 1/t = exp lim t→+∞ −x/t2 1+x/t −1/t2 = exp lim t→+∞ x 1 + x/t = exp x; (12) the interchange of the limit and the exponential function after the third equation sigh is justified by the continuity of the exponential function, and the fifth equation is valid in view of l’Hospital’s rule. We will extend the exponential function to complex numbers by requiring that equation (11) remains valid for complex values of x, that is, by requiring that (13) exp z = lim n→∞ 1 + z n n , valid for every complex z, Various properties of the exponential function can be established with the aid of this equation; we will confine ourselves to establishing the equation (14) eix = cos x + i sin x, called Euler’s formula.
Given a fixed real number x, we have (15) 1 + ix n = r 1 + x2 n2 cos arctan x n + i sin arctan x n according to (6), (7), and (8). Hence, by de Moivre’s formula (10) we have (16) 1 + ix n n = 1 + x2 n2 n/2 cos n arctan x n + i sin n arctan x n .
The limits as n →∞of the expressions on the right-hand side are easy to establish by first changing n into a continuous variable t and then using l’Hospital’s rule. In establishing the first of these limits, we proceed similarly to (12), so we will give fewer details. We have lim t→+∞ 1 + x2 t2 t/2 = exp lim t→+∞ ln(1 + x/t2) 2/t = exp lim t→+∞ −2x/t3 1+x/t2 −2/t2 = exp lim t→+∞ x t(1 + x/t2) = exp 0 = 1; 5 the first equation used the continuity of the exponential function, and the second one used l’Hospital’s rule. As for the second limit, lim t→+∞t arctan x t = lim t→+∞ arctan(x/t) 1/t = lim t→+∞ −x/t2 1+x2/t2 −1/t2 = lim t→+∞ x 1 + x2/t2 = x; the second equation here follows from l’Hospital’s rule. Using these two limits and making n →∞ in (16), we obtain lim n→∞ 1 + ix n n = cos x + i sin x.
Thus (14) follows in view of (13).
Writing −x instead of x in (14), we obtain exp(−ix) = cos(−x) + i sin(−x) = cos x −i sin x.
Solving the system of equations given by this equation and by (14) for sin x and cos x, we obtain equations (5).
6 |
7719 | https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/lung-blood-flow | Skip to Main content
Sign in
Chapters and Articles
You might find these chapters and articles relevant to this topic.
The pulmonary circulation
Pulmonary blood flow
Total pulmonary blood flow is approximately equal to cardiac output and is a topic which belongs to the field of circulation rather than respiration. We shall not therefore consider pulmonary blood flow in any more detail than is necessary for an understanding of the changes in pulmonary vascular pressure and resistance. Pulmonary gas exchange is, however, greatly influenced by the distribution of pulmonary blood flow, and this is considered in Chapter 7.
Methods for the measurement of pulmonary blood flow are outlined at the end of this chapter, but at this stage it should be pointed out that most methods of mesurement of cardiac output (Fick, dye and thermal dilution) measure the total pulmonary blood flow, together with the venous admixture (see Figure 6.1). On the other hand, the body Plethysmograph measures only the pulmonary capillary blood flow. In this method, the alveoli are filled with nitrous oxide at a concentration of about 15%, and the amount of nitrous oxide taken up by the blood is measured by whole body plethysmography. If the mean alveolar Pn2o and the solubility of nitrous oxide in blood are known, it is possible to calculate the pulmonary blood flow on the assumption that the alveolar Pn2o equals the arterial Pn2o and the mixed venous Pn2o is zero. The method may be used to measure the instantaneous capillary blood flow, which is found to be pulsatile.
View chapterExplore book
Read full chapter
URL:
Chapter
Volume 2
2022, Encyclopedia of Respiratory Medicine(Second Edition)John C. Huetsch, Larissa A. Shimoda
Blood flow heterogeneity
Within the lung, blood flow is not uniform. Given the unique characteristics of the pulmonary circulation of high distensibility and compressibility and low intravascular pressure, it is not surprising that pulmonary blood flow and vascular resistance would be influenced by factors independent of vascular smooth muscle tone, including both gravitational and structural factors. For example, within a column of fluid, the weight of the fluid exerts a higher pressure at the bottom than near the top. Similarly, in upright individuals, the vertical height of the lung is approximately 24 cm (at FRC), and while alveolar pressure is fairly uniform throughout the lung, intravascular pressure is increased in the bottom (dependent) portions of the lungs due to gravity-dependent hydrostatic effects. This pressure gradient causes progressive vascular distention, decreased resistance, and increased flow at the base of the lung.
Based on the relationship between pulmonary arterial pressure (PPA), pulmonary venous pressure (PPV), and alveolar pressure (Palv), the lung can be divided into three zones (Fig. 6). In Zone 1, hydrostatic effects cause both arterial and venous pressures to fall below alveolar pressure such that Palv > PPA > PPV and the alveolar vessels are completely collapsed, restricting blood flow to the corner vessels. Under Zone 2 conditions, PPA > Palv > PPV and the alveolar vessels are partially collapsed, with the driving force for flow being the pressure gradient between arterial and alveolar pressures (PPA − Palv). In Zone 3, PPA > PPV > Palv and all of the alveolar blood vessels are fully open. Thus, blood flow is driven by the difference between pulmonary arterial and venous pressure (PPA − PPV). In a seated or standing normal subject, Zone 1 conditions are typically absent but, if present, typically will be found at the apex of the lung. Conversely, at the lung base, blood flow is greatest and Zone 3 conditions dominate, while Zone 2 conditions can be found in the mid-lung.
In addition to gravitational forces, body position can influence the extent and location of the lung zones. In a supine subject, Zone 1, if present, will be in the ventral portion of the lungs while flow increases to the anatomically dependent (dorsal) regions of the lungs, resulting in Zone 3. When a subject lies on their side, blood flow increases to the dependent lung. In certain cases, alveolar pressure may exceed vascular pressure in the non-dependent portions of the lung. For example, regions of Zone 1 and 2 can be created and/or increased during hemorrhage or hypovolemia, which result in low intravascular pressures, or during positive pressure ventilation or forced expiration, where alveolar pressure is increased.
Regional blood flow differences can exist even between alveoli with similar vertical position, indicating that blood flow heterogeneity is not entirely dependent on the effects of gravity. At the microvascular level, the branching nature of the pulmonary vascular tree results in structural heterogeneities within isogravitational planes, producing variations in local driving pressures and resistances. This heterogeneity in driving pressure creates gravity-independent differences in regional pulmonary blood flow. Since regional blood flow is influenced by both gravity and vascular geometry, all three zones can exist in the same vertical plane; however, the lung apex still contains a higher density of Zone 1 physiology while the base contains more Zone 3 physiology.
View chapterExplore book
Read full chapter
URL:
Reference work2022, Encyclopedia of Respiratory Medicine(Second Edition)John C. Huetsch, Larissa A. Shimoda
Chapter
The Pulmonary Circulation
2017, Nunn's Applied Respiratory Physiology (Eighth Edition)Andrew B Lumb MB BS FRCA
Pulmonary Blood Flow
The flow of blood through the pulmonary circulation is approximately equal to the flow through the whole of the systemic circulation. It therefore varies from approximately 6 l.min−1 under resting conditions, to as much as 25 l.min−1 in severe exercise. It is remarkable that such an increase can normally be achieved with minimal increase in pressure. Pulmonary vascular pressures and vascular resistance are much less than those of the systemic circulation. Consequently the pulmonary circulation has only limited ability to control the regional distribution of blood flow within the lungs and is affected by gravity, which results in overperfusion of the dependent parts of the lung fields. Maldistribution of the pulmonary blood flow has important consequences for gaseous exchange, and these are considered in Chapter 7.
In fact, the relationship between the inflow and outflow of the pulmonary circulation is much more complicated (Fig. 6.1). The lungs receive a significant quantity of blood from the bronchial arteries, which usually arise from the arch of the aorta. Blood from the bronchial circulation returns to the heart in two ways. From a plexus around the hilum, blood from the pleurohilar part of the bronchial circulation returns to the superior vena cava via the azygos veins, and this fraction (about one-third) may thus be regarded as normal systemic flow, neither arising from nor returning to the pulmonary circulation. However, another fraction of the bronchial circulation, distributed more peripherally in the lung, passes through postcapillary anastomoses to join the pulmonary veins, constituting an admixture of venous blood with the arterialized blood from the alveolar capillary networks.
The situation may be further complicated by blood flow through precapillary anastomoses from the bronchial arteries to the pulmonary arteries. These communications (so-called ‘Sperr arteries’) have muscular walls and are thought to act as sluice gates, opening when increased pulmonary blood flow is required. Their functional significance in normal subjects is unknown, but in diseased lungs flow through these anastomoses may be crucial. For example, in situations involving pulmonary oligaemia (e.g. pulmonary artery stenosis, pulmonary embolism) blood from the bronchial arteries will flow through the anastomoses to supplement pulmonary arterial flow.2 It should be noted that a Blalock–Taussig shunt operation achieves the same purpose for palliation of patients with cyanotic congenital heart disease.
View chapterExplore book
Read full chapter
URL:
Book2017, Nunn's Applied Respiratory Physiology (Eighth Edition)Andrew B Lumb MB BS FRCA
Chapter
Volume 3
2022, Encyclopedia of Respiratory Medicine(Second Edition)R. Naeije, A.J. Peacock
Pulmonary Vascular Pressures and Blood Flow
Pulmonary blood flow (CO) is driven by a pressure difference between an inflow pressure, or mean pulmonary artery pressure (mPAP) and an outflow pressure, or mean left atrial pressure (LAP). From these measurements, resistance to flow in the pulmonary circulation can be approximated by a single number, pulmonary vascular resistance (PVR) which depends on the ratio between (mPAP—LAP) and CO:
In clinical practice, measurements of pulmonary vascular pressures and pulmonary blood flow are usually performed during a catheterization of the right heart with a triple-lumen fluid-filled balloon and thermistor-tipped catheter introduced by Swan, Ganz, Forrester and their colleagues in the early 1970s (Swan et al., 1970; Forrester et al., 1972) Fig. 2.
The procedure allows for the estimation of LAP from a balloon-occluded or wedged PAP (PAOP or PAWP). The fractal structure of the pulmonary arterial and venous trees allows for a stop-flow phenomenon downstream of occlusion and de facto extension of the fluid-filled lumen of the catheter to same diameter pulmonary veins. With injection of 5–10 mL cooled saline into a lumen opening into the right atrium, a temperature change over time in the pulmonary artery is registered and CO is calculated from this thermal dilution curve.
View chapterExplore book
Read full chapter
URL:
Reference work2022, Encyclopedia of Respiratory Medicine(Second Edition)R. Naeije, A.J. Peacock
Chapter
Ventilation, Blood Flow, and Gas Exchange
2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Frank L. Powell PhD, ... John B. West MD, PhD, DSc
Distribution of Pulmonary Blood Flow
Just as for ventilation, blood flow is not partitioned equally to all alveoli, even in the normal lung. Both gravitational (topographic) and nongravitational factors affect the distribution of blood flow.
Normal Distribution
The topographic distribution of pulmonary blood flow can conveniently be measured using radioactive materials. In one technique, 133mXe is dissolved in saline and injected into a peripheral vein. When the xenon reaches the pulmonary capillaries, it evolves into the alveolar gas because of its low blood solubility. The resulting distribution of radioactivity within the lung can be measured using a gamma camera or similar device and reflects the regional distribution of blood flow. Subsequently the distribution of alveolar volume is obtained by having the subject rebreathe radioactive xenon to equilibrium. By combining the two measurements, the blood flow per unit alveolar volume of the lung can be obtained. The distribution of blood flow can also be measured with radioactive albumin macroaggregates and with a variety of other radioactive gases, including 15O-labeled carbon dioxide and 13N. Functional magnetic resonance imaging of the lung has been used to assess distribution of pulmonary blood flow.30 This noninvasive technique does not expose subjects to radioactivity; therefore it can be used repetitively and shows great promise for the future.
In the normal upright human lung, pulmonary blood flow decreases approximately linearly with distance up the lung, reaching very low values at the apex.31 However, if the subject lies supine, apical and basal blood flow become the same, and now blood flow is less in the anterior (uppermost) than posterior (lowermost) regions of the lung. Thus blood flow distribution is highly dependent on gravitational effects. During exercise in the upright position, both apical and basal blood flow rates increase and the relative differences are reduced.
The factors responsible for the uneven topographic distribution of blood flow can be studied conveniently in isolated lung preparations. These studies show that, in the presence of normal vascular pressures, blood flow decreases approximately linearly up the lung32 as it does in intact humans. However, if the pulmonary arterial pressure is reduced, blood flows only up to the level at which pulmonary arterial equals alveolar pressures; above this point, no flow can be detected. If venous pressure is raised, the distribution of blood flow may become more uniform in the region of the lung below the point at which pulmonary venous equals alveolar pressure.
Three-Zone Model for the Distribution of Blood Flow
Figure 4-11 shows a simple model for understanding the factors responsible for the topographic inequality of blood flow in the lung.32 The lung is divided into three zones according to the relative magnitudes of the pulmonary arterial, alveolar, and venous pressures.
Zone 1 is that region of the lung above the level at which pulmonary arterial equals alveolar pressures; in other words, in this region, alveolar pressure exceeds arterial pressure. Measurements in isolated lungs show that there is no blood flow in zone 1, the explanation being that the collapsible capillaries close because the pressure outside exceeds the pressure inside. Micrographs of rapidly frozen lung from zone 1 show that the capillaries have collapsed, although occasionally trapped red blood cells can be seen within them.17
The vertical level of blood flow can be influenced by the surface tension of the alveolar lining layer, as discussed earlier. If measurements are made on a lung immediately after it is inflated from a near-collapsed state, blood flow reaches 3 or 4 cm above the level at which pulmonary arterial and alveolar pressures are equal.16 This can be explained by the reduced surface tension, which lowers the pericapillary hydrostatic pressure.
Zone 2 is that part of the lung in which pulmonary arterial pressure exceeds alveolar pressure, but alveolar pressure exceeds venous pressure. Here the vessels behave like Starling resistors,33 that is, as collapsible tubes surrounded by a pressure chamber. Under these conditions, flow is determined by the difference between arterial and alveolar pressures, rather than by the expected arterial-venous pressure difference. One way of looking at this is that the thin wall of the vessel offers no resistance to the collapsing pressure, so the pressure inside the tube at the downstream end is equal to chamber pressure. Thus the pressure difference responsible for flow is perfusion minus chamber pressure. This behavior has been variously referred to as the waterfall33 or sluice34 effect and can be demonstrated in rubber-tube models on the laboratory bench. The increase in blood flow down zone 2 can be explained by the hydrostatic increase in pulmonary arterial pressure down the zone, whereas the alveolar pressure remains constant. Thus the pressure difference determining flow increases linearly with distance.
Zone 3 is that part of the lung in which venous pressure exceeds alveolar pressure. Radioactive gas measurements show that blood flow increases as one measures vertically down this zone, although, in some preparations at least, the rate of increase is apparently less than found in zone 2. Because the pressure difference responsible for flow is arterial minus venous pressure and because these two pressures increase similarly with distance down the zone, the increase in blood flow is not explained by changes in perfusing pressure. Instead, blood flow increases down this zone because vascular resistance falls with distance down the zone, likely because of progressive distention (confirmed histologically17) from the increasing transmural pressure (intravascular pressure increasing down the zone while alveolar pressure is constant). However, resistance may also be reduced by recruitment of capillaries.
The Effect of Lung Volume on the Distribution of Blood Flow—Zone 4
In spite of its simplicity, the three-zone model of Figure 4-11, based on the effects of pulmonary arterial, alveolar, and venous pressures, accounts for many of the distributions seen in the normal lung. However, other factors play a role; one of these is lung volume. For example, under most circumstances, a zone of reduced blood flow, known as zone 4, is seen in the lowermost region of the upright human lung.35 This zone becomes smaller as lung volume is increased, but careful measurements indicate that a small area of reduced blood flow is present at total lung capacity at the lung base. As lung volume is reduced, this region of reduced blood flow extends further and further up the lung, so that at FRC, blood flow decreases in the bottom half of the lung. At RV the zone of reduced blood flow extends all the way up the lung, so that blood flow at the apex exceeds that at the base.35
These patterns cannot be explained by the interactions of the pulmonary arterial, venous, and alveolar pressures as in Figure 4-11. Instead, we have to take into account the contribution of the extra-alveolar vessels. As pointed out previously (see Fig. 4-10), the caliber of these vessels is determined by the degree of lung inflation; as lung volume is reduced, the vessels narrow. In the upright human lung, the alveoli are less well expanded at the base than at the apex because of distortion of the elastic lung caused by its weight (see Fig. 4-5). As a result the extra-alveolar vessels are relatively narrow at the base, and their increased contribution to pulmonary vascular resistance results in the presence of a zone of reduced blood flow in that region. As overall lung volume is reduced, the contribution of the extra-alveolar vessels to the distribution of blood flow increases, and zone 4 extends further up the lung. At residual volume the caliber of the extra-alveolar vessels is so small that they completely dominate the picture and determine the distribution of blood flow.
Vasoactive drugs and interstitial edema can modify the contribution of extra-alveolar vessels to pulmonary vascular resistance. For example, the role of the extra-alveolar vessels can be exaggerated by injecting vasoconstrictor drugs such as serotonin.36 Under these conditions, zone 4 extends even further up the lung. The opposite effect is seen if a vasodilator drug such as isoproterenol is infused into the pulmonary circulation. With interstitial edema, the contribution of the extra-alveolar vessels increases, because the edema creates a cuff of fluid around the vessels and thereby narrows them. This is thought to be the cause of the increased pulmonary vascular resistance seen at the base of the human lung in conditions of interstitial pulmonary edema,28 in which the distribution of blood flow often becomes inverted (e.g., in chronic mitral stenosis).37 Under these conditions the blood flow to the apex of the upright lung consistently exceeds the flow to the basal regions. However, the effects of interstitial edema on blood flow distribution are still not fully understood.
Other Factors Affecting the Distribution of Blood Flow
Because the topographic distribution of blood flow in the normal lung can be attributed to gravity, it is not surprising that, during increased acceleration, the distribution of blood flow becomes more uneven.38 For example, during exposure to +3g acceleration, that is, three times the normal acceleration experienced by someone in the upright posture, the upper half of the lung is completely unperfused. The amount of unperfused lung is approximately proportional to the g level.
By contrast, in astronauts during sustained microgravity in space, the distribution of blood flow becomes more uniform.39 Because it is not possible to use radioactive gases in this environment, the inequality of blood flow has been determined indirectly from the size of the cardiogenic oscillations for Pco2. Cardiogenic oscillations are fluctuations in the concentrations of gases such as oxygen, carbon dioxide, and nitrogen during a single expiration. They have the same frequency as heart rate and are considered to be caused by differential rates of emptying of different parts of the lung due to contraction and dilatation of the heart exerting direct pressure on nearby but not distant lung parenchyma. For oscillations to be detected, these differentially emptying regions must also have different alveolar Po2 and Pco2 values, and this happens when blood flow and ventilation are not uniformly distributed throughout the lung. Microgravity almost abolishes cardiogenic oscillations, implying greater uniformity in the distribution of blood flow and/or ventilation in the absence of gravity. Interestingly, because these oscillations can still be seen, albeit to a much smaller extent than on earth, some inequality remains, indicating that gravity-independent mechanisms are also present.
Although gravity is a major factor determining the uneven distribution of blood flow in the upright human lung, it is now clear that nongravitational factors also play an important role. There are several possible mechanisms. One is that there may be regional differences of vascular conductance, with some regions of the pulmonary vasculature having an intrinsically higher vascular resistance than others. This has been shown to be the case in isolated dog lungs,40 and there is some evidence for higher blood flows in the dorsal-caudal than the ventral regions of the lung in both intact dogs and horses. Another possible factor is a difference in blood flow between the central and peripheral regions of the lung,41 although this finding is controversial. Some measurements show differences in blood flow along the acinus, with the more distal regions of the acinus being less well perfused than the proximal regions.42,43 Finally, as pointed out earlier, because of the complexity of the pulmonary circulation at the alveolar level, including the very large number of capillary segments, it is likely that there is inequality of blood flow at this level. Reference has already been made to the possibility of recruitment of pulmonary capillaries based on the stochastic properties of a dense network of numerous interconnected capillary segments.22 There is also work suggesting that the distribution of pulmonary blood flow in small vessels may follow a fractal pattern.44 The term fractal describes a branching pattern of both structure (blood vessels) and function (blood flow) that repeats itself with each generation. This means that any subsection of the vascular tree exhibits the same branching pattern as the entire tree. Were a picture of such a subsection to be enlarged, it would overlie and match the pattern of the whole tree. Just as mentioned for ventilation earlier, repeated branching of blood vessels has implications for how blood flow is distributed independently of gravitational influences. The greater the number of branch points, the greater the likely inequality of perfusion among alveoli. This implies that the finer the spatial resolution of the method used to assess flow distribution, the greater the amount of inequality likely to be detected.
Abnormal Patterns of Blood Flow
The normal distribution of pulmonary blood flow is frequently altered by lung and heart disease. Localized lung disease, such as fibrosis and cyst formation, usually causes a local reduction of flow. The same is true of pulmonary embolism, in which the local reduction in blood flow, as determined from a perfusion scan, is usually coupled with normal ventilation, and this pattern provides important diagnostic information. Bronchial carcinoma may reduce regional blood flow, and occasionally a small hilar lesion can cause a marked reduction of blood flow to one lung, presumably through compression of the main pulmonary artery. Generalized lung diseases, such as COPD and bronchial asthma, also frequently cause patchy inequality of blood flow. Sometimes, asthmatic patients whose disease is thought to be fairly well controlled show marked impairment of blood flow in some lung regions.
Heart disease frequently alters the distribution of blood flow, as might be expected from the factors responsible for the normal distribution (see Fig. 4-11). For example, patients with pulmonary hypertension or increased blood flow through left-to-right shunts usually show a more uniform distribution of blood flow.45 Diseases in which pulmonary arterial pressure is reduced, such as tetralogy of Fallot with oligemic lungs, are associated with reduced perfusion of the lung apices. Increased pulmonary venous pressure, as in mitral stenosis, initially causes a more uniform distribution than normal. However, in advanced disease, an inversion of the normal distribution of blood flow is frequently seen, with more perfusion to the upper than to the lower zones. The mechanism for this shift is not fully understood, but, as indicated earlier, perivascular edema causing an increased vascular resistance of the extra-alveolar vessels is thought to be a factor.
View chapterExplore book
Read full chapter
URL:
Book2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Frank L. Powell PhD, ... John B. West MD, PhD, DSc
Chapter
AGEs in scuba diving and in DCS-like problems in breath-hold diving
2018, Gas Bubble Dynamics in the Human BodySaul Goldman, ... Kenneth M. Ledez
5.6.2.4 Pulmonary blood flow
The pulmonary blood flow Q(t), defined earlier and needed in Eq. (5.12), will be determined by integrating its rate equation, that is, by integrating:
(5.17)
In Eq. (5.17), Qs is the initial value of the pulmonary flow at the surface, and Qo is the minimum pulmonary flow under dive reflex activation. As shown in Eq. (5.17), the pulmonary blood flow rate involves depth and time constants (KDR and τDR) (Box 5.3).
Box 5.3
In Greater Detail
We have found that for linear ascents and descents, that is, for dives with constant ascent/descent rates, the pulmonary blood flow Q(t) can be obtained in closed form by analytically integrating Eq. (5.17). For linear ascents/descents, the depth has the time-dependence
(5.18)
Here is the constant descent/ascent rate. Constant descent/ascent rate breath-hold dives can be described by a piece-wise continuous function, with each stage of the dive (descent, rest or stop, and ascent) captured by the general form given by Eq. (5.18).
By using Eq. (5.18) for z(t) in Eq. (5.17), the integrated expression for the pulmonary blood flow is found to be
(5.19)
Eq. (5.19) was first derived in . By using this closed-form analytic expression for the pulmonary blood flow Q(t) at arbitrary depths in Eq. (5.12), one very significantly decreases both the computational complexity and computer time needed for these calculations (Box 5.4).
Box 5.4
For the Math Mavens
The way this equation is implemented in practice for the full dive profile is as follows: operationally, one replaces t0 → tk, , and for each segment of the dive profile, where tk is the time at which the kth segment begins. Then is the expression that gives the depth at all times t for the kth segment, and is the constant ascent/descent rate of each kth segment in the profile.
Eqs. (5.13)–(5.19) provide all the functions needed on the right-hand side of Eq. (5.12), so that the arterial nitrogen pressure at any depth in a breath-hold dive can now be readily determined.
View chapterExplore book
Read full chapter
URL:
Book2018, Gas Bubble Dynamics in the Human BodySaul Goldman, ... Kenneth M. Ledez
Review article
Includes papers from the Symposium on Developmental Transitions in Respiratory Physiology at the first International Congress of Respiratory Biology, Bad Honnef, Germany, August 2006
2007, Comparative Biochemistry and Physiology Part A: Molecular & Integrative PhysiologyMichael B. Thompson
During development, relatively little blood flow (∼ 10% of cardiac output) is required to sustain metabolism of the lungs. After birth and in series with the systemic circulation, almost the whole cardiac output goes to the lungs as a result of a major decline in vascular resistance and pulmonary arterial pressure (Mortola, 2001). Pulmonary arterial pressure falls from 80 to 30–40 Torr within 10 h of birth in calves (Reeves and Leathers, 1964). At birth, blood flow through the ductus arteriosus, and the right-to-left shunt, decrease (Reeves and Leathers, 1964), resulting in an increase in blood flow to the upper body, brain (and lungs) and a fall in blood flow to the lower body (Mortola, 2001). Closure of the intra-atrial shunt occurs in both eutherian and marsupial mammals, although the processes are slightly different (Mortola, 2001). Interestingly, the ductus arteriosus does not close in all reptiles (Burggren, 1976) and the part that it plays during parition in reptiles remains to be investigated. The onset of ventilation, inflation and oxygenation of the lungs also has a great effect on pulmonary circulation by reducing pulmonary vascular resistance. The volume of the fluid-filled lungs of the foetus is greater than the air-filled lung in neonatal mammals, so the fall in pulmonary vascular resistance may be due to an increase in lung recoil pressure. Together with the formation of surface tension, the result is a five-fold increase in pulmonary blood flow in humans (Mortola, 2001). Additionally, lung ventilation stimulates the synthesis and release of prostacyclin, which has a powerful vasodilating effect on the pulmonary vascular bed (Leffler et al., 1984a,b; Velvis et al., 1991).
View article
Read full article
URL:
Journal2007, Comparative Biochemistry and Physiology Part A: Molecular & Integrative PhysiologyMichael B. Thompson
Chapter
Respiratory Effects of Anesthesia and Sedation
2008, Pediatric Respiratory Medicine (Second Edition)Daniel J. Weiner, ... Julian Lewis Allen
PULMONARY BLOOD FLOW AND PULMONARY VASCULAR RESISTANCE
In contrast to the distribution of ventilation, lung perfusion increases from the upper to the more dependent regions of the lung, except at the bottom-most region, in which pulmonary blood flow is also diminished (see Fig. 26-6). The vertical distribution of blood flow is similar to that in the awake state and is related to the hydrostatic gradient.
Hypoxic pulmonary vasoconstriction is diminished by up to 50%, but not abolished, by several inhalational anesthetics; this may not be the case with intravenous anesthetics such as barbiturates. Loss of hypoxic pulmonary vasoconstriction, while somewhat of a protection against rises in pulmonary artery pressure engendered by alveolar hypoxia, leads to regions of low /.
View chapterExplore book
Read full chapter
URL:
Book2008, Pediatric Respiratory Medicine (Second Edition)Daniel J. Weiner, ... Julian Lewis Allen
Chapter
Monitoring hemodynamics, gas exchange, and pulmonary mechanics during veno-arterial bypass with extracorporeal oxygenation
1976, Physiological and Clinical Aspects of Oxygenator DesignF. Lemaire, ... G. Atlan
CONCLUSIONS AND SUMMARY
Some accurate measurements allow a better understanding of the physiological changes occurring during extracorporeal circulation and may lead to an improved mode of perfusion.
The actual pulmonary blood flow may be measured with the usual dilution techniques only if it be possible to rule out an upstream aspiration of the indicator by the drainage cannulas, and a mixing of the different blood flows before the sampling site. Injection in the pulmonary artery and sampling from the left ventricle may obviate these difficulties.
Left radial artery may be used as an index of aortic arch oxygenation, and left ventricular under 100% oxygen ventilation, as an index of lung oxygen transfer. Intra-pulmonary shunting () has to be interpreted in relation to the pulmonary blood flowrate, and calculated pulmonary vascular resistances.
Tests of pulmonary mechanics provide a sensitive index of the condition of the lung. The measurement of static compliance offers a reliable method and is easy to perform under extracorporeal oxygenation. Data concerning the dynamic compliance are more controversial.
View chapterExplore book
Read full chapter
URL:
Book1976, Physiological and Clinical Aspects of Oxygenator DesignF. Lemaire, ... G. Atlan
Chapter
Volume 3
2022, Encyclopedia of Respiratory Medicine(Second Edition)R. Naeije, A.J. Peacock
Pulmonary Vascular Function
Pulmonary Vascular Pressures and Blood Flow
Pulmonary blood flow (CO) is driven by a pressure difference between an inflow pressure, or mean pulmonary artery pressure (mPAP) and an outflow pressure, or mean left atrial pressure (LAP). From these measurements, resistance to flow in the pulmonary circulation can be approximated by a single number, pulmonary vascular resistance (PVR) which depends on the ratio between (mPAP—LAP) and CO:
In clinical practice, measurements of pulmonary vascular pressures and pulmonary blood flow are usually performed during a catheterization of the right heart with a triple-lumen fluid-filled balloon and thermistor-tipped catheter introduced by Swan, Ganz, Forrester and their colleagues in the early 1970s (Swan et al., 1970; Forrester et al., 1972) Fig. 2.
The procedure allows for the estimation of LAP from a balloon-occluded or wedged PAP (PAOP or PAWP). The fractal structure of the pulmonary arterial and venous trees allows for a stop-flow phenomenon downstream of occlusion and de facto extension of the fluid-filled lumen of the catheter to same diameter pulmonary veins. With injection of 5–10 mL cooled saline into a lumen opening into the right atrium, a temperature change over time in the pulmonary artery is registered and CO is calculated from this thermal dilution curve.
Methodological Aspects
Fluid-filled thermodilution catheter measurements of pulmonary vascular pressures and CO are accurate but may suffer from an inherent lack of precision (Naeije et al., 2015). Errors on the measurements of PAP with fluid-filled catheters as compared to gold standard high fidelity micromanometer-tipped catheters range from − 8 to + 8 mmHg (Pagnamenta et al., 2013). Errors on the measurement of thermodilution CO when compared to the gold standard Fick method range from − 1 to + 1 L/min (Hoeper et al., 1999). The magnitude of these errors is not dependent of the absolute value of PAP or CO being measured (Naeije et al., 2015). The estimation of left ventricular end-diastolic pressure (LVEDP) from PAWP is near-accurate—with LVEDP reported on average 3 mmHg higher—but is also unprecise, with errors ranging from − 15 to + 9 mmHg (Halpern and Taichman, 2009). Accurate but imprecise measurements are suitable for population studies, less so for individual decision making (Naeije et al., 2015).
Fluid-filled catheters measure vascular pressures with an external manometer which requires zero leveling. For this purpose, the best reference is at the cross-section of three trans-thoracic planes respectively mid-chest frontal, transverse through the fourth intercostal space, and sagittal (Kovacs et al., 2014). This is close to the hydrostatic indifference point, where pressure is independent of body position.
Pulmonary vascular function is affected by changes in lung volumes and intrathoracic pressures. PVR is minimal at functional residual capacity (FRC) with the glottis open. Lung volumes below or above FRC are associated with increased PVR, because of increased alveolar vessel resistance at high lung volumes, and increased extra-alveolar vessel resistance at lower lung volumes (Howell et al., 1961). Pleural pressure is slightly sub-atmospheric, on average at − 3 mmHg at FRC. Pleural pressure decreases at inspiration and increases at expiration. Pulmonary vascular pressures, systolic and diastolic right heart pressures, and diastolic but not systolic left heart pressures follow the changes in pleural pressure (LeVarge et al., 2014). Expiration may be associated with dynamic hyper-inflation, increased FRC and positive alveolar pressures which are transmitted to pleural pressure. Respiratory swings of pulmonary vascular pressures are illustrated in Fig. 3. During normal breathing, pulmonary vascular pressures are therefore best averaged over several respiratory cycles (Kovacs et al., 2014).
Adequate measurement of a PAWP requires attention to wave morphology. For the estimation of LVEDP, it is recommended that the tracing is read at the nadir preceding the c wave, or use a QRS-gated approach (130–160 m s after the Q) and omit of course the V wave which is systolic (Naeije and Chin, 2019). However, for the estimation of the outflow pressure of the pulmonary circulation and calculation of PVR it is preferable to average the reading of a PAWP tracing over the cardiac cycle as the V wave pressure is transmitted upstream to PAP. Omission of the V wave could markedly over-estimate PVR.
How to read a PAWP taking into account cardiac cycle and respiratory cycle variability is illustrated in Fig. 3.
Sometimes a measurement of LAP or PAWP cannot be obtained, and a total PVR (TPR) is calculated as:
Since LAP is not negligible with respect to mPAP, TPR is larger than PVR and this difference may be flow-dependent. Thus TPR is not a correct characterization of the flow-resistive properties of the pulmonary circulation.
Pulmonary Capillary Pressure
Micropuncture studies have shown that pulmonary capillary pressure (PCP) is higher than PAWP (Battacharya et al., 1982). Thus PAWP should not be called “capillary” or “capillary-wedge” pressure. Estimates of PCP can be obtained from the analysis of PAP decay curves after arterial occlusion, as illustrated in Fig. 4.
Pulmonary artery pressure decay curves present with a rapid first portion, corresponding to the interruption of flow through arterial resistance, and a slow capillary-venous portion corresponding to the emptying of pulmonary capillary compliance through venous resistance. The extrapolation to the moment of occlusion of the slow component of the pressure decay curve defines PCP (Cope et al., 1992).
Based on measured distribution of resistances in perfused normal lungs, with 60% arterial resistance and 40% capillary + venous resistance, PCP can be estimated from a simple equation proposed by Gaar et al. (1967):
Measurements of PCP by single PA occlusion in healthy volunteers show an average PCP of 10, ranging from 8 to 12 mmHg (Maggiorini et al., 2001).
Limits of Normal
The limits of normal of the human pulmonary circulation derived from invasive measurements reported in 47 studies on a total of 1187 individuals, of whom 225 were identified as women and 717 as men (Kovacs et al., 2009), are shown in Table 1.
Table 1. Limits of normal human pulmonary vascular pressures and pulmonary blood flow at rest.
| Variables | Mean | Limits of normal |
---
| CO L/min | 7.3 | 5.0–9.5 |
| HR, bpm | 76 | 50–100 |
| PAP systolic, mmHg | 21 | 13–29 |
| PAP diastolic, mmHg | 9 | 3–15 |
| PAP, mean, mmHg | 14 | 8–20 |
| PAWP, mmHg | 8 | 2–14 |
| PCP, mmHg | 10 | 8–12 |
| RAP, mmHg | 5 | 1–8 |
| PVR, Wood units | 0.9 | 0.2–1.8 |
Naeije R (2016) Pulmonary vascular function. In: AJ Peacock, R Naeije and LJ Rubin (eds.), Pulmonary Circulation: Diseases and Their Treatment, 4nd edn., pp. 11–24, CRC Press: Boca Raton, FL, chapter 2.
The Calculation of PVR
A vascular resistance calculation derives from the physical law that governs laminar flows of Newtonian fluids through non distensible, straight cylindric tubes, originally proposed by the French physician Poiseuille and later put in mathematical equation by the German physicist Hagen. Poiseuille showed experimentally that flow was inversely related to the fourth power of the internal radius, and confirmed previous demonstrations in animals that arterial pressure remains high in arteries down to 2 mm in diameter while venous pressure is low. The Hagen-Poiseuille law states that resistance R to flow of a single tube is equal to the product of the length l of the tube and viscosity η and a constant 8 divided by the product of π and the fourth power of the internal radius r. More generally R can be calculated as a pressure drop ΔP to flow Q ratio:
The ratio of pressure drop to flow through an entire vascular bed accounts for the resistances in series and in parallel of the individual vessels. The fact that r in the equation is to the fourth power explains why R is exquisitely sensitive to small changes in caliber of these small vessels (a 10% change in radius results in almost 50% change in resistance). Accordingly, PVR is a good indicator of the state of constriction or dilatation of pulmonary resistive vessels and is useful for detecting changes in arteriolar vessel caliber due to changes in tone and/or structure.
Effects of Age, Sex and Body Position
There were no discernible sex differences in pulmonary vascular pressures and body-size corrected cardiac output reported invasive measurements in 225 women and 717 men (Kovacs et al., 2009). Aging is associated with an increase in PVR. This is due to a slight increase in mPAP and a more important decrease in cardiac output leading to a doubling of PVR over a five decades of life (Kovacs et al., 2012). Body position affects PVR through associated changes in systemic venous return. In the upright position, LAP, right atrial pressure (RAP) and CO are lower than in the supine position. This results in a de-recruitment of pulmonary vessels, and mPAP remains essentially the same. Accordingly, PVR is increased at rest in the upright position (Forton et al., 2016). When cardiac output is increased during an incremental exercise test, body position does not affect the slope of pulmonary vascular pressure-flow relationships (Kovacs et al., 2012; Forton et al., 2016).
Limits of Normal and Definition of Pulmonary Hypertension
Currently, there is expert consensus to define pulmonary hypertension as a mPAP > 20 mmHg, pre-capillary pulmonary hypertension by a mPAP > 20 mmHg with PAWP ≤ 15 mmHg and PVR ≥ 3 Wood units (WU) and post-capillary pulmonary hypertension by a mPAP > 20 mmHg with PAWP > 15 mmHg and PVR < 3 WU (Simonneau et al., 2019). Data from large population registries show that a mPAP > 20 mmHg and a PVR > 2 WU are independently associated with a decreased life expectancy (Maron et al., 2016, 2020). A PVR at the upper limit of normal of 2 WU can be modeled to correspond to 20–25% obstruction of the pulmonary circulation at a normal cardiac output (Mélot et al., 1995).
View chapterExplore book
Read full chapter
URL:
Reference work2022, Encyclopedia of Respiratory Medicine(Second Edition)R. Naeije, A.J. Peacock
Related terms:
Nitric Oxide
Venous Return
Blood Pressure
Blood Flow
Lung Vascular Resistance
Lung Artery Pressure
Heart Output
Perfusion
Lamb
Lung Circulation
View all Topics |
7720 | https://arxiv.org/abs/2401.16153 | [2401.16153] On the best constants in Khintchine type inequalities for martingales
The Scheduled Database Maintenance 2025-09-17 11am-1pm UTC has been completed
The scheduled database maintenance has been completed.
We recommend that all users logout and login again..
Blog post
Skip to main content
We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate
>math> arXiv:2401.16153
Help | Advanced Search
Search
GO
quick links
Login
Help Pages
About
Mathematics > Probability
arXiv:2401.16153 (math)
[Submitted on 29 Jan 2024]
Title:On the best constants in Khintchine type inequalities for martingales
Authors:Grigori A. Karagulyan
View a PDF of the paper titled On the best constants in Khintchine type inequalities for martingales, by Grigori A. Karagulyan
View PDF
Abstract:For discrete martingale-difference sequences $d={d_1,\ldots,d_n}$ we consider Khintchine type inequalities, involving certain square function $\mathfrak S (d)$ considered by Chang-Wilson-Wolff in 1982. In particular, we prove
\begin{equation}
\left\|\sum_{k=1}^nd_k\right\|p\le 2^{1/2}\big(\Gamma((p+1)/2))/\sqrt{\pi}\big)^{1/p}\|\mathfrak S(d)\|\infty,\quad p\ge 3,
\end{equation}
where the constant on the right hand side is the best possible and the same as known for the Rademacher sums $\sum_{k=1}^na_kr_k$. Moreover, for a fixed $n$ the constant in the inequality can be replaced by $\sum_{k=1}^nr_k/\sqrt{n}$. We apply a technique, reducing the general case to the case of Haar and Rademacher sums, that allows also establish a sub-Gaussian estimate
\begin{equation}
{\bf E}\left[\exp\left(\lambda\cdot \left(\frac{\sum_{k=1}^nd_k}{\|\mathfrak S(d)\|_\infty}\right)^2\right)\right]\le \frac{1}{\sqrt{1-2\lambda}},\quad 0<\lambda<1/2,
\end{equation}
where the constant on the right hand side is the best possible.
Comments:16 pages
Subjects:Probability (math.PR); Classical Analysis and ODEs (math.CA)
MSC classes:42C05, 42C10, 60G42
Cite as:arXiv:2401.16153 [math.PR]
(or arXiv:2401.16153v1 [math.PR] for this version)
Focus to learn more
arXiv-issued DOI via DataCite
Submission history
From: Grigori Karagulyan [view email]
[v1] Mon, 29 Jan 2024 13:35:09 UTC (15 KB)
Full-text links:
Access Paper:
View a PDF of the paper titled On the best constants in Khintchine type inequalities for martingales, by Grigori A. Karagulyan
View PDF
TeX Source
Other Formats
view license
Current browse context:
math.PR
<prev | next>
new | recent | 2024-01
Change to browse by:
math
math.CA
References & Citations
NASA ADS
Google Scholar
Semantic Scholar
export BibTeX citation Loading...
BibTeX formatted citation
×
Data provided by:
Bookmark
Bibliographic Tools
Bibliographic and Citation Tools
[x] Bibliographic Explorer Toggle
Bibliographic Explorer (What is the Explorer?)
[x] Connected Papers Toggle
Connected Papers (What is Connected Papers?)
[x] Litmaps Toggle
Litmaps (What is Litmaps?)
[x] scite.ai Toggle
scite Smart Citations (What are Smart Citations?)
Code, Data, Media
Code, Data and Media Associated with this Article
[x] alphaXiv Toggle
alphaXiv (What is alphaXiv?)
[x] Links to Code Toggle
CatalyzeX Code Finder for Papers (What is CatalyzeX?)
[x] DagsHub Toggle
DagsHub (What is DagsHub?)
[x] GotitPub Toggle
Gotit.pub (What is GotitPub?)
[x] Huggingface Toggle
Hugging Face (What is Huggingface?)
[x] Links to Code Toggle
Papers with Code (What is Papers with Code?)
[x] ScienceCast Toggle
ScienceCast (What is ScienceCast?)
Demos
Demos
[x] Replicate Toggle
Replicate (What is Replicate?)
[x] Spaces Toggle
Hugging Face Spaces (What is Spaces?)
[x] Spaces Toggle
TXYZ.AI (What is TXYZ.AI?)
Related Papers
Recommenders and Search Tools
[x] Link to Influence Flower
Influence Flower (What are Influence Flowers?)
[x] Core recommender toggle
CORE Recommender (What is CORE?)
Author
Venue
Institution
Topic
About arXivLabs
arXivLabs: experimental projects with community collaborators
arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.
Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.
Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs.
Which authors of this paper are endorsers? | Disable MathJax (What is MathJax?)
About
Help
Contact
Subscribe
Copyright
Privacy Policy
Web Accessibility Assistance
arXiv Operational Status
Get status notifications via email or slack |
7721 | https://math.stackexchange.com/questions/478375/mathematical-formula-to-find-adjacent-items-in-a-grid | Skip to main content
Mathematical formula to find adjacent items in a grid
Ask Question
Asked
Modified
12 years ago
Viewed
2k times
This question shows research effort; it is useful and clear
0
Save this question.
Show activity on this post.
I have a 3x3 grid of dots. Selecting any one of the 9 dots, I need to find out which of the remaining dots are adjacent to the first dot. So, if for example we chose the first dot in the first row (dot 0, numbering them from 0, left to right, moving down the rows), the adjacent dots would be dot 1,3,and 4:
[X, 1, 2]
[3, 4, 5]
[6, 7, 8]
and if we chose dot 4, all of the other dots would be adjacent.
[0, 1, 2]
[3, X, 5]
[6, 7, 8]
Is there a mathematical formula to calculate this adjacency?
Ideally, what I'd like to do is extend this concept to find a formula in which a user must select 5 dots in total, always picking an adjacent dot when available, and not permitted to reselect a dot, such as this:
[X, 1, 2]
[3, 4, 5]
[6, 7, 8]
[X, X, 2]
[3, 4, 5]
[6, 7, 8]
[X, X, 2]
[X, 4, 5]
[6, 7, 8]
[X, X, 2]
[X, X, 5]
[6, 7, 8]
[X, X, 2]
[X, X, X]
[6, 7, 8]
Additionally, calculating the number of possibilities available would be useful.
probability
graph-theory
recurrence-relations
Share
CC BY-SA 3.0
Follow this question to receive notifications
edited Aug 28, 2013 at 18:37
mheavers
asked Aug 28, 2013 at 18:27
mheaversmheavers
11555 bronze badges
Add a comment
|
2 Answers 2
Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
Two dots are adjacent (ignoring diagonals) if they agree on one coordinate, and differ by one in the other.
Two dots are adjacent diagonally if they differ by one in each coordinate.
This can be expressed by the inequalities
|x1−x2|≤1and|y1−y2|≤1,
where (x1,y1) and (x2,y2) are the coordinates of the points in question.
Share
CC BY-SA 3.0
Follow this answer to receive notifications
answered Aug 28, 2013 at 18:32
Nick PetersonNick Peterson
33.1k44 gold badges6060 silver badges7878 bronze badges
Add a comment
|
This answer is useful
1
Save this answer.
Show activity on this post.
Let c be the number of columns in your array and n the dot number that you are looking for neighbors for. The dot to the left (if there is one) is n−1. You can check if there is one by checking if n is divisible by c. If it is, you are in the left column. The dot to the right is n+1 and there will be one unless n+1 is divisible by c. The dot above is n−c and there is one if that is positive. The dot below is n+c and there is on as long as that is not greater than the total number of dots. The diagonal ones are combinations of this-can you see how to do that?
Share
CC BY-SA 3.0
Follow this answer to receive notifications
answered Aug 28, 2013 at 18:31
Ross MillikanRoss Millikan
383k2828 gold badges262262 silver badges472472 bronze badges
Add a comment
|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
probability
graph-theory
recurrence-relations
See similar questions with these tags.
Featured on Meta
Community help needed to clean up goo.gl links (by August 25)
Related
4
Most likely number of matching cards when two people select different numbers of cards from their own packs
1
Eigenvalues of periodic lattice Laplacian?
2
Tough Probability Question: Visiting 4 Friends, Finding Probability Mass Function
2
Sampling with replacement with multiple attempts
4
Probability of Adjacent Birthdays
Hot Network Questions
Setting minimum scale when rendering vector layer in
Is neutrally skewed the correct interpretation of a box plot with equal length arms?
How do computer models, languages, and algorithms relate to philosophy?
How to do Lucas' quest in Lil Gator Game?
Switching bodies to get everyone back to their correct body?
How do ordinals manage to descend finitely?
Is pre-Greek Minoan?
How to make Rsync newly write all file perms, ownership etc. only?
A week after the Putin-Trump meeting at Anchorage, what's the quick elevator pitch on proposals currently publicly on the table?
Is "from the rising of the sun, and from the west" a valid translation in Isaiah 45:6?
Relation between the truth of P→Q when P is false and the principle of explosion
long division symbol between number and fraction
mesh shading problem with "flat shade"
C++17: shared_mutex protected DNS cache (rare writes, many reads) return-by-value vs decltype(auto)?
Where does the voting occur in a 3 Computer system?
How did Henry know that Jimmy was going to have him killed in Florida?
Why do winding numbers replace two-chains in the proof of Cauchy's theorem?
Can a minor armed attack justify a large-scale military response under international law?
Understanding the practical application of Intel's _mm256_shuffle_epi8 definition
Memorizing inconsistent key names in pgfplots
"Intermediate value theorem" for manifolds
Shebang apparently calls wrong shell
How do you formally indicate to repeat the second ending of a volta (or ending)?
Why does he have a phobia on these dates?
Question feed
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. |
7722 | https://www.hhhh.org/perseant/libellus/aides/allgre/allgre.190.html | Allen and Greenough's New Latin Grammar, section 190
Heavy Construction
The Allen and Greenough is still under construction; so some links may not work quite the way you would expect.
DEPONENT VERBS.
190.
Deponent Verbs have the forms of the Passive Voice, with an active or reflexive signification: -
PRINCIPAL PARTS
First conjugation: míror, mírárí, mírátus, admire.
Second conjugation: vereor, verérí, veritus, fear.
Third conjugation: sequor, sequí, secútus, follow.
Fourth conjugation: partior, partírí, partítus, share.
INDICATIVE
PRES.mírorvereorsequorpartior
míráris (-re)veréris (-re)sequeris (-re)partíris (-re)
míráturverétursequiturpartítur
mírámurverémursequimurpartímur
míráminíveréminísequiminípartíminí
míranturverentursequunturpartiuntur
IMPF.mírábarverébarsequébarpartiébar
FUT.míráborveréborsequarpartiar
PERF.mírátus sumveritus sumsecútus sumpartítus sum
PLUP.mírátus eramveritus eramsecútus erampartítus eram
F. P.mírátus eróveritus erósecútus erópartítus eró
SUBJUNCTIVE PRES.mírer
IMPF.mírárerverérersequererpartírer
PERF.mírátus simveritus simsecútus simpartítus sim
PLUP.mírátus essemveritus essemsecútus essempartítus essem
IMPERATIVE
PRES.míráreveréresequerepartíre
FUT.mírátorverétorsequitorpartítor
INFINITIVE
PRES.míráríverérísequípartírí
PERF.mírátus esseveritus essesecútus essepartítus esse
FUT.mírátúrus esseveritúrus essesecútúrus essepartítúrus esse
PARTICIPLES
PRES.míránsverénssequénspartiéns
FUT.mírátúrusveritúrussecútúruspartítúrus
PERF.mírátusveritussecútuspartítus
GER.mírandusverendussequenduspartiendus
GERUND mírandí, -ó, etc.verendí, etc.sequendí, etc.partiendí, etc.
SUPINEmírátum, -túveritum, -túsecútum, -túpartítum, -tú
a. Deponents have the participles of both voices: -
sequéns, following.secútúrus, about to follow.
secútus, having followed.sequendus, to be followed.
b. The perfect participle generally has an active sense, but in verbs otherwise deponent it is often passive: as, mercátus, bought;adeptus, gained (or having gained).
c. The future infinitive is always in the active form: thus, sequor has secútúrus (-a, -um) esse (not secútum írí).
d. The gerundive, being passive in meaning, is found only in transitive verbs, or intransitive verbs used impersonally: -
h=oc cónfitendum est, this must be acknowledged.
moriendum est omnibus, all must die.
e. Most deponents are intransitive or reflexive in meaning, corresponding to what in Greek is called the Middle Voice (§ 156. a. N.).
f. Some deponents are occasionally used in a passive sense: as, críminor, I accuse, or I am accused.
g. About twenty verbs have an active meaning in both active and passive forms: as, mereó or mereor, I deserve.
|
7723 | https://en.wikipedia.org/wiki/Coxsackie_A_virus | Published Time: 2002-02-01T01:50:06Z
Coxsackie A virus - Wikipedia
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigation
Main page
Contents
Current events
Random article
About Wikipedia
Contact us
Contribute
Help
Learn to edit
Community portal
Recent changes
Upload file
Special pages
Search
Search
[x] Appearance
Donate
Create account
Log in
[x] Personal tools
Donate
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
Contents
move to sidebar hide
(Top)
1 Viral structure and genome
2 Replication cycle
3 Diseases
4 Signs and Symptoms
5 Outbreaks
6 Pregnancy
7 Transmission
8 Prevention
9 Prognosis
10 Treatment
11 See also
12 References
[x] Toggle the table of contents
Coxsackie A virus
[x] 8 languages
العربية
Čeština
Deutsch
Español
Français
Italiano
Polski
Simple English
Edit links
Article
Talk
[x] English
Read
Edit
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikispecies
Wikidata item
Appearance
move to sidebar hide
From Wikipedia, the free encyclopedia
Virus that causes hand, foot and mouth disease
| Coxsackie A virus |
| Virus classification |
| (unranked): | Virus |
| Realm: | Riboviria |
| Kingdom: | Orthornavirae |
| Phylum: | Pisuviricota |
| Class: | Pisoniviricetes |
| Order: | Picornavirales |
| Family: | Picornaviridae |
| Genus: | Enterovirus |
| Species: | Enterovirus A |
| Strain: | Coxsackie A virus |
A transmission electron microscopic image depicting virions causing acute hemorrhagic conjunctivitis, primarily caused by two enteroviruses: enterovirus 70, and a variant of coxsackievirus A24.
Coxsackie A virus (CAV) is a cytolyticCoxsackievirus of the Picornaviridae family, an enterovirus (a group containing the polioviruses, coxsackieviruses, and echoviruses).
Viral structure and genome
[edit]
Coxsackie A virus is a subgroup of enterovirus A, which are small, non-enveloped, positive-sense, single-stranded RNA viruses. Its protective, icosahedral capsid has an external portion that contains sixty copies of viral proteins (VP1,-2,-3) and an internal portion surrounding the RNA genome containing sixty copies of VP4 viral proteins. This capsid mediates cell entry and elicits the humoral immune responses. Enteroviruses have a depression encircling each fivefold axis (canyon), which is their binding site for immunoglobulin-like receptors. This binding can trigger viral expansion and release of its genome.
A complete genome analysis of Coxsackie virus A2, A4, A5, and A10 strains isolated from individuals with hand-foot-mouth disease showed that natural recombination is frequent in the virus's evolution. Its strains in China were related to strains in Mongolia, Taiwan, likely to those that circulated in Europe, and form a distinct lineage from strains imported from Japan and South Korea.
Replication cycle
[edit]
Replication of the coxsackie virus happens through contributions of the host and virus components. The virus enters the cell where it is internalized into endoplasmic reticulum and the Golgi apparatus. After viral un-coating, viral RNA is released. Ribosomes on the rough endoplasmic reticulum translate the RNA into viral polyprotein. This polyprotein is processed into structural protein P1 and non-structural proteins P2 and P3. Via the virus-encoded proteinase, P1 is processed into the viral capsid subunit proteins VP0, -1, -3. The 5'-non-coding region contains sequences that control genome replication and translation, while the 3'-non-coding region contains polyA tail needed for virus infectivity.
Diseases
[edit]
Coxsackievirus infection on skin of adult
The most well known Coxsackie A disease is hand, foot and mouth disease (unrelated to foot-and-mouth disease), a common childhood illness which affects mostly children aged 5 or under, often produced by Coxsackie A16. In most individuals, infection is asymptomatic or causes only mild symptoms. In others, infection produces short-lived (7–10 days) fever and painful blisters in the mouth (a condition known as herpangina), on the palms and fingers of the hand, or on the soles of the feet. There can also be blisters in the throat, or on or above the tonsils. Adults can also be affected. The rash, which can appear several days after high temperature and painful sore throat, can be itchy and painful, especially on the hands/fingers and bottom of feet.
Other diseases include acute haemorrhagic conjunctivitis (A24 specifically), herpangina, and aseptic meningitis (both Coxsackie A and B viruses). Coxsackievirus A7 is associated with neurological diseases and can cause paralytic poliomyelitis.
Signs and Symptoms
[edit]
Coxsackie A virus leads to a number of diseases, however the most common signs and symptoms that appear with infection are fever and flu-like symptoms, mouth sores, and skin rashes. People who are infected may present with a mild fever and sore throat, and a general discomfort three to six days subsequent to exposure. Painful mouth sores (herpangina) may be present in the back of the mouth. These sores usually appear 24 hours after the flu like symptoms begin, and may blister, causing further discomfort when eating or drinking. A flat, red skin rash may appear, commonly accompanied by fluid filled blisters and scabbing. Rash will commonly present on the bottom of the feet, palm of hands, and other areas of the body, and persists upwards to ten days.
When the symptoms are incredibly severe, some may require hospitalization due to dehydration caused by inability to swallow food or water without pain, or seizures and convulsions can occur due to high fever. Signs of dehydration include dry skin, unintentional weight loss, or decreased urine output/darkened urine and if present should refer to a health care provider for intervention. Other serious complications include inflammatory brain conditions, such as viral meningitis or encephalitis, which require medical intervention. A professional health care may need to monitor if the someone infected is immunocompromised, or the symptoms do not improve within ten days.
The diagnosis of this disease centers around the appearance and behavior of fever, rash and mouth sores. Outside of the symptoms, age is also taken into consideration as the most common age of infection is under five years of age. A healthcare professional may choose to confirm the diagnosis through collecting samples from mouth sores and skin blisters, or a stool sample may also be ordered to rule out any other causes.
Outbreaks
[edit]
Since 2008, coxsackievirus A6 (CVA6) has been associated with several worldwide outbreaks of hand, foot and mouth disease (HFMD). In Finland, the initial HFMD case caused by the CVA6 lead to its identification of being the responsible pathogen of the outbreaks in Europe, North America, and Asia. Coxsackievirus A16 (CVA16) has also been linked to HFMD.
Outbreaks are more commonly seen amongst children (those seven and younger) compared to outbreaks rates amongst adults. Due to this, there are outbreaks within daycares, summer camps, and early autumn.
Pregnancy
[edit]
Serious pregnancy complications due to hand, foot, and mouth disease are rare due to its limited data. However, HFMD is concerning if the mother contracts the virus at the end of her pregnancy. CVA16 infection has been associated with third trimester massive perivillous fibrin deposition leading to intrauterine death. It has also led to first trimester spontaneous abortions. Overall, there is limited information about the Coxsackievirus A strain in pregnant women.
On the other hand, there has been some reports of the Coxsackievirus B (CVB) with regards to pregnant women. Contraction of the CVB are not associated with a higher risk of spontaneous abortions. Although, complications at the end of the pregnancy carries an increased risk of stillbirth or HFMD in the child. There have been reports of congenital heart defects and urogenital anomalies within the newborns of women who seroconverted to CVB during pregnancy. CVB is responsible for up to half of all individuals with pediatric myocarditis. In the past, it has been stated that newborns who have contracted CVB have a 75% mortality from myocarditis.
Transmission
[edit]
The Coxsackie A Virus is a highly contagious virus that commonly causes the mild hand, foot, and mouth disease but complications may lead to more serious diseases that can affect the heart, lungs, muscles, and more.[citation needed] The modes of transmission of the Coxsackie virus is primarily through contact between people, respiratory droplets (fluid from coughing and sneezing), and contaminated surfaces. All age groups can become infected with the Coxsackie virus, however it occurs most frequently in young children under the age of 10 and in who have a weakened immune system.
The main ways the Coxsackie Virus spreads are:
By direct transmission (when an infected person coughs or sneezes into any mucous membranes of the face i.e. eyes, nose, and mouth)
Fecal-oral transmission (virus in the fecal matter of an infected individual end up in the mouth of another person)
Via surface contact (when an infected person touches their face then a surface, the surface is contaminated with that virus. The next person comes along and touches the same surface then touches their face)
Via airborne transmission (when an uninfected person inhales respiratory droplets of an infected person)
Although adults are less susceptible to infection, it is still possible for an adult to get infected with the Coxsackie virus. If a pregnant mother is infected, there is a 30-50% chance the infection will be passed on to the infant.
Prevention
[edit]
There is no vaccine to reduce the chances of infection and spread. It is critical to use non-pharmacological interventions to reduce the spread and transmission of Coxsackie virus. The best and most effective strategy for prevention is adopting proper hand hygiene, avoiding contact with the infected, abstain from touching mucous membranes of the face, and sanitizing frequently touched surfaces.
Prognosis
[edit]
Some of those who are infected with the Coxsackie virus may have complications that may lead to more serious issues. Complications include stomatitis, meningitis, pulmonary edema, myocarditis, pneumonia, and possibly spontaneous abortions.
Treatment
[edit]
Treatment is dependent on the disease process initiated by the virus. There is no known cure or vaccine against this virus.
Most Coxsackie A virus infections are mild and self-limiting meaning the infection has the ability to resolve on its own without requiring treatment. Symptoms of a Coxsackie A infection tend to dissipate on their own within 7–10 days. Treatment tends to focus on supportive care where the symptoms of the infection are targeted but not the virus itself. NSAIDs such as ibuprofen/naproxen and acetaminophen can be used to manage the flu-like symptoms, fever, and any other pain the infected individual may be feeling. Do not give a child aspirin as it may increase the risk of Reye syndrome. Fluids are recommended as to decrease the chances of dehydration. Mouth sores will make eating and drinking painful and may potentially lead to loss of appetite and refusing to eat in order to prevent mouth and throat pain. Severe dehydration may lead to hospitalization. Additionally, topical oral analgesic medications or salt water rinses can be used to help numb the sores and ease the throat pain. Since Coxsackie A is a viral infection, antibiotics will have no effect on the infection as they only work on bacterial infections.
See also
[edit]
Bornholm disease
Coxsackie B virus
References
[edit]
^"Details - Public Health Image Library(PHIL)". phil.cdc.gov. Retrieved 2020-07-31.
^Ren J, Wang X, Zhu L, Hu Z, Gao Q, Yang P, et al. (October 2015). "Structures of Coxsackievirus A16 Capsids with Native Antigenicity: Implications for Particle Expansion, Receptor Binding, and Immunogenicity". Journal of Virology. 89 (20): 10500–11. doi:10.1128/JVI.01102-15. PMC4580203. PMID26269176.
^Zhao Y, Zhou D, Ni T, Karia D, Kotecha A, Wang X, et al. (January 2020). "Hand-foot-and-mouth disease virus receptor KREMEN1 binds the canyon of Coxsackie Virus A10". Nature Communications. 11 (1): 38. Bibcode:2020NatCo..11...38Z. doi:10.1038/s41467-019-13936-2. PMC6946704. PMID31911601.
^Hu YF, Yang F, Du J, Dong J, Zhang T, Wu ZQ, et al. (July 2011). "Complete genome analysis of coxsackievirus A2, A4, A5, and A10 strains isolated from hand, foot, and mouth disease patients in China revealing frequent recombination of human enterovirus A". Journal of Clinical Microbiology. 49 (7): 2426–34. doi:10.1128/JCM.00007-11. PMC3147834. PMID21543560.
^Lloyd, Richard (2016-03-30). "Enterovirus Control of Translation and RNA Granule Stress Responses". Viruses. 8 (4): 93. doi:10.3390/v8040093. ISSN1999-4915. PMC4848588. PMID27043612.
^Mao Q, Wang Y, Yao X, Bian L, Wu X, Xu M, Liang Z (2014-02-01). "Coxsackievirus A16: epidemiology, diagnosis, and vaccine". Human Vaccines & Immunotherapeutics. 10 (2): 360–7. doi:10.4161/hv.27087. PMC4185891. PMID24231751.
^"Hand, Foot, & Mouth Disease". Sarawak. Archived from the original on 2008-05-12.
^ abcdCDC (2020). "Hand, Foot & Mouth Disease Symptoms". U.S. Centers for Disease Control and Prevention. Retrieved 2020-07-30.
^Yamayoshi S, Iizuka S, Yamashita T, Minagawa H, Mizuta K, Okamoto M, et al. (May 2012). "Human SCARB2-dependent infection by coxsackievirus A7, A14, and A16 and enterovirus 71". Journal of Virology. 86 (10): 5686–96. doi:10.1128/JVI.00020-12. PMC3347270. PMID22438546.
^Saguil A, Kane SF, Lauters R, Mercado MG (October 2019). "Hand-Foot-and-Mouth Disease: Rapid Evidence Review". American Family Physician. 100 (7): 408–414. PMID31573162.
^"Coxsackievirus". www.uspharmacist.com. Retrieved 31 July 2020.
^ ab"Hand, Foot and Mouth Disease (HFMD)". Princeton Emergency Management. Archived from the original on 2020-10-20. Retrieved 2020-07-30.
^ abcdefg"Hand-foot-mouth disease". MedlinePlus Medical Encyclopedia. U.S. National Library of Medicine. Retrieved 2020-07-30.
^Ramirez-Fort MK, Downing C, Doan HQ, Benoist F, Oberste MS, Khan F, Tyring SK (August 2014). "Coxsackievirus A6 associated hand, foot and mouth disease in adults: clinical presentation and review of the literature". Journal of Clinical Virology. 60 (4): 381–6. doi:10.1016/j.jcv.2014.04.023. PMID24932735.
^Bian L, Wang Y, Yao X, Mao Q, Xu M, Liang Z (2015). "Coxsackievirus A6: a new emerging pathogen causing hand, foot and mouth disease outbreaks worldwide". Expert Review of Anti-Infective Therapy. 13 (9): 1061–71. doi:10.1586/14787210.2015.1058156. PMID26112307. S2CID31843865.
^ abDeeb M, Beach RA, Kim S (2019). "Onychomadesis following hand, foot, and mouth disease in a pregnant woman: A case report". SAGE Open Medical Case Reports. 7: 2050313X19845202. doi:10.1177/2050313X19845202. PMC6498765. PMID31080597.
^Guerra AM, Orille E, Waseem M (2020). "Hand Foot And Mouth Disease". StatPearls. PMID28613736. Retrieved 11 January 2021.
^CDC (2020-01-31). "Hand, Foot & Mouth Disease Complications". U.S. Centers for Disease Control and Prevention. Retrieved 2020-07-28.
^Yu, Weiming; Tellier, Raymond; Wright, James R. (2015). "Coxsackie Virus A16 Infection of Placenta with Massive Perivillous Fibrin Deposition Leading to Intrauterine Fetal Demise at 36 Weeks Gestation". Pediatric and Developmental Pathology. 18 (4): 331–334. doi:10.2350/15-01-1603-CR.1. ISSN1093-5266. PMID25826430. S2CID24791832.
^ abcModlin JF (June 1988). "Perinatal echovirus and group B coxsackievirus infections". Clinics in Perinatology. 15 (2): 233–46. doi:10.1016/S0095-5108(18)30709-7. PMID2837356.
^Sharma V, Rohrbough SN, Goessling LS, Brar AK, Eghtesady P (2019-11-19). "Abstract 17090: Perinatal Coxsackievirus B4 Infection Leads To Preferential Damage Of The Left Ventricle". Circulation. 140 (Suppl_1): A17090. doi:10.1161/circ.140.suppl_1.17090 (inactive 1 November 2024).{{cite journal}}: CS1 maint: DOI inactive as of November 2024 (link)
^Dancea AB (October 2001). "Myocarditis in infants and children: A review for the paediatrician". Paediatrics & Child Health. 6 (8): 543–5. doi:10.1093/pch/6.8.543. PMC2805590. PMID20084124.
^ abcdefGuerra AM, Emily Orille E, Waseem M (July 2020). "Hand Foot And Mouth Disease". StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing. PMID28613736.
^ abcdCDC (2020). "Hand, Foot & Mouth Disease Causes". U.S. Centers for Disease Control and Prevention. Retrieved 2020-07-31.
^Giachè S, Borchi B, Zammarchi L, Colao MG, Ciccone N, Sterrantino G, et al. (July 2019). "Hand, foot, and mouth disease in pregnancy: 7 years Tuscan experience and literature review". The Journal of Maternal-Fetal & Neonatal Medicine. 34 (9): 1494–1500. doi:10.1080/14767058.2019.1638898. PMID31291796. S2CID195880613.
^"Coxsackievirus Incubation Period, Symptoms, Treatment & Rash". MedicineNet. Retrieved 2020-07-31.
^CDC (2020). "Hand, Foot & Mouth Disease Prevention". U.S. Centers for Disease Control and Prevention. Retrieved 2020-07-31.
^ abCDC (2020). "Hand, Foot & Mouth Disease Treatment". U.S. Centers for Disease Control and Prevention. Retrieved 2020-07-31.
| Taxon identifiers |
| Enterovirus A | Wikidata: Q2725073 Wikispecies: Enterovirus A CoL: 39YKK NCBI: 138948 |
| v t e Skin infections, symptoms and signs related to viruses |
| | DNA virus | | Herpesviridae | | Alpha. | | HSV | Herpes simplex Herpetic whitlow Herpes gladiatorum Herpes simplex keratitis Herpetic sycosis Neonatal herpes simplex Genital herpes Cold sore Eczema herpeticum Herpetiform esophagitis Herpetic gingivostomatitis | | B virus | B virus infection | | VZV | Chickenpox Herpes zoster Herpes zoster oticus Ophthalmic zoster Disseminated herpes zoster Zoster-associated pain Modified varicella-like syndrome | | | Beta. | Human herpesvirus 6/Roseolovirus Exanthema subitum Roseola vaccinia Cytomegalic inclusion disease | | Gamma. | Kaposi's sarcoma-associated herpesvirus Kaposi's sarcoma | | | Poxviridae | | Ortho. | Variola Smallpox Alastrim Monkeypox virus Monkeypox Cowpox Vaccinia virus Vaccinia Generalized vaccinia Eczema vaccinatum Progressive vaccinia Buffalopox | | Para. | Farmyard pox: Milker's nodule Bovine papular stomatitis Pseudocowpox Orf Sealpox | | Other | Yatapoxvirus: Tanapox Yaba monkey tumor virus Molluscum contagiosum virus | | | Papillomaviridae | | HPV | Wart/plantar wart Heck's disease Genital wart giant Laryngeal papillomatosis Butcher's wart Bowenoid papulosis Epidermodysplasia verruciformis Verruca plana Pigmented wart Verrucae palmares et plantares | | Bovine papillomavirus Equine sarcoid | | | Parvoviridae | Parvovirus B19 Erythema infectiosum Reticulocytopenia Papular purpuric gloves and socks syndrome | | Polyomaviridae | Merkel cell polyomavirus Merkel cell carcinoma | | | RNA virus | | Paramyxoviridae | MeV Measles | | Matonaviridae | Rubella virus Rubella Congenital rubella syndrome ("German measles") | | Togaviridae | Alphavirus infection Chikungunya fever | | Picornaviridae | Coxsackie A virus Hand, foot, and mouth disease Herpangina Foot-and-mouth disease virus Boston exanthem disease | | |
| Ungrouped | Asymmetric periflexural exanthem of childhood Post-vaccination follicular eruption Lipschütz ulcer Eruptive pseudoangiomatosis Viral-associated trichodysplasia Gianotti–Crosti syndrome |
Retrieved from "
Categories:
Enteroviruses
Pediatrics
Hidden categories:
CS1 maint: DOI inactive as of November 2024
Articles with short description
Short description is different from Wikidata
Articles with 'species' microformats
All articles with unsourced statements
Articles with unsourced statements from January 2021
Taxonbars on possible non-taxon pages
Taxonbars with multiple manual Wikidata items
Taxonbars without primary Wikidata taxon IDs
This page was last edited on 30 November 2024, at 20:27(UTC).
Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
Privacy policy
About Wikipedia
Disclaimers
Contact Wikipedia
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Search
Search
[x] Toggle the table of contents
Coxsackie A virus
8 languagesAdd topic |
7724 | https://math.stackexchange.com/questions/2920032/inclusion-exclusion-at-least-and-exactly-arrangements | Skip to main content
Inclusion/exclusion, at least and exactly arrangements?
Ask Question
Asked
Modified
3 years, 3 months ago
Viewed
2k times
This question shows research effort; it is useful and clear
6
Save this question.
Show activity on this post.
The question wants to count certain arrangements of the word "ARRANGEMENT":
a) find exactly 2 pairs of consecutive letters?
b) find at least 3 pairs of consecutive letters?
I have the answer given from the tutor but it doesn't make sense to me.
Let's start with the base case:
S2=(11−2×2+2)!(2!)(4−2)(42)：All possible combinations for 2 pairs of consecutive letters are known.
S3=(11−2×3+3)!(2!)(4−3)(43)：All possible combinations for 3 pairs of consecutive letters are known.
S4=(11−2×4+4)!(44)=7!：All possible combinations for 4 pairs of consecutive letters are known.
The equation for exactly m conditions：Em=Sm−(m+11)Sm+1+(m+22)Sm+2.
The equation for at least m conditions：Lm=Sm−(m1)Sm+1+(m+12)Sm+2.
Answer for (a)：E2=S2−(31)S3+(43)S4.
Answer for (b)：L3=S3−(31)S4.
For (a), I don't understand why we need to multiply (31) with S3 and (43) with S4?
If we have S3 that satisfies the requirement for S2 (as three pairs would include two pairs), then wouldn't E2=S2−S3?
For (b), wouldn't the answer just be S3 since we it contains the combination for every triple pair?
I don't understand the given formula used for Em and Lm, mainly the combinatorics part because it looks to me that we already handled that combinations in the calculations of S2, S3, S4.
Could someone please explain the formula and why the answers are as such?
Thanks!
EDIT: I got the question from - start the video at (4.35) for the question
combinatorics
discrete-mathematics
combinations
inclusion-exclusion
Share
CC BY-SA 4.0
Follow this question to receive notifications
edited Aug 31, 2020 at 11:11
linear_combinatori_probabi
1,63011 gold badge1919 silver badges4343 bronze badges
asked Sep 17, 2018 at 9:26
thatguyjonothatguyjono
27122 silver badges1010 bronze badges
0
Add a comment
|
2 Answers 2
Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
Terms used in the answer:
Ak: The set of elements has (at least) property indexed k.
IEP for Inclusion-Exclusion Principle.
Exactly-none IEP: exactly none of the indexed properties got selected via IEP. i.e.
⋂k=1mAk¯¯¯¯¯¯
Long answer to understand the two coefficients:
Let's define Sm: It's a shorthand to simplify the formula of IEP. Assume there are n properties in total, and I is an index set, then we can pick any m≤n:
Sm=∑|an intersection of m sets|=∑|I|=m∣∣∣∣⋂j∈IAj∣∣∣∣
2. E0 gives exactly-none IEP: (Let S0=U):
E0=∑j=0n(−1)j−0(j0)Sj=∑j=0n(−1)jSj=∣∣∣∣⋂j=1nAj¯∣∣∣∣.
for each Sj, the coefficient is (−1)j:
E0=∑j=mn(−1)j−m(jm)Sj,m=0=∑j=0n(−1)j−0(j0)Sj=∑j=0n(−1)jSj.
The (jm) disappears in E0. But for Em, we got more than that. So maybe there are more than one exactly-none IEPs in Em?
3. My attempt to explain Em:
E0=∑j=mn(−1)j−m(jm)Sj
3.1. Observe the first term, j=m. Now Sm gives:
Sm=∑|I|=m∣∣∣∣⋂j∈IAj∣∣∣∣=A1,2,...,m+A1,2,...,(m−1),(m+1)+...+A(n−m+1),(n−m+2),...,n
3.2. Now we calculate E0 once for each term A... as the universe. So we have: (The notation E0,U where U means the universe defined for the calculation)
E0,A1,2,...,mE0,A1,2,...,(m−1),(m+1)⋮E0,A(n−m+1),(n−m+2),...,n=∑j=m+1n(−1)j−mSj=∑j=m+1n(−1)j−mSj=∑j=m+1n(−1)j−mSj
you might have many questions at this stage:
Q1:
Those A... might overlap some of the others?
Yes. But if each E0,A… work, no overlapping. Because Em means exactly-m properties when the calculation is complete.
Q2:
Each E0, ignoring those (−1)j−m, have 1 for each term. Now you're trying to convince me that applying it many times will create (jm), a variable for each term in the resulting Em? By intuition, if you apply it, say 5 times, you should have a 5 for each term?
Yes. I don't know how to explain this either for now.
4. Now, the strangest step, I cannot believe it too:
Let's try to calculate how many Sj are needed when j is given. This is equivalent to finding how many universes include each of them. So here is the term:
(jm)
given any j properties, you choose m of them, you find one universe, i.e. one left-hand-side on the list of 3.2. that includes it on the right-hand-side. This explains the mysterious (jm) of Em.
5. On the other hand, The formula of Lm(Count the elements included in ≥m sets) is:
Lm=∑k=mnEk=∑k=mn∑j=kn(−1)j−k(jk)Sj=∑j=mn(−1)j∑k=mj(−1)k(jk)Sj=∑j=mn(−1)j∑k=mj(−1)k(−1)k−m(−1k−m)Sj=∑j=mn(−1)j−m(j−1j−m)Sj□
So Lm does sum up the coefficients of Em.
The rigorous version of the explanation can be found on the linked question: Combinatorics meaning of Lm=∑nj=m(−1)j−m(j−1m−1)Sj.
Share
CC BY-SA 4.0
Follow this answer to receive notifications
edited May 8, 2022 at 9:55
answered Aug 27, 2020 at 7:53
linear_combinatori_probabilinear_combinatori_probabi
1,63011 gold badge1919 silver badges4343 bronze badges
0
Add a comment
|
This answer is useful
0
Save this answer.
Show activity on this post.
In order to analyse the relationship between Em,Lm and Sm it is convenient to have a closer look at the sets which form the building blocks for these numbers.
In case of problem (a) it boils down to show that according to the formula of Em
E2=∑j=24(−1)j−2(j2)Sj=S2−(31)S3+(42)S4(1)
and we also want to clarify how to derive the binomial coefficients (31) and (42).
The Setting:
Given the word ARRANGEMENT, we consider the set
X={AAEEGMNNRRT,…,ARRANGEMENT,…,TRRNNMGEEAA}
consisting of 11!(2!)4=2494800 permutations with repetitions of the letters from this word.
We introduce a set U of properties
U={PA,PE,PN,PR}
A word in X has property PA∈U if the letter A occurs consecutively and the meaning of the other properties in U is analogously.
Given a set of T⊆U of properties from U we define
E(T) as the number of words in X which have exactly the properties T⊆U, and
L(T) as the number of words in X which have at least the properties T⊆U.
So, for instance ARRANGEMENT contributes to E({PR}) and L({PR}), whereas AAEEGMNNRRT contributes to L({PR}) and U, but not to E({PR}).
The numbers E(T) and L(T) form building blocks for Em and Lm. Since we have
Em is the number of words which have exactly m properties from U,(0≤m≤4) and
Lm is the number of words which have at least m properties from U,(0≤m≤4)
we can write these quantities as
EmLm=∑T⊆U|T|=mE(T)=∑T⊆U|T|≥mE(T)(2.1)(2.2)
Example m=2:
In the case m=2 we have according to (2.1)
E2=∑T⊆U|T|=2E(T)=E({PA,PE})+E({PA,PN})+E({PA,PR})+E({PE,PN})+E({PE,PR})+E({PN,PR})=(42)E({PA,PE})
where the factor (42)=6 can be taken thanks to the symmetry of the words with its respective properties. For L2 we obtain according to (2.2)
L2=∑T⊆U|T|≥2E(T)=∑j=24∑T⊆U|T|=jE(T)=E2+E3+E4(2.3)
The Setting continued:
Now we take a closer look at the quantities Sm. They are calculated for 0≤m≤4 as
Sm=(11−2×m+m)!(2!)4−m(4m)(2.4)
and their meaning is:
Sm is the number of words in X, so that for each subset T⊆U of properties with size |T|=m we take the number of words which have at least m properties from T⊆U, i.e.
Sm=∑T⊆U|T|=mL(T)(2.5)
Indeed, taking for instance m=2 in (2.4) we have
S2=(11−2×2+2)!(2!)4−2(42)
where the factor (42) represents each subset T⊆U of properties with size |T|=2, whereas the factor (11−2×2+2)! in the numerator represents the number of permutations of the other letters besides the selected two pairs. Since there are two more letters which occur twice we have to divide by 24−2=4 as they can't be distinguished.
Sm is given in (2.5) in terms of L(T). We can find a more convenient representation in terms of E(T) as follows. We have in case m=2:
S2=∑T⊆U|T|=2L(T)=∑T⊆R⊆U|T|=2E(R)=∑R⊆UE(R)∑T⊆R|T|=21=∑R⊆UE(R)(|R|2)=∑j=24∑R⊆U|R|=jE(R)(j2)=∑j=24(j2)Ej=E2+(32)E3+(42)E4(2.6)
A derivation of (2.6) in a more general context is given in this answer.
Answer of (a) and (b):
We find in the same way as in (2.4) to following identities:
S2S3S4=∑j=24(j2)Ej=E2+(32)E3+(42)E4=∑j=34(j3)Ej=E3+(43)E4=∑j=44(j4)Ej=E4(3.1)(3.2)(3.3)
These relations enable us to represent E2 in terms of S2,S3 and S4. We obtain
E2=S2−(32)E3−(42)E4=S2−(32)(S3−(43)E4)−(42)E4=S2−(32)S3+((32)(43)−(42))S4=S2−(32)S3+(42)S4(→(3.1))(→(3.2))(→(3.3))
according to result (a). Similarly we obtain with (2.3)
L3=E3+E4=S3−(43)E4+E4=S3−((43)−1)S4=S3−(31)S4(→(3.2))(→(3.3))
according to result (b).
Share
CC BY-SA 4.0
Follow this answer to receive notifications
edited May 19, 2022 at 7:10
answered May 18, 2022 at 20:27
Markus ScheuerMarkus Scheuer
113k77 gold badges106106 silver badges249249 bronze badges
Add a comment
|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
combinatorics
discrete-mathematics
combinations
inclusion-exclusion
See similar questions with these tags.
Featured on Meta
Community help needed to clean up goo.gl links (by August 25)
Linked
10
Combinatorics meaning of Lm=∑nj=m(−1)j−m(j−1m−1)Sj
4
Combinatorics explanation of Inclusion-Exclusion Principle Exactly-m Properties Formula Em=∑nj=m(−1)j−m(jm)Sj
Related
0
Combinatorics: 3 heads in succession in 5 coin tosses, using inclusion exclusion
2
What's the probability of getting a pair of king and ace with the same suit?
4
When drawing 14 cards from a set of 52 cards, is it more likely to have a full house or two consecutive pairs?
3
Identity for a sum of product of binomial coefficients
0
Why are my two answers different for this same combination problem?
3
Contradiction in arranging words in two different ways
3
Arrangements of a,a,b,b,b,c,c,c,c in which no four/three/two consecutive letters are the same
8
Arrangements of a,a,b,b,b,c,c,c,c in which no two consecutive letters are the same
0
(a1,a2,a3,a4); (b1,b2,b3); (c1,c2). The total number of combinations with given condition
Hot Network Questions
How do I fill holes in new pine furniture so that the color will continue to match as the wood ages?
how should one interpret a cert who's Issuer is different from the DirName of the X509v3 Authority Key Identifier extension?
SciFi story about father and son after world is destroyed
Why do these two lines have the same probability of intersecting the circle?
Find an explicit real number that is not in the set produced by the convergent rational series
How Do I Talk of Someone Whose Name Appeared in the Paper
How to balance research and teaching responsibilities?
No /dev/sr0 when booting with customized kernel in qemu
Splitting a long introduction into collapsible elements easier to read and translate
Example of two topologies with same Borel sets but different τ-additivity
Why does the Apollo LM's Cross-Pointer display have a bulge?
Best LaTeX environment to recreate this example
Other than the tank PSI, what else could cause a Whirlpool 3 stage RO Under Sink Water Filtration System to have low water pressure?
Activate the Laser Gates
Film about alien spaceships ominously hovering over Earth, only to form a shield to protect it
Would weekly seasonal changes still allow a habitable planet?
Does 我们员工的休假已经达到了150多天 include weekends as holidays?
Simple present tense of "wear" vs. present progressive form of "wear"
Will NASA send Juno to investigate 3i/Atlas?
Why does this Association show its Head, and how do I make it evaluate?
Which passport to use at immigration counter
A friend told me he is flying into Birmingham, Alabama, on September 29, 2025. Is this the correct ticket?
Can you get valid SSL certificates for hosting local services over a domain that you own?
Fewest cages for unique Killer Kropki Sudoku
Question feed |
7725 | https://www.math.utoronto.ca/mathnet/plain/questionCorner/tetangle.html | Question Corner -- Angle Between Vertices of a Tetrahedron
Navigation Panel: Previous | Up | Forward | Graphical Version | PostScript version | U of T Math Network Home
University of Toronto Mathematics Network
Question Corner and Discussion Area
Angle Between Vertices of a Tetrahedron
Asked by Lee Lude, student, Michigan City High School on February 5, 1998:
Given a regular tetrahedral with a point in the center, find the angle formed from this center point to two corners (next to each other) in the tetrahedral. In chemistry terms you are actually proving that the the electrons in a tetrahedral shaped molecule are 109.5 degrees apart. No where in the proof however can you use the 109.5 degrees. This is the angle that you want to find.
This souds like an assignment question. Our goal here is to help you understand mathematical concepts (especially ones that go beyond the standard curriculum), so we would prefer you to ask "Here is a question I have been given, and I would like to understand better how I should go apart tackling such-and-such a part of it ..." rather than simply asking us the same question that was asked of you.
There are several ways to tackle this question. The simplest involves using vectors. Think about the four vectors from the centre of the tetrahedron to the four corners. You know several things about these vectors:
These vectors all have the same length l.
The angle t between any pair of vectors is the same, and from the theory of dot products one has the fact that u·v = |u| |v| cos t= l^2 cos t (where u and v are any two of the vectors).
By symmetry, the vector sum of all four vectors is zero.
Combining these pieces of information, if u, v, w, and x are the four vectors, you have
0= 0·u= (u+v+w+x)·u= (u·u) + (v·u) + (w·u) + (x·u) = l^2 + 3 l^2 cos t
and therefore cos t = -1/3, which you can solve for t to get an answer of approximately 109.5 degrees. [ Submit Your Own Question ] [ Create a Discussion Topic ]
This part of the site maintained by (No Current Maintainers)
Last updated: April 19, 1999
Original Web Site Creator / Mathematical Content Developer: Philip Spencer
Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu
Navigation Panel:
Go backward to A Sequence Describing a Bouncing Ball
Go up to Question Corner Index
Go forward to Counting Obtuse Triangles in an Inscribed Polygon
Switch to graphical version (better pictures & formulas)
Access printed version in PostScript format (requires PostScript printer)
Go to University of Toronto Mathematics Network Home Page |
7726 | https://pmc.ncbi.nlm.nih.gov/articles/PMC106833/ | An official website of the United States government
Official websites use .gov
A
.gov website belongs to an official
government organization in the United States.
Secure .gov websites use HTTPS
A lock (
Lock
Locked padlock icon
) or https:// means you've safely
connected to the .gov website. Share sensitive
information only on official, secure websites.
Primary site navigation
Logged in as:
PERMALINK
Structures of Toxoplasma gondii Tachyzoites, Bradyzoites, and Sporozoites and Biology and Development of Tissue Cysts†
J P Dubey
D S Lindsay
C A Speer
Corresponding author. Mailing address: Parasite Biology and Epidemiology Laboratory, LPSI, USDA, BARC-East, Building 1040, 10300 Baltimore Ave., Beltsville, MD 20705-2350. Phone: (301) 504-8128. Fax: (301) 504-9222. E-mail: JDUBEY@GGPL.ARSUSDA.GOV.
Abstract
Infections by the protozoan parasite Toxoplasma gondii are widely prevalent worldwide in animals and humans. This paper reviews the life cycle; the structure of tachyzoites, bradyzoites, oocysts, sporocysts, sporozoites and enteroepithelial stages of T. gondii; and the mode of penetration of T. gondii. The review provides a detailed account of the biology of tissue cysts and bradyzoites including in vivo and in vitro development, methods of separation from host tissue, tissue cyst rupture, and relapse. The mechanism of in vivo and in vitro stage conversion from sporozoites to tachyzoites to bradyzoites and from bradyzoites to tachyzoites to bradyzoites is also discussed.
Infections by the protozoan parasite Toxoplasma gondii are widely prevalent in humans and animals worldwide. T. gondii has emerged as one of the most common opportunistic infections in patients with AIDS. Toxoplasmosis in AIDS patients is considered to be a result of reactivation of latent infection, but the mechanisms of reactivation are unknown. This review focuses on the structure and biology of T. gondii stages (tachyzoites, bradyzoites, and tissue cysts) in intermediate hosts (humans) and the resistant (oocyst) stage outside the host.
BASIC STRUCTURE AND LIFE CYCLE
There are three infectious stages of T. gondii: the tachyzoites (in groups or clones), the bradyzoites (in tissue cysts), and the sporozoites (in oocysts). These stages are linked in a complex life cycle (Fig. 1).
FIG. 1.
Life cycle of T. gondii.
Tachyzoites
The term “tachyzoite” (tachos = speed in Greek) was coined by Frenkel (73) to describe the stage that rapidly multiplied in any cell of the intermediate host and in nonintestinal epithelial cells of the definitive host. The term “tachyzoite” replaces the previously used term “trophozoite” (trophicos = feeding in Greek). Tachyzoites have also been termed endodyozoites or endozoites. Aggregates of numerous tachyzoites are called clones, terminal colonies, or groups.
The tachyzoite is often crescent shaped, approximately 2 by 6 μm (Fig. 2), with a pointed anterior (conoidal) end and a rounded posterior end. Ultrastructurally, the tachyzoite consists of various organelles and inclusion bodies including a pellicle (outer covering), apical rings, polar rings, conoid, rhoptries, micronemes, micropore, mitochondrion, subpellicular microtubules, endoplasmic reticulum, Golgi complex, ribosomes, rough and smooth endoplasmic reticula, micropore, nucleus (Fig. 2 to 10), dense granules, amylopectin granules (which may be absent), and a multiple-membrane-bound plastid-like organelle which has also been called a Golgi adjunct or apicoplast (24–26, 103, 145, 176). The nucleus is usually situated toward the central area of the cell and contains clumps of chromatin and a centrally-located nucleolus.
FIG. 2.
Tachyzoites of T. gondii. A dividing tachyzoite (arrowheads) and single tachyzoites (arrows). Impression smear feline lung, stained with Giemsa stain.
FIG. 10.
Transmission electron micrograph of a tachyzoite of the VEG strain of T. gondii penetrating a neutrophil in mouse peritoneum; note the moving junction (Mj) at the site of penetration into the neutrophil and the extraordinary early development of the tubulovesicular membranes (Tv) in the space between the neutrophil and the tip of the tachyzoite. Am, amylopectin granule; Dg, electron-dense granule; Rh, rhoptry; Nu, nucleus.
The pellicle consists of three membranes, a plasmalemma and two closely applied membranes that form an inner membrane complex (Fig. 3 and 4) which is formed from a patchwork of flattened vesicles (121, 136, 145, 176). The inner membrane is discontinuous at the anterior tip above the polar rings, at micropores which are situated laterally, and at the posterior pore at the extreme posterior tip of the zoite. Polar ring 1 is an electron-dense thickening of the inner membrane complex at the anterior end of the tachyzoite, which encircles a cylindrical, truncated cone called the conoid, which consists of six to eight microtubular elements wound like a compressed spring (Fig. 4 and 5). Twenty-two subpellicular microtubules originate from polar ring 2 and run longitudinally almost the entire length of the cell just beneath the inner membrane complex. In addition, two inner microtubules terminate in the conoid (Fig. 5). The microtubules are like a rib cage and are arranged in a gentle spiral. Individual microtubules have prominent transverse striations (121). Between the anterior tip and the nucleus, there are 8 to 10 club-shaped organelles (164) called rhoptries (Fig. 3 and 6 to 10). Rhoptries are excretory structures, each consisting of an anterior narrow neck up to 2.5 μm long that extends into the interior of the conoid, and a saclike, often labyrinthine posterior end (up to 1 μm long). Micronemes are rod-like structures which occur mostly at the anterior end of the parasite (Fig. 3 and 6).
FIG. 3.
Schematic drawings of a tachyzoite (left) and a bradyzoite (right) of T. gondii. The drawings are composites of electron micrographs.
FIG. 4.
Schematic representation of the apical complex of T. gondii. Modified from reference 51 with permission of the publisher.
FIG. 5.
Transmission electron micrograph of a negatively stained apical complex of a tachyzoite. Ar1 and Ar2, apical rings 1 and 2; Co, conoid consisting of coiled microtubules; Im, internal microtubules; Pr 1, ring 1; Pr2, polar ring 2; Sm, subpellicular microtubules. Reprinted from reference 121 with permission of the publisher.
FIG. 6.
Transmission electron micrograph of a tachyzoite of the VEG strain of T. gondii in a mouse peritoneal exudate cell. Am, amylopectin granule; Co, conoid; Dg, electron-dense granule; Go, Golgi complex; Mn, microneme; No, nucleolus, Nu, nucleus; Pv, parasitophorous vacuole; Rh, rhoptry.
Although tachyzoites can move by gliding, flexing, undulating, and rotating, they have no visible means of locomotion such as cilia, flagella, or pseudopodia. The functions of the conoid, rhoptries, micropores, and micronemes are not fully known but are probably associated with host cell penetration and creation of an intracellular environment suitable for parasite growth and development. The conoid can rotate, tilt, extend, and retract as the parasite probes the host cell plasmalemma immediately before penetration (19). Rhoptries have a secretory function associated with host cell penetration, secreting their contents through the plasmalemma just above the conoid to the exterior (123). They contain a proteolytic enzyme (140a). The micropore is a cytosome-like structure formed by the invagination of the outer membrane of the pellicle (19, 123, 124).
Tachyzoites enter host cells (Fig. 10) by actively penetrating through the host cell plasmalemma or by phagocytosis (15, 54, 95, 120, 123, 125, 147, 152, 160, 182). After entering the host cell, the tachyzoite becomes ovoid and is surrounded by a parasitophorous vacuole (PV), which appears to be derived from both the parasite and the host cell. Soon after penetration, a tubulovesicular membranous network (TMN) develops within the PV (Fig. 8). Some of the TMN membranes are connected to the parasitophorous vacuolar membrane (145, 149–151). The TMN appears to be derived from the posterior end of the tachyzoite (150). However, convoluted tubules, structurally similar to the TMN, were observed at the end of tachyzoites by Nichols et al. (123), and we have also observed similar structures in in vivo-cultured tachyzoites (Fig. 10).
FIG. 8.
Transmission electron micrograph of four tachyzoites of the VEG strain of T. gondii in the final stages of endodyogeny that are still attached by their posterior ends to a common residual body (Rb); note that several host cell mitochondria (∗) are situated close to the parasitophorous vacuole (Pv), which contains extensively developed tubulovesicular membranes (Tv). Am, amylopectin granule; Co, conoid; Dg, electron-dense granule; Hn, host cell nucleus; Mn, microneme; Mp, micropore; Nu, nucleus; Rh, rhoptry.
Tachyzoites multiply asexually within the host cell by repeated endodyogeny (Fig. 7 and 8), a specialized form of reproduction in which two progeny form within the parent parasite (Fig. 3), consuming it (145). In endodyogeny, the Golgi complex divides first, becoming two complexes at the anterior end of the nucleus. Next, the anterior portions of the inner membrane complexes and the subpellicular microtubules of the progeny cells appear as two dome-shaped structures anteriorly. The parasite nucleus becomes horseshoe shaped, and the ends of the nucleus move into the dome-shaped anterior ends of the developing progeny. The inner membrane complex and subpellicular microtubules continue to extend posteriorly and surround one half of the nucleus, which eventually pinches into two. The progeny continue to grow until they reach the surface of the parent. The inner membrane complex of the parent disappears, and its outer membrane becomes the plasmalemma of the progeny cells. Tachyzoites continue to divide by endodyogeny (145). In vivo, most groups of tachyzoites are arranged randomly due to asynchronous cycles of endodyogeny. However, rosettes are occasionally formed due to synchronous division. In rapidly dividing tissue culture-adapted strains, T. gondii within a vacuole may divide synchronously (14, 140), but this is not the norm. Rarely, tachyzoites of certain strains divide by binary fission (63, 139). The host cell ruptures when it can no longer support the growth of tachyzoites (Fig. 8).
FIG. 7.
Transmission electron micrograph of a tachyzoite of the VEG strain of T. gondii undergoing endodyogeny within a mouse peritoneal macrophage to form two daughter tachyzoites. Cd, conoid of developing tachyzoite; Co, conoid of mother tachyzoite; Dg, electron-dense granule; Ga, Golgi adjunct; Hm, host cell mitochondrion; Hn, host cell nucleus; Id, inner membrane complex of developing tachyzoite; Im, inner membrane complex of mother tachyzoite; Mi, microchdrion; Mn, microneme; Nu, nuclei of daughter tachyzoites; Pl, plasmalemma of mother tachyzoite; Pm, parasitophorous vacuolar membrane; Pv, parasitophorous vacuole; Rh, rhoptries of daughter tachyzoites.
The rates of invasion and growth vary depending on the strain of T. gondii and the type of host cells (1, 101). After entry of tachyzoites into a host cell, there is a variable lag period before the parasite divides, and this lag phase is partly parasite dependent (1). Mouse virulent strains of T. gondii grow faster in cell culture than do “avirulent” strains, and some strains of T. gondii form more rosettes than others (1). Although T. gondii isolates have been classified genetically into types I, II, and III (90, 94, 148), there are no appreciable structural differences among different isolates of T. gondii.
Bradyzoites and Tissue Cysts
The term “bradyzoite” (brady = slow in Greek) was also coined by Frenkel (73) to describe the organism multiplying slowly within a tissue cyst. Bradyzoites are also called cystozoites.
Tissue cysts (Fig. 11) grow and remain intracellular (Fig. 11C) as the bradyzoites divide by endodyogeny (64, 65). Tissue cysts vary in size; young tissue cysts may be as small as 5 μm in diameter and contain only two bradyzoites (Fig. 11B), while older ones may contain hundreds of organisms (Fig. 11E). Tissue cysts in the brain are often spheroidal and rarely reach a diameter of 70 μm, whereas intramuscular cysts are elongated and may be 100 μm long (27, 36). Although tissue cysts may develop in visceral organs, including the lungs, liver, and kidneys, they are more prevalent in the neural and muscular tissues, including the brain, eyes, and skeletal and cardiac muscles (35). Intact tissue cysts probably do not cause any harm and can persist for the life of the host without causing a host inflammatory response.
FIG. 11.
Tissue cysts of T. gondii in mouse brains. (A) Tissue cyst with three bradyzoites, each with a terminal nucleus (arrows). Note the thin cyst wall (arrowhead). Impression smear with silver impregnation and Giemsa stain. (B) Three tissue cysts with well-defined cyst walls (arrowheads). Note the tissue cyst with two bradyzoites, each with a terminal nucleus (arrows). Impression smear with silver impregnation and Giemsa stain. (C) Intracellular tissue cyst in section. Note the thin cyst wall (arrow) and the host cell nucleus (arrowhead). Hematoxylin and eosin stain. (D) Tissue cyst with numerous PAS-positive bradyzoites (arrowheads) enclosed in a PAS-negative cyst wall (arrow). PASH stain. (E) Tissue cyst freed from mouse brain. Note the cyst wall (arrow) enclosing hundreds of bradyzoites. Unstained impression smear.
The tissue cyst wall is elastic and thin (<0.5 μm thick) (Fig. 11), and it encloses hundreds of crescent-shaped bradyzoites (Fig. 12 and 13), each approximately 7 by 1.5 μm in size (118). The tissue cyst develops within the host cell cytoplasm. The cyst wall is argyrophilic (77, 153), but results vary depending on the silver impregnation method used. According to Sims et al. (153), the cyst wall stains intensely with Bodian protoargol and Palmgren silver but not with methenamine silver. The lack of staining with methenamine silver indicates that the cyst wall contains no glycogen or other polysaccharides. The cyst wall is composed of host cell and parasite materials (64, 65, 153). It is ultimately lined with granular material, which also fills the space between the bradyzoites. Some bradyzoites degenerate (Fig. 13), especially in older tissue cysts (129). The cyst wall is only faintly periodic acid-Schiff (PAS) positive (Fig. 11D).
FIG. 12.
Transmission electron micrograph of two T. gondii tissue cysts in the brain of a mouse 6 months after infection with the Me-49 strain. Note that the tissue cyst on the left is younger than the one on the right because of the differences in their bradyzoites. The bradyzoites in the tissue cyst on the right contain more micronemes and amylopectin than do those in the one on the left. Additionally, the contents of rhoptries in the bradyzoites in the older tissue cyst are electron dense (arrowheads) whereas those in the bradyzoites in the younger tissue cyst are honeycombed (arrows). The cyst wall (cw) also is more branched and prominent in the older tissue cyst than in the younger tissue cyst. Reprinted from reference 46.
FIG. 13.
Tissue cyst in the brain of a mouse that was inoculated 8 months earlier with oocysts of the VEG strain of T. gondii. This ultrathin section of the cyst shows approximately 110 bradyzoites (Bz). The tissue cyst is surrounded by a relatively thin cyst wall (Cw) and is situated within the host cell cytoplasm near the host cell nucleus (Hn). A few of the bradyzoites appear to be degenerate (Dz). The area within the rectangular box is shown at a higher magnification in Fig. 14.
Bradyzoites (Fig. 3 and 12 to 16) differ structurally only slightly from tachyzoites. They have a nucleus situated toward the posterior end, whereas the nucleus in tachyzoites is more centrally located. The contents of rhoptries in bradyzoites are usually electron dense, whereas those in tachyzoites are labyrinthine. However, the contents of rhoptries in bradyzoites vary with the age of the tissue cyst. Bradyzoites in younger tissue cysts may have labyrinthine rhoptries, whereas those in older tissue cysts are electron dense (Fig. 12). Also, most bradyzoites have one to three rhoptries, which are looped back on themselves (Fig. 3). Bradyzoites contain several amylopectin granules which stain red with PAS reagent; such material is either in discrete particles or absent in tachyzoites (Fig. 3). Bradyzoites are more slender than are tachyzoites. Bradyzoites are less susceptible to destruction by proteolytic enzymes than are tachyzoites (93), and the prepatent period in cats following feeding of bradyzoites is shorter than that following feeding of tachyzoites (48).
FIG. 16.
Anterior end of a bradyzoite in Fig. 13 showing the apical complex. Am, amylopectin; Ar, apical rings 1 and 2; Co, conoid; Im, inner membrane complex; Mn, microneme; Nr, neck of rhoptry; Pl, plasmalemma; Pr, polar rings 1 and 2; Rh, rhoptry; Sm, subpellicular microtubule.
Enteroepithelial Stages
Cats shed oocysts after ingesting any of the three infectious stages of T. gondii, i.e., tachyzoites, bradyzoites, and sporozoites. Prepatent periods (time to the shedding of oocysts after initial infection) and frequency of oocyst shedding vary according to the stage of T. gondii ingested (37, 48, 80). Prepatent periods are 3 to 10 days after ingesting tissue cysts, ≥18 days after ingesting oocysts (37), and ≥13 days after ingesting tachyzoites (44). Fewer than 30% of cats shed oocysts after ingesting tachyzoites or oocysts, whereas nearly all cats shed oocysts after ingesting tissue cysts (37, 48).
After the ingestion of tissue cysts by cats, the cyst wall is dissolved by proteolytic enzymes in the stomach and small intestine. The released bradyzoites penetrate the epithelial cells of the small intestine and initiate the development of numerous generations of T. gondii (Fig. 17 and 18). Five morphologically distinct types of T. gondii develop in intestinal epithelial cells before gametogony begins (47). These stages are designated types A to E instead of generations, because there are several generations within each T. gondii type (Fig. 17). Little has been added to the structure or biology of types A to C since the original description by Dubey and Frenkel (47).
FIG. 17.
Coccidian cycle of T. gondii. Reprinted from reference 47 with permission of the publisher.
FIG. 18.
Enteroepithelial stages of T. gondii in a small intestinal villus of a cat fed tissue cysts. Note the heavy parasitization of entrocytes, containing schizonts (type D) (small arrows) and male (large arrow) and female (arrowheads) gamonts. Hematoxylin and eosin stain.
So far, only late stages (presumably type D) have been studied by transmission electron microscopy (TEM). These forms multiply by a specialized form of schizogony (Fig. 19). As in normal schizogony, the nucleus divides two or more times without cytoplasmic division (61, 133, 144). Whether daughter organism (merozoite) formation begins after four or more nuclei have been formed is uncertain. Before or simultaneous with the last nuclear division, merozoite formation is initiated near the center of the schizont by the development of dome-shaped merozoite anlagen; whether one or two anlagen are formed near each nucleus is uncertain (61, 144). The merozoites eventually move towards the periphery of the schizont, and the schizont plasmalemma invaginates around each merozoite, forming the plasmalemma of the merozoite. The merozoites separate from the schizont at their posterior ends, sometimes leaving a residual body (47).
FIG. 19.
Transmission electron micrograph of type D T. gondii schizonts. (A) Intermediate schizont showing several nuclei (Nu). Mv, microvilli of host enterocyte. (B) Mature schizont. (C) Longitudinal section of a merozoite. Co, conoid; Dg, dense granule; Mi, mitochondrion; Mn, microneme; No, nucleolus; Nu, nucleus; Rh, rhoptry.
The schizogony observed in T. gondii type D organisms differs from schizogony in conventional coccidia (Eimeria species) in that in T. gondii merozoites are formed internally and immature merozoites do not protrude from the schizont surface. It should be pointed out that many details of schizogony of T. gondii are not clear even in type D organisms; nothing is known of the ultrastructures of T. gondii types to A to C described by Dubey and Frenkel (47).
Piekarski et al. (133) proposed the term “endopolygeny” to describe the division of T. gondii schizonts in feline enterocytes. They believed that merozoite formation began after two nuclear divisions. Vivier (175) had earlier used the term “endopolygeny” to describe observed divisional formation of more than two daughter T. gondii tachyzoites in the peritoneum of mice. We would like to propose that the term “endopolygeny” should not be used to describe schizogony in T. gondii but should be restricted to the merozoite formation in Sarcocystis and Frenkelia spp. (51). In Sarcocystis endopolygeny, the nucleus becomes multilobed but does not divide into separate nuclei. Each nuclear lobe is eventually incorporated into developing merozoites. The number of merozoites in Sarcocystis schizonts varies from 4 to 100 or more (51).
After the asexual development (types A to E), the sexual cycle starts 2 days after tissue cysts were ingested by the cat. The origin of gamonts has not been determined, but the merozoites released from schizont types D and E probably initiate gamete formation. Gamonts are found throughout the small intestine, but most commonly in the ileum, 3 to 15 days after inoculation (Fig. 18). They are found above the nucleus of the host epithelial cell near the tips of the villi of the small intestine (Fig. 20). Female gamonts are subspherical, and each contains a single centrally located nucleus and several PAS-positive granules. Ultrastructurally, the mature female gamete contains several micropores, rough and smooth endoplasmic reticulum, numerous mitochondria, double-membraned vesicles, and wall-forming bodies (WFB) (Fig. 20B). The double-membraned bodies are located near the nucleus and are probably derived from it (67). WFB are of two types: type I and type II. Type I WFB are about 0.35 μm in diameter and electron dense, and they appear before type II WFB (67). Type II WFB are moderately electron dense, less abundant than WFB, and larger than WFB (1.2 μm in diameter) (67).
FIG. 20.
Transmission electron micrograph of T. gondii gamonts. (A) Mature microgamont. Fl, flagellum of microgamete; Mg, body of microgamete; Pv, parasitophorous vacuole. (B) Zygote in the early stage of oocyst wall formation. Am, amylopectin granule; Lb, lipid body; Nu, nucleus of zygote; Ow, oocyst wall in the early stage of formation; Wf, wall-forming bodies.
Mature male gamonts (microgamonts) are ovoid to ellipsoidal in shape (Fig. 20A). During microgametogenesis, the nucleus of the microgamont divides to produce 10 to 21 nuclei (47). The nuclei move toward the periphery of the parasite and enter protuberances formed in the pellicle of the microgamont. One or two residual bodies remain in the microgamont after division into microgametes (Fig. 20A). Microgametes are elongated and consist mainly of nuclear material. The anterior end is a pointed structure called the perforatorium, within which lie two basal bodies. Two long, free flagella originate from the basal bodies and project posteriorly. A large mitochondrion is situated near the basal bodies and just anterior to the nucleus. Five microtubules originate near the nucleus and extend posteriorly alongside it for a short distance.
Microgamonts have up to 21 microgametes. Microgametes use their flagella to swim to and penetrate and fertilize mature macrogametes to form zygotes. After fertilization, an oocyst wall is formed around the parasite (Fig. 20B). Infected epithelial cells rupture and discharge oocysts into the intestinal lumen. Five layers of the oocyst wall are formed at the surface of the zygote (67). No extensive cytoplasmic changes occur in the zygote while layers 1, 2, and 3 are formed, but type I WFBs disappear with the formation of layer 4 and type II WFBs disappear with the formation of layer 5.
Oocysts
Unsporulated oocysts are subspherical to spherical and are 10 by 12 μm in diameter (Fig. 21A). Under light microscopy, the oocyst wall consists of two colorless layers. Polar granules are absent, and the sporont almost fills the oocyst. Sporulation occurs outside the cat within 1 to 5 days of excretion depending upon aeration and temperature.
FIG. 21.
Oocysts of T. gondii. (A) Unsporulated oocyst. Note the central mass (sporont) occupying most of the oocyst. (B) Sporulated oocyst with two sporocysts. Four sporozoites (arrows) are visible in one of the sporocysts. (C) Transmission electron micrograph of a sporulated oocyst. Note the thin oocyst wall (large arrow), two sporocysts (arrowheads), and sporozoites, one of which is cut longitudinally (small arrows).
Sporulated oocysts are subspherical to ellipsoidal and are 11 by 13 μm in diameter (Fig. 21B and C). Each oocyst contains two ellipsoidal sporocysts without Stieda bodies. Sporocysts measure 6 by 8 μm. A sporocyst residuum is present; there is no oocyst residuum. Each sporocyst contains four sporozoites (Fig. 21C).
The ultrastructure of sporulation was described by Ferguson et al. (56–60). The cytoplasm of the unsporulated oocyst (zygote) has a large nucleus with amorphous nucleoplasm and a distinct nucleolus. The zygote is limited by a unit membrane with few micropores. The nucleus divides twice, giving rise to four nuclei, which are situated at the periphery of the zygote; at this stage a second limiting membrane is formed. After the cytoplasm divides, two spherical sporoblasts are formed, each with two nuclei (57).
As the sporulation continues, the sporoblasts elongate and sporocysts are formed. The two outer membranes of the sporoblasts become the outer layer of the sporocyst wall, and the plasmalemma of the cytoplasmic mass becomes the inner layer (58). Ultimately, as the sporocyst develops, four curved plates form the innermost layer of the sporocyst. The plates are joined by four liplike sutures (described below) with a depression on the surface of the sporocyst wall at the junction point. Sporozoite formation begins when two dense plaques (anlagen) appear at both ends of the sporocyst. Each nucleus divides into two and is incorporated into elongating sporozoite anlagen. Thus, four sporozoites are formed in each sporocyst (59). A prominent residual body is left after sporozoite are formed; the residual body is enclosed in a single-unit membrane (59).
Ultrastructurally, the oocyst wall of sporulated oocysts (Fig. 22) consists of three layers: an electron-dense outer layer, an electron-lucent middle layer, and a moderately electron-dense inner layer (159). The middle layer consists of remnants of two membranes that evidently were laid down between the inner and outer layers during oocyst wall formation. Treatment of oocysts with 1.3% sodium hypochlorite (Clorox) removes the outer layer. The oocyst wall contains a single micropyle, which is relatively small and is located randomly in the oocyst wall (Fig. 23). The micropyle is a 350-nm-diameter indentation that consists of three layers that are continuous with the three layers in the oocyst wall. However, the outer layer of the micropyle is thin and moderately electron dense and the inner layer is electron dense, disk shaped, and slightly thicker than the inner layer of the oocyst wall. Although the function of the micropyle is not known, it might represent a permeable site in the oocyst wall that is susceptible to the actions of CO2 and various enzymes that allow the entry of bile salts and trypsin, which stimulate the excystation of sporozoites from sporocysts.
FIG. 22.
Transmission electron micrograph of an oocyst of the VEG strain of T. gondii exposed to excysting fluid (trypsin and bile salts) showing one sporocyst in an early stage of excystation. The oocyst is surrounded by a fine reticulate veil (Ov) and an oocyst wall (Ow). The sporocyst wall (Sw) consists of a continuous thin outer layer and an inner layer composed of four curved plates joined at sutures (arrows) containing an interposed strip of electron-dense material. During excystation, the excysting fluid acts on the sutures, causing the plates of the inner layer to separate and curl inward (double arrow), releasing the sporozoites. Am, amylopectin; Co, conoid; Dg, electron-dense granule; Lb, lipid body; Mn, microneme; Rh, rhoptry.
FIG. 23.
Transmission electron micrograph of a sporulated oocyst of the VEG strain of T. gondii. (A) Oocyst in the late stage of excystation, showing several free sporozoites (Sz) and collapsed sporocyst walls (Sw). Box B (see panel B) shows a micropyle in the oocyst wall (Ow). (B) High magnification of portion of panel A showing details of the oocyst wall and the micropyle, the oocyst wall consists of an outer electron-dense layer (Ol) and an inner electron-lucent layer (Il), which are separated by an electron-lucent line (El). At the micropyle, the outer layer of the oocyst wall becomes electron lucent and extremely thin (arrow); the electron-lucent line runs continuously through the micropyle, and the inner layer of the oocyst wall becomes slightly thicker near the micropyle and is continuous with the micropyle, which consists primarily of a curved, moderately electron-dense disc (∗).
The sporocyst wall consists of two distinct layers with a thin, electron-dense outer layer and thicker, moderately electron-dense inner layer. The inner layer consists of four curved plates (20, 58) (Fig. 21 and 22). At sites of apposition between two plates, there are two apposing liplike thickenings and an interposed strip in the inner layer (20, 159). During excystation in the presence of bile salts and trypsin, the sporocyst ruptures suddenly, evidently at the sites of apposition between plates, releasing the sporozoites. As the plates separate, they curl inward to form conical coils (Fig. 23 and 24). The sporocyst residuum consists of amylopectin granules and lipid bodies.
FIG. 24.
Freshly excysted sporozoite still within an oocyst. Am, amylopectin; Co, conoid; Dg, electron-dense granule; Go, Golgi complex; Im, inner membrane complex: Lb, lipid body; Mi, mitochondrion; Mn, microneme; Ow, oocyst wall; Pl, plasmalemma; Rh, rhoptry; Sm, subpellicular microtubule; Sw, sporocyst wall.
Ultrastructurally, the sporozoite is similar to the tachyzoite, except that there is an abundance of micronemes, rhoptries, and amylopectin granules in the former. Sporozoites are 2 by 6 to 8 μm in size with a subterminal nucleus (Fig. 24 and 25). There are no crystalloid bodies or any refractile bodies in T. gondii sporozoites (Fig. 24 and 25).
FIG. 25.
Schematic drawing of a T. gondii sporozoite.
The ultrastructural differences of some of the organelles in tachyzoites, bradyzoites, and sporozoites of the VEG strain of T. gondii are compared in Tables 1 and 2; the VEG strain was isolated from the blood of an AIDS patient and is mildly virulent to mice depending on the stage of the parasite inoculated (42, 49).
TABLE 1.
Relative numbers of organelles and inclusion bodies in sporozoites, tachyzoites, and bradyzoites of the VEG strain of T. gondii as determined by TEM
Stage | Mean no. (range) ofd:
Rhoptries | Micronemes | Dense granules | Amylopectin | Lipid
Sporozoitea | 5.9 (2–11)4 | 55 (40–78) | 9.4 (5–15) | 7.8 (3–13) | 1.25 (1–3)
Tachyzoiteb | 6.7 (2–11) | 25 (19–38) | 9.1 (5–17) | 2.4 (1–6) | 0.6 (0–2)
Bradyzoitec | 5.5 (2–8) | 75.5 (36–112) | 2.7 (1–5) | 21.8 (7–38) | 0
Sporozoites were freshly excysted from 34-day-old oocysts.
Tachyzoites were obtained from the peritoneum of an IFN-γ knockout mouse 8 days after inoculation of tissue cysts.
Bradyzoites were from cysts in the brains of mice at 8 months after inoculation of oocysts.
Numbers represent means that were obtained by counting all organelles or inclusion bodies in 20 longitudinal sections of each type of zoite; ranges are given in parentheses.
TABLE 2.
Relative sizes of inclusion bodies in sporozoites, tachyzoites, and bradyzoites of the VEG strain of T. gondii as determined by TEM
Stagea | Mean size (nm) (range) in:
Dense granules | Amylopectin | Lipid
Sporozoite | 208 (175–250) | 356 (200–460) | 388 (200–550)
Tachyzoite | 244 (133–334) | 201 (103–333) | 224 (150–400)
Bradyzoite | 181 (167–201) | 358 (192–603) | 0
For the source of these infectious stages, see Table 1, footnotes a to c.
Ultrastructural Comparison of Tachyzoites, Bradyzoites, and Sporozoites
Sporozoites, tachyzoites, and bradyzoites of T. gondii are similar ultrastructurally but differ in certain organelles and inclusion bodies (Tables 1 and 2). All three zoites have similar numbers of rhoptries but the rhoptries differ in appearance between the zoites. For example, the rhoptries of tachyzoites are uniformly labyrinthine; sporozoites usually contain both labrynthine and uniformly electron-dense rhoptries; and bradyzoites usually contain uniformly electron-dense rhoptries, some of which are folded back on themselves (Fig. 3 and 15). Tachyzoites have few micronemes, sporozoites have an intermediate number, and bradyzoites have many. Dense granules are more numerous in sporozoites and tachyzoites than in bradyzoites (Fig. 3 and 25). Amylopectin granules are numerous and relatively large in sporozoites and bradyzoites but are few and small or absent in tachyzoites. Lipid bodies are numerous in sporozoites, rare in tachyzoites, and absent in bradyzoites (Fig. 3 and 25).
FIG. 15.
Bradyzoite in the tissue cyst shown in Fig. 13. Am, amylopectin granule; Ce, centrioles; Co, conoid; Dg, electron-dense granule; Ga, Golgi adjunct (apicoplast); Go, Golgi complex; Im, inner membrane complex; Mi, mitochondrion; Mn, microneme; Nu nucleus; Pm, plasmalemma; Rh, rhoptry.
Attachment and Development of T. gondii
In general, the attachment and penetration of host cells by T. gondii zoites (tachyzoites, bradyzoites, sporozoites, and merozoites) appear similar to those described for other coccidian parasites. The mechanical events involved in zoite attachment and penetration include (i) gliding of the zoite, (ii) probing of the host cell with the conoidal tip of the zoite, (iii) indenting the host cell plasmalemma, (iv) forming a moving junction that moves posteriorly along the zoite as it penetrates into the host cell, and (v) partially exocytosing micronemes, rhoptries, and dense granules. Zoites of T. gondii can penetrate a variety of cell types from a wide range of hosts, indicating that the biochemical receptors involved in attachment and penetration are probably common to most animal cells. Host cell receptors consisting of laminin, lectin, and SAG1 are involved in T. gondii tachyzoite attachment and penetration (100). T. gondii zoites can, however, enter cells by means other than receptor-mediated penetration. Speer et al. (160) found that after passing completely through cells, some sporozoites of T. gondii carried an envelope of host cell membranes and cytoplasm but were still capable of penetrating other cells. Zoites of T. gondii can also enter cells by being endocytosed (160).
Recently, a great deal of research on the formation of the PV has been conducted (96). As tachyzoites penetrate host cells, they are surrounded by a membrane that is evidently derived from the host cell plasmalemma minus host proteins. This membrane is destined to become the PV membrane (PVM), and a number of parasite proteins associate with it, including rhoptry proteins ROP2, ROP3, ROP4, and ROP7. ROP2 is located on the host cell cytoplasmic side of the PVM, suggesting that it plays a role in host-parasite biochemical communication (2). Within a few minutes after penetration, tachyzoites modify the newly formed PV and the PVM with parasite proteins and a TMN forms within the PV. The PVM acquires pore structures that freely allow charged molecules up to 1,200 kDa to diffuse bidirectionally between the PV and the host cell cytoplasm (142). Dense-granule proteins (GRA) are secreted into the PV after tachyzoite penetration, with GRA3 and GRA5 localizing on the PVM and GRA1, GRA2, GRA4, and GRA6 associating with the TMN. Collectively, these modifications establish a parasite-friendly environment within the host cell cytoplasm that is conducive to parasite replication.
To date, there is only a single report involving the expression of parasite antigens in T. gondii sporozoites. Speer et al. (161) found that GRA3 and SAG1 are developmentally regulated; they are not expressed in the sporozoite or the sporozoite-infected cell until 12 to 15 h after sporozoite inoculation of cell cultures. The first parasite multiplication occurs after the expression of GRA3 and SAG1, indicating that although the sporozoite is competent for cell penetration, it is not immediately able to establish an intracellular environment which can support replication. Therefore, many of the proteins necessary for parasite growth appear to be down-regulated in the sporozoite, perhaps due to dormancy within the oocyst.
Most of the available information concerning the interaction of T. gondii zoites with host cells has been derived from in vitro cultivation with tachyzoites. There is limited information on in vivo zoite-host cell interactions. Recent studies with sporozoites have shown that T. gondii interacts with host cells substantially differently in vivo from the interaction in vitro. For example, Speer et al. (161) found that in cultured cells, T. gondii sporozoites induced the formation of two types of PVs. Type 1 PVs formed first, were relatively large (20 to 30 μm in diameter), had an indistinct PVM, and contained no exocytosed dense granule material or a TMN. After 12 to 18 h, sporozoites in type 1 PVs actively penetrated into the host cell cytoplasm and established type 2 PVs, which contained exocytosed dense granules, a TMN, and a distinct PVM. Parasites multiplied by endodyogeny in type 2 PVs but did not multiply in type 1 PVs. In follow-up studies in mice, it was found that sporozoites passed through enterocytes and goblet cells of the mouse intestinal epithelium and entered the lamina propria, where they infected all cells of the host except erythrocytes and underwent endodyogeny to form tachyzoites (52, 158). In contrast to the in vitro studies, sporozoites did not form type 1 PVs at any point while they were in the intestinal epithelium or the lamina propria. Even though the sporozoites were just passing through ileal enterocytes, they still formed PVs that contained exocytosed dense granule material and well-developed TMNs, which appear necessary for parasite multiplication (102, 106, 149, 150). The presence of exocytosed material and TMNs associated with the PVs of sporozoites in transit across the intestinal epithelium indicates that parasite replication does not always follow the formation of a parasite-modified PV. After entering cells in the lamina propria, PVs equivalent to the type 2 PVs were again formed, but here the sporozoites multiplied by endodyogeny. These findings therefore indicate that the type 1 PV seen in cultured cells might represent an anomaly of in vitro cultivation.
Bjerkås (5) and Sibley et al. (150) suggested that the TMN was formed in association with a transient sac-like structure at the posterior end of the tachyzoite. However, as stated above, TMN can develop early near the anterior of tachyzoites even before the parasite completely enters the host cell (Fig. 10).
DEVELOPMENT AND BIOLOGY OF BRADYZOITES AND TISSUE CYSTS IN VIVO
History
Lainson (104) reviewed earlier literature on the development of tissue cysts. Levaditi et al. (107) apparently were the first to report that T. gondii may persist in tissues for many months as “cysts.” However, considerable confusion between the terms “pseudocysts” (group of tachyzoites) and “cysts” existed for many years. Frenkel and Friedlander (77) and Frenkel (70) cytologically characterized cysts containing organisms with a subterminal nucleus and PAS-positive granules surrounded by a argyrophilic cyst wall. Wanko et al. (177) first described the ultrastructure of the T. gondii cysts and its contents.
Lainson (104) provided evidence that cysts were formed in mice as early as 8 days after the inoculation of tachyzoites. His illustrations of a cyst with four organisms on day 8 postinfection (p.i.) and with 20 organisms on day 10 p.i. provided the first convincing evidence of young cysts of T. gondii. Older cysts were up to 60 μm in diameter and contained approximately 3,000 organisms.
Jacobs et al. (93) first provided a biologic definition of cysts when they found that cystic organisms were resistant to digestion by gastric juice (pepsin-HCl) whereas tachyzoites were destroyed immediately. Thus, cysts became important in the life cycle of T. gondii because carnivorous hosts could become infected by ingesting cysts.
When T. gondii oocysts were discovered in cat feces in 1970 (27, 72, 75), oocyst shedding was added to the biologic definition of the cyst.
Dubey and Frenkel (48) made the first in-depth study of the development of the tissue cysts and bradyzoites and defined cysts biologically and morphologically. They found that the cysts were formed as early as 3 days after inoculation of mice with tachyzoites. Cats shed oocysts with a short prepatent period (3 to 10 days) after ingesting bradyzoites, whereas after they ingested tachyzoites or oocysts, the prepatent period was longer (≥14 days).
Ferguson and Hutchison (65) reported the first in-depth ultrastructural studies of the development of T. gondii cysts.
Dubey and Beattie (45) proposed that cysts should be called tissue cysts to avoid confusion with oocysts.
Structure and Biology
Host cells parasitized and prevalence.
T. gondii tissue cysts occur in many organs and cell types. Most observations have been made with tissue cysts in the brains of mice, but the acid-pepsin digestion procedure and bioassay in cats have shown that tissue cysts occur in many extraneural organs. Tissue cyst distribution is in part controlled by the host and strain of T. gondii. The distribution of tissue cysts in different organs of animals fed the GT-1 strain of T. gondii is shown in Table 3. In pigs, dogs, cats, and rodents fed oocysts of the VEG strain of T. gondii, more tissue cysts were found in muscular tissues than in the brain in cats, dogs, and pigs whereas more tissue cysts were found in the brain than in other organs in rats and mice (38–40, 49, 110). Infections with T. gondii in experimental animals appear to be essentially the same as in some naturally infected animals as reviewed by Dubey and Beattie (45). For example, Jacobs et al. (92) isolated T. gondii from 18 diaphragms and 9 brains of 31 seropositive sheep killed at a slaughterhouse in New Zealand.
TABLE 3.
Persistence of T. gondii in tissues of animals fed the GT-1 straina
Species | Day(s) p.i. | No. of infected animals/no. fedT. gondii | Reference
Liver | Kidneys | Brain | Skeletal muscles | Heart | Diaphragm
Sheep | 97–173 | 6/8 | 5/8 | 4/8 | 6/8 | 7/8 | 6/8 | 32
Goats | 335–441 | 6/6 | 3/6 | 5/6 | 6/6 | 6/6 | 6/6 | 29
Horses | 33–476 | 0/13 | 1/13 | 2/13 | 2/13 | 3/13 | NDb | 34
Cattle | 256, 267 | 2/2 | 1/2 | 0/2 | 1/2 | 0/2 | 0/2 | 31
Pigs | 38–171 | 2/8 | 2/8 | 8/8 | 5/8 | 8/8 | 4/8 | 50
Elk | 77 | 1/2 | 0/2 | 2/2 | 1/2 | 1/2 | 2/2 | 53
Bison | 28 | 1/1 | 0/1 | 0/1 | 0/1 | 0/1 | ND | 30
Coyotes | 49–84 | ND | 1/4 | 3/4 | 4/4 | 2/4 | 2/4 | 28
Dogs | 166, 167 | 1/4 | 0/4 | 0/4 | 4/4 | 3/4 | ND | 33
Pepsin digests (50 g) were bioassayed in mice.
ND, not done.
Size of tissue cysts and numbers of bradyzoites.
Tissue cyst size is dependent on cyst age, the type of host cell parasitized, and the cytological method used for measurement. Young tissue cysts may contain as few as two bradyzoites surrounded by a distinct cyst wall and measure about 5 μm in diameter (Fig. 11B). Tissue cysts in myocytes are two to three times longer than those in neural cells (36). Because tissue cysts are most numerous in the brains of mice, most observations were made with neural tissue cysts. Beverley (4) measured the volume of tissue cysts liberated from the mouse by grinding brains with a mortar and pestle in saline (0.85% NaCl). Tissue cysts grew uniformly up to 10 weeks, after which there was considerable variability in tissue cyst size, perhaps due to a second generation of tissue cysts. The tissue cysts were up to 58 μm in diameter. He also stated that a tissue cyst may contain 60,000 organisms. Van der Waaij (174) made an in-depth study of the growth of tissue cysts in mouse brains in which 100 to 500 free (liberated from mouse brains by homogenization in saline), unstained tissue cysts were measured in the brains at 4, 8, 12, 16, and 24 weeks from mice inoculated subcutaneously (s.c.) with tissue cysts. The tissue cysts grew uniformly in size up to 12 weeks p.i., and then the growth levelled off. The tissue cysts were up to 70 μm in diameter, but the mean diameter of 100 tissue cysts at 16 weeks p.i. was 42 μm.
Ferguson and Hutchison (65) measured tissue cysts in thin (≤1-μm) sections from mouse brain fixed in glutaraldehyde. They ultrastructurally observed 140 tissue cysts with a minimum of 10 tissue cysts at 11, 21, and 28 days and 3, 6, 12, 18, and 22 months. The tissue cysts were up to 20 μm in diameter at 28 days, up to 30 μm at 3 months, and up to 50 μm at ≥6 months.
Dubey (38) measured tissue cysts in the brains of rats 72 to 75 days after feeding the rats oocysts. The brains were fixed in 10% buffered neutral formalin and sectioned at 5-μm thickness. Although the tissue cysts (n = 224) were up to 50 μm in diameter, most of them were approximately 30 μm in diameter.
In addition to these reports, Dubey (44a) has never seen T. gondii tissue cysts larger than 70 μm in diameter in formalin-fixed paraffin-embedded sections of brains of hundreds of naturally or experimentally infected animals. The measurement of tissue cysts in formalin-fixed, paraffin-embedded sections provides a standard means of reporting results. The sizes of the tissue cysts vary a great deal when unstained tissue cysts are examined between a glass slide and coverslip, depending on the homogeneity of the brain suspension, the amount of fluid, and the pressure applied. A highly flattened tissue cyst that initially passed through a 63-μm filter is shown in Fig. 26.
FIG. 26.
A highly stretched tissue cyst estimated to contain more than 1,000 bradyzoites in an impression smear of brain homogenate from a rat 14 months after infection with the VEG strain of T. gondii. The cyst wall (arrow) is barely visible.
The size of the T. gondii tissue cyst may vary with the strain of T. gondii. Although there are no firm data, one of us (44a) has observed up to 300% variability in tissue cyst size at 2 months after oral inoculation of mice with oocysts of different isolates of T. gondii. The tissue cysts of some isolates were only 20 μm in diameter, whereas others were up to 60 μm in diameter.
There are no firm data on the number of bradyzoites in a tissue cyst. Most of the information is from reviews (4, 137). In a large, flattened tissue cyst illustrated by Huskinson-Mark et al. (91), 990 bradyzoites are clearly visible. The 60,000 bradyzoites in a tissue cyst mentioned by Beverley (4) appears unrealistic.
Separation of tissue cysts from host tissue.
Cornelissen et al. (21) described a method to separate tissue cysts from the brains of mice after suspensions of brain homogenates were run on discontinuous Percoll gradients. According to these authors, T. gondii tissue cysts have a specific gravity of 1.056. This method has been widely used. Tissue cysts can also be separated on discontinuous gradients with 25 to 30% Percoll (7, 132) or a 20% dextran solution (79). The degree of success in purifying tissue cysts depends on the host tissue and the amount of blood contamination. To minimize tissue and erythrocyte contamination, the mice should be bled out before the tissue is harvested and the brain homogenate should be passed through a 90-μm wire sieve. Tissue cysts of T. gondii are smaller than 90 μm and pass through the sieve. They can be stored in saline at 4°C for 2 months (41, 93). However, the mortality rate of bradyzoites stored at 4°C for prolonged periods has not been determined.
Genetic regulation of tissue cyst numbers.
Mice are often used to obtain tissue cysts of T. gondii for experimental purposes, and the most frequently used T. gondii strains are Beverley and ME-49. Tissue cyst numbers in mouse brains vary depending on the strain of mouse, the strain of T. gondii, the route of inoculation, and the number of organisms inoculated. More tissue cysts are produced if mice become mildly ill but without obvious clinical signs. With the original Beverley strain, outbred mice inoculated s.c. with tissue cysts developed mild illness (weight loss during weeks 2 and 3) but survived. Hundreds of large tissue cysts were seen when the mice were killed at 12 weeks p.i. However, this strain has been passaged frequently in mice and has become more pathogenic for mice. Similarly, the pathogenicity of some lines of the ME-49 strain has increased since its original isolation. Therefore, to prevent mortality, one may have to use prophylaxis (sulfadiazine sodium or sulfamerazine in drinking water at 15 to 100 mg/100 ml of water) when the mice become ill. The dosage and duration of chemotherapy should be adjusted for each T. gondii strain, stage of the parasite inoculated, and strain of mice; there is no standard formula.
The number of tissue cysts in mouse brain is genetically regulated (6, 16–18, 117). Brown et al. (17) reported that the tissue cyst burden was regulated by mouse chromosome 17 containing the class I gene Ld. More tissue cysts were produced in congenic mice.
Tissue cyst persistence may vary with the duration of infection, depending on the strain of T. gondii and the host. In one experiment with the ME-49 strain, the number of tissue cysts recovered from the brains of CBA/Ca mice at 4, 8, 12, and 16 weeks p.i. were 3,720, 2,158, 3,133, and 1,538, respectively (62). Therefore, the optimal time of harvest should be determined for each strain of T. gondii in a given host.
Suzuki et al. (169) compared tissue cyst formation and mortality in two inbred strains of mice inoculated intraperitoneally (i.p.) with tissue cysts of the ME-49 strain. At 3 weeks after inoculation of 20 tissue cysts or more, nearly 10 times as many tissue cysts were found in the brains of CBA/Ca mice, all of which survived, as compared with BALB/c mice, two of which had died. After i.p. inoculation with 80 tissue cysts, the mortality rate during acute infection increased to 75% in BALB/c mice, whereas all CBA/Ca mice survived. However, during chronic infection, 50% of CBA/Ca mice died between 2 and 6 months p.i. whereas all BALB/c mice survived. Therefore, the optimal time to harvest tissue cysts may vary with the strain of T. gondii and the strain of the mouse.
Tissue cyst rupture and reactivation of latent infection.
It is well known that T. gondii tissue cysts persist in organs of infected hosts for several months and perhaps for life, depending on the host and parasite strains. The localization of tissue cysts also varies with the host and the strain of T. gondii. Although more tissue cysts are found per gram of tissue in mice than in other hosts, the fate of tissue cysts is difficult to study in mice because mice are never completely immune to T. gondii and new tissue cysts (Fig. 12) are formed even in chronic infections (4, 46, 62, 174). Even tachyzoites are present in the brains of chronically infected mice (52, 62). Because T. gondii tissue cysts are small and tissue cyst rupture is unpredictable, Frenkel (70) studied tissue cyst rupture in hamsters by using T. gondii and a related parasite, Besnoitia jellisoni, where intact tissue cysts were not associated with inflammatory reaction. Leaking and ruptured tissue cysts were accompanied by marked inflammation. Microglial nodules were common in chronically infected hamsters, and in some T. gondii antigen could be shown to be present. He proposed that some tissue cysts rupture in chronically infected animals and that the released bradyzoites are destroyed by the immunocompetent host. However, in immunosuppressed animals, the released bradyzoites are believed to reactivate T. gondii infection.
The factors affecting tissue cyst rupture are largely unknown. T. gondii tissue cyst rupture has been documented only rarely. Ferguson et al. (66) quantitatively studied tissue cyst rupture in mice chronically infected with the STR strain of T. gondii. Only 2 (0.27%) ruptured tissue cysts were seen among 750 tissue cysts examined, although T. gondii-positive debris was found in eight (1.4%) glial nodules. Tissue cyst rupture was documented in a Panamanian night monkey (Aotus lemurinus) that had been vaccinated three times with an attenuated (ts-4) strain of T. gondii and then challenged orally with tissue cysts of a complete T. gondii strain (55, 76). Glial nodules were found around degenerating tissue cysts. Degenerating tissue cysts were found in rats inoculated orally with oocysts of the VEG strain in the absence of tachyzoites or formation of new tissue cysts (38).
The mechanism of formation of new generations of tissue cysts in chronically infected mice is unknown (Fig. 27 to 29). Clusters of tissue cysts are found in mice infected with certain strains of T. gondii (163), sometimes in clinically normal mice (Fig. 28). Whether bradyzoites leak from intact tissue cysts is uncertain (174). The rupture of tissue cysts and subsequent multiplication of tachyzoites can lead to fulminating toxoplasmosis even in chronically infected mice (Fig. 29).
FIG. 27.
Section of the cerebrum of a patient with AIDS. Note the large area of necrosis (large arrow) and several small satellite areas (small arrows), probably due to tissue cyst rupture and subsequent growth of tachyzoites. Intact tissue cysts (arrowheads) are also present. All the black dots are T. gondii. Immunohistochemical stain with anti-T. gondii serum.
FIG. 29.
Encephalitis in the brain of a mouse 87 days after being fed oocysts of the VEG strain of T. gondii. Hematoxylin and eosin stain. (A) Coronal section showing grossly visible areas of necrosis (arrows) in the cerebellum and pons. (B) Necrosis (arrowhead) in the cerebrum and numerous tachyzoites (small arrows) in a meningeal blood vessel with severe meningitis and a group of extravascular tachyzoites (large arrow). (C) Four tissue cysts, two of which (arrowheads) are degenerating. (D) Numerous tissue cysts (arrowheads) and tachyzoites (arrows).
FIG. 28.
Cluster of T. gondii tissue cysts in the brain of a mouse which in life showed no apparent clinical signs. Unstained squash smear.
Little information about the mechanisms of relapse is available. Treatments with corticosteroids, antilymphocyte serum, and anti-interferon (IFN) antibody are known to induce immunosuppression and relapse due to toxoplasmosis (3, 78, 81, 82, 119, 126, 162, 165–167). The animal model, route of inoculation and stage of T. gondii used for primary infection, and criteria used for evaluation of relapses are all important considerations. For example, hamsters are more corticosteroid sensitive than are mice or rats, and orally administered corticosteroids are less effective than parenterally administered corticosteroids (74). Chronically infected hamsters initially immunized with the RH strain died of overwhelming toxoplasmosis following corticosteroid treatment (78). Similar relapsing fatal toxoplasmosis has not been induced in mouse models. Odaert et al. (127) examined the brains of chronically infected mice 6, 9, and 12 days after oral administration of dexamethasone. They reported seeing more foci of necrosis and more tissue cysts in cortisone-treated mice than in controls. However, there was no convincing evidence that relapse had occurred in this short period following corticoid administration (74). In experiments involving chronically infected mice, Sumyuen et al. (165) found that there was an increased mortality rate after immunosuppressive therapy (cortisol acetate or azathioprine, or azathioprine plus cortisol acetate) but that the number of T. gondii organisms in the brains and lungs of drug-treated mice remained similar to that found in untreated mice. Similar results were reported recently by Nicoll et al. (126), who found increased numbers of necrotic and gliotic foci, but no increase in parasite numbers, in dexamethasone-treated and chronically infected T. gondii mice. However, many tachyzoites die in the necrotic foci and do not infect new cells.
Miédougé et al. (119) compared the clinical course and parasite numbers in mice injected i.p. with tissue cysts of the Beverley strain of T. gondii. Starting 50 days p.i., group A and B mice were injected weekly with anti-IFN rabbit antiserum and group C were controls. Group B mice also received antitoxoplasmic drugs (pyrimethamine and sulfadiazine in drinking water). On day 68 (i.e., 18 days after treatment), five mice from each group were killed and homogenates of lungs and brains were bioassayed in tissue culture. All the mice remained clinically normal. The T. gondii numbers were dramatically higher in group A mice. All five mice in group A had 856 to 1,337 T. gondii organisms/g of lung, whereas only one of five mice in group B and no mice in group C had organisms. Because histologic examination was not performed, the results may be misleading because rupture of a single tissue cyst while making tissue homogenates for bioassay can result in a ≥1,000-fold difference in T. gondii infectivity titers. The authors also reported that there was parasitemia 53, 61, and 68 days after infection even in mice not given anti-IFN antibody. Thus, the mouse model does not appear to be an accurate representation for relapsing toxoplasmosis in AIDS patients (3). One should also take into account that clusters of tissue cysts of different sizes, numbers of tachyzoites, and reactivated lesions occur in chronically infected, nonimmunosuppressed mice (52).
Although it has been known for 30 years that immunity to T. gondii is cell mediated (71), the precise mechanism of relapse is unknown. Clinical toxoplasmosis in AIDS patients is thought to be due to reactivation of a chronic infection, probably mediated by CD4+ lymphocyte deficiency (82). Production of the cytokine IFN-γ in mice is considered to be the main mediator of immunity to toxoplasmosis in both acute and chronic infection (167, 168). Mice chronically infected with the ME-49 strain of T. gondii died 10 to 18 days after administration of anti-IFN antibody, and depletion of both CD4+ and CD8+ lymphocytes was needed to induce mortality in chronically infected mice (82). Depletion of CD8+ lymphocytes alone induced mortality in some mice, but mortality was not induced in those depleted of only CD4+ lymphocytes.
Lesions, similar to those seen in AIDS patients, were seen in mice that died after the administration of anti-IFN antibody; they consisted of necrosis and infiltration of neutrophils associated with the presence of tachyzoites (82). Because a few tachyzoites might be present even in chronically infected asymptomatic mice, it remains to be determined whether the reactivation is initiated by these “dormant” tachyzoites or bradyzoites released from tissue cysts.
BRADYZOITES AND TISSUE CYST FORMATION IN CELL CULTURE
History
Even though Hogan et al. (89) were the first to report the presence of T. gondii tissue cysts in cell culture, little interest in in vitro cultivation of tissue cysts was shown over the next 25 to 30 years, and few reports appeared in the literature (88, 97, 98, 114, 146). Hoff et al. (88) demonstrated that in vitro-produced tissue cysts led to oocyst excretion in cats, indicating that the tissue cysts were biologically the same as those produced in vivo. A renewed interest in studying the biology of tissue cysts and bradyzoites was evidenced once the importance of toxoplasmic encephalitis in AIDS patients was recognized in the mid-1980s. Improved methods for the induction and detection of tissue cysts were developed in the early 1990s. Many researchers are now using in vitro systems to investigate many exciting aspects of bradyzoite-tachyzoite and tachyzoite-bradyzoite interconversion and the development of tissue cysts.
Studies concerning the development of tissue cysts and conversion of tachyzoites to bradyzoites and of bradyzoites to tachyzoites are not always clear-cut because of the terminology used to describe stages and the gradualness of the conversion process (85). For example, is a PV that contains tachyzoites and bradyzoites, as indicated by immunostaining, a tissue cyst, and if TEM demonstrates a tissue cyst wall surrounding what are structurally tachyzoites, is it a tissue cyst? The results of observations in vitro should always be compared to what is already known about tissue cyst production in vivo (74).
Evidence for Tissue Cyst Formation
The methods used to determine whether tissue cysts and bradyzoites are present in vitro are widely varied.
Light microscopy.
Light microscopy is probably the least definitive of all the methods used to demonstrate tissue cysts in vitro because it is impossible to distinguish bona fide tissue cysts from large groups of tachyzoites. It is somewhat easier to identify free-floating tissue cysts that originated from host cell rupture and are floating in the media in cell cultures, but confusion with groups of tachyzoites still cannot be ruled out. Phase-contrast microscopy may provide a more definitive identification of these stages because the tissue cyst wall is phase lucent (178, 179).
Identification of tissue cysts in cultures stained with Giemsa, PAS, or silver stains can also be highly subjective; PAS and silver stains are superior to Giemsa. The age at which tissue cysts become PAS or silver positive has not been examined critically. Immunostaining with specific monoclonal or polyclonal antibodies is highly specific and is the technique currently used by most researchers (see below). Quantitative studies that use only unstained or histochemically stained cultures must be interpreted with caution.
Acid-pepsin resistance.
The tachyzoites of T. gondii are more susceptible to digestion by acid-pepsin solution than are the bradyzoites (93). Resistance to acid-pepsin digestion was used to determine the numbers of bradyzoites in a cell culture in a modified plaque assay (134). However, this method cannot consistently distinguish tachyzoites from bradyzoites (see below).
Transmission electron microscopy.
TEM can be used to demonstrate tissue cysts and the localization of antigens in bradyzoites and tissue cysts. The matrix of tissue cysts produced in vitro is often absent or greatly reduced (113, 116) compared to tissue cysts in mouse brain (65). Free-floating tissue cysts that originated from host cell rupture during processing for TEM are most likely to exhibit a lack of matrix material. Tissue cysts that form in vitro are generally smaller than those found in vivo, and the identity of the host cell is important (87). Young tissue cysts many contain odd numbers of organisms (Fig. 30) (89, 154, 179). Generally, it is difficult to obtain quantitative data by TEM because of the small amount of material used for examination. Immunoelectron microscopy has been used to determine the expression and localization of bradyzoite-specific antigens in vitro (85, 86, 172, 179).
FIG. 30.
Transmission electron micrograph of a T. gondii tissue cyst in human foreskin fibroblast culture (HS68) 6 days after inoculation with the VEG strain zoites. Note the three bradyzoites with honeycombed rhoptries (arrowheads) and a thin cyst wall (large arrows). The small arrows point to the plasmalemma of the bradyzoites.
Only the bradyzoite stage of T. gondii will reliably produce oocyst excretion in cats. The prepatent period is short in orally acquired bradyzoite-induced infections and can be used to determine if tissue cysts are present in cell cultures (88, 109). By using TEM and by bioassay in cats fed T. gondii, it was determined that tissue cysts were present on day 3 p.i. but were not infectious for cats until day 6 p.i. (109). Repeated passage of T. gondii in cell cultures results in the loss of the ability of bradyzoites to induce oocyst excretion in cats after about 40 passages in vitro (109), a phenomenon similar to that which occurs after repeated passage of tachyzoites in mice (74a). Functional bradyzoites of the oocystless T263 strain of T. gondii have been produced in cell culture and used to orally vaccinate cats against oocyst excretion (134).
Immunostaining.
With the identification of stage-specific antigens and the subsequent generation of bradyzoite and tissue cyst stage-specific monoclonal antibodies, methods for closely examining the developmental biology of T. gondii tissue cysts became available (9, 10, 13, 86, 99, 171, 178, 181, 183). By using both tachyzoite- and bradyzoite-specific monoclonal antibodies, it was determined that tachyzoites and bradyzoites could both be present in the same PV (10, 155, 156, 179), indicating that stage conversion from tachyzoite to bradyzoite is asynchronous. Immunostaining permits accurate quantification of specific stages and is important in studies designed to determine the ability of agents to induce tissue cyst formation in vitro.
Host Cells and T. gondii Strains
The types of host cells and T. gondii strains used to study tissue cyst and bradyzoite development in vitro have been widely varied (Table 4). The majority of studies have been done with fibroblast cell types (mainly of human origin) because these cells are easy to grow and usually survive for extended periods as an intact monolayer. Most cell lines will support tissue cyst development, and the type of cell line probably does not contribute greatly to the presence or absence of tissue cyst formation (109, 116). Most strains of T. gondii will produce tissue cysts in vitro (113, 116). Strains with low replication rates, which are less pathogenic for mice (i.e., VEG, ME-49, Beverley, Prugniaud, and NTE), usually produce more tissue cysts than do the more rapidly dividing, pathogenic strains (i.e., RH and BK) (83, 114). Before the use of methods to induce tachyzoite-to-bradyzoite conversion, strains of T. gondii that were less pathogenic for mice produced tissue cysts with larger numbers of zoites (both bradyzoites and tachyzoites) than did highly pathogenic strains (11, 83).
TABLE 4.
Combinations of host cell types, T. gondii strains, and infective stages that have been observed to produce bradyzoites and tissue cysts
Host cella | T. gondiistrain(s) | Proofb | Stagec | Reference(s)
Human fibroblasts | NTE | MAb | T? | 83
Human fibroblasts | ME-49 | MAb, TEM | Z | 178–180
Human fibroblasts | Prugniaud | TEM | B | 22
Human fibroblasts | Prugniaud | TEM | B | 173
Human fibroblasts | WTD-3, GT-1 | TEM | Z | 113
Human fibroblasts | WTD-3, GT-1 | TEM | Z | 112
Human fibroblasts | ME-49, GT-1, RN-2, RTH-1, WTD-1, WTD-3, WTD-13, WTD-20, D-45-15, D-45-19, D-45-21, D-45-70, D-45-73, D-45-74, D-45-76, D-45-78, D-65-5, D-65-10, D-65-12 | TEM, APD | Z | 113
Human fibroblasts | VEG | TEM, PAb | Z | 108
Human fibroblasts | T-263 | APD | B | 134
Human fibroblasts, Embryonic bovine lung | T-263 | PAS, cat feeding | T? | 135
Human fibroblasts, VERO PLK | | MAb | T? | 170
Human fibroblasts, Verp | RH | TEM, MAb | T | 155
Human fibroblasts, Vero | Prugniaud, 76K, BQNC2 | MAb | B, T | 156
Human retinal pigment epithelial cells | RH | G | T | 122
HeLa, human(?) fibroblasts, retinoblastoma cells | RH | TEM | T | 89
HeLa | Beverley | TEM | T | 114
HeLa | Beverley | LM | UD | 98
HeLa | 76K | MAb | T? | 171
Human fetal lung, MDBK, bovine monocyte | GT-1 | TEM, cat feeding | S, B, Z | 109
Human fetal lung | Beverley, Prugniaud, EA, Jon, Ore, Sau, Tre, squirrel monkey | TEM, LM | Z | 115,116
Monkey fibroblasts | BWM | Cat feeding, PAS, SS | T | 88
Mouse Fibroblasts | NTE | MAb | T? | 86
Mouse fibroblasts | NTE | MAb, PAb, TEM | T | 8
Mouse bone marrow MP | NTE, RH, DX, BK | MAb | T? | 83
PM mouse bone marrow MP | NTE, RH, DX, BK, 561 | MAb | T?, B | 11
PM mouse bone marrow MP | NTE | MAb | T | 10
PM mouse peritoneal MP mouse fibroblasts, mouse MP | NTE | MAb | B, T? | 9
PM mouse bone marrow MP | NTE | MAb, TEM | T | 84
MDBK | ME-49 | MAb | B | 105
MDBK | ME-49 | LM, MAb, PAb | B | 154
MDBK | Beverley (=RRA) | MAb | B | 183
Mouse embryo cells | Beverley, S-273 RH-cyst III | MAb | T? | 128
PM mouse astrocytes | Pe | TEM | B | 97
PM rat astrocytes | Prugniaud | TEM, MAb | T? | 172
Vero | Prugniaud 76K, BQNC2 | MAb | B,T | 157
Rabbit corneal endothelial cells | RH | PAS | T | 146
Human fetal astrocytes and neurons | ME-49 | MAb | T,B | 87
Human fibroblasts | Prugniad | LM | T,B | 138
Human fibroblasts | RH | TEM, PAS | T | 23
MDBK | RH, Martin, ENT | MAb | T | 1
Mixed mouse brain cells | BK | MAb | B | 68
Mouse brain glial cells | DX | MAb, EM | B | 141
HeLa cells | Beverley | LM | T | 98
PM, primary; MP, macrophage.
MAb, monoclonal antibody; APD, acid-pepsin digestion; PAb, polyclonal antibody; LM, light microscopy; G, = Giemsa type stain; SS, silver stain.
B, bradyzoites; T, tachyzoites; S, sporozoites; Z, mixture of bradyzoites and tachyzoites; T?, listed as tachyzoites by authors but probably a mixture of zoites; UD, stage not determined.
Methods of Tissue Cyst Induction
Most strains of T. gondii will spontaneously develop tissue cysts in cell culture with no manipulation (22, 109, 111, 113, 156). However, the number of tissue cysts produced spontaneously is small and manipulation of the cell culture system is needed to increase this number (11, 83, 155, 156, 178).
Early on, the use of T. gondii antiserum in the culture medium was used to promote tissue cyst formation in vitro (88, 98, 146). The results of these studies are difficult to interpret because questionable methods of identification of tissue cysts were used (98, 146) or tissue cyst numbers were not quantifiable (88). Popiel et al. (134) achieved an enhanced rate of tissue cyst formation in cultures treated with 5% rabbit anti-T. gondii serum but attributed the enhancement to a reduction in the replication of tachyzoites. It remains unclear if anti-T. gondii serum actually induces bradyzoite and tissue cyst formation.
Tachyzoite-to-bradyzoite conversion can be induced by applying external stress to various types of infected cell lines. Many investigators use pH manipulation to induce in vitro stage conversion. Both acidic (pH 6.6) and basic (pH 8.0 to 8.2) manipulations will lead to the conversion of tachyzoites to bradyzoites (155, 179). Most researchers use the pH 8 treatment. Exposure of extracellular tachyzoites to medium of pH 8.1 for 1 h increases the formation of bradyzoites and tissue cysts (180). Temperature stress (40°C) and chemical stress (sodium arsenite) will also induce tachyzoite-to-bradyzoite transformation (157). However, sodium arsenite treatment did not result in the production of tissue cysts when examined by TEM (155).
IFN-γ may act to inhibit tachyzoite replication in static cell cultures and permit the spontaneous development of tissue cysts (97). Although IFN-γ does not cause increased expression of bradyzoite-specific antigens or tissue cyst production in human fibroblasts (155), it is highly effective in inducing tissue cyst formation in cultured macrophages. IFN-γ treatment of murine macrophages induces the release of nitric oxide (NO), which reduces tachyzoite replication and induces the expression of bradyzoite-specific antigens (9, 11). If IFN-γ induction of NO release is inhibited by treatment with polymyxin or anti-tumor necrosis factor, no inhibition of tachyzoite replication and no increased expression of bradyzoite antigens occur. The highest level of bradyzoite antigen expression is found in macrophages where the NO release is manipulated to be 40 to 65% of maximum (11). Exogenous NO supplied by sodium nitroprusside will also inhibit tachyzoite replication and induce bradyzoite antigen expression in murine macrophages and in host cells with nonfunctional mitochondria (12). Released NO reacts with iron-sulfur centers of proteins and therefore reacts with several proteins involved in electron transport. The activity of NO in host cells lacking functional mitochondria indicates that its effect is exerted on parasite mitochondria and not on host cell mitochondria.
Mitochondrial inhibitors (oligomycin, antimycin A, atovaquone, rotenone, myxothiazol, and carbonyl cyanide m-chlorophenylhydrazone), which affect different aspects of mitochondrial function, will also induce tachyzoite-to-bradyzoite stage conversion (11, 69, 170). Studies involving host cells with nonfunctional mitochondria indicate that the inhibitors are affecting the T. gondii mitochondrian and not the host cell mitochondria (11, 170).
A common mode of action of the pH manipulations, sodium arsenite treatment, 43°C treatment, NO, and mitochondrial inhibitors is that they can be associated with the induction of heat shock proteins (HSP) (9, 180). The relationship of HSP to these stress inducers is being actively investigated in several laboratories. Bohne et al. (8) have cloned and characterized a bradyzoite-specifically expressed gene (hsp30/BAG-1), which is related to genes encoding small HSP of plants. BAG-1 is located in the cytoplasm of bradyzoites, and expression of the 30-kDa antigen appears to be regulated at the mRNA level (8). Weiss et al. (180) have shown that a monoclonal antibody to the inducible form of HSP70 reacts specifically with bradyzoites and recognizes a 72-kDa antigen. If quercetin is used to inhibit the synthesis of HSP, induction of bradyzoite antigen expression is also inhibited. The roles that HSPs play in bradyzoite development and induction will be an active area of research over the next several years.
Current Knowledge of the Events of Tissue Cyst Development
Sporozoites, tachyzoites, and bradyzoites can develop into tissue cysts in vitro. If sporozoites are used as the inoculum, the appearance of tissue cysts seems to be delayed by several days (109). Ultrastructurally, tissue cysts have been observed as early as 2 days after inoculation of a mixture of tachyzoites and bradyzoites (111), and several TEM studies have documented tissue cysts by 3 days p.i. (109, 179). Tissue cysts present by 3 days p.i. are not biologically mature, because they do not induce oocyst excretion in cats.
Immunostaining studies of cultured cells have identified a small population of tachyzoites that express bradyzoite- and tachyzoite-specific antigens (10, 155), which most probably represent transitional stages. In the absence of any induction treatment, tachyzoites obtained from the peritoneal cavity of mice do not react with bradyzoite-specific antibodies, but by 24 h in culture, a small population will express both tachyzoite and bradyzoite antigens, and cells containing only bradyzoite-reactive organisms are present by 3 days (156). Maximal expression of bradyzoite-specific antigens occurs 4 days after inoculation of NTE tachyzoites in pH 8-induced cell cultures (85).
When mouse brain-derived bradyzoites are used as the inoculum in the absence of any induction treatment, the bradyzoites begin to express tachyzoite-specific antigens as early as 15 h (156). Mixed populations were observed at 24 h, and cells containing only tachyzoites or only bradyzoites were observed at 48 h. The observation of groups of parasites that were reactive with only bradyzoite-specific antibodies indicates that the inoculated bradyzoites can produce additional bradyzoite-reactive stages.
The source of tachyzoites and the strain of T. gondii may influence the development of bradyzoites. According to Appleford and Smith (1), a small population of organisms from peritoneal exudates of mice contain bradyzoites as revealed by staining with a bradyzoite-specific monoclonal antibody (Pb36).
Methods for Tissue Cyst and Bradyzoite Isolation
Few data are available on the isolation of T. gondii tissue cysts from cell cultures. Soête et al. (155) isolated pH 8- or 43°C-induced tissue cysts of the RH strain from Vero or human fibroblast cells by scraping the cells from the monolayer and rupturing host cells with a Dounce homogenizer. Low-speed centrifugation resulted in a sediment that contained tissue cysts and free zoites. Bohne et al. (9) used magnetic cell sorting to isolate pH 8-induced bradyzoites of the NTE strain of T. gondii grown in L929 fibroblasts. A surface-reactive bradyzoite-specific antibody was used to purify the bradyzoites from a mixture of bradyzoites and tachyzoites. The final product consisted of 95 to 98% pure bradyzoites free of host cell contaminants.
IN VIVO STAGE CONVERSION AND FORMATION OF BRADYZOITES AND TISSUE CYSTS
The period needed for stage conversion of T. gondii varies with the inoculum, route of inoculation, and method of examination (Table 5). In these studies, tissue cyst formation was sought in the brains of mice. To circumvent the process by which the parasite reaches the brain tissue from the site of inoculation and is then converted to bradyzoites, Dubey and Frenkel (48) inoculated tachyzoites directly into the brains of mice and found that some tachyzoites converted to bradyzoites between 2 and 3 days p.i. At daily intervals, T. gondii-injected mice were fed to cats and the feces of the cats were examined for oocyst shedding. Cats fed mice infected 1 or 2 days previously never shed oocysts with a short prepatent period (<10 days). The results were not related to the number of T. gondii organisms inoculated, because the cats were fed homogenates of many mice inoculated simultaneously by 4 routes (i.p., s.c., intramuscular, and intracerebral). Dubey and Frenkel (48) also demonstrated tissue cysts by conventional histology with special stains (PAS and silver impregnation method).
TABLE 5.
Summary of in vivo formation of bradyzoites and tissue cysts of T. gondii
Inoculum | Routea | Earliest bradyzoite formation (day p.i.) | Evidence | Reference
Tachyzoites | i.p. | 8 | Histology | 104
Bradyzoites | s.c. | 11 | Histology and TEM | 65
Bradyzoites | i.p. | 7 | Histology | 169
Tachyzoites | i.c., i.p., s.c., i.m. | 3 | Histology and bioassay in cats | 48
Oocysts | Oral and others | 9 | Histology and bioassay in cats
Bradyzoites | Oral and others | 7 | Histology and bioassay in cats
Oocysts | Oral | 7 | Bioassay, histology, IHCb | 52
Bradyzoites | Oral | 6 | Bioassay, histology, IHC | 42
i.m., intramuscular; i.c., intracerebral.
IHC, Immunohistochemical staining with bradyzoite-specific antibody.
Recently, tissue cyst formation in mice inoculated orally with tissue cysts or bradyzoites or with oocysts or sporozoites was examined (42, 52). Tissues of mice fed tissue cysts or oocysts were bioassayed in cats, in mice after pepsin digestion, and by immunohistochemical staining with a bradyzoite-specific antibody (BAG-5).
After the mice were fed tissue cysts or bradyzoites, bradyzoites penetrated through enterocytes, entered various cell types in the lamina propria, and divided into tachyzoites by 18 h p.i. The infection disseminated to extraintestinal organs, with parasitemia occurring at 24 to 48 h p.i. Bioassay and histologic examination showed that tissue cysts formed as early as 6 days p.i. in the brain and other tissues. In general, the detection of BAG-5 antigen paralleled that in the cat bioassay. The acid-pepsin digestion procedure was unreliable because it gave inconsistent results with tissue cysts produced during acute infection. Evidence for direct conversion of bradyzoites to bradyzoites was not found.
After the mice were fed oocysts, sporozoites penetrated enterocytes but developed only in lamina propria cells during the first 12 h p.i. The conversion of sporozoites to tachyzoites to bradyzoites required 7 days. Thus, there was a delay of 1 day in bradyzoite formation after oocyst feeding with respect to that after tissue cyst feeding.
In both bradyzoite-induced and sporozoite-induced oral infections, T. gondii organisms were not seen in histologic sections of the brain until 6 days p.i., and at that time individual BAG-5-positive organisms were seen mixed with BAG-5-negative organisms (Fig. 31). Whether individual BAG-5-positive organisms seen in the brain had migrated from other tissues or were released from tissue cysts could not be determined. The stage of the tachyzoite-to-bradyzoite conversion at which the tissue cyst wall is formed in acute infection is uncertain. In chronically infected mice, a tissue cyst wall is clearly visible around two bradyzoites, both of which also have a terminal nucleus (Fig. 11). Recent studies by Sahm et al. (141) suggest that bradyzoites invading the host cell secrete the ground substance of the cyst wall and that during the invasion process the parasite determines whether it will form tissue cysts or tachyzoites.
FIG. 31.
Section of the brain of a mouse 10 days after it was fed T. gondii oocysts. There are individual (arrowheads) and a large group of (large arrow) BAG-5 positive organisms with different intensities of reactivity. Numerous BAG-5-negative organisms (small arrows) are also present. Immunohistochemical staining with anti-BAG-5 antibody.
SUMMARY OF IN VITRO AND IN VIVO CYST FORMATION
The recent development of stage-specific antibodies makes it possible to study in vitro conversion of bradyzoites to tachyzoites and in vitro conversion of tachyzoites to bradyzoites. Soête et al. (156, 157) inoculated bradyzoites onto MRC-5 cell cultures and monitored the appearance of bradyzoite-specific (B+) or tachyzoite-specific (P30) antigens. At 15 h after inoculation, the organisms had both B+ and P30 antigens. Organisms began to divide at approximately 24 h p.i. and vacuoles containing two doubly labelled organisms were seen. By 48 h p.i. some organisms had lost B+ antigens and were considered to be tachyzoites. A quantitative analysis was not possible after 48 h p.i. because some parasites had ruptured and reinvaded other host cells. These authors concluded that not all bradyzoites transform into tachyzoites, since they found multiplying bradyzoites (156); these results were different from in vivo transformation of all bradyzoites into tachyzoites by 48 h p.i. as determined by the cat bioassay. The same authors also observed that cell cultures inoculated with tachyzoites from the peritoneal exudate of mice had completely lost tachyzoite-specific markers by 72 h p.i. Bohne et al. (9, 10) obtained similar results in cell cultures inoculated with bradyzoites. The bradyzoite-specific antigens declined from 100 to 15% from day 1 to 3 after inoculation of cultures. Conversely, the bradyzoite-specific antigens increased steadily from day 1 to 6 in cultures inoculated with tachyzoites.
By using stage-specific monoclonal antibodies, bradyzoite-specific antigens were detected in the brains of mice on day 9 (127). Mice were examined 6, 7, 9, 12, and 14 days after being fed 20 tissue cysts of the 76K strain. The brains of the mice were stained with tachyzoite-specific antibodies (P30) and bradyzoite-specific antibodies (P36). On days 6 and 7, only P30-specific organisms were present. On day 9, groups of parasites were labelled doubly with P30 and P36 antibodies. On days 12 and 14, the number of P30-positive organisms decreased whereas the number of P36-positive organisms increased.
BAG-5 develops early in vitro and is probably a small HSP-like molecule associated with development. Weiss et al. (179), using the BAG-5 bradyzoite-specific antibody, found that tissue cysts had formed by day 3 after inoculation of human foreskin fibroblasts with bradyzoites of the ME-49 strain; it appears that some bradyzoites formed tissue cysts directly without conversion to tachyzoites. Using a monoclonal antibody specific for the tissue cyst wall, Halonen et al. (87) found tissue cysts in human fetal neuronal culture beginning on day 2 after inoculation with the ME-49 strain. The development of bradyzoites in vitro occurs in a series of steps, whereas the development of mature tissue cysts is less frequent and requires a minimum of 6 days in culture. This is supported by studies of antigen expression as ascertained by the various bradyzoite-specific sera available. Most bradyzoite-specific monoclonal antibodies and recombinant polyclonal sera available are expressed after 1 day in culture under conditions which induce bradyzoite switching (9–11, 155, 156). The late-appearing antigen is not seen until 5 to 6 days of culture. In addition, results of feeding experiments suggest that even at 6 days, not all of the observed tissue cyst-like structures are mature, since only a subset behave as tissue cysts in animal bioassays (88, 109). Thus, to some extent, the data on in vitro cystogenesis mimic the data seen in this study. Early on, one would expect tachyzoites that are in process of becoming mature bradyzoites to express BAG-5 antigen (day 1 of transition), but full biologic maturity would take several more days. It is clear from the above discussion that for in vitro work, markers of mature functional cysts are needed.
While these in vitro studies are helpful in elucidating stage conversion, the results do not completely agree with those of in vivo studies. The 15 h needed to convert bradyzoites to tachyzoites in cell culture agrees with the data obtained with mice, where functional tachyzoites were not found 18 h after bradyzoites were fed to mice (42). The 72 h needed for tachyzoites to completely lose their tachyzoite-specific markers also agrees with the 3 days needed to form biologically functional tissue cysts after inoculation of tachyzoites into mice (48). However, when bradyzoites were inoculated into mice by any route, the minimum period needed to form biologically functional tissue cysts was 6 days (48). Unlike in vitro studies, all bradyzoites had converted to biologically defined tachyzoites by 18 h p.i. (42). Therefore, biologic measurements should be examined in the context of parasitologic and host factors (74), and caution should be used in interpreting in vitro phenomena as having biological significance.
RESISTANCE OF T. GONDII TACHYZOITES AND BRADYZOITES TO ACID-PEPSIN DIGESTION
Until recently, acid-pepsin digestion was a generally accepted method to distinguish tachyzoites from bradyzoites and to recover T. gondii from tissues. Biologically, bradyzoites are resistant to gastric digestion and thus remain orally infective whereas tachyzoites are often destroyed by gastric juice. This resistance of bradyzoites to digestion by gastric juice has been known for over 36 years. Jacobs et al. (93) found that bradyzoites can survive in acid-pepsin solution for 2 h or more whereas tachyzoites are killed within 1 h. These authors digested homogenates of liver, spleen, and lungs of mice, inoculated i.p. with RH strain tachyzoites, for 60 min in acid-pepsin solution and found that tachyzoites were rendered noninfective to mice. Direct microscopic examination of tachyzoites in acid-pepsin revealed that they were immediately damaged; they became more granular, less refractile, and ghostlike within 15 to 30 min (93). However, Jacobs et al. (93) tested the infectivity of tachyzoites for mice after digestion in acid-pepsin for only a 60-min period. Subsequently, most other researchers accepted that tachyzoites are immediately destroyed by acid-pepsin. Unlike acid-pepsin, tachyzoites survived in 1% trypsin for 3 h (93).
Sharma and Dubey (143) quantitatively studied survival of tachyzoites and bradyzoites in acid-pepsin and trypsin solutions. They reported that bradyzoites survived in acid-pepsin for 2 h but not for 3 h.
According to Pettersen (130), the destruction of tachyzoites in acid-pepsin was due to acid, not to pepsin, because no differences were found in survival rates when tachyzoites were incubated in acid-pepsin at room temperature or at 37°C (pepsin is active only at 37°C and at low pH). Pettersen (130) also reported that tachyzoites of two virulent strains (RH and 119) survived in acid-pepsin for 20 min but not 25 min. He thought that bradyzoites were present in the peritoneal exudate of mice inoculated with bradyzoites of two avirulent strains of T. gondii. The peritoneal exudate was obtained 4 or 6 days after i.p. inoculation with bradyzoites of strain DUE; the organisms in the peritoneal exudate survived 90 min in acid-pepsin at 37°C.
In a follow-up paper, Pettersen (131) proposed that bradyzoites can be excreted in the milk of mice 5 days after i.p. inoculation of lactating mice with 1,000 bradyzoites. The evidence for this was that milk treated with HCl for 60 min at room temperature produced T. gondii infection in bioassayed mice. This result disagrees with the conclusion reached by Dubey and Frenkel (48), who found that bradyzoites were not formed in any tissue of mice until 7 days after bradyzoite inoculation.
Popiel et al. (134) used the acid-pepsin digestion procedure to quantify the development of bradyzoites in cell cultures by using cell culture as a bioassay. Tachyzoites of the T-263 strain, obtained by a 2-day cultivation in cell culture, were killed after a 10 min digestion in acid-pepsin. These authors used the same concentration of acid as used by Jacobs et al. (93) but only 10% of the pepsin used by Jacobs et al. (93). Bradyzoites produced in cell culture survived acid-pepsin digestion for 30 to 60 min. Popiel et al. (134) concluded that organisms that resisted 30 min of acid-pepsin digestion were bradyzoites.
Lindsay et al. (113) compared the appearance of acid- pepsin-resistant organisms with the development of tissue cysts by TEM in cell cultures inoculated with the RH strain and a temperature-sensitive (ts4) mutant derived from it. Tissue cysts were not seen in cell cultures inoculated with these two strains, but the organisms survived acid-pepsin digestion.
To resolve whether the inoculum used to test the effect of pepsin digestion, the strain of T. gondii used, the method used to obtain tachyzoites, and the source of tachyzoites affected results, Dubey (43) conducted experiments with extracellular tachyzoites from the peritoneal exudate obtained 3 to 9 days after i.p. inoculation of mice with tachyzoites. The following conclusions were drawn from this study: (i) tachyzoites occasionally survived acid-pepsin digestion for 2 h, which was not due to protection within host cells; (ii) the strain of T. gondii did not affect the susceptibility of tachyzoites to acid-pepsin; and (iii) even extracellular tachyzoites were infective to mice orally, but the infectivity was dose dependent (the infective dose of tachyzoites by the oral route in mice was 1,000). Therefore, it was concluded that one cannot rely on oral infectivity in mice or digestion in acid-pepsin as the sole criterion to distinguish between tachyzoites and bradyzoites.
FIG. 9.
Transmission electron micrograph of a mouse peritoneal macrophage containing several tachyzoites of the VEG strain of T. gondii, one of which is escaping from the macrophage near the top of the micrograph (arrow). Note that the PV is no longer evident and the tachyzoites are free in the macrophage cytoplasm.
FIG. 14.
High magnification of a portion of Fig. 13 showing the cyst wall and part of a bradyzoite with an active micropore. The cyst wall is approximately 0.25 to 0.75 μm thick and consists of a parasitophorous vacuolar membrane (Pm) that has numerous indentations (arrows); the rest of the wall consists of membrane-bound vesicles and moderately course granular material. Note that the micropore of the bradyzoite consists of an indentation in the parasite plasmalemma (Pl) and of an electron-dense collar (Ec) that arises from the inner membrane complex (Im); also note the presence of an electron-dense layer (EI) of material on the surface of the plasmalemma within the micropore. Am, amylopectin granule; Mn, microneme; Sm, subpellicular microtubule.
ACKNOWLEDGMENTS
We thank J. A. Blixt, K. Prokop, O. C. H. Kwok, and S. K. Shen for technical assistance and David Fritz, USAMRIID, Fort Detrick, Md., for taking the photographs shown in Fig. 19 and 20.
This work was supported in part by the USDA Animal Health Funds Grant No. 192380 and MONB 101406.
Footnotes
Contribution J5148 from the Montana State University Agricultural Experiment Station.
REFERENCES
Articles from Clinical Microbiology Reviews are provided here courtesy of American Society for Microbiology (ASM)
ACTIONS
PERMALINK
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
Cite
Add to Collections
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894 |
7727 | https://www.mathway.com/popular-problems/Algebra/260523 | Solve by Factoring 2x^2=4x | Mathway
Enter a problem...
[x]
Algebra Examples
Popular Problems
Algebra
Solve by Factoring 2x^2=4x
2 x 2=4 x 2 x 2=4 x
Step 1
Subtract 4 x 4 x from both sides of the equation.
2 x 2−4 x=0 2 x 2-4 x=0
Step 2
Factor 2 x 2 x out of 2 x 2−4 x 2 x 2-4 x.
Tap for more steps...
Step 2.1
Factor 2 x 2 x out of 2 x 2 2 x 2.
2 x(x)−4 x=0 2 x(x)-4 x=0
Step 2.2
Factor 2 x 2 x out of −4 x-4 x.
2 x(x)+2 x(−2)=0 2 x(x)+2 x(-2)=0
Step 2.3
Factor 2 x 2 x out of 2 x(x)+2 x(−2)2 x(x)+2 x(-2).
2 x(x−2)=0 2 x(x-2)=0
2 x(x−2)=0 2 x(x-2)=0
Step 3
If any individual factor on the left side of the equation is equal to 0 0, the entire expression will be equal to 0 0.
x=0 x=0
x−2=0 x-2=0
Step 4
Set x x equal to 0 0.
x=0 x=0
Step 5
Set x−2 x-2 equal to 0 0 and solve for x x.
Tap for more steps...
Step 5.1
Set x−2 x-2 equal to 0 0.
x−2=0 x-2=0
Step 5.2
Add 2 2 to both sides of the equation.
x=2 x=2
x=2 x=2
Step 6
The final solution is all the values that make 2 x(x−2)=0 2 x(x-2)=0 true.
x=0,2 x=0,2
2 x 2=4 x 2 x 2=4 x
+2+2+3+3+5+5
(
(
)
)
|
|
[
[
]
]
√
√
≥
≥
7
7
8
8
9
9
≤
≤
4
4
5
5
6
6
/
/
^
^
×
×
∩
∩
∪
∪
1
1
2
2
3
3
-
-
+
+
÷
÷
<
<
π
π
∞
∞
,
,
0
0
.
.
%
%
=
=
Report Ad
Report Ad
⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
Please ensure that your password is at least 8 characters and contains each of the following:
a number
a letter
a special character: @$#!%?&
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Targeting Cookies
[x] Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
7728 | https://math.stackexchange.com/questions/4686316/every-graph-is-bipartite-if-and-only-iff-every-cycle-is-even | Every graph is bipartite if and only iff every cycle is even. - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Every graph is bipartite if and only iff every cycle is even.
Ask Question
Asked 2 years, 5 months ago
Modified2 years, 5 months ago
Viewed 703 times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I have a problem with this proof:
Let G be a graph with no odd cycles. We will consider the case that G is connected; this is sufficient since if we can show that each connected component of a graph is bipartite, then it follows that the graph as a whole is bipartite. Let d(u, v) denote the length of the shortest path between two vertices in G. Pick an arbitary vertex u ∈ V and define A = {u} ∪ {w | d(u, w) is even}. Define B = V \ A. We claim that the partition V = A ∪ B demonstrates that G is bipartite. Assume for contradiction that there exists an edge {w, v} ∈ E with w, v ∈ A (or B). Then by construction d(u, w) and d(u, v) are both even (or odd). Let Puw and Pvu be the shortest paths connecting u to w, and v to u respectively. Then the cycle given by Puw, {w, v}, Pv,u has length 1 + d(u, w) + d(u, v), which is odd, a contradiction. Therefore no such edge {w, v} may exist and G is bipartite.
How do we know that P u,w P u,w and P v,u P v,u are disjoint? And that v v does not occur in P u,w P u,w and that w w does not occur in P v,u P v,u?
graph-theory
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
edited Apr 26, 2023 at 3:18
Sarvesh Ravichandran Iyer
78.1k 9 9 gold badges 88 88 silver badges 161 161 bronze badges
asked Apr 25, 2023 at 15:38
user394334user394334
1,717 12 12 silver badges 22 22 bronze badges
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
This is not a mistake, it's just a conflict of terminology.
The terminology I consider more modern uses closed walk for the general concept of a sequence of vertices and edges in the graph that represents going from vertex to adjacent vertex for a while and eventually returning where you started. A cycle is a closed walk with no repetition of vertices or edges aside from the first and last vertex being the same. From this point of view, the proof above looks odd.
It is also not too uncommon to replace the terms "closed walk" and "cycle" by "cycle" and "simple cycle", respectively. This seems to be the terminology used by the author of these solutions, since later on the solutions mention an "Eulerian cycle". So there is no mistake in the proof; it just proves a different statement (for which we do not need to check any of these disjointness conditions).
Going back to the first terminology, we have here a correct proof that a graph is bipartite if and only if it has no closed walks of odd length. It is also true that a graph is bipartite if and only if it has no cycles of odd length, but to see this requires an additional step: showing that if a graph has an odd closed walk, then it has an odd cycle. (The converse is immediate.)
The quickest way to see that is to let W W be the shortest odd closed walk. If W W is not a cycle, then a vertex in the middle of W W is visited multiple times. This lets us break up W W into two shorter walks, W 1 W 1 and W 2 W 2, with the length of W W equal to the sum of the lengths of W 1 W 1 and W 2 W 2; then one of W 1 W 1 and W 2 W 2 will be a shorter odd closed walk, contradicting our initial assumption. So W W is actually an odd cycle.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Apr 25, 2023 at 15:53
Misha LavrovMisha Lavrov
160k 11 11 gold badges 167 167 silver badges 304 304 bronze badges
2
Do we have to have that in the walk P u,w{w,v}P v,u P u,w{w,v}P v,u that no edge is used more than once? If yes, how do we prove this?user394334 –user394334 2023-04-25 16:33:04 +00:00 Commented Apr 25, 2023 at 16:33
Do you mean in the original statement? In the original statement, that is also not guaranteed; we just have an arbitrary odd closed walk, which may repeat vertices and may repeat edges. (In the argument in the last paragraph, that follows for the shortest odd closed walk, because repeating an edge there also entails repeating a vertex.)Misha Lavrov –Misha Lavrov 2023-04-25 17:50:40 +00:00 Commented Apr 25, 2023 at 17:50
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
graph-theory
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
5What is the maximum length of shortest odd cycle in a non-bipartite graph?
9If a graph has no cycles of odd length, then it is bipartite: is my proof correct?
2Number of odd cycles in non-bipartite 3-connected graph
3Graph with no even cycles has each edge in at most one cycle?
3Planar graphs where every face boundary is a cycle of even length are bipartite
1Prove that if G is a graph with no even cycles, then every cycle in G is an induced subgraph.
0Help formulating a conjecture about the parity of every cycle length in a bipartite graph and proving it
Hot Network Questions
Numbers Interpreted in Smallest Valid Base
How to solve generalization of inequality problem using substitution?
Overfilled my oil
Is there a way to defend from Spot kick?
Is it safe to route top layer traces under header pins, SMD IC?
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
What's the expectation around asking to be invited to invitation-only workshops?
Why include unadjusted estimates in a study when reporting adjusted estimates?
Is encrypting the login keyring necessary if you have full disk encryption?
What happens if you miss cruise ship deadline at private island?
In the U.S., can patients receive treatment at a hospital without being logged?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
How to home-make rubber feet stoppers for table legs?
What is the meaning and import of this highlighted phrase in Selichos?
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Discussing strategy reduces winning chances of everyone!
Should I let a player go because of their inability to handle setbacks?
Who is the target audience of Netanyahu's speech at the United Nations?
How to rsync a large file by comparing earlier versions on the sending end?
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
Proof of every Highly Abundant Number greater than 3 is Even
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Why do universities push for high impact journal publications?
Making sense of perturbation theory in many-body physics
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7729 | https://flexbooks.ck12.org/cbook/ck-12-interactive-algebra-2-for-ccss-2nd-edition/section/6.4/primary/lesson/characteristics-of-rational-functions/ | Skip to content
Elementary Math
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Conventional
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Probability & Statistics
Trigonometry
Math Analysis
Precalculus
Calculus
What's the difference?
Science
Grade K to 5
Earth Science
Life Science
Physical Science
Biology
Chemistry
Physics
Advanced Biology
FlexLets
Math FlexLets
Science FlexLets
English
Writing
Spelling
Social Studies
Economics
Geography
Government
History
World History
Philosophy
Sociology
More
Astronomy
Engineering
Health
Photography
Technology
College
College Algebra
College Precalculus
Linear Algebra
College Human Biology
The Universe
Adult Education
Basic Education
High School Diploma
High School Equivalency
Career Technical Ed
English as 2nd Language
Country
Bhutan
Brasil
Chile
Georgia
India
Translations
Spanish
Korean
Deutsch
Chinese
Greek
Polski
EXPLORE
Flexi
A FREE Digital Tutor for Every Student
FlexBooks 2.0
Customizable, digital textbooks in a new, interactive platform
FlexBooks
Customizable, digital textbooks
Schools
FlexBooks from schools and districts near you
Study Guides
Quick review with key information for each concept
Adaptive Practice
Building knowledge at each student’s skill level
Simulations
Interactive Physics & Chemistry Simulations
PLIX
Play. Learn. Interact. eXplore.
CCSS Math
Concepts and FlexBooks aligned to Common Core
NGSS
Concepts aligned to Next Generation Science Standards
Certified Educator
Stand out as an educator. Become CK-12 Certified.
Webinars
Live and archived sessions to learn about CK-12
Other Resources
CK-12 Resources
Concept Map
Testimonials
CK-12 Mission
Meet the Team
CK-12 Helpdesk
FlexLets
Know the essentials.
Pick a Subject
Donate
Sign Up
6.4
Characteristics of Rational Functions - ALG II
Written by:Sean Regan
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
This is our newest and most engaging Algebra 2 content! If you would like to revisit the previous version of Algebra 2, please CLICK HERE.
Common Core Standards
Focus Standards: F.IF.4, A.APR.6
Additional Standards: F.IF.7d, F.BF.3
Lesson Objectives
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.
Rewrite simple rational expressions in different forms
Agenda
| | |
--- |
| Introduction: Bright Lights Revisited | 5 minutes |
| Activity 1: Vertical Asymptotes | 10 minutes |
| Activity 2: Horizontal Asymptotes | 10 minutes |
| Activity 3: Oblique Asymptotes | 10 minutes |
| Activity 4: Using Characteristics to Graph Rational Functions | 10 minutes |
| Wrap-Up: Review Questions | 5 minutes |
(Students cannot see the purple text)
In this lesson, students will derive the formulas for determining the asymptotes of a rational function. This lesson will delve deeper into the algebra behind rational functions. Take sufficient time when presenting the rules for determining asymptotes; many students memorize the rules without understanding them which can be problematic in higher levels of math.
Introduction: Bright Lights Revisited
Much like there isn't a specific parent polynomial function, but instead a parent function for each power, there isn't a specific parent rational function.
In the section Rational Functions, you use the function @$\begin{align}B = \tfrac{L}{4 \pi d^2}\end{align}@$ to find the apparent brightness of a light source based on distance from it. The numerator of the function represented the luminosity, or light energy, of the light source, and the denominator represented the surface area of the light photons as they traveled in all directions in 3 dimensions. The denominator takes on the formula for the surface area of a sphere, @$\begin{align}4\pi r^2,\end{align}@$ where the radius of the sphere is the distance of the light from the source.
This function has a parent function of @$\begin{align}f(x)=\tfrac{1}{x^2}.\end{align}@$ Not only does the structure of a rational function affect the graph, but transformations can affect the graph like they would any other function. Use the interactive below to examine how the structure of a rational function affects its graph.
Inline Questions
Given the graph of the function @$\begin{align}f(x)=\tfrac{1}{x^n},\end{align}@$ what must be true about @$\begin{align}n?\end{align}@$
@$\begin{align}n\end{align}@$ is an odd number
@$\begin{align}n\end{align}@$ is a negative number
@$\begin{align}n\end{align}@$ is an even number
@$\begin{align}n\end{align}@$ is a decimal
Given the graph of the function @$\begin{align}f(x)=\tfrac{1}{x^n},\end{align}@$ what do you know about @$\begin{align}n?\end{align}@$
@$\begin{align}n\end{align}@$ is an odd number
@$\begin{align}n\end{align}@$ is a negative number
√ @$\begin{align}{\color{white}n}\end{align}@$ is an even number
@$\begin{align}n\end{align}@$ is a decimal
What kind of function is @$\begin{align}f(x)=\tfrac{1}{x^{-2}}?\end{align}@$
linear
quadratic
cubic
rational
True or False: It is possible for a rational function to have a range of all real numbers?
TRUE
FALSE
(Fill in the Blank) The function @$\begin{align}R(x)=\tfrac{f(x)}{g(x)},\end{align}@$ if @$\begin{align}g(x)\end{align}@$ has a degree of n, @$\begin{align}R(x)\end{align}@$ could have up to n vertical asymptotes.
How will the graph of the function @$\begin{align}g(x)=\tfrac{60}{4\pi d^2}\end{align}@$ change from the graph from the parent function @$\begin{align}f(x)=\tfrac{1}{d^2}?\end{align}@$
stretched horizontally by a factor of 4
stretched vertically by a factor of 60 and horizontally by a factor of @$\begin{align}4\pi\end{align}@$
stretched vertically by a factor of 60
stretched vertically by a factor of @$\begin{align}\tfrac{15}{\pi}\end{align}@$
How would the graph of the function @$\begin{align}g(x)=\tfrac1 x + 2\end{align}@$ change from the graph of @$\begin{align}f(x)=\tfrac1 x?\end{align}@$
shifted up by 2 units
shifted down by 2 units
stretched vertically by a factor of 2
shifted right by 2 units
Which of the following equations represents the function @$\begin{align}f(x)=\tfrac{x}{x^3+3}\end{align}@$ when shifted right 2 units?
@$\begin{align}g(x)=\tfrac{x-2}{(x-2)^3+3}\end{align}@$
@$\begin{align}g(x)=\tfrac{x-2}{x^3+1}\end{align}@$
@$\begin{align}g(x)=\tfrac{x-2}{x^3+3}\end{align}@$
@$\begin{align}g(x)=\tfrac{x}{x^3+3}+2\end{align}@$
@$\begin{align}g(x)=\tfrac{x+2}{(x+2)^3+3}\end{align}@$
Activity 1: Vertical Asymptotes
A rational function is undefined at the vertical asymptote(s). Since a fraction cannot be divided by zero, the vertical asymptotes will be at values that result in the denominator being zero. Find these values by setting the denominator equal to zero and solving the equation. Recall that a rational function can be defined as @$\begin{align}r(x)=\tfrac{f(x)}{g(x)}\end{align}@$ given that @$\begin{align}f(x) \text{ and } g(x)\end{align}@$ are polynomial functions. The vertical asymptotes are the zeroes of @$\begin{align}g(x).\end{align}@$ Since it is possible for a function to have unlimited zeroes, a rational function can have unlimited asymptotes.
Example
Find the vertical asymptotes of the function @$\begin{align}r(x)=\frac{x}{x^2 - 9}\end{align}@$
The first step in finding the vertical asymptote(s) is to set the denominator equal to zero.
@$$\begin{align}x^2-9=0\end{align}@$$
Solving this equation will result in the following zeroes:
@$$\begin{align}x^2-9&=0\
(x-3)(x+3)&=0\
x = 3 \text{ and } x = &-3\end{align}@$$
Answer: The vertical asymptotes are at @$\begin{align}x = 3\end{align}@$ and @$\begin{align}x = -3.\end{align}@$
Inline Questions
What are the vertical asymptotes of the following function? @$\begin{align}g(x)=\tfrac{-x}{(x+1)(x-3)}\end{align}@$
@$\begin{align}x=1, x=3\end{align}@$
@$\begin{align}x=-1, x=-3\end{align}@$
@$\begin{align}x=1, x=-3\end{align}@$
@$\begin{align}x=-1, x=3\end{align}@$
Find the vertical asymptote(s) of the function: @$\begin{align}f(x)=\tfrac{4x-1}{x^2-16}\end{align}@$
@$\begin{align}x=4, x=-4\end{align}@$
@$\begin{align}x=8, x=-4\end{align}@$
@$\begin{align}x=3, x=-4\end{align}@$
@$\begin{align}x=6, x=-4\end{align}@$
(Select All) Find the vertical asymptote(s) of the function: @$\begin{align}y=\tfrac{x^2-4x}{x^2-2x+1}\end{align}@$
@$\begin{align}x=0\end{align}@$
@$\begin{align}x=4\end{align}@$
@$\begin{align}x=2\end{align}@$
@$\begin{align}\color{green}{x=1}\end{align}@$
Activity 2: Horizontal Asymptotes
Horizontal asymptotes represent a value that the output of a function approaches but never reaches as the x value increases or decreases without bound. When observing different rational functions, notice patterns in the values they approach as the input increases and decreases. Use the interactive below to look for these patterns.
Inline Questions
(Fill in the Blank) Describe the end behavior of @$\begin{align}r(x)\end{align}@$ when the degree of @$\begin{align}f(x)\end{align}@$ is less than the degree of @$\begin{align}g(x):\end{align}@$
as @$\begin{align}x\rightarrow \infty, r(x) \rightarrow\end{align}@$0
as @$\begin{align}x\rightarrow -\infty, r(x) \rightarrow\end{align}@$0
(Short Answer) Set slider a to 8 and slider c to 4. When the degree of @$\begin{align}f(x)\end{align}@$ is equal to the degree of @$\begin{align}g(x),\end{align}@$ what does @$\begin{align}r(x)\end{align}@$ approach as @$\begin{align}x\end{align}@$ approaches infinity?
2
(Fill in the Blank) To find a horizontal asymptote, you must compare the degrees of the numerator @$\begin{align}f(x)\end{align}@$ and denominator @$\begin{align}g(x).\end{align}@$ Use the interactive to help you fill in the blank:
If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is @$\begin{align}y=\tfrac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}.\end{align}@$
If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is @$\begin{align}y=0.\end{align}@$
Find the horizontal asymptote (if it exists). @$\begin{align}f(x)=\tfrac{1}{x-2}\end{align}@$
There is a horizontal asymptote at @$\begin{align}y=2\end{align}@$
There is no horizontal asymptote.
There is a horizontal asymptote at @$\begin{align}y=0\end{align}@$
There is a horizontal asymptote at @$\begin{align}y=1\end{align}@$
Find the horizontal asymptote (if it exists). @$\begin{align}f(x)=\tfrac{6x^3+1}{4x^3-2x+3}\end{align}@$
There is a horizontal asymptote at @$\begin{align}y=0\end{align}@$
There is no horizontal asymptote
There is a horizontal asymptote at @$\begin{align}y=1\end{align}@$
There is a horizontal asymptote at @$\begin{align}\textcolor{green}{y=\tfrac32}\end{align}@$
Use the interactive below for more practice finding horizontal asymptotes.
Activity 3: Oblique Asymptotes
When the degree of the leading term in the numerator is greater than the degree of the leading term in the denominator, there is no horizontal asymptote. In this case, there is an oblique asymptote instead of a horizontal asymptote. An oblique asymptote is a slanted line that a function approaches but never touches.
To find the equation of an oblique asymptote, divide the numerator by the denominator. In the previous activity, the equations for horizontal asymptotes were derived based on the outputs approached by the function as the value of x increased or decreased. A similar approach can be used to derive the equation of an oblique asymptote.
Example
Find the oblique asymptote of the function @$\begin{align}g(x) = \frac{x^3-4x^2+7x-10}{x^2-5x+3}.\end{align}@$
As the value of x increases, the value of @$\begin{align}g(x)\end{align}@$ will approach infinity. Consider the leading terms in both the numerator and denominator. The value of @$\begin{align}x^3\end{align}@$ will increase faster than the value of @$\begin{align}x^2\end{align}@$ by a rate of @$\begin{align}x.\end{align}@$ Furthermore, it can be stated that the value of @$\begin{align}x^3-4x^2+7x-10\end{align}@$ will increase faster than @$\begin{align}x^2-5x+3\end{align}@$ by a factor of @$\begin{align}\frac{x^3-4x^2+7x-10}{x^2-5x+3}.\end{align}@$ Dividing the numerator by the denominator will produce the oblique asymptote.
Use the quotient @$\begin{align}x+1+\frac{9x-13}{x^2-5x+3}\end{align}@$ to find the equation of the asymptote. The remainder can be removed because as the value of x increases, the value of the remainder will approach 0, a result of the degree of the numerator being less than the degree of the denominator.
Answer: The equation of the asymptote is @$\begin{align}y=x+1.\end{align}@$
All asymptotes thus far have been linear; however, polynomial division could produce a quadratic function, a cubic function, etc. Consider the function @$\begin{align}h(x)=\tfrac{x^4+1}{x^2-1}.\end{align}@$ The quotient of @$\begin{align}\tfrac{x^4+1}{x^2-1}\end{align}@$ is @$\begin{align}x^2 +1 + \tfrac{2}{x^2-1}.\end{align}@$ Since the remainder approaches 0 as the value of x increases, the result will be an oblique asymptote of @$\begin{align}y=x^2+1.\end{align}@$ How does the equation @$\begin{align}y=x^2+1\end{align}@$ relate to the function @$\begin{align}h(x)?\end{align}@$
Use the interactive below to explore this relationship.
Activity 4: Using Characteristics to Graph Rational Functions
Using your knowledge of asymptotes coupled with your knowledge of intercepts, you can quickly graph a rational function. Recall that the @$\begin{align}x\end{align}@$-intercept is the value of @$\begin{align}x\end{align}@$ when @$\begin{align}f(x)=0.\end{align}@$ Additionally, the y-intercept is the value of @$\begin{align}f(x)\end{align}@$ when @$\begin{align}x=0.\end{align}@$
Example
Graph the function below after finding the asymptotes, domain, x-intercepts, and y-intercept.
@$$\begin{align}f(x) = \frac{3x}{x^2 - 3x - 4}\end{align}@$$
a. Asymptotes
The vertical asymptote can be found with any value of x that makes the denominator undefined once the fraction is fully simplified. Since this function cannot be simplified, set the denominator to 0 and solve for x.
@$$\begin{align}x^2 - 3x - 4=0\
(x-4)(x+1)=0\end{align}@$$
The vertical asymptotes are @$\begin{align}x = 4 \text{ and } x = -1.\end{align}@$
The following are the rules for finding the horizontal asymptote for a function of the form @$\begin{align}r(x)=\frac{ax^n + ...}{bx^d + ...},\end{align}@$ derived in the previous activity.
| | | |
---
| | If n < d | The horizontal asymptote is @$\begin{align}y = 0.\end{align}@$ |
| | If n = d | The horizontal asymptote is @$\begin{align}y = \frac{a}{b}.\end{align}@$ |
| | If n > d | There is no horizontal asymptote. |
In this case, the degree of the numerator is less than the degree of the denominator. The horizontal asymptote will be @$\begin{align}y=0.\end{align}@$
b. Domain
The domain of a rational function will be all real numbers except for any holes or vertical asymptotes. These values will result in an undefined output. Since this function has no holes and the vertical asymptotes are at @$\begin{align}x = 4 \text{ and } x = -1,\end{align}@$ the domain will be all real numbers except 4 and -1. This can be written using set notation as @$\begin{align}{x | x ∈ ℝ, x\neq -1, 4}\end{align}@$ or interval notation as @$\begin{align}(-\infty , -4)\text{⋃}(-4,1)\text{⋃}(1,\infty ).\end{align}@$
c. x-intercept
Since the y-value of the x-intercept is 0, the x-intercept can be found by setting @$\begin{align}f(x)\end{align}@$ equal to 0 and solving for x.
@$$\begin{align}0 = \frac{3x}{x^2 - 3x - 4}\end{align}@$$
From here, multiply both sides by @$\begin{align}x^2-3x-4\end{align}@$ to cancel the denominator, leaving you with only the numerator. Another approach could be putting 0 over 1 and cross multiplying.
@$$\begin{align}(x^2-3x-4)\cdot0 &= \frac{3x}{\cancel{x^2 - 3x - 4}}\cdot\cancel{(x^2-3x-4)}\\
\end{align}@$$@$$\begin{align}0 &= 3x\
\div3 & \:\div3\
\hline
0 &= x\end{align}@$$
This function will have one x-intercept at (0,0).
d. y-intercept
The y-intercept can be found at @$\begin{align}f(0).\end{align}@$ For a point to land on the y-axis, it must have an x-value of 0.
@$$\begin{align}f(0) &= \frac{3(0)}{(0)^2 - 3(0) - 4}\\
&=\frac{0}{-4}\\
&=0\end{align}@$$
The y-intercept will also be at (0,0). You could have figured this out without finding @$\begin{align}f(0)\end{align}@$ since the x-intercept was on the y-axis.
e. Graph the @$\begin{align}f(x)\end{align}@$
A quick method for graphing rational equations involves first graphing the asymptotes. Once these are obtained, you just need to know the direction in which the function will approach the asymptotes. To get a better sense of how the graph will fit the asymptotes, plot the x-intercept(s) and y-intercept. To get even more detail, plot additional points.
Another quick approach to graphing is to use your knowledge of positives and negatives. Once both the numerator and denominator are fully factored, substitute an input of your choosing into each factor and just record if the factor is positive or negative. The value @$\begin{align}f(-4)\end{align}@$ will be used to demonstrate.
@$$\begin{align}f(x) &= \frac{3x}{x^2 - 3x - 4} = \frac{3x}{(x-4)(x+1)}\\
f(-4) &= \frac{(3\cdot-4)}{(-4-4)(-4+1)} \rightarrow \frac{(-)}{(-)(-)} \rightarrow \frac{(-)}{(+)} \rightarrow (-)\end{align}@$$
Since a negative divided by two negatives factors will be negative, @$\begin{align}f(x)\end{align}@$ at @$\begin{align}f(-4)\end{align}@$ will be negative. A faster way to do this is to graph the factors in the numerator and denominator individually. Each section between x-intercepts will be positive or negative, depending on whether the graphs of each factor are positive or negative in that section. If there are an odd number of negative lines in that section, the graph of the original function will be negative. If there are an even number of negative lines in that section, the graph of the original function will be positive.
Use the interactive below to explore how this would work on the function @$\begin{align}f(x).\end{align}@$
Inline Questions
Find the @$\begin{align}x\end{align}@$-intercepts of the function @$\begin{align}y=\tfrac{x-2}{x+7}.\end{align}@$
(7, 0)
(0, 4)
(2, 0)
(1, 4)
Find the @$\begin{align}y\end{align}@$-intercepts of the function @$\begin{align}y=\tfrac{x-2}{x+7}.\end{align}@$
@$\begin{align}(0, -\tfrac27)\end{align}@$
@$\begin{align}(\tfrac27, -\tfrac72)\end{align}@$
@$\begin{align}(2, \tfrac27)\end{align}@$
@$\begin{align}(-\tfrac72, 0)\end{align}@$
What is the domain and range of the function @$\begin{align}y=\tfrac{3x^2}{x^2-4}?\end{align}@$
Domain: All real numbers Range: @$\begin{align}-\infty < y \le 0\end{align}@$ and @$\begin{align}2 < y < \infty\end{align}@$
Domain: @$\begin{align}-\infty < x \le 0\end{align}@$ and @$\begin{align}2 < x < \infty\end{align}@$ Range: All real numbers
Domain: @$\begin{align}\color{green} -\infty < x < -2, -2 < x < 2\end{align}@$ and @$\begin{align}\color{green} 2 < x < \infty\end{align}@$ Range: @$\begin{align}\color{green} -\infty < y \le 0\end{align}@$ and @$\begin{align}\color{green} 3 < y < \infty\end{align}@$
Domain: All real numbers Range: All real numbers
Identify the vertical asymptotes, horizontal asymptotes, holes, and @$\begin{align}x\end{align}@$-intercepts of the function below. @$\begin{align}f(x)=\tfrac{x^2+x-6}{-4x^2-16x-12}\end{align}@$
| |
| Vertical asymptotes: @$\begin{align}x=1\end{align}@$ and @$\begin{align}x=3\end{align}@$ Horizontal asymptote: @$\begin{align}y=\tfrac14\end{align}@$ Hole: @$\begin{align}x=3\end{align}@$ @$\begin{align}x\end{align}@$-intercept @$\begin{align}x=2\end{align}@$ |
| |
| Vertical asymptotes: @$\begin{align}x=1\end{align}@$ and @$\begin{align}x=3\end{align}@$ Horizontal asymptote: @$\begin{align}y=\tfrac34\end{align}@$ Hole: @$\begin{align}x=4\end{align}@$ @$\begin{align}x\end{align}@$-intercept @$\begin{align}x=-2\end{align}@$ |
| | |
--- |
| Vertical asymptotes: @$\begin{align}x=-1\end{align}@$ and @$\begin{align}x=-3\end{align}@$ Horizontal asymptote: @$\begin{align}y=-\tfrac14\end{align}@$ Hole: @$\begin{align}x=-3\end{align}@$ @$\begin{align}x\end{align}@$-intercept: @$\begin{align}x=2\end{align}@$ | correct |
| |
| Vertical asymptotes: @$\begin{align}x=-1\end{align}@$ and @$\begin{align}x=-3\end{align}@$ Horizontal asymptote: @$\begin{align}y=-\tfrac34\end{align}@$ Hole: @$\begin{align}x=-4\end{align}@$ @$\begin{align}x\end{align}@$-intercept: @$\begin{align}x=-2\end{align}@$ |
5. Use the sign test to sketch the function @$\begin{align}f(x)=\tfrac{(x-2)^4(x+1)(x+3)}{x^3(x+3)(x-4)}.\end{align}@$
| | |
--- |
| | correct |
Wrap-Up: Review Questions
| Image | Reference | Attributions |
---
Sign In
Don't have an account? Sign Up
Forgot Your Password?
Having issues? Click here
By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy
Student Sign Up
Are you a teacher?
Having issues? Click here
or
By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy
Already have an account?
No Results Found
Your search did not match anything in .
Save this section to your Library in order to add a Practice or Quiz to it.
(Edit Title)46/ 100
This lesson has been added to your library.
|Searching in:
| |
|
Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for:
Go to the Table of Contents |
7730 | https://www.onlinemathlearning.com/fractions-worksheets.html | Fraction Word Problems - Mixed Numbers
Related Topics:
Fraction Word Problems Worksheet
More Fractions Worksheets
Fraction Games
“Fraction Word Problems” Worksheets:
Fraction Word Problems (Add, Subtract)
2-Step Fraction Word Problems (Add, Subtract)
1-Step Mixed Number Word Problems (Add, Subtract)
2-Step Mixed Number Word Problems (Add, Subtract)
Objective: I can solve one-step word problems involving addition and subtraction of mixed numbers.
Follow these steps to solve the mixed numbers word problems.
Step 1. Is it a problem in addition or subtraction?
Step 2. Do you need to find a common denominator?
Step 3. Can you simplify or reduce the answer?
Solve the following word problems.
Mark ran 2 1/3 km and Shaun ran 3 1/5 km. Find the difference in the distance that they ran.
Brandon and his son went fishing. Brandon caught 3 3/4 kg of fish while his son caught 2 1/5 kg of fish. What is the total weight of the fishes that they caught?
For the school’s sports day, a group of students prepared 21 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres left over. How many litres of lemonade were sold?
Darren spent 2 1/2 hours on his homework on Monday. On Tuesday, he spent 1 3/5 hours on his homework. Find the total amount of time, in hours, that Darren spent doing his homework on Monday and Tuesday.
Brian has a bamboo pole that was 6 ¾ m long. He cut off 1 1/4 m and another 2 1/3 m. What is the length of the remaining bamboo pole in m?
Lydia bought 2 3/4 kg of vegetables, 1 1/4 kg of fish and 2 1/3 kg of mutton. What is the total mass, in kg, of the items that she bought?
Kimberly has 3 1/2 bottles of milk in her refrigerator. She used 3/5 bottle in the morning and 1 1/4 bottle in the afternoon. How many bottles of milk does Kimberly have left over?
A tank has 82 3/4 litres of water. 24 4/5 litres were used and the tank was filled with another 18 3/4 litres. What is the final volume of water in the tank?
Try out our new and fun Fraction Concoction Game.
Add and subtract fractions to make exciting fraction concoctions following a recipe.
There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics
of fraction addition and subtraction or challenge yourself with the insane level.
We hope that the free math worksheets have been helpful. We encourage parents and
teachers to select the topics according to the needs of the child. For more difficult
questions, the child may be encouraged to work out the problem on a piece of paper
before entering the solution. We hope that the kids will also love the fun stuff and
puzzles.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
Copyright © 2005, 2025 - OnlineMathLearning.com.
Embedded content, if any, are copyrights of their respective owners. |
7731 | https://www.urch.com/forums/topic/137132-cheking-divisibility-by-3632-and-24/ | Cheking divisibility by 36,32 and 24 - GRE Math - Urch Forums
Jump to content
Urch Forums
Existing user? Sign In
Sign In
- [x] Remember me Not recommended on shared computers
Sign In
Forgot your password?
Sign Up
Browse
Forums
Blogs
Events
Staff
Online Users
Leaderboard
More
Activity
All Activity
Search
More
More
More
This Topic
Everywhere
This Forum
This Topic
Status Updates
Topics
Blog Entries
Events
Members
Home
All Activity
Home
Cheking divisibility by 36,32 and 24
Cheking divisibility by 36,32 and 24
By MikeNoob
June 21, 2012 in GRE Math
Share
More sharing options...
Followers 0
Reply to this topic
Start new topic
Recommended Posts
MikeNoob
Posted June 21, 2012
MikeNoob
Members
7
Posted June 21, 2012
I am only familiar with divisibility by 2,3,4,5,6,8,9,10,11 how do we check for 36,32,and 24 ?
Quote
2 weeks later...
DaveTutor
Posted July 1, 2012
DaveTutor
Members
2
Posted July 1, 2012
For divisibility by any number that you haven't memorized a shortcut rule for, consider all the factors of that number and pick out a pair of two factors for which:
a. you know a divisibility rule AND;
b. their least common multiple is the number itself
Here are some examples for those numbers:
36's factors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Well, 49 = 36 and their LCM is 36 so you have something to work with here. Check if the number is divisible by 4 and 9 and if it is, then it is divisible by 36 as well.
32's factors are 1, 2, 4, 8, 16, 32. However, since the only prime factor of 32 is 2, there is no pair of factors for which the LCM is 32 (besides 32 and 1) so there is no shortcut divisibility rule for 32.
24's factors are 1, 2, 3, 4, 6, 8, 12, 24. Well, 83 = 24 and their LCM is 24. Therefore, if the number is divisible by both 8 and 3, it's divisible by 24.
Quote
4 weeks later...
mxplusb
Posted July 23, 2012
mxplusb
Members
47
Posted July 23, 2012
For 32: well, you can first quickly check if the number is divisible by 8, because if it's not then it definitely isn't divisible by 32. If it is, well, I find dividing a number by 2 is relatively easy. I would do that twice, then check to see whether the resulting number is divisible by 8.
Quote
shadoWizard
Posted July 24, 2012
shadoWizard
1st Level
307
Posted July 24, 2012
mxplusb said:
For 32: well, you can first quickly check if the number is divisible by 8, because if it's not then it definitely isn't divisible by 32. If it is, well, I find dividing a number by 2 is relatively easy. I would do that twice, then check to see whether the resulting number is divisible by 8.
mxplusb, kindly explain the rule that you have used to solve for 32.
Regards,
Quote
shadoWizard
Posted July 24, 2012
shadoWizard
1st Level
307
Posted July 24, 2012
Quote
32's factors are 1, 2, 4, 8, 16, 32. However, since the only prime factor of 32 is 2, there is no pair of factors for which the LCM is 32 (besides 32 and 1) so there is no shortcut divisibility rule for 32.
DaveTutor, I've never heard of this rule that when a number has only one prime factor, then that number's any pair of factor's LCM cannot be that number. Please elaborate on that.
Also, to find it for 30, we have factors 1,2,3,5,6,10,15,30. So here LCM of 2 & 15 and 3 and 10 is 30. So which pair be used?
Regards,
Quote
mxplusb
Posted August 1, 2012
mxplusb
Members
47
Posted August 1, 2012
It is not a rule, just an observation.
32=822
Therefore, any number that is divisible by 32 must also be divisible by 8
so if it is not divisible by 8 it is not divisible by 32.
Also dividing by 2 twice and then checking for divisibility by 8 is exactly equivalent to checking for divisibility by 32.
Quote
4 years later...
SpaceExploder
Posted April 5, 2017
SpaceExploder
Members
1
Posted April 5, 2017
DaveTutor said:
32's factors are 1, 2, 4, 8, 16, 32. However, since the only prime factor of 32 is 2, there is no pair of factors for which the LCM is 32 (besides 32 and 1) so there is no shortcut divisibility rule for 32.
Actually, there is a rule for 32, just check if the last 5 digits are divisible by 32. This is reinforced by another rule that states that you can find if any number is divisible by 2^n by checking if the last n digits is divisible by 2^n.
Quote
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Reply to this topic...
× Pasted as rich text. Restore formatting
Only 75 emoji are allowed.
× Your link has been automatically embedded. Display as a link instead
× Your previous content has been restored. Clear editor
× You cannot paste images directly. Upload or insert images from URL.
Insert image from URL
×
Desktop
Tablet
Phone
Submit Reply
Share
More sharing options...
Followers 0
Go to topic listing
All Activity
Home
Cheking divisibility by 36,32 and 24
Contact Us
Cookies
Powered by Invision Community
×
Existing user? Sign In
Sign Up
Browse
Back
Forums
Blogs
Events
Staff
Online Users
Leaderboard
Activity
Back
All Activity
Search
×
Create New... |
7732 | https://artofproblemsolving.com/wiki/index.php/Absolute_value?srsltid=AfmBOooP9-QYHPg3l6wpEDOVQy4IK8Ik2yWkUyndaGtQgD8dFnDNVTyq | Art of Problem Solving
Absolute value - AoPS Wiki
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical CampusVisit the Virtual Campus
Sign In
Register
online school
Class ScheduleRecommendationsOlympiad CoursesFree Sessions
books tore
AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates
community
ForumsContestsSearchHelp
resources
math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten
contests on aopsPractice Math ContestsUSABO
newsAoPS BlogWebinars
view all 0
Sign In
Register
AoPS Wiki
ResourcesAops Wiki Absolute value
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Absolute value
The absolute value of a real number, denoted , is the unsigned portion of . Geometrically, is the distance between and zero on the real number line.
The absolute value function exists among other contexts as well, including complex numbers.
Contents
[hide]
1 Real numbers
2 Complex numbers
3 Examples
4 Problems
5 See Also
Real numbers
When is real, is defined as For all real numbers and , we have the following properties:
(Alternative definition)
(Non-negativity)
(Positive-definiteness)
(Multiplicativeness)
(Triangle Inequality)
(Symmetry)
Note that
and
Complex numbers
For complex numbers, the absolute value is defined as , where and are the real and imaginary parts of , respectively. It is equivalent to the distance between and the origin, and is usually called the complex modulus.
Note that , where is the complex conjugate of .
Examples
If , for some real number , then or .
If , for some real numbers , , then or , and therefore or .
Problems
Find all real values of if .
Find all real values of if .
(AMC 12 2000) If , where , then find .
See Also
Magnitude
Norm
Valuation
Retrieved from "
Art of Problem Solving is an
ACS WASC Accredited School
aops programs
AoPS Online
Beast Academy
AoPS Academy
About
About AoPS
Our Team
Our History
Jobs
AoPS Blog
Site Info
Terms
Privacy
Contact Us
follow us
Subscribe for news and updates
© 2025 AoPS Incorporated
© 2025 Art of Problem Solving
About Us•Contact Us•Terms•Privacy
Copyright © 2025 Art of Problem Solving
Something appears to not have loaded correctly.
Click to refresh. |
7733 | https://www.education.com/worksheet/article/evaluating-expressions-order-of-operations-with-exponents/ | Evaluating Expressions: Order of Operations With Exponents | Interactive Worksheet | Education.com
SKIP TO CONTENT
Worksheet Generator
Subjects
Grades
Worksheets
Games
Build a Worksheet
More Resources
Roly Recommends
Log InSign Up
Subjects
Grades
Worksheets
Games
Build a Worksheet
More Resources
Roly Recommends
WorksheetsMath Worksheets6th Grade Math Worksheets
Worksheet
Evaluating Expressions: Order of Operations With Exponents
When evaluating an expression, or finding the total value, it’s important to follow the order of operations! Give students a chance to practice using the order of operations to evaluate numerical expressions with exponents in this sixth-grade math worksheet. As learners complete the multi-step problems contained in Evaluating Expressions: Order of Operations With Exponents, including problems with various grouping symbols and exponents, they will gain an understanding of the importance of solving problems in the correct order.
Download Free Worksheet
Open Interactive Worksheet
View answer key
Add to collection
Add to assignment
Save
Grade:
Sixth Grade
Subjects:
Math
Mixed Operations
Exponents and Roots
Algebra
Exponents
Order of Operations
Expressions
Understanding and Evaluating Exponents
Evaluating Expressions
View aligned standards
6.EE.A.1
Related learning resources
Order of Operations Puzzle #1 Worksheet Order of Operations Puzzle #1 Use the rules for the order of operations to write in the missing parentheses that make each equation true. Fifth Grade Worksheet
Algebra for Beginners Worksheet Algebra for Beginners Want to know the perfect recipe for some spooky math practice? Witches and algebra, of course! Worksheet
Dividing Decimals Worksheet Dividing Decimals Children learn how to divide multi-digit decimal numbers in this supplemental math worksheet. Fifth Grade Worksheet
Base and Volume Worksheet Base and Volume Calculate the volume of each object using the base and height. Fifth Grade Worksheet
Place Value Puzzle #2 Interactive Worksheet Place Value Puzzle #2 It's a riddle whirwind! Hold on tight as you use math to solve this place value puzzle -- are you up to the challenge? Fifth Grade Interactive Worksheet
Greater Than or Less Than? Comparing Fractions Worksheet Greater Than or Less Than? Comparing Fractions Covering a variety of fraction skills such as converting to like fractions, this worksheet is sure to challenge your fifth grader's fraction savvy. Fourth Grade Worksheet
Hungry for Math Interactive Worksheet Hungry for Math Figure out these foodie math problems using lots of multiplication and division. Seventh Grade Interactive Worksheet
Combining Like Terms #1 Worksheet Combining Like Terms #1 Give sixth- and seventh-grade learners a chance to practice combining like terms in algebraic expressions with this one-page worksheet. Sixth Grade Worksheet
What's the Angle? Worksheet What's the Angle? Once students understand angle basics, challenge them to find missing angles in this exercise! Fourth Grade Worksheet
Autumn Word Problems Interactive Worksheet Autumn Word Problems Welcome the season with autumn-themed word problems. Your student will have the opportunity to practice addition, subtraction, multiplication, and division. Third Grade Interactive Worksheet
Order of Operations: PEMDAS Interactive Worksheet Order of Operations: PEMDAS This fifth- and sixth-grade math worksheet is a great way to give learners practice using the acronym PEMDAS to follow the correct order of operations. Fifth Grade Interactive Worksheet
Metric Length Measurement: Word Problems Worksheet Metric Length Measurement: Word Problems Familiarize yourself with the metric system! Convert units of measurement in this series of word problems. Fifth Grade Worksheet
Sign up to start collecting!
Bookmark this to easily find it later. Then send your curated collection to your children, or put together your own custom lesson plan.
Sign upLog in
Educational Tools
Learning Library
Worksheets
Games
Interactive Worksheets
Worksheet Generator
Lesson Plans
Common Core Resources
Support
Help center
Pricing
Education.com For Schools
Get a Quote
Give Gift
Redeem Gift
Contact Us
Connect
Blog
Tell us what you think
About
Company
Careers
Press
Reviews
Privacy Policy
COPPA Privacy Policy
Terms of Service
##### IXL Comprehensive K-12 personalized learning##### Rosetta Stone Immersive learning for 25 languages##### Wyzant Trusted tutors for 300 subjects##### Vocabulary.com Adaptive learning for English vocabulary##### ABCya Fun educational games for kids##### SpanishDictionary.com Spanish-English dictionary, translator, and learning##### Emmersion Fast and accurate language certification##### TPT Marketplace for millions of educator-created resources
Copyright © 2025 Education.com, Inc, a division of IXL Learning • All Rights Reserved.
Create Account
Step 1: Who is primarily going to use this content?
My Child
I'm a parent or guardian
My Students
I'm a teacher
Myself
I'm a student
First time here?
Education.com is the world’s largest collection of educational worksheets, games, videos, and songs.
Pre-K through 8th grade
Create account Keep exploring
Unlimited downloads
Guided Lessons to cultivate a joy for learning
...and more!
Upgrade |
7734 | https://en.wikipedia.org/wiki/Electroscope | Jump to content
Search
Contents
(Top)
1 Pith-ball electroscope
2 Gold-leaf electroscope
3 See also
4 Footnotes
5 External links
Electroscope
العربية
Беларуская
Беларуская (тарашкевіца)
Català
Čeština
Deutsch
Eesti
Ελληνικά
Español
Euskara
فارسی
Gaeilge
한국어
हिन्दी
Hrvatski
Ido
Italiano
עברית
Қазақша
Kreyòl ayisyen
Lietuvių
Македонски
മലയാളം
Bahasa Melayu
Nederlands
日本語
Norsk bokmål
Norsk nynorsk
Oromoo
Oʻzbekcha / ўзбекча
Polski
Português
Русский
Simple English
Slovenščina
Српски / srpski
Srpskohrvatski / српскохрватски
Suomi
Svenska
தமிழ்
Татарча / tatarça
Türkçe
Українська
اردو
中文
Edit links
Article
Talk
Read
Edit
View history
Tools
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikimedia Commons
Wikidata item
Appearance
From Wikipedia, the free encyclopedia
Early scientific instrument to detect charge
For the 19th century conceptual model of television, see Telectroscope.
The electroscope is an early scientific instrument used to detect the presence of electric charge on a body. It detects this by the movement of a test charge due to the Coulomb electrostatic force on it. The amount of charge on an object is proportional to its voltage. The accumulation of enough charge to detect with an electroscope requires hundreds or thousands of volts, so electroscopes are used with high voltage sources such as static electricity and electrostatic machines. An electroscope can only give a rough indication of the quantity of charge; an instrument that measures electric charge quantitatively is called an electrometer.
The electroscope was the first electrical measuring instrument. The first electroscope was a pivoted needle (called the versorium), invented by British physician William Gilbert around 1600. The pith-ball electroscope and the gold-leaf electroscope are two classical types of electroscope that are still used in physics education to demonstrate the principles of electrostatics. A type of electroscope is also used in the quartz fiber radiation dosimeter. Electroscopes were used by the Austrian scientist Victor Hess in the discovery of cosmic rays.
Pith-ball electroscope
[edit]
Pith ball electroscope from the 1870s, showing attraction to charged object
How it works
In 1731, Stephen Gray used a simple hanging thread, which would be attracted to any nearby charged object. This was the first improvement on Gilbert's versorium from 1600.
The pith-ball electroscope, invented by British schoolmaster and physicist John Canton in 1754, consists of one or two small balls of a lightweight nonconductive substance, originally a spongy plant material called pith, suspended by silk or linen thread from the hook of an insulated stand. Tiberius Cavallo made an electroscope in 1770 with pith balls at the end of silver wires. Modern electroscopes usually use balls made of plastic. In order to test the presence of a charge on an object, the object is brought near to the uncharged pith ball. If the object is charged, the ball will be attracted to it and move toward it.
The attraction occurs because of induced polarization of the atoms inside the pith ball. All matter consists of electrically charged particles located close together; each atom consists of a positively charged nucleus with a cloud of negatively charged electrons surrounding it. The pith is an insulator, so the electrons in the ball are bound to atoms of the pith and are not free to leave the atoms and move about in the ball, but they can move a little within the atoms. See diagram. If, for example, a positively charged object (B) is brought near the pith ball (A), the negative electrons (blue minus signs) in each atom (yellow ovals) will be attracted and move slightly toward the side of the atom nearer the object. The positively charged nuclei (red plus signs) will be repelled and will move slightly away. Since the negative charges in the pith ball are now nearer to the object than the positive charges (C), their attraction is greater than the repulsion of the positive charges, resulting in a net attractive force. This separation of charge is microscopic, but since there are so many atoms, the tiny forces add up to a large enough force to move a light pith ball.
If the external object (B) instead has a negative charge, the positive nuclei of each atom will be attracted toward it while the electrons will be repelled away from it. Again, this causes opposite charges to be closer to the external object than charges of the same polarity, resulting in a net attractive force.
The pith ball can be charged by touching it to a charged object, so some of the charges on the surface of the charged object move to the surface of the ball. Then the ball can be used to distinguish the polarity of charge on other objects because it will be repelled by objects charged with the same polarity or sign it has, but attracted to charges of the opposite polarity.
Often the electroscope will have a pair of suspended pith balls. This allows one to tell at a glance whether the pith balls are charged. If one of the pith balls is touched to a charged object, charging it, the second one will be attracted and touch it, communicating some of the charge to the surface of the second ball. Now both balls have the same polarity charge, so they repel each other. They hang in an inverted 'V' shape with the balls spread apart. The distance between the balls will give a rough idea of the magnitude of the charge.
Gold-leaf electroscope
[edit]
The gold-leaf electroscope was developed in 1787 by British clergyman and physicist Abraham Bennet, as a more sensitive instrument than pith ball or straw blade electroscopes then in use. It consists of a vertical metal rod, usually brass, from the end of which hang two parallel strips of thin flexible gold leaf. A disk or ball terminal is attached to the top of the rod, where the charge to be tested is applied. To protect the gold leaves from drafts of air they are enclosed in a glass bottle, usually open at the bottom and mounted over a conductive base. Often there are grounded metal plates or foil strips in the bottle flanking the gold leaves on either side. These are a safety measure; if an excessive charge is applied to the delicate gold leaves, they will touch the grounding plates and discharge before tearing. They also capture charge leaking through the air that accumulates on the glass walls, increasing the sensitivity of the instrument. In the precision instruments the inside of the bottle was occasionally evacuated, to prevent the charge on the terminal from leaking off through the ionization of the air.
When the metal terminal is touched with a charged object, the gold leaves spread apart in an inverted 'V'. This is because some of the charge from the object is conducted through the terminal and metal rod to the leaves. Since the leaves receive the same sign charge they repel each other and thus diverge. If the terminal is grounded by touching it with a finger, the charge is transferred through the human body into the earth and the gold leaves close together.
The electroscope leaves can also be charged without touching a charged object to the terminal, by electrostatic induction. As the charged object is brought near the electroscope terminal, the leaves spread apart, because the electric field from the object induces a charge in the conductive electroscope rod and leaves, and the charged leaves repel each other. The opposite-sign charge is attracted to the nearby object and collects on the terminal disk, while the same-sign charge is repelled from the object and collects on the leaves (but only as much as left the terminal), so the leaves repel each other. If the electroscope is grounded while the charged object is nearby, by touching it momentarily with a finger, the repelled same-sign charges travel through the contact to ground, leaving the electroscope with a net charge having the opposite sign as the object. The leaves initially hang down free because the net charge is concentrated at the terminal end. When the charged object is moved away, the charge at the terminal spreads into the leaves, causing them to spread apart again.
Gold-leaf electroscopes
Condensing electroscope, Rome University physics dept.
Electroscope from about 1910 with grounding electrodes inside jar, as described above
Homemade electroscope, 1900
See also
[edit]
Electrical measurements
Electrostatic fieldmeter
Faraday cup electrometer
Radiation
Footnotes
[edit]
^ Gilbert, William; Edward Wright (1893). On the Lodestone and Magnetic Bodies. John Wiley & Sons. p. 79. a translation by P. Fleury Mottelay of William Gilbert (1600) Die Magnete, London
^ a b Fleming, John Ambrose (1911). "Electroscope". In Chisholm, Hugh (ed.). Encyclopædia Britannica. Vol. 9 (11th ed.). Cambridge University Press. p. 239.
^ a b Baigrie, Brian (2007). Electricity and magnetism: A historical perspective. Westport, CT: Greenwood Press. p. 33.
^ a b Derry, Thomas K.; Williams, Trevor (1993) . A Short History of Technology: from Earliest Times to A.D. 1900. Dover. p. 609. ISBN 0-486-27472-1. p. 609
^ Elliott, P. (1999). "Abraham Bennet F.R.S. (1749–1799): a provincial electrician in eighteenth-century England" (PDF). Notes and Records of the Royal Society of London. 53 (1): 61. doi:10.1098/rsnr.1999.0063. JSTOR 531928. S2CID 144062032. Archived from the original (PDF) on 2020-03-27. Retrieved 2007-09-02.
^ Sherwood, Bruce A.; Ruth W. Chabay (2011). Matter and Interactions (3rd ed.). US: John Wiley and Sons. pp. 594–596. ISBN 978-0-470-50347-8.
^ a b Kaplan MCAT Physics 2010–2011. USA: Kaplan Publishing. 2009. p. 329. ISBN 978-1-4277-9875-6. Archived from the original on 2014-01-31.
^ Paul E. Tippens, Electric Charge and Electric Force, Powerpoint presentation, pp. 27–28, 2009, S. Polytechnic State Univ. Archived April 19, 2012, at the Wayback Machine on DocStoc.com website
^ Henderson, Tom (2011). "Charge and Charge Interactions". Static Electricity, Lesson 1. The Physics Classroom. Retrieved 2012-01-01.
^ Winn, Will Winn (2010). Introduction to Understandable Physics Vol. 3: Electricity, Magnetism and Light. US: Author House. p. 20.4. ISBN 978-1-4520-1590-3.
^ a b c [Anon.] (2001) "Electroscope", Encyclopaedia Britannica
^ phi6guy (2025-07-03). Coulomb's law | Electrostatics | Electric Charges and Fields | NCERT Class 12 Physics |. Retrieved 2025-07-04 – via YouTube.{{cite AV media}}: CS1 maint: numeric names: authors list (link)
External links
[edit]
"Pith-ball electroscope". Physics demonstration resource. St. Mary's University. Retrieved 2015-05-28.
"Computer simulation of electroscopes". Molecular Workbench. Concord Consortium. Archived from the original on 2022-07-03. Retrieved 2008-02-03.
"Pith Ball and Charged Rod Video". St. Mary's Physics YouTube Channel. St. Mary's Physics Online. Archived from the original on 2021-12-22.
Retrieved from "
Categories:
Electrostatics
Electrical instruments
Measuring instruments
Historical scientific instruments
Hidden categories:
Wikipedia articles incorporating a citation from the 1911 Encyclopaedia Britannica with Wikisource reference
Webarchive template wayback links
CS1 maint: numeric names: authors list
Articles with short description
Short description is different from Wikidata
Pages using div col with small parameter
Add topic |
7735 | https://deleahora.com/conjugacion/conjugacion-del-verbo-preparar.pdf | CUADRO DE CONJUGACIÓN preparar Ger: preparando • Part: preparado INDICATIVO Presente Pretérito perfecto simple Pretérito imperfecto yo preparo preparé preparaba tú/vos preparas preparaste preparabas él/ella/Ud.
prepara preparó preparaba nosotros preparamos preparamos preparábamos vosotros preparáis preparasteis preparabais ellos/ellas/Uds.
preparan prepararon preparaban Futuro Condicional yo prepararé prepararía tú/vos prepararás prepararías él/ella/Ud.
preparará prepararía nosotros prepararemos prepararíamos vosotros prepararéis prepararíais ellos/ellas/Uds.
prepararán prepararían SUBJUNTIVO Presente Pretérito Imperfecto yo prepare preparara / preparase tú/vos prepares prepararas / preparases él/ella/Ud.
prepare preparara / preparase nosotros preparemos preparáramos / preparásemos vosotros preparéis prepararais / preparaseis ellos/ellas/Uds.
preparen prepararan / preparasen IMPERATIVO Afirmativo Negativo yo — — tú/vos prepara / prepará no prepares Ud.
prepare no prepare nosotros preparemos no preparemos vosotros preparad no preparéis Uds.
preparen no preparen K vos: preparás Copyright © 2023 DELE Ahora deleahora.com |
7736 | https://bcrc.bio.umass.edu/courses/spring2019/biol/biol312section2/content/draft-po-ratio | Draft: P/O ratio | Writing in Biology
Skip to main content
Writing in Biology
BIOL312Section2;Spring2019 Brewer
User login
Username
Password
Request new password
Main menu
Home
Image Gallery
Perfect Paragraphs
You are here
Home » Blogs » aspark's blog
Draft: P/O ratio
Submitted by aspark on Wed, 04/17/2019 - 23:53
The P/O ratio is determined experimentally and is a number that relates the amount of ATP generated per electron carrier. This number can vary depending on the physiological conditions. P stands for the moles of phosphoryl groups consumed to form ATP, and the O stands for moles of oxygen consumed to be reduced to water. The P/O ratio of one NADH is 2.5 ATP, and the P/O ratio of one FADH2 is 1.5 ATP. FADH2 has a lower P/O ratio because FAD has a higher affinity for electrons and donates them through the electron transport chain at complex II and contributes fewer protons to the gradient. Per glucose, 32 ATP are produced. 4 ATP are produced directly while 25 ATP are produced through NADH and 3 ATP are produced through FADH2. The total number of ATP produced can vary from 28-36 depending on the P/O ratios of NADH and FADH2 dependent on physiological conditions.
Post:
Drafts
aspark's blog
Perfect Paragraphs
Each week, post your own Perfect Paragraph and comment on three Perfect Paragraphs. Suggest improvements. Don't just say "Looks good."
Recent comments
Comments6 years 5 months ago
Feedback6 years 5 months ago
Descriptive words6 years 5 months ago
Sentences6 years 5 months ago
Suggestion6 years 5 months ago
Suggestion6 years 5 months ago
Suggestion6 years 5 months ago
Feedback6 years 5 months ago
Feedback6 years 5 months ago
Feedback6 years 5 months ago
Blogs Updated This Week
afeltrin
alanhu
angelinamart
aprisby
aspark
cbbailey
cnwokemodoih
cslavin
cynthiaguzma
danielnguyen
ddoyleperkin
ewinter
jhussaini
klaflamme
kwarny
lgarneau
lpotter
mqpham
mscheller
nalexandroum
ncarbone
rdigregorio
rharrison
sbrewer
scasimir
sditelberg
This page is maintained by The University of Massachusetts Biology Department.University of Massachusetts Amherst•Site Policies• |
7737 | https://zh.wikipedia.org/wiki/%E4%BB%A3%E6%95%B8%E5%88%86%E5%BC%8F | 代數分式 - 维基百科，自由的百科全书
跳转到内容
[x] 主菜单
主菜单
移至侧栏 隐藏
导航
首页
分类索引
特色内容
新闻动态
最近更改
随机条目
帮助
帮助
维基社群
方针与指引
互助客栈
知识问答
字词转换
IRC即时聊天
联络我们
关于维基百科
特殊页面
搜索
搜索
[x] 外观
外观
移至侧栏 隐藏
文本
小 标准 大
此页面始终使用小字号
宽度
标准 宽
内容会尽可能占满您的浏览器窗口宽度。
颜色 （测试版）
自动 浅色 深色
此页面始终处于浅色模式。
资助维基百科
创建账号
登录
[x] 个人工具
资助维基百科
创建账号
登录
維基百科Discord、IRC、LINE、QQ及Telegram等各平臺交流群歡迎大家加入。
[x] 开关目录
目录
移至侧栏 隐藏
序言
1 術語
2 有理分式
3 無理分式
4 腳註
5 參考資料
代數分式[编辑]
[x] 10种语言
বাংলা
English
Español
Italiano
Македонски
Português
Română
Simple English
தமிழ்
Українська
编辑链接
条目
讨论
[x] 不转换
不转换
简体
繁體
大陆简体
香港繁體
澳門繁體
大马简体
新加坡简体
臺灣正體
阅读
编辑
查看历史
[x] 工具
工具
移至侧栏 隐藏
操作
阅读
编辑
查看历史
常规
链入页面
相关更改
上传文件
固定链接
页面信息
引用此页
获取短链接
下载二维码
全部展开
编辑跨语言链接
打印/导出
下载为PDF
打印页面
在其他项目中
维基数据项目
维基百科，自由的百科全书
代數分式是指分子及分母都是代數式的分數，像 3 x x 2+2 x−3{\displaystyle {\frac {3x}{x^{2}+2x-3}}} 及 x+2 x 2−3{\displaystyle {\frac {\sqrt {x+2}}{x^{2}-3}}}. 都是代數分式。
有理分式是指分子及分母都是多項式的分式，像 3 x x 2+2 x−3{\displaystyle {\frac {3x}{x^{2}+2x-3}}} 為有理分式，但 x+2 x 2−3{\displaystyle {\frac {\sqrt {x+2}}{x^{2}-3}}} 的分子為根式，不是多項式，因此不是有理分式。
術語
[编辑]
在代數分式 a b{\displaystyle {\tfrac {a}{b}}}中，被除數稱為分子，除數稱為分母，兩者都是代數分式的項。
若代數分式的分子或分母中包括複數，則稱為複數分式。
簡分式是其分子或分母都不是分式的代數分式，若一個表示式不是以分式的形式表示，則稱為整式，不過只要將分母設為1，即可以將整式表示為代數分式，帶分式指整式和分式的代數和。
有理分式
[编辑]
主条目：有理函數
若代理分式的a和b都是多項式，此分式稱為有理代數分式，或簡稱為有理分式。有理分式也稱為有理表示式或有理函數。若有理分式 f(x)g(x){\displaystyle {\tfrac {f(x)}{g(x)}}} 滿足 degf(x)<degg(x){\displaystyle \deg f(x)<\deg g(x)}，稱為真分式，否則稱為假分式，像 2 x x 2−1{\displaystyle {\tfrac {2x}{x^{2}-1}}}為真分式，而 x 3+x 2+1 x 2−5 x+6{\displaystyle {\tfrac {x^{3}+x^{2}+1}{x^{2}-5x+6}}} 和 x 2−x+1 5 x 2+3{\displaystyle {\tfrac {x^{2}-x+1}{5x^{2}+3}}} 是假分式。假分式可以表示為整式（可能是常數）及真分式的和，例如以上提到的假分式可以表示為
x 3+x 2+1 x 2−5 x+6=(x+6)+24 x−35 x 2−5 x+6,{\displaystyle {\frac {x^{3}+x^{2}+1}{x^{2}-5x+6}}=(x+6)+{\frac {24x-35}{x^{2}-5x+6}},}
其中第二項為真有理分式，二個真分式的和也會是真分式，有時會將真分式的分母因式分解，再將真分式表示數個真因式，其分母分別為原分母的因式（或因式次方），這稱為部份分式，例如以下等號右邊的即為部份分式
2 x x 2−1=1 x−1+1 x+1.{\displaystyle {\frac {2x}{x^{2}-1}}={\frac {1}{x-1}}+{\frac {1}{x+1}}.}
因此等號右邊的稱為部份分式，例如真分式積分時會先進行部分分式分解，再進行積分，稱為部分分式积分法。
無理分式
[编辑]
無理分式是指分式中有變數的幂式為小數，像以下的分式即為無理分式
x 1 2−1 3 a x 1 3−x 1 2.{\displaystyle {\frac {x^{\tfrac {1}{2}}-{\tfrac {1}{3}}a}{x^{\tfrac {1}{3}}-x^{\tfrac {1}{2}}}}.}
將無理分式變為有理分式的過程稱為有理化，每個根式為單項的無理分式可以用以下的方式有理化：找到所有幂次分母的最小公倍數，再將變數用另一變數的幂次取代，使原來的根式都變為新變數的整數幂次，例如在上式中，幂次分母的最小公倍數為6，因此可以令 x=z 6{\displaystyle x=z^{6}} ，得到
z 3−1 3 a z 2−z 3.{\displaystyle {\frac {z^{3}-{\tfrac {1}{3}}a}{z^{2}-z^{3}}}.}
腳註
[编辑]
^Bansi Lal. Topics in Integral Calculus. 2006: 53.
^Ėrnest Borisovich Vinberg. A course in algebra. 2003: 131 [2015-07-06]. （原始内容存档于2014-07-05）.
^Parmanand Gupta. Comprehensive Mathematics XII.: 739 [2015-07-06]. （原始内容存档于2014-06-28）.
^Washington McCartney. The principles of the differential and integral calculus; and their application to geometry. 1844: 203.
參考資料
[编辑]
Brink, Raymond W. IV. Fractions. College Algebra. 1951.
| 折叠 查 论 编 分數&比率 |
| | 除法 ＆比例 | 被除數: 除數 = 商數 | |
| 分數 | 分子/分母=商數 |
| 代數 長寬比 连分数 十进制 二进分数 古埃及分數 黄金分割率 白銀比例 整数 最简分数 精簡 最小公分母 音程 百分比 單位分數 |
检索自“
分类：
初等代数
分數
本页面最后修订于2021年10月13日 (星期三) 23:27。
本站的全部文字在知识共享 署名-相同方式共享 4.0协议之条款下提供，附加条款亦可能应用。（请参阅使用条款）
Wikipedia®和维基百科标志是维基媒体基金会的注册商标；维基™是维基媒体基金会的商标。
维基媒体基金会是按美国国內稅收法501(c)(3)登记的非营利慈善机构。
隐私政策
关于维基百科
免责声明
行为准则
开发者
统计
Cookie声明
手机版视图
编辑预览设置
搜索
搜索
[x] 开关目录
代數分式
10种语言添加话题
选取内容的语言变体
确定
保存您偏好的中文语言变体，以避免显示简繁混杂的内容，提供最佳的阅读体验。 |
7738 | https://www.wikihow.com/Multiply-Two-Digit-Numbers-Mentally | Log in
How to Multiply Two Digit Numbers Mentally
Last Updated: March 10, 2025
This article was co-authored by Joseph Meyer. Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University.
This article has been viewed 57,012 times.
Using the standard algorithm to multiply two 2-digit numbers is sufficient for most purposes; however, its multiple steps can leave you looking for a quick and easy way to find the product of these types of numbers. If you know your basic math facts and have good number sense, you can use a number of techniques to multiply two 2-digit numbers mentally. If you are familiar with the difference of two squares, you can modify your two factors so that they fit this algebraic formula. You can also manipulate the factors by using the distributive property, or by doubling and halving, until you come up with two new numbers that are easier to work with.
Steps
Finding the Difference of Two Squares
Using the Distributive Property
Joseph Meyer
The distributive property helps you avoid repetitive calculations. You can use the distributive property to solve equations where you must multiply a number by a sum or difference. It simplifies calculations, enables expression manipulation (like factoring), and forms the basis for solving many equations.
Doubling and Halving
Expert Q&A
Video
Tips
You Might Also Like
Expert Interview
Thanks for reading our article! If you’d like to learn more about math, check out our in-depth interview with Joseph Meyer.
References
About This Article
Reader Success Stories
Aarna Tyagi
Feb 10, 2017
Did this article help you?
Aarna Tyagi
Feb 10, 2017
Justin Usoria
Jul 2, 2018
Quizzes
Do I Have a Dirty Mind Quiz
Personality Analyzer: How Deep Am I?
Am I a Good Kisser Quiz
Rizz Game: Test Your Rizz
"Hear Me Out" Character Analyzer
What's Your Red Flag Quiz
You Might Also Like
How to
Speed Up Numeric Calculations in Mathematics
How to
Do Math in Your Head: Easy Tricks for Mental Calculations
How to
Square Any Number
How to
Multiply Using Vedic Math
Featured Articles
How to
Stop Instagram from Suggesting Adult Content
Will I Get Together With My Crush Quiz
The Easiest Way to Clean Your Room from Top to Bottom
Wondering if Someone Likes You Online? 11 Important Signs to Watch Out For
How to
Increase Your Self Confidence with Positive Daily Practices
How to
Fix Painful Shoes
Trending Articles
How Rare Is Your Name?
5 Different Types of Butts: Find Your Shape
How Will I Die Quiz
How Many People Have a Crush On Me Quiz
Labubu Blind Box Generator: Unbox & Collect Your Own Online Lababus
How to
Know if a Person Is Interested in You
Featured Articles
How to
Use Castor Oil to Promote Healthy Hair
How to
Apply for an Internship
Tips for Trimming Long Hair Evenly (Your Own or Someone Else’s)
How to
Whiten Teeth With Baking Soda
130+ Sexy, Sweet, & Seductive Messages for Him
Something Bit Me! Common Insect Bites (with Pictures)
Featured Articles
Do I Have Common Sense Quiz
50 Cute & Flirty Knock-Knock Jokes to Make Them Smile
How to
Calm an Aggressive Cat
How to
Get a Flat Stomach in 7 Days: Exercises, Diet Tips, & More
How Do You Know if She Likes You? 15+ Signs She’s Into You
How to
Flirt
Watch Articles
How to
Remove Hard Water Spots from Glass: DIY Tips & Tricks
How to
Sew a Button
How to
Cook Egg-Fried Rice at Home
How to
Purposely Shrink Any Type of Clothing in the Washer and Dryer
How to
Become a Vegetarian
How to
Make Your Own Deep Conditioner
Trending Articles
What Is My Mental Age Quiz
What Is Your Soul Animal Quiz
Quizzes
Follow Us
Don’t miss out! Sign up for
wikiHow’s newsletter |
7739 | https://pubmed.ncbi.nlm.nih.gov/11700123/ | Mild conversion of alcohols to alkyl halides using halide-based ionic liquids at room temperature - PubMed
Clipboard, Search History, and several other advanced features are temporarily unavailable.
Skip to main page content
An official website of the United States government
Here's how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Search: Search
AdvancedClipboard
User Guide
Save Email
Send to
Clipboard
My Bibliography
Collections
Citation manager
Display options
Display options
Format
Save citation to file
Format:
Create file Cancel
Email citation
Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page.
To:
Subject:
Body:
Format:
[x] MeSH and other data
Send email Cancel
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Add to My Bibliography
My Bibliography
Unable to load your delegates due to an error
Please try again
Add Cancel
Your saved search
Name of saved search:
Search terms:
Test search terms
Would you like email updates of new search results? Saved Search Alert Radio Buttons
Yes
No
Email: (change)
Frequency:
Which day?
Which day?
Report format:
Send at most:
[x] Send even when there aren't any new results
Optional text in email:
Save Cancel
Create a file for external citation management software
Create file Cancel
Your RSS Feed
Name of RSS Feed:
Number of items displayed:
Create RSS Cancel
RSS Link Copy
Actions
Cite
Collections
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Permalink
Permalink
Copy
Display options
Display options
Format
Page navigation
Title & authors
Abstract
Similar articles
Cited by
Org Lett
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2001 Nov 15;3(23):3727-8.
doi: 10.1021/ol016672r.
Mild conversion of alcohols to alkyl halides using halide-based ionic liquids at room temperature
R X Ren1,J X Wu
Affiliations Expand
Affiliation
1 Max Tishler Laboratory of Organic Chemistry, Department of Chemistry, Wesleyan University, Middletown, Connecticut 06459, USA. rren@wesleyan.edu
PMID: 11700123
DOI: 10.1021/ol016672r
Item in Clipboard
Mild conversion of alcohols to alkyl halides using halide-based ionic liquids at room temperature
R X Ren et al. Org Lett.2001.
Show details
Display options
Display options
Format
Org Lett
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2001 Nov 15;3(23):3727-8.
doi: 10.1021/ol016672r.
Authors
R X Ren1,J X Wu
Affiliation
1 Max Tishler Laboratory of Organic Chemistry, Department of Chemistry, Wesleyan University, Middletown, Connecticut 06459, USA. rren@wesleyan.edu
PMID: 11700123
DOI: 10.1021/ol016672r
Item in Clipboard
Cite
Display options
Display options
Format
Abstract
[reaction--see text] Alcohols were efficiently converted to alkyl halides using 1-n-butyl-3-methylylimidazolium halides (ionic liquids) in the presence of Brønsted acids at room temperature. The alkyl halide products were easily isolated from the reaction mixture via simple decantation or extraction, and the 1-n-butyl-3-methylimidazolium cation could be recycled for further uses.
PubMed Disclaimer
Similar articles
Spontaneous product segregation from reactions in ionic liquids: application in Pd-catalyzed aliphatic alcohol oxidation.Van Doorslaer C, Schellekens Y, Mertens P, Binnemans K, De Vos D.Van Doorslaer C, et al.Phys Chem Chem Phys. 2010 Feb 28;12(8):1741-9. doi: 10.1039/b920813p. Epub 2009 Dec 21.Phys Chem Chem Phys. 2010.PMID: 20145838
Nucleophilicity in ionic liquids. 3. Anion effects on halide nucleophilicity in a series of 1-butyl-3-methylimidazolium ionic liquids.Lancaster NL, Welton T.Lancaster NL, et al.J Org Chem. 2004 Sep 3;69(18):5986-92. doi: 10.1021/jo049636p.J Org Chem. 2004.PMID: 15373482
Thermal Resilience of Imidazolium-Based Ionic Liquids-Studies on Short- and Long-Term Thermal Stability and Decomposition Mechanism of 1-Alkyl-3-methylimidazolium Halides by Thermal Analysis and Single-Photon Ionization Time-of-Flight Mass Spectrometry.Efimova A, Varga J, Matuschek G, Saraji-Bozorgzad MR, Denner T, Zimmermann R, Schmidt P.Efimova A, et al.J Phys Chem B. 2018 Sep 20;122(37):8738-8749. doi: 10.1021/acs.jpcb.8b06416. Epub 2018 Sep 10.J Phys Chem B. 2018.PMID: 30130967
Nonsolvent application of ionic liquids: organo-catalysis by 1-alkyl-3-methylimidazolium cation based room-temperature ionic liquids for chemoselective N-tert-butyloxycarbonylation of amines and the influence of the C-2 hydrogen on catalytic efficiency.Sarkar A, Roy SR, Parikh N, Chakraborti AK.Sarkar A, et al.J Org Chem. 2011 Sep 2;76(17):7132-40. doi: 10.1021/jo201102q. Epub 2011 Aug 3.J Org Chem. 2011.PMID: 21774556
From α-arylation of olefins to acylation with aldehydes: a journey in regiocontrol of the Heck reaction.Ruan J, Xiao J.Ruan J, et al.Acc Chem Res. 2011 Aug 16;44(8):614-26. doi: 10.1021/ar200053d. Epub 2011 May 25.Acc Chem Res. 2011.PMID: 21612205 Review.
See all similar articles
Cited by
Microwave-assisted Mannich-type three-component reactions.Leadbeater NE, Torenius HM, Tye H.Leadbeater NE, et al.Mol Divers. 2003;7(2-4):135-44. doi: 10.1023/b:modi.0000006822.51884.e6.Mol Divers. 2003.PMID: 14870842
Related information
PubChem Compound
PubChem Substance
[x]
Cite
Copy Download .nbib.nbib
Format:
Send To
Clipboard
Email
Save
My Bibliography
Collections
Citation Manager
[x]
NCBI Literature Resources
MeSHPMCBookshelfDisclaimer
The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited.
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
7740 | https://scholarworks.law.ubalt.edu/cgi/viewcontent.cgi?article=1289&context=lf | %PDF-1.7 %ãÏÓ 45 0 obj <> endobj xref 45 37 0000000016 00000 n 0000001352 00000 n 0000001456 00000 n 0000001477 00000 n 0000001863 00000 n 0000002128 00000 n 0000002398 00000 n 0000002672 00000 n 0000002950 00000 n 0000003215 00000 n 0000003490 00000 n 0000003769 00000 n 0000003919 00000 n 0000004462 00000 n 0000004951 00000 n 0000005466 00000 n 0000005884 00000 n 0000006272 00000 n 0000006728 00000 n 0000007181 00000 n 0000007611 00000 n 0000008258 00000 n 0000008614 00000 n 0000009271 00000 n 0000009915 00000 n 0000013691 00000 n 0000013869 00000 n 0000021703 00000 n 0000021875 00000 n 0000022398 00000 n 0000026508 00000 n 0000026690 00000 n 0000030473 00000 n 0000030646 00000 n 0000054765 00000 n 0000056263 00000 n 0000001036 00000 n trailer <<7E9AC00205A4B2110A00A084872BFC7F>]/Prev 427609>> startxref 0 %%EOF 81 0 obj <>stream hÞb``¢VFé �abO ÞÞ®}¾wÌE³ïÒx2-ëͶW«Ì×ni-{Öe©¾î´CGH40p(µ40j�P»([ö> ÄúDøX·°/à ¿ÀÁ¡á @¬ídíØR'|ðo}<´a®-S ÿÿD×4øÄ)qw{®aj5àz ÅÁ0Ñ©ÕÓA á%¶@$¬ÃÀÖ|H3Q@�,ìB endstream endobj 46 0 obj <> endobj 47 0 obj <<>> endobj 48 0 obj <>/MediaBox[0 0 612 792]/Parent 42 0 R/Resources<>/ProcSet[/PDF/Text/ImageC]/XObject<>>>/Rotate 0/Type/Page>> endobj 49 0 obj <>/Border[0 0 0]/Rect[284.652 646.991 540.0 665.009]/Subtype/Link/Type/Annot>> endobj 50 0 obj <>/Border[0 0 0]/Rect[81.0 624.297 129.672 636.309]/Subtype/Link/Type/Annot>> endobj 51 0 obj <>/Border[0 0 0]/Rect[81.0 609.891 171.06 621.903]/Subtype/Link/Type/Annot>> endobj 52 0 obj <>/Border[0 0 0]/Rect[510.324 617.094 549.0 629.106]/Subtype/Link/Type/Annot>> endobj 53 0 obj <>/Border[0 0 0]/Rect[243.264 230.364 407.7 242.376]/Subtype/Link/Type/Annot>> endobj 54 0 obj <>/Border[0 0 0]/Rect[145.74 211.794 214.836 223.806]/Subtype/Link/Type/Annot>> endobj 55 0 obj <>/Border[0 0 0]/Rect[123.813 144.141 295.173 153.15]/Subtype/Link/Type/Annot>> endobj 56 0 obj <>/Border[0 0 0]/Rect[124.576 86.5415 178.768 94.5495]/Subtype/Link/Type/Annot>> endobj 57 0 obj <>stream HTMÓ@ ½çW̱\\Û3©Zv RøÚUÑ.¥ø÷83Ù63jyö³ý<B@Fj$7§§.0 >Æ;}qÑõ¿O»nõïyýìDXÆÓ ÌF\±&±A\_¾¾tËdݧn=t5ÊFtjÄÇ\#G^x dZ¥¾uË ÅB8ì»ÅçwføÞµ×Ìê·øqø!fTC» H«óӯùÄh+ì9îÍ:£\suä1úúüù|x) BSÅ1 ,6Uäs]]µùX¨<Íô뿦íÇa|Cö�Duߥí0Nì2Q@ÁNaC|ÌÖ¥&óßg@T¬ÜsdBÒÔ¤# ÁM>{D\_IhËaÜà1Át86ùÙÛy;"¤²Ý^×6"ÏDÒ>ÎÔ3B x[5|(ú;MÕpºhM;Û XØ�B½J:ʹuæé"¨ >U²RqéðbU!~]@gØ\_
ÙôÀÍ}=]"ݳf+vÓÝË¿ùÊ�º#rîÄ槩¦bþ 0�H endstream endobj 58 0 obj <>stream H´ÍN#1ïó>¥qÛ«Q$3Zå Ð!+H$òü8c¢½Ûæ2Ò|.wUwO.N?̯K³ý38GÖK¶üÌö÷p±'¤Þïg�¦,¢È«yÊ Í]UEª)Zmà4H)j§©xA9íó6SÉä¢öG-,@Ù§ÿvíðs;$®ï={rðõÀÍ¢ô¥ÇC-ç(¸®ÜKÕA¾¶îVÜÿKZÝò}«£-JÄþ|äêV »¦xX'bî÷Ô@¯ÀÓ4UÃ(±µ±ùÈï,:U¹°Þ
çioõm-~áK Û5}îÜÖzàê¦^ìV@TSqÝ@ÒFÄ%ãN&×Û4:¶ ~Õ6Ôõ·®-ózÕ¾8hË×ûÁsºtgÑf½(#<63 $òºF^µÖ¤/åu;®<×ÿ ¥C¸A¨´lßb×Ñj©|ÌVPh3P°KVÊÃÌËn89æçÉebI5ï �U å endstream endobj 59 0 obj <>stream HTÛNä0 }ÏWä í\HiRÄ#Ð?ØÝAB Õò²¿¿N¢i:Hl6¾ãäÚ
;¿ZÔÇÎ?Í5ßbAAácsäxãA{>4+Eq%ÄÅmØÐC¢m>¾ë¶A ä|ÇqAÀÖµÐ0vp(ùËàÆRC<´±÷Iuç¼üDeFm îÈ¥7PËÔaäUäI5éÜ"ÐÈ!7KÍ7aê˺hùge.~üx7Wl¿Í£¹ Ú#H:÷iØ¥3ù
¥qÆ jÈú.:K´b¯å2³:»^S´/|�ñ;Lë ãwõDåná\_@ö¥çAß¹y¸�ÉûKÃU¾@út¯û±:ÿ=¬TÈï¦~¦:I{oæYCu!×
¤5-ܵB¤êYÓ\ªÔÿ\_æ´¶ K:wÄ!T¸WÛ1´yÏ a¤¥A£Í=^RRu$.ÝàIç ×°²½tÊ%îK[!Å£S¿ÒxÈûnòì÷ÃUú\ìÎÉõ:kFûO�¾)X endstream endobj 60 0 obj <>stream H¼TË0 ½SX¶;´ZÏP%ìiû/CfÀ0ÒK$^l¿!Bô=æ§_º/xAbÎç´ßýúÛ1ª?ß>Æ3@pà Ds# JyÜÀc«g ¦A:4Ф'6¼²á27iÁ©2G#@rnç8©!Xvìy( u&T"Mnj0Æ{÷Å (xQëÄZç °MsÆ
?ñb<Ã&Ãñ#ÞsT²6²pµ1�ó73×ÍÌÚ 3Î .û'7±»'7k¸ç¤¦¸æ´'%Þ4Ov²ßñ÷¡¡1ÙUAÔXûù µñGú&6D» u'[Yn¸È}Ï}McJ!ïÁ¿ <üS%ðRöfØxoïJécíƵTK 1ïHe:4CeúÚæþO�° endstream endobj 61 0 obj <>stream HÜSInÄ0 »Ï+òA-Û@QÐ$/'ôÔÿ? ØÎÄ2è¹9ä:IÑ.hÏòø¾}0}.0À×0(é¬y;ÀÄy�
øÎö¦JL Ð>^ÏÐÊé$y$0îlu8ÈN¯QÇ- ¹A]Êó/©8;êó4»v3HTÕR!ü¶©(9ÞrÕÝyãð²ªýjªM"#¯H·)>¾L'Óv{Q>stream HtAnÄ0E÷9EN@KU¥Ìdr¡j¥.{ÿEØ7Ydñ ¼8Kí÷Ïé
3/|ý÷ Dq§ôPJîZ!�ÅØGÜ!î±BåoûtÙ'¥\kczìÒ«ÎL ¯'§Æ´¢fd@\_«=l¼Ô:ûl-¦zC ëÑ>4Í¿ßq¹v¦×V-³ö
FXÈÁç:4PcBF 7Þ¨RBêÛù©uKÕ
SË< òöì|6Y!'í]¶ÌýìIk¦·gøË´yÄJ
ÙÃ/¬öÖ8V(~Ñm[J JðmøTÙÓH|×Äxh¸\_ÄÊc+]£BºO¬5ª<õÏí/ñ¿;£ä¤³d»éZuß áá@ö(ÎÙßÅ·u9ë v4ÿkeøµ' vV¼bÁ{ 4ÿ 0� E endstream endobj 63 0 obj <>stream H¼SËJA¼ïWÌqçÒv÷¼Ac6'ë4Q1\þ½Y±ôb[ÓÕÕUc! óO÷Ý%ñ;1zð5ä!Ù(!�Á8 ¤ °$£eðÎL¸jà%H§äÀqóM¦ Ãv"Ç +Ò÷Y1´çu¡¬CðÑ´~#8lçë03ÊÚ_N,ÙåM ÉQÃóöÙ Æô®PfeÛº\kl ø¦vË{:-QF¯ª"°¡M|²ÆÍbJó°ÈÆn1vÆç (à ÎO.VF¥:°éú;]ô%+Lì\_t)%±0ªßêêL¤×%MGça¯©h$ðº=èÒ%+cê?Õn£ë² E ?Í ¬0»?ÔïùóiÅA x®8dú>YØ~\_y"{y¡ºÕ¥¥WëÊHzõ¡VzÎøËóÅÊ÷osLdNWjsÜó³êK�)- \_ endstream endobj 64 0 obj <>stream H|AnÄ0E÷9EN@cKU¥ÉLr¡j¥.{ÿE©H5JMßÀçû%/Ï Ðyÿ^qE¢åmÞ¿¦ -)ÎèûôÊ·.dÈH£ ] ÎÅNV¼õ¶åyA÷ÅL¡¼ ë>-ûTé)(Ô\ýDE±ÖØÏQk0¶¾w%¶2 «[.-ì"´±2ñå|~V3êó}²Øæc>X =X¼¦a>%Öx3 fiõäÆV¿x^¬¯:õÏÊÔMN³½¤Ô2NÄ ¥çp I.BÅÀPTvÕ®© n2p®¸» ¬;çËÈv"÷XZï\%õÑusTçær¾ðÀÈgßWÇoN@5ß9r¦¼zee:øä@òf7ò4OXYêpñvûçÑÿêðS&å"J =Vª¸Î? �hØöÔ endstream endobj 65 0 obj <> endobj 66 0 obj <> endobj 67 0 obj <> endobj 68 0 obj <> endobj 69 0 obj <>stream x½Xilçyþٽwv»KÒr¹\î"Š˶HvìDÒJ\+\\ëHÒ =Ü$îDqAì&M+h&mP4 u ¥hÿ¤E¤hú'(ºBí6Ôöýf#uårfßïï{Þç{¯YB.ô)Dû½°ÖÚªg ç«áðÅÕËõLAþüÇVÚ¥»Ã{f ÔXõyÚÐXYë]/} g¡ýhhµ{¡
ÐþÚðxïÿ§? -~|ÔÓ
r!ÚÌEÎi¨ßþÏWï}ëÍ÷n㯾ùùîöüéí{?ÜÆïmÿóÏñÐ=e[.CQ ÿ.9BÎ !´F(Rj4kúx1^jB¡G5Eò¥B©ÞlÈÏD½\ÁåW<ªÃlÙY¢µF½TÈ+ù¨fhF§)&üÌO(oìÀø+×.E~\$´2B°×¯ýÛË á®R=Pó6ÔP9ñ{É KJ?;¼?Î^Ýl¤9O¡(¥Äk\_oha® F<éßãØeLïÝ¿ýÖc¿vïdAÏ¢iým çÑS;:MãZSÈ~À.?�?\á×ltztg´àhA vá$ 'ͳflÀdYL¼1t»´çÒc�:)lU(M2Ay(j~éö\LQH®r¸¹ñ|Ï)OPÎÕÜÛ·Êy
Øx½ýïÞÅmP×�nÁ23:f Ðo_îN±u3ó08~IàRn ¢ÔÇ&ÁªöBª2b øÅ9BC'c#qÆD$@þÄ¿E!Y /Á(
VÔÑØ_=YÝòÕ7J4W(skP!¦#ØÖÔÄl 8á.%Í_ åBÓ~U8ÊFÅ©ÔÍD¤ý)xm /e7(å¨ú&8,0ïÑÎ gZLñFs@á¦Á-XøøUr Ñäó�³ã|û4
Þ ¸lýt,(cìÉ¢yÔÑ!é;QøÇ8¬§)w\¥D$«Ù wS0ø¾~8V<^£QäFæj%¯»
àÌ]Pó±CUÀÉçÀl¬éwBÊpïZË!nz �È×ÆÈÇcTJ÷À!ËOå ¥7q.ö üÍkcÎ58¹±
sôÒQ?cæ@Ø´bÚ6õãßA[ö¹7bÑe4×mvä]·xÛ9i¡X H+KÒTÓØFbG÷ÆýaHnéµ3! P¤c@$#Çëònî=ZvO¶ yHõÌ73qÌ=þ$>¾¿Tà ¹Aû[®ÁdLWÁj£J8}.Åf>ºßtÕk2ÖÛ×|ÛácVXÜ¡p#úd6àñqOöR§a©wbÕO Vù(k;ÕËheäïÕz {·ÝwtûÌì¤Ï\_n kB ¿rõÙM®½²wpо¸tÉ]}öp\_øp5'W5ôÐïÛÿCi´, Ç>k²Àxî?¥$ã%nR¸@\AQzŹ=É'<$|Pøî )|çV2M&ñ+ôÍûahØÍ<å/»cAÆ#OÀ,m-OPẤZTSæuÃ:o³Å/£QYK;ñ¤¶Cj²Ò¬¥ ¤VàH¦ÓºoÁÔÄ¢©ìÞJçðü¬3cX\U8:ùôñR®¿úXë¢æ
u3;ªÕ~<0Z\4°0:u Àã@Ý·\þbjM¢È3í¿ vù#ü"j@1X%J;\ËS´Ò¡Jt§þ³?%ø/¬ZJ»ëË!Lç±BÓÜ®KZ>�ý1O42ܹ}UKLÆ´ÀW¬2|RÅÔÕü"èqm\£Èd(«¿g\zêù¹ x~ÜTýYÿçäø²B¢;ué.fEr@ ÚàwgIì<(!ÉBq23O3(w¸ÊëwMåáÄ+
æ/,üh:y@ _,PÏ úg= zêîFÑúÝ h�¥h©Dñ¹ñðÍAB"ãçñÙ @¾¿qá~üuÆùjÒoòúR2eLêøN_Ç?vΤÞ4RqÜVªîTä¶J4¥Ji¥QKòrú)8ïpÓhÕÝ,Èp×¾¨ò gz >úù×\_Uïû¼Ó(¾©È ·ñ³ EaI
Ò)êßÞ{oáǧ^? ´6ò ¶óu)cÿlæ.òÐÊî;î}{绿zï7ø[tCÖcíl¸ß¿÷º#ýÕí?ào½ïå;ß;¬BC24ÒDaòòï¡<~ àGPGà Æ,"D
þë(?L<üÞpñþ{ä ý·ñýwµÚÕåõÀõKðN¿a¿ ó×áU~®wR!¿¹RpõàÞÜwòCÈë
ë6B¾¯Éß6,ôQ¼BÖúíüö0¿«O vul¢]Tô=G&¡¿wd ¼ÃêÌÑ^üç,xYAgÈÙ
ô´#{P~ÅýhÞC;ÇÆØì ëÖR\
^udØþÚ)Tï?pdº$£^pdýÿîÈ ºMìBÚpdªÑeGö£ôÏlð¨¬n!ÕP>MÐê¢ó¨ òIt·Èkðm¢CùnòÞ²Æ�¥|n>&::è"Z-«Õï6̽êÌD·ÌZµÚ4ºçÛæÉ[½öÚyhýBws£»Ùêµæêªy¢sq¥·ehoµ7¯B'6aÿ.¬z ¤.tl®wÍc -BsGÝU¸Á>U¸¡ixj? °$ûyûéûUªccÓ ûLË%aÅ{_øÄÃÓN[ n]n>´ý1 q§G~¦.sÁ{uw¬/rt H½ ëÉ9Ëл «ã±çAªB ^BN·7·:ÝuÓ줪ÕénïBwýªlU&ªÓkËíno¹²Ú9_±Fµ6ñÖë>áë0[põ¬c_²L¢òeèëÒ_ÎpäÊëÖª+Ös§ ÕÙ; =ÓÞ}zGçMwÓZù9Ø:Öìc-³eö6[KíµÖæe³»ü¿ÙY7{+móÔzG¶Nö __ín]Ù4/t[ïmvÚ[°þ GÙMØVBµ=-Qìl³Ù3O®vÖη.@ô¨¿TE׬OÅ¢Ò^§b©±c= ozÐJ¯·±wtôÚµk]¹Ð]íÝØhÿ2«6xf ¾+Öók ý=FWÛ[«ÞÚêûüËþ1õ]þîûþ¿ÿI/ endstream endobj 70 0 obj <> endobj 71 0 obj <>stream xÝzit×uæÛjé}ªÞ»º½ 4^� HA\D"A& I-E²#[Ù²dÇN$':²cYÆÊf¼kbÉËÌdlÄË£±5ñq¤d"YñôÜWÕ�IIOòsð^-¯îúÝï¾nBt;¢õ÷Z9¿fé^8óaïÙ
[Î<öhê0þ4BMs³3§_Ï´"[PuNÈA¾ó-0o[\½yöù1ÿÕò©Dçw#-Ãüû37Gï÷ sÃ\_Y=óèÿ-Á¼ðóËVë¿Þö,¿F\>ì0ùP&l±Úì§Ëíñúüª¡p$ëdª)Éæó-m
öbG©³\Aÿ?ýø/ü\yQ$çìÈ
P'ᡵOóàDzjøÕ7Xê+\_ÿ Þ÷ÄW Ϭá»ÖÖ¿»±ö½\_ãæui/CP?êäC°FVõ(9q6íÇ=Xó ×4Õ¢ìÃLBhaw9øÔJ¬]¶0²Ò|Þð?Õ$&¤\nÅ¡úöTÅ0 ±Ñ¡z>N~|ùÊÕZ Ãò\_ÕT1U Þ»ÈøÈì ÓØÍ"»Æ+ØöØÚnìcßùðÿyùíc2ש:ý°×ºký¸ÎfIûa>vêì¡Q¾Løc¿ËL}ó#¸øóaÇØê§uê¿®X¿Ë#ejUCóZµ¤Å0$)£´h!.bqc<Ùa{={Rr;#Dj Þ'¦L
úëè£hû£VÒTÅ/©d6S)תgU×í+.×;ïrG.ËRýnþ\Kýü0y?UED)zA%üY½&,")½Æò$ôµC¢Hó¤ÿ µï¯á;É jF( '3©¤ÊÀ¹]ª2(Ieø¬VÕDÅo(ï Z¦ÓeJó>ÕE)ÓTBHöxÈFJ>$¸(£,yàޮ̢Ne¿úÝ®paCHý×øyrÀ·IIÌÀË:KÜ N"Z+UkUUåÄòI ¦RÎ|z¨yph¶oøè²+XT½ß >"¦ìBKÓÒ¡�ën?°
hcØK§ÆO^ ·miS
XXø´Gg1oy{ïÒê..O¦þü °ÇWR5p{E4Ýb¸ZÈë,U+UCª! ²ý´wOc|oµmç¾î³»(öÓS²»´ Ø®¸ÞYmïííô£ääD×®^LqëJe:ZpÛ©,µáª¦ mݺÒ,ôuY$Û yªÿ3þ,¹©(ÍMÕZÕ4!ÃS%ð!a9?ó½~LÈÁBÿ±s¥-'f¿õ4³~3÷==
Ún²<ÿ(a¸enÏíÛôß=vÏk=DYÿ[Å¢DNÙ=49 >¢¨X? 6¢<ê2ìbÇ ¯îÇfÔjIÑÈMÃ,Ô"I»6t¸Ðñö-÷Þp¬}Áätwµ+nÙ&°þPsIsrhêÏeÚFJÛNëj¨b÷ï,ä¥|kO±ò@ehí¯CSµTy7Ç&½þ5ü,9(f¤xÙ4i/&MÕLïífV|þ] xÃÞÁþcÝÓsÿÍUìÙÛÓp2q´UV¾÷!¸4?^9˼wöþHßî¾}û3Z1pq/äòçñYäD1(t×g³ïMÉ}S«Ë ÙÌ;£97î9;bäy0àÅ.§;tò¤xÜ°>àõ½hí5L+%Ô4·j,^ |ñÎR(~'I%AÁ27þ¦:KGgÉ5¯ñÈ[ÞºÞwlgqn,Ä8Pôwïìo¾ù½Ïî¸0®8ÞÒ!AÞà9=(æãç·´æÜG¿Ð±Çkר³eW!Ó~Ó¹÷ݵîéÞñû;õª±·?sdzå±[G³#é,ÁCû2WZÝÖ¦dõ£9ßMQÉ:±mrúËõ5òN2úÐÜðÓÉ#ºÔ©jÞÎrÉ Æ5-Ä|'çU(ÃI
ëÊó$QkÀI'µÙeùÝIU¯R·BÓ^èÏéø~ ¶+þ�xDO÷ B¯Y·wY!ò!)ÚõõHû1+ëwMÕ(üÓ¨ ÊÑSå^ÅåÚ8ÿ;ÿ4.Bqèïyå®ÿxÈ;P êAcW%%®%wO0#¿Ð55k«q§Û:®Í<ã ( Ð)¦ÒX"r:«v¥ ¿³{Æò,ìKò8@rÊï© æzöcý#'4Ke>sø&¶HÏÔ°×þKÕ Ã$HD"øCúóéKÊ@»ýÐeØ#·ýÍÈðÝ=׶9 \¯Õ_!É-ôá@CêjÆx~5#$5)mÛ»pó=·¦õ¨¸r§gtñg&öN].Ý#; ã¯ö´9Õo¿ídh§MH³üÈóÚN)¼75í²À3Ä´g×k ? ÃÑ¥
×5C{ÖáëC¨ÖO «s%áüg Bm©$ÛwÇíD¸CY%A20M²¥ü;n)+l'óÉ}NJõ¼0[Ü~ÓÄ=õ§%ÐÃ{uI $æ A ª¹-øqpp\_>¤ T4j\µú#M\_|6ò»5tn8\\_CÀDßN§ÍX\l&ZI7Ä¢- \Þ¶AÈäîJ¨åXD?î4éñ;ãªXmdjPÍ©UçO P£æA-æóG~ú@»Ëå äsÖE¢ _÷µòq¼2äÝp?cymÈsãÌ:Éîc 8
ÇÖc¯~W viøÇð¢D1kUÝ .Á6CZRy½îÿG\Ú(Iå±$è+CàB3Ú[ýêã!]e"Èì�¿2ï¹\sMÕ°Zsd¯×~¨\_2Á/ïÛY~|¼.>EҲŧéR9åpÊ?É( úòL.[ªëOcÊãÅïw´/äzj"$h"" Ìôú8Ü4ÙÐDÊ2ÑÖY¿ÇÆãøQ�±õæ¢!5Ü$Ú£ødÒå6yØ^}w@eÉd˵°åá·ÅÐuö(mÒ%T·0æåb}ã 2§bëhsÌ)¡J:ק?Ù¡pìlô@±¿¿µcKä¢W"Ô<1zPÖ¾ ûxð(-%,h±Î'_p
U¨¥N[+A£Á�}¶£©ëÕ(\YçÏá ´úJ^E]~ Ú¦ÝÍî©+ÄN%cN-º(Z]ÒV@§' C³#ÙÌÜÍëC§7ô9Í.álhÐP2=´ÔÉhñ=)g4
=þÅ
ÅAçfu¾)<ñµ¸¢ÀYA°&¿#P[ìàCäë¯àÏ�îv@?Ñà©e®"@-(ß¹ ´ëkP8®_? ·5õ·líÚ5¶ëüæsÌ"CÈñh¹£)í¨õäÍg&ûEÅ¡O+åBokËÔÀøQOëq(,ûýJ.Ûpu}ö,Ä-Kq&LÔç _\ÿ.Ì]ãRt8Êr!8 ]aÎÍÿ3']e&0$&¶{ì ·ûðpªÒ§²}AÑ¥S¡l2Ú LÅ4�Ý9~C×÷�JIH2àX5ôÄ¥ñ¶-ΦÝ]÷ßC1l&R��ñôýiÛ:{&GqÎB üYý ò>Óèõ y5H/ðÎvüV¼£@]¤LòòpL42¿®Y´w9)Q/ú,Vµ eìMǶAö¨ª?|KÒ¹Ã÷zÔhC@& ¡Áõ§¸·ôDBÌj^@$½½oý[S]¢Ø7/åãô¯?®ÿ~C£²Mµ!º¯TóÖMz¹o:C�:MJDÊ|ÄU ÌÜGû[~ÓE í'äb\p/Æ Ë²HUûµìàDqOµr³5îJZÌHMñù¸fªÉLÒoZÿÆ[oc¬oý+�ôJ0)«j¸G¤PôsÝxë$YÿÔeºÓøÄöDX9dê¯áÿA5´ïdÆûL óø7øºÑã34½ýU2[8üÚ
Þ�añ}¾)e ¢)ÑKY;ä(Þß 'çP>lTÊè͵eÕÑC=¢©J¦/|«7'Ð I
éNGZ²´[¨ÄðtF&ªË¸¯üø ÓÈãÐá@ôa³±àÈ pª\ßÒ0ÝÆ'QìÀ®òcÚ³ÜÞÒz8~0oW°û])pßÁDÀvò ð äwcMz?Rîÿî%?
CU3R!h#Þ§áÞÎ.U´¸þ uRÍ{vÚ¸ô~7ø:×´)«Ù¨h¼Ç6~#\]ñãÛÓÇ:³³C}
¦ �´JT¬tUªéüþ²°93²eübo³Ëáe[b\Ù»eok¹(ªà÷|ýUü°QŨhª¹=ïLm¼¼Ñ LBmy# @?F68k¥ß@Müc\+G½Bj"¸-ßMêg ¸êõe)eG!rà0XÆþíB´e¬§-!óÊçÝ6k;Ü LVªÓí#¡4cC¢ÈBBØFÜ= ÿé+ªÓÑ·FsûÆeçîÔ¡Ókøn°¡Ú¸N\]kÉFëÃ0|L9Äwª¼Ã
ùPf¥øÑôîB×tg×à68°Dz¿&²tËPÿ¦"wºV:I©ÛÊ»Î\éILW%-§GÀÖþ\+îë¯@É~×<Ä}êÎpl¼ ¡ å+§²Ó1öh¸o Ï9ÇÛ0ÕȳFuU y¯ÛöçüTì²&Ü z:Ûª>à¡!I
tÀã³a,æ%ÞaÎDpD>Ê{Àú$ElÀ¡¨Ç.Z0µOë³´0Jò^ÿÌC$Ó@ãW"1hr$z°Àúkø9ü!Ðaà{fz@Tp;]2·06\æÕF¾VÑMµýhaÓØÉ6ºlMýa¶åüÄ=²]sÌXîiïq²7[Q¶vÕzvkUL8¶EÝ-$³Å¨'\´ nílôzk,~ÄgU±H,YÙNR"[«-D²Õ°Í×C-MÁ®[ÑF[<L2î ÷w=ÝÓÜé÷XòÙ¨Î}ÊÀ§/O'ÌZ]©¹ÔØ:ÑüW&m$¨� ß¾}2Iø?¾E²b@ÉñîAì°e Ò Û[²Dmv V¡¥9m'ÄÒEH+<õ+ÖBEvr¨D»WÿG\{îz«GEH A[ê>T·
-®/GAÀÌÊ^¥x ÒÐCñ|òÔ×Ðß=Ȫ ¿AÌÈõ½ ÌÁÍM[ü±E ²?¬½ £8 @½ U²¥áäà Ʊ¹&Ö¸"{eûÚYÄxgôoQelsPü°ÿJJ[X¹OÄú/ñwÁ'LÖØëbè¿Çr Ô)B½)ÌÖãú¢À}êi¸.óH¤¡{î X0ñ-ÞÃÅÕÖë²kÄæ ¹ÀQHP¤Pb帬9�¶KD¢Ð½8cx©³g!S·àV Qnû½>÷TÖëbÙèA"ãáRH£ qN¬ÿXð©;6fö·øé !-ÓP§|UÐFyKè9%I·;¡e¥vgXÃødÁËw¶ÅµQêµERZ~ð7±¡ zD ÷v¾ðíêc4·ÌØS8!/F\�NÝØ h¼Zóo¤ÖHU¹6A6{ع&®ë 3zkH dèí){´ Ö'ûjÛRD8°¥{[myPQÐ2ÀsN4WÈçÛZ ¹Ýa°òǽ]¢P < >snoÎ9°lw÷CZ´×BLöxkêm6=9ÐzClÚSý=oonôÉ¡Æ
&,Û?¤x$èLý"¢+»6Ç7÷çfêÌÏÀÇÀ×xqÃþZ£{ÔÞbÆ&ß^1 VSTövõ¥BÄêNÕ"ñðN°MÔîá¦ÑéBÚJ°×ÎËíOD·an9O)æ÷ n"ÅUS ÷Ür1Y0kÄÔÃùOÓ¡' QÁÆFmþ2Ôf sNvuSïK×̦ÏÜPë½ñÐ@2ïï˪ÝøC·ö¤ß5}®7u&NN/)©£ÎÁº¤.à/á?A];|D¹uÔïäpªpÿæöhÍl]¯!N0ÊÈ (ëÝÍíRÇÒöfG=-Í\ëØ.è\_iïvBì8]\_oËxûbÙx!ÀÇöY! ¯ÊX\_¡2~ÓçÊ:ÐÉLeVÁ¬¼øáZHE]t9þ,î 3Gwµä &üÃ[L&kl ¾7ÃÀ^e#V¨U±?é LKG!¬>-Ç#òh P TGbYìòôd4Q%o
ÅónïÌ L5kE5pß6{Àë£Tµ³H(ß÷ñøÕÁÇ5ðqÁøôe³6úh³¾öã¡«m·¦<(Ê!_,ÔÞÒÛzx¸¯:=ùrj$Þlï-íÙOºñ½á=ÀÔú\v½
kê¯}Jx»f
)-Ö+ üë5$I5>«üÄ0zP�ý>Ì¡H3»l )ÑQHBCR[ûCN,V! RvÖ¢0;4SÆã_ÖCB´µßídüõj�Jpþ(4qylõÄÎÝÔ :ßBdCåR5SFíïìÇм(8ÛÒPSñXjÔóU¾e¢M(µ1Ju]$Ä)ãyÁKôwÇe·öå�Tb8å÷þÜ-(TõYlmðôÀ]ÏG ³Aÿûs¨§ÿ¿W}º¹·glã5dCÜ+¢´ñ}{»±-C ÃÁl#A§É»$B»TÏÅ+'Ô MàĨ¤¹ã[½®ÇáÞçqyø&A5vçPh[è¥wïh²@ÿd¶òT rzÛµmkù-ÈLy_¯ÿ<�ò¾¼¥J7ìv<ß¹x×Þcl±ò�"øï=ð ·¡h[¼ØådbÙÆvAAi2WE$b(Á å /ÖE©ÇêóZºzµ}§ò[pvÛ :)ÔÐ1Á~:fÖ}yGæ£N,92ѧ|ài&ä@иþ{±/¯èÀ>ñ_}:^ÝN;ßäsÏpÄ÷86;yþ@ÁJ£�d$ImÔ¾Sd;lLqóO éß2 ÛC
ÛdKJ Ë;å? hø&Ó0sd?gx· Ý¡zAʬÿïæ@ØÉ»÷ÇöR¢ùõ ¸ºBægòTT�¨ Å?ÇÉTø¦Ù,x®ðªýÆ$}=£µ¬¾Äü'r ܬßUÀ~)¿§ïÅ®úßAnk²ß[V$,80çq¥W± tY¹,Z}¾J>n3j- O917ø{«iÀr'/¾ÐÐðò{=Á®ÙÿWþ&9.óæÍ»AaÌÏÇ?OihÕB+Ó!ì¨%RÆ/ ÚÎ÷Fö^¶ó²'| Ymf/ÐÚ²Ñ Ppë^Õ;&S±;ZJ>îv&; \Ýeëb8D%Ï\Äû¬à¯À0ÙÁ]"ua!+a x¸õQÒ;5v\_«[c©Ñ3xcç#Ð3|éM=cµxiþýV¨}à$iýÃ!ÅjîC3µÃ¤ÄÑ´ù=«ÐýMMòPå@s{0L]© 6lÅF=!)�élß@�@úþ%{!TlÚÚÇq©µù)Gîg[
Ú!è2¶Oú Hê2×ÆAü¯Å#'ZÖ Jn ´ø$ÑIMe[�¥»A[;-²ÅçÞ,2¿»VÿeüG'\½¯#ý?<^<¼úçëëï^¦ga1¿¿ÉãëC/ÎúÚç
ç¯þ«Wá÷çù§¨¹tút�ß@üjÁßG³pDàÈÀ£h/ݪd¶þ2éCnò(ª? á!"ä$?DDr
ñK(O¾T²\ÿºPÿ þ}Xëï!{ByüOpü1"rã?G¡Wèe$ñkHÁ?{~F~|Þû_F6ü'õ7Èmõã§êoà;Q¾4ÒÜ ýv7Ï|Æøa8þ,ãÆúú_b½p]
ÓpÀ\ü$BÒÎópÀyË÷²Þ ê¿wû?"äóN8çþÝmã\_ò|!ïípÀu¬ã>RJ®ò~ Þ©,8ð¾Ð» ãY" Kto#°^îÓÿ¡Ì°^êÿáÅ\¾<ÈüØÆwð¶nú5�öÙðu ßýçÆ }«1¦à7OcÌ ¬ßcëÆßlEä"m±\_5Æä¢Wc ÒÏ7ÆNÔÄæ~Iì¼3 L7ò1ä2z¢1&ðÄó1Eeô½Æ¡nm´5Æ"Gc,¡ÏB c Ñ#± èݱí¤eA'Û>tT6W éhFËèF4 ãýèè>Wa¼uKè]Kòß3ƵÓÀLùs ð££I4΢9¸rÁÍßY¸÷RãNôq½T,ÖôÁÓË7Îêûo¹°:»xA]:µ¼r~yefuötA\XÐ'çÏÎ^Ð'g/Ì®\Þ°ï_U'´ 'Võñ×EA¸&gÏ^\A¼°¿;P<>Ð122i{óÓ
bGGÏàðÐH¼ msÁßòè[î?hè~l² Ïé× 4öÝ8ÃzÀjË·Sƽ6¯Ùò«ðs°¿ç ]UoÏ÷t�^a,øàìÊ
ùå%ÝTdb?=sË«§.ñY¡RìY97»¼z¦°0c ÎtT¥Ê¿]Á«N7>Ǫ§°áOsË ò¿/¸øÊKƪsÆs6¿ym?Ì;Í·/ÁÙvãyÝ0Ôá�ÝXù"\å²Íw4AÑWWfNÏ.άÓÏü+¡©Ï/é«s³ú¥y>Û¿ 'áñ¥ÓíË+ú2\YÑO-\_\Z]½Pàai\»¯å¢r-g.ÃkVVõý ó7Î> endobj 73 0 obj <>stream x]Íjã0
÷~ -;bë¿
Z YÌ°%954¶QEÞ~ttBfó IÇßïm\_¯yÚDû3/á61NsÌé²\sHbH§in¤q Û}UáܯM[.o-ó¸4»hÍËoák\ô¥iäò4ÄÃcY¯ëúÎiÞD×ì÷"¦±}ë×ïý9¶^{<IJ?m·Çrçßß·5 U×2aé²ö!å~>¥f×u{±{{Û7iÿíYÍ+ÃÞû\Êr´ëT·/¬È ¬ÉOC~[r�;r{ò~ª¬%ø\3{²däú®H®ïJä>stream x½X{l[×y?¯ûûyÉ+Qäå¥HZ|¿)Y,É-E#KqøµÐ2m)DEb[§IP4¶Ð¢Ø´Û° MH³¶+R·º¡K ]6¬"]±bIȺZÜw.)ÇÉúÀúÇxqx¿ó×ï{s.!}ÑÎ+ÛÎ3aÏÅõ/,_{æW@¿³²ÚlëÌ0B®@Ú0ägé2BîÔW7Ú¤DüwPúùõÖJ¡k þ,ÔW7l¡uü BÔõÍÆFóÓ¡'@=py«µÓî\A%¼¿#!èEô;øý BÍö|ɤä=rS +,fGÈát¹×çW(ø»×þú 9_%×aС;ê¦Þ¨7àÆQ+v{Ãø_>¿ûÅßý8¾ýù'×®Oã§Pg÷¼{ý¨3»uø|Mw®ã ¤ÒÅUQ21C%jõZ}ó?R%]FMõzÒc«ûª
2#Çjé1\0O»g«ùÉ2¡¹éäüʱjvjÒÛª }¤8glüHaôú«Õü¾©¥Æt:?ÎíHP¦ó þ{À@ ·]-a\\É!ªè5!K¼1Ã#LÁ?Z#̾ TFFNò³xufªZ:© ã¯ó§fÆïƶî|ºCÆff4»µ=¹ãÿ"¹eÐ(G¨Úó O^RËÀT$ú]4)QÃ7®ä+,P×[
ÛÖdÔ¼Ãí#©ò$ e®eϪ.Ùñ£sSµ3àÛ#1¢Jø¶åÏÜrôÄb"@lj@¯OýWmª[j ¯,\_=êÅÇï9x2è0 c6ÛwdrxäÞë?Á?>1lÄ4ªØdh ¡èÄ@-þ [s ÿNÈ> þ~?h\p \Xsg-h:Ø°ó ðªÒðD¡Ä0Í
¡çÅÊdÆÙL5óCçί l[ìZ²T+«¦HÌ@05¢l\¾?àñr-ð·Ü1Y%©jDû¥g;Þ=DIºí±ä롳V¶Èá4³úldnJ$Gvÿ8ñÁy¬±S~/ÝLåø¸ö ÔÌgKÿ$ÿJ.¡É=¼7Á5Ñþ÷zæ^Àï Ü-!¹eþrvøNª;ãr\VlÊÄA{2JèxÑGõBT³~ýjUÇeL5YT.\_¨yõÖXjøØÚ½/Ä\>=$GAÔuí¯Ò?$;ÁS¸"vc:Ùy?ÿÅQ
gg¿ c©îIé'É \NêJWç!±ÿQ±oÁ¯[±BÁß ±VLë÷\²-D°Õ÷3mÞðØTËÎfxmñÚQ¹êÃ:3|8«×Áþ³zH8\äïÅã3o_çx?¬(Å36÷ùýê<Ûp}ûCïf¼ÿ9!p6îi1Ï ÍdnjJf¶Í¤ÇñæÜ!ÁÚòhÙÁÐdE ÉbY;,Ú]ÙúÑ£Çû]ÿÜê|
|s{sÕkU3ûÌ$4a#7²Cäæ Èj×\1à Dè2g Rm"KwÈ ¾©h>ÙÓ= :¿@?]º±3K1U·g?3ñÅY,ÄqÝr N-MÅðÑ îjÕÈ"$Ørl7çÅ , á?ï3çwÞ YæÞ²'({oµ»ßs¯¹yMnÍX7¥~ C2º¹&f¤õyj#nV{(±+ÅÕ«ñ>Åcvo ú¼O÷(!ÁÐN ªÊ
lÕKi·£_²d
Åp1ÖbçyÞq|~å6ù ¼@Ò¢Û6T<}T;pçøm\(ÜÖªOÞ:ÓY\\v9>¶êê4ÑydÞÌNì«{rÏ3Á¶¯µ{4±~»Ïí¤Gor¡> ±øcFB·ªºOPK ¸ ÚÜL.)LĤ"RÁ o¼ræ6B6³î<0ѨFåæйNǧ=rc¹ßÉÀ'Ss¼.8ñÜ8íÜ\ê3wé¤!þrqúÐþr¾T.G=9
¸{z±vì©DlqôðÃ@x·óзñ\Q÷,Ì=²xÞOÓUÓ»$ã&×÷DsÑHOx4à²a±¦H5Æ,4#bdÁ &·@v#ÕôHÀaÍË"Ó®fj~)LRFZ]±á%gAÆ X� ?gvãÊÄaßø!¥· ûÍrlìü÷ã" ªY5Ü ×Æ Ý)a\öÇúRÉP®VEeÀOé¾pÐgD#©5Öì.%g! éª/®3Ze¨�xbg Àcz?5HÝl¨y+&pZ/õWªtãPÜ%¬bÐ[wQlN)Ñ¢YÍ.6j'¥ TBîVìHÀN>~¾
nØoIÌSX(¤>Øi
EÉ ØJ<ÉLx«Õ%xϯßÄ_q{vVÛ h±TM¨=¤ïèJB2ýª^»Äw $|Ì£ùLT\ý't»ÝÏĪC$vKLav_ } °(p-Í>WÕ$c1ª'¿ îÓò§Á ¥yò¼s°(xmÉê]¢ v²ùr²UḠ?¸á¾+|¨÷¥Ü�YÕÆlèhç%<ÖGÄd¿Ï,MØìp2±Å5×)ÔÝáøÄlù85ÄY«^-¯D$ê:þSu %x-ÉþÀüÉK£l|c B>¼ °ÀE#+WSa¦@9prucÄÔ«òî¦Àoü ²?,.¤kØãpÉnKÞu6 "Ãê°õj¶À±YIp1MÎö¡·;÷¢ÿÝþuT"ßBqü(?RxEqÉØbø{À{ Ó(?ÒùfoÅ^y VÛdA( j!uAòsÄ"$¹ ¼|Ê5²Ü åëY§¡ü�!ÛãÙ' @?ÇGù÷Sj _FnÔ=ì}¸!o'n|d©£zÑK= }¿GSÍÖ£èfºG h?þn=ZBgÉ=ZANºÙ£H£/öhI/!ÍÁño8æÆ°w=×£ ²¡¿éÑÐ?÷hB KP/÷h
ëѺF=ZAaz´G[Q^íÑ4K_éÒÇÆFÑëà©5 >ÓBçPèeô ÚAm 7à£9À¼Û Iþß0ÛΣ9n-¡5tBËYk» }/÷z¢/è¥B¡®OokêËî´;úÜæJk{«µÝh7ÏçôÉõu}iíj{G_jî4·/Á Û°~ f]ªí;¸ Ô,×å×�k7Ö×à]Õ ð_;þ$FShƤº³tçȾwh1W(G'§§fFùä0wvo¶ß:îýï0EÞÆ ÒßeÔºÇáÏ((«}/ßhË¡ªÙºË\ùx ÀåKuûá[Zmnוּ6õ®Ë(FW[íÖæe^ËU £KÍVûBn}í\ 8ÅZ¡Tý?J÷®!ÖL#7 ´Mw8oºJèKÀkÞßÏ¡øÌ欫æ¸ãP[»Ñ¶lk®ÅWßnÞ¯ZZ5µ¯3ßÛÙ;×sµ½¡··çíKzëoqG}mSo¯6õãk¼¶Ü& ß<omë-hÙÖWZ÷m¶·×;0ÿROØmXCå:Ü�Nà¬B3,³ÝÖ××6Î5VÁ#¥ ïòè~óɪìÎ3Åض6(o 8hµÝÞÚÏßÿý¹[imäÛn5Ùò° Ë3Ço�õ×ȯ7/6Ös«íõßw<«¡ÎÛPãßÆÿ÷ï�g«©& endstream endobj 75 0 obj <> endobj 76 0 obj <>stream x½X l#çuþ¯9È!9 úÉÿÿ>ÿÁï¯K"ØêBBE_Gkîà õÕð;ÿñÒÞoóå½_ç_úæKä«»§ñ»{ßÛÅ?Üýþ¸OÚ5AFç¨
Îð5j%] %1̤+åZuÆãr»GÛ§+äS\Ëïs»8°ÓðÜ/ãix.þÑ'ïloܲú£V}ÿv[©ó6¹\Gihäp&J¡ 6RU-äÁ\¬6"JL¥®p~îÚÏ=~v°'E\JÁ;yî\_9wnùZî æ%Ä¿;sé{üé\_ïÁ"Í,I¡W>¿üô@>xücr½ÁÎ.þi¢i°þ©²Rý(!\_Ö5] JInQ\Á gd¸4øQ|7ò¯TÓz&Î\_;DL±³(÷KÁ$áQe3LW7 g"úRt>wãÕ\± Q¾ÞK§?Î2óÞ^&$»1
EeuGÆ\_ùsO´¸0%}ù¹Ê!'½óù¸éË2÷¦¼o<áÎW¶i}-&CAÝrQ´]LóNËE¢ !Ä}�K7&®S¾ówµ)I]'ûû0Á=!]×6ÝË9!¬¨î øJ}³ÒaæwaB°»÷yM(»+eÈ=¦õËkeJF'¼$3¢ÿßôù¾/Ám>$aV\_þ^o0(:ÿ (nåØá³Ü÷Þλxï 2:¾3v�MÊLà¤ÜRÍæj¦> é®TNñ[¤«TßöG+~wWø<,ú±5¤}ëéy°g9Ñ §ç ªé 1zNÅD8#«T ¦' ñ©½Ã¢ÊhW´d!te"ÆHÖ#
ýÑs Îý#¯¡ ZZ9åÁÜ>îDµådãtñèA°Z2}Ð"n&¥©ÁÌ
ÈÙX2òbÆtg !f¨é³á\¿1É(2² Þ)çfÆTQÄ_ÃånA0¢çwwÜtÅ¢äþTÆHIbwjpqz·{pO B}¥v ¨#¥}j¹Àd0&E ÈnTrßK (-úúÆ4
Y Ê>HÏ¡á#NÖG>0=qåÐäAò-áXé0!]ªhÓÆôÑÚàLÈ;ÎøÒÙ±²Õð×é¢ xfa¨WúåîÝ>[\Mö
ú$2>\B¨3c³Cþèî#GãA¯sèBsSÎEwgý©!yþXAåÙÞZߪføxX3fÛL(]{"À4! ¸Bª,SyRÕ0ãNØK° O rH ¢Ca·è¥Ýx,c %Ã;ú¸è]T-QýÄÃÅ°Ìx,$°í EÎ;æ.²¤m&ï¨ÅHSÃ¥bʨKCO0%�Óظ¦ÒLtx³ÿ3»£Ï8¡±Å÷ÿ&¥!M$A?C~ò¢ ;¸íÿõWFͤ
®,¼xºÞA¼!{ôM¾éÙ¼}\~BÎ$®T¦P É øîêŲ«èã8êËcRyæÃAF)c!Ý.bÁÄþ> û|L>ÊÄ¥.»¸V%eÔH÷|ò³\_ðvÀ+Ʋ/ιkèW# Á½/
¢¹_}=ß-RÁÚ|]VNy!§¾8»G¾Ú
Y²päøiAÐ(mcxÆFÈßø¡ SnÎÒhÍèª7nÒjdب�j]åB°êìÄbFbp $#üÔßé (SÉ㥾sÑÏ "5}/!¯ôÄsz>ÔQ/ %%a7߲̺ÞuýEðSEQ^E÷ë§UÍSUËéwçïÍU\SwcÉPmñøP8;wÖ«£½S÷2QoêÒÒv8ûå ¡0Ú5«>ób|&·af2ÕܤÍOã}ïFBMÒ{¢Q'Å~ÒSáÆþJ¡êò2@ª;wCÓ©^ FODöó�B/rG &
@Û Bg¿Õ$6Ë?¤)à{À1afìß$Òæa;dÀHÝrÚ(î?7^îÐãö{ú±Áé#> ðcÆý¨ËG³Ý±TqÒï9é¦uGo´'ÇGÓCGö¨÷Ð{PéÀ�G($Zªk|570U2cvóSnÝÄVg¼>XTûPkA( xDÎÔ$&²Å¨ïÆ£úÕ¤{Hÿ@ø®Ø³~3É xѳìI,z¿@Ä Rç-üü$:÷ ¿UGíC{zÿÀbvë&[3±Ïdëôe= ¢÷T>ëòNßS&îó3ì ¸á8xº0(ù³£Çópxv&DPÒ/éL84åk=]Ä°©+Bøgþ8é§Ä,0qPVO>òZLàïS¸óoÿÄÏàO¢Q80ØøÆ-ãåwdßöýÓfó ä[ͤÍTò$FsÊTX).SÇ�Óéxp6&\Y¯;(:Ô®QT2þy¥.°Õý+ïfÜÐSÅ"£@äî±7HbðäÓzÌç ÔzÍ
ë×?u¿:þRè?qøßlL¦÷ï{ e:Mr×z1\_òêÞ·ÐBªsq÷9¡|Ç 3B~~¿×/û÷÷Ñ4y ð¿ ©"o¢^üFçüÜwPÿ¼M¹Ñ"|¹ñ· õáòá\¼v\_G¾Ùùw{ý3öõ8\3¾ æ°rhàÝ_¯ýÒ>øàx!ç ¸¾ò0B.(½®à}ÿ/az¡áäCV¥ÚÿßàØ\_Q¼|àk mà¡¡¿´e$´kËÅñI[fàçS¶, äÛEÀäQ[Ð}tÞH¥fË ³-{P/»iË^$³WmÙÜì/,~öhÇÌÍ Ó.c4^²e+½eËMÁÊ- ¨ÍE%K¶,¡ß'ÏÚ²õÐß±eèßٲͰ¼-{}Á}¨}ÍÁN'û{ô20¤ ð4.£º / hÎ ´wÍ/+0ºÈóߺ9vÞøskð5àDº®¢&lÜ0÷=½l
1y¹u©a,ÜÜn7Ö·ÙÖÖfk«Þn\ÎkkÆüêÕf{Ûol7¶®A' [ ¿«ÎÄïgtèÈÛ×Xu&om´¹q¯cXÝÞ\«C?7è\zжÍÆÕ×ê Á´üÑ(:/'MÉRi)¾Eáê¹B±86ybêä× MÅzÿ×+ýÏ/nÃÔ,cÜfîÄi¿ÇÙ,¼bνv0Cst=�ëñ9W w V½ °æQî%[Û« Ãrnn
X³Õ^im\ã\¥0¶^ Ñj_É^AO±Z(UþÏþ~ëUGu¸Ú&ã.läÁ{�úZàÁOÇY¾ò¹jÓ|n Z«c Y3-íÐ77LÜf< såaÛ¶jÎÎÙÜ[Ý6êF{«~¹±^ßzÀh]ù17V7v³a,m¬òÖB:áñËùÖÑ-c¥õàF{kµ±ã¶Ý2¿pzêNAÍVÛXX[]¿T_zÜ? .äÑuó3¡´ÖÉn¬ÃXÀÛÔl·7çóׯ_ÏÕ¹Ù¹Öz¾}s³ñÓ¬ yÖÕá3_éGëȯ5®Ö×rÍöúÚG³þ ³ÑKÿ;²þÃÖ:ø#¹óiþ?õG?ÿ rS endstream endobj 77 0 obj <> endobj 78 0 obj <µ!%'6³'¿!-<´$21?FHJ»(/HHF¶.1QSGJ¾+9D¶2 =F9F}B>NO\ RPOM½0@¼4 yDJ½3-¾5(·?«=JDNTVYVVTdSU¼?È=;Æ@1MVPX]]\_][»KU[ÀJ ¯JWÃI\x[\\§OYlaÆJ2ºT eecdehÀQ ÈM@[cÌT2^fkkoÍT8mljÆ[bhÇ[%ÑVCÊ[0~lmgljnbrsvusq}rrÈf"ÄiÑa@Êf-Ñd9²hqot¨luovwz|{{{¦pwz{z}Õo?ÙnI¤y}Ïw.Ç}Óv6µxÏ|&×wAÊ \ °¨áMÞJVÚ:Ò Ö4ÝCÕ-¥ àMÜA°¡§ºÜ%éWåNßA£¥¢¯¢£Þ8©¤£· £Ø¤Ú£,äL·ª««®«²«Ê¦«ç¨Bó¡lç¨Jë¦Sã®9¬¯äFä´.¿²³¹´³´¶³í¯Qò²^æ¹@ä¼3ì¸FÀ»º¼¾»Ç»»ï»QñºZÕ¹½Å¿¾Ñ¼¾ðÄPÉÄÃìÉ?ÑÄÅÆÈÅðÈJÎÉÈÓÎÍÏÑÍÌÏÝÐÑÖØÕÜ×ÖíÙÛãÝÜÝßÜêÝÝèáäáòååèêçîéçïòîöðï\ óöòý÷öøú÷üÿû)]/Filter/FlateDecode/Height 240/Length 23161/Subtype/Image/Type/XObject/Width 245>>stream xì½ \SW?~»daä»±¾ÿKP;Ô|afµRø-¾ üUa)aº58VJµ¬ã¸¤+Jø ;ÐaµÅ®ÆSburë_ÅöÆ·Dk¢!ÉlÒdî½ 7ÿsîÍÔÙÙýþÏLòrïýççí<Ïs<ÿþÒD»ÝÃÔr¢ÿÚõ"´YÍfUÉå¢z/5Ë(ª2ÍVIüµñy=LZL¸RV({FTTTdxÀ_3²áår5n&ÿ?ÀuÚe3iUq11È)&&&_,×ìÄÿ\äÕöÔ§Ær¼¨¢æ§¥¥W¢ðb\û NxdÔz ªµþì.§A)ÎgDDD.,¯it(pÕ9 í´Lj¤¹¦¼(jZ>¢@$W]î¿ÊÃ?6½¬¾<>|xLLZu£T¦µåñ ¤Íf{leè± þk³$9ìñPiÑ¥ÿ&3 ~5:§ZÿG0¦ÌLÆA8±iEb¹ ·:].Êe2ÖIöïíÞ¹%>ú÷sþðÍ7Á >årY\R'la±Hn ÿÚ &&· U/,Vû¨ÕAÐùøÁW:YV¶¦lí¦5ñsçÎÿB?_z}ù7]µ¦lgçùKþDº<.Òj×ç, -hF-ÿ}§9í¶DéÑÅùB ·�ÆÙ4è±ßìþéúµ+ãçÌZÈMLä¦Ì/;kÎåÿ{íÖïþÍ1¥ÆævÙ êa6¼ZÄZNþ÷èV?NÊXáo56XÚ¦K-y=Rbò´£ >6'5uùÊM;v>p {H»V^�ø&Rý÷å4ÓéÝÚ©a·C/ÿùOôÊ«K¹¡�.°?yÿ_ôê«k¶ü\ú 9ì¶kdp¹pbj6ú¿nš²É42bNdB=·n×ãþÃo¯Z:géÒqÙ»@0楯�–ÏYUöËó&’¢mYN˜BQ9͆ÿ³ÜÖÄÉiÖØÜ..ß²~Éë³A zÙø£þ^4{ÖþÓá~»ËCXäФJ\m´Èe£ò£%#T)ÊvÿÜÛññ³¹Ü³¡ÕwíÚÛÒ°¯Ò?$oæå®Îú_£?Èå&Î_YÖùHÄo$ùÀ²Êj©¿:·Êêp$:½Fn"ë·¬Z»}£aEÒHлnï{ %u_V±¯Þѵ¯æóRÞmÈÇ°ûW>6^m y=ÖÿjÈU2Xéfi<·%~N°Nʨü�©¾àÍ#§ªü/óY´oèöæ$VOçÖÝdßüðÞ@Ój^anF¦w¾³ ø3yVüªÃ¸}PV�nG Ô%dTæµ9_l�ȱ²×ã½f?K©:òÙààÍC)|AShïÓ·Ý.ñ.ÛÜ6Ýݽì÷>Õéºöv·Ð×Þ²ºÊ¼ ßrãg®9øàÏ.«,-A$Æÿº9îÖKÓ9È´"¹Ãmùb÷ÊøE, Á~Vüíͪ¯úûOåá6 /³ pç=x·þuF§»øÉ÷NäÑ~ÊSÝ%àûü¬Ç¿å"TGVãÿEÚ¶+˧!|± û¯\ºÈ÷
{Ö1¿¤dì9;¤;±µðHSÖÐZt¿Êô¾}¹ùüæ+ºkUUuÝí+x£hóÛWX·Úÿ;gùÎKv· �~å¿ÙææùáHX9fw¤oÇÏüc!ÛÙy,Ì÷¾~øðdÓg²F£Ø«ú¹§obÞOÙz}àòn]&Ï«ã uwZN´¼ôÚ¢WWýwS÷ÅÀ©×üåu¶Û$r!XÍÎþÝ«úµTRVISwïÞ\¯ôì¹6¤» Hm\Ñ}]ĬҪ¾ë hþÖÛð¸üe¬E,½Zî ~²®pï
îͼ�%ÎZ\ö;Ë\ÍQú/=Å eÚ '¥ÿíúÅþÍË+mï;»÷Í %¯Á§èÍ]\_ Íc
ñ�óÚï=ì® 7¿Ú®a~Ãàà!ïüxÞ×7§¼¶ñìà ¯#Ýì¥ëwã¢ÔõÀI9þ¢Éc±aHt¢ÍÌe9»zk+
22¾ð2öÞê+Imò [®ýóÆÍÝÿO{!3/r¥Ó
¬÷n=¼X±z¯îÆ[#ÞI¿é<å¶tÄ"aÑbÓ_ÎÑvã¢8³°Çá¶ýûúx¿¯,¨kï=ýûmoǬ֡ë£Ö43¹[O¼páÓ3í «YÝt oðÖéÑö¤ÕÝOUfdä À^üýë]NXBì/¥²ÝÆÚ$üPëÁÁWçú °Ì÷{¯´ìº2ÔW2tFÝÐÀ¬�X7vÀOY·îÒÒÒ ßó¾þ7µ?þaÔÛîèîhmøÏÛÝcÞN5sæ �uX>úñ³)4(çZ=I~ñöÔy ÷W+²²tí ¶ÝÕ]¬Êcp @ÈÿgJVVîXÌ|°¢o¶m;Ñ{ëÆcÞ$.]²å7VÚÜÆÉ9þºC"MÀÏX3'Ù'P ½u|+X¥}}> ë}¯òË¡;§Bx>¥4^\Å7¯=ì®JÉ|÷àct¤äÔUGSÎ Y;fl%5¹\,g¨V2í)Ø\Þ_ Ümób-´º«7¾lÙxFn"CÉÉð?ø/ø ÿYÄ A|~nËàP _zõfpàóùyoø0û>I(³Ã¨Æçm§¸äQÀ½Ôýñ³½<¼ßVÉ°þ½Í×ugóg÷î/-ÁLîÊ×ãWmÚ´óàG<|ôèÇ;vìø¸lÓªñÉ!"«÷|=t±Ý·}|µ5¼¶úî#¹¯yoµèõ²K$¥)KOd~®¨)ù ÉdÿK½|)l¿ÐWÐP¯¥´ ¶×ÜäJL^¾ü'ï|¼c{çùþ?Þ7?öÒÿøÕùÎwü?ùñòå#BkÒ Wç,©ô®= ,ª7oÆØ,¯ÿÃ¥?÷y¢¦QHD
ÖE[ïÚu-wtw·1RPycàr°Ý´Ì÷fêÚµnVãX®ìÞ½víröZÌ¿y»Îô5d%e¬>5tç}ïÅøún µgÍ©9¯H¬n»Bü¹)lK°<+ÔnÇù5½$ø°éôÀÐé¼ àNáéÝQA¥Ë7½³½³ó«Ç¤Ë=á »\¶ûç;ýÎOÖ.MõL)løôHÝþã·N{Åáªöã{Ûö¯£Ä5ºM×@¹<ÈÃ6 XÏÿdpÛ¤k0ÏJ\_Ák
§´äx±ùÖÐ ÇMÝ´»ÅmgÒ °áØçÿòöJÿªÜÛ}¶¯¯É{Ѽ־îÊÕMº»>K
UÜYÓwß÷ØêãÈrýsáµUpÔöck~ÌÃm7uW+@ôêny?Ï;§SSüÎѯ|;e«|Ôß¹}ÓR6Ȳ®ªrsf´gǯ\;±yc÷мó x n«4 f{3À\þ}OÈøp&r¦Ôç̺3§öúÞK\µ½óÒ#;õ4ã>L;Lß;¸6Õwe8mÝÕr¼e\_[o[0þAÃ[rg¯üÜmÄþìËÕæv¾²~ôú,?}¿¬;¡ló lAéÏÖå²³:uå;ç�áî.Ç£óv{u@ðVïõ{×%åÖ¿8 ó
!¼Ïñ=$/~[ï1Õ_³ñYe8%OC^¨À(Û¿ýp¶ÑËÖ½ÿ¦Olë¸Qé)¬Iç&¯ýå¥GÏ!ÚmÑÛ´±\ø øy®\m�vI漧 ìËôÅ2Þïòù¯;{æ¿è ½¬g »óN6îr¹-o¶^¹Þ½Êh 63ܺ²93ØLþÁ¦£ýùï¼ßY¶¹3EÕ&0É2N>¼½9 ¼ó>õE_ßùÈ¥+Jf{i6#]EÇÖÏòòòý3¿üòu,hÞmÀ%¨¼Á¢¦wéÇýöççÝnJ6 T%¥¯¥ð¸Ë.?¼�ÍÍl½ò¡O¼þ/&¾Af<ÏåǪ̈ëÛc+}fbÕ©ãïæ-{·4Ïç8m»¨{xûgìÉk7î·=á45µ'"ôïþñR?gKöÞê.ñdUöén·yµxbüîG.¼ÃÉ?uàÌüç()I[?ÛwËýgßõ²Üë(h»ó°ïP!» ï8g"'¼¤´0ýÔ~šìÿåJŸÔXÖtU7tÜzAÉg÷†nù@óæ,þ?6·42ˆ^ãS¢¶Icf‚Pœ»ˆË{¤¯$cuÉž¶OÙµþæ6oÈãó�ëzùÛ&d¢Û¬HDÙåJ{jäzüÅÎUàöWPuîÓ>û<üÖ}»2ÁÝMH³_=f# N
á©0»äóð1ÿá§ñ¹¦kÝMíÝ×o鯷2Ãù^0ðýþå¯.¥[ö"#ÀÒöNñ¹®~æÛ×.lº>pýÓ7qÎªÏ {s-y� 9Z_ö²ÏÝ3ºòÚà×w�ÝlÙÈL²eð_îÒ5ǬO�bQ©³«íÀaA¸)î$'üú°õÒÁåÉ3ñÖBFrf¤´ÝºÙÉÒÜ_éwê#¸§ «$ u>zÙèR® ènÝüäÐgKü·J^~ðþ5n4A pCvÄDhAk2cJËO|÷£7±¦© ±fä¶ÞºµoÔ¦HÜ·ï»Í5HºzÊÚNù=¤íñ³F\Sð^Óñw ó23><ëw-·üFÿÄùJH+nEG5kdBù»,8ª4Ñ´·NÚ¥$#eï Ý½Ö¼Ñ :Ü×wZÝXv§bªV8¥ÄHÔ¹×çð2(%)++) ܳ´Å :1õßMb9ÅÐlöØ\89Ê®Úr!m² 2Ô®7NÚãêßîÏÒJY½çöÐC«yѶ¥¯ç£g>2Mlf/B½ëþ¹ìjx×43ÄËà} )ôbþÌò1K7Ú=6Y|±ï2øæ5LêqÜHÁ=£'8 Ô£ÎÿÍ2;ooÛ ðlÝwßvÐì#m^¥)&Ö£hhNº¼¿óUv³ò ×eú®/øÙº¦O÷ÃÕ¼ê q2ÖÙwY,L&ð.mAP\ÇMäÜvfÊÐA N·Ã+Ûù˹л930xv5/^ʨj=syFnü¦Kn[9NaYS2¨¡]¶ÃKØW×}±¯}obÕ:uñìö
ts-J9n ẵµqaN¹V wçGu¸qARJ}/hz¿ùî¶]í2røý£6¢+/´Þ²HÉ»5o³ip½g[ÚÚNß½ú{æ/Èàæ2gê~ç$/J«þ"4NwÙhi ¹eÇ�Y\_T¤Ë)÷Èä·á+á ªNõ6úfo;ÓÀ2?~ñï<Q£äã É-%;g2VIFÉñÖ¼Ó½þ;Ë#º|{¨åLÃ2Ê>ì¦(7íc¡Zkw»ÝnR«6Ó@ËppÚ%¸÷UÚ(ÉcNÁÙ,ID?8º³ÊÞÁëg> Än¶ö¦+%ÎyûOnøŵÛÆ6¤!áõòw/Îe·ÎÏÝ |ö.ÙÀä1úð9v&V°(
ó\ÌƪñÞÖ,,ÆpLÞÑ(×[æÚby}¤Ì±é f¯ïE7£Òy2\_EÖÆý|2£wþ'ûúxóvn©:18tÄÅ»mö®X$KÃÌõhýßÃëfd}rçÎïK²2W¸ÜwrßÖí¥¬È\Zö|&µ
es\°úhò^îÝc2gXI§¼È'G/²æ!åi±HDºÔ(úí[ { e¬+,\ö=¯y²§O§Ó þX4û¥snSP¼e´ Hî.±{1{¥×n麷VÐ énYÇ mòÁþ1RíqHç-àfáE¢ùááQù5÷qµ±ó0UoÙùªàqfACIípÆÊ«åZMOy8çqû¢ßȲ1¼ù¯ñ6Ô]Õ/w\_÷6kîOï»Ð@"%P¨ÎZ§À@þáUV[6}väêîËCmgÚÛ{oß¼ÝÛ÷®¸K·÷Vý.Ïo4mDÖJ;¥¸Í7-ÅÑÍibJ/ N ®YfM¯S&Üb×ÊÊ#0Ä!oRÃ.²IùI
§¯ Ýëݶ"·®û#Ò%"|"æaE42Cî2ùüɳ2öüpøꬬÕ÷Ú[¹>w|5j± ÛeRH,!ÝÐLTpüttðà[ÌS=Áz¢º¢õpÿ¦å¼ JÙz¦nðxif&ã<ó¹É«~=Ú$:"S±ÖdÉVyd5 eé ÅB¹Ý^OQÓ´C®"aJ¢L¹Ñ�d|q½OKÑN-¡ÊZX&qEïS Þï½v[§;±'Öo4ù³Öt¶8ÕO°Pô0éØ=×8ð& äµÞº|òVf.uåÑ)¸41ìlä ójËsÒb9ltìhj,Gb;zP¡6z^ÎôµV«Õ¨B&) zL.ÕFG§UÌ+¶ôHùÿ©¥@=WPéÊ0ºYÖëÚï Ýó&ê0ÄøM{bê ý^Bdë]ýëgù yö)«.ê.Ôey³vÇã 2 Bó;0T.acW) c k³
r¡<]Ô%ÍÎÖJÄBHåBD+ÀZ3ÚÍÁ3Ï [ã1©ÖµãüZ¸ OØÈCÜYKörNXíD¦2-@8RÒzpq"ô&y_wåáÕ:æ
äF~ /¨v2 ÃDXèd+ÓÃEj F«#ÒÒ
R%nÆ1Hf(¦ÕjQýféîؤóÃóQWOz°a;¶>[æ®ãM0{ïÏmò:s×ô»ÕéHrE# õÃV½9û~{PSJnËÕ&µ9ùQƧUã"XD]=hk'Q6W$®¥ôôbQ3ãz Á6hzP(Ifá=Ò³
¶+ëEòò°m°úxy"ßûFD¥uyðgò$pÃj&H×qÄý§ìÞpülCÀ®åòóZJñ[yþÛ_SÅ!µµJtC©S%Ì4 ð8(°DÓD ´3©$ô©g²@´n³jÔsÒvæÇEÒt êþvÆã¬5ÚsægÌCÏ]sX=Q»ª]Ò$s]ú>ÛÔ©Z·"Éï³.c,±ä¥?¥¡¥PDÅÔÊQÜ2J´ÉºÜzäÄËÓ2¥ÎlÜ@eeöÞ};óÃ1´RV3
HG°0&ϯePgN©:y½aý¢ýÙ.Uëw�)!['òy\Þ-u<%£²W§ènI{ IØ©ñç\Ú$Z"#H¾Þñ[Ö½Ê{oÝêm-7uC½òr«N¤B¦QfÓn¬Gc ¤ípPÕÑóã²1¹ h%NÃ8ÁAé1¥Q·VÃpR3 =® ²t çR»~÷¯²îÜ;¹9©zøAgËælDPZËp]¶L÷ @?;~U7tç˧¿Ë¬èµ¿i¸z: f
iÌ´S]ôkLz!4e&�ÚE¡]js¸<s¡Aq?+3ÀôVKéCáÒ
Õ2bWØui=³¡¹îìñ ü~Y¼Ä5ý¤ øcËzhu2£¾XãO:µ¡{·¬.µ}¤öÊgWÈõ.h(biAxHê÷-0Òëת1T$5ÆÀôö-dÒ´r¢Å´ydVÉã£p?Òp$¸éµ×U»hñQ9=,¼y kàZ/¶[z&Þì^~fæûgt ÞåäÈuaXpÄéDN©"#,µ µR¥ï%fBE�³Lú¢ "2x» HoþC´SJ²ÓcGæ®?1®uÞ[> xÛ¡'¼hæ¦ûDó´°ò1:ÐDu ß/èûR6·^kcLäíõ%ScH&-ÛÅé»Ç$®úC&fSÞ,VÈä(j¥±ØÚµJÁ8Å.� aPЦ·QÇH×ñkÆÝJáùåö÷ø+6·]gÒ~0ó é¸Ñ^¾[ÌÓºÎdô|ai@p3K?-eª0Ræ()ûeèiÓ²%ɺT �,½Ûõ n£Õ±°9UlvvBtL1î Mp"Ô.±ÿ"ä#gÁpÿÞn Ömk»<4Àä3¦N?lµä¿=ZÜg© ¿°¡µ-/h TP·ëÇ<¥]Oª%JÜê±Êõ¾ù�ãùîyc&B -wÓåÌ®¯Ã sÃêÊ6G) Ç16ÇnYáºëJÖÕ9~e@7tM¬½þ>¥ä Åæ\_¢ ùþ·YÐ}ÛVÁÅÌEº½ $Ù=gb&R)j4Ùqß·i+ »ô)jjݸ¢UZÝaVû#ø<Ò6ò±ÑØõ7×áì¶Ó§O:uãáÐÖô¢WÑÆ$}¤×A)æ! ÃÇ^<uz×åÊVOÝ1zESh.Unn#¢n¬¼Þ WS´0[ n¿SfÖÔv5ÌZAÏÇè 0¹Áº]#\×µcçlÍH(eàB£^·íÌ2îWn÷¶Ø¶ïäkûsÙXá¢ek1!ñ¨Ð® ¬MѬº¦ZëVLkHÝ=ѦÚ
ùE]Rí¬v2 &eù=¥ÇPT®ÆmV«Qåå]&ÔL1 =4&9ÊC©¤Í"¡¤YÔ1vO2èf|´Tr}±Ñ7Ï^9TRÒ}gèÆ!_ÜgÑ˯\¤Èü k gÔ£/c2}ùËefVµÖÈÊ(Ä_}{-°@Ã"jÍ£ßzB#³cåR % h¢ìT®IÄÙQÍV¬-h\ ¥È\¡R¡bÌÒÚÚêüü®ÑïXEÍ_ÝÒTUrzàb??;kú%6IájÛt°/òÅñ7··®DMxKßy<úöxÚ 0¼Ù1ÁÝY&»Ê%Ù±E"á¼ÈÆúiE=f«U©Ñ#-] jH#<4\kÜF3�4éFë zÆÎ"g-DF¯\ò<ãl¥T¾Å+9{wè¶ i8÷ n¬<ø ²($ÍðçÃ>TÀ\ïÀU)¾¸"wÕ1æq4^4AF[Íì#Ñ
1¢f¡U6¦×« nÐ ]ê(µQö9DQ&-´V0cFìü"q(D@VÃF~ÑÒÑ·¥0ü̬>îbU® +¿'x¤Hñ4x\ÆÈn¸!àqSR\ª|ɾ©[uR,µaáoXq æºiÈ´¥5ê¥ùaQåÍò®útU©Ö,ô¦1 ÅhÊBx}þè° Bz¤2zlá·GÙ¼3C7[ 7løåMÊþáW2 )zZk>gúj ÝIï½Ör¤ÊÓÇ«¼<õç!bne\ÎDÅ1Ú@hF3ä!HdWj 8.&aZTTdÄ´fB©VÆø@ãF�Ún·ÓÜZ/ ëc¦Õ¸ïpÿFU¿÷éëWºÚ[víÚµµr3S}Áý»£.ôàSöj5¡Ô|Ò;Mq<,#5JÍÞ·�Ú¬èHôÖÚr´"¢]#ÔL×ñ§7¥Ä=ÊüZå8EÃÞaO\<4ºÝ¬õßPÀõ¥?ÔGD©Gk½ îÁqk¼ÃÕzdé¡#'¤ nÓÄ ìd>ÖzÓrV§5Hk
R!còFï% hÚ4~rº6[JMôx½MÛMu )+E0FNr!Rá³4ê2ªñRÞ8´|G
È,n¤g§CN ê6èÝ�4SÁd2nÜlÄ41ùè°Y$34Þê´o'uq¦Íc¨j?ÊÈÈËKðÏ{å¨ ËF¼#jKCbÔäçñã^ÛzÕ³9ü¬9aA§x1"àäàf¥×²h´¸Z%q¸µ"¹vªYÐ$j"ÄÕ 9,ÍL|l~zÇ8ñW×Wï0O,nKê]w~«ÏFo¬e!aO1ÃÃЪq@;AËl! h#Lz\~IxA'F6¬³þ ¡TlÛ®¡eó¦T¸6MÓao'ÓÈ#ÙÄñì!º{rpÇAÐï=6½¸&)2;á^÷Vf& ó�½éRhÐÃXãb\ÑØÍR '=ÛóÚÞI<^nCÃjôck¡ií $ÍL|elWnR1&ø; ·Å,ϯ/æä§5Ѭ@Ã]>$A% ¯0Hä\TSmrJ ÜôoDtl¹H¿ epÏ.]O¢ ÙÉäm»ð.ÝãIMíôì²?Ù+çAc±Hű{fȦà[k>©ÜL¯FÓ8a^Á»£¢ÃHPAC¶òÈ
Z´NkEÓvØàÀÙÈ�ÌQcA;kLÁðÃAõyüÍ°é ßÛÕiŧ èYþd¶zeÙ@«sbrv zÒÞQÞ,±Ûð® PXýrD´ýêB³ »XãÆó}')ÅuÙA»]aS´ü$ÊëYÐY½ ;ÿ0 ùÛ®xAÿÑZAoY ¦wfßP÷wÓã.~ûI ÕL.÷
_Ù1Q-.×EEÓIµ0¸/
¾ÆB7)TêÄÉ1,�A´5é99óã'Sl4ÜÉl±g¹Ð@?ªNÞd£g½Ýo {À¸¥Ó4»m díÛJËÍ-Ù¸5B¹«¶< 4ÃÌ)ò àv©ÃD'ø(¹Sa#¤µÁÎ &ÁéF)V§ÿÍ9mZȧá!ësj+Ê¥¸I«c\õæ9Q¹"##cuå¾ÞÁL {Öú~k='\AK4½¢íÖͳ§Nÿ>uêT+Óø»\T$t$¹ñz¡PX^ïKbµk(MÅ\KwÊ4 µ~¤fCI©ÔäôÅ,H³k5÷ÅÙõ(ÜÔäZGÐÌôP®õ¨««ké»;4øI4)Áf4õx º÷Ì÷«öïo¿8 ølÿ2^è]¬±ãkdAkËÜR¦-�´Õë¹Ó"$�t¸ØX.n(XY5¶F)iã.X¡I8fL\ äëÝ¿Í\_×záúí@×[ÙãÙëû]Á ]ÖCÐ
-Û2^{®oèá×ulöÊÎÉvåq©ËçåHar ¼ Nù@wÁ£4<6wp¾6¥ÑT»Ïö¦J6ØwqSmÒr¿Ë¤\¾óðá×'vy·¦F&YÐïWAozÅ¡>Ý@w7o|é¤A{Üò"(bmÍNè[jYÐ hAÎCѱ@A¨¹,¿ÖÉ$¸)»J/ÖÓ&î3^�÷]3ò\ëê\ó|ùï³Bf+|òÚoêt'ýw§�Úãfã·6n3ø2¦¥3 Ãåaµv|>U.-või@Þ hþ6Ó~4oARî¡{º®óví@{ ¦5¸5³ óÐH¾I¤|ÄJyÁ¬ÅD3¢ÓïôÛ^Nûþök :ݵ_å¥øCM«´ÌnÊ-l [XÌN7ÈÃ#GeúÉ°4I#ÖÍsÍ®é@¤Ìgq,ý£ ½ÒØæÛN»íù xzÐ&Le´0çb ¹¸Ó¨ +ªXÈy>ÓÛwd¬/4pÒ+ÑÐ; ëÛ³©T--eqb¡ãEQV¦GãÔkP¹§ü 9R»£i×µÄÊ˵LAÇëq¬fFä¬)Ò°We ÞÊÈxÍWNβHoR«ÜÜJÆ }±Èvõ=Ô]lH+ ·¤É úÆI0¹ ADÍ U $ƸTiå@¨DitZòÚb)a@ \�z¡\I¹L¸dòí\BýÕ6ÓóÄ¡o~ 'ÌZÅ¡2÷H£°mh ïÌwKß=òiïÅVôÑ)µÄ7a¸Y+ïÊÄB íÛº¢3XÐ!V¨KÝØ(WCRéI]aPÉ^ª
á%Ú¤dñhøØzÖ ½zóÊ3Ç4¬+,¬lø÷1«ìÖÑ^´ÈNݾ}åxwoß7oÞ¾Çv®^òDÛ{ÄMµ2UÊeR#.Ö×w$Ñbe,ãKÔ8ÜXmáq8|ºæAʵapWC̦zÆ.ÝÃYÐ7>ÒëäÆ×w{aKtàOû½,D°!èSgölÌZ]²uë®][[úö>h5X,f3I»3øG7³ 9Þ ¡ ìÀå&7è´´.ùÐ¥DS�Í:z¯�º9 »70¨ÓÝèí "�h=qNï¼÷}ÊmAJR%ûÇòSqÜ0ãj-mS¢Z-f¦ÄB <¯0N#íR;Q îã³Å æa¦ÉiRåQ¿{7)¢Ï±ÇM%UëJö^~~9ÅZúA{½¬7ò^Øæ +r·¶À 7uçÓ(Ær#1°[²ÇãØ÷f"<1F6\V°våj¥Ô(xp´#ü!«¢Gi¡Ý8f°jج¯=j)\Òº´Òt02Åj¹òÙlo['»/}ÑSKøÖ¦ÕÞ¢»mø<_VØ.-ô@ÑÐÐDå ßã#8éÐÞR¤Jj·a¸E¼0æò:dRÇmWw¤s8ùBéÙ Aÿí´ùÐÅ[º¯÷&%eí¹««Í< ýé!LVÃfÍ1² {=yå7S× ^Z< @G¡nI¦Ô)íÐhp[zÍß{}àòmo&åº;¤»ºò2~U?Ù1Í×ê<·L7AVv Ë KJVøS¥3 ÐfÓèx@!Üÿ (sDc =.Siç4»Ü8¤hsEö3ÆÚ×ö\8^Ïxæm8ÐÞ^ÊN×öÍþMyã<$ß@ìþ.L\VÈã/{«tS{÷§|N öü$ÐÀÛÀG¦·¼H®Ç½ÛV´]Ojº0¶§Ó0M;ðÆ¢g3¾ÉK è¼A¢ÕYÞ¾lYkDÎÞÛ$¨ß¼ mêN:{¶» ЧÇëØoÿãÓVyY&BÐåÞ«X¬ÅZ¤nÒê åUntS¤ÃjPJÊõ8¤ozÓ
G�,àÍ|ÛjÎF"¼óJC"eÄ9&¥ªîêõ³-{÷l] ¨¤é³÷/¬:?åvÙAL!B¦µDÊ+´´±¶Ã ªõz=©p¥Db$ÕF«J.zº:ÔÏyøÁn_Ô¨÷ÊÎÇld¡7ìN×r"%ä&Žêì§
|Ëçs¹¼u'ÚNÃã¥T=´è ½c´°h'L.U-0ÄzÔ§ÃP)0TÉæêôùV~FgÚC÷3 }ý Á¼èÕ_ÊH±oË@Ì¥ÇLÆf{] ÜÔp ¯ýõ%ˤU(l^ÐÑÀ ÑxASN¹Ho©í Ãp«µØÓ³;0LñlLf¸Ä(¡ w ²g-9OÊ"¡´26¬ÜüíÁÁ[ËÞË ugZ ¦þIèÁÈiP ,è®hNB8cÐ2Án«nÄq%t6îðXäfºº¢C.=>ÿÖ\_0exY¾ç= 3ôûÍ5a¿5f2OK{ËómôyßÞ ÌÉ|©¡(> 4ªñFæEDÔô|Emuɳm© æÙ$Ý,¶èÉ´Å{=zinPòYïÊÜçÏ^ÿ'Sù>jICb®K/&òF.þÕm·½GNÝ&6á8 ºgG%snRSËMsy´ÌJ4;0Ò\º#v^XûQ~xÓûïÀ§]¶çúÐ×ÛvU ]Ò³v
/¤ûwDI-%/úq1
¹Üd¦7·ûñE î=Hï(19é;=tÏ)á!4uØY¿£Ûß>çC &ØwU7¤ìíÞWY²b#ÓiÖôN"©ö'P2§Öf]371,«ü íêÐôS{~Ð]1Ò¥ÚíÁbr Òð2kôÚFiz>§:¦®q\_]>}ò²îáÐÐ˧[uÃί.îwJ8@�J= I0ؾ }Ê[÷ìk;}õnðV«÷H§ÐÔ8±Á®èE'®§=Æ´(©îZZ+" P'ÜÇîÎVÈæS'úÁöTÈéÒ³mKJöÐéà®åÙ\ ¥¯l"Óºü£KsÔu~&Ôq{ún\½qÒÐÃË ¾Í;§fA+Ø Gúú³3°ÿCÊÞ¹
ÛNôl8î+G]ÊH¤8hÅÔJåÛÝPSgu^þ¤eóæ7%eøô÷ãÿ2hÜ Z AÁB\Úádò(H5:lÏ 4¾
þl¼pasOPzê[�NËØFmÑ÷¿"9ô 4µÚõ»\Xÿp|[n&ô¯©ìÔSt:ôz/ht^¢(²L�@7Ú¨fæ°2á=eíRé h×¹×Ùväí×÷däí¿p³·.{Ü4'N RÍíîe6òî×Æfºåþ¬Éà!+O/>½Z(ÊPXÈ©0»Ý¨X>AÝùäÉæu6VoînÉã«ki?É«< e÷ìé¤0 þ<)àÔzL?z Ì,XKù_#OXM\ùë)Ú'm�´$?I´7i+ëhÓØ x6Uyº3aFÑl³ªÌ ¾ÁËäåååVê{ 8µ7s8å#ε;ypº¯[FÒ¨rÔw¦Ñ0áJµ4G_Î@+cT4å=ìÇR¬Ïù2Òc?ÐcËÇ ZõéxN¢é«¬iEËFv¨4ý#2su¾×/uyô϶ÒÁñ ¯rz?) «³�ÛT¥i£gãæ£A¿.RüD»ä~ÐÄ«dr U¡2;Ól ~ö
êY9íºT6ú½ÏÍã&N/û5 ÑÐZDÊÜÞ!)±l\�Z�P/Èj 5K ´Ýua ÚÔ,³>ÛXÑ;ØSò!æfÏì´¡ÑÀä;°¤A\L¶¼ìÑc¤BòÒcS7¢£1�:¢\i%\Y©%¥J¡-å°1ÕÆÃ\ úúÃÇ]Ñ©sW=°9¡Ï¦,Êo·xVhî¢EÜÀà½siò¬&T�è(9á4¨Tbô¨ÄNÒ[ÎädK¬ÒgMvV0xd~0ÃR_>hòÇéñM 9áõÖÇ^ö²äø¿óïÌôÇK=:ùUMJqh¤&»VSÈ@[-Á -É%°K®ó«|<ýï¾ó!ßü¸ç.¹û¼%vjÔ3tl¸ó?ÉY÷7ày¿ó¢W´¯EYAGd'Iµxº!QÒ4kѬ@³+°g:IxøÑvï¹ ^ú¸Ùß¼4ëë?Äã~wûcS9'RzÚjðfç£õs}éÓÿÉú¿l[ºct»qÉmAÙêW4åÛsËõv¦m�mxõgÿ9$³¾Ã<ñßÇõùoå3ã\Srm ¿Ä0.Ö¢¼µ$ß ,Ç&ýtnLjaAÃ?ïïöcmô<Ö�´Ílbqè^¼ÑW¾ó�§ÿ-ûÄ/@O\_=idܲTc6##¿Y=A{w®SG7ÆHe³7õ#BýïR.³ SXqà¸ØDµÏr<øãø%÷w¿ãí\_s\_9fSÅ!QcÛÃzÉ&@ 3½ã4½ßéÍÆÌpWìó=~ÐéE¯Ñ¸D«ÔL¡Ìl { ?é>9ø¦·4÷»DpáøK^½÷1륿e®0\_ÚùíäÙ#6° aU,h¤iL.²zµ(á8G:LÜýe½èE{Ñ3=]¸³gû³T0þì@g»§³kd«²f½ãIVHp?èrÄ ³ó dIR[ÒõO.¢M¾æ cçÎOüw³|RÜé?zDÖr&piu:é¡°õÞD.7#cw¤ßÅM,{0)Ô¤JƤ0Ø|TõiÊÕFy#¬Ò£!Øoû|iðóq9å{bÀê9¯#aÍ´|¢åck«úÛ ]=ÔÎÊÐòàe¡#îJ4©¬-RÛq ÛÓv]z{\ûSÀK|yǨ°¡+ýÙh{°~ÎhÓ·§¤ø¢ÔImb¨Ý f÷éeÇ-0²¢0 c<5hJ³=1d�eÏÜ\_²Xd!:ñî C
$àäùÅñ£A{-pø#yÇdô©WAФî=¿áÆÙý:»ÕãjÞÐzu2í漩'%¼Ò)aîß'æÝbiöst>2M2¨ Hr#ÙùJ¨~ÖyïÁÝòÅ&¬r9ø§>lF xÐ>RfFÐvæh§jÜÓÆ!úþAÏY¥o
xP^K[ÃCu'e6»ï'�µå|¼ý¥±áAëö,æ�n¿x)ò hM2MÄuSn;îƯÀܪÀ)X«%à Ê=é¥í~ðKop#ëÀ¶ÍYàëG'¿¼¤ìøP´ ¢¤QÈ.òÁéc@ï¿®»^·ÌëxÕ\SÝiØÃÇå&Åð@» ;Á; §Y.zÔ©K& I�z²Btß?ê[Ï)ï^»s¡!/¨¸òdÖÌ\_[±lG8Ù}Cy'Ýäè|e´¯ûrHwÁ{Ô
¼vb#ÅÜ$:,¦µh2©T¦4D¤MM£öE2èPfÙb·àj Ó\O"ðÙ¿úòNé¾þdsFÐæþà¥ßZEa:® ·àÿjÿÓÇ3gײ,|¾y²20©e£Oô ZYpQv¹X®PÈb}n3\dVÊ5á IÒ3á'¸2yÿkËÖCCWö¦øaϹêDO 5
ÜKgó4»QÖ¿¼h}_Ø|×C÷Ú ùzbòO.M ÍÌr ªéê)1L©P(PC $aîÊzJÔ´X¦ÑÆ Ma
]ýGSð½O!xmã7OoÛÈ÷vxýûW:Ú4dÚÖZòFòü}Î<�ù¼yä|O^{lüx�ZM¹N§Ûf4ÚÓ¤]ÕÑ!HËëQ;9ìq9-(îo×DQl\¡ÿ¨>²ÿmò¯å6]<ù«:ÝõÒR¦Cêì¹ÿ§©óBöÊùÜh0Yͦïø'øûu÷ZF)»9usÛ®hÄP qÛMºÑFa±X£V.3fµ Z0-NØÿ:à8¿#uÔS9UõÞÙ¡³ yàÏÄî;:"8ÙÔÒ Y42MJ<Þ=}6\kìÅjy ë¯
|fW\ƾ[Cׯ8ô«Ó°T°,)hèÅGëEdÞ$Ír£QGhtÉØ(».05È!)uÓoM!Ä°YÍãZiÊö ¿+ ^ιíwºûõ[m¬ûËýÝÝìdZ×ÓD&lÍQHÄfÝÍD³Þïî®Ì ¼öãoȱÜ2µ¤].×$ÛG®û~|þ_ì½~{pðÃýä93ôÀÌÏiõSKñ¥q#u<>8¢Îø ÷lûãduðR·n3²Ô3¦$Û7/OálÝËßÐÝ×ZYÒÐvºï4óh³f¾ÝoS¤!§ÌÒ¢ d¾|¼\.~JÓ
ÏJï4eF÷÷Ääÿø\OÆ¥á?uîHe\Inî»Ë¼ZÀç׸°7wÅu
L«Ôék.ßü#,|êú.¼Ìù)ÃëÌÊðòöûݺÄÛc|>yÛcÉí¸ôÑZoÃÛ¼Ö½ ¢]?ß¹Çò¥»ºúòZÕrܺ¶Ùeûzçc>[AÑ×èÉ/Öò3¹w¾xCw·.�ìw:ïãOOäéü;ËýÀy¥ïx^^çu_ùly�øüÈ$DNxÞÏdÈÑ ìQ¡ÉqéûÓyG|~æ»m½g?9=põ@üå}ÜäU»ÿãÑs=̬p [ÎÑ!h¢h@� #íja,u©´È+�0ñ0 Xê{¢OÔJǧVh¿³&hfüAí±Äð½
ÀL&[Ýd|µL-I5ÿ!õª]èHܦw¥&Í_c~ü?V¬¹%tMk/DX] )K¢æÙM,¾EÞ±'OÖñ=館ûfH«[>¡677?kÔiG¤_W~vha¥U´}×Uª&&bZdÞMС¸7²ûtCù°mç?s2û/mA¡ÞìÒOÝJÄî°ÜUÍÀÓ2û±ZäìÏ;¯¨Ö@¡PöìÈ=vìÓ¶ïÊ p2l²ÏÁçÌÔÔ¸R)ýkZnnøJ¤U¿Äí�-%'åÐ['òh�@+òXÙÆü£Gõý®p0éB§ÅRúFmáú¤
YÖ{8íG¡:ðÄÒ©î G�Äݱ9 HÙóΧ%Ù%µb©txxD£Ñ(Ré¶ÚséE%ÇöZüõ$2Ãkú®·kç�4Nú»ê²æú,Ä£Bñ£/hÁ®Ãn(oî½øÖé;pÌåÍUÊsÑ~OÙǾyÖ~C,!,45¯]¦î(@rv_Ý´wµ!t Ê{ïåææþüûÞ]Ü.ÇbqyM7kÎ6ËD¬
@D,["õô´×[È Þ/ qaDU+6 Y CHaBþ4Z¼ßZ|ÊEB ©ÍA"4ýìSQqü cóÿ°8b
# rNM¯ºçRrü1Ó$ÆÄøOZz{ÿUf91ÁH!°|S»ÔR¤H«~êÞ~¸Å{X¢«5¦Ig:ÙÔEg¬dÙ/Ró:3 R@¤çå¥Ìo6F}©ÕßÏôB¹&üÈVµNû'WÚ÷ÚsHYµ®iÚ´ÄÉ;E)ÐÀï
B6mùLe.#&y¯Ëx»zÇ]Dèeö ý 6] êZßvA¹í[¯ìy?Èå~TaëÞÝî8·0ïªe?ÆÒ¯Bg5·³cI',Ø�LäÄÅ1YdD£ rKcõ)Æ´OÊêPR)»ÉË=5|_ð¹QhëBo¨ZÓnñ µ¾{?ÌSZu ôXx�ξ[!õ±JTF&Ò¯ª:«;Î.vb "³¯
/q¸Jc×\ÿª¥ñáú<À)ÇÊyå7ä½WXûqhxÿÍáimºûî(Öui¨pÃ+h_®B?q+=8ÐVñzr-{ØØñÕ¾åÔà#êQóÅëͼXk<ú:ÎÕ§Óél\Õe#Z¤¼»êÿê%ªïDåÉBC°Qç!åV8ô ÖY°!ÁåÄ×CÊÊèÍþX¼òxØÍV=¹²dqÓtº3+Qf[æ&rHÒ-jY5ͦUÚeÙO]}$hÿü,}áÔ©P\_ô ñ8øM¡ tÞYQÕÁÐ\_ ûU ¼åg²ÍÇoB·ü êdn\_)?°îþ aPIt{A £i ú ¢ÊPjù únß"PУnþ@ç{{êOÒó:ðµixËå±ÿÓSr+cºõM4 ½¥J;«»õî&L¤Íg[I²V«¡ÕB�8ËÊÏX Hò?UO VZFe6«§"O)¶®Fb6
WBC µaM«¡/ åYØ/$bqv?~¤þÈr2[ç#Þ©~TǶÒAYËXiWdjö
ÝÅUM9L²=EÀm .N@C\ox4Çt!Y ó]o»¾ö)í!eí¡p¿@$ø"»+¾d¦¥JÔO:®ß¸aO÷ðÔªê¡8ÜAh"S$|¾Ï6?lôSž°½yÍæ«
ñÙiØÆ|«Æ òdIwÒ±DhÎCõ£3å¡gUVóã æýº5TImºY¾ð=\_tö=½á¹?Ò+¥~ëy°+ÓòH@:Eå¡°�üË6" q$êÎdàpÖ#Uo¹%¿wq@bqëf¦äYh](% 0<ñÜ päöûh´Ç~ÁKºø®·Ø·6îu¯ÖÖÃÌl
´²Æô¨êÍ $N×ÔÈTQijÃ
õà K¯ìÑòÃÓ:?¶~Tû'À¨$¸¿ÁgÆÅEXLäÿm4´#xªf±ÐÔT¡Lvùø¸8GRÏXR2Á§×ë Ú¿gx¹Às·o iV'=G´íPÚ= ºìhL ~E±iìuGyÔÚ§òNÕÃi¾Éc.þ:Á¦}&�_ ]CþÆ%?´êÎ3ëyHS
´»róͼoì|ucà.ûÔíæõyû q¦29íß~{ãÆçGè²+dãÆW³k3&èßÛæ {Mïþ_\Ë65 WØ¿u~X ÁýñHoñð¦ådc³K äådåØÇûb£µ)AzÞpØé·å¼33¼k4ÙðÒ´Íu[iFéGÅEéÁXl8eI½Áv��XK@ÚPûê7:p¢V:´}§c\òJ¨z¶ÖCÙÝ1n0kpóáÞ3 ÿJ+ÓÃ6Á ¬·mÖJ ÄR �x¿q¨R<® m7 1k··ÁÙ\_a;b^¯¸á ëÛÝ;éüIH\¤uç²Ãüq8ÊJLP#)8!X¿èÄì¢:±T«u#w.övíÚ{?wHïÔM;ÎÄô¼«êmOXÝ®I
UÝÚIG¥mµ%iïì ÁbýüýpxGø×^ ôÃBöî=TRÙödþÑX ô}\_ÄpÐ[òcúuÉt:&íýKÛ|hßB~ÿؤÁd2á{·¾øsQö;£v. ÆÎéÙ%°zGu d4L>Æø=oÐÑÖþµ¶ û90oîo=½ßéåê¹5)ßúL£A#M+ßKÅ·ÚjÛ¬?\¿zP§yÖÍ/}«'òÌÜ|>%6üªöKaÒ4äy¢-eç®Þ ù¥×ºî ½Ñh´»¤Y£Ñd¦Amw \?ie§oÆ¥~Ýzæ8׳m+7?ÁÇÖ¥íîî\_ʽÆç nÛ àóùÜó?Hðrw[hçp8¿Uþ©¯õÀüì´¶OÀ-Íðöðpu³tª^YÝ\=<¼ ¹>~vÍ'}ÿ@íýÖªÓAAA¼ ì0öò zë4·¡õ¾æ
S7dý¢5tc}}}]·kU
6V}Éo¸Ý\_Ñè ¦~ü endstream endobj 79 0 obj <>stream xíÙ¡M�EQÕ!:@®ÐShB ÿ!Jð\ý?Í9>stream xígxçÕíر|vÑ1ÆØÆ0ÕÓ X
"Do¦ÞAHè¢.z½SE!DQï½ ÞµÚ6ßlpÁ í®fæLyîëþ+¹æyÎÎî¼ïË0��Å£Íe¢û1ûm6õ?��ÄÉòdüÛFLìh&ï2£WSÿ���Hö!å÷òCÛ0É.LI(õ¿ ��¤Pż>PþkC&vsÑ«¨ÿ���¤@ÚÚ·ÏÿÒÊðØ¢ þ·��1z-Ö±üòÒ!ÿþµEKýï�� > î0P^öáéFFý¯�È NSß»WèåwDξ}Y7g¬\æà2iRÒ ýú½0ÞÆ&¶k×Æ´kÝ¢ÅíÔéå'¡oßÿÛäqãÒfÏN_¼8ÓÝ=gïÞ¼S§Øÿ»GJ#"4iiºbêk!Mæ3S~ÿBÌI^Ȩ"©ÿ�4Ðåçݹölööíl§N8hPlÇ UôÎ;AúH þóÃjÔjÚ4®{÷¤¡CSgÌÈX±"gÏüóçÙyÇ>]QõÚ|Ã\0{¦à 1�ÀÐææàÄ÷êÅÔ¬-[ ¯_פ¤P×bs9S~ÿ©¥=ó|3£Í¡þÛ��£×jUAAyÇ?wqIè×/ºE Y>zðgðûïG6hÀ^º4ýû=Õ¯6êdêyÎgÊï?µ´bÒVãG|�$¾¤¤øþý¬mÛÇnÝýìMË23äÃÙ'»ÄÁ3׬)ôöÖåçS×¼¤oák ü×FL FMý§�Bÿë¯ìcH¼MXõêä«@ÙËÎ^|¶l!ØrPwñèðù)ÿ,ñ@ê?�ð:zµºøÞ½ÌõëlÐ øÝwÉC¾-GdÃIÃgmÜÈ-u¿¼¢GB WÍ=¦þËP:êýûS§NnÑ"ø½÷È)[,¶dláØò±E¤î£ÿ%yÁL1ØÏPÿý�( ½¾øÑ£Lw÷~ýðldKÉ-+[\¶Ä ¦+1üN3Sþc̦ðåE�@îhsöíK:4¼V-ò�¼Ê8qàÀì]»h\_r/ÿN!oá��¡ÍÊÊ;~&Ì0vSý@èõa5j§³ióæqßxùÞôÓÁl~¾×ærY�>É£F§³E·osßv SéGC mY�ø#y'O§¡ÿú^Ëulj³ oêNêÕ÷�¼.//øý÷ɳ ÓĹ﹬Ãô³[ã'0êTî/�Ò$¶S'òì4çàAî.z�ýàÜsõ">S��¡ÈX½<» ~÷]mFÇÝVkØü|ðd=Süã+ÔPÇ¡1mÛrßmiëé\_2Ió ¿ëÔ!O0(6Ó.å¸ÏôZ&¼uæ bØ÷LÎY¯�Ò!eDòb³ä1קëÞ¥O{!û
Q'q| ù¿ýFPT׬ÉýIô9/°!øí(]qqÈGçÉcÆpÞdLÈ·ô!Obô ¦$ãë ¸³²"Ï1(óÏá¸ÃrÎÐg;¡Á ÛæëK9¾ª�¬MÈs Äà>ÐrÜa±#èÜH[¼l BiL yA÷ã\·W¢¥$Xbüæò4b0ÓÝãÞJßFæ¢,@¤ÎEfP FGsÚYz&¢}M<°�¹Sxý:yAr#êÕ㸱Ҹhê˨"9¾à�½ZúñÇäiM1ãÆJ^DÝb6¸a £ãø² ììÈ3 ÒZèåÅeKéJ ÇWç¶øɨӸ¼ò�ì]»È3 ò÷¿ëK9ý?÷}\KEvøæ^àò@&%%¨R%òdT&ôéÃqKÅO Ïji4Ñq\�èjÞ<Ù ì\ͤÉ0¼ÝDÒ3¢SÄõ�ñ|þ|òd4VªÄ>¨rÙLèóY¢77² ô)òõ¥7Ha¥%ÇÍÙ>%müDFÇqQ�½VjaAoPx/XÀe'Òg² ïbXàI2<ß ðûùqÙF©+éY73|dÉõô$Ï7(°aUª0:îÝé5LX{ú4 Sm>g@@´YYÁï¾KrPH ã²ò½éCX~FXál/ QbÚµ#O9(¤¹GrÙ@³éX´dò8Ýè��AH\_º<å¿÷6;³îÑæ^
%_ìbøv�éP@tP0c:tà²{²Ò§®ìÎh²¸¬�¼¢×׬IuP3V®ä²ybÓG® ïÌqY8�ø$yôhò¬¨ ä¬oT1ôa«39g9«�|wêyÖA ¯S˾y¾>i&~^R@üþûäù6eîºFoØÿQá S³3³ÊÀ)'çäÏÒî¾E¥ð¥ SwßjÀ/çäÉÈ 8ku Ô>Há«Fõ1Ô�1¡/) ùè#òô|6gg±>BáèÆÆrVe�¸ ÞÚ<ý z{sÔ#z&'úüo4´5Sø£BÀY[¶§äÜÐO>Ñk8Z|]ü>9a7cr/pSk�:><�!ç&öïÏY¤,¦MX ÝUÙ yBnÍÙ¿æÐ2¡m¨'Þ1 mölò \úÎ;ôtn#ï}TBãMrôjnJ¹Þ¸A;£[µ¬9§Ðç$4IÃ%5��¦£W«C?ù< !W¦»ºrÓ Ãf ä! M5º£åî´h�L'ÁÞ< !W?zÄM[dî§Gh¶:6�Àt²wï&OBÈa5j0z=7mÕ>¡ÙVD&pÓ �&55¨R%ò<7yÔ(nz¢$>a ûQEpÓ�Htäy+nÞ©SÜ4DÚjúH7¬=SMK� Ï, ÏCXAß_ÏA7èµLØ÷ôy91´5SÀAW�Å~~ä+hlçÎÜtC/}B iÅqôæ�F¢Ó
U©B°"f®]ËM3$΢AÈ!Lá]nÚ�ãH:<aET
qÐÚ|&¸9}BÎÅX{ø0y\B³øòKnú û}úA4?núòÐfg¿÷y6BóL:>B}?1VÄ|÷y6Bó,¸r
(¥=È·+@(Ò-#ÏFh!û^¥ o¢O<(±r LJ?&Ghñ={rQ=Ñ>î 0´Äæ-wôúðZµÈjööíT¿ð}ÐAÁ|¾ <Ç%OHh\©¹¨ý|ú ±ÀòN¦IhQMrPx]aµ5yÖAÌØÃAÃ�ºà> ÏIh¼Ï8(|îyú¬¼ÛÉ:A·�±]»ç$4Þ"\_\_ª7:î ß6b²sÐ\�Hæºuä9 4ô\_ÿÒkµ-¹:Õ8ô¡ù(�¨È£iàÁ nÌä^à O�0/¿$OKh¹qPïHúÜ<ÜÉ»ÌA�PR¦L!OKX®ÁｧÍʪh±ésòdps&ßH� B\ºD°\c¾ûb§,¥>Èì@)ðá C�¨0ú>"ÏLX¶ËW¸Ò¥Lh[úôòac�DC¼ yf²-yú´¢eÎó¢O?ȹ!-qX0Y[·g&,ÃðÚµ9(süú�Üj8}>Þ�SÔ ä± Ë0yܸÖXex)<!¶c¸��¸'²Q#òäo3ÿܹ8Ë>!~ÇpqëÀ isç''|£Áù®°°¢îG+Ã~TÑ\Ü÷�ðE¡·7yx7×½{E«[N+Ã;1ª.nz�xD¯Ñ~ò y~?µqcE«¶> !'²¥4;�ÞIè× aÅèaØS�é yþ<èwÈS¾jê¬Ykú<4²'£Éà.@P¢¿ýI°"F÷g´yÝß�Mñ½{äA _зoE+3>¡ÙF4wq©P!üé³aü$Ãéi�Èï¿'OTzîr
ª¼>¡©ÆOaô{Ó�ñ±|9y¢ÜÛüµrzZ~hÿÅûiY&þV«+1ÖDÐ$¦3z ?÷4�<}Jª wMÍÆÿúiÞK+[9v²ÉuÏ%ïHµF[~ s/Ð'$4É$GÃÎl�ÈðÚµÉsUÉÚ7ïýêLyÕo/ºþäµGáe ¸qô! 7ÉÈèGçªb}RéÚ?ÎxÛLyiÞ ¹ì;ëó¬ôµá¢NcÑç$4Ò¥ô�#ùEª£×üÙ¤êÛª?y´VÏÖ.w ¼j]ûEV»|/ä÷'Ýô9 4u ì]6~Yuô3;1U³Û¬Çï¼K®ÊtRýN&Í6ºÜÉã|qÐOôQ ñùê[�y ªÿ»®Þ}¶Ù3åü.Ô Ë°}:�0¿ ØA.û[9Q?µ@³rg/Í.V=û¥O±,E´62¹ ÑhuG¯ùýû»Vë>çá»ï'rÕ°»Ùųd�ul·ÜØ°[�R£¨¤ÔãìËn&/Ý^å+ò¤Ußthv¥ü/¶£OøG0¹¨³�Ó§û±õ-åp¼pd£äI«ý{e³ËÔfÄdúð4¸ïM�@^aëK/׿snýN©ÃV!:ÕmcvÖoìEð53>Ô ±ä©Ü<½¾°§iòRöó3yÞ\ÁNmW \V)w[ÐG(|ÕK¦À:$�0RõãÞõ,á{¼ýüL·²×÷ÏVîá^즦Pøª!-¢Ô9@ù¨J5ì4ùßMÊ°së!ä+{7VÿÆì;Ð>EáKCZ1EÔQ@ùüæÔrô!§É «ô{ïÏ!O]y;°©yÕù¬÷ì¢ÆôA _Ú)¢ �ÊáAH¼©Ûªpë¦êõÉSWÆ>«Té.ÓÌ+Íä
Cè¾0ô;¦$:-�(ÄôQËYT-<'ibM¼2öØ'5Ì.Íísè³²ýÀ¨¢¨·¢\Õ¸yz·OçÖí2ý,M½ruF½æÕÅrÈ} uBÖ°\:3�x+G¯ù7¾|¼êÉ«g¯\móÝHó²|}ðNLi,uf�ðf¢3,ÜK>Aþè¬/Ûg¯,õþËß-̪
Õ¼ïôªp#º1êdêØ�à ¨J5®{.U³u&oô»v#ÈãW®ü´©yùiü/ôªp#z0êêä�à ø<úvôjòÁQ{8Üþà¯ä ,?{[ÚW};¢U%ÙÑdP'�¯\_4aõ1ò7»qMÍÆä ,3¿ónÍn³Ì¨EMÛ9yÒçªbêÍh2©Ã×9ëóìÁËÈ
Ú7ëM2sß¿êWÑÃésU±Fõa4YÔáÀÿ;Èeù0ÉÚ?ÎxRéòã¿éb^-®èDÊ4z £Í§Î�þS7°0úç§ä9,'vøÅ\|ÓæYCútU 11P¨ÈÊ+¾ä ùh0Û)\_w$ÏaÙxù¯ÿ0¯ Vô£OWû3£+¦�þ˵Gáúõä¶l?<e£KæU!èjúU±£] u�ð;¥£Ç¯x¹«\¯}ø1yËÃîqý;H°J3~"£/¥N3)(Ѽ@ý¯�\ÞyÊ&òYÀËk7'Ocøà½÷«ucÆõߺÕ>ceüF¯¦N3ñÍéáæÛÜéÆ»ÔÿÀ gïÔìµ|phÏ}ÉYn«ZÏ\_ÍÆ!ã^3úU Ó½:EÌA£Õo¸ÕbþMv °¶^èVHý"¿Hõ˪£ä#skvðÎ{ä,uG4úÉ?pæHúUIó½:HÌ!)»dضG/¦ÉKm~ Öê¨ÿiÀLG$6¾<ÿyrÅçä,iÿô§z'qåÏêJ´ 1ÉI¢åʳçß/öym ¼pãhê0CWÕè5<ùùóoºÇ²¤=óU͸ì\_ØÍR=Æ1Á²aôÔAb2%jíü!o&/lá|óaLõ?@©F;cãiòÌçÛF?'eI;÷vf\öY®éÃV ¦®@H-èë~¿òn¾yÅü
H¤eå[Ïö |a¼ô×'³tíÐv¸×üÁ
öôy+{Sݤ8Pù&¶^è]î@yá!Ôÿ^P>÷C]NõéR§%y2KÔ;ïT¹©¼ÍÉôy+{o¢É-VÏô 4r¼ôʳçÔÿpP{Wµq"Ïy!íÞr y8KÔõ5qÁ׸÷¦\y±:HLæq\îO«î:PX;,öIÍÁ¶�bD£ÕÍÙr<á
·Z÷9Þ{<¥h¿f½L½ÚnKº2VjE¯gÞIhµÀØï»þèýO©ÿð:y
V³¶Ç;U¿"ÏgÉù´Ò;uºN7õR÷<>uekC&ë0uÆó<ÕØfOyEDtrÈOûåÛõ hÉyøq©ìëN¼rµ}:KLãVhF§¥·+>PX¿\_짢þ¿ X¡õ;M ¤hÉ9åë¦^çÚ½gY Oåu@©F·ê|¥3Óä¥sQÿYÀp¢Võr^Òh¼çþ^
<¥¥eËö£M½ÈPg¯ nÌä^ ÎL+´7bùÞ É þã§<¶¬çÄy_´%Oi éõáÇf\ä[g~ OÜɽD%&pÈ7±ù?Çm·¾ù%XIZ£æ~<ÆEe§6CÈZB.«ÝÜÔ+ÜdÐ4] æv 4cò½©ãÄXØ´sØäå'¦º×pê?Tqª5à Ïp±Y¹ï?$Ïj©hÓÞÔ+¼du\_úÁÍ[Ôqb,OsÌZ~bª-o>Ë¥þsD~ÊzyÓÕ¿!ÏjIèÿî«wmÒµµ°u³}ËÆK¦àu
N¯÷¸óíN?À>ëîj°¾dåuº<ºEë &6äq- wU®kêµíöË/ô9,CZ2E©ãÄ(ÒóT¿ì~,Ø4y©ÇõXê?]þdæv¼<·Åì]¦=«T<±ÅïÝL½¶»wXÓG±< iÅùSÇQÜËì¼å'¦ÚjwôsÉ#©ymÇ#mñ{üä-~uoÒUn;7ûASú4¡mb ¬Ðêô[½bZø}×·ë1õe-Ié¹Æ¬!kI8ë«ïÉ[äþö7 S¯êÏsGЧ± ý)ÀÖî ÅC¶<$&/½ð$úbÈäÌf#d{ò/ç¶k÷3yhÜù·6õª^:Ö>¥nX{¦$:NÊçã´ö®·È§É »®¸S¨r.KÍjþ³yPKHæù|ðWòܳ]Z6éÖï7Có ËR\8P:2ªê8)¢ÒrNû%Ñýrõ
éõ-%OiɹºfcòÜ÷þü\=æt=õ§ÏdIÞ)¥rMÎï½îùù£xǦQ\_9Ýdø ò|¢öÍ{G·hÝT¾©×óÙ¶ô±,]#º1ê$ê8)ËOZ. ü9¾lÇïÆõ%53Oá{×WÄÚ?ÎxRéòô§CXt1;HËUÞQúXÌüÒ {OrÅÞ!=§ õصäÉ,iý³6yzÐg\Õí2Õ¤+¹yKOòd¨a×[O^µM£ïzpßð¬.ËîÏ c´Yãõæ\_ÔaÒòLº¿îHà"ôÄÇÕMºUÒï5'g)|Mý~3Øk8r§V'º0Ôhõ~·;nÄR\_6éQP¬ê6} y ËÀo¿Cà"töWíMºf"g)tµM=û//ãäu'tz=uºüĬ¡[ÏSmã]B}ñ¤DQIéO3{ <çz}ø1yÍví~6éöü<%§ÿÅv\_ØÍzíJ:l=G0¿séiZ{WòasR\_?É Ñê/ÚOÃrrYíæä.\}>ø«
)°nYªÇ8&Ø4\hÿy×Ê W¾N2|ö+°ì?Ûà^¯ëv<e¦M {ò«j51éN\_4<¢¥åís>í5§Kºëü]ª K)è³^ËOLuØÖGbúQ¤°ÈX~Vï6;àÝ÷È\<Ú5ïcÒô;ß<¥%¤÷éjõ,k °V±vº|Oèý¾Ø>äØz!_§ý ï`¼W\Ç®àLyÜ]¹.yÄÇï¼[ûÇÆ_ºC§êéZÞ8ýCMÛrÊ ?³s nÅJnzÆÁgäS[û¬»§ÕáYåÍÜzUÍÖ<{åê¸?¹H<ð¯ÏLºt«Ö÷&j©xáhê6&lwÓpèò¤t!~ðÍéáæK>øðÜ#Q¯$¥"21ýËþÉWÆ6ê8<ÌEÄú¿nnKÕðüÓÊ Û_WÈãk±~Ãå(ÚÓOxµûJ_Zt«~hIÏ)° y÷Ò_ÿIçb°y±Æ_4ÛIcȳZ;нyÙþØ'e ß&½å'¦zðNWO¢«Ô8V^|Þ<ÏɽòÑ?Lºh{{ǵø=¼×üòéî§8Ï+Ï¿XªËOL²ãÒÛ%8ZÅ^¯çv;X¿\7Nÿ@ÚbvëVÎÊ ¿ì¿8.5«"©"³å'&¹×p®Y°óE?WòU#õ vB·WùÊøkÕdÐ4] ~««Ý{óÑ¢&m(V©ÍBv¾ïzMv¦æ(tcI¶g°=_wHìlÔÃøkåºÊ<·E+{qøëÒñ«)ãrZu<ÕÉ]z/¥Û3äéªXÏþ_Uòl'¦õ;M2þB
ßhEÝt[_¾»tß
{FNg8í÷[ù.?1É n¦(ïQeï
{乪dç}Ñ<ÞI<÷÷Æ_¥.c'G·Õ6X°¢�]ZÍÖù^p\¹aòyŲݬX¥n;nyW=öI òLv~ÞuºWfÞ²äI.UI
Ë¥E¯×÷rØA°\þë3òÌçÕwÞ«Ùm1ó>³J4&vS}-ÛLØ6Îõ&"_Y:çp õL¨(®>"OKhëw"}^Ýkñ¹bêÁäÁNe²¯e«á¥ÝòƳ~#Ï^¹Úbþͤ¬bê±>é9\_ö\_LÐ[|?<öyu\®F^»¿~Oí$Æù´l>d\aþ{ùÉ5òà·ë.FROó½ü0yTBãõúðcòäçÏF?õ³å©ú@úxÞØ[ß64ª÷ªØº|3åyÞ\Áö®>
\InÕrán0yHB\ZÛ<ùyòÒ\_ÿiäEX¹Î<Þ
7ìzëݲskaù°¾H=L&·°Ç£HNëoû?O.ø¼1WÀj^OKòØkmê÷AÒrVNX~"¼6«ïêtzê)a³6!OHhªÕ»Íö÷ÏäùÏ?¶2ê )ëãÈ^¼Ú|Ýo&I¿LÃòodPO ¸}½$ê®ÊuÉósï¿÷AÕsùóìþ<ä
Ôÿb»/ìz¿sk \ÛßwÑ9qïêAa,Z®Ã¤ äÙÍslÉG�çnúµ1{sò)èྫÛXXcù ½Î72¥ñRñ®ówÉm£ãÉG�çkleÌß>ÆqyÎ æís>í5Gø«f·¬ÑÌóä Y7\_.?ЩÉÊ+ú¢+y0xáoÿ"ø§?}Õy1¸×ÉäQ/·ÎüÀ> ÞZuok6ËOÄb×w4Z±ÿRïèñ+y$ ºàóVäCO\Í¿ºÁéÚ@E|ã4Á@ù÷é'GÉS¾¦È©K«jãD°þØjù àÐÙ\_~gÌ\_½p¥=yÚ àÅc«ÛõºÖèçÖdÎEòütÚgÔs£,,ØK°Ví1÷þ{Ï®lßn1uèµÖäÏ·çtx XX9}1zw3,?«ßοù·µÍX¹²èÎ]~>u?Ф¤ä?ÿØu
õlê=çýþz¤ yòóäÖ6B,?¨xgª©ïÚßÉÈ)ø´ÏBò�ûCae¨¦MÓ ½½Mú KxØçè wgn:ÓtøÊ?þõìg>mDþ|¸Ú½üïý¸ÒpÚïxOòlf{ì^õú; v^ O?È{8ÜyÿÃ×ùðÃ>}²wíR'¥M",þùS>¶s·W±þ}!Õ%ÈÃ]WÙ Ö\5û¹5Æò;Ãúî4W£W9\_,@麾FÃß¿Ýúøã¤áÃóÏ׫Dú\»©ä»0xÑ~ÿË2ÜE°bñã¿OûÅòÉké|#)þ çÈsòç f=c;vÌ=|XWTDÝk¼¡Éb22¶äõ ¬ì'L{TééÒê)ò0\¹Û;ö^KÍ®nëL{[±á¸wjF.m Jqì´ f« l8Ëu 0Rk�N?ý6ܧ½§»">ȹ&ººú¨¤T,/ ûر Þbì@¾È¨³,+Nû¯i
Tw^dbúË_9¡ ¬jã4qÍñ'üåtL¾7;|R©æYÃqNÃèj}6ñ+yôA$2Ê(cæ 1P ËO½Èsò§Í?mZÓ²±§½ d8cãéÄôÅOøä³ãª6é0ïV©l3ÿëGÈ `Bðw^÷ºV# K¾y$LÁ&º?ùyUÕãÆgä»[jدl<ËOêß"¾±²òjõ^@Ðlçx<L¸mäC¾7Ñ|°-ÊSÃ|þóöæ8í¾Á÷yÜý/¿HUÛÚê¶Îû/oU þ¾Ô°s~
JìoÆGÃT±uùfÊ òà¢uÊþ§üÝ.= ÏLX .÷ å¯äcB,Í(ÑÞ-ð3Pjaù ,ÏÖ ½ UZî§o!Mø6;MÙ®¤Ã³¤EÑc&ì{3Jȵ6õûÍà¼[,pú 4ÚÁ¼lWBðmu;RªÖðQwÀçL½I%È«Í×ýfrÞ-å'Ó°ü«Ë)^¶r.Ä¢ÕÍó^º"&~%àR»/ìfqÞ-qoêÓO v^vóý$UêÏí'|ÍÊV[Oßæ¶Ö\_ôj&ɱÜòðwuûp=Pð}4×$Ô¿@ð5«÷É/ÛBHßRÆ@¹sîûÚ½gsÛ-Õì5òh®{¼ã¸½læl'Pøª5{-ðzÆm dìzã@¹uæO{Íá´[ §ýÎÃòh¾ãw?æ°÷ÃG(|ÕÏì\|ÅpXb@CÖ&¨Ñ«åæjõär àôÈz«Ô:®ù«ä) \_Ê>¡ø<ª¸Ì/ÊÅc«ÛÎå°UjôsÃi¿+¹:¢K¯×[\E¤ð
5zÍÇ@Yì@9¤KuÎÅOÓ~ñs<äÎÍW£9é÷'IäA \_XÍÖùò=ü(/C¬båÀUTííÚÚiò2sø¶Gt»ËîäY Y«X;ýz'ÕV{{S~¹÷Ûù7 Jº°Z¯×7¾7/¾ýiæ6U)6T)yÆßwm|=ËO @Z:ßÈÌ/5»·ïÇ'ªm8tyZÇÛì�ñÃÞzÕlËmê}5õyÎ@Eéhþ¾÷ó¶ýJªJ¶f¯Ñ)&G¯ùÑ¿/?qÄò(´k.D×Òz|ñEí±ëÜÆ36~ccTééÒê)òlÊt¿yý¥´ÎÞ|ÛCÑZÏöx1j XÛ§ýB:[/ô.Õ³ñ×Ã×ÉsU±v´¿Ëø´ìÿ[ÓO 8 LÈ3£fm#VeúKLr&çé$Êo¾AÿzqÚïôsäa!ë!ßSÛ8;¿¨µyº\Óý+ô8svÞÄò(ç 2µÏø<%Ve:r'¡$Mq©¶÷º{äIá Wß5µ'¯;A® ´Þ%ϳ±¼ØÎ7ÉÃf²òQ¯×3xyÀ\ÐÓÞOø % uÖ\$O\_x#ØÏ¢ÉÓUu=À\_P¢ÖÚ¬ñ#Y7\6ádÀuGo¬Ò¬k¿(5Ó×ó¢xméL'Ùiì?.³|»íÌþÈ #AäyáwnétFm^TRjÌæuC»NÛ¬ÃVöÀ8² K¿\_ìC)Ʀӱ·Gg¬¢´°r¼Çw9±Ï'(&@'åã½bH(Û~ù%ål¾ç7?ò¤Uk\_&|\ºÿ)y°@%Sv\_}aokËFCçìµ°r I2m8û(
<[ 2=äøÆÌÌ+$Ï[%ØËaÀiZ«ë¼ì6y¼@êv>=yûi4yÞ\ÁS7q:0à, I²ÿéÒãìò¼½õ.)Õh bÓÈã\ÐÞëî½±!§¹$\Ùë¼ý¼À9ÅòJ³Õï7øØuÚfòȽøuðÊ©Éä hröë{ájuº½G®¼í8y#IÎ�åP¨Ò¶^èM0PiÞÊ~#ÓÉ#Wö®?v"f²qðy@¥yò~òk}]YøÖÊ1.5$d¢¸ø$"¨7]n3n-yüÊÒ\_V¥Î� tr8úùÏÛ!çöZr~ïxøªW¦u´<~eiÿù{¨³�\_ö\_L~;ÈÒék+FºNÛL^YÊ^XêÚÀhu:¬AãÉ9[ÎRWXÍÚF^Yj9Òº¶�0y
ä÷\÷pÞDïy;ÉK#Këôu¡®-�Ldb:ù½ W±^à X°¼4²ÔÊQUª¡./P:~±ä÷\]yëßÀX·#䥫 éÔåJç?ù W·¹C]^12oۯ䥫^Ã¨Ë §ù W] .¯qó¼F^¹ºû¼uyÒ¸æ8ù Wñ¡ñìüõ.yiäª ÔXÏñ ¿äªXuyÅÈiï'䥫£¢./P:M¯ ¿äj\juyÅÈÍòÒÈÕ¶ãÖQ(Ü ,xäÍ¼ê §É䥫U¬JJÕÔÊÅ//óe5[g½^O]a1C^û$2º@¹ìùÍü«õ-¥.¯H)\)%¯=tõu
r½ù,ù- WÛýo¶ß ¤çOìÁn~üi3g;uyÅ zäOÛ¹h<@F«ûÌÎ
ü«Ã¤®°xéÔ/.»/HÆn9
Þà³"¯»H]añ²û<^8äÑKñ½+ !~'åS,(¯aä±õ-ÅÚ( 0q©Yä/o³ò¨,^HÞ>H¤.2PØWqkÙª±)¯®>|ºÈ@Y Xó[y´ÃDwê úIÆö¹º@A°köZ@Þö2vë~ê"î3¶IÆV±v·¯@0nøc³q~uò8O]d±3f%N¥ç×Ó·R(çíçÉ^Þ$úrY¼÷2yäí°Å¨ N¯o0dyÃËÛ~ÁÔu;.? /¼Þs>Nðà3Sx7$.ºÎbÇ?,¼L²÷øõ�ê:ùã°õy«Ë[öó¡F«£®³Ø)V©+ãubÅ»"o´:Ý7ñÅ¿v´ºÎÒ Õ5äÅ·5záë/À/¾ÏbÈû\öNX}ºÎÒä2OòbÉÞç|©ë äÌ/«7¹ìÝtu¥ÁªC×È%{¿å·/²ópf«�^N]jipán0y±à£°êRy²ý¬/y{+Á´¬|êRKldÓÜORÈö)¼½eo½K¨ë,ôz}¾8½w?³sÉ/RQWÈp,½çí¤.µè1»~ áãÞÔ¥rïØ£óvìôe37!/l8ty©FK]m b3«X;7¶ô¼òºÚRbÏo8DX ^ó§®6s·í¼@ÈM%/Bü~;WX«7NKÏì\´:ìÊb:½¾®ý"ò)ÄþÔrõáëäͬívQW[zô¿¼p ±ÓxT$3·°¶ÝBòfVn^ÔkðG@/ÜÅ) B¸î¹DÞÆÊñf�¾[0ÛO£É §Û\_¯Ã£ 0ôOûà!E «X;ceÉVµÁKùë@ꩲû"y+ǰŽ¹t²¼|ÊÑrä\U©ºæ@zD%eT·u&oå8wË9êKG_É˧(Ýݤ®9C]·®¢/§½w©ýqúêÊË~Øø^;° x#¹
% ,#oQm¾8ÁãìòRÓM'nQáK7§2}@]|9A^JeZÍÖ9 <ºþ@\x^yHÞÊôëAK±)W´µ¼ Ê´í¸uÅ5uýX?àá]/\ñ1à@RB±¹xZ£Å|»@ÝòáêýPò\V +ÇK~!Ô-�èYºï y7\VÃ[Ĺx3 K«÷O^VÅZÛnaH\u�JNßæ{v²ºä^5¡µåè5¹
%Ô]�h?QàgZ±g8ç¹\öáµ~øv4^¡ÔÊ1.5ºäF±JOJä®:tº huºÁö7í:m3u#ÈQËîùͺpLs?IÞrpË)êF'çnVµqºþ(º¬?v¼ß
cRz.u/Èbº¶¾þ¢÷3; èêv�ürüz�fäÍ{ÌÜJÝ rfÌÊ#ä%¬ÍF¬Äg's70ûD³w¨ÛAÎ\¸L^bø¦ÃW\lY[«7(¬låOÝr¦TùÜ~y¡á [YS@ÝKD&áÖs<¨;Bþ\_}¼Ðð¥Ýgl-,.¥n À ±)8ETî:º)ä×0òBÃWí4e#Î3A1©õ-%o'øÒj¶ÎϳñÅïèôúf#V¾jÛqëR3ó¨[Ï£°/ú¹7|ÕQËQ÷
RXuèy¹ák¶³oI{Áqø EÞð n ¥[ÅÚ¼ð5[^]$Çí§Ñuúº7|Íæ?»a{=!è²¼èð~Ù±ï³êî�Ʋû¼>ÓÕ¯Sw²ÀBÑZÝÖù8¤=ìg'óäÝß(;èSð¥°h´ºC¾ÑÊVØõNÌj´×'ïø6ºì£î%²dïeòÒÃ2êz Çx¸Ô¬&m oXîS·KÍ®=îÄm«1kBqè°¸ö(üËþÉaáË5Zu§(Þóv7�,ÛÚv Ïú<£îøÅuÏ%l5,~ì½LÝ,ÊåÔM(,Ù[¸ëB©ZCÝ/Ê%#§^£3¨ûE¹huº£pR¶4l3níÓÈdêQ"'o>®%qä2Oê~Q:;Îù·4Òª6Nn×ØOÔ]£ Ux¿KZ'RwÒ)V©¿ß¥dÇq©YÔ#üb[]K^nh¼ØÙ^$,Û
¼ IÖè5}`)Õh©{GdåZ~¼ÊÐT/ù
P÷0SÀfy?@Sýiæ¶ðçÔí#7®? ÇÆÝR´Ý/ëôØàK4Ls?IÞÐ «Ú8ÍØx:ErAH\Ó.òBó
ZqàêG¾à^ ¿À?¸ÿ}NZLzNA¾X#Ë÷sûEÇ®í1së }×wÞ~~Íáë{³î½pMî:÷Åîæé5oÛ¯ãÜö¿§ë´Í-FÆ!>Ð;OÙ¤ï}à¬;z¼
ð¥wFSç"0U©Ær$6GáKR"¨(çn7Öè5?.5:ØÎÝNÞNB
í"eCt vºÚȲbuÎæ~¼© ÕóÊCê\Sð¹ý"ò¾Ð.S7ãýaùá~ì&ykA¦
£ï³êüÜ£Õép=Pçn9G~/cS±WP0 _æåÍÒ}WÈÛ B¨½QgàRµ¦í¸uä!½cV¡< wc° >Wë \WHv@ fm:CÞrB{ÄË:çpä©[AÞuBYjï¼:äÐ\½JÞxBùYÛnaóê\æIÞ~B¹ù¤u¶ò KXIÞBÙ8ÐeÛ°(¿ XlY !äÄúfäP§ fÉÞËä!ºVXáX4Z]·é[ÈB(ign:Cf@,Ħd~fçBÞBÚfÜÚb:É8péy[B¥hu[ç§ÉÔDǨåÈB(97÷¦N/ F2ó YFÞB Ù{ÞNáÞÆ ÿÊØ^Bhõ,ÁyP6Oú7PüV³u¾GX@ìèõúÑË·+Pän9
=XQ~7~=yÇBE+ûÉ:¨JÊøÜ~yßBE(ûýäIR@b\ò Áq×d?mF'gPç$±ðÙÏì§MêdRE§×÷¿¼!"qÉÞËÔ±¤MzNN²örØ¡Ñê¨3 Hðç_ôs%ïg!¡&m((VQ§ wcª÷OÞÕB[GC@Vºù¯A¨@kÛ-ĶÃVô"oo¡Vµqé'ôzýÄ5ÇÉB(û/Þ§ gÔmÇä}!Àù;~£ ò KÚc70åî¥q0 ¨¤¯,&ïy!Ov²;z! MÅXP²%·°:cNÁZHef»_ÖeæR§ P(BkÛ-$¿ h9rÖ6Z|DÕè
%öJÞæ?»%¦ãpy@Ï ÿìÜ¡¤m0dYLr&u�ð;¿ùUµq"¿/ føÕÅ!qiÔ)Àÿpüz@el¡Ôü¢kt u~�ðv¿}&!öYxçi4ur�ðVN{?©fëL~§@˵Þ%Øp+÷Cñ&"·áÐåá Ï©Ó�£¸ý4ëV #WÅ¥fSç�&à[§¯ ù½!|Í£× "O"ê XB~A_Úa¢{zNu6�&añÏ]N~AY»MßSPL �T¸´o/#¿ T¸?Nß\_D�p@jf^ÇÉÉï)ëGJÕê$�3 K,ÜK~gA¨@Ý<½ô8±ÈV7kÓòû BåXÕÆéÀåÔ·>�<qö¶P�?³s¹þ(úwÎú<ÃB{yµñ°A1©Ô÷:�q?$KW äÉ6à¸F 4ØQìG)ò»Bi5k[fJ$+¯ÈÞy7ù=¡ltÝsI£ÕQßÙ�¡ÕéÜ<½ð«=ôÓ> OÝ|B}C .ß ùÜ~ù] ¡Dm1juÎjà¢2Úý²üÞPr[| ¿HE} : U#yß¡JÅÊV{c<�o½;Ø{¤µùÝ ¡Èý¢ëÕû¡Ô·,�à\_Èýß³Ö¶ãÖ
Æ¥Qß©�HÁöß¹ÍÊV®{.aa�Ìàè5ÿOûà\{·ñ°>O¢¨ïK�$LDBz§)8{yãÜæPß�HÖ°.?ÜCÅZ×~ÑYgÔ7"�²AH¼åÈUäw7k=Ç#>-úþ@dæ[|üPÙgsÃÏñ-õ¹z?´Éplh en×icq� �BWXèñ+v²ôÓ> =ÎÞÁö�\_lë±kÉ�BµsÚE}o PJJÕn^ÕlÉ£� úeÿÅG¯ùcó.�È NéÏΧ¾��¿SªÖ,?pµzÏùäá�¡IÖ¸äØõ�ê�ðR2ófl<ßî¡$¬m·pÃqïbú¾�ÅãDë9äáÛ´°r°úûú^�Ëå{!#ÝÈÓ×´µ- <úþ��LI©zÃqï:}]ÈcBÖæ?»õy7»�4iYù36ÆÐÏì\Ø7ìê»�À áöλɳͪ6NÓÝO%¦çPß��îñKä²VÕoC£¬k¿ÈÍÓ+- {?�Þ lP°qAYP´6¾Òãìbu·�¤lh°ÑA_PTv¾å¬Ï35Î�ªTsàòÎ+[9X°÷êýPì¬�¨8é®{.}=h)y¸Am?~=ûÄS@Ý��¹QªÑõy6Èev½µz/±ñôÈ$ê¦�ÈäÜ Ç½À¯-2ôIö_¼WXBÝe��eQªÖ¾õ}l©nëL°Öµ_4iíñÛO£ñ �Rõå{!Vû´ÏBòl&ùõ ¥¿úÅju:ê>�ÿ¡X¥~ñ \Dîçöfl<Q�¹
%G¯ù³Ã¥üÌÎ
}dKÕê��IÏ)8å?få¯,&OTÅj9Òmöæ³ý K©;��¸!.5Ëãì;§]Õ{Î'YÙ[×~ѨåاÅçÙØ � gUjïHG_ÿìF½rÒʱóMn^O"ðC �@ièõú°øçû/Þ°úåÈUä,E+[9v´aîs§n>IÎÀA½��ð;Åöù
ýmç´«F/|?öV¿ì¿x˾ ǽýbÙ>êº�ØaçËÍv¾ qÝo9ÒÍÊ<É fëüÝøõcVÞrÊçAH<¶�À¦hh\ÚYgìa?¢[\%ã)SÅÚ©Ý/ë&¬>Æ>\¾G}ù�@æ¤eå³2ìçöYÎô¿§åè5]hùi
íǯgÇ<Îï½pÏ/(6{m��5:½>)=×÷YÌá«V¸:~õ1«YÛ YF>5^XÕÆ©ùÏn½çíæ~rÝѧn>y=ä�@R°³&3·0"!}¸èìyåᦷ\v_ºþäP×ìÜi3nåÈU¬õ.ù¢+kåò¾Xc^ü7ÙõË>kØÎÝ>béÁÎ,ÛeëéÛG¯ù_½Êèä >/ProcSet 32 0 R/XObject 2 0 R>>/Rotate 0/Type/Page>> endobj 2 0 obj <> endobj 3 0 obj [4 0 R 5 0 R 6 0 R 7 0 R 8 0 R 9 0 R 10 0 R 11 0 R] endobj 4 0 obj <>stream HËnã6÷æ´lJ¤Du× :@E]Ë6k %'ÍÛ÷PòwÄÈEV?çò¾ì²hg¢_¾E»Sôé¡Ë(¡?D"Í¢"±vmôÓ׿¾}ÿóçÝO¿ï¾¸ùrq>QGq:|ÒêªnèÝ>-Ýiå%';ìéoõóyÆ,
)ÒLô %àoÝ÷ªïkÝAVH _�Q@ÜÄÑ-ònåOºL½¿dÔV\ÒM¤tAåÇóû·ù¶¼@L=ð¤¥: ¥5ªUÝ Ö8A°dq222 Ëè{ö9[�.ÐW3I,"/F"^j5¼ïaI ØpVF-W9$Ô½{Êée%Uç"Ã;ºQUÿy;î0Ç¡U¦y[«Dpdcju×ù/§«¦®ýÏyl «¶ÍhzÊ5n,
{ãS}ÕÞÃz¥³Þ]ÝäI¸P?êrîòÛ«M$0äF¿îJvþxÆé¥>^«f³õÁÖÍÇÚÕ_ô(IH£¯Ï>íPÁs¼Ö'Ac·wMè;rhñÈqS¾súMÝ"à°ìÈñ¶î4NãfªßÖÎó~o?É�4ê¤Ø~ gPn¹½ ¶DÔÁ7O5Û+ÈMcå4aæB×u_-{5¼úëkx¨ú."ÊØ)¨BØìΦ3·bç4]¦(®SÈx1rË'Ê1$³ºê3<·»êwJ·wÈþm]ÖôÅé,7A8�´îÔ¾ç@Hñ&ªÝÚ&VÉáU/S×·B¡Ív$J§Ò-t+7h3ñÄGYR¤\_CË0c#"]¸ ¦À¼û6¹Y¬p¾Ó\_Rt6]>ĦOÇ¿´6ZhYÕÛºä¥�¤ÁjGXÀZ½ý'À�zèÌ endstream endobj 5 0 obj <>stream HËnã6÷ænRW¼HÓ¢³ëÎ@7³:VK)%»ÈÛ¤¤Ã_&ãdl}$Ïõçáýö ³¯Û/wõFe¹ýaöWälóí)»oóÀ_Û»?¶w"Ûì÷ïEƲí>ûrÇòM½¬ãvý{Zöï¹Ú±í»i{û¶pe \¿'@.@Å8êçæ83lC6æ0¦}>A´H5Ý!d +s@Îc«ÛþV௨ëMáW¶¾´c³;¾?Ū |n/z}f¹�È´&FÃ%0cÄÐxÒ«éw$¸D«6? Û É9aeî~ïMdS!h»alºGý@þQÕpãR»«|²Ò#xìGÝ5Ý·BàÎǧ[¾°²M0ºioj ©<2Õ·³dK¾u©&G«" è'õaøªÍø(õK¯0k¨¹%RQÌÖ5Ö´M&ô
ùç »ØÅôû2 }Iu˳´Ñõ=®mÁ§%uðUµ7ýé¯õJösW4± ß} ¡(j|¼~Ú®q§3MH'ëÚÃþà%¯ý0èa ÆÊP5nµÿ Qüºp¡²%ÀKßv#)f(;%sÐÇ0ªëZb©þ^0)Òæ½NÿðzqS ÎQ{IVóÕIY%"&¡V%jI;ÊÄ.®kÀH÷¡M8¼Ó ¶KEkuN:\è¶ëÉ.nzsüé.2P2Okû éEËzoVFP½¼äÓ]¥8èÖÄç±"\_Y4j£1qé©wæ4üÙwgÓE¢S2.ØÅã]æB5 t�q´¸xíIMþ~Ú,¶¢@ȲzXÆÙ¼ áuþÉÀÅÆNÊ/û¶á¤Æä·-Å@È_� ï?îYø¯äÕ«¶íí"Eî3ôN«¤Êx¢|ÝD?>|ZJ¸ªf¥µµht BÓ!áVH÷&®wûïÓkÇ«ñ<ö&^wζإ°ÃB@JÜp¿Ü4xª¦ûü#Û£ßv2³ªØ=Î í´ÃA ùþsÄ4 &DÏÞÏY ï!Yy¢ïÓ©ÿÀUÛѤ^]Ç6û> ßãñË᳿,c/]³r_Kÿúwä£5¼SP8Ë5×·CQ(d1+
®¢¸'"Qîäèûz E¡ñTv+SB¢à«³s»rN ñLMX·åa¨ ©à¥;á;8;¡®ß'K�|F@,¹¤ð.pc9u%@à« yü,Àys>/óÝÞìéE àÆ>öÕe¦{kåÕê¬+Ek·ONÑ® 9æf\¥Öwï4}ÃüCò »(Ü%Wy"V¯T¸ùpþ¢M¢ÛE à:º¶ôG{dg7úÔçóDý/À�Äý7 endstream endobj 6 0 obj <>stream HËnã6÷ú¬ £¤HIËôÌbÐAá¢o¶UÈ+Ê òD}ÍR²Dþ Y'ï¬G<·ÿµIî×ÿpGHJÌþHä4¡Yd}JVç®=·]\_·Zöjb³´Y�£ìTúÉaa9ìë.´óÍüº¾ûi}ÇuüøOH²Þ'£åb>Hí¿eVã¹oö²Î´gxÌÏíN9éG'+Ó
3:îOü»Àñ¢.ªpäI&®§ª4wÏ�Ú¬HUäû4
¿\_a³fýä¥MÎÃúpì£bfx12ÅÈôíÐt~^dðü[k2Æ&ÿv6Ø|©Nÿi.²i^r(Û¶½ÒR÷&R]4Pém{ѽêèÞKD1×òÙ«®üE±ÜÆ8Vï]¸?<|©fD Û-O:yW ý¸vï{ZѨl"�ú ë}ýùÙ¼¥l\®ÔÛHCsÌ˱nTèrY�Ón·ók!Grô}£#R:±yÇ ±uË®¥Ñõ.7¨d%0ÝØZÕ¡Ì�j»ÐJEVúº B8@èªGØÕN5rPUs¬Ï¯ÕLoØ˺1aÄÌzFê]X@J\/ÑjGÄYÚÈFúmTÀz[;LÓý¸¿ÊF¦»énîVYʯÑdðxV }C§öYíbc3µå¥çÍ:x É+ÿÜT1Q
wÄ#PTöoêÒE@øÇ/ÒÁ·Rï}ÛmÁ}Ȳw8ïaÊÁKêHóehÀ\T¤8Øp#iÈÄÏÛV¿#.îD06µ©w0:|[± Z3r´KÑíT °SvBj=vw4ZÛYïvV;¿Ç!jõ²Ô¼yjg:Ø�p pf2«rsH�qvK@dñ×ÄóÎdsÁÿ~1~fC_Q+Vz|2J÷¡¢Q»y ÌÛ©Ð>stream HMoã6ïú «¤¨ãè=½øR Y¦c¥²èÕG²ù÷¥(óÊ òp83W÷yÑû_îØ.ùc»§i y°¿÷wGd3!v·sY8H ]«É%% ýÅìï~ßßß¿}ùç)0Y(æ'w»PÚsªóz
¸
" Õµ)\_Uë02FÀåµîΪ]¥9ZÓ'ONÚ1FR¼ ¦|>w[ï0Þ(ͦ8? =²-3ZÿÓ£SGûZ¤$NÙXÐç]RÔCÔêëϳNǯÓÕ2ùÆ Ð×Uþvê+\t Ó·ÑS&C ÷ó/p.7eÍM¼nßÚïuCÙç' =Ëa(?R�®Ü=ÿz¿U9útÉ}Y3Zwë71ä°Ùá7ù°5öÚ÷¾là9(%Fò)¡j�(´Ùå¡ïg\ l26í� r,B·ÝZAOñ©¶çCºÈI¸ÚÁ|¤P<5bè1q�÷^E4LÜLM½õ@ÿ 0�kõÉ endstream endobj 8 0 obj <>stream HKÛ6
÷ä?Ù4U¤$JZ¶@ t]/³¡ez¬F]ÿ}©Ç«dÌI{Ï}èéãÓïǧ4:ѯæèÃS³(1ÿHÑ鯬<:vѧFEÿ|x"$& HÅmh^þ9�²@}¡)3@¤aJê J ¨Zô¼×_?©¯¸-fËùÿâ;ê¢Ëùg7©Pª½}LîD-d1¿[DQ@N
%ÄYô8o¯JiÔÐ<_µ:¼)fNñò 6S9ìK.<ɬXOßywkÅ!ÅÍÛÉb2?ÔªIí;Ïöf<¥½Ó¨M{».ϵOåzs M ÷Ú(]�v&ç¶tå/b«Ê»ð$D¿_gÀÜÄðe_0ñgùjÛ®ÑZãµüTÆi¶þöhaêRO�±½�@Y ïCq¬{ñý&j v/]¤ ï,ÞÊçû{AÓµE~Å@;ä°NÅÐDU¨Z^ù©Ñ|2b iÏp¶¾úü¯jL0qÔÓkÎVÁÌjN-öZ÷äÌ£ô¸¼7¶föÁYè_A;@y¤¬¡¸q¥à¦9@²uÇÜC²N:ë%¦¸S.[kÎÕk ìl®ãëkÓÑÜ/1ÏÄ £Ò{BårA)¦¸ð®iïÛä3¼æ2GÑëö(±õAKÏ
ybäf{C'¨é÷Ry)ªø!àß½°X]e°Ô!ü¢Ã{Ez VBÊqã&õúµ_ ¥ÓjÉÊxñ\_ãm°iÆôÚ¾}Æ×סPúpÏiÀ §u¶1ÍL,ÌðÖ¢LÄö
z95Ãk]äíbØLÓ!nìýVê· ä´@(!¾2$À³ÓÈ{gÒXlfÃ
ïñ°¹Õº�Ãý°í²4Ë ö°¼ÀsL-BÅ:L>¶DãM¦^EÀéta!Z%^§y¿»W=¬ÄLæô¡gûk#F5ò Ü67= =×^9º¶ÿQd ·°ï(ªrh6õgf¦@s�O@n4/¸NP�D-;7ÃQÓÔmwD¶ (Ùþ6gb·nÔõx3ßPq@8£�A¸ëÑeHà{Ì¥�7ýö¸t]7ð3kuMó7#¯ÅËüÕhkaóÕæ¸+?´$ôRos X÷ÑÝSHÜf2^»»5>|§VÉªË »Àp4.²×y!Í·¡MÑtËýÝYÌöê¤f¶àk£¯û¦ô2h¾\Û¸·³T˱7Ébl¥¹Øç¹µ uJ°¾ 8bvöî§&¾2¿¤É:7õ¹Ì4.ÇMãRm5g¦$ãÒjÉÌ0ti\|T|lÁ %�çF)® y»ª ¯½Ò²¯ÈÒÒÑ �§ endstream endobj 9 0 obj <>stream HMoã6ïú= à¨üÈ£ûu(°À¢ëöbÓ±²dH´ÿûR\_¡Åx·rIR3ï ßK{H6ß~z $Õ q?$"INt'còhfBhJ=¢0Òª°eSwò´aB!¶iÿÿ±yøuóÀMüò§Lh²Ù'Ã)òy×2É´rÇê½mqî̼9cªÙ/¾.F@ E\Ôæ="b¢ÆÔfå\IrÄxÍÐ.á]EçRa¤nì}t R.¨Q+Sû8¹Ä4çÎ]ºÜ?÷×p,®³$Ëò Kÿé 8éw§!E½Å¡©#ðÕù%\_®3]ç²(X¡p!+ÇPÖïEûÑçJCFÑ"Ü¡À ü(
}¤Rèt ÎÖ>=8gkX{ åè)BE[Ú¢¤ÍvhË#¯×ûÙ&p
rX\9¤GºSsFzÓ@OÅ BdýMÇàn CØëÕ¨Ïu
M8Gȶ9·vu?d çîýÒs½3mäjB¶UÐ#GÄ碽V¾Ë>Ó6'ÓZÚ×q¿ÝóýxpêÀåsi}6cŹ±×[»0½½zæó.x"ùÍv î×| ~'ÅÀ\_gôP£÷Ï8Õ¶ûà ^¯ÓGlvw½½xf]×iDþ~=óõ3ÉæªD½íåê\L\\ÞQT; oEëã:@OümNÕ½x5î¼î�L×ÿoS¥Ë¦.ùë,¢H\_Ý3òõ|:¥/OMÜ$á©Of¿7»uû\_¢@kÿ¿êÿÿÿûzKýÿöýý45oAÝ6ÿ×ßëýõ¿·]þú×tÐvúßúß]ßék÷ÿ½W]U×ôßÿþëÞ¿vûwÿýkýo'ýq}ßi}õ·\_W·§ÿtÛ¥×úÚûÿíbBÏÈ0I]+îa4ҿ׿þC5íéAÞ·p¯¿ÿ·ÿöÚV«kim¥«^ßò D5a¶È°pdy[Kàa í;Úa¤m¥ %x4¡a¤~¶¯Ó¤ÈgÒÿàMǶÁPdpÑ!TÅ]Bª(øhbÃBTRv " d á°a¤¦ÒAÛH4w÷i' euÛiÚ¶© ÔV½õpNôÅ4ÅEÃaàÑMH?£�à $0¡{·´Ðal&A¦Xaoµ±±ôíI«_MtbpÃÄC Ôzah0A i¦XäÜ«²Cr#§ÃM0 ¬?Í5µ»mò.ñ\ÀÈE###¢ÃdIBá"S9SÐh0ä\ãØUTd[ð'ÉÕ"":!¢-#azNN""?ÒOü%¯õé/ïÿlÔd¾E\t#Y¯þ ÌÌ2) A~º´Ñ]°©£cUFÆ´lh×ë¬Lì$A6ÚA°áÒ©6Aýu´0ÒmßÁDé>Ò´©ÕûIûÿ¥½«iimÕ¯V¯^7ë{ 0××÷ëøuõØ0¬0pËàÿÿ"!ƵbBÞ"ãëþ_°WÈyB?øo«pÓ åÖXÿúù 2f$¯ôçR®ød.¯íÿwñ÷×ëϧ§uémõ}¯ÝÚ¯W¾þ¿·i'°¬3éf´m)ÙýÇwL0îÛINÓKcjÖ AÃbPi¨oe@A 'à ¦Pè4A )Ý5òå?þZÉ3#^ZÀÆv ¼à§~ e ¹à x kÐá0ý DJCb¸AÐ@Ö Ò}ôפ÷ô^¾¿_úõÿÔIèoæ8.I"tèGQshE8ιÃÍEÓú á4ÌÆ\ Í a jx°kDXxL&1éøL4gNCûÛMh&ôJÚ%n"êÑ;q+ŨD¡äHÞ´úM¤á£úI°ÚVTõôõ°Ý«².3M².d$Óé»^½uúzªé§^[«½kýïÎ:ýzÿkþíúJÿºÿëý þýà ӧÿÛÿøÿ¨®øþ£¿¥ÿ ½õÿÞÿM÷ü²1íü
«ò å4¿Iÿÿê¿ü!û}ÿÿ}ÿú_¨ß»÷¿û×úÔ;ýþëý÷»mº÷~ß~·ûîlÔºÛõnÛÿ×½¶M¿³û ÿì2ÙëlÑ}Ù¢Å4ÈaÕ¤ìgbûä;×d-ï,Ñü,1LP Ý AÇw±Ö3!hÉhA Ä; aÃ5m0®Á¤ÚL'm ¦Ê »4ÍçabÎôWh gaKtáá
¾} BÖGp G'" á¿ÃNþE ÐvKß¿ÿn¯þÖ¿ýÓýéÿÿÿï¼èqØzËYOKÉ ×lÀ¾h)ÐRqHc(B!Ó6hÃÉGÆf2b( Y8 Å$Ft? 0AéL&L&pa|aOv¡xUï°>4]za0T×#v'¢O ¢qD% [aa¤õ_@ EÛFî½ÉÅ¢ã¥7::'moÜ:mL=Eö½ÒnÉt$ø^tkaåÄJÒzÒuüGm+ôÞº¾ôôñýµºvn¥ïþÿxéº×i¹\j·Ç'ý×ÓKìõcö0ƯZOÞ5]ý×íëøcÿ7"O!(/ÿ·Zþ¾ Bô¿ä,ƾ¿ðpq°pü|õÖþBðÈÀ¿á¨/õûûaÎ?ý;KÁ~¯aÈ!rWï\_·ûþÛC¾+ÿ}´~ë,/]þÜB7ÿ\_»rIá¿ýGòwþÔè=ík¼>ýÑ8Ø<ÿ§ØÒþºÝþ£ÿûüÿz¯îýýu¾êÚ_ßù½ÍîÙÒ ßn½+w_~Ôþÿûûÿ¿§__û¯çíׯat½»]ívû÷ %·WõgÕýE¿ëôÇÅ0ÕÒ½{a
±Ý2Í ý½oµcm+YÇÚKéïi²vÅ ÒÌ8ÿÝÚ°I°¶BÅEE:PB86¹ÃPÚbPà CµÔá§pXP©¦&×wí5Ðk AÕ{ L&ª4Ê)2 !DDDB"""""#ÿüÅ$M&EY¡vH:
2ùW m0áÃÎÖ²ùtmD2#¢9Fh£Z/Í£4uG#y©¡Í":)ädH2]e£-ÝçeM ÑÙ1ÌÆGDxçäB">aóEò
(AL .~0Ppg
64Ñ µ°Á0EÀh»¨0ÀÁAQ²±Ñy ÐE» ¨¸£s»Åh'aa»ètn-Abé°àéI¹ùÓÏÍ®A°c é6 Èñ½HèK+Ðt HÇ%³DK_KÚMV÷IéôêzNTþ KWz´¯O½té=]?îþïCNº½?^Úë×ô/«nÆ¿Åq»»UÞmzþ
oé{×÷ÚÿßÝ_^¿ýþÿõÿà÷ß©È.ü'ñ«ûÿ¡¯ÿÿã³×û+ ^Ú@/ëöõÿÿýo¿í BWaO:øDñïË«ÿÿ&cÿÿ¾LIõýåÁÐVôôõÿé´¿ÿ_ÿîEËÿΤÝgB¯JÕïÿý}ïßýõáõÿ®jÿýûoûIÿ{~õûôÝyô÷çülWö¿é1ÿvëï?ßdÇékkþ·êׯïk¥þßuj¸]¥í¿i]kíë¥av°°°Øki5fÙýÚVGûÿfkJÒt¬'i6°½DãëØøg(î %Ç{ +GzÄî±û vÃS (Ú (ØÜ0¬(oLáPئ¡Å5 b°@é(b pب¨LSPP®!5{°¬&Õjçp°×Aa0§ \5ÚÜ0ª0N hÑ0DB" ÁÐDDFY²½/T¿]!õ×b¡ú ÇÊb¤WÎêÌí̯PgÒ2 ɲÈÊãc; e(B\P ¥Xg!Ó aKuda¬#¿ó7»áÁÁ |Éc;H3°<ì]ïgf¯AØ>@ÙttH. nj¯çv¾«ó²×Mÿ§µäoÊÍþ×öý±ÖF²ýwáZV±þÿ®·ôv5?úßoíñ×ëþç_ÿ÷ÿÿçzÿÿýzÿ$òbî²%JÊ daúG#AHy: yñlôh'þóRάd)ì¤3ª<ÉNNGAFCG"päù¤l¢å<aÉfß{ø@ÉÙ#¸ Cè ò03A
Be0¸}¥ÁU¦°T-ÖT¹ªhÜÕB° ônóä#6 08à¾0(ͳ16hØ2®4¨Cù¶f2ÈbÏ AÁA
· [pj\a4ÏàÓ BEà¡0ªv61®®¨Z7 UFÊ7=S ïèÙb hÙbxaÖ4 Ot:zðtÄ4laTB.èØ]°WMBå¬3"á¢>0ÁB )PpaÈPlaUtÑ¡£u l\7¬NÜ1;´n«(z ÃCcA:NÌ1ÁÁêTO#¸H:M9#æétØmë¦Ã:Aó#® ®É<¯ÿGtáØè:A´oº µh4±ÐAѱÇQhÙÕEÃEáH8c JÆl$è'Ñ812>a¦Ã,ôIé6;%mX/§KdîNíþòx´½ééô ôÛO]ÿïÚnáÒt'NeYÉs§Ó'ì8dy i6n]Õ´Ðzzu°i'Í4ßOîýü!ýǧKéÿëKÞ§þÒÇÅ¿í|hW«W¿ûM÷ì¯WKÖÚÔZ§Wdîÿ·»n¶NëÓ¤ßôëÓõÿxÿÓêôúý= ý~ÿÿõïýkuÿ{ëú\_Ã$ ëK®Æ¯ÿÛ¤ëK¡þÄzútºq{«¡ýþ¿ëÿÿÿíßõýuª\_M{K{¬/¯û¯õÖÈNþ÷ö®£¥ú×ýWßÿãøÃëýýWOÓ\_¯ð\_¯ê;¢ëõÿñvýÇþPÿúßëmuá×õë©ä[í×öðaõí%t¢:dCþ=| ññÿþN×ïõF?ÿõÿú¯äÖÚXD®ºXD£ÿ÷üÏýa¶ ÿþÁÿÇëü|{!UÿÿFBxuÿðZäuÈ9ÿýJ?ýrÓº@ÿÛIÿùb?RÊ ^ß OÿÿÕ{i~?ÿöÿä,?\_Èr28[Gÿë·ú$ú'åËÿéÿûa$þüWÿúÿ¶º_¥¤¿ÿ×ÿúX?ÿü7ÿC·ÞòÄe[@¿ÿå'ݯ×ú ÿÿÿAkÿûÿü0¿ÿÿ¾+ý¿§ûÿ¿¿×ðÞþÿR6?ÿÒ¯ûÎK¯ÿÐX6 ÛÝ?ú_ßÿÿþÿû_ÿ³íKÿßÿ»Òÿÿº_ii?·ZYåëÿÜ?ÿýÿøzïþý,^ÿÿ÷ïÿú_ÿî ]µ× :ýïþý/ñZ¯Ýû÷Ýñéw_Þ¿ß{íÕ}ïºÿÿÞ¡ÎÔ¿Öÿé[_ÿÿî¿uþýýmnþï~Õß×ì÷Z_ma/ÿýÞôýñÞ¯Þ{¯öþöû÷ÿ¿ßwÿv¾ÿðÖöõ°¿k}û}^klì~ÚP zäSìøûKûk¯Õ¥{ïýÿ¥
¿ÿnþØVÿ[ a[4[fÿêÃ.Òïl÷ ¾bqö¼û(ìJ{g®DÎÃ9&ë»OAÄîîÙ í¢ @úBF:Úö
ýöøgøeÕ{¶f¿íöîõÒ¿Ç°Ïñ±)Û ,Áú}ÝÄîØí°z Ù ¨0ÈW á´±@´6( t60´aCP aPJÓAÛi1;µ $È£v >DÊl'lÒbqëiÞÁö·mCcAAÃdX ¨lVÅ àÁ8l(t6!0AµvÖ¾á
iÛXa+^[í5XeÊð! )Oøa8lB¨Å1 M46(6( qP¡Ap¬TÈÝD jÚ ¦m[áauÿ]L¨(xPQ ) «¢èØB"" ëB"""8ã2Ùíh5ì& ÖÖá
'ßöA
qBB ¶DèCh'þ""""""4"""""""$'T"}ZþºtÿûÕ.´«_þºT´$³¾×¶¤MükJI JMàaLB«ah2mEZäåq@AB; d ÍDAÈd 9 òÜv ¥¢lU²ÌBiÚ¨ i´÷; FB£Sµ&à òÖeO¥0aNaÐxU³¦ºúm¤á5»ðt¨}0~É×}ý&Êÿ¿áä&L- ZÚé?_÷'£!µ¨L44)þ©z×üuÃAqÇk×ëõýGõ+ùÞê=ÿu÷úßF-.Jþ²Êqý\´+Dc)Ï)kîõd1ëÍ ÑçNeyç²ÊSÏ^j24ÔgA Å$ )8¤ù BFldØÃE @gÝ Í é°" ï¤=L{ ] UEØag¦gfHÙ>g62Ê3l<á: CÎDò|á\á
:#7 B3 ä}BA 0 &8A¨0¶pPD8ô^0¡?EÄ'ºVwh7ï^tµAEö00®<)ÁOa<&g¡í 0>A52 ÈA FØ ôá¢Ç(x¢c¹ ¢a ÆÆ©£sFê/!¨¾è7§¦a¤÷A6ÄÐÖÐAÒ[é¯Hø|4¤ jÚOÑyô-8´nz °ªEã¾jÍVÑxí ð©Ç >º,¤Øb{iÒzti=aÒlúzpòG¤Øa=:N ²¥¤ô º'Mw¥|¸²w«® Rpd² %{Èñ¥4¯'¶táXÝ,81à &v)0¬B4á4 0 øÄDDrÆE)£ °;ÉdSä±q ¨Í«ÉFtÐc5
¥Çe±ZÉjû;-P ä3 ¬Ï Ð&UFv3"2Zö0A60H¨/¼a¬·é üÙ)JP@ïØAZM¤Þ ©7 éA¾¨h0ö êíîÍ®d.hÖ¯ÿÉÖý?îÝ:I¥h$¡tõJ:Mþûÿ ×ÿN¿ú_¥C÷ZþºWÿQØXiPU£»§Izº®¾ï÷ÿÿÿããøý]t¿êÿÿÿþ½t½-ûJ½Æ¿[´¿RAãÿÿôFj%(Ù«6D4p¤¿ÿÕ/öõ~ ñ;%ü¤_úák&|a:]¿_Öýa¥úA ùÚ!Dá ãdC!3 ø¹ è£("¹²!¢tpeáÑDiÍÐbfáFDóC:O£s¸£s¦Ð½4³¿Ñ²^õý½_l1=ú.(AÔ%AágÐgpL§$>0PO0 #2 ÁA?I´tÁã'ÆN¯Ô'[$¾þ\TaøFî qh¸nª|ô\B£sBÑyÕÕA¦¡UB-Ú/¨0²Z£cÕRm'í+ð5ûÕ/µaë^¾pyôØd»: ÐMÈ[¤Gy,%e' JÈ4 R'Æ ÁÁòv0sc}t'ßë#ÍZò<}CKö4÷ßé¾àî´ß¼vúpf¥+i´z û6ï%¸MÓ«ôõúéצ×ízoÉQm_á?±0ý'Ù=ßíÿÿëWÜ[èWÅ÷Þú~ëê
.Û÷¦Õ¨AÕºoþµªoýº~þ¯ÃZ_°´þÐ÷ë÷÷_×÷ªõãý^;÷ããØCXû¯ñh}è¿aW ûÿëÿü/þ»ÕüþÔ?ú_Xoÿß÷ëÕþì~x ©D7ßöäÓÿÿïäá><MÙ JýÈ@ÿò$!ÿú!Gÿü±#£úüä5¯^_IõÛúëõúàßë{ïðÿéßÿ°jý¯ÿúMþû«²r(ÿÞkKÿ¿ÿ÷ûèòiè/¿Ãû}/þ<¶þér2wÛ¿ýûÖX?XÿÝ$A¾½ëwÒ÷ÿþ×ô¯ø7_á¿ÿô®ÃKX/·ÛõúÜÕ[ׯUô¾ý»þÿ___ú^ÿçS¿òÿ_égBý]þßZ·Wÿ«}[×Þ=oÿô³?¯ú®¿¿ÿ׿úO¤ÿáÿeô·oïµÔÑ]t½îµ¤IÒt¶Û´´÷ÿÿ߯ÚùÍëüzôÿüçþßÖí&׫ýÐzNúMÕá/+«gøiwµïz_ÿ÷¯ûý/´ºÛýwý.õà ¢n ¶g«÷¦¶ªÛui: MB 6m! êÚÿi/a}¿ ÚþöÚV¿×IÛ×d(ö±2¡ ÛئF: ئ+'¸ bÛ }OäHmñ8ÿa¸¶E¶Ý/küJÒÇ{g¿Jý\¾Ïö BbjIÅÚvÕ!{A«eÝjOç0 6 \¨ÐA°Á8mC0@ØÁÈ°ð°H pàÒ8H0B¨±"8pjaa6 Þ0!eqFÕáí
§l+jÚPÄ$î!'QTWl%¶í°bC1¸O¸k9pQpB Áb;(º &¦ë#£"""#¯é&ÛKÿMí-ÿÞßý_ôô!îÒOíb©Ø¢"M¢ÉZK!L¿:²
²xíOÜ9]ó"¡NŦvTs'«Ù®j3YÆ4ÉFOªÉ»ÇjD°QÎÔe¾¹'¿ü( Áh¼mª£=,<ì(Ì M
OUTðÁ8T]´li Ý$BgyVDYA´ýuýªAÒm4¿UwÃËyL0{öûÿ×E½m'ÿi%ê<®TwRxXýt½ÿßz¦ÒÛÕWê ÿx3½>Ô¹«§VÙ(íìÚ§ ÝêßèÏF¨@#ã]U+Z t
ZG¸A:ªZ Õv)iÒ¿¢#Î 0§ñ 2a3ä0AlCÃ032 ÏA°@ÉÙ§¡Þ¸CõÓ÷[¯Mü!¡þ^®ñÐNªÖé)ò¥;Ã'5¥Îôx¡Z\ï餹P":WK¿Ië´pEáCp6
æÈEãØM]´]jM4Ñ°¦¢°ÁWz]«Wö×ÿ¿úÿû×ÿ®¥çǤºQÚWJJ«¬tyá.\Ó~ëú\_òC pd¹¤2<¥ñ5Òpȶ&Aé½ÚH7Aѹ'±4PMÅ&'¶ª7¥ÿ«ëýÃëý?\_ëµ CÞ»Ó/Pø:ÒXþòáÇôºÕ(ªÒX¤iwZºú 8hztiºe¤1m&ØL&ý&÷ªÒëiÞÛ6°@Ë8Nl. ïñûû!Ö?:ëøÁþ"6¾zªáRZIE}i$¾5ûuéa}#gõ¥½:ý?Õ:ëöªûï '§¦ªÝ+¿ý¿×¥þÃnò: DoþùþúY@tXJñøxI% uIi$J®½×Õ&ÿë¾¾þ?½UØ»ëÿwëûÐÿÒòÄÿnÒeòÄP/ÿ( úþL»ô ½ ¯_¥ªRº[®ºW¤×ÿþ·ü?õ^Õºúÿÿ¯ûëÕ´¿Áدýÿ^ÕûåY|WñJ©im¡ ë
WÄ?Ý~R\_ûëÿãý&:n£þûýõñõäázúÿÿüÔa
ýÝýþµZá´¸\º¯¯à«éWJj m/\_Òô·ÿÞ¿uýtì°½/þ¿®¯úé_¶¿ï5_õÿþ¾¯÷5§éÝ(-/Ò
úÒ¶ª×±_i^þ¿ýÒýOöäÆt¤W9 "WÿþëùªK¿ÿëü÷¥~ßý뾺Z[ÝúÒQIz]aôªé^i~Ú¥é/ÿÿá¿kÔ7[|°m_ÿØa/ØC¯ûÖßÓÿôÿUí{µ¿K[ë«Ø_úÿê¾ÒÒ õÚý-$ÿÿáÿ÷:éY&I?ÿìB¿ÿí.ÖÒ¾ÚOÿ«¿o°¿k¶f¬-/lÿúÖ ii//þ¼0¼§JýoÚDú¾hÒèÑÿýýWÿáïoÃÿ^ÿßáuaÕOõûIcÃ={ å±8ûS´Ø0N'w¢¶-¤µõúEæ>³?õ³5K¥ ¢ Ý%ëmPr Û¥D©íWÒþýOnþ®k]/]¯£ ~ÿ 0BcAþAÁlP pØ @áC0@Ú6!0¡ TVélÿÚIù}RÖÕ%#Ö®átm$¯¸ïVÖÒÿíÿ¥ÿï÷±Wuÿÿ¯ÿu¶!PaÚkTþa
Ú©c ¥º8R²¨
¥Û m¥¬TU¥°°ÁÝ]-¥ ÁÔ=[ {¯ý
ýoÛí}¿¿µïn¿ý½×°ÚA
ôx4""""Å1]Å%HBb¸LBb´Ø¾Å\ua«kblRÓÓØ}§È.LzþÒ³×M´¬öÒôArdLÛ ÷Ýx#B
¦Ú¤áSL\ \6$0DDDA68mAAý 0a&0±ÃXÃA; pð¶4ÛJÿKðÁBb""""""(ÐÄqݵ[m ¶a; $é
Óa4ì1IÃNÚ¤µ!7JDDDDp &W&W¡ !· !fµ)Ê!e$" 0ÁWõúM¤éUQ³±ô¿ÒtÕZOý×Ó{¥]^¿I×÷ñ×ûö¶k]}jÖ) &Úô¼0¤aÚZ[Ý-XF@hFL%m+HDEA+a&H0Cb0B%¹¡3´wã-Õ¨UÔ·îáî4Ç"©c;³#YÔÍDgÃw£-c¬t2&\83@ÀAX ÍиP as%"23²%gblÙÝåqkP°ð8Th0Tk~T]êexR64$ÊAëúê uOI·°Té T i0ÛOÕGÚúÿéA7h¼\O¢òIæNE½&Òt¶ï 8iWÿõ£][á|÷Û´Òº&ýþßýÓ¬&Ô«ÿÿÕ-:cýj£¤þ»~¿ðKÿú£ßÿÿ¯õëøÿþ«îwÙÔÈeÑè§2#'<áLê.l: "á1FP&C#7dã®ûGú·×úIkíÿM.«òº#Ä<0@ì afa0á0AðL Ïa|ac|gLpÏNizûcýx¿øÿ÷ÿûÿÿH§]´]/!¹ ]B.ÞMº FÆ¡4XïL ª/¡P Ú¥£gÿKÈa×ý%WÛK{û£ûÿÁÃÃÒmbg íÏÍ'î'¶ãA8sGO &ÁIÒÿï ûþÈ:|4ç³3f¢$#PdmF¨¥³Yÿ~yÏå fFðÈbý>¿ö<'½tí>û®×]kôum ¯\_½z ÿ¿õUïküZhØÁ}ÐTM3ÃCCEÀauúA\ p-Ù±ª.¸L"n.Lði ú¯«Æëõèw¦Æìkúûzoñ×üRÿÛI úòp×ù;uú°ÿE<0lxh4ÄÃFÆÄIA6Û«ãA:ÅoT0
è¸q»ÿ«VØ×ý/ýíÿïö¿Áþ\_Ö©z^ü Ý]Øa+û°Aëú d}§ :NwÝ),i7ñÓƯ#FáIs¦ÒxAÁ<«ýÜÿös_!è=29
òL_ÿøßds!v.¿ûKÿKñtµcFwþµnNýôÓ{lÙ;êêúßAÕÁa:\$¬t¿Û¦Kÿã³!Aþ>#ðVBñÄ}ÿ÷Òðkÿ¤é/û¤ÿÿ"=ÿ[UÐî½cÐÐõi}?\_~ëÐïõ¯ãÖÖßÓ »ôAÇò\_Jíõ"zG8{ÿú òÆvÙp\_½ª\_þý¥¸kPÿÿ\_ú¯ÿ\_íuÞ¾¾õÿþÇêÿ¯åaºù1b] ßÉÔXúþÃ~êC!}Pº¯ûK½úQ¤ÿò+ý´¿KíkzëT×ß뾿à×ÿÿò¬ßïÿVýú^¿Ñ¨¿Á´$Cï×ÿkú¤½ÿñ×ÿzã_[ûµß¿ ô?þÁÁÿÿÖ¿ÿ¯nÿÚÕ/èÁ´¯úÒ×õÿÈt}/#zýN½WðDuÿúL"(ÿë_óZwÿïüú}'_§Oïÿ¿Õi~_êô«uëòÆvÿÑ}þ]_úîXÍ<´ÿëäÄèÚýüÞ÷÷¿éÝi]õ®·þÞtJÎÌþþ«ú¯Ó÷_¯ýKïccë]ÿÜAÕ¯÷ý~߯ôÛ®íÖºVîú}õm+þªAoê¼%Õþún¿K_þ¿¿ÿ¯ü:ÿ»_Û3VºßÙÛ3_a/µ³ýþýÕwȸé?ÒïZÒuÝo D>Òé_'ó7_ûÿÿ¾ßë½þû_ûuÿxwß 'Õ¥"^Ø©ÁîwvB +Û»NÛ 13» lPOAà «t¶i"rx×¾Òcíª^ûKýï|WÿûÚïÓ¯ïýþ@×ÿì(à B¢60@Ã4 UE0jAÛP ðB÷÷M ºM¤½¶½4I´ÿu¿úÿv¦u{ÛîÚ½·~¿ÛW¾L&ðÁl)´vHA§§GAØA¢|A !>00¯m%Aº±;°a¥ $´Il$Ä$¬ LS\_³5õý
mµMý[¿»µÿ¿µý±(F¼ÕÙ sÀP@Á? CÄ-m𠮪l0k§ñ8ô×rbSúk¸g)Ú°ÎTÚIȶ»ICÿlô"7°¦ËIQ4àÏa0CB `E! pØ @ `Å5ø pج àØ @Øa6Æ£ª0ÃÄ[»ÿa¯ðkkü0O + V× íl4Q'ÄDDD¡BgØ20[kÄDDE'KÕÕ!m.ëIº4¦IC¥¨aB2WiËLPà!ß NÈ1H1N£%êF¶K ÈÞk Ä]DtxË¢èò(Ë2A µa0¸4"."0¦Jq+&ÀC4k`ð@à©£[¼ J ³¿2.f\`¿ôÒ~ I´+Á¿ y`òÉY(Û%?çúW¥¢o7M ½¹Ó{,x:\·ìªS8ÁBÕ{{wìOZÚIíéT>Ìÿ²O2RÊ.¡Ô´ÃþN®=:½i{×þª¿]Áÿÿç뺣ÿKÿZþO(ýúýq÷ßþý®7·ºÿÿ× \í+OæN3ZòF³]×~S¯~éƱÇþøÒ>vµÁ "0ConD;þ»\ ×¾µkæåú¿è¥/ëáh\úÐPª4\_nµ×ï÷w:üȪW:#¢<ÍdFP<è2B8GAóeDh9 S¨Êÿúª6B5ô¯GèUFÊ=ô»ÿÝúõ£Køkù±òq 1f|RDm` Ì0A¢'pa0>A@7 1ùڢÿôwªTÊèuIѪT
ýÒÒ×Þ¨=¿´¿LÌ&< ôÜ Ï (P|/7©£c¸n &Î-ô,(Fê@¸AçaBól(QÐBt)ó'¼ÙÈ2ÈO3Xò|ã#³Ìä2Ê!'!Ñæ´ºÒzñ©æ´¼}vºOKö±¯óv¾/@.á
²Æ|A6ÛH6H=$àÈñ `ÉvFÔ6ºOxaB&ìðÓ<0
8Háj²&B &@ñ .Hg|kúU¨Å.´Ã¿Û¯ô½K7ð¯És\&ÝjaÒ .I¸A×K[jnÓÓIui7ôÕ á=4o ªFæÊ70Ýé^4l^5T.ê@PABz64¤ÿ\_¥o÷õúöB¡:ê£i};«ïÝ{'zk¦û®µéúëiý×ûM °b~`ÄøÒm'¤§VÛKí H:MÛA8;àè àØ<ÙÒ}$´´¾RI%ÿK¯KÚûËäq=ÿ¯±uQèRõÕ÷[¡ïWß[Ò÷û÷%~éè:¾ø3S¤Ú[i2uþ ïÓÞ©ëðKUÙõð¶¿é^ù$èGÖÿ}Ýÿ\_}jþ¯
zþ¿«Ð«ÓtÝ\_¿WÖÛ¥ÓÓ®·Ö¿¯O]7¯Z¡ .\_¿þµô¼{ä+ÿÿ~èïÁuúµþ?øÕ2;ÿÐØ=}tõïcêö)wWýãÿ×·\_øýU×·Ò Ißþ¤ÖCý/×d!þ \_Ûïúÿø"1ÿÿî½ý×ÿد֡¯úþºý}ê°Zôu:T ºýÒßÒë¿wK\_ßa#í¤þ½Ô±¿òÆiÕÿ¾LC®ëÿü
?&7þ-&7ÿÛ®¿: ².Ê¿¥ÿ÷^>þ½.ªþ«ÿmô½ÿ~ëõÿ\_t¿×ü¨kÝöÔ×õ¯ÿýþ Áþ¿ÿ¯?¿kû zú½$F-ðm¾é%ÿýû÷ïú\_íÿÿßðßɹ0ÞX$SýûK¯Á½°ý÷]%>µÑÇë×\_Ú¥M¤Q¥õ¦¥\_ù¨þÒ¾¿\_Ý{ÿ\_õ×Ýýßx n¿\7 ýwûýè×ëH£î«KèõëkmM#ÄtGü\M¤¸K¥ÿÞ¸Øvþ×û¿¥½é^êêÍ\_×zëüèzàúuù&VÿþÿùÔ©BJGóö\_´õðGìõd}&Òm°D7I$íAjþ{}Z×ïúºÒ{ußïþÕÕþÿ«¯ÞëØ{FÿaÝ~ý¿ÿK°êÅEXJ; \Ra(øi6n»ÓDom, `B`Áÿë^|.Þé}·tÝþ ½¶Õ-/÷õ¤ù¿%º ïÜÕ¿ûþ½®¿G¿ªb&+ Z¦!B¨ÅAjlt 1 ÁE$ëõµ§Û"Xù)Z[l3Ü3iH¸Ë¸ßr)q)í¤ÁÃ]×Õ¥ii·ö\_ßßö»ß¯ð¡0¡5\B¨L&$AðÓ´ÏÖLéX аRw§iD§~Äãµb7m$±±°[ pÚ=¨6)¶ÚV×Kw¥iZûß·ý¯«k
°¿00CDDq¡@ÁSPýà ¨t¦('l&Ða{ ðÐka Ú¦ß %W\_lõ{gý´¬{gûJÏÿ±iÖÄ¡ö¶º¥½i
´Ó^Õ0PQ@Á0° ì(¨¦Pa بa&0ñ¦b¡Ó Û!!î¨q;Äï»ITDDûb!BÅE[A¨¦MíÚ AÃi0AÐL5mC¨0ª0B) )v0@Áh0 &©Pa2á2AJ¡
®ÊË &S jL{ $¢8""""¢""""""""""""5\_¡é~º\n¿§a´Ëd¥DK4¨Ë"¨å¤K3QqAdÈ|©äù&Ë(íS;©(N"»MRPå½Dxë#°¢;0:]´"ð6ÑgIwULûùn³þW-I¤ ©Û ÙYýNÄ¿°àòÝPS·\\4ÚI6é6%JÝß;&;Oçd'¥ $®éµ\s %þ.Ãxë¥ö»õá?Åoþ®¦EI¥ý-Ý&Õ%ÿÿ\_î®]ô½ÿë×þÿÅÅWí%í÷"¥AUÌ£jtäHh¢'6F°ÿý/ÿt»ëô¬Ù ì Íd1& "xffr±Ã<3ÃL&fH0½ùȧd,eÎD?ëì4ï£d"ª£d&©£s Ì\ªônZ.çj¨y\D$"B!»ÉÃ$ѸÐ\Ó$ xCÃ3Íè)0 fÁ¦xAAÿb³Ú¿¸tªNÝ óó OtT `é6ÏÐÈò';,Cd!apg §$yÁü"c´Bd&7 "í¢!0¶ ¡4la¸];S"\kÉtKåZÿ]¬ ý|4ÁDÛ%þ°IÁK²QÒôÚ¹Ù@Ó T\©øaQ¹ú70a
A´l,L;AÝ ämRpÄñÁ(¶m8C82Ø3PsFiÐE8!Á¦lgQ ó ÌR@¤
óc:(ó|ý
Ú¿¿úôþîôûoMкßû·½;m]='FÏ»A&ðêߤÜä²M²GK h'JÁ° ]pj "ᢠEÀg pa4A~ù3A¦\jph±á8hÊa8d øÏ àÓ?NG+ÿÇúÄ\_\_éÿë\2]¦Ã,Z¦Ýè7ý2W\_t¥pjjÚØz~äóº¿]]¨ Û6 Æy/ ×EßP ¢á Ð@ØX°qzÄ4ýÜ&ä(ôÿáÿ÷ëÛ\_׿ôêïëÕWú´4ÿÛiUûú»Ö¯ôÞ¾ô«IÒÃ#úM%Í ààÉcª^%Í4´t$ò:åpdºÑ%{úONÞõäDPþB-Þ\_ﮯûÛoëÿë¯×}j
ë¥Óläî·»M¤ôé?AÕvééÒä¢uV 0¦é·éô~ WOÁïì0\_Ý}õýþ¿KÃÿô«KÓô]{ZýcÿãókKÐÐý}×\_]~½jô>/ºþëÓ×Uþþ?þÑ1°çÖû×ÿ \_ßûë²"9 OoãøÛãýÿßÿÿ÷ºþ¿þ×ô5\_cú\_ÿ[¥úíþ ÿh/ÿý[üAýÿð~ ú\_òEkù ÿD×ùb=}þÁ|{×ÿßÿÿx,¿ÿßYäGÈï÷ô¯KÁ÷ó©U{×ÿþïõûíö+ýü±X7RbõßÿÿÙ Oÿÿõé~ºX8ûãëÿB!ÿýWøðëuëêïþÿÿß\_ø/ßÿ¿ÿÿwÿëÖ9ßÿÿ^®ß×ôØZ\_úõúúí¥çÓÿUÕ÷÷ªõçÓÿÚõßçQúÿ¥ÿ¶ëÿþ½ýÿ©1ÿÿ½´¾»þXÀXnXµy1\_×ÿÝ7IwV«÷ÿ}}?~¿¯Úÿ\_ý~ßß¿ûïþ÷ï\_ÿûÊïþ¸×ûôCû÷Þø¿þÍ¥«ûúÞë·^ÚV·X\_m~ÿSûúÿ¥ÿÅ?÷uÿÝûÿëð~¿ÿð[\_ú´°oÿÿ×÷IÒKM°¶2?gÿ´¬Û©®ÚQ(~±8ôö%?·\_þ×]cÛ^Þõ÷ý[ÛÒ»ýßîkO÷\_ÿÌ3×[Ûûûÿõ¿i°Òa¤U°Á(Úi1±¦pÒ0AÃNØH0ØI¨4 $pÚÕ×k¥¤Úû\_íÓvûfÒ²þöáM¥"àg¥þ·Ý½¤ëö¿¤¬}÷ßêÿ½u×rpF)¼1 ¨qTAN(0®ì1M1 Ô1L-Mwm\_ºPÈý
øa; ú°g-Xg-mȸö+iFlvÆÃKb·¯oKJëÛ\_×®¿o\_ný÷´÷ö]Úa I0AÊt `È a0B ¡Á9h!¸iG¸pLNáÄî ôD SLVÄ&aba îÝ¥öf¾Oü5þ÷ýæýªðÌ×Úë"# á)¦¡ÅU°i¦ÐP[ -UaîS¦SB@°`Är¤©°`¶UbqÚ§r û¥NØfýÇÚL3 qì0ºÄãÖöÒ}A! B#ÃLP@ÁÅEBb@àØP8v< a ¸V)
) aÄDDDjA
¸a0¥kp×¾Þ×xaa×4ðö»NÛKТéDDDDDDDDDD^è'í~iWýé¤ô¿¬D´
4¿ÔW´ÛiøJX`
"%Ê;3±FTÊZ
9Ù
å-¹ÊãtðïaÉ»²Á0E}Eàî]BñĤEyÞv Þwëß\_LCÚa ó²¢ á?%=ý|«ý¨ÿÒûÔL§^ý~M+´ka(iWîú]D\qÇ×ùÞGe7ÿ\_¿ïõU,çÈE<?9g²4YæJ e9b!º¯¿ÿË!TOæÆtè!>ÉãHØC3tÌ\;xC!FcÁ(xAh aÿYÙ[UÎ3[ÒÖYxSá0SÃ8?EÃL'¦ utéé0ú`¡á4la460¼y¼öR2#Æu 2¤ë623ì(@ CÏøÊ#ï0Aïéê4l"ûø` ¼£cÑ\_x°h\(Þ¢qÊl1C¾zVêê`é;(rèKªÝ0A% ¢#4DgfdhÍúuÒ¯ÝUü|zú}mz}¦ÿºÒxAÁç#PK¡ü-'m&ØMÓá?¸3tØ3Q;k[²wIÅéë±aº¢ø5Q½/jMÿª×õÚøûm×u߯]j=íÓ[§º·ôûÓ§Õ6ÿëkCZÖ¾Gul1Ó| é6 °bhÍÐÄöôëýÿo¯\\ÿï8õþ¿ïLæ¿ ]zWÓ~4?cN·øûýÿØôú{ÿïøAºmNjÓ.«aj\_þ=ÒÞ126ÿKIëÿÿ?|÷ÿ½.ÿë×õÿ]çëýþµéÖôºw®ÿû×ÿ½>ÁpDþß}¯ÿýrp\_¥þ¿ükûûÿÿ"+ý«ä+þ£ïô4ÿKëÖ=~?òu\_÷I¯t ÿ®´¿í?ÿ/ñ;,ÿÿª1²þßÿàÿêÒIÿ¿×ÿýG¿oýøÿÿ÷è/K×í}ºúÿ¿ \_ÿýRéºÿØÿèÇÿ¾/õî^ÒÐõþ¾+É"2ÿëÖÚ]uéwû«ikÿ÷ÿ~>ïÿøo]Ûÿ~¿~àÿ¶Ô/ÿû¿\_Õ¿ø]v ÿïÿ×ýý+¨¿ýëßµ¿ÿù ~ ëþt/¯¯Á¾ÿÕþºòc?Úë,gJûíëýö»ÿþÇOýÿý¿Kÿø?`ýuáÿûü>þ}}ÿwÖ¿ÿÿA}á½knýzz½}Jý¿´½í¾½wÿßwûüæþ×Ù×þ¿uúÿ}þõÿ\_Y¬~Û[\_½» Ú·ví~Ý|3û Òÿúÿ9÷·Ò×ÿÿÿûµ®Öío×ÿÿ¾øxïaqûM 9DãÖ%=12)1U\_¦ÁÚÛa$í Áݤ«¿ûýõi7\_¯ijß[uµ
Øk~ÙûÛ×v½¯úæÙ"E¨0ìlS Aµ
»0Æ8lP@á±LPb~Ý/ûöÿ[ 6ý\0}ÈØg»Y)5ÄãµÇr$SÚWþýoÿ¾½iÛ ]jí
°R=Vû 0¥L&a{F4¿NÒØ`¶U¤°á¤BL tlpÐTÁ·j@ÚÐ@á±]+Û^ßïð¿o`f a`Á´ABÚÝ´80Á@±Q P@ø "V Å|'PÓµ¸jÃI³ÿil3ÊÒGÛ3þýV"#µ &1 5)C 0« ,ÌÙèU9PW"t @Í!r¶H
1¦ÃI6-´àÁ+ØíS5ú !0Dþ"2ÙF3a 1 Ú SA ¨aÔBDDx@Áa` +)Ðal¡Îè k®"""""""""#¯ëªü·²ºq;)@Ëh©«+
vPpS¹ÊªwÒêòH®Ì C£Ä=ÊäÌ\"`«}ÒÜË"7UÜ<²ò¹.vT¼É'¤T«@àáüÈOLÈÊ B{Au]{ûFB«2,þªñÿÇ\_"øOO¦:ýým4ëúú\?ÿ¯?ßÿS°rÊUÿÿÿëÿûÒ ¿LqP²d3!ÝÐe[ 0¼ÙÌDo3Q¨ú!pßߺþ«ÿá2x H°dìÀ¥6Í8\¡ w ÑaL@ôF POÿ¼öSÅ9Ñ,3©ùÆk,ã¿ëÑ à0L&>4D(Ê-Ø$g 4vѱѰ0\Ѭ0UÜÐ.ØPªvBe0]¬\V8ÍeÉH¦p¶h)8GAJ3o/¯=tõè} ÿûÐÿø:MOàÌðAÒ}^²wIºuÙ;ZTëÿõ¾×Û»H¾aB/h·qapaB60FÀª7A0æalB.Úÿÿÿßén½ÿßÿßî¿ý~¯Ó¥¶·¥´ø«Ó×C·öºN/øûúõÐAá7¤àÈî4z cIºI°Å$ôuãAØÐO;µþ¸ak÷¨àÈ»¯[Ç[ëñÿMÿWÿ·µ×Ó¥uõkþü?öÿV«.¸AéÚuvêFõpÉÎìþ ÇMYÓødoVôC¼ '÷ÇÿªÈpÿÿ÷×ÿÿ߯ô¿õÝÿßÿ÷× ðq¯Q÷þºúI °zþÓªbm¯ü}% É.£v~Òï¡`Átó¾ØHÉ9!p8I°dq¨O0Ä\BÃX¼È¸Î)P3@Îtd " UÄfH:ä³)rA5 `9Ëå;#¦(APÁ$ 8PT
°A¨('qÙn,G®¶4Ö& aSÐ-7 ÁB
Ñvè\.òº3½5S²ÊÃZa ©Ø\#Iià ª H úºi»ª¢ÝhÏéH 9¾v|ÞÑ¿-ó{ 9¿Í£2sFh¢ã8Äë-é\ä&røp÷¼!Òdn¦ØMBéé¸.%Ó|.P¸[ ÖÒéëJ2,YØëÞº\ÒJoãûªK¾8øãïßZéÐO÷jÒiß÷ÿúYÝ×ÿÿékétÚøý¦¤E¬\þºñ}\ÿÿÿºÿUׯþ¸®ÇJµI±Îÿþ«ûÐÓýÚÿûñI%ðCüçþaÃJ¾ÅW¿ëÿõT½i/WKìIiøý%Ñ,dlÿ" ÍdPDM"pÍdDÎÍ dÙG~HÍÆóf|¿ê½zUFËußôÿ¶+î³µ¡IÙH;4zl ax \A ðA$ " gÈ&ªÇFB"R×ýz®T ÖôïûuÝRKA ! "î ?pU» ôljL%T6
ÉѹÅna5
UFæ4çaa(`FAôHUe2|Øüè2áHè¬S2>BÌêN^u2!ÿU (ºzIþoÿ76þÖ«»Å
ÐAá ?Å MÚ Om' Èí¤ØdyIº¨">ÒlíNÃAÏ-Ã0P<3D ìñAA¦320AA¥ê¼Ã¾K+i/ôß𱪽6 di;® ³üIºzwuxAöZ´ðk´ë1àÌ`!d¢h=ØMÄ\Za Åós¤,(PÇi' 'qì0©¤ú ,CiPzÒÿúFr}UÓÓ\_ûÿ%kÒÖß®´ºë§¦é¿úzz®t ô,ã¦Ã#È:MÁæF?I°È¶Ñþ Om¦§æPAÕáI]RÃuW\_ÿ[a ßµZñ±ýZÖ¿oþñuuýÞ
}úêí5tÒ¶o ;´Ø8.ØO· Òw¤¦N º&Ù.úÕV»§ÿýý? I?¿¿÷ÿôWû¯\_×ïýu~õº¿ýôþ¯O]=v¯ÓÓ¿Ói>[¥êÿþ~+í/\_þõ\]ÿ\_#Büÿ»ÿú÷û¯§ªú¯Çõ®
¿ýv%ÒKÿÿÿ¿ÒIYÚÐNþB!ýãÐû¯~iëëÿÿë¿ýúÿúõõÿZÿ®¤ëéiÿÿê½F?ýÃÿTAýÈÇ,«í¤+§ÉòÄ\_úÿÿý?ÿ\~CêÒ+ýkjùÀ©zI}~éÿu½õIh¾õáÛÿäêÁÝzØIÝkýõ¿ÿÿÿÿ\_æïüªûÞ5ô½$²%ÒýZ\_æ¿úôUÿ`ëþï~¾ìR\_\_ÿ¯ï¿ÿÿ¿ÿÿäð\_í´%¯Dת^jI%©t®µûóRó=-}k×ÿ526¯ï·\_ ¿¿¿ÿ®ÿÿÿÿ\/ÿô ûÖLg4^ ª¨&Ým¯fnÌ":mKöûý%K3¿ýÚÿûý@µýÿû\_÷ÿÿ×÷ÿëÿiWÿ¿¡ô¦$ºK´"Ga
öÒõ\$é>I-:Wü÷õ÷ßé}¥o½î¿ÿÿúÿ¿®½õ×ÿõM=Wí-´½XÚa; 'I Øh&ÒÓA}v«ß\_ýûm¶~¶Köï׺[í¿ÿö¿½\_öþ¿ßûö¿ø" 2 5ç\)&( `¨f @D( `b²@IU}{û¯¿öÌÕÙÿÿ\# &þû=|¶¿û\_ÿëÉîÿþÿ\_Ø£W¯ÿ"aÔùõýõaa©K»É»'iÊ2oeÝmr²4Õ5üki2.Vø0VAði´Ã¸b AÐ@ 0Aöõÿ{këºUþÿk{x\/¶oÄDDFDDhDDDDDDDDDGÄDDqi:\\ Á§Ó ð»W ¤ì4©´ I§v Å %½WnþÒM>D"Xgÿl/¶}Wj¼EGìÍ}i=B¿k
5þA 'iUìD¤dCb°A"66( Áaà ᥰApßc;ÝáÆõ¸iÛq°ds½RíDDDOÄM£"""6aiìBAß±VipÚ´¥jjLP áÕ$Ý¥ÎÕ0B"" ÁÖ A &VCL&PáO2Aè0§3q#©PS= a>©SªOÄDDDDDDE¬DDDDDDDFº´¿T©¥úún´«¤¯WÒút¢´-´¬BÒA
W\UaB- J"YK&[´Cb
ɲE¸Ôd)AÃs"ÓLÔ2ÜÔ Ðåp{»
\¿ÓÜÈ·/oæ@ÙMº¹xIààõSµv '¸ÿÕpèêùbÑðpá=ÿ¬ÈK×ÓÿûØ`°aã\_éþ+ÿògçeAwN³°¿ÿýÿöVü/ýõ¯?¯QýuÈB%ò3HÙ.fÕ,D 0èÔy²5 2\d\dçB!£j2 ë8gB:úÿÿúuù¦
0L fdhFxL ÏA 4ÁAA¢"¸MÿDc#h¬".¿ÿ\_¢ñ
T]¶Ò ¡iªz60ÃZaSDÇh¸£sZ6=0¦ "c´lùÃ5DP2ȾpÈy@
Fpe3)DaçÁ"/<þEðdÃ4ÿï8õýPM± HÙ x&Ój'¶aí AÐAÒm$1=´n`Äöê'ºH=á0äfAAÙ ø<<"á$E¸g¨tþ¸a>ð<5YØPÈy±
0!Áç ÍÎEÿ6f³#fDJÍt®²a×ØdnÕÛJ`A°Áë}(Aëa¦ÒupeÑ6O83!n»jp¢ù./Fí§A7¨P .4ma(0¯¨EægÆM{4..ô@xy£4AÙá ïæÆA3çÏf!´UÚ[»]7«éõ«×étÛýc×öÿ¥CAÃH éI¾Gl:NÌÒpÈù¤àÆÉS\5nú Ø1 Ñp' E°A \.ØEÛB!G] éõü Ñ¡¦GøDÜ0<0²þ¾\í¤=é[õ¶ø¾+ëûïÓØYuÝý+¦Ã-XNþÐnJ\ÛM«rWkÓ×Wß¼Õð%ͪmibxh'AÁa«Å C0«Xá Û0I¾záûk¶+]éÿµÿýWµÿ]NÞßÿJúzèk§úëqvÕ!ñ¯·N·P},áV×a] êØ2é¯zI°Éu§i&Òl1ÿ¬tk¯øùÝ¥ñùãò¿ù@`nìWßþÆéë¯ýÿïéxõ×mS×ZWO'Ý¿Ká=tØiµUzvJ?¿N ßõZ\_õེþhÿaþ¿ß¯ÿ¿¦¾Ú\_ÚèïHÿê=Sõ~¿tÿë¯Úúô÷Ýv·O×òa!¿\_òe]´¾X<ý<±"}ÿÑ<·üZòcÿöB+\_ÿÿñ\_ò@\_ë^ßúýúïß\_½Z½Wÿë¤ôô8ÿI ÿê×ÿôö¿ý}×°ÿÿÿ\_¥ê!2Qµo
úÒöëÚ¯²èÇÕ«×ßíÎô¿ÿ¯Ò^׫úÿý/ÿõ]ÿJÁÿrÒvåÿðÝü±¿ÞLlu¤zðDQ÷÷ïÿÇÿñßZßñÿ÷ÿÿú;ºú«}´¿úþ¿ëÕÿÃÿBøoÿðÿþµýá^¾¿Dó^ëú\_ËrÒãGþC¢Ýté/ÿ¥¯¯ö¯kkÿ¯ÅGÕþÚ÷ÿëýN¿ù¨ÿÿ5××WÝ×úíÓÿA}jßë!þXκûòÆ}öëÿÿtÿî¶ö»ûzßÿwýÿß¿ýë{ÿ÷ÿþ=$¾ýwíÿ¥íÿÿÿÿißïKi7kߥõý×a{ûû ÿáûóéþ¿Ï}¤ºé?ÇëLë\_ý.ë¯ÿý´þ¿JÿûKlÍW
þ[ ÿý0Ï÷Ú°Ï-)ÿì.²\¿am/ß[Ûúë÷«Jý£¥îÞªµÛ\_ý¿úÞ·ýÿ×ßÜS¿i[ ¶'viÄ ci¤á±§q;¶-±§'vpôlLáÃKûM~í.»û v¶}ÒþÙµü/ ¯¶ÅZLWöÿÿ±U×þñ½~a IL Ø^h6AÃa Ú Bi °Ó 6Ú °á\S ºöÌöé0ÏM¥ßÚP n¡g.D?ON+ÝÇìD§ì/ë¥û½ý®ß-ïo®µ´¶WXL Ó Ù1ÔùycË &e9Fe! bB" "&DQÞÈ£wLSi AÚQ@°Èå @Úð@ÁÅBÕC ¨á§ØMmV^×!1ZÿhDDDDDDDEÌ!!ªZV#úÿëIWë]^ Þwä©AéÔÈ·R²HAuBÊåçÈê¨Dl(Un¡Í ¯ÌÇfI YºÄi)\§ÿbÉC R`èûx888pp)¤ªvhÖ[²¹\_äçuyÚvS¯hPR¼Æ¿áS]UF:/ÿÇë×U¯ÈKÉ÷¿¾¿uéWi?¿þý?Xõ2'Uÿþª¾¿Ô.uÉEçf¬¬D±÷Yìd óèdÈo:Pÿþ¡. 3Ö`\ØCbãl Éã<0D'0á;P@ÁBð@O0A`3a ¡þ«ÿD'l0¥Ç ö: (EÀ`a0·M#Xa tEÛEØa¹4k ØÒ vD/ &a0PD&Í%MYs5 C"c" æÌÖEB6ffu3ª9ò?ýPA¸ ëÇ°N¨&aÄôaA'@¬Lôh ÚÆxѾ¨ ÜL=aNËÉvhÁ(¡È `00A03$Aȵ2`5 `40r8>vó5 ê#AhÖf¶Uù ;ù\láä|÷ Æ0×Ò¦AÃ&pM«i\乤Ø21é:¶Êº,@Ý82\Ù-v@ਸLp`©
MÚ.áp¶]¢ñ¨ML'¦·aUBa>vXA4@ÌÈA@Á§. äIä^5\Ë2$pdÛ3ÐdrÏtx³©ÓrQ\_
ú²w[i«¦ßKªZ°DxõV4öéà A¾ AÑvѹi¤A é6it 144Òîl2;ÃFæÃ¥6zEã \5°©£]l\võË ÈBTXáA&|"cxa0ô×õý?h0«¬2^öÒl/é:Në@És i\'Ãã4aò`ÇNÒ ÒõÿÒ¤×úú³5¯ÿÿµÿÕ|ÿ¯ºìzuz~ñ±¯uz}õô¯ì^ÇõNÉÝ'§KIé½z¾Ö}:O°Èß»M¯#x3@Ë¥ÿÿöþ?×´¢+ÿÿ¿ãð\_Z¿ìkþ¿W»[ÿýÿKÚ\_ÿOдÚý×í«ëU±~¶¥p[kÓoÿÿ¯\_ûò{è/ÿèÿÈ"8"Qý^ë ßÿȧ"PIG]~¯ÅþC¼è'ÿ×õ}ô¿oÿ¯÷ÿmw}¿ÿÛ®XÞß.½k²Ìå÷Ô±@¿þ½¼0^¿þ Cø×ÿßëó@ÿÿÿ¼0¿ý
ýé~ý\_ûKÕûÿþ¾¿ÿÒûÿ¿ÿþî´¿ÿöè"Oõÿà !Hmÿÿ¯Ë^çôNÿ§þ?ëã¯ïßÐOö£öë¬ÀUß¿ö׿ßÿ¯ÿÿÿßëÿïÛÿÿîX6ÛþÿÿôøA×êBy¿EoÒýjtþºn¾»ÿþÿûi]§ý½}ÿúÿýy ZAzÿù@ÉÉ2$ÿÿ¿ú\_õÿåùuÿå÷ÛKù:ÿ¤¥íÒ¿]×~ßÕÔnÝ/÷ß¾÷]þ¿ÿx=þ÷õÐ8=ÿ¿ÿwÿú\_ÿÿ¯ÿ÷´¿ú
ÿýúÿô;ÿéoÞ]ïÛÒþûuÛû\_î¿w¥÷\_¼s{¹«siûÿÿ^¾¿ÿÿuÞ¿Õwþékÿß×ú\_½¥·ÃKªlÏ¿n¶ÌöV˯lÍ7Ùuí³êêD¶¾ü2>ÿ©íÿûôWûÿúûK÷oû ªÿÿêî¯ßV\ÿßׯþßoý6Ò¶ »Ó`í¥LlO;¸29¶QklvŶØÖØÚw¥õ÷kúé]?¥õ~Ú^õýßßýØ\_û·«ºé}×ýÒôéikév ±@c@è6( ÁÅ; ¨t0Å1M@´Ä&8mX á´ßvÒlÏûM¥i{ õ®ÿir&i³Òü0\_ µºûJߧÿµí{õmíêqá
\xa~XaÇAËL)¡Êô `LØ e`":`è0-ÇÚLXlNø¦Ò´¡¨a(i&à 6Aá±A 8`È@þí¥°ÏölÍ:ÃJØgûIÿ¶Ìý{JÞ+úÛÄDDDGl@b`ÁÔ&) j\0bvÅ' =¯l05i
k0A¶lqPdsV+iuß bø0[J ='""e\ BÚÚ &ì.Xá00 Á0A[xL'lP ØP¨AÄ&\8`)`0@`Á(¦(=§a°Í¼<ÄD! [&±ÃAﺰ\_ Bp×ÖûÕüa:MÓ¤áé°dvé6nÛI°ÈìJÇÕ}pápÉt$$&õ#¶¤ÛlM°ÐaDÐOoºw lZ ½Ba,ZØEÜ A°64/ø¯ÿþÚÿûtºûûu]kõÕ®®~é< íû2,á]Ó¤ºé;'}}éÒt^g ÒoIà °eÔ'¦ô1?0cVÃ#æÇ"ÏÈèGtjH$Ø2;ÈCûª´¿Ü/;´¿ý~Ö¿ÒD(ÿß¾à÷®ëõºõJº~ºé¯\_n±n
ºÇúë¯Ó¯Õ=V´ézO äïVt¹; jÓi:N º&á÷aûûÿê ÿÿõñßÿÿ×È]1\_}ÝûïÿuoªíÿuÖÿÿ¿\_·TêÚ¯MЫ«{xí¥ÓkúÞ½~¬?Ò÷ÿÁ»¤ÿ붿$9Ç2?¶²ptÿaÃøp}k¶µõÿ¯Ö«õ
ÛK õþæþ»ûõþ÷ýzÿÿQßýÿí÷ÿü.ÖÒÿúøB?ýPÿÿä$ÆÈIKÿ;µÿûÿö×$¯Êwúÿ\_ÿn»²> ÿõïköÿUÿÿKôºÿô¾¿÷ÿÿÿ¾¿ü6ÿý/ÿúÿ]w¬ï\ÿÿ×î¿×ׯÄ}^¼~ÿèzú îûõþ½/ÛKõÛKï¯m-»ëü8nXß¾Ú\_ܱõû]éºÐ/¯ÿdÇ ë&ÿÿ®¤ÿWù?ÒþÿM}O§÷\_õÿبÿïÿýô¯ÿ<>ÿí~ºÿþªëAè/ÿüG^ÿÿ{rÄ\_þ÷å[züNüËׯ\_zýÿïÚ®Åjú^Å]Þ¿êçaEd¿ýßÿÿÖÇ¥í¥ÿÿß½þÿëÒýSúõZÿÿî×ü{ï÷zíõëûkzÿÚ·\_þÿm+×þÿõëÿúþÿëëÿÿÿîÿîÿÞé~ïÿý
øe×V
þÕ¯µþÕ¢:´½¦Ía]Ý#²³WOÿ¼~ÿÿߣßþ+\_õµ×ÿ½í¥÷þ¾ºW]ÿÿï®Äîí&?m+bw}ÛÝbqö1)ÛÃ.î!]ì+¾ÐIÞ©þÞÿÝkµ\_ý7ôO®H\_ënºû{{×ûïþøýÖ׫ëÞý/îßý¨qT6 @Ú8
¨ l(0AÔ6" Á±MCbpaG.½/ßÞ»N¿¿í? eð×¥_k½
°·ÿo×ýZûít»Ý+íêÿû®õÚá5á
Ûá¯}¯éð«Ú &¶´ûb½³?í.Ò÷öÈÿȦ颵øa;V'ÄãÛK"÷³þÝoû{kim¤JÿJý¶Ýv÷¡ÄDDDDDDD\DeÊÝ¡ Á(a+I6ÒNö8zi°Á$w§U»A¸AÈ@àÚ¨6[NØöI§Èûá{ mì0g®D¢Ù; ÷ òºGìÿßUB1A6( áÖt´plPL ááIR umZíaV[aÐA´8lPAµ°b´ãcm(ØÓýva0à ßÃM8kh07³>ta4UHÈðhYA"B" C Tá
ÞÒýmp 1H&mR´h; ¯¡ B Á4!
-ÁlC V a[êñÅD\Dkþÿþ$W\_õû]SªIga¤z$¿ëôKêUI®¿Qª ,Ô©\¥½Åhì zô wÕ èb3¢MøAÆM«"¹AKHÑåu©W¨89nËp²8çeÒä~W¬(Ýdd ÖÛäÊú}¸y6êv®UTpº~a×ÿÊê÷Ëq2Ò ÈÆdOÊû£È«ÿiêµ á?#/%¶µÿã×Õq¯\W£#¡çEÙNÍfC#¡Le!vRdtEbSÄÎ#/=ò:_ÿ××ô§dÌà Ê ( Á0@Á8AP!T@Á øCb Áà ßÿúýuèk^1¦0£pjFà//!(¼¨FàÁBxL#cXL&A An "D 63 Õä gb/åB""±³óÙÈ !u9ØQQÇÈ«DS$ªý;ïÁ'YI#é6 JÆtÒ ÐM¤¤'·Û 6íÏͦí¤ÚÄöÒlì©ÁÄ' Í>þx@ ÐBñ çò/4f ià0@Ïñ2ç dt3äv2Bvm! h"¢O^N!Ð!AFÆtÑfc%òÙyÃ5 2"'>JRèË&¤ô&Õ½×õ}[zv º}(AÒlj¡0.EÜ"x\&Ò6ÓP7ÃHØÁRFÆpEÛ¸¢ñª¨"ÍÐ
£pt[´!!ìÝ
FÆvÎ 0D8ð{]
à0»Ó3à 1ÁpaLAH3ä"Ðë×êÝ6¯×Ók_Õ×MÕ8¶ÝV¯î¸bxÛAÐAÒtoèÜà $lOl1>4 &ÒM±¤(Ú6AÐA°phá$ônG'c 6léi/ hx°T?-ª7°:ÿÿéþÃìa¯ÿ_þ¿ý_° ´ÁzOÓé¤l&°Iþl2QKY½:ß²wjpf¢½; °f§A8dîp1àË=- Ð37¼: i6 dg%a\Á&éÿÿKú¶D¡=(DGzþE¿ä1[¯K÷³ ku¿ô¼]¿]¦®«zj½ Ó×M¥ÓïCã«Ç§÷èiµá_ý:O«¥eVúpjn¶nOV{Y÷_ÿûRÁÅA²ÿÿæÿßýöºìÓKkýúú±ö×ÅoþûþÇ×ýÿÖëêì|_úÿ_ñWuÝ/ÿôì6ÃÃoÿè§'J´ßù<+a_µëóìúë´wu®È¦?ïÿ"$÷; ùÆ¬Ú «ÿöú¼=Õ%ýúþ½_b¿ïþ²qhÿÿkô_{ßÖ?¡²É6|}~¯úÇÈu¯ÿ¶ÞvZöBñÿ¯ì=IqÿüöæaW /ÿëS¡) èkÿ÷ú[¥ýÿ"pÃÿJÿ¥&3}þé7¢;(oùcú¦ÿ¢yý°ZäÜã0V¿äÇ&ûÁ{×®¿ý:¯Î3ûú¾æÓÛOÝ_ÿýßúú¿òÁ¿ô tKÿ¾ý(CkÿÿçQWè/ðmúõþÿ ø@e}ýýxKþë«Íï~jÞýÖ²BûÿX¯¿êÿl«/úím'ÿ¯Ò}ò ÿïÿÈÇõJÿ)Í/ß_ÿuò\ ÿ½wÿñÿýûK®ºoúî¯ ûÿ«ûפÿáþõúþ·V;Kßû{o~××áýoïþí{Áý-ÿ¯þ¾ô /ýV_´&Ò×Û3ô¿Òøfkû _iY~Òf¿ÍiúV8«þÒ¾£¼Þݳ?ÿªù¢õw÷þ·þÿýÿªúªiÄC[´¢iCJÒÁ(iC \0ß{«w¿[m(auïû¿ëoýzÚ}ÿﺵÿ·®¯ß÷Ýéokºÿë¿Ò¿o]ø aáÓLU1QQLTTP Ø8t86( Ô6 j¾ßýø]õûI´ïÒ¿m½l/õvÿ~í¿¿î׺úí{ý×nÒúyÇGRÇW¸h4P igï \uð¯ /§ Ô¨0´û=ZR&[=ñÕ0Î?¸h&×Øg-=sÃ9AIÇk"CV ¿»On0m¬ÿÛ"æDË ôÇ"cKJü.Ùþu÷im®""(³ÄDFZôB"""4"""!vá¤àc"Ào±AT0¦Á aA±AM0HVÄ+PØA±LazxMd$ö¤·»¸a5[öº+m ëM8l(¦µm^ì%Ó¯a\Bi[ &a5¨ijÖÚb¡Za NÚ6i« ÅE ¨( a @ãú"@e BE!8à
¢_DJE:'Ķ°""&BHA gxaOÁ¥8A¦ ö°^á°´½$DDG(!eúø¸Ú]%ÿÇû¥¯ØidØ°×Ã:ø¸a_;òl|ÉaDÌÆdµ-æErJ¼0pp¦JÌì ¥ò¡S%£HÄîI¸:vXÉ+Ï2?LíR|)]t0éý<² ÇUýuWßþÿ¿×þifSÈ^dµ|Ò̼îµÿT¿úí.¤QQMefë"Úí¥ÿ|~¼}¸©®Nf³;æF^ku¬òÍH¿)´kõÿÿ×Ò¬0 ÁL%2s.eÁ?vHR3 NÁdñ¶mÑ £Á5!O00B "Oa?¯ÿõàªêB.Ú3Áül B/(A!Ç
M3b" ¶\.Ú!nal&v.áéÉÃD$̦^jGvÿ!ÿa8;Æl ðl#} Øb{{ÄðÁäm$a¢hh&/6t ÃS²± N)<È îÌ è!Fmñ±I#>!ÁS$#bf!ÁIq Ä' Q³ ΣflÀȦlRFaHÎ è)6ò)6YÔ~yÌ¢§æ¦DþJ)6¾Õ´YÕ¤õmdxx@Û]aKË8'Vè:WX2Á§dö\xi
· ØÕ!1 ÑnÏaxiB©¸N)Æ>aÍB!m\a ¾¡¶^AAPd<4-Ùò LpÏ¢#°>A4tw Ñ°$Ù 0AÆÙñ Úu×þIin®¯ éݯÿëzzÒÕ¿~Ý ÞEå¤7
vèÙÒ Ó»4ÉñAQ±»l6ºÓñxOIP@j'£sB|Áù0°ÁAT4Bñ¬3c pÑpÑ µ8ýãÿ×OWý¯^»«¯®¿ªúA°cAI°bÒl9.tØc JçM#$E¢pb~tÛ¤Óì¡Ë¢páêO¥ ¹¬.'æ iÁùÓÆplº¦ø('wIØØP± ë×þ½}\_þ¿ý{öÿxõrQI×dî©ÒÙ;´ÞJ4ôAÉ]=?IkÉ\\_mÚm]´½NÓrQ[§Iä¢rzØAÁé°e Ú¶1 ì¡Ë¢l8Iºx":xúÿÿ\_ÿÿø28;õ/"#Þ½Ót?õ¯CXô?ûlW§¡Çéûn¿]=:NÓjÝííéuO\_'^¶NÿàÁ½ÿ¯ÿÿõ»þþ¿þÁû^ôºú¿ßOÖÚëþÚ½õ±þÿ§¡úÿ]t¿½ñé½[¡¦÷ÖN¯ÿ$9Ü£ÿÿß×ÿE{¯¹?·÷×ý¯¯[õúÿKÞëÕZzþ¡ôhðÿû×Oûú¿Z¿^¿Zö;ýÐÿþ¾éÿÉûõ¿oë\_®:ûó@ûïæaIb·$þ"6ìïúþ¿é:×ûöG þׯk«÷þþ¿ëí/ïõíoy ¾÷ÿ®¿¯ðDoûIÁ¿ùxDïük¯Ãõý§\_ÿþÿþ?ÿ¢¤-?ÿïþµÿÿßøoÒïï#,fý½ÿÿé¹c\ÿ¡Ãÿß÷¯ÿ§ÈtùFAÃûÿû¿éúÿÿûí½è~ÿÿÿ¯Ð^¥ÿ ¶ÒÿKþÓù¨~ýwõuÿÿÿËÿYb2Ë¿Ûõÿ]ûm}ï´ëéîûþµÿÿ¿þ«ÿúý}ÿ¯ý~ÿý>ïÿÿûÿÿþíØ[\_mµNÒ¾½=tµÛßþÛׯ÷ÿ¾ÿïõÿK¿JK¿]}Ñýýÿ×÷\_^´¿ÿþÁý³ûiX\_³üFÙm[m}²ëí+ eû3V]ZPÈëÛ3M·][iwö¿ÿ÷¥þÇëú±Ûÿ¿ÚMüÓþô»]/ÿ{÷ïÿïÍiøpÁXvÄôíb®6Òd ;n\'w -´¢ÓN A9¬,¥}~Ý¿û}®¿[Ö¯wøUþ¿õkõ÷K÷\_¿ßÿÿ~ø ÃØP 0Å¡Ó8t á5 ¦)¦) §bSLP m¥°Ý\_ýö°½½ÚÃK Ó¶f»µ×lÏVýÿ·imöuýÕþÒºÛÿ]½¿Ê#²´GA \i{8ú &4 aA÷ I¹¡4A kv0L&h0½v'Ùé6Ë3m¸0ÙùP{b ÜSïØaRor$0ð m(0uÈn¾m+.½³5þöÞÝÕ¨"""""""""""""""""""ÃAÅ Ú Ã§bB :cbj¡ 86(%pj(0A â0@Ý $ãm(Ó¸Ú÷"fÚÁµ1á·°M
m^Jjí§i¨A
nA
´_зíBíá&Ä(AM bAÁ´Å40Ä Ãv!¤)¯ 8Y÷åz¡PahDDDDDF""#5¦HX BB iG@8":s¾g(|ì0
ÓXñ¾ºïZëÿúëékÝ=uøAõ¥\_Îüy)ñ ÐA¥ ¦8I°¨AÄe¤¶3²Ò+oR¹� (,=Ë"A¹ÚÀÈÔ~2YÍ=çòpäß0RSLà2Ô(&l¬·Wy¬ìo;)ôöN~igeìø!¨R PRZÉIò2ü/øUðp~á8D± [ia0 "!�þ¿ýòuGt5ZÒaÁá¨@®M;ûÅÿð¨Z é×¢1Ú ´´H0ÓFæ×Çü|qúOé=s|%8×ðIåå&é·ÿ¯ÿZª®ôÝUú¥]Òu×6GR\¸Ðd2²!3¢7èÐd4sÏ¢¹æk2Éf¶AÞ{"×z_]K®©/_jÜAæA ¦M|r ä%Ex ÁA
3æ40AÙÔ!üøð@Áûê´´=W^?õôëÖM ¨ØÓ¦Q¸0¨0ÌkÓFÆ1á"mPa4Ða¡¾FÆd³ùª&DBò5öj¯&"¡9©U¢2!#Yjô@Ë£ñ¢'ÉÑäf¤i.:ó»¯KJÿßäĬa')'Fî:¤ØbVRlöXÚM éÔ$ Oyù'ºµO ÙÙdld1 JN g3FR°@Á0BÏ4eIà@Ó( gù @Ï40MjðPMA4gÚ"Yhk6GãHÎ9ÌÞßO]m!ÿæ䡤Ý7´Û¢´]2èëé6é°eÍl&÷a6 ±øAÒlì\apÁSEÜ"á¢á]¶» A¦acBFÆ D>6BAU5A^0UÚÞpZH$.ªcýB¨U ùÝ,ëïý~µé¥ëá= Ý== :ô뮽oOMÓ»}Uév& AMa Ã;FÊ@Û«¤ÜLô 48;à O ±× u£ET.¡Rôo¤x
ª«Iý ×wÿÇÇúzµºúÿ÷«êëµÓé÷%{I°Áî¯Nê'ì;IÒ )6 m9tONÃ6 ¥ÈÇ\}ø§=s½Qß. ¥Öw¿¼ Ý[]WXøÉÄ÷¥þ?t¿þ5ÈbþóýzuéñôÅ}Wý&ÇúzÿW±úzn®ýÔVUÿZ¥´±0ù»Ú]:ñl:¿ýÕÔÿþ¿ÿ\_ú|Ð?ökÿßýýû××þ¿þ·[ýÿú¿Õ/¯^\ê½bº o^å5Y0,rW¿ÿäÆvÒësÿ#r´á¿ÿ×þÿðz°pÖÿÿÁ\_÷þ¶EGýÿÿ¥I%ÕuéV]%I ¿µëJ¡6 &?ûþõÿ×ÿ}üüû¥ßñü
ñÈQÒ÷È0¾¯ÿú¯þßývÿýR×õRÝ÷¯ÿttÈÿi÷K¯ïëý\_ûÿõÿ½Òÿú]tß ¿ðßÿÿßÞßÿÿõõÿJª¿ªúªÒ§û]uLº é§ý~¿ût¾ï~þûý"c8y1 /JX®ýzÿõï×Òëõÿ×¥Z¯¯é$½iwï\_¡·ºBºÿ}¯ûõ×Þ÷o[ÿÿÿïä1¹V7oÿôýÿÿ×\Éÿÿÿ¤Kפ߯êÒ\_öþ©iôÿ¯¤·é7ìgOÿµ»ýkûwÁðKÿëý¿÷ÃÿZWþëÿGÿ¥IV½\O^Çùºëtûû4[ví't+ýûm½wöýýÒêüò|×½·ßþ½íÿ½\_Þjß\_÷Jûßõ¥é%¥uúèî¿Kë¼ý\z3°Û öÄãÖ $¶Imò%®Ã90Îvaqéò"}}¥®¿ßñ\_ÿëÿ\_~þQlê´µIR¯ô½/¤^´¯/Öþ½þ[Ú C8 % @é¬0bØa\b AÃi¬ pØa+¦ÿÛ齿ް¶¶½«u^ÚZX[JÒ뻶J\_×I2?ùªO6uî³ KK0©T.k¿jOýZF¥«^VL&øV¨Am\{M(4KµµºØb¡ úûI³þ¶i³5~¶Ã Ä¡ìN;äKÚWÈÛáXîJ Kd\Øa+XôTiq¶°ê¬Tkåô1ú]l5Ou}QÈ ¡X ZC©DDJ1a@B""""" TwpÖ8a&=8q;¶¡°¡ AÃx0A Á&!CQL0 b
¨]%°Á&Ò«AU|V¤ÃItqýÝZZa&ÒÇ8q5vvNÚ¶«M¤vÐb¢W 5a5TÕk
± \UBøK°Á( @Ú6<¨Ã 0I(IµþXé
Ï4 Ê¡ÑF"!!ODB L0C,¹dÌ8B-A&aBaU0h0 ¦ªªªbW¨{íb¼LWøDDDDDDDDDGaDA ªô¿Ð¤"?ÿKô¿¤¯þ¼ï¤¿ëüA4áaM §)ÔDDe¦Qá¢l:¡Ù3 ËHS¿Ü¡Jà±/®ÁîÃJÔ®xîÜíaTU7R_îôÚ>ÊH¥È¦pSXg2 \ìk ã½zdhE Q Î&2B5¡iQ#Ñ¿ó¾ó½g;TGeZM¥ïÜ4vA &¡0Ï�î"Ô. ÖÓKª
¨6ðÁ ~½Uô]°Ò»tr(äc<=¦ãþþÿ´ça]~iùöÛÕ'H ÞsÌ!¯_[jÚ÷¶ ÓJM:zêé½¹¦ûä'þ«¯õ×ûb8õºnÕ]-&uÓë´Ö»û¿Mö¿¿ú}ÿHãÿtúÛñNÖ#A äJ3CÍÇóbôO3P«#¨É1|ÙD¸¤çAÃ:.õ!ÏYÇÿû|ïÿN«ÿ~¾¿ãþv¨0»33ÄwdqA
"d "cAò = 324 ¨°D=v¯ßþ«}+®Ò]Ò¿ÿA(ÜõP¸aÁ8´nh6a¤707©¢ÇhC^ÎÌ!nô¤6Aæ(" ÈÃ]LÑÈÐgDx¤CG,Ó1öæh eÃ5 è2å S¢6Tä|:]ÿ\_äA&Jø=8H:NÌ v¨2Rl1:A´X1>4GZmÒoíp@Í !g øÐ
tpA ¶GÁ ³fÆ3cDGa|h5°¨´¥º ¯iWþ½à3HØÉdñ¦msdHÄ óqÈ\Ùÿ¤û´Ü½ºÚ¬0t¬C µméX2èû´àÍM=¤á²ßXL\$á5§Ñ£¤B/T/¨BÃ5ê Ñ|Ñ»ôoqtè»ôêV+×ú§.ØE¸©[°SÃ8J7 ±3ÃLÔ0EÚÖéè}zëë¦Õýþ
/w^±é¿Nµ½Ûù²Å; FËÛ ú41+(&ô×H8b|¤ÚNH:MG:Uè$-¯®õ_H6ôÕ
Ñ»¨Eáé}ý{ý×ÿí¯ß_Ö+ÿ¿_ãï®Óa¦ÃNo0(dVÐmpÉDëßÚn¯ÜÒ~ÚoºÏZµI×uýÖªôI°d±î k²9ä:NºOãø5ÓZëûßÿÿÿïþÿêþútðõÓu}un½º}×ÿ§Iÿõn~¯U× õ¨jýZ¼ õû't¡6þÉG©s Á²\
Oã¿ô¿ëø÷ý_ãè'û!þö´>:ãïU tÝpëû~ñµënÖéi¯ñz½_Øýôþô:õÖÃú¯Ûÿ×úÿÿÝRìû×Óuúÿ«î¿¿×{jð«ßÿNÒÿFÆÿD}%ÿõßÿÿc_íÞùb;ú_¶ü°/üÏüÍO¯øoÿ´³ ¿ä+_¯øþÒþB_Oÿ¡kãK×IÿÚIRúÓKäTÿàÿ¬ÿöÿدÿ_¯ýõ«ù¬_¯'¶ÿÿú÷ÿaÿ°v¿×ý4¾$WAÿÇI/î?úCÿd!¯ù©ýô¿þþÿöÿ¯Û]þÁ÷þÒÑyõ'}ûÿùb=¥¥·Wí×_z1sKÓÛëéRûjC£°ëä'ö·_Ñô÷¿þóÿÒÿÛÿú÷ÿVýýaëÿ¯ÿðnëÁ¿¿®¯éi}-~«IËÃwòÃ¥ý®ëv±þï÷ö½þ¿æ÷þºû®¿÷ÿí¥¥O\_!ýÿ½ý´¼¤?¿éRÿü?ÿðb¿ßµô¾¿\_íëÿ¾¶þ××úëØ/ÒëÞ×ÿ[ÿü?WAÿêõ¿¥¥ï^¾´ýø?ÿø?°º¾Òa°ÌÕ¥ïýÙ
íí¥~Ù¢ÛmnÌÖíôµýÿo×÷ýë§ûô³Ë÷ÍÿîµÓ·ûJ׿¯IR¯WóÉÿóý>éSI¡°Áb®ëîAqö Õqì-Ã76d\ n'ȹ´¶Ð^Òzñ^¿köºWÛ¿Ò´þëÖüé9Lº#ät]Zî¯úI.f»«õõ»ÿ¿Òû! Å4b¡7 8vÐ hÄ,plP llU1@ ê á÷Åïký¯Ú½ZMßé 'ÿ½¿ý»KêD[a~'ÒÑ'Aíÿý®ßÿúþ 0è4Ó þíµ´A¦L& puà áSW´ívØw¥ $Ã9TDñ{J (i,$õ¸g9°ØêÚMý¤¯ºL3ÿºV»fjËÿÿ§¯Í°£R""""""""""8¨fÙx@g xA±;°60 %¢N AÁ¢0I @b@ÁÆý êÔ 0a+'ZÖ8a&4íÓ#wià ]È1×3Ý 7V¦éµ jÓM ÒÕA\T&!1 Ú°»ða%,·µ$9Ý5ÛÄ'°A/á0Å0 ئ ÚJÒßÁÐDIñ ZC)DO¹iñ5ËNUI,iè ÌðX~oõU»i"8Ð#ÄlGjDGýÒK]+ÐIÒô®¿êÚÓÿ¶¾½%Òí'I/êÝWVÚM×¥µµ \Rþét(0Ä"Юwú]ºtP,4¨8@±L)à¡Ð0B"1,3»Æã¿ÄRÔXAI""ã"Ù00»ÐOORn´v²ÕÃàö\eòã\a c-"P
%)\)Ø,o¿úC"h Æõ&\,,ÙPg\_¦Md ZÓ}¬©¸@áö0ÁØhàéæj=>¡ÐX@HÃÚxA´v¹Ø'ú£²ÿ]¿H9ÖîO§ ´v5¼þôþºZ¤ÿ·NñU í/ÿþÿú[\_ÿR¸y±Å!RAP2tH 6flêfò=YÃ:y 3[É%JCy«µÒý~µw®¿+Ìh00 Ì Ï d hZah3Ä£0C Ùr.Á 2ñ·8ùßЯ\_}¾ýMÚ7QºÍ¨Ù æ©¡h0 k l !Ú¦ vDÞ 5L"ÍtBá°0P»ý¯Ö¥mkeM|áÉÌ1« JÆi>é6¤: °ÄöÃ#±=áoQ=´ A°Ä«ntãc¦Úiéà OH76X;ÿXl:ßëÖ×]ÿÿíÿÝ/|©tÿÒw0,R´Ý&«§§ÓN»'uëëðjÒÁ4òQzh8d¢¶¡Òl>Û6:+¦Ã%Ý÷Bðä$ÈJ?ûò¾?#ëÿÿ" ÷ûé/ý{]jü\uõµöÕë¯þÝ 7j"öø¿öÐý=:ß\_é=;«ryÞÿa0l:ÿ¯> ¯¯üÿ×ìõºß÷bl¿^5øôôõ×ýú××Ø~>ºÕu¥WC½¯c¯ýxO׶ܱ»ÿýËuò~MÊ×è¾»Köÿu¯Ô_ÿ¯þ¿¦½þþøøø2498/¾¾ß¿[×î½^·ÿÁ°|_ÿôýGö«öÈè%øuÿu¯½Vþõ·öé¿U ý/]!ÔY I '¯þ¿¯û"#¿ÿ¯¯ ²LÁÿ¯úí/ÿûï E~uÿuÜ_ÿÿíÈ ÿß&1I¹C°D·#ÿí×þÚ\ÿþR« á÷~ûï×[½ÿÿö]_{¯uÿ ÞÝÒþýI÷ÿº m Þ?ÿú]øoÿÿJëõ©1îjTc}Òÿþ+û×î×Ðóÿëöíàûkÿ¿ÖÝÿU¾ÿïK& P_ýûÿWéßÿÿþ×õõ}uÖ¯¤Ý¯¿·ÿ¿½û²íÿ¿÷·Û[Vÿõývá¿ÿÿ´¿öüÔÿúÿ %¤Ú^Ù¿JÞ¯û°µiv¾?l/¯ý.Ýk\_¾fköÒû[~»ÿþöùäô¤Fýïÿ¥¿¥¿×ÕýJÒ´a;JÕm.Óø 0fÛ»a¸"?l¶Äã½9ûÈÕ$ÛKß½¾û~ÖõÞ¶ö¶þ¦}ëßî«ëõ±õ}1翵ýëÿÅ!Å1HTB0
¨Å5 86)
0°Áµ , Û °¦Ñúàým3JÿlêÛ:+JûÿAÿo¿ð½&Úö·kÖ½½}Úëü±Ða0¡0Ma
ÿµò¡jà ×Ûkµ©6!'¿ pÃ$WöÜ2ÿjØTþÒo·LN;lþ¿/°Î[g´£ù,2?alºn·°ôÿKkõ×azÄDDDDDDD\DÖX+aq3j Á ÒlNîØ0JD±8ãqÛ¿NÛJAñ±¶¨laÓ´ï´plP la4JB;xÜq0î6Òº[iWi[i.im÷¤"! ãBài
LkaL5ý0˜M0ƒ4DK"šˆé–8">">ÔDA˜H+aƒ°ƒL,4vƒ ¨XbŸ¥ $õ RµißTê"""""#ˆˆˆˆˆˆˆˆˆˆˆˆˆŒDDDDDDDD^Ya•LDA‚Á B~«""=t½/]E/ëiA\^[¼“iE|'à U-LB…\DFMÉGè'¤D²Ó;#;RÊéµ †w2;\u;ÊáÊR¨pêÃÙhRŽÄ;½¾êdXd#2È8¨ÉO2:$òRl§fF�¤ÀS1J¦j3[:W2ÎƲ|×äÊë!2¬ë¯0UUS ˆH!˜A‚p}> Á¦’Jd©k†µUkVa´\0$ ‹‡¨Tk mØ2~Æ@à¡„NAõ[\_Ç]GQM-ØA$ ‚ Ö M ˜i7MïX\O:rº—é))\_õÿÍЂM&ÒIô›–Š¢Þ“i:MïÚòßV•ÍøV’ì&vSÿü'KÝ%…¥p©½ëWßzÇI½é¾42¾¢Moÿý[I442õœxéZõý~—úªO @Σ)"03$FÅÉÄó:@³löH6?!D
',y³!¢G gDpú©ßé%«ôýÿý¯«ßOÕg\6ð<4ÐgvL eA#" e4B!83ãAÈ Ç àÏDgÆc¤¿×þÿëþõÿÎò^ýAݶŮ6øEæ,& ¢î,ÒÕB¤|7ê¿c¥ÿiGC"/S©vð@Î 42|ÙYFOf¡¤^u2")òÒÛJK®ÄüúnE¡.i6Áé©,r\I6IÛ"Ìaäw4j' JÆ5êõ\_¤ýÿ¨A ·@ÿßj"C³ã<3ÄÄh6«à<$Èi"(! <Á"l¨gDpÕ۪駧ÒÁªt¶Nú°P ]-v^mðeÕ´æC꿯ê¿ÞÃK} °Õ?ÑxþÕT\4l¢îÕT& ªvBnÕBñpÁ0 A¢ 8(A¦>H0&afAÛ¤.×Ózý{пïý¯OOÐôõúMÓ®©¿ÌßI±T¸T±;¾Õõûá ::MC§ 'WPn4llXMb.áÎ-4-Q¸5OØõkiÿ÷[×ÿéèVÿÿÿÿ ÿïªý\_[ }é[öö«·NtÓ8eL½úÃ'4Ø2XéÃ%Ùè2< ÓaØ2;à 8b{A\_ÿúßÿ×÷¯¯ÿÕüh]\_¡kUí¨aÓðUÓÕß
a¬t¿«{Jß«\_¶«§i´&á4ì&zmèmB{ýGü×ÿ¯ÿÿÿþUþG/ÿuéûI\_ؤë£ÿOõU¤þþ=~ú¾×µÓÁÓIþ¿Áÿúõ\_ÿÿÿÿó!Ê«øMÖÄÃÒü1hwíÿï×[!£[JiÿÛ¿½zýýGëzÞ¿ÿýËTü±~ÚOÿÿÿÿÿûÓ§ý?½/dp ëðËáµûØZM¥; ®þõÿÒä1uÿëÿõÓAõþ½ñ__ÿÿÿÿûüþI5 ÁA÷ÛÿzúÆ"7ÿ-ÿÿÄ~ ¿ý-ÿÿû¯ÿõûõÿÿþ¿ÿòOÓ<?ïbDcïûh#úßzîÞýþ¿í$kß!Ü/ÿÿÿÿÑÿ¾½ÿwÌ3ÿß_ÿ_÷ÿóÿzÖ¥}väÆ~ëÿ]_ÿ}¯ì0?,G@¾×ôÝúÿÿ´×¯ ÿýRûÞ×ÿû^¾F9Cým¯ý+ë_ûþÿÖ:]íÿ§ÿV)oý/KÓ^ÿÿôµþÒ¶ú·¿ýßmÿþý¯¸!ýÄïI"é[\z^û¯_ß©ÿÒöõ÷v¿ïöÌú» ßa.ÌÒW½SûÝ&úÿí+ võéôÿoN¾¿¥ïíÞ»\_õ}~íý?}^}é¾ÿúßýVÿ¿ö \·íx0V|a÷ ¿ù!ÚPÎm¤"gD½¥ÿôßפH´¯¿oë4]Öho ¤»\_µ´½~Õ÷ÿý~ïÿéëØ m0Á88 b¶A±aBÓLS $pn !76@¥}zW¾At-/nÒÐ{k קÑö
mõk÷ú§Wwú÷¿¶õÿö¾[°¿áa
MxajÚí4A
{ ¹P ØV«½Òu[ot½$¥JÒoÛ³ý½®µ¬ ¸{ba(¿3Ü)^EÀÏÿazµû´¿ýëÄDDDeÄDDf¤DDC%3HHCI´vÚM¥ B $Ú¶MÇöH~F 6\(666)@á¸A8x@bvħ§l¡§°f&ÒûO»Kú') ئ\Á¨ é0A1 aª>·¤ á¦AL+[i4íõM¦8u áàÅ ÁA°@ttAʵ6XM<¤ CéF0Lü ÁKp
¦"£!0FÃB!RîNÓk 'è0 ßÛ B½kèDDDGDFbh#¥é7ékÒô¾ãÝi´.º[i68ý6 ¹\_´ 1ý´\_°Á!1 2ÒJ#±vW,v7T)ØÖv7!±>{å§Êáü89n0BÐ.RÈMqKr ßþgcy,h,¯6F P¸P¯ÖB^Çm4é¹ßýëè í ü?þ¿êÓ~[
ÿ¢)ÙWÿïþþD4ªÌ}ÿíÿuñ ç ÔyV~{)ÈïFu2"7WGnºïëÿö\_ v ÉãùñN30¡F0@Á'2p~3 ÌÄ<31s!Q RB9bÿÿüí0D.¥ãCݱ$þQw»© MÈ\ !t ü"Ü=0&áLÖ0EÏ 'üíN\×\Oyú©ø)7¤LttlLaïa|üg4D/fa0AÁ|&DAL ã t] ¡2av| Ï fdibÏ Ö\_BLsYæE Löj{$F©éºWêÖá?kýtÓëC Áªõzíè6lmB¨MSý
´ ÐEã&1èÜÕAT'MÐEÅnoYèaªi^=0Q4^2 aQ1ÃT[³ð AL Ï 6-ñî¯ßñÖ¼Wÿÿëm\ºý}ëúvA¡±3¾E °ÖI;M$ìsó@Òl8;2-¡£u¦@è Ý' i8:NÒl1=¹º êVh&ã@a áEãFúèØаþ¾×ôëðÿÿ÷ÿöÿÿÆ ða o[ÓÕ·ªl'¤é:»àá0ÓwI:õ¶NôúÛ´Ø3$.ka=6ÐpËIZJÁÒt¤Ø;ÒOV i82;ÿÅÿ׹ݥõÿÿû i&¿ÿÿôP¢:.únúªõ[ û¯WòwvýÕÓt"ý6ôÿ·OZM¿}Z»'uuzûªoy(I6ÐpfÙ[ð[ý=ÿÙ ÃëúïòpÇñ¡×èxB"Á¿®¹uÖÛUVÛ¯¡©\_èzñï+¯¯¿\_Uøõî«O·CÝkT?MÿNÿÿ ¶öpÛúúG\ô«ÿèíL{Óïÿ÷]}~£ÿÿïKÿð¿ý§ï¿ß\_\_^?õøë×þ\_ýrÄ}´°Û, ý¶×´ oI©:ÿ&3ÿú}¤!ÿÿîHK
ïDh'ÿõÿFþ¾¯íÿ\_ÿõ×ÿ~ä1ÿìBÒþ¥ÿÿø·¯ÿ÷·M×@ÿýßúÿÿÿ¨/ïZ\_õ¥! ãÿúîÿß}ºÁÿð׿K·þïý{ÿ¯ÿþ²bRcDkÿþÝ×äê\_'ßÿù1"WþÒoÿë^ôz¯×¯ô¿ñZçwþ3ý×þÿ¿ÿõÿé?OÖ{ÿ¯è×Ðÿ¯ÿû¯
ÿú\_üºòc.Éýþ»«ýü\_ú\_ý¬SwúWÖý?ÿ¿iz\_^¯ÿúöß/¿ÿ¿Iÿtþþ÷ïr ]ý°½Õýô×íÿ¯§Ö¯é6ÿ¯þ·¯ÿVí÷¯n¿þµ¿þÿþ¿ëú\_ï×ÿïÝ,ÿ3ÚȺ´#þÙÿ´3]ü3ëó \-0ÏöìÑ}¤ß¯ë¥\_ñ±W¤ýoÿëÅZ÷\_]ÿý÷\_ø¨¯ÿ\_tíßm¤ÇvÁÞÕÚÁ¦´ m:#¨Û°Áb "nú×®³EönÇvuíoßWooÕëÚ¯÷zê¿¿õwÞÇfûþ LSLPA áqÔ:b©bÅkk¥aM§í×ßáu¶ÖÖõÖ×°·µ[ õýíý«öõö¾Û}·î "! Ð0è00ß'Dï&= ¦Lãö÷ m0· )PapWiZà 6û]ÒÚa©û ßVþ6{ åmÄãöÒµäPdL å¶Æ¼M?·TÙvf½³E¥ï¦"""""""""""""t´)Çq )iSNÒ®)Ø 1;È&bw á8láíA±V(6=0Ãê0A äG áÿvÚI°aX3 ´ÁÚVÃ=®È¦å>0L(t}î¦
m!5IÝ°ºkaC axaí4Wµh;iª6!Å1LìTHla«æE£i|±Á3:i&ð+᪠R "t a0DEȺ¢"83ì anËyFN@ж °[j¼DFÄDDDDGÄxýkI%ô¯Ö½\_î¥Ý/^¾Ã ~¥¦!÷Ij$´Y:B7 ëZcj9iÎÔë,Õ,¥\gf¹MrÎ,A2¬Ú± )AÎÎÊäni=OÊ«ÁT¡-=Bül%\:j:aÁ×<puoönï»õ2©~º úþwz\_ïý »çdå;½¯ý¥ÿV¼éÿøÿøÿ\_Uï;\ò$oÅ:õÿþÿýgbÃL&|hO /ÿÿÿúø¼hÙÿÿù#7 CÍ"oÎ2$o"ËïÁÁ+ è:#¤$³«'EáGÎù S3#³ùñÈb#|!¥ Lfb£!æÆj%%ãHÀè ù ið'fl Á ¤H4tZ3Q¤HD:³± ¦ÌáÓia<àÎ &háȳ> ða4NDv;O3ã 3XÎ MUa0Bð(¨TLøÓD#ôDöfAQ 2 ÈF:Vÿ¯Nwø°ÆFæ¾~áa¤k£{in ¡=:-Ø°a"áÅ]øº@Æ #±4gæM Ø7ѱŠ4\_ª,"ñ¨T^B.Ú¨5Úé]Ã#¶ÒAºI¿I¶¶H:M%{«¤è8AAøAÝ>% ÌrQ[m'Ypá¤/ÓJ}$ tòàè'A6'¸Fÿÿÿ«ï=6 ÔØeÑ?îß ïL&éüBõé?Õ<'Kéû?NîðÊOMÓrWiºj'd¢ ÇMÓízºN &ÒïÐþ¯ë_§Õ¿JôÒõºÝowÒñmµQouÞ¿Ktøëþ?Xô>·WCÓþõ´Ý:¿ôýzÿëÿûë¥Óõýøÿ×û¯ú÷ôß_~¶ê¾¿ÿ±\¯û^Nµÿ¿û}~»z^¿ÿúàº0﫯óá?ºù¯¬/þ¯ïÿÃþµõëôúýcÿ_ÿßÿürðW Nõ_ׯ½¿þhô¯ýt?ÿ~ÿ~ïõë÷ÿÒõÿ}[{ð[ÛÁ¿÷~׿÷Jÿÿø"7ûõÿÿ°y&ÿ\_÷þ¿ßÿûí¥ÿü×ôOH¾ÿ¯MÖþ¿ýß´¯þ¹1^Ü6Nÿûÿö¿þ¿ÿý¥ïÿø~½j^ãþ»êý?ÿKÿûëÁºÿõÿ»SKÿûÿÿïÿòLûô¶Õúï¶ÿýïJµïÿõ®ÚÕÿ÷æ¢áÿ÷m°Õ?÷þ·\_oum/¿]ÃýýEE/^¿ý×ßý/¿¥ÿ¯õü>¿ÿzÈDÀfæ%?¿Öû\_ÿ¨¨í^¿Íkÿú}ooÿÝýú~ߺÚZûõ»}êïÿÁ m4Å5¿ÿ¿Ý}ôÖ·ÿwû~\_´¿ÿK®ÿ tµúíuK{^\_µëݦÚþÝÝnºº^ÃWK}¿¾ûJª×Û ÿOm/äL×±(|00½ÖõßÛ¯Þ¾Ú°ÍÉþÃ9HÒîÚOÄãí-=r´¤K°ÍÜJw¼JzìNûî3Xdr plí h6¡Ä&&ÜzvÈøg(½&êUÙ{ Ø l0PA ¦ plTSAÆPßj@àÚ pþ¾¶´ «¦Ö6(:869 áÁ¤Å1;´áÄîK{ V
j¯|1 ÓÚ¦¬0¶M[\_¼D0D0D4,÷s4òÇ ôk +i<0WL+@ì5e¤¥bÐB!Á"%×!Ð 3 'Øa4ÁùÜã¦qøKý/ôºìïn¢%iKL¹ò02n.GgB/;
áÕòЫɱW7÷°àå¼Ììi¶v(a8Loåº}¬Ìµ þÐTðÓ¸tº¹nþ=n¤sdhà¼zÿ+åä.]NÑ%"sãñ¯×ÒúXªK
ÿ¿ÿÓÿ¥ÐRÜh27 FCÍÊ2|Ù|Éã<ØΦj:¨9 3QäN¤NÿÕkßð àaIL ÏL ðÏ ÓL¸Ða ÈH&fA<&4BYiÿÿ¯Å^=$njLwñhØÃp°FÆ7
¼azjz7xBÿùØ[+ÞþòpƤ8d[àáå'tå Ü uI´1=´nzM¥A¦Ø d2)tGHÎpSÃódjHGòB²)dSA,2ÙH ê!pÅ Ò62äL"á hEA¬ êpÍDK Ê 3[ÎÁ²hºn»þÛ ý¦Õ¶Òh6ºN º-^z}'¦ ð²'aB øÓ>A§¢&S2\#øè BháAgÆN¸DÇL"áø&A Î Ñ"rh040 dì Á3f<ؤ uiקKþéën©ºt½[K~·õµ§¤_AQ¹MØEÄ'§ hØÕ-tnh¼h¼ 0½Ì&4[½0Ä0TÐA° (µFçXEÛ \¢í ÑpÁT"Ó\0Oÿ¡þÄú×UXûbÿ÷7¤Ütyú!íîÒQ=ÒpÆNô§H:H:MÒN pb{ A§j¸2XÁÒmè2;ãKGH6"ÅìOmc&ØÕZAÑ¢íkÿü5ÿÿcÿÇ]Ó÷Úºðf§ÛH=;A¶ e é6ÉCû\_z°eéX3Q7 öáIô©7Oû é7N Ã{Xm&à °dcÒl2^ÒpÈÞ ºh}øøù 'ÿøÚÞ+ãò$'ÒýÕk¦ëKÛ§i½µ}סë§é\_Ié¿êëvëôý7u¿ôßO]?Ó·O¶ÓÕ¯ûuþ¿°ÉOÿV¿ÿÁ|Ì?ö·Mþ=tÿ]\_××i\_Cúûøÿÿ]Ö·^½oýzzznºëÄuäÆQ2hþL#ùuù:¸D³Ò'÷IÒ®uÛýµÿ÷õXjþ¿ý/u¥ÿëû0¿÷ÿþõýµáô?¾ù Tÿ{§×ÿÁzh/ö+ä-1ò'\_õün¿!ÿß³@¾ÿýÛ\_Xÿúãßýãë yÔ/ÿàéõ÷¯ÿ¯ºWû µaôþÿÿþ«÷LÒÿ ß×¥Dþ½áÿzý×þÁ «ëßyôÿÿGWÿ[ìÅôúÿí¤¿&)ú#÷ÿ,g@¿Ûòc?ö·òÆþ¹2½ònU]´¶ÁÇ÷ÿ×ÿÿµ¿ý¿¥úû¤£ýú þ¿ÿßò \_ïÿ´÷¥ýê¿õÿ÷ûþ¨ðÜ»{{úýí¤Ý¤Û~ÚÚý
ý&û×ëÿí/ÝWð§½~¿·¿ÿ}ÿï¿ýÿ÷ÿö¸:]ïÙìÍ-¥a\_´´®í¶×\_×mSßéý}-wÿÿíïþ»úö·ÿúNö¿u}~~·ïX=zkl0]°·|N=n%ÖØg%áM=È·q¶á£°½÷U¿ÿ®ëûÝw5nõÛïíþ·¥¿ßÿýkûû¯õÏ'÷ pØÁc4 aC8mC8lRL8Å1ìm0Ù h94ïݾÚë¥ÖÇûé¯kÝ®ßÚý±ëÝWíþßÛþu¤ßlQ«ï^úà à 6)«Ok¼0z 4 &«ð×lPoT»°¶ÚV«ûÝ{t·«³5Ò
öêöýÕÿlÍ}õzì;K´{ðø[g """"Ë&]eAö¯,3JÄãã C/d\ì3VÉ>ÒØföÒq
´Øl3ØÓ¾ M; ß"i,ol÷ %ºP äSªõ®"8@ SL66EP ß6( rà 0Á @ض0A°ê pØß´pðÅ4a P áìP@è àØ Øk ·¤á¤ÄãÙrÙ©:Q a[ ZêÓ ma^Q1 aA
áû°« a~ì.ÃA
´¤Ø
MÚ¥4)
üía3ìl°f)F Á3¬"R eHÔªãhDDDDDD¢ àù§@Á0CÃ(¤"""5ñÄqèDDDGúÿ×ôªýWKÿz¯Uÿ¤iKýt¿´¯Ii~´6\«×¥ô\\¼a"n¥¢4?HDu8%^wÃ0ª¤ AÚt[æ"#ÜÔeuS25dÊ EÅ"2·'îé«Êå=ØjWNdØÎÉ]þNõ ÙË\_vaZì¡î«ÿcÅã×ÿÊôéÊ¢A ÿÿÖð3GkQÙL§}äÿ\_ëøªJ6GBÊHd[(ȲlÏe?æ¥QÏÿ]õú Ó<ö|P3Fl a!
0'Æ0Lx "ÒSJÊÉØ"!°aÆHê×õýE£cpV» "ñÁa4]°¨Ö±¨(A.Þ ¸L"Ã
xFÀÏ .à<0_ªÿüá&Ò &Ø!Çh&ѾA¸Øv}AZ64/ i&aôv-³°¡ !åB6Y Ó6y óQhɤuUÜ ÝgïI&ë 2]I°® t^K´ìãvJÙ#Òl m.aøIêgÈøÏvxA¸AØ ÁH>!ÁC#62α®tÑ8ÍGú·éë«{µº·OòWßt® |pö¯ÿ½]7ô^5H "§\0|'Ï? D'xANDPDX2: 'Óþý×µu¯MÐÖ}XÕoÕ]=¸ûP@é6;£cBgj ª6,0HÙFöXM &EãP
é"b~×ô¿þµßñßx{ÿëý×]ôÛ¸2~Ù\'WdsMÓÉuè$&é&Ã"Þ~#¶IÃQÚ75Þ?ÚH!ÿÒ¿Ù´¡ëôø+ßþÿ¾¿Õþ=[ ¾¬5ûÓýkaNm^,F<gON ½oÿãÿõ#¯ÿÉx/ýl
'_êõÿêúúO]}%Ó¿Ó´Ýnêðôþÿ ¶|:Aÿÿëÿü"7ïàßÿ}F=ÿ·ÿÿÿØû¶¸é7ãÕý^=cýI§þEÿÿËôïô º _ÿ'ýoõà_ÿ
ÓéÒ¯Úézößÿ×ûÿ¿ÿÿ¯A~¼/ÿÁÒ×úÿßÿ¹0¦ÿøü'}ïþC¯½6ÿõ}ÿïþ?õþúÍG_ÿÿ¯ïôþÿ¥ï÷ÿþþ ôüT{÷ÿºßöÿôõõÿÞy?¿~ÿï'ßÑ1 ^ßÿÉXDóÿÿà_Öýïëöê÷ÿ¿ûU}t¾úýÿÿZþý´¿êôëþ¾û«¾¿ûÖÖïï¯Ð]}¯ßÿj½ÿëÒúßúô¿õë¯úÿô¶Mòû]Øgÿ_na[3ÿm´¾Õ¤¶§
´Þ¿¬+ì0»zûÿÿîöÒ»§¿__¿ÿø¦6=Ø0VݧØIß[dG?4N6![/ %í#§OÇýu¯×K¿_î?_ÿß¾¿éxLªb84 1MA±@4PÒÔA P@චÁÔ»«\_õÒÛÞµßmëoõק\_ûó°V Ó¸a·}ØMa
îanÖÊp°a;[ ØBwÃWy6ðÈêÕµìÍ]
ö®ûÿ]ïºÿ"""""%DDDA$b&x´¢qà Ü\N=I°ÁdLi18ýv¹´¤^ õ|"+m¥Èíañ± «AÃi
بPÁLm¸ N"; ¦è;TÞøvÆ0M=;Aáa &½Ü4Ä&AÛ Ðpè&:Q¢ fÔ"B" "&É m45DDDDDDD}/KÒuc¥¨
yEc!µ©NDج꤮U3gdã±Ø°p¥sDW|¨ÎÆÔ0pýÛNdqØ¥FFg&3²èÖDA¤)n)_ýýC µtA&,$¯ FȯZ;#+ßÿåoww2Zj¢v$0UÚkÚp«ZáÉ[A tß~ðôÇÄu}°»µù£Ôíj;)æ@«HppÿÿÝ{KOµ ôLyÚWÿÿÿ¿ÿ¯×ñUùíÿþésßÿÖ¿àDæ¢0Gê!¨9 S©&UOç43j OäHÊ $"ÎáÅÊÃófj/öý'ýïþ7gxhz PBg paa3Ggr#°a<ù< άYÕ~jYÿÿ÷ÿè ß
0ÑqE㻢ò±zé£a0ÝPª©ônqh¼êD=ÓEÛÒ¡ w@§³F'¿Ì ê!C'æÈ \èDáHè;%DE ô#áàôØAI´táäX{'¶îø8<øÒt n Lí(Ú ¾¡9 ÁÓ¨O<)±tú"fDÇ ð Ða !<3Ä 0DRFÅ3c#Bë#¨È6ÍFU¢^ÈÁ-0W§KIÒmáðdÁ¬82ét²,ô§WØ àË8A×Õ>ôþï Å µTl!.4MèÜ^5L&a» ¢áh
gAÈ
È jgvG@>!ÁB 0DZ².02CÌU¿ët'ÛJ®ú}ÿßÜ}n¾§I¼ pú-þþÝ'ÒÝ&ÁÍNÄÐçè@¤Þl70ÄÎÖÅ It\_8´na0Z70]¸°4]´v±¨&MBz "pïïÿÝzëǯÿìÇýêñÿ¯ðI+ä2S>õ?pàé?©¥=;ë'0«Ó«.zv82öª4¹è:M%&é)´°Æ ²$U a$ñÍ,"òmA ×þ×þºðýÿþúÿÿxkÿÿý>íö¿ÿkoÓ¥uÓÓ´Úé]-µoÕÖnJ4/ONY2Î[¸aÁ®ö´Ø:N PAáçM½%kÿú\_Çìÿü>ë: ézLQÿ».ï|\_úéõïú±èwºt[§ôôô>·ONü'®NôÝ?VôÉÞºu}ºtýïÿ\_¿ûÿÿ¦ù8\_ý¸7ÿãÛúôôõ½ý¿Õ\_ÿõkÓúëî·B¿êþ½·]cÓk½%ÿÿô¼\7ÿý¤Ãû%¿ÛÿôA6 µüï üqÿÿõä Oøü'ýý¯\_ûþÿÿ¯m/þ×ÿ{\_ÿÿ¿ë ßÿì0ÿôÿ'²[×äáéèÔýWÿý|úü×þÿÿÿü|uÚ\_Ù|Cè/^uÞ¿ÿý÷ïÁþ¿Ø¢&?ô¿ÿ:ÿß¹\_ Ú
þF.d7\_þ¯¹4ü"yßÿÿÿ¿ýuþ÷Qäh/~Lú¿ÿ÷ÿûÁÿÿÿÿÿÿÿ·]hö4JÿÕ:ÿýÁ½ÿ× ¯ÿúÿÿOü¢G+G׺ÈxXD¡uà£ýºm.ý}{í_²_ýÑ?ïÿoóßÿÕïÿ_Uÿ¥ÿìW¥ÿÛõ_úý{ÿOþ¹b·¶¨@¯ýϯèÔµþÿ_m®ÿ_ýÿ®¿×ïÿßÛ}ÿëÿ®ÿßìýÿ®·ÿÿÿÿ_ÿ¾×éúÿÐ_[v¶¶¾Ú°Öí-{Kýzýuö×oÏ?þßë_j¿[Ýz¾qûéß÷_×ÿý?¯õî¿éÿúOúPÈþ¢PöÏí
é;KþE¡Ë^Ë« alúý/û/÷fknýõû^Õº¿ûõ_Òo«ÿ]ÿZþõ÷¯½Ýý{i_ÿz^¯Úì4Á¸L' jLîIbi î"¤\L;ÝÆXia5¸ÓØ«l3¢Ç¼|v}+ÿéj¿§m¯ª÷´ïí¯÷ßîê?ÿ×ú½êÃé¶!¬0PqA 0Nm0¡°¡´6( à°A Û{Ý6lk|µ \¾VÏ}H¸ý{ ?{KJ¯n½µþ¯¿ßi´ð¾Úë¥ÔSgk B !3Dt!eE hÐÙN¤Ó;ÛaIh5°ÖL¡Ì:h0
¸kpï6¤W7]"8m aá LLáÄ4±;¶6JØïݧÁ«d\é{uȶëlý}VÙ{\_µ°¶jöfDDDDDDDDDDEw /i¯Ú¾¶»M¤Ób·8p¦!4@aCá1@à°Èä8l0@Ý[l%P«¶Ä§lN> m-µ b%0B ½ê0A6%B aa28àé# ÖôÐama»bt¶@á´Å;P áÔ8
&éÄDDDQ´D!a çt ö l/à a<¨A[^ø?ëþº^ëûa+^ÅuÃà qbÁ
1Î8"?ÖYdn;ȦBu&R²|¬S¹Æ\DMªê¹.ê
+
ì¬ ÉÔìFaÒIÃÒGªv/©¼ìYÙQb¬É-äÒHǤûý]´ :jR]-fUØÜi!"=O BgØVN)j/:,äFD(d ¨?ÿõHïuý13¹± 6 é16hmBÆ}ÝðTnqgfA Ï äH%7C è2@ì ó§. òóG#C$FÈÖ"P2M"#'"\ÍEQÍ^Wkÿ7NʱTØ2]¦ÒlÃ%\i6ÈÆÇM¤òXÁðpcVÃ#ÎÐǦPD>Q²A£sFÇiÁUØP¡0P6(T\0Qè²o3Ã0^3 L 30PAA;0y£B¸ .&&l¤tÌ31k®³¨]HfuwOMî/WM®¯%a:¼Õ´ qÐuÒjý;%Øt&ÃÝ'ô Òm'H6ì[è Ú¸Tt\cEߨ¼ªat^0ت <"pÁS BñnÓ fd¡#¶¸^ }8÷ª¾é7WC\_VÐÿÿ]¿¥Ðï»ýl&Áx2ÏIÐAÁééäÿN°áèd|Á&Ù\dsNl4| éÃC&ØÚ ¶5tmF¶¡FìZ-Úª/7Ö8é]°cý~¯[ëÿOñ¿j?ÿ\_ÿ«tÿ\_iv¯Nû'uo§§º¸kúzt Ã¥°#nÉSMG¦Òya4Ým|$nlÁ¿¯ÿþºÿë\_úú~5ûþ¿Õv=t5ôÝ= \_ÖïÕ¯§þõtêìÃNÓmX3Sý6»#OWTåÄW§¿3 Áéý{ñÿü]ÿðÿÿý~öþµþªõþìEéë×Ø×½7]«q}/zÚnu»Wà¾ÿX{{õÿÿ\_ÿ×ÿÿüB¿×ùõr)µÿ¨5õû}ï]Ö·ÿÿý8Ö?ÚZ^þ/ú¶¾¼°?ÿüøDt¨£ÿþLgØ8ÿþ?Õ¨?( QÿÿýÁQÿÈ?ûïÿ¾¿Ãõªé×þÿ¯KÖïÿþ¯øݧÿÿúnBÿä åSßl?Gÿÿì¯ÿÿÝÿÿµÊqÈCò@ûpÔþ°o}ÿþïÿÿÿÿì<±ý¯þt å{ÿþDóÿý¿ÿþ¿wÁöþéº_¥Ãÿ÷ÿ÷Tïþ·ÿß_Ð7þÿÝ1ë ÿúþûAÿÃÿÿý.?,QøD¯ÉÙöîÿöäp%_=ùÑ_îõÿ÷ÿßþ¶ÿþÿÿïõû~¾¯¯5\_ÿó ÿþ¿ý}ù(Ð/¡§KþäëÕ¥
ëúûµÿuûK÷m¿Õ¿¿Ï'}ûýJDÝJþüÍÿ¿ø~¾~ÿý/Á¿J»ãÿmëí+\_ºm/¶ëï´¶Íuÿwÿavï׿ÿ·^õõ׿½wªÿ×ÿùíõïûªÿÿ\?ÿí¾Ä&%>Ò:Úû åizÛt¬0¶"Dt¬3 Ü¿±8ïr_ÿþÕ¸k½ëÿW]¥ÿëí¯þºõzÖõ¿ýìõú{ÿ~Ýý¨qMSÅ1@8lP@ãa X86) aj la´gþáþÈþÙýD¶ioavÌÿ» m÷ÝúíZÿývé4®ÝëõVÿúì[WkßB pV ë°0«i &Ô5jIÚv|kq<¶M%N'v ßÒ^DzlþAzV×µ´¾Ýÿá=°ûjhöÛõ³ J0B"%0!\0DD0B)pÚAÐi6
8l(±@ 1A ©é0Û! :zpQVÚÄ&édKm¯ÆȸÏz\_ {?ß[ïfjþ"""""4 nØNM;4ÖÎ=¯ /ál/¨¦[i4AÅB©ÈǨ;bwh6<nrްħÞË,"&0B%aS¬±ï°¡Xj CJ×l-¦B à ¤ÃPè6Ô0ëÌZ Z@B"!aa0CL+8ði§}ÝÃM~¿DDDDDDDDDDDä¯ãcïÿõ~ðþ)ü á4ÆAÙ\²ÎXìTè¨S¿A52Ì:©3j)Dß#,Àã·Ër¹U6i4Ðgf ì¤
F%6Þy$T×Srêge@ÁÚRiW¤eW¿àèÑN}ëá AD¬çe$B".l¨È;µkTý¦Aº0 0éa0@ÁT¡3 äÇëÄGW3i<ìèýÐ iôl
PTkaH-Þ®«¯þ:êº
ôéí Ñ.¤ØIAP@Ý$ßúÿý?ÿÝ/ZtMê1é6Iôr=ÿÿ¯ÿ¯^Òý-Õë¤-'ZVÌ Èè¤uY#9Y¨f:$"ÈÄLÌÍ 'EV#¿õÿþßZXýõè9Ã3$ Á<àÍLhYA ¼ØDE óùÃ4a4 pfÙó d3aùÑäMÿëÿ£¿Û¤µéQÞÚ70PM A
êôv4,(EãPD=ØNÌ=p»P·
k 0B<ÀÈbæÆu(°AÙFmHò¼ÙBfãA3eË!ùÃ5£!ÒûÇ\_õÿZû 7Í!N÷ÃA¥cA6;¦ØµÃA ÄAãA7ì Ø´?ìàÓ ¢á¸Ñ!B ð & ä3ÄyákxL ÈÆ2xÁN²wÌÑÌÐRp²42ò«P¯¯]%÷uKdÄ'dèÐ6ÿAzM6K Y#VðwÃKµ@áãA/êÐèÞÓX°ÀÕQ±¡iè¼a0\ ÑpÁBg¨¸ \uS3ã:0uX@Ê @áéõö4¤ÛÓoMÓ[ï\W®µn?ÓkUÉ\_þÒéÕÚy4é¼=$ôÒl2<Þh;"Ãçç Ö%&ض4oAôQxÕ=LxEØk©a8þê¿{ããûýý¯ÿ_ºú¯BïOßôðÚXe ÔáQWicéXdnÕ¸ àÍKêìrÎOôáóaóÒ4ÄöÕÚÓ±ql×kßI/\_ÿ\_º]}~ïÝi×þ¿ôÿþëÓÿû¼!«¯ÝµxMV÷´ýXz¦ÕØO%umæ[>®t[ AüýøAÿJ®ëåXÐOÿýG¿Ç뺷8þÕ©þÚ\_ÿ]\_úÿØÿïé·Oêëÿ{m×ài^ê÷®pô¾¸íªUßà¸/ÿuñÿÿýi6±¬oÿýwïõÿð~¯ÿÿZKé}÷õUÕ>üq«ßWK]nÞ½.½å.×I%|"WÿÝ~D£ÿ\ÈþÚÿH#!{Kû\_þ;ãþB×õÿùÔN¿ô=õüØO¯\_ÿûýiéÓ~Æýü \_ÿ¥òu#ÿNòb2Æ»ÿ¶þ¿õ°ÿûÿ\_\_(íu÷ÿúúßÛ¯ãCÒü=ðÙé%ÿ¯ÿþßÿëÿß{¤ëþ¿±\_üÏ,fðßÿOüXÏ¢yïÿ³¨ÿÿÿÿ¿ïoí½Sïÿ z®þ¥ïþ¾÷ïîýûuøºúÿüÿU_ÿô÷ÿB¿þCpºÿ÷¯!Ëþ¿÷ ÿø¤H´ßÿúÿuÿëÿülUÿÿèÀ{ýÿüÿ÷ÿëúÿ·®ÿÿßOËßíõÿÕ?¯__ýí¿ÛÿcõدÒë×®úÃýÿÿÿë~¿þ«ýw¿¦ÿú~¾«ÿWÒý´½{_ë¯nØ[ÿªÛ¶ÿýêïûͶõïýöÿ¯~ëNÿõëßÿ[ïàÖÈþ¿ú:£õá
ì/}ïöm[ ¶ødö3[ úúOáûô¯Ûêýû__ýµï¯¯ûëKúé]÷Ýì4ºÿI!al4A±;´ØÚkj]±±8øvÆ·mW G4Õû¯i?ß}ÿ×oo·ïø]¿ýí.×^í÷mõ®¯þ¯ßt¯¥ö)°ÕAÃAàÓ iÐ@á´{ì): á¶0a½³äH¤ÚZ«"ãØg½³òò,ÒäLðÏöMº´·oVMtM« [ïüÍõ´¯ðthF Ó]U ,tðD~·uAØM pÕ=4,0½ÿ¾ø¦0AÃb!A AÁá0±A0aÓ¶8
ÚR áUm(a°ôÆÃ=ÿØ7ûº ÄDDDDDDDhDDq¡ñp í
vÓJb=¤ì0I°¶Vb@Ú8mP aQP à %!\8 & ïiöi>W¯ÐAb ÎZDA ½ a? På:(ÁAmBö
Ä ÕL Ø0N0D¨¬ JKôµÜDDDDDDDDY-C-ÊZA Á ?YPPö¦Ë# ð?é}ÆDDDDDDqq´¿®[þ½m/ïMªõÿÿ¢l:K;(JéM&ÒB´)ÿi7áPÁ%üS¥IÐb}«HDE5¦OA ÄhIØK2Ü3´eS2«àðÈP#%µ'ëe¹&fW>v]CÏ8tg ¤ vRÀî¡ËH/twH}ÓtAË ¯ªEÜYgAðk?ð×꺩®I2Kýï¿Ôw\_Ö¦tÒYÝr½úõÿׯÅv¯ý¯¯}~R/ÿô·ûÿúDS@Èy"lU6F¡ ²0 &r%&(&è4P00L&<0!v ò]gSìæTüÿ¿ ÍQnа5Ç/L(L"ñhã
¼za0©^=0¢îdñ·¡ } F É2@¤Ü¥@¤|Ù8¤HÔèF¢!ÈÇäWR<¤Ý #±=°}Úå Ä÷I¹úºM±ÛI°ÄøæÌi6Ú TB:p|.Ã
Ê ðÎ BLøÓD#Ó ¢&S3$!0l0Càk\Óh?éSkm0KÛYËÃ#xe:[MÓ#[ Á"áÚjJmcã¾Å¢áÆ/7BxAá[Ñx.Ú.à©]´Bl+WÛÝö¿¾'¦½ºÛiÞîÓp«{þdsNÉ0pt2.a¨:M¤ûÜ=:A6m' K ߯¯÷¯ïÞ±ÿèzz}ÇÇëìUÙ;ZªÃTüÒvonJ2ϧðf: ·¸Iµm^2VÙ-ªõ¦º^ÿúéëÿUÿ}ÐíÖ¯uô4õBúÞý?õÖýZMº|y O_n¯ßøÉ:þ?(CþDÙgÿþ¿ÿÿ±ë¬U¥ÇÿÖ ûþµÿôõýyðÿükí¥ÿý/Z×ü?øu}ü:÷ûòÅè'ÿn¿ÿü¿òbÜ"WäÇ¿ãïýÿýo4 ú,ÿ! ëã�ú]õýÿÿÿ _pZýzô?êªÒûÿþ°DQÿÓý°ûÚØ?úÿzUÿé_ÿé÷ÿßäÁI{ÿÿÿz'Ýy1öõíÖÿÿï÷ëíÿ×ÿÿÿô¿áÿÒÖ¿ÿôÿàÞÿ7÷ÿÿU§ýíöï÷ÿïò!?[~ëÿÿûþ ÿ@÷ûeÿî»ô»_cÿþõÿþôµÿýïÿú¿¤µ÷þ^°ÿÿooµ÷ÿÛ¯ï ÷ßÝ¥k·ý÷ñ]ý¯¿Iùç·û[5'÷ü~ËõÖ¯m%öÏQ8þD¶äLíᬳÿÈÒêßz¿û[õú¾ßÿÚ½÷^ 1@²,0AÁ Ü' °H b1 Ý8lTBbga60:A¤ÙwWi^ÿ -¬/vÝë}0Ööý}SÚ¶µNîélR§m0»v:µUh oäLiHO¶'w&>DËgsö±ºMí/n0EdLÙÃ,ðPaÓ0æÐ e9A©)Ý·S¾qÌè0·i
Ó±T8Ô8©ÃãA«AØNI®á¥ om(8""""""""""#µ°jÿi8M]´ &+h á Pئ( áÅzÊéx Ͳ² D00 CLЩÇ)Ða00 ô4#3Êi_ÐúKZ_¥ïêZªÒZXÐõé ·a]' Æ£-#8î|´Æw;ììÈÉIÔÿ
PRn©Áæþd-þ]ðàáÓ+
KÊÍþý¿]?;ôFõÿÿ +¾ædgj½iýc«JêvS25EÜ_&á(DJÁäõåa ±E" HG êGTpüè2Rdñ!²²±k 'Ü5t'Ah xÜOl1¤ÚQun0Á0! tXáLv|iÓ9ÙáAAÁ¦G a0ADã e }][Üz¿¦éÒéáÃ$ú°en°ôØ=8zyø: Úá@Ì"n|(EãM4Á´^viB«¦gD&«ñú÷õýzzzu~[º¼}ü¦Ù;«%ZNE¶h:O¤Ø28hÜÁã¤m'zÚ/+¤Î0µÿ÷[±ñ÷§ú·ÿhjñßýz®Yá}6¶uÒé°f:è=;6 ±¢TºM¶A \ºÿÿ×Ðÿ믥ë½cÿ½?O¯½=kêÝ?ÓÐ=_ºK®%àËÚ¯¡ý _Û÷µKt øÿÿþú½==ÖÖ?ê¯[tÓÓý?üÖàãûúõøÿä2½}ÿÿ÷õzýêñCÐøÿÿrª'´BaÈ!£ÿ_ÿ_ì=õô¼{ü9ǯëúÿ׫
^¿ëð
¿ÌC ËýÿË/ý½¾LsWÿß_®"¿ã|'ÿÆÑÔ'è´z¬Ö;A/úÿÿþ~<#úò7þCÿ¥s1Hïý?õÿ_ý\®ÿûÿ¿òTY¬ÿ#ÿËÿ,§ü²ýïþòB'ä ?¿¥þ¬Ñ;û÷ÕÿÿuÝ_¿ïïÿ¿õTÿ¯h/¾ÿmn÷õ¿¿×Òý÷ò6«3N÷õïÿ÷Ý»ÿûúißû Þ?'uiõÖßýö¿ÿ~ÿÿÝ}¿Óëÿ÷Ú3þëìN={DÿkÚMÛ ßÚ[Õ®ÞõÞ·í÷ÿöëoWjô¿ÿ¤ß÷ÿv ãà78x@(píXqÛ î!+ ¤NÛaÜ0ÚWiH¾É[þïnê¶ûKþ×ïzm+Õ¿¿íaÒm't Ó 6) Ãbb qA± ä"fDË öÚW°Ïì6ødtÃ?úa~Ë«/Ú¶f¶ì+w·üD Á²gÈa ¦¼¨[5ì,5á4Ø 6ÒáÆÈ'0[vÄîÓãm ¥q8ø3a°Î_"l(ËL14ê0ºM¤Ã aìBi @Ú86bm
hq§bá±Dõ1Á XB Á2 Ñ4[L\§i á¬4Õíd Ò;( 3HþPra¾Èþî "çíøP´ßݯ½ÿÖÛ ¿þ?ÿÿÿ\_ cÏÕÏΨºD3ä0þhè!Ôd0¹ B?F ¢ ©àzPAPAÒ#¼Z¨DX}"(í¡5 gË"¶W ¬}'§µIúHm~Dý"VÑ.ÉöbÁJãv¦2lkóJ¤H×MåÐKt½Òn' ¤Û¿ÎÕYÙD[Õt½àÈËФ¯t´µ÷zÃ"[AAS÷þÿÞWûÿÿVéó´e"}\_^ÿ×ZOö¯¯§üÉ)¶µ ¥ïÿÿû?0Æÿÿ5£°¥´ÜfßÿÿüL>AæAïþ¾µ çjÚge2÷ïÿÿx66=ëõÖ¿MR÷éÿíßmûÿ½]oa®ÿßîÜsQ߯bÈ)̼õÿÿÿØ
ÿ×í¼D7½y60 gÄ8( ÈD棲
ÅXɲ>Ψádæó|þlÌÙ¨´u/ÿþÞ×ÿþNÃëøMEØnÒ.PÃ<3ÃD pa0Aá0<0@Ï ðhf&a|dtGI#q¼/ÿ·°_íÿGp|¤%?áàO Û4M_P&ü(T6ê6Ç HÜz¾ 0¤,&SECávNËð¤ì3ÃO0)ÐR(ódj£7±IÅÎ(Ðd<ÙyÐR2#C5 f¿¢ÁÿÿU1Ù¯cáMºTé:¶#G &éÐNé$Ødv&Èb|ôÛT Ä÷I°ÄöÒTa£ÕlD&Sv!na3 · "ÙáLÌh&DðÏaa\gË·¥÷õÿûjûÿôÖúZNø4 NººMè%ØAÛMé6ÝBi´a82ÅË¢mXÐMÄXX´ ´ab/Eä a4l-;Bѹª£cõH/oD çûµVêÿÿöëë×n¶ëïÿ¾»ýoú¯S¯O½a=&Á>²G%Ô&ÙऺŠlt#ÊGl1IÐ päØzß'ÞCÔCÆ»^½ººÿýûÒ]zõìk¯I~ºýúßÿõmäî¯MXzun¬n¹+«¥a¡I¸ONÓ,lÔ¶ÉE&Òt;Ðm§w~ÚÛÕ¥ikÚ\_þ÷Çëÿûk\_ßVßýÿÐ÷Óôß«ûÐÖ®ëÖï 맧[KK®mM1@±QPa&¡
I6ÒNÿ÷ÿÿýä/>;ÿúÒëÿIÿþ¾úú¥×÷\_ÆkëݵUmTSÄ$(lPAÃÿô¤=?ÿû»
\_Ö·ëïÿÿÿ¦¿ÿÿÿëÿõN¸ý@ACÝÚdÇM &Úÿý¾]ë×Ãˬ±Xÿôëüοÿ×ÿÿÇÕ?@ûëçÄDFN:0!tÃ3
ÿé}ëøo¹ÿïßuÿúÿýÿò:¯õÿþÿÿþAÄDDDDGûïÿÿü;öÿõß÷ÿÿÿÿRÿäÆm=Ñ1Ñ<¯ÿÿÿ¯þ+½ÿû§Ý^ÿëý'ÿÿÿÿÿ]ÿWô¿ÿ§ÿÝc}zÿÔÝ»v_ÿÞ¿ÿÚ_ÿÛÿ¿õý%Kÿÿßÿº{mmuöÿý¶ïô»¦+¾ÿ!ßû×ÿú_þÿÿÿÝo¯¥÷J3í[« wÖmí·ý¥¶¿íþÛ¯ÿÿ{ÿ¾í{ÖÿWzÿúa4Øi; ħvN=Kq¥jÃ9C9&óÒü0Ó_ÙÿäNR&Rû_ÿ¾¿÷ëþÿ½ë½»¿i Ðp©MC SP~ÅEE0v( ÁÃÁ éØ0·\_߶ÚWì>kmþþÕt¼-°&Pà a8aN 'aa0% ÷ Ãé§l+¶¶[äLìÚLa{ ÜÚR&qíÅüM·[jÙå òa
±B"""""""""%"bæ.D£4A@ lS B a Ø @Ðl ÃA1lq3a8#¸aSTÚ x°ÓÐanÖ¶ÒlB¶(4já
¤Ð0B 0Acô""#éz×Ç¥É.C·Qeª;gdîzPdÎÆÌìHÎÓÐa·)ïAbA{Aáô÷RVE¸CÕVv-
ÖTþìR2S!{þðíõ^ÓÚô¸7ÿêÇÔù_úÿÿWÿ®;ÿºÕR¾Hd_òÿÿÿýãð@Ï3CÍ#"Ly R|ÙH¤!3PÉÆCÊ2tqçB)ú:gè29G"è³EÆDD6fNÉ|íN§,\_ÿÿðPIþ&Lg ghÍp ðÏÁ 2ègØN ØCÙñNÔÉ0Lã3C S£à¶<ÍpÃÍ(ðCèÌyE(ͳ1ëü ~Ñpü0A¤]Ñ|TnhXOL Ѳxа¼"ø=tÑvÁT"Ë·^
¦UD èبBa تu¢°ðV´[°1ØDǵÐi%Líg8GB:#kÀ§ófk"3!"çc¬ú5¢:þH'þAÐOI6 3ü1=ºH:M"Å©°àáí¤lj°a%|stÃðA¸Ã!6Æ ÐhAPA¿ a67 ÐAà A¨°ÆªxÎÕ3 ¦ áÕè0"f;0Á'i<ðCb!l2p
FqóeÿúnÕ°fi¾ØAÁkKa×Ißa82æ8dý>»M=6ÉE½ú°v0Öozd©|UÒ~ÛWR<¤Ø1¤ÞjDÞ Ò/B!îF¶ ¨» 4\3cP
8N 0pÎ &A Ïûî-þ«×ý==z}=:ZßÓ½l¼¿}Ö×]<'é^Ní4êÛ'tÔ=\_ÑU6«ët¶NéaíÏÐÄöÐ@îÒ6©DÐÒ ±êv-Þ¨° -Ôn,\-Ü\_áÿÕ ßMÖ?øý{ÿþ¯VÚ¥^=6µt=?Ðý·Z¿¾«þ¿mõ^Þ;möØNk®Ã&\YÁ:N/ÆaóIÁGMB%Úupd±ÈÄ?ÃÔÛ×õ__áý Oõýú¿ÿý=+õÚ_õ¯ßÅ×ÿc^[u×z}ÖÛTýÖÚ'êÛ'pj×nôTì'üû
ý/~?' û"(ÿÿÇ\z;´¿$ ÿÿñè}×þýëìF¯8ÿÓõ]ǯ»úJè~úz¿î¼[ßíø"+ý¿þ¼ø:õÿþ»¦×ÿûKÿÿßÕ×GÒCø;Vÿ£õÞÕÿ´·ÿÞ¿þÇÿÿá¿Dûú_òÆtO>¡¹c?ÿú&3ÉÓkî<ûÿÿ$9Ç(<çGKßêAÝmÈ ë/®¼Oô¿þ5~¿DQÊSi§ÈKø-}HcÖÿÿú üþÿý]j¿¯T¿´¿þKo×øÒ&2YzWü̹?< ßÿÿêßÄä-p?É@_ø?¯íúë×^k=~ÿ¿þô¿úýÿÿîý/ý¿òí½ÒúÿDóêéõ$[þ¾LrDoH"Qö¿}þ§þ¿®ë¿^ýïÿúûý½´t¿Ò¿þÿ¿zIßëÿ}ú Ýÿëîºíz¡é½ß ÉøDù~׿×ã:§×ÿÿ?¾û½ÿñQ¾¿íýûîzÿøwcÿþ¿í~ÿÿïõKòõõuínµÕv{}·úÿí¯wëû~ß/í~3¬ëþß|~¾Çwûëßo¥õÞ¯ÿÿÿþê¾ÿÒþÚZPÚ¥§¿°¿}[íéZýïîÐUß_ ·_÷êû××ÿµýÞF=î>¿ßKý'íõüÿúûjÄã¸õsm)Ã9|";VÏKÃ9¶¶ "Xg¤Óaë¯VûÓk¯¶Øgþ®Ò÷lÑ6h¶Òonz×}§Zþõîµ»½RWnÿ_´
5 0ìl0Å";7bpØØaÄÎpÝÓcß0@áÄîa]¤¶ÃV-ò&4i ÃUVÊ«á³ÕZVºm÷WýÝuÿ
½ÿÿûMVÖTba-°;j) ´Å5atÐ aêí« Ä @ÁplP lP Úi@ö60§p Ø A±]°½C<©´¹;gû m¥a&ÌþÚµl%i6h¾ì/Ì"ø¤ Ú6D0D0 ² "|DA5"aàkQ !'ÉÑ´Ò¿A
í0½ Á ØÕjØa&¨ÄîîØ·m$ôö8¥0U)Ç¡ Õ^D0->"¸ÐB"""#B"""'ò!']i¨¦(m ±TÅÓL(t
1AL=%þ"#RÇDèD0N Êär¢Q°P¶ ÚÚÙN
xi®îªIRX""""2ÌÞÕ%¤¾×ªJ´«þÖ)U¥éTRQÿå¸ã,æxZÒô´
ôÿê³±%ëÎøqÓÒ ê""3´µ õdC"YÚ3+rØÒ+S²©Jè\;á¦[ lezM}|2µ)¾³$5%ã#<²¾¸È:?úÿþ\ÿÿýFvX²UÙçU×P8Ödÿ£ëçea4 à¤\0)#0a0-' ñÄ"vP) S ò,)>HÓ8d?2v0 Ù£'æ31I>GB:dèáFj,¨Dèó)3j#QÈègB!HD¿Óë¢m¨O¢Z4Pð84 &hA0/8@ >4D[Y/A¢à0PPÁO þ³¨Õpi Ì &a #Ä2\ ÁA gØ Í 3µH<Öë÷î¨ØXì0Qj¨ Ü^|4\_Q|ÖР軣uQxB- ¤Û'ô?¢ãá -B/CL"Çk¦ÆqUMPa£ep¶¸\60¨3°@
hÁL'f [5FÆEE5 d!Ó6JuéÑÆC@Ψ#Ñ!daÃÉ §?@ÈùMÈñôØ2XEÇNn´A<'ÒpôÐ2<Ðt:MÓ ûá&ÁÁ=ºÿW~ýn°«ýõ×^8¥ÿ\_\_ýÿÿÿ¹b;·ÿ&¤ç >éÿÿùc|? Oí¥ÉøD¯ü]ü6ÜrÐoÿÿçÇÇäL'ë÷¯kþÿÿþ«iÿýä1oýAx>þüÿýÿÿèýß^ Ál7ÿúíékà£ÿÚÿò(ë¿þ¾úõ߯¿ûUä°ö!}}×þéioÿÿÿÖ¾H"@ò¬ÿÿöY1È~çþÿ¿D¯}ýÿý¥ÿû¯¿ëúSéÿöz¾û0Ïþ¿~}ÿ÷ÿ]/û¿ÿðáðpo¿ÿ£¨ÿWô¯ÿ _ÿ]°Á/ëìf¯KzÖíÕî½Òü~õ¿ïªúé|Wÿïÿõtos{٫ͦ¾¾÷Á½ú¿®ýëÿÒÿûþýþ¿@ºö½^ºþ¿auúÒþû¯»_×OßooµþÛ_ô¦÷ýµþ´Ï§ïWõÿ×÷ÿ}ÿþeí/á
÷á ¶ÒIý´¯lÑYþÁïÿ´ØgûVÒµ°ý|/öf¡
þÒ´/½+þ«u¾ÿêÿÛ~þ»¯[¾/ÿAS ±;°º¨¥m(cM;M&ÒMX2çlL=m+DçiA bqÃMbqÜNîÖ ¯lN;µ¶MÇ~Ü8MÒ}¤ÙÇ}¥¶}Ò¯[y «þ×ëý}m»U_÷øA ØV7¨@± aAÃb ا
à@ÁÅ'm0®!¡µ aÐ"@Á à SPÁ â¨0J$ $pÛI°ºR&3ÿ}ßa[a[iw·]
nÛ¦ÓûMu!a©eF¢p}Ú °·½ü0B ÁÐa0ÇÞÐ0·ÃOû5ûb
A
¢¶);m(Ü5ØÐxANØvÈ'6Òûm+lvÁm¥îDÍò íLDz¡cÄDEìDDDDDDDDqiã«#ª:±\)¨q =0 ÐA±@ÁÅÅ1±MCA8;¤± e:AÉÔ>a4 Ö]a
µL+Ý¥]¿êõø)8!ú¥×éb!Uú×¥ñô^z+Ñ.é6^ ¸i66XúÄ\í+ Á \ °DDYñ0Á(£-êDUqRÝ c-ÅR6Es6W.Ê;24Á5PR½d ÃÛ¦ÒGÃçk2êd»!QÝÇgÉ'ÎÅÕ¼5
ÃØw5K&;Á¡#±!RÿÓ¢bÌÓîõ檾º J¸\_;GáÁ pptÃQÇovû»þîíßÒ\_ñÿU}ÿõu¿ÿþëíNÁc¦T"]ó¢ûÿÿÿÿÖ e 0B"( åL¦ÆHè RT!ÔRqä8Dù|ðÏ¢pÍeYÃ5 £ÌDã ã³@æÈê2\D4r4Ó8f¢#,óÔf³:D"è¯ÿúá0AL"à=? p&áÓ.3Ã80-L0@Ñ!"d32|E»·0! Ï (@Ï@ ù' " 30¤ì"<# Ó5èD1Íè)0!@8¿ßÿ¾ÍѺ EöA´6¸EçðK aª \£sTaV-§ª75L"ñ¨H!eÈ5¦&¶ 3Ã6(L"Ü
Bä &á&áèá¢è×Hê §3RAJ3©3Êjë:Ä"5ϪAé°d¹¤-Ó©.úM¤ØAA88pb~nÐpÈñ¤ÕDùIÁí±¤Øb{i<p¤yI°b|a&Áí¤ÛM0¢ga&/Ag~Å Ö46ônÙبCÒ"øgÆ|°C7 ðAC30@Á¼ !ñçÆ!à:f2DO(]=:Z·ü'ú÷_z{§iÖ¦áIêì éXa¯ Ù(«pÖ(.'« XÃ%yÉs_ÒnÒv·EàzÚÕ.ÃÓ»jxOÑnþÑ1à ÐaA3ã<4ãÖ¿·×úëk×ôÛ¯OOÓ¯MÐzºuw«iºt¦Õº}ê[KVö¼'©dÔ.§Iúl0°A6ý_Q¹é:7^i £[¬ °L0 è»Å^4\Q1Úª/(¼Ôƽ5ÿúz¯¯ÿÿüU¾ûúmuè~ºñõ§ýþ«÷Û÷^þõôé\8= Ý'öAµAý&ßi&ÐAäxÒtx@Ø8=:Mð_ï~¿¯]Þþ¾º×iëéýWþïuëºÿ¡§ÿÿÿìûú¿¥}úûÚº»Õ÷õ~«¤ôôÚO[º½|ô;Fÿqÿ_î}ÿÇÿù¿¿Æÿÿa}×µ]úúÿí'¯uUºÆôé_zºû¡WÓë^]}ëõo#ÿM¥ÿÿ×ÿÿÿþ¯þÿÿëüÖÔkûÿß~!?ñ÷_úþØýk¯ØÿÿKu×õµô þL¿þþïö×®ÒþD,ÿÿïüærþAÛÿÿÿõ¯áþº0ÿ¿ü$ð}þ¿Ã_öÕ}}ÿîÒú þ¯ëýúïm%ù0¯¿å õú~·^X¥òu_ûKù1WüÉ5ªU]y ²¼È@×Õó»¨ÿÐÔÙcõÿvºÿ¿ï__ÖÅwëZýKþµûÿÿ÷ÿþÅwý«umß~Ûßëö'éµ¥õÿå8úÿ¼VµþßûiW°]îëÿoúKWí¥¯ßÿì/ðþµëÿ=»þX§ÛDóïmí¥'Wù:¨ý¹µ×îö)?ï®ïÒÖýï¥ß¥þ©oé}[ÿÿ³ }Ãý¿·_õõ¹ _×üA~¾¿_× ¾¸!·¯·µ×Jöúé1_¯÷¯Õ¯_íëw¾ßlQ¯ÒnÞ¿ý/ìûÒVÚ^ÿߥ¨;×ÿ¿zë_ÿ´¾Î¥{°¿l35ïß Í=[vÒÿïîÒÛ]&׿í-é4¿Áíýoýí~8¿ðúõº÷êÆÝ]þýwAû ð'zÛ^ M'°ÚL3 çaa«OÀ©8a¡õ 'K3°ÏPÐ]¤Ã&ÿ²ë« 0Ïì6ëÿîý³E¤õ·OÓ¯ç¾ý'µÏoõßQ¿é[¯¦{ûU pØÃá0b¦66&pAýRºBbáÅE&à !LiŵÄ&Xf+ -ú|êØ0¶¶°ÎOÝZ§k
ÿ««oÿ_ëjÛRwkÞ¿»p { ÚOÔ4a4ÕÕ'vÛA; (@°lT&ma¨@b© @Úé&(1±A80@Ð@àØàL0H Øê+ Ô5µôI°+ nÛV¾ú »µµÒ¸ÑA0 ËpDl0-ep@:CD0RÝ2§ LpÓT¬§00®ý;v½Ä Å+ vÒbqÚlNíÇí«k +Õ>4öá{rbqÚPÒaØg6;½³ÔDDGFDDqDG¡¡¢ñ(Ác@½¨aA4)H&)¨a 2#l-4Å5DvÄÎì$Åu éb&ÄDZi¦»mk ©Ü ÓZqL.í^Ó_AÓPêS_ÿÒIÄA>Ìfmà ah a ³h0L¡ÐjaÓA
ÒõÿKÄDDDDE!ú×úT¿÷ÿÿ¤ºZ_éýoÿWI.F °ùßIZU ÓNî6CAÐa Á:b¢""'f¨Ajò¹+Dv¥äÞª W ÈÎÔ í²(02º¢2¡ò¹ôKréx:a pøtå¹NHx:M¦æ@Æd3±";Y9¿¾ûå¹Z-úWÉw¨S>2T"/ oÜ/ùØTv SnRzOê×Òk¥ÿUVè/îþ?øã¿þ>dCÿó¸ÿÿþ8ýmÿµþ¾ýÿûK^±üyÙig]Ù¯:²b_ÿ220gC%±¿þßÿøA³g#Fi ( D èLæ9 wù²:AæÙ ò©DP!"6D1sdCÉóLfÈÌÍ d ±þf¡H°¤gR:#u2éNr$ã\Ùçÿ¬"í¨" ()Á§]´\Qp.Ùá¨SÃ<0D#ôBd ÔJxA¢èáA ðÎA£pa((a3ã B©tDÇ &pÈè Ï&3B ¢' NÉ<0ÙuAÿý ÛÓNÒmøøXѹª^"î‹FìCUM&Ð@áðàìZ.Û¼Z"‹æXEàqa4!„ô^uTô.ÚæÈ‘&¶f)#0GÌÔF æ†{#¢RCG9G²4gTxÉwu,êdR!")},21àÌzAÜ2_Òot&ÁÁÁÁç#§]IsIÁü&äZi°dºÈî ±«¤é8w º.ÈÉr<¤ÚI½àÈíÍÐd[ëO'ºQ0í&Ñ p<4È1ªgAS&|a¡r5�ÌÈ Ï@óÄ0AÙ¡ 3 NÜ ×þÕõÿÉçéÁß[dîûZMÓ;Ðt.¿§®á:é:N ºxAé¸N ÇÓnì&Á@é6±z7UB/ZhØ.!Eàqxa4Tõ»h» ÑvaBa´Çn«IõÔlwÇëÿ¯è×ÿ¯¦÷Ýjºl[§®½µ¾ýºë«ôûéúzà¤yÚ4 { ´Ò6 ÞÓSeÄÃ½Ò ¤$mmc´]/=ÿ_~½u×ÿþÝÿîÓØíõý?ÿÿ¥ÐÓÿãÓþ?ÿø0PI¶J4ì ÓtÚO×~ê ØÏ]˦éÃiÇ[uÿÙàà×ÿÿý?_ßíxïÿû÷{wëë÷ÿWõþ¿ýÕ:ÝS¯Oj¯¾ý=;Múßôôôû×ÿý{¨6B×ÿÿÿ£É_ä!þÿkÿUñÿÇÑÐ_öD×_üO¦ßëõÞªëô§¯é÷ÖÇñ×ÈPþÿ ¦ÿÿÿø$ÿûõÿÿÁÇÿõÿòpÿð}ÿñ_¬=/ÿÿWþÿ÷õì}ú¾,Goÿ¤L½»ÿÿÿÉK+ïíÿÿ¯ô~²bÿ©7¾ßý¿ÿ¹ëý�±ïÇ[iîµþÈ t°¾#¥5ÿ¥ïî ÿÿÿ×ð¿×Á¿úZ·ÿ
OÿÕýÐîýp~¯úµ_ðú¯ÿÿí?ñÈr$ù0/®ÿýÿëÿûý/úó©kûÿú®½{ýõz¥ÿ}×þ¿ùò75;ií¨Dï¯üþÖÐ"7¿ÛJÿ»5 æÓ¿ÿõÿÿß¿íoë{Ýýw÷ý¿ßõÿÿÿøoøCÿÿô÷ÿ ÷ Oè¶w8á u«ýz·CMCïOTÚ¶º]6·¶½jáÜ ôì&Á$]-,á¤é6ÉP´>lÑ0é?Õç¤u_¥U×ÿ ®ÁÁþ¿ßÿÚëþ£ßÿ[ÿö?ÿî·ýVõú½]ÿ¯ONð¿W 5¯'vªäS¿÷}_(ÈQ õ¿ÿÊÈYø¯ô¿UÿûqÿÿÿÙ\_üþºCÿÿÿЫõû×ëûi7W«íк½ Oïÿ®Ûÿ믿Øaýÿÿý÷ß?ýÿø·3 \
+ÿÇïÇç¥üè5~þ×õWëKþÿì·¶¨Ýÿru¯Û@¾÷oä¿ßÿ¯ýþ¿×à¾ßÿÿÿÿÿñÞHÿzù:ÿìC¯úþ×KÐX?ÿ\_ømía°
ßõãÿü±ÿåú'ßo®¹ cüG[þBPÓú¯ý¿þøÿÿU½ÿ\_ïÞûû÷:OÞt"ÿ¿¯ÿÿÿÿþ\_ß&'õDÆqåÿ°eõùýu\_ÿ°~ºKÿ÷\_^¡ÿ¬8ÿÝÿÿÿ¯ÿþ¿ÿÿ÷ÿÿûï
þ¿ü²üÇïþÝÕºÏw½vï¿ÞóÉÍý÷¿wûïß÷ïúßÕ~ïþïÿÿûÿÒÿÿÿûëÿ\_¿ß^»þínµÿýßW»¯ýuÒ¿ÿµUÿÿ\_4M]uï×ÿÿ\_ÿ××ýýw¯]yнuV¿ém·k¥¯ëk¿úë~½ï¶þÛÿûié\ßæ·ýÝuvõoô¿ö¿Ûÿûë¿ì0îÇil4¶ÉrbS°ÚQÈn£ï´®¡
aäKi0Ï(i½¥ +·i6í·¯w
¾×nÛJÛé´»m¶×ºí~þ¿nÒý¿ºÓèú{R# 6¡°a&B8n8a)Ã8@Òbw ¦I ªm¤Ä㶶«pf&ÒNûdK ÛX0A1(wÈÒ¥¹ öÙôÚL»µöîÏ««.gûð¿íÿý_ú·ïÝÔ0]l&êÃl [ÕÒÅO¸¨¦¦Ó@á± ¨lP@ MC1@ÅT0@é Á¥Å¶ØØÛ E°Ç|t¤Lì3Ókd~öÿô¿]b""&p Á¡ÁDA !Ä¡ Á !.á Ó°pi«Ü4\ uiT¶®Ã ÃÐi;bl'bLP@áÈ áîà iÃbJ;¶4í´¨½p"" d@Á¬±aM¶M) M¦!8 á° paB¸ÐJ;0pÁ"èêÉ¿atDDDDDDDD}uõÇúZQÿÿÿÿÿòÎÒ'hFÄ\R>G=DGÈüìÀÁdeȾ]Ì"|º1¢:0ÄfDpØ]Dp¤|ÆG28˲ùtf²SDDDDDDDDDDDDDDHd 9W8å9+ò5 eh7$9!ÉFäO)Êà ðx8är9sD"""""@Ðv,"""""""""""%"'Fh¼r#¢;.ù#Æ2èÚ.ùt_(GDs+(áòã6Ð+£yì¼_#¢;.ù##ØkÃÄDDDDDG)¡5;0©Èà
ÁHÀT.I¸dÑf2:=Dc<ËWÉÉg8åTUÊ v7 ¶ cØçaÉQÎ9ܱÎ9PUGDtGÈæGDt]È2GDvo#¢;.Ìdt\Bè">/ProcSet 32 0 R/XObject 14 0 R>>/Rotate 0/Type/Page>> endobj 14 0 obj <> endobj 15 0 obj <>stream x]Ksä6¾;¢ÿ¢/kG´5ÄcÛÝécv=:ìFô
Q]l×C.VI£¿H< "U:ÖG $2¿|�úåáu÷p¸ûÛÖwîáùîÃÍ}}WÙ?QÝIu×Ô¾»{ØÞýø¿_¾l¾¾øáï?üø ÷ ÔwP÷îÁýÛîÎ=&àUªxÙOÓ0Mã~w@Ò"eÄ?öoÃëpøÄàjMpûãzy¨t¤¼oÜca;<ÃÞ/"®güûçñøó´îãî%Fµ-ºß=H&@.xDgr�᫯§î%à©Õþ´;Ú~ï·(r8ÿömÿ@×:Ï7ieLô¯ý¸é7C9s¢®,> À·õ°»^¶ 3~\¡E#F;̺u«B1] éjÉWj³AìOÓq8Ü/ # Ï+EÝáÖCùvã^wýñDÅ1¶µÕ~ÏQ[ð3]ã~÷Õ3êVXíîßÞ9½Tè'l°êsÜ Ó²° ]2ÓyúÕºlÝ9 ã80p;wÒ°{F«@ XsçÓ¹OV¦<Ñ4gwY? V7AZ:|±6
ÝnãÇ
$êgEÕuÄZáp|Ow¬Ðʽ¢¼Ç4»iO6aê8¦eQE33EõfÊs=̬ £ïÑMé0¬ÕÅ7%\íq\×®!Ç RçTKÆqÝK1aád|qÚnÉî´ä÷~³)]õ&Ì×t\_ΨóòeÇìGY9ÄLÉÈVgþqv«Wq©Ã /«8ù:LuÓÍÜ4¢ðmxËXJ 6ØNJ![é�fþ´G øv¬£&f#ÍöÀ¨Ç·aÙÝÌdÔ|¢?ÒvU$/ýáÈ
V8X[,XZªr?iÓ~Ã86!ý×V§j'Ó»êÆá]ïwhÝÆ:ïàuöÌ«¹D�²öÁ�Àz=l=Üñ[m3 ?¿hϨ ¡rØ©0þcò¬Ú8PªÝûò\I:WÈ!Ü^.Ýx¦þeØMýÚÐdØà}M7wnÁNõ81ÃxHéÒçnKXúúãôõ§rY:1ÐUAA,cH¯¤°áiMàKx[H²mí,ªñ$ V°û�^èf 9õËÏÃêX~>4ã0~[¥¯nÈ«Eóùý÷Ë$j³ 0HK^ï Tí~7é²£L#î!¿hR·WòX d)¿?ÝÇÑ¢2XË �? BuùPD|gTv#ò© ?vªô-}ó½|çÌ% PSÕALGøYûú£èêæëO÷imPGê~eUu°OÃÖ4H¤XòÒñLhAípmàGm8 ¦d¶îHÌnb SîÒ YNo¢0æÌ9üݹ²wÆo P¡&Æá)ÕIHïû÷3saAWÎE¶²\´S¿/¯G¾óFÐc¤3ÖiSÿD óú78«§=¤ðLYA:LôLB¶ðHR¼r椥 ©Êã°êÓe¨éÜ'kÖP^(åªà,ɵI ÿÃj®öÊjr$wÆAüÇ·ýÓPêxÛ:ßÌ\ÈúTÐ�ä]à ëÙ:·6ÊÁH0oئÈD¶ûÃñ[ÿmøTyãq ý¦¥ÓÅt´KÆ9ÿ¢à-úÓ°/-¿ñ¨Ã>aDÁ] Õ¨¼È¦Ó¡Çø>3å¥ý821´ ²5rªxIªr/é¹T¤PÚ¡ÌÜ~¥÷«aÙîwÇõæB¬ESó&èõ)¹Ò/µÅ tý# ³dP#\_ ±
AÆðñmÚ<Ý èÆ×ý+³&Ðgég)ÒèÃÇÝêt8°ÃFÉýøM¸±so:¦5gsmܨcå
5Iu °_¯&v20acë7ÈØs
Úá0ÞÛp)ʼ¬Ó¸7= -,Óåáj2\
©�2Tˤ¶iÊþX~R¢wÅÇã@Ù~¢aª¨ÉùúM¡ÐÝix[Ó§¼¹Sè,ß?2ÁWè,S&CRóí!!8MÖhòYÚ ÏyëñÛÈ76á0ð°ßnûiÎ7
^óÖM�¿á=gãá´xÖY© cþ©ºG¾y^I:ÔÏ//KLM3:Pj»øÆülêOEàm¹,ÒUɱ|RaÈæÆ@¥£M: ."² $FÖùúÓ²-mè\ oüúã_§ý¥àµÇÍèѶ¢�}ñ XIÔoxÈ»¹qg)8M$wg ¨ÒÑ÷^HäîCaõ÷þð¾I´úá^ _³±àÙDî©Ýþkð¯a5rHh+ÒAÙ&[Út¡øi³õ£1ô2
áû°JD°!,¬3Io\_TÈ3-/z©·f¼jóÛì S fÉPq0¿dP2ùZ÷ Î5®(ìGôøPIHN
hv È3CÚ8s %åÎüi¸À8iãB\_¾Éò vÿ+ÖíÜÛöx"ùnÿ:~)x hçó¸b¶dãqAÅ®®u¨;
HÌS¹
lh$pQ¹Ýå2Ò¶Þlãz|ß»N¦]XÁKªY±ÁÊá"YçbpÑxLÉÞòlH:¤g±8"^úñéÎa¼ÿ³L!´µû=¤!ùÄÈ ¥"¼Oå¿Äêd js'S½dXfð"ÛäS�Ù4§OC Là\_k)¿NýÀ´ \ éõMpZ®!Å\Çh)þgí4G/Søïã²b¤]å@Ñ>gÉvt íX¹jÍéÔl¬Öón6¤¦¢|é~´Ï½ '¸Vë5z´ÁĤGW¿ÛJ%F9¾ß(ñ;÷´¹!ù(¾yܦTDÞ\ Îziɬ{Tw4h ç·&§áI!B0ú|¢ÈÆOPÅd}ÛÈ~wÁѹÃú}9:];D ¶ÊtÇ� .ÛÚ³ëm$$RKíþjk+°?>cÍí!ʳóo Íu(£w¾wÞåQ\_M+ËûqWØ4¨z ,¢Ü\Kkê2TY¨� Pä¾õ§ v¹°¼vÈ4Y3dKoëtͺ}29MFÄn¼wÝ s.
µ�mZ4®Ëºq @LN[b¨¤ùòf@ÌÃgêQk ÊÎ ²ªy§m9Kè
búli\_[ã óBpÖä§$ní¼Nx¥ÿdp>'sí^O6ûÝ·ëÝÀz ×Û Ç,¯Ìɼ¹èqëQùêl)¸1³e+Èæo&öwptÚoÊ$[ñ0ñYKä]óÊ<&5$ðë.fHÉ8R�øû¯$çq¥þ)eç)è~: +5ð{Pë¬w�³Õ�)ìIªh6éw&+3KZ·w<öÛëÃÌ&¦ò^Ȧrìö»£A ù\²æÀòÜ6i©k�¼àò§¿êܹ ÿÇ~WýD½I.#5<64'MÿVîdÙ¶R¤Ò~«AXkzgH Kt-E¡«È+¦sÙ^©5õ//;9|ϪìZåñ
ÔÛID¥$hoXI8G$ì3U è"L.ù\M ÖÇ?PèJ¡;ÓJÏ{Æ1+8Ñv÷¤Ä|%«ini«¦Ìºµ;mïà]å 1\nJUiòP(ÿ©8¯6§§«Ë Qý'¡1ñ8²V8?G7cÛ8ÌLêsäWr·:màéN
áÈV©©Îªô¥©)fÖD·i Ê50ÛÈDæA'&ÓHD»ÐòiYçd&³½$ܤq^úñÀ.¡Ñ¼.!ÒÙÚÝþ:Ю$ºÑjÿÕúÀd®qJé8:PÙöl.Qj6_"Æ>¿ç½X DÌÿ×°àu.%%´C %:éýnÉïÑ\iQ]^/¤ÊÆ!¸)óljB³²é ¯óvìôBj_¨¼ 7¨¶iS
¤Ëí®¤éôο.òhrK(û{Üö\c«r"" -ûïa''ØÕ)¶!9ÒÒ&nmùM X¸2³S#ø,9ºðɳájÐÎÜbLPhdë ±h~¤ ýD¾àû%çä+½§©B$0y¨ésºP÷+!pÐ VÉWwmÇMHQº4-E)ÊÿwejTÖAü^ö ðøuÓ¿ÍçU»gBkåá ßÕBò i¾á¤ñÚè2ÍéÌáÔÅÓ8m6:ߣ-=Īñ_ð;|<©QáN´4¬G[àçgzÈ3m(áßffÚN÷WPp® 17WQ1Êe[LnÓ ¯½yÆB�ØB#£b¼<»mTbpyTÈ#öwì¦ IþmG(iMëߦ8¬¢pôá\P9HÙÁZ×nü+ÏøA�Î ò´ÃÌzf6àsª²ñ¯ ñä8ì$oÓùñäU6ÃðJ-8YÈ4)\nÅ
òIÆj³Nçjãñh+Ó~.k aê£� 2 W&ͯ$tÿÄÀ=£TC gNWÐ5^u¶%tyåÝ1ß4êÿË0¬ÖùÐb1=t£µCÅö±Ã¸¡Frù´DHF-
ÖZk$H µ¡þGy¬Èy\_vIã!¯0.'sJ§üع·Ñó4pj ÓäæËÌÙv¸ÖU)vë\C[$9-q& |QÖÚCNpØ ¯9í¾¦#×PéDBÜ çÀA& µË¡OËgÞtÌyL\_!¡ÊR& HÖ±$ûÆ6±+8Ä cMó#æB.ÓÙ""GðUÒ¥VmJ©À/Lá rE Á'#Üi%[Î+)© ݲÀU¬f>í¦pÎTcðOK7.%Äë~sÚûÃÈt©ªqÈYÇÙT8
n~})ÑÕÅxF09�5§¢Uå'HÎ%¤ÏnzªÐyöF-e{ß ë¨Údky>mì»ã-V·p9X¹Úp /ÆÓi¸/çò ]ðË�¥à!FTñ×WÑ÷ÞFõ:8nQã5ÚAgkB;@áøD£Ó¶å2p!iþäY±AÔý ¦8çTJðU�øæ ¸NT åa·§Ãä®ÊÍ G>¢ÃÆlø@Ñ[¦.¿Äáj÷{ÈßÀheJÞWiõ5DX;Hãbyápó"mnig$ÕWQf¡&Ýúͧ²éÀ÷»¹ÒØ ®Éî® d¶»IaÅ£bïĬU(9j~wò¤,[ÔÖs[P«0þÎÓ°iùkÉLïRò¾#tõ]#"GøÖnHoKG0'µò"íîª Rº!Ö\_7ôM4þ2}ÏÝì(/tÁ'5i\_gh-÷éiäÓ§ZùBns6¨ /î|¹Î(w sò lù7ÉÕ+w+cÇ\ ¥\=×´UIú]©ÏÑ\j=¢¤Ö.ÖUËDQÒH¨'7mrGÜ1_ ·ü5ñ¼!K5´û^ë;SÓ°¨"«×'qý¯m.5 \ãý|ÁS[@lÑÒ|g»ªcÛ:¬ô Ûñ´eÒzF5èG¦FìI õè R7º¢a¶iKÛÄßC´/üæÁ@iÑð£,}IÚÖ´'ÈÅ
j6X¦eÊHÒÀÈäÓ´ ¼D}¢kHuì¬'f\YÜZ ;þuá,sS D\굡^ Ò³%êg2«º!¸2C»Û@�áÀÀÀyr°« v)}"©òaÄ
L
°,ÍèôÃÕ»È÷wÚÐõ ~wôK³ýÍÔæü5yðñÚXÊÉÏ´ye¯@x~ì´Xè[K]ÃEÊØÅ521¸3Å ¤Ð"BóÛn°îBhÑdúYÅÓSxgvìÊ\~¤'4Ó¤ÀEÆáçÍH é$ö¨{H#»×W·Ð![æÙl9å±Í[²}«Ûº8ߨ¡ aggd5ÀåL0 ÎØ´7<ëNHTÿÌ£Ñ~h±¿çÚØhÃ�m
ÌPÝÅÑL+vO¶ËJ/%Ï]åÉã.. ÜçÖ´ó ³qCúÙMavÎWG.Ð~adíÄÕÔãÖf?MÍné3géBFP+÷Äý ã½0!]Χ bѱÎÛ /àÝ´¤Y³QÒÜî5õo0Ä{Êo¸pZfêÝ s½ltLÆÝùkæo©èpÛ©[ó,éÞ·ß2fÒ(ÿ¢ü¶cº- z÷×µ¼"za²kµC´Hrø0Í÷sw Æðñý
=P\_ù÷Ü%mn 6¿¸VÜ,^÷kàe:´ô(è�]ÛZ;XYOo²á@Tæ4èà&¾|ñþtîéV Ø?~KZj vÏ\Ut]Ü¿á~lF¬OØüènæ§[).¨&Úî_%çZ@bjåWµÈÍ2POÅÉߥ@öwípºHÍ,É ·!j7³h&ÔcEìÚÃgN¢FÇmåXÓäî±Ñ³(Öþ=çº,¨¥ß¿ùîTfVZÿN>q÷F}Ý1Ù gãÒMÕ³MÅÚ¸tå©\ EDp½@z�û9ÏFÐPÇ\1KMC,5%jØ ¾|¼&×#Qä¼4F |¥;U´î
üÜ6jbIvÞñïcÓ×ÿ'&0ï¥@Vc"9« g\cøñÄþc ðJYrn ¬®î´ÆúHDê4ø83Ó�j®2Ó#ôRÝ&\Þ w+²Çñ5ééDeXê ×F'ø¾pÝ\)ÛëÏ«bAwüÇ\ß«Ýv%pV
©üîÞÃÈkhÖdñ °5agù(r:{Ø %·¡Q+¸ºÏÄIÛÉOãæ¦Éÿfúy o¸8íªö Cãæ+$Ê®Ë/=K£4ò#;'N\:ĵ \®L(á\Ã5Ú.]83û"K9:èDÛ³ìEç0e»'µÿ q8G&Må\_T\Ô9Û 9Ï{óIÍFqPV\f! ºjÉì}½PEQHmÜrzUÝ9\8·EÚ³\$ (0ÒTD�éµ°ª¸pÞ¥Mr£Ãð|"µYÖa8ͼ méC&¥'XOB%nÄGR-´¤;ìë¬Ôдäka2C|Pë .¡áLiÛ@gft%L¿}
ÔfÛÿÉZ(£Zj¢í¾>õetR Åï25!¶"WNs �¢ð÷ 4þ÷ÕHÿKMHúWú¸¿õ7Él,/C1LNÁE2 ëñPÐÐîÌTuÿ¦á]çE}%Ù|á_c?Ùr¨$ëO7¹ùó¦YR3J¦UÃÕ
ðf»rCXÛ µ3�¿ßÖÉ®ð´Ò¤ýKBSåbk¢ñc$üOüõÁ´ê¾ö+Òtø¯¬¶þöe+åÝoûÿóáÿÕ|¼ endstream endobj 16 0 obj <>/Filter/CCITTFaxDecode/Height 3328/Length 103907/Name/Im22/Subtype/Image/Type/XObject/Width 2432>>stream ÿpEÿ (Gÿÿÿÿÿÿÿÿÿÿ-°kÇÿþv$gTl)xù0ÇN æ´0@ $Ã¥ÓX-µÿ ëôq¶ø÷KÿÝ}%êÚÿä!ÖééدÒÓ ÿù¹[í}ú à»Kî?û¯úý}?ü~ÿÒÿ·ÿý/ë¤ÞºZÿÿë¤ßþI¿Òþ¿þ¦ÿôª´ý%JêÃ0@Á EärÇ(¥ÄDDDRº]. z DõO ,´2»ÿÿÿÿÿ 5IGÿÿþv4dpäpÐ_0DtaGø>£ZÄDDDDDDH,@ðhqÈ8ä(ä)ÈmXç
¨ä±Ì9CÊ(r'r¬¡ÊM)ÄDDDDDDDDDDDDD@DÎPåhTC©M¥) ÒDDDDH-9(r;Ç9MS ¶C
9\S7""$2hr!8äì®(r ²\ìQà&GB8iÀð±+©æ´s.fdvGdtldvGÈñ#ÀÛ70d|EãGäp%ÃÀÌ3Àõ2ÑÄmÈùtp0G¸x6EÙÈñ/#0 """"dVÄaDx6!¼À!2àl¹ØÐ5Ùt]Å#_3qÈ$DqÑGÃ$ È9 9Ç<rÇ0æ·<9NPåTr¦P)Nt,tؼDDDDDDDDDDDÙDs¹NPåS2iE½h8ñtGÍD|#¢>G"80¸R8l#.Èá ñqGÑtr3#ê"""""""$ÝQQ¦a2GEÑ"&3#ÅÁDDDDDDDJèFÚ.¢ù2<\B8@"""""WHÌÈè:/ÀÌ """"?ÿÿÿü¶Ærõ(ÿÿÿÿÿäØ\;hÉ(¤vI©"
¶©JPÎÔvZ3V¬Ùc,
+kû¦ª¡T·3µfÊÀ¤ÛKþÁéÚÇepXU$7;+G Fó#0±ü©/·÷GÙªñ© ÿ û^B¯HìbýÃ%Ðpd&2%áÜ/ÿ¥ê£FEgÉÿoÁÁAg/¸ªv_¿ú¢_õ÷þvXÐCùñfÈgÖF¢:®#23ÿýúÇÿëÿóµ¥/üØd Á$'OÅ:£
çVR×ÿþ¿ÿøÿצá5A£sFèEä#c»¢xÀfc øD%"ðN)0ódjôkÊx":ß÷ÿÿüg;ô» bW0bV4W§Iº éú6øF°Ú.ñ§A§ PÎAS#QewmH$F§ÿùÐÍWüuôÖ©è:O]:ºZX:M¹ÃT $ÁazZhÜD?SÓ ¢Þ ?]2Aá<þxR .UÀñ döQ<£(HÍ2:5eãúçAÌÑ@L<<§DO B¬¹³5½Jvu-E #¿ë¦öýt¯÷ÒºxÒmúI¾©°cñ¤üLøÚ.Ú¼¢áøaxM¥t^Jªa¢í¢á£ÁQô\3Ã8ÏñeÊu Aú
D'!<<&xPL&fa ÉÚ ÌÙ£3a0DR dª"ÐO_M/|lÕëIôõï^J4é6d Ð<'¦Ã }Ú¤¢|i7ÏÌ0°A6¯ám_ % %^¨;ýº¦"C½;H&=0ª¹áBjlSbBg
0¡ û¼ãÿßáßØÃþ]uýÝk¡§[}á7ºÖíët ³§I¿}=^I@ÒovHù ò:R #ËÛAÒíá£_¯ zð;p¬XDá¬Z'µN0¨ø¿ÚMÿõÈàáÿÿÛk«}+Õ^ý{ßý~êõkôÛ}Udï\_ÁÁ+ß«,éðWOþ\!çëi$êóã ¸dW¥æDîMáçÿ¥ÿ»È\_
9ßþüîÒ´õ÷¯úûQúéÿ]\_¡\_ÿÝ]ßN·ÿãí®ôÞïiwûÓi5Ðo}Þ²YIþá7÷ßÿ½¶ä= ÃaýÿüzQþ¿þ ëíÿÿöþ÷uÿú\_ú￾ýXãÿÅCÓ]ô×OÉ{¥[Öÿ´®¿¥Á°ÜFûÿ÷IùÑÔÿÿd,ùô¿CÈbº0þ¾µÿ\_ïÛGz\_ßq¯òÿ»ýyOÚÿpÅ|z÷\_¯×ëþø9ÐyH@ßÿÿ}¥,%eÿÿà]·Á¿ÿÿºm%þ9 Eþïýÿûjhö ÅG ?ÿÈyúÿPïý«¤ÿë¨8;Áþ¿ÿKûÿ«×ýÃz¢ûí¥ü<õÿònaÊ ;ÿÿý-_ýáñ¼>¸+íõÿ°ßΡ7XûñÅWVqy_ÿöÒ·Kÿÿô³ ÷Kÿ_ô«}ÿB!¿_ÿö÷¥ÿ¹ÉGÃá¿(ÜLÿÿü^þê¿Òëï«ûÿî¯þý_ÿ¯úýµýúÛô·ü.¿ÿý+ÿ·A~á¢WðtJÿÿÿÿÉ 'í¥µö¯ÚMßÖõåºìm±OGÊu×ÿþ{û_úÿ×Ò÷_Ã÷ÿõÿ¥ý÷|\¡Ç¥÷Ö/ýz¿ ÿõîZðWVÙÿí/O㺺÷´µõÿÿûþ6ëß³Z~¿ýWbÿ÷Ö?S?ôúëgTft¿ÓùÔ¿Iÿ±[!0PÖLv´¶ë»Um'ØgúlÍ~¥zékûz®ÅúµÒÛ¯ß÷Þÿÿ·ýÖ½ßó;uÿî׳ºÿíßðÁ5_AQ )¨¨ @Ø0H8x@Ð6ÒÚbí(i2/ûJõÝ
þý¿X"?ºý¿ý¯½};ÿÛvõï^ýRïïöÒÿÓïþÿ"#,À a&&áPbí&b«kÅDîá¥lvħÞÓ"çaøabºöÌûJË®ð¶¶ï¯Òo ¼Õµbä%³Æmïü\_þØ^×{oþ""""""" ¸a4 Tîb8pPÅp ?60¿ÛI5Á+J-YîäKm¥ÿÈL á´Úq¦»kqÚQöÿÝEñêÙÅ!\ñQfAªa8a¯M-°l(nìP@Áb¦M Ú47;Ý©Ã0±TÈÜ:p\0oÈ£pä#Aà !4í[ýF T &TCL /õi;eDA
8éÝ ÓXi
µÓ°ª¼5÷Ä'´Ù § pýb""""""""""""""""""'D!"%ÐÃDCBT! )ÜPåfÓA Á2Î?êKúP@û+¥\q 2¸ÈÈMJú&3jo$WªyJfF£&ÀqاôzCùnfAo<9oë/ êg8ÉdaJqMQ}n;A°ptÓôNºB®ÿðiÃÞ>õ¯¢§çd/è¡gz´ d÷ áL¢-êçeHÜh2Fr4s)û¯üqð¾8êv\I@ Í"Iá8ÿÿ®¿\_ª£sTÓØÑ4nÁB Ìaf³Ê°±ÆtÍÇ¢|ÙFêD½Uÿÿ\_Á,UÒl ð
L(H#sXPTåß ñ
.3032!b0@ÉÅ;%B¾w¸ah @ÞD>ÌB60 ± -ÙÁ
D ø°pD8ôn \hì2¬iF¶|jN aUaqeÈa040@ðAB·zÿþ¾»èVÿýÇ¡ëÛ¦ééº~Ò÷a6¤Ðzo§'æÏ=VÁíÒM¾sóÝïÒ éoaBAÐAäé;à mT % ^5A[¶i L"ì0T!l¼^3¯¯×Óá×ÿö5õÿZãÖ.¿Ó¥Öûûp¦÷Wì«àÌtØbqé:¾ V Jý6 T:Ú¦ÕÒÒ²,éÃ%W4äcI¨N®%ev2;ÃIØ\tlýøÙQýÿ3 "×þõÿÿ¡±§õ||}éþºIéõá:ÛOÓÉÝ^^JØ0
ÖºuÿVéÇJênJàЫԹ¦ÚnÖAé°aÃ'?¶åü¦ºÿß«ä?ú õú÷ßÐÿÕÓúþ·ý]ôÝÝ=u\_Bë×ÿÕº½?NãûºëÓÿM5¯[^¯'ÿ'Vù1ÿúáù¤wÿäð\_)S¨YýußÿÿÕÿÿëõñÛUÿÿ}¥ïõºüoþ^úí+ßÞ¯ÿõÐ?ÿ]oÑ<ÿÚA¿®AÃð]aïÈÿÿÿ
Ýõûúú;SµÿKïÿÿýwþíÒö¿ûëÿ½ßß_ý1ú ÿl$ð_O;Aª àß$ÿÿÁÿý Ð&¿ûÆÿ¯ÿÿ^ÿïýÕ29®àÿ×ð{þÿÝ|?ý}_û_·¥ú Ä>Ô]ðÿJÿÿÿ ç%ÿývÒÿýÿëð^¢?3Ú_Ç_!Iÿ¿÷~æ¿Òï¿ Óúÿì%ÿëúì÷½è/¯ýÿÿþM8/ÿ~¥þÿõïßþ¿ûH"yo¡®ÿùûÿ}þ~g~ë]ÿüãõ_×Áõßþ¿ÿÿÿ×ý/ëµÿWI.ÿÿûÿßýµAvÒ õþXá¿êÙÿöÍ =¾Òî¾î¿_ðºýúÿ¹Ù~¯Zõõëô¿¿ß÷Kÿý¾·ýÿëµýX
Xb¿
·_ÿßá°¶»%?³þP^ÿý¯ÿ¯Úÿë¥Û¯ÿV¿ýuÿë¿Õÿþ·¿¿ÿÿuá]}WÿÿÁ¿àbÔ0 x
mªÜ[KáaÿðÈèßj»ka~ÖÿÛý}tº÷ûÝuïþ¾¿öÒU½µÿ¯íÖô Ñ«õÕ¥ýþhÿwØA«A¢P lB êAÅC´Ó¸8ûØö%?» Ý¥ +3M¯ku¥÷
ûÿ¿õô»ôøímûÿztµJõÓþ¿§®"""" as8T^[jÁ0íCPu vÄîÙÚX0A6y[ #ì#Ó»[.ºûzØfk ¶éZ\_Ó
mXaV÷KîÕ~ëµ×»b¿nûû^""""B1Á³ÔGì0·ÚÚîÚ¶ QLla &Å\TBOdAË)ñw¥Ü-°±8ûïf ºxìJ|Nïd\Ú\_"c¯Ý.Ë«µA¹¢ÒÕÿºûÞ¿¢é &º &!B &aZ ¨p¨ Á¦¡´lP pöcaC86AE1VÇP¡µÒ xAA;{I4íÇ ÅO^È
»í³ú|/Z ÄDFZC!"Á2K)c
» §jà v a5C T&¬5ma«M¤Ø¤6a ÚÃb¢Pt0@ÐlPAÝA (w]'ÄDDq¡gX9vC- )c Ó0B! g":aÐa8i¶á¨Mn½ÔS: ÿ±G]"(Ç°¸" a7ÿëô¯¡ú_¿JôÿÿÿZ×ÒöëÿUôÛKÿK_ê6 ¹ýÕ¥¬C_~ÖAM|ïÚV%:å006aMQR1íJõåzYD"¦ dc)ÈíC2öeJ d63'eR½ÔéÃ^v,ÁJî3!vWd&esC_»î¦r
)]@Ë~fC¯þyðàÕ®K¼> ãÿûþ°ª¸8ÿÿûæþd^÷!ÿÿþ¢¾Ö¾WY¾ !N] ¡¬6f²&_ýcÖúãþ¿E»FÆpSA¢çaTgCøCbæC(dªÔòÿkúÿ¨@¤ït.ÈBª 0>A ø ä´I dlSZάÈëúÿû«¥n dt±¤ìOwhØÃ0´Ñù4 ðngôDÈL Éàæ>Çç ÖdµÿÿÿÖý85Ðze
Nõº¹úíOÏѱÅè ØZ #c84Óü 3ÄhÏ@(¨9?eFQH: èyìyÙf¡3" È\Ùì/ªôýW½ZôÕÓzÓté;Ò¸~KºM½ÔOtz5°°×i].ÃèÀÐh¸ à )á¢$O xa&&Ï ¡Ì á¢'ÓAdæ #340<2Dat Ö!@ÎQseç¢0Ȳ7 èWÇêÿ_þvêmÛÖ·éé½-×ÚX0Aô÷\O ÚA°Ú ´gh Þ6pë£cEóXEó ¡µè0 ô70ª¸0¡0DǨUL' Ï B Ìh0¡8,pÍ d@È ëKÿÿÿÿXíзãtú÷Õ:·Oµm¯ÖÚ!¥ºG4Õ$lma¨~yÓI¸":XÒlRlÚ8v&Nî=Q²Å£t @Üp B
á0.x$;÷÷í~ÿø+izÿÿÿúýx¿×N»i5jÝ=;BüVöÁEawP@ØtxÚpxÒð Ý ò\Á RuÒt VÛ]u¿TOÿÒ÷ý/õïïIÿ}t¿×éûûïÿ÷ ÿM«ÿÞ¿]ÿþÿ/ÿþIÊÉ%wV½kêÿV=+Õ\_úIv8¯ÿý'ÿí>¾¿ßòL¿¾´¿ÿÿÿïÿë»^ ÿþÐú#iDãöäK{ÁD~¿°µ¤Þ¾ë~ÝûW÷£;÷ÿõ\_ßéÿOÿoÿÿþ¾·ÿÿþÜ1LSPÁAÐR: ÜD·Äãø4¬ÿaj¯am.÷®ÿùÓÿtJþÿþÍ[ýlmÿßþ·ïÿë×ûïõëôº0¯vÐvÚA86¡¼SN÷ÝÚÄãá¶{ê÷4\_ÚÓ¦¾Ý¥ví¯ßÿ¿ñ\_ÿuû÷ÿõÿ½êº÷Û½¿"" $'ÿau(p ¨tÔ064(&ÆÖ°a{ì"Çô¾í¦öí&þ÷
ÛÒí.¿ÛÕÒ¿öú½Wô´×ï÷\_ÿýSdhDDqráì/¶¹Ý&ÜpØ @áÄ(mzj0Xg$ C9m±;û#í®Ã=þÝ{ÿ}]HØgý³?Ví´·\_m»î¾õonÖ©:T"""""""HMS(ra
û^î0Ãb¸Å6 lq;4°ªÕ.¬3wl0§lUÁ[l3~:SöáØg½&Òû#û¶f¸g&;¯"""""""Ó · &ôIí\ ¦ 56!pÁ 8lSA @Ã6: pi4hI´ãµÇ"8a=iPÆ1KPaOÿü0¼ì.0@Ê+3>Pð 3}å@[ »ÃAÅ1A Aà a0?Ò""""""""""""""""""""tÍl!hYC °ÐVÉm!¬DDDDGëÛI.Ø$vP>
\aqNs²^hÎÔvga<)ÜÇ s²¥¶YËòÎ'&HÈHfU3²¬
äØ_Ô:x) ÊÉɲqìÖ+þW\{¦éþ§j¯^MÔ~BÿààæG@áïý+ê× ¼·|ïMws»Ý¼&<ÌÇ#©K=ýÒꪰ{EÛ85Ð0AàsY 2ZÉ 0Q5KÿÿÿÕ(Ú ½B7pÑyÓ» "íA ø 33; 0Ëu(Öúÿÿø2ΤØ10õnK ôÜO è&ØÚ6=4naAB/ |.Ã64Á4QHͳÞlGd2 &Éò1g×ÿ×ÿ§ø'®½´[VÃÔ§IîL"í "áKgxpdt\30@È30@Él\À¤J ò<áåFj2y:$#W#µ(Øɶf!"63 ÈhÝç2è»8ÍfDdæJVÿý|W§éºÒÚ½ºtôt¶/¨ ÜL; úéø!è¼i¦0MúhÐE»¸ ðÏ xA¦GDtu#Äa<ñÙXÐg¡pÏ ñÓ/Çw ÍL dÃE0ȾBaI±ÆF³lÌR) L4Èbýþ¿ûuoéǧWÿzäóAÒ°²,ãñáÐA°Ä÷bxE! Ó Ú@è ߣqaµT.8H"Þvðª7¦Íp
aBå0°ÁUØ\iÚÜ -Ã\ÉB&8TBrî ßëÿÿïþÇWÅ-[§ üg° &tßÄ®jôé6D¬Ð81=´|8:Ø1?0cI°ÄùK4ºXéÝBA¶&xMaZ6ªF¡ÞxÑy vê#ã\ $
QäH^¸\_ÿ¿ékëè}þ½éövê¯Ò¿Ø\láI÷i'Vá;'zvi7%e,á6}ab~Óß%Ô5m! ôôÐAé&þ¨HO"aüh~Ié/äB6¾4þ¿WþõãëM¯¾þ½>þúO[·ºÝk´ÓõÓü'[¦¦Þ©µÙ(ÓÓé6¸3Q83RÿñäÅä7òdôÿØ>?\_ú÷ÿÛý¯ë±\_ÿ¦ÿõ×K¡ÞÔ{\_éº~¿Õ'«¡[\_ý½/ûïz òê&3¢ûöþÞ"5ÿÿÿÿ÷
oÿßÿÿJ;ÿûõ÷õ½W¯¯øýz½xºþ8ÿÿýzÎ¥×ú^ýømdÆÿÿÿëë( ¥èr\_ý{ÿÿý}ãÿ÷µíVö¿×Mþ¿ýïÚ×ûý×K~ÿÿýûÿüÛÿ÷߯ßýëý_m/Xªò@_%ÆÈC³ _»ü÷ýþ¿¿á¿ûÿþ¿¯¿ô ÒêLoú¶ÿÿ¥¯ßüå_¿ã.¹p¿3vÞ½yð«ÃÁîÞÿ®úOýZùôýÿÿÿÿõþÿðo÷¯ÿïÉícUT"¾|±e@½wQNO:&9C« çDÇ+iPÈÿÙxuÚéZVÿ¯}köÿÿúÿÿõ¶¾ä ÿPÅõþÿÿ¿·ßÿú ë_¤¸7 ©\lzlmì#ûI³>û_~þ¶òÝú_}×þ¿Zï׳'úýþÿ¯ú_éWýþÚÿÚ_e¥h8a àØ«S ö%ûJýM®Ûÿÿõ÷_ÿØ®¬'µôþëÿ[ÿ÷ÿÿ]]á×®áh¦Uà ì(qP AÃ
và 18áÃI ÖÛKÿ´¯û¯ï×ÒÿWÿßû[Ò»KKö(×ß¾ÿûú]½}¿¯"""" a3 &m]à Ba\U1@ì3 ´»3ØAX,2:MÒ¿û ÚMû}ªÿí_jÚ· -êÒoð_ï¿õé¾í{´û0!¦(6Ü0E1;ÃI}mǵ $Ã9ZM¤Ã^~ý¶F+ -¥ $Ã=ðÈûÕ^Ù¶gÿa{[Ûíéx[sEav´9Ç ¦Ú 1@ 8l(|Bba&Á(ÁØ á±¦l^'vÅl,B Bcqklí ¾Ä§µL3Ã9C±û ÷{"8äL b"""" è0F¬e:iÚz0Óºv0a´66 بXb¡:zh pØ @ØÁÔ>) b0±ÈÝS"ÀnÇq6""" ¥\5@XPåjY3)%¦sº /à Úe8A
µaD&Ú A""#ãÐ"""$5B" Áz¥¡Ò_-ô¿¥éRT¯üzN¤µÁÒó¾ÄéM B´aU2+DF¦Aã+¸\þOðêM´½£Ü·UÈr³;¤Wµùؾ¢pS¼^±b@¯ï
£±»pwÔqtdX8ÿªÏdþáÁɱc°»éZ\®aäÉy_×Þý|W¬'á__ÿÿëµNÿ03¨ÈLáÿú¯ñ¿¦3YAæhÉólÌR¬\¨2DSTBýñþ@vL"ç<5¸ "Ü9° ; 0) d<Ø΢äGDtD ?ΨÊy[ÿÿ\Òl1=Ѳ¸}PNA°¡0z70|.ÙÁ¦gÆ^"gÈ ìÑù8
A83 <ÛS §B5 j"ÉTAÄCRX¿ÿÓ®nYáÁ&Òo&ÕÛwFÈa6©!ÇBå "c³±Q¦. ²'aa0@Í a DCÍè)#7AѲ5 gB8Íq¿!¡ÄäH&gTq"ôk2´_¯näî¯Nb.«l¡Ë¥°Évaä$¤ëh v-0ª¼!,ÖÑ°ñ¢í
H"á¢á0Ï (DÇ ñ Ó.3ä| xAÆ e ÀäL Tc3CC" 0A3Fi0DTD4fc$fÈùóH£(FÌÑ óÄC#¢:_ëñÿÆo¦ÒêÅë§Ý¦ëzI¸AÃ,öHún2\%s«9p©¤i:M £ce'vÍ ¦õ@i¦qÕSO ¨0T\0Ú ^0T-áB.ØPSPá¦ÖÊqPpÏL ÏÁá Á0¬ABþë×ÿ_¾:Òº¯V=:_®ÓoÒ×
NrW¬êËõý8 4rGºÆĬ ÁÁÃÀñ=½gçGP·I7NÌè w Omòu¼lL;Fäzj¨ ÜZ¯Å Ñ» jwÿÅÿO¯ñÿ÷Å´½/ýoõv§öþ«§[§Wí6»0Äã÷K ·ItØIðqw tá 8I°bW°cI°ÈìJÆ,Ý' Jìý ~ëÍ×ñ¯býþø"úÿÿÿV?ú¯ÿ^#Ó¥i?MÕÐ×NUÓMÓû¶]ÿOOÓ¯òy«¯éÙ;®iÿiß´Ûtü±§¢w¿äuù.)ôï¯øßÿ½ÿºýtýwúôþ?ï]õãÚí=kÿxî®tô){\_û^õÓëïÿAòÌ?DóÞ¨ç#ûiÿ
ÙW÷û¢FÈ
\_ÿ¿ÖµÿûÿÕÖúÿ¿®ºïúßÿjÇqú½u~ÿoè/\_@ºÿDÄù&ÿú$ýj ÿã¿ÿ÷Á|ðL¨Ø2__HùN?[ÿÿÿûë~ÿ¯ÿþ¿×A¶¯ÿ,¿¯Ýn'º!%ÿþAÿÿç@¾úÉqßÿ|ó \_¶ª?úÿ ù Oÿ¿¯þÿ¯¿ÿßßï¯ø7ÿ®Å ¶+ ^¿ëòÄeÿÿÖþ´¾X'ÿë ×JëòEyûß^ÃðDQÿÿºõõû¯ï\_ú¯í-wÿÿK \0²ëÿÿÿÿÿýüt ¼¨ ÿü7ú¿Ëü±ëü7èþÝö¾½ÿ¯ußëîýÿýÓÿsU橺ûÿ\_ßÿÿÿ\_맪ýýçRþ¿¿ÿõ¾üÐ\_ûöÏ^Ø[/Úíí¥êÛûk÷ï¯ú¿ÿéR¥·ÿ¿\_þþÿý-ýݽÿ¿\_ûÿÿÿÍGÒ\_ð@ÁAÅáÄîãbqû +ip0öí+/¶f¿\_mí{ý¿ïzõµVÿýµ¿÷ÿÿÿÿÿ§Í^¿¶¿ßÿ÷úïÿëßÿ¿øaµ ª)´ ¶A1¥-ñ)Ú|30Ï~ȹ²ú¶f´¿µïUïð¿jõýíûzÿׯú·»ÿÿþ¾«ÛzVÿÿ~yV¿ëUpC¦^LR
m1 ¦(pÚpl{N|0[K[mÈA| ¿³ý¥a}µ´¯Ûïî×ûK×î¤ÝéX\_l/ÿêýú÷ÿ¯÷ék·ÄDDDDDÁh^ü00vÐA´a pØ D-7ö\LNîíݶ"çrÛ8ÓÛ=ÃÃÿ°a~D8ôlõ}
øí¥ j×ÛlÏÛ\_ì/ÿh+ ÚÄDDDDDJ"e9Q a×áá«pÔUCm ج4 1@±A w½m¨PVÅC DðVvLJwq)Ú{^ ,3nôî'¿iA¢qêÄ¡Üqqc°ÁÈDé÷ ' Ý
Øi®îã °¡M{a
va\¡Ô0AÁ±\!Å pÁÔA E5 Ú CzÁ4%B:±eo &T)Üã#áìãðÁa]ö¿¶ M®©ib"""""""""""3mH¤£4ëéñ«_ÿô¾¯ü´Õ¥}[IÒý6 0ZÅLWÃ&P ; !Èl=KqLå¸ËuéÔ@åt0¡KrX·Jg}ð§eY]®[© A S·7gj§}BeRôeÖ-üÉbS²ã Ñ?.Õaðpö®ÿþ+(äüªÑö°¿þ©
ý?;ÎÅt¿F±ç".ÍK\_¥úñ×iB®×±Ã00@Ì)ÐdÙL~kûÿþ¬Wñá -úaB.Ú!<· ñ fì wy²:çê3©ÌêÍLèëÿ~Úõýsõ£u ÚaPECL"
àhÈM a 0AÙ!22'aB "(21zëëÿÒot é7%i''Ì l1<5 Bs¢ ¦¹
M¼¢îv.ác¸gÆ<(A °A É !;(Ê èdÆTdºÉbëÿëÓkÓ¯V5m7N ²eYíp¢V4'¶Ó ÚNè ÃÓFàÑ ±roØæÈ]°1ÃTBpdt;4Ó0PÁ24!iÄù¶|d<ÙÆN<À§8;sdu"S¨¤Ê3LáG#´{4G ÔDh Êxÿ=¤G_Æ«¦ëW×Iû§ÅÒ°q§^tI÷W§Ü:N JÊM
@ÕÒ}PM£[Ð m¡Qn"Ûi]aJq "nÍðÏPièïÑ E "Ü0L3Ã8Ag ÁÁf2v(3¨CÙñpA@¼}G×ÚÅÿÕúzm¦Öÿêý}nKe^Êi81«tKIéI'TnoH'Iѱ±hÝãAõÑnÕ0FÇa :EÄ"ãêaµL"ìXôla¸k¡©BÖBavÁ§ß Î .ÆN"ÿWÿÿýÿõï~+ïïONõß'ºoV÷j']ËvÐJÝ7MK¡ ààÉs§4lÚ:MÈùõ"¢|¤ÚMt4ÛT1=´:AµH76bÕPMé A
^bø/úükÿÿu÷ûûoÿW_ÿýýzöÖµ÷¦ëéÿä¯NÐt[ï@n[IÒßéØOÉNÈæ°dýÓÉvF\íæ×%ÐtÈóDÓÅÿo«àý?Ü~ðpýý'èvî¼~»\_ü]´±~¿ö©Ý+^º¿ÚuéëߪmקW\Vé÷§ÓrQìjWýïòc8Dñþ·ú×öÈ\_!xrd-?ÿò{"ü.¯þÅw×ý==:û¶¿úÚý¿ÿúOWÿWëÐÚ[×]ÒÿÖ¿Ð\_ÿÿ,\_Ý6rÛ\_ÿÖ\C4 QÁçA8óPêˤ0\_\_ø~úÿZÿý tí×ÿ¾þÿOOÖꣷþÿKÿÿÿðÛrboÿÿ&9UÈ8è(Ý0x'®H ñ
ÿþBÓ¹¨OãûKÿ×ÿ\_ìº\_××ôwÿÿþõÿÿþ¾µ¯!@ò¬þ»þ?&Éz/¼Ü"yÑÊ 0ßÿöàÇþ¾ÿõúÿõÿ\_ñÿÿô¿ô·«ßÛÿÿöûáÁÜ6ÿÿy&[¥ø6©Wè.Ùeè/ÿðßáïòr¯ò¿]´¿´¿ä å!a¿_ûi_úíÛÖý·ÿ_ý}·Òó~m>kMߺÿÞÃú_¥ï¥§cKÿüè?êÿún»ÿÿÿý°ü±Aÿû¥ÿ{±í^Ý¥vhÿ·_Ý÷[ýþûï]æûýÝëÿ¹ßõÿþúUöÿëÿïýZþÅ\_÷é~½þÿõb
lwÁm¤È¸ o?ì7¥ûjuÒöïþëï÷óèÿöjõ׿I+Í^»÷þ}?ÿÿ¥iu½¥êõøf1¿ëÿÿý´µßᦠAÁ)¦0ÅXNØä\iHXg(au}¥fkô×Ý-¸zQíá¦ïÿÕ·ß×m\_\_\_ïßï¿î×A}×þõ±\_ýïPàACè0NIÛ Ðpê¦Å1L0429à CJî´¸gg«×³5k¶h½p¶ÝáR{Ò°¿kûþº¾Û¥Ý{ÓÚLWá[u¶ýëþ?ýb""""CB"" ^P£ TTP Øpl4llT0&Ã#¬0¶"À;03ÜÒ¶ÿð¾ÃOûJޫݤù
{fûnÚ\_ýÝØ\_»µé¯ï"""HH» a ,5xb0Ðj+P m6ì{<$Ãb¡Ú{£iÚjÄãøïÛ=°Ï|W°¶ ÜÃ<¢qðM!ìúm)Ã?ßkí[×ÿ·ñq 0
T |5ÓVÓ ¨P÷j)¨a ¤ a\ Áj pØØÒ
XAÁ±Mãj! ÃÆÚ[ mÇkq§ %»§¦Áø2ª8àDA"ÁL ÊtÌ8"?´×»
í} à <0Ði®ª!Û´Å4Ä @á°¡bàµAØÐ=@¨FD ÌÕËN#a Ða3 è0spÖÓ»ò1á
õHEqq¤dDDDDDDDDDDDDDoÒJ©}\ßIê)j½Rÿº]B\_iùk,SÁ'XñÃ)ÚÒa;UÊãÃ&A¼DDåo&XÈAey¹ÙKÓìeuB,F@¨/¨pîv¨¼È¨dXfBæ[ß&ã<Ãá¤ò¨ÎÀL°¥k;'§@º´ ·î ý¡é¦ÿëÿþî þ¡ýoþÖÑ\/$§óOÉ|ìò »ÿ2S5ÿUãUÚVé¨Wz¹÷Ð2: Á'é:æÌùB\_ÿë?®>8÷]¦xÔ"ì32M3AÐ(0@ÈlRmuÍÖD¿¯×ÿýWÖ¢VRl"òn,'¦Fà¦@Ô"à31 Áhª|!Áóã3üØû6dc; d#/ýýQßðf!7Z:MKøÁX'¶aá±Taé£sP{ ßÍM=@Á40C"Y¶f)Fpd2:"r>)"^"UbT/þ¯÷ïý:¶º½BziÒvpd¾ i81+V {L PÖ ÚaaB )00¡pÍa|g&fd!£> à¥")s"y³" ðGÄz{$D.u<ÀÎù ¤Éáå Êu\Èc& ç%)2¡.Sºÿ×\_m?TëtôÜUºtx3!ýÐ;%Ic¨ØI´l'¶}ô6é\:Õª4Bi¦ g¡-ÙMB&8g"ó°@ôE»5L"cpa"Ã.a °0A0Ír.x gÌÑÙ'hÌ NÐ!üø¦Úy~Úê)?¾þÿ] ]7ô;Ó¯N ÄôôÛ Òpc\1+°a°ÇO?0b{aØÅܱޡ an,&Ô& £t"Î, z/(oLÁmE
Ãpa= 8®¡]°TÖѺà¨
¨ÿÿú¿Þ§[¯Kºß~o}ÒvNé;MîìR}áIÃA.@àÇMõMKøÒl:N©7%Úl: Õ°b{r=$Ú:¤ØA¤á¡¢z$;Ç{
AØwNh0VXX¿ÿÿþµÿïý{{¯¯ÿã×]t¾é¿W¶©é¹+¥n«´Ó®pf§]tëû¥peÒ º5t.iÒi&ØN º Ç¡áÙCJ¶ ëjÿAÙÈóõ)ÿý}ÿÿñþ·;Òÿÿÿ_ÿ¡±®º§ë¡Wº¦öºÒoýzÒûöéþªµõºuºn¡;ÓøúWMÓöÚî Boýÿ}µÿÿÿu÷ïÿ©tG@ ¿úõÿõïýþß_Ý_ûïñ¯þízý×Þ¾§ÿÞ¾»¯Þ»ßÿõëÿÿË/ßÒÒÿþ"3²¿õä{"ëÞºý-kõþô¾økÿõÚõøþÿ÷ÿþ¼1Õ~»ë÷ÿß¿ÿý¤Þßÿüçr¶XJÿýx/Ã"A÷ûÿýÛ¿ùMÿÅr±ÿ÷Gí.ÿïzýømÎ>¿ëï}ÖëýÿߥúÖ¿B0¿þLb'aJÿõÿÿ¥ÿþÁ×ÿ°ëÿãÿÿëïÿô-¢;ÿûßÿÿÿïû÷Kô«¿ý/ÿïKðt ×õÿþùb?,gÛ¿ùpÜuÿ¢(å#öwÿÿÿÿ·¥¥ÿ¿Ûµõïºý-ëKûئÿÿúÿþª¿# ÿíÿÿuÿÿ ÷þ¡êÿéúÿ\_¾¿ÿÑa¶öÿüý·¥ßýk×µÕ¿b¿ÿuþÿú¿×ý}uÿþSÿçB¿ú¿ÿßÿÿÿ§ô¿\_ºvÁL3t4½n¯´Õ¥éáië¯ûÿÖßUüÿ[ïÿõn¯ûýÿûÃÿ÷÷ïïÿúÿ¿ëÿ¥÷Ú8lSØH à §mà[iCA0Ï[d}ÿ»ûû×Û^îõëZÿK¿\_ÿóDôÿùÏÿ¿ï¿ÿúÿÿóÛé7ÿÁ"A½ Á Aa¢¤ã¤DçÈÛ?¶á®Ý¥
þµöÖ×íw¨Ûý¿¿»¯mú¿ÿÛ\_®ëïÿ\_¿Ò}Øúú""""!
)Îéë B\_N6 Ó¶ '¶Ä§{ å Maÿð÷ ý »Úm~ÿ¿Òû3KíÿêÝ®êÞ¿þ¿¥çßzÓKA¦«¤öPaðÁb¢¢¡iÔ0½=µµN\Û.Êûl÷ò$R-þà ö¶ùò$[?18ëJ˯´on¿^ëüþ.#¨0IMnZµ0®!qMC8mC 6$Ó Øim A±@Ã&(é8¦ÆØI{a$î/XÄã¹-¥~i=,wõª@B" "?=@ iéö«bK¶Km1L&+l0¸ a¤¦iÓLPAÁ´GBÐ@Ø Aà ,U8@ÜpZÄDDDDDDfÚB,@Ða2c !A !a¶M2ÇR Ó8ðÖ_Lñp×á
é|DDDDDDDDDDDDDDDDDDDDDGêÿú_¼ì4¿] øA¾MíWIñOaNPà©6DerX-T·#±»hÑ w¨ÌI?Þ+/2!~ %©nJ½upùÚF"]ên-D_°@É H#¡PdB/ÿõP°P.°¡ÃLe dLL tG ÔD©eÿõýÓAÐAѺѺyEØg¸hL33 HÌ3AJ4Ê"|Ù3QÏÍÿ_ JöÓîi=8tmAhÜ ÑAaì"cÎ ðÏA ì0) ~gQ#"r:£ÑNõ¿ÓúN¿¤î®.i7GI6 On¨ ÚT06B-á¼]½0D&S2\ ÌÔ32@2xÃÏpeIf/eÓþ¾¯N¿ZôðVäòéÐtÙCÔ&Òj Jê Þ%|i8bV:IÐNÄööÒl1+7ÿáÓÁé?®ÿöqî»ÿþ¿ûÿÿzºÿZjêäî:Án°Im7Oi4ÓÓm83:NpZ¦Úpf- ÿá°Ü± þî»üG×!ÿú÷ÿþ¿÷^»®µzÇÿ§zÚקïK^קõ¿^Á°ïþ»ÿÈ£úaÿ^¾¿¡ÿûþGFûþÿÿúôþµÿþ;ïøÿþÿòyHd2ÿÝ/þYEùeß÷¢u~ÿ¯×ÿ®"?Òÿ¿ð¿þ׿úÃQÿÿ½/¥tqnÒýÿûÿÿßýÿÚ§äÅ?ÿ!ÈíÿïýõÿÐëvB)R¿d%]ÿó~æÍúßúý¯ÿÒ½úýÿÿÉú_Z_j(Ú¢ë%/ÿáöÿ ~ÿÚM¯ïoîþ½ý¯¿O¿ûî»úÿ¿ÿÿßÿÖÚHÚR [òeß÷ü7&]ÿ~Òë+ý-é÷ÿ½%sVö½ßÿOÿ××\_ºÿþÄ$¶!?H/ÿýA¿ûßÿñ ´n=Z^¤½¯uïýþ«ÿ÷\_Þ½®¯ý}ýì$¶ÿÚ^½yw×E9ÿÿà a+ Å&4¶Ò½ºVÒ¾þKoJ×öÛ ¿¥ÿw\_Ý\_·ÿÍ\_?Jÿw¿ýî¿ü±Ó¨ÔTP Ø AÃl ÚKl,B¸g+XÛ ÷Gí+I¿ßKn´ýz^®¤¿u¹?½¯ÿ¿ÄDdu3[a8a=±I±@ÅB P vƸÛJ aäLì3Ü0Ý+Jÿ½:éÿ[UnÒÕÿô÷4võÿ÷\DDDDD! "> ±ÖQ
iÝ°ÃiLS@Á±Å' (¦[»îÒì¼-Ùm'Z\_nÿInÿé¿í×"HÄDD ÐaJtÓû ¨LP mz§¶Ç Ç ÇÈ/@à D¡ßöÃ=®È¸úlöÚ[v¬3ò Ñ|DDDHÌDAà0¥ammA lP mp¡Û\á áð[¤ &0îà |FhDE¡# e=# Á;MØj é>ØWmD\ ! îLBÿ""""3R"¤eªDD(DÑ8(A /úAè'z¡ÿ뮺_ô«Òëÿ¿úPþ]TSþÖýFvßZZÄi«a&ÒRY%Ðp \Tìb1u;RØ«%he|0ºa"e:êVvR2ÅðEã è4lkga00²ÒKDb;¯ ×ýúz}[Më¶ ³õ4¼,ÔOm&ѳ½' 6Å^B¾ "¤\ ÌC=Á(R%äDh2ÊC:£F<£ÔDHÍQ¡QÃ)È ÙñÿÇõ¾[dï¿MY]ái=6. ¥¤ê AÖ\a4&ár 2c´DBv¡ Ì È2ÚONm' ¤îìNúvm_Òÿø"WÿFÿÿ=ýþ /ÁÚî¾äØfGGiH¡SzõôþîÝWûÓÓô÷× ¯§K]
¶ô_ ^ÿõ#sÿÙ Ãþ?£²¥8B¹¨ì,g®òïn·ÿÝoÇmW±×ÿÄ^½Òoõ×ß µ\_þpßÿÉ¿°ä:>¿oS±¤).)3Ù¨÷ ÿÿ¯ú±\_ kþ»úMúï¯éwýoßÿ
úý ¿,gäƽÐ"¨4C«{þfKøÿÿ:¨Ùk}uò!½dQû¶¾ÿß×ÿ¯¼7ÿöÑ ïÿaü:ÁZ\_ïàÇ¿B{öó uøúë¯uúïWß'ÿþ¿þûûUIÿýô¿ÿô ôLCDó{ÿ°ÿ/äÜ >Ýzí}}~ÿVîÿÿÒ9÷ÿ÷ú\_µký%·ÿÿþíà ºÿðoôýЯ®D·ØFd}ºì/º\_\_Ñ÷ÿ÷ýnß½½Vº]cÿ¯ëûÿþI\_§þu/Ò¯÷ïÐAÛq¶±)÷ÚR&mÿþý» k¥ý·k·ÕUö¿Õ}ÿuþûß¿øÿÞÕ](i8l+iA¨a ,'±¤©§Èìÿal ½³?Û¾¹ÞºÎ>wþ«±ÿëú¥y«ô¾ÿÏÿöïWÄDD ' k) @0AÃp بÃÈäÁfæÒµ¹!Æ,!ÔgAMdHDÏûÒûJÖ»Ûý뿦Ûÿ¿_÷µôÿ½X¨2P#Z &1ý°°ÓW8ê)A bi ÿáVBa=AÇ£63 ¥|uàÉNNé9 ä<¨] ÿfkî×ðªßÖØauõ××Û¾¸F"!°ÌàþA Ä+VÒÑ+jª;L"P,&P¶ ¸ àÔ ïòG ûÓ;0dq Qj»i5¶ mD:NDq)él3Õ¬0±8ýþ¡
¸a~äÙþJ¢"""" ÁõM0¨'NÐM²}Qpü0J©î¸LXMa0L a ð8tÃ#@é¨a&Ã8u B\0»¶'vpØØÅÛJ×vPä¾ÁMt é'¯VÆKé6 pÿAÁXû§FèthÙá0iÝÃ Ú BÚkÛ (
V¬ ap6 ì4U×ôGD½?_õµjòwÛéé÷Ii¶'{×N °"""$ >¡DA8B"Q "Ð0D í¥ö×ëïo¯¿{×ôýýp¯ou~DDDDDG¿ÿÿÿpÇúÒèGý/÷Kqë¯ZîÿÒ¿
'õï×Ûÿﮯõýkuÿ¾H ÕCΡ#~<êô¿ô¬üÿKìý¤hu ¶Á©/X"(ÿýü+¶ ÿé~¿Ø¤JìriÞÔ±X'ïÿµpßü(ÿé}«õÚ«ÿýߥþÝ ]-ÿý/þü%ØX;÷×[þ®·éù)"Ð/kzù«èºç\=úþ¿é~¬R×ë×þT+ýý®©_·ß¿þßÿëíÖ½w áW¯Il 9|ÖÌ ¾Û~ý~(êÿÚãþýDjí¯q¡G;Vû3^߯ëÿ
ëûxËA\x lP@ÚÙÃؤ$aXaSá¸Õ¾Ø]ûZá
qíl%¦Ú¶ÃLSÅbb0?L'aH8ÝÖÄãÓ¶'+'a m§ &2P»a+^ÐA á Ü&Ô0@Ü8P ¨3µ¨ìW
#1(ºDDDDDDiDDDÖJ>HtONYÙ ÎÆÈ®°¹6+XIOÐЩù\+»ëªÁÁé´ÕzõTè^ÿÿêçìío;Jß;2]»ÿKÖ©uÒnĨAA!YÌr$çAÑ̨"ý¥ib¿ÿM-Ú¦xÁO ÈLpÏLe ¤ÔS¢9ÐÛ4s%ýOûaXa$¿þè Ø1Ú è pM ª£pg¡xh032 ÌÍ04\À¤3GS"&!1Eq<\\_þ·Zz® $ôÁ&Øѳѱ£qé#sFÆ&EÛA¢ªh·r óùáB ff0Á0HQ)}Oþ«ßÌênGl27Ó§\1¤ØI°b{½Û AÁAPh#sEÛ /#¨Èüã5 d dάe:=NµUgk/Z¿ï¯ ©ÚOOMÓ.¢½7 I°Ù aÓr7]ÒVmÝ°P E Ñ!>a0 ó3F$Hò1dt 8ÙFuÐPDYÿßد½õ½=kôëÓt°q×WVÃD.AѽðEã ¨·hØBÑgÆD-¤\7T ÑpÑ^á ';+ÿø¿ÿý?õûkíõõ×ÄöéÒmb{i6 6©ÃÅXUA½ÓAÅ A°º74^«ï¬>:ßõÞµ÷ÿþ¿=&קYt[?B'¤ôØ3méÁë%Nø:X1+©7# 4AÒxDóÿ»\_ð\_\kÇ!ǯßÿý¼ãÒ×û2;éקþu}^Ótàúÿ½:Ø44ë¤étôÿá¾LezJ>÷]!ÿÿjüRa?ÿë}6½ =\_WãÕÿ\_õ¤ý=k{wÒßÿýßÐ/Ð-ÒåÔJÊZ ÿÿVûÿÿÿ~\úÿïö?cýþ·ãÿÿßý/Ð[ %ô=òc:ú&3ÿ'©!ÉPßñïü2>¡úÿká¦àûÿþ$"9k\_KVý¯ëb½úÿýöÒnªÒýøªÿ²ü
#² [ÿÇ!|CëðÿçÓöö¿®ÁÞÿõÿ®ÿfAyÙWü¯þIÊ×ÿ!Èx"QßØÿý=»\_czÿÿÿ@»¿ïÿ¿Kïß ®ÿúü!ÿåè/ÖßÃ}ÿòzäÆ7Ø\_µµ¿m^¯\_Õ-þ¿ý/cb¿ëý¿ÿ÷ÿZ ×£X¾ õëôø68ÝíbPý)þí.ýªöß{ÞýmßOÿýÿßÞ·}Ø?ÿÿÉ2°¡a\Ƕ±8þPù RÃ?¶f¿³5÷¥ßoOÿêúÿ·ÿõ÷÷÷û§ûððÖöÅ4ñMb¨Ãj;aÎ 5¶_dZà ç"OMöf¶öDÿºþßn·ÿ¯o翽íÍ0A Ê[ ý
mÅ( ¢éÁ{;>."!ÍbdPÏdµ#YÑf¸Y(]ûwWÿwßúíþý=~ÝÞñ¼§+aî^Ôîa5»LP h7ZÓ¸gEÏ& ù 2vH# d^"Éy:ô+Ó¡ÙNÈÃ3¶m+#þÃ?í·V]6gý
îßJ½/½?ݽ(dDDDDDDDA:j-à Á¨´oL ¦±65L& D(L ëE « L1Ãva ÁÝlm¤Ç{ jÁÛiG ñ8ýdLßúöÚL¯l3Ó ò{µ¤"""/¥%ÕÁ\ÒlÚN©6Ú66&®EÜ0 x´Ð lA "ì°.ÈN5A8l&) A lP llBÃ860@Øop $6K±Q±ÃJ¾½7MÓÓ.ØOAÃSJ÷ ·i'H6,±¤Ø1é<ýIMD Á4CLã# e8DtM¼5û Uq v!5v V¾ëuë[«¯[§úlj}~¶u¶J Ë¢eÎôËqs§#¡D¡ ¾êµÿÓúÿø½=ÓÝníítôúôÚëؤ""";Jç}\_þ¿ÿþ¿½î¶#U¿t?é6ÿõ¿ãÿÞ/CðWé\/Pa?ÕÿÿöÒ¶ZU¢ëÿZÿ' ½ùFù 5ÿÿÎü0Eº¶m[þLg¿Ñ1|Î+ýx"WÒa#uÿÿ@½Dm]ÿÿÿ \_þï ¢ûÿûjm)6AV
é{ÿôÿéôÝ#XÒ¿ÿíÛJ¢Å7ÿïÿ¿ÿßÿißÐb« Véx¯¿ëïþëJ¶Ò(O×ývͯ±Ù'OÿÚû÷ëýþ±õ\_ïú\_=yØXÊàýf½ºûÛö×ûJ×¾×û\_KUnæGÎÌ ì]½^Ã[ÒÛa¤Kgÿ°¿eÕµk« ¿×®«Ã§ÓÅ $6)ØÐw»¾.!D§}¬Jwé¶"Yð0EÏC»Õa
¸b¼0©86¡0¡¼SP6@[Ä ByÙîDDDA a UGRôÔ±Ó]S ^ØÓp¡»W©Î;Z×;®ñÄFDDIØRCÁ¡°a)Ýì¬õñ\_+è¼EFDW;£R¼à¤Ëÿø\_IÒÑöG:ÿ]/KHí׿ÿÿKÝB'u%w6GQ¤xί&\_¥ú]aU7tBÜ ¨@AC0i0DT ¶lÍdFÙ®ÈwöªãëCÐõÓFçL& ¡ã.E»1ANê)8òB(Gêd»R,´´·ÿõ¬ÄùI°ÄöѱºIh Ú8-¦xÑ®.¸a5.3A¡FÄ36Y)éÿûðxN»=8bÁzé?ÛA6tvta£d"ïúP!Á32;aa#,#¯ÿëÿÛÞ¦äñzMø3 ']'JôINNñÏÐÄ·aB/¨»hX ÒbbAQ28Èb# <À΢yz(6~t D»È
e"RÈEë×W¸øÿïÓïÚ·«¤é}4ÞÛàËHÈò" F¸LðÖD'a¢è0»"ad~AÁhadt}fÞ3æfVhÁPÁIÎ^¾¿ì«ºúîµþ¯xN"yñéTz
Õaë£q), hÜÑc´laBa0UÓ "z6 D/ÿð\_×úö4ýkkT?N=zXý«¤Þ¤©lÙ
¹¤àÅWíêa &v«ÄðÃ] ÜZ ÿ3 ür¯½¬úÿÿïãÓÕüúÉDuä¢õ.83+.Òt,aàð¸vPå%Ù×äŸDïþÃõ½î=õÿ(]+2¸º¡ÿº¯n^Ò×[ÝØOÓ~/«tàÒ·þ¼ ¿åÈaÿ_úø/ùÎ+«nSØ®Úÿ^ßãÿþû×_C××ý^ÿúÿ²_å~¥ßl+ÿ#¨|· ]i?úõÿôýÿ¿ý>ÿ錄òÿÉÿÿÿ¨/ýý§ä\;i¿Çÿ×úÿõÿ~¿ûÿÿÿwþÃßß÷ú¯þüJïxv"Wÿý~¾ëý[ÿÿ÷¥ÿÿþÝ¥ÿÙ©äM;ÿuõ\_÷þ¾[N">ÿÿ&%Ý-ÿÿõ\_ÖÿõÿzX\_û£7{·oÖúﯯû×ÿ
Ì/m.ÿúøa-ÿÿÿþý¥ü³ý¥kÙ
¶h½}}mVÞ®õÒÿïþ¿ ÿÿÿ÷áïþÿÿë¯ý¸AÛûa
Ò«#ÞÃ?ôÝmÞ÷kiaÓ\_¿ó-pÿ¶ÿëû2{û¥ß×ÿÿýÿi4UPÁhlpiq¦Û VÃ96 òaØam+ ½·¯ÿÿ¯ý×îþýïÿÿëÿýÔDA
,p^Õ°¢©4¢al$űÄξÓ7±(¥ uÍÄl~lÈÃ+q E×úZñ^¿÷ékÿm×ÿ¥ß½oë(ί£ &ka1 a é¨0AÃi¡-Ã4 A§ ¡3AÄñDlæNF¤z"H2¦|ký®ûuß·ëÒ{uzÿö¥ö½o\E!C-al÷ 0M µTn ½ ] ¨ÖÁS lg|ið30A a0"320@Èqù±iD§wú\ËädQ i6Ú\_eý;JÙn»»\_B"8M)&Ö'¶}BMÅ£gÓÓ ©
L"ÞBA£XgÆ>ESPÁ @à ÁA±HNÓ¾6LsImøÄîãVÒùuÖÙ(¤àÌBulÒl.a àààÉñ=°cFæ OmaÁØì qXAl&½± Ø[a °i¦)Ø8t tÔAbìK¥ÓÓ÷Ó×'zéÒþM#}; Òt·Ã\.²DÇMÐ(pB"%"A Á¡ìãÃL¡ËÌ:3 ÔÃGL/h1 Ö·¿Zô+×ÿ[Ó´é×VýÝ]8xN¾"(Èþ?Õ~ßÿô?¿ô#ý?ÿ«×îI6?ß\_û#¯ÿ\_ÿ¿üWú~ÿÿU¯ÿÿÄ|uÿ×µD¾²#ÿ÷ÿÿ®ýÿÉÁUQ ÿÿWC×Çÿÿ¥ÆðÿÿÕ&_ÿýÒ¢yþLr?ÿíÿ¥Aÿÿ~õÿÿº ¡ÿÿ×%5ûºßÿûïÿ×Kþÿ®¿2&Ò¿úWßÿÿ]¯ÿ¯«þê÷þ¶õÿïÿÒµýû¦!L£±Óuº\_míýÿú]þÝêÿþ^·2$K)" ÈÔJ²+U¥ß"ÿlÏìÍ^öûIp¿wzi7\_ûcõ R Ï$ÆkV ÓV jÃ[Íñ8ý>KÞÙí³ý°NßÙéú!XbAÅ Ãcj a 8lllL8hB´¹ØY}ÃX8:B!ìØMvÊt ´ì0
p Ð àÃÙlv¨ÎÊr¯ýþÍTSÁ2Ü ¥AÝ8kØ#²Á©Ø©ävM×ÿþ½ºª . W¿ý¨\:< p}§ÿÿ×ÒÉü=¯ý\_ªó°DvV¾v Êwÿÿÿûþ'\ìP ~R@¹²5:k÷õÿõIF¨4lhWMáè0|ÉÅÌ Kä(Ȥ]㿽Aîl,XAí4lxE¿ ÓÑnäÇhe[0ah"a # Ñ3Lò8ÿZ°I°e汤útlzx@ ÁTB60P!lðÔ Ï ãZ¹ôFÿ®o§¦éÒo >ñêúMV)&Ã; oPw¦A æ$DÅ'!HͲqMC,Ùeófk".D¢]è«ôV½êzuú{éáÒl2èQ é;82Àí lß ô.E» ¢'pÏ BAá ü Á§A" ¸ D ¨h#}ÿ¯XõëtÔ'[§é¹(ï ß´h ÜXM êwFæpôðÎôѱ[ѹCáv¦ !+5É1ëûÿÿÿïÿ¾8ýz½ týI{IÒl.²TM4tñêÝDöM ¤ÃTlO <\Aøÿÿ¡ü}~õÿ¦¿ÿ¡ýµ§«^¬>ìZûÓé|'K IÁDØa§ m9b ?ÿÿ¯þNøúßãýë×ú,jzzz\_®ºoÛéöë[׫úq÷éÿÿÉy1ïÿmUÕ|ÿ̲YoêvÕ ¸ôýÖµ¿ôß÷õþÿ¯ßúÐ\_òe;wäÆðDQþÚAJÔvPr%ÿêÿ}tÿ×û\_ÿÿûÿÿ¿ýþ×ú¯ÿ~çü:h;þ8ºÿøÍþ?xfÅÖüåù \_àÿ>ïúÿïú\_þõïÿè/Ý'Ô4èAÁ\_ÿÐ"Oþ¿/ÐÁ/Ù õÿWÕýúÿéoÿÿ¿×]äÜJÜí^äûÿ&.XÿËN þYäÇ(Qÿ&98"xÿoõý¯¥ÿßÿýÿ÷ÿëþí-ò¢éúÿ«Ð\_ÿuãè/öÿj»ueÕ¤Ã3^ÞÚ¯k¥ý×ÿÝíõ?ÿÿßÿõßöÿ¿×ùH/ãNÛJ. » õÄ¡ÿ
ÞÚûºúüW¯þ·wýoiþþÿúÿÿ8lBj)AÛV±)ñ[ åº\_gÿ²êë°·ýýo^ýßú[þ÷ýÿþD'þvP jT Ãh ÃPШ8@b;ÁS¸¶ÒعÚóffd<Ù³Yl¯ÿ[I»þ×Û×û#?ï¿ÿKûÿ+nÔ gÊs¼jTa
aA Ó(AB-Ã>H0A Ð Ph'ä!².dÍÓñÎ<ÍDL" fD_2]oiYlÏ¿n¶îÖÐ[ißíº[§ÿ_ÄDDDDDDDD 9CGA#á¦×vèÙAha4-0Æ&Kpj|a8g"8ðA&30äïÁa8aWô¶Írcá¿Øgûº²ýþ\_eÿþ""""""""4ÈìOlí¤ M ê ê@Ú¨L0 ÓFÆ60¡0qè¸EÃTByn1LPAÁÁcb¤G( ØðMXÙÆliÜNín;K[õ°a=[ Ãó4 JçVÒê%cI·IÒl1=´°º6º¨ ܨØb´L&ç$à 4Å48mC6¢8zK§ëÚÕá;MÉ]m¦ÒvJ'§éÒl5Npe#=tÈÞlDDDIpg((LhD0A (pé zpÓ ¾¸ÓõõOBÓ¯OõN½§«{ø4©6¸hR±ªÿÿ¯ôÿïô;kÿãõþþêú÷ÿßúî¿þ¿øÿ¯ûÿÿßÿÿ÷øÿ_òè/ÿ¯\_Ý/ÿÿÿÿõÕºõ¨4#ÿø×\_^¿ÿÿÿÿòc?übáÿ޽ס_ÿÿÿÿÿéoXÿ'WW}__ÿõÿOíú½Çÿßýö¾¿õúýú_Ð?ÿûÿQ·ÿýÿëµÒ¿oÿ=»ÿûëë2#»¥ÿí}4·«Kø£_÷ýÿëw÷̦L3i~½½{iwÞßà¾ßÓ÷ö»×&ë[¤iÚiðÒí¶ÒL3ß"WÙþõ³ÿß[ Í6·k[ÆZ Å&p Ða@Ã@a¢£Mýma¤ÇÜxí ħÈëÞK²Ü8!(BÎr §¸aoV
0]h qL @á´7Ô¶?ÄDDDDE°É $ ¥ç B:0§Ã§ maÖbngd¬·|·ge×ãô"""""""""" ÁËXPRºXõ«A?;þ±Ãåz¬_¡ñOÕ-Á¤Ã§¿®¿¢n²«yßõÿ®J/
kyÙHKÍÔRB9Q ò£úßÇÇõª.!.h±Ùò A Á(!Q"
Ðwßéþ0 0ºa ªH·£paSLðÐh·gh±Ø@Ð2:0@40²@J62DN)(AgS"%Îÿ_ëý<uå[ a'bgiºh ÞB¨ÜT!x¸vñaA0ÚhWýýlMÖÈÞIXa%t1+¡+%u'Fæ¤Øbxj Åèݸn " ¶ ³ ÉÛ4"òÃó§2¦Í}¡~Ý[ôêÝS´§¦÷§Yé; ätMÉcIÐAÕµA6Å¿R';>Ja;.r!Á8Ã5o®ºþ}ÿß[§éÿIÒlt¶.L ^4la U Bå0¶ l'D Ó<5¸ ðÓpjAùA=gÌùÿ×èw÷ßë±ýé¾õéÝ.$Òm'PàÄöÖ' ¿A6otõAôÐ'IB5h¾é^5 ¼"l+ãô¿ëÿ¿ÿÿcNºýX2è{-&pf@AµlzÝý5lÍ&ÁºnÓÄ÷I°àĬ×ÿýøDuÇòZéÿðñíÿr¸¦dE«Ó«k×ý:\'Ié¾½=:»'paM| ÚM¶Ói0_þLgÿãÿÁÿßù%Ýl/ W84ÎÀïè¯ÿ¡ñúzn}Ðîºôÿ¤ë\_\_¯¿¯åá°çÿÿýýÐÿAå-©Ù1^¬\_ë°ïÿÿÿï¯ÒýºÛþ½ÿý÷ÿÿàè/ïÿÿÿ!ȪJ«¦©è|rëÿä åàÁ<ÿÝïÿ¸k¯OþëÿþÿÁ´¿ÿ\_ü¿Y1"ßÊÑö)TþýqgÃûë¯ér%ãÿ^¯ïÛýõ[áÿßÿ¯ÁÿßPa?|ìb¦ù11ÿ½ÈQÊ9+ÿÿÈ á{}0W÷Û¥ûû¯í+\òéÿµüò¯ëVge¶IÿÐ7ÿòu
ú]ÿ,GÒßäÆWt¿~Þßÿ«}ÿuÿÿ»ÚÿK©kýgÿûü/ÿúëü.µ^ÃJÚLÏý¶öÿX\_uýëÿ¿ïÿÿVýýÿúý×ÿÿ¿mp¿Ø
aZaM/K§L3ëaaþëî×ûíºÛß×ë{ÜÕ¿ï§ßÿÖþ¯ýo×û,p¡MÐ@ئ(ÃbáƤȸá¤ÙíÒ²ýNÒPDïþ¿öÿë÷þ¿õw·ÙÍý÷ÐCQ eZ ÷]Å5 ±A á4á¤Æ÷ ;Og¼ØÊ2I gLãÈÑy=ÈÖFÕõ®Þßÿï··ªõÝiú¥zéDDDDFc W ý Å'iÅA "c/L"z, Íð²# Ì'eÌ2vAà#A ± d1J3dt6?!yÔ¤L åÙþÒþË¥ìÍvf¬/ -´¡¤Ù¢ÿÒm&ÿ¾""""""" Á Ðan bgkÇx'\_¡0WèØ\iéEà N-Ù0>4Ñ pMàÈèo ØÐiR ;ì0dsV q)÷dt-´¡ dC¥¾Ò{OÑö§O¸0Aä0y0sFÐ:MÛØÒÄÎ'wôtn \7¶´n ISA0Ô0AÄP lTSX@àÁ(ÆawÒÒn¬¿ ºÁúxAÃ,äcÒl0L Üpû'Ñ6. 2Mf hD2¦]2GÃÉ &òÇ aB -¦¶!JtI§ ¿ë«ô5ÿë·´ôÛ¯Z¶ .Ótí83Sø2éqDeïXB Ì©ê=µýÓúûÓô4õôû½z^ôÿþýd?ÿõ\_ï¥÷ÿ¿út+×ÿØþþ©?÷ÿÿ¯ÿõúÿýÿõÿ¿ÿÿÿÿ»½ÿëÿÿògøü Où_éjëþ«ëþÿÿþÃÿüÿö¿ëðAþÿßÿÿÿëþÿòÄt_?öý[ô¾÷Òë×Kýú_ÿÿý/ü÷Ký(ö¯¿ÿÿíÿïþ~¿¥_àÿþwøI´¯ÿߤëïõïëþy;Ýwÿþþí °Þ^ïOþ¿î¿ý_×ÿOü?ø ÁLT® 7i{}i8_µïõÿëíý¿ëïÿº,"é )NA¥;®16¿"e8a dL·ZûuÿÞwWþµíßî¢"#íNÃ+±Îá) @áÐ a»tÛI7Um.ûÖøg6Òd\ ô¼Jß"_J¬ºº¥a{¨"f0Ðb»inqÓPè Ø A lP a4@àLm¶( í«îUbÛJiw þ"""" '@å8Dt-ý
§íñ4Å4AÃ
iÃk]1E¸DLS"ò·GhXuDDDDDDA<ͨ e
A[)Ù;ÕýTBy-þ"""68(PS¸ïS¦¿tÐ">ߤÚÿV?·è!páÿÿÿúEqX©Ú^RÒ¿þ¿m%\_ì'Óó²#eä1\_:ÿôõÿÕa(A¢á;°AF s:Î¥È#¬òö×wþ¶©
Ñ»¢ÀÎ Q A¢!Éóà2 y²58f£:M¥ÿ¯øMç¤Tn©é¢í
4na¥}0UÐxAXáá0A¡D{ÿÿ¼'\þáK9#]ÜtM°°ÝáØôL&i\ÆÅ' S S üÆtFÌÌò@ΤH+6gTyyì2'襮/·^÷ÕÉ\phAKéðÁ¤ÉsòĬÍÔ£c- ¡Î B.D'0"{C"ñO0AÀ[Á2]ÈEÃA: QëûWhzq÷¦ß¤®×§ôôÝ\_Ónú6 A´]hÜ.!U$^4-<"ñ¢î·i°·L#cAÚÁ4Ó· àÓ6¿÷ÿ¯~«ßï×^ëi=?î OTÚ5I6à °dv'·¤H: Ú ±8í&õF¶Ð0¢ÓAçe!?äM}]~¿Þ;JºQú¿||G;ZPJêØ3S¥5:Ng%nOÕ×¼¯L¡Ë8@éįtÒK¬¤Ø2XáA¿ýõ÷òïûïÿ_;U"Ü«$«v¾µú×ééÚkÕ¿^v¢ôêÝ=_§^{U@ø">ßî¿#~ûÚM¤ÿèwÅ ÄT&Kò Õz·½~+u¿];½tßÓbíÖÓtÝ?AÆ û{¯ ^¿ÿÿ¯ÒY y%¤)Ø.F wWþ¿_¯ÿW¶ÿïëô¿ÿÿõÿÐ_þÿïòeOÃGwOÉ[úþ5ÿ÷ûþD9m/×ð^«õÿ¯ø7^ú\_û{i?úý%Áa´ ;+³²Ü¥'Ü=ÿÿ\_\_ÿÿìÿü~f¾ûýäë÷ÿý~´¿ÿm½s©¥®&Fî¿ÈÿÿW\_ÿ¿ý¶þÒþBPNô¿¯ö¿oÒ÷Kÿ¿kýëÿÖßÿK¿µÿ®ÿ¿ÿúÿÃÞ¯¯Éx[òÄ
ÿÿuÿÖþ×ßÿ]sɪéÿÇÇñ×þÿÿÿûÉ$ÿÿIz_ÿwü{Kî»û íÿýÿÛúúö¿ÿ××ÒïÝÿuÿáÃôÚ_¿¥ßýÿaC\4©6Ò¹DHDºT×Ýÿ·ö·Úÿýý¿½ûÿÿÑÍͧÿÿÒ÷¯ßûWíßu ôºxN°î'YÉu&ÁÁmÖ@Á8a- "uÎ:dÇÎ: -ÚÝuÕý×_ú»jµº½>étÚ^ôõòwÄDDqÄFDDDDm¯éþÅkúÕ}ÖÿýÛ¯Ö¿öÒþ46».¾×Ñýô?ßî¿ý÷ׯzþ¿ÿúÒýWøë'÷ÿý-(òÅÉ|ÏþÞßþÿç?]¿ÿÿýߥ×÷ýItJÿÿ!ÇúZ_ïOÿ¯û__öÿ@½åÿõýÑÝ÷ÿîú¿ú½U{ÿÿt Jp÷µ}ÿ¿K÷þ¿_ÿÿA¡ñÝûÒ»¿öêýoöïKÖ×ûðLä×m´ÿýwÿ^ý{ýI¦dó"ãìõgýä\ ÿöaYtÝ5^ÿMÚZ»¯»~"5;ÁNþ6XAÃbPíG$î8¶Òß[ÍëöÃ=ØJUCáÿlÍ}|HI9ÜY'aI6b pÚiè8aÆ80@á±Ä&'v6®;X0NúCO̬R\_:1!hL¡Í ÓL&Óa/íB
i¡´8a@#
2";ÕTª{jËÀÐaAt"" ±3AC !eD">[
;Aæn©´¡£cÄDDDDDUAJÐëôÚ&ïI½áôÿõü uIõÿÿôÿõäK!S°µÿëUhãû×U]çbð gAJ3o!t%9óëëÕÒÖ¿Õq -ÙTDM2>ILÁ0.ªiÆxCèôie#'T¦´Q¿ÿÿÓAÕSÜ.!aT(F²é¢¯AU<(PPTBCèÒ ð£guß~»ðbW´4m'WIÒJ¡]ªÒzAUÃÔ&-.ÚKþ°DNÉÜsé82Æãäì) ~Ï:aèøIiéV¡V¸in(Üh3¡äÁÇu7:çLÍLdºìt]iµéúuøA¯aSÔc£Ítò¢ ;Ô$ÜùÁÑÇ£?Kb¬"Ü9=0aa°PDOfÌ@@ʱITF±fq;'Ì3#djÛÆëñ×ÞÐuKê+KIqC _Ixè.qAª£cOØѺp^.Ú!1é\4kÏ58h·áN BgÆ ¾¿ê/_UòýÃKúKàøIb©tµ¶o4£sÒm]$ljl6Ct ¤XFîÅ ÿT\8°º7\
¯¿ý7^êæëôºK´¬2ÉzÝ>u.°eÑ82èm¥ÓÉu' ÒÁÉØ1 %ÔuoõÔWý¯ûþ½zÖÚK¯Å$£;;[{ôÕ¯»ï¾î«{{Ó¤ïúôêÒ\_åòuëþé}.º÷ªªê´6 vT!ÇjA ¯zý ãÿcÿÝ}Ðõk½>ÿC]]^߯ëþ×ëkÿÒ×\_õ\_Òü;à )ß2>a20R gQÄJÿ÷ëëÚ\_cïýv?Òÿÿbëiiii}ZZHH°82vðpÚá30ÀR()/þ2yÊú¸V+à×ëþÒíéý+ï¿Ã ýk]zþ¿×JI Þ=¤ðPtu0 ߬٠H+¡ä0Á=ä->j®?ý\_üã?\_uézI~½/ôé?·Aúx/õèØÿäÓ¾Ç(sÔ(¬8X"7û ÿÿaþHû¯¤÷»þ¿ÿ^¿ÿIi}zD!ÝôósyooÖXýªMûôøB-èÉÑ}åhÿøoôO¿rr´¯îÿKì£ÒKK=\[K\_õ]R~¿JºC ×XB³Î©?¯¥îTKý.ìÿþú uü0 úí³=»ûý5®ë^´5¯®PôÈzïë×ïþÿõÿø7ÿû£òÿïüÿ÷üBaÍU.äB}cü%ëikåý$ô·[\_è ¾¿¤)}zèïÑÇÕ£îÕû¯Íiú÷ÒUwzýÕ|?é}o¦(0@Ø ÆÐ"¶8u¢u¶WÚúIZZVãI=ÑØ"ýzþqÜïÿ]_×}mïïõûý&ìÕºÿ{Þyzÿ§~[ )Ê¡Ðga0M{§mhBIb©j!a í&+UÒ;+ d<Û4£<Ò3Ssò@¦hÿ_JúßKêëï]»¯]ûT¯ûÿúï××jì\DDDDDDFi±¦ U ª
± K ÌEÇ ÌÉA¨L(TÕ:ùÅïY :úXémî°Ïö³E÷anÍë¶ÞØaWúµí{v| Á0aBVMÊØ}4[´Õ
P¨öÕ5P¡W°D]}øZ¬¸×¥í(i&¬29ÄîÓÁm&E8ù
Jl3ÝìJìJ}ߤÅí,°Ï&Ï÷ÒÄDDhDDAI| ÆzI3Òyñê~Rý´´]ÒkOA°¡´Å5 [ ª pêA &DÁ°¶.:ëÐü'§diJzZMÒOëHïÖÿ\ ¯izJÅYËA©Þ (L&h0A¯pa=ØjÓ¾®ØPÓ /ô»î>ôùt4×],_©y÷¥IëëÜDDDDDDDDDAX!>ÄDDDAkýXÕ½ Þ
.®ÿª ý«Ö-$¶ÄDD¥ëý0ª«ZÒI.÷~é´µû/tBÓ ÇýWþ_ßúT?ÒXF·Òaø"1õ_Zêµþ©_ºþ»®ZMÁà ý~X%$¿é½W¤í/E«Ñ²>ø}!Ð_þ¥ý%ª_¿úèDE¤é¸)þõ×¥®úUöÿ¥ïÿû×ÿ«õÕ/V:òReÖÎé\^YQ6Zëõõ¥\_Òý~GD¿Wô4&m"¡ÕGÿí²1}VM®© %uÿkK¥ß21ÿá}åÿ0ºµ'¬éåõI-à©/ko'û¬%¿;ÔRÜi|0õa×\%m
×JGÃKBÛUþ§zXTAý[\21IhΣ(d³"ÙP9'@Ãb¡Ãb°A-6b68
UÒÃ/iXwðÒé{Xx h\h· a8$HL¨ çÆKÄk!æ¦BG£VKYq]µNuP¢¡S©¨U®ÅÔ<6 :ôÛ ]pT6¨:Ðú $¡A gvyhDDèÚB ÐhUU \s
Ûízq :S0F!oÑoEã§D¢qÓOѱ¢í ¼8DDDDDG¡Á \p×]Áàs
fçÒMAPA°ÛeýÆhDwãú½>útüáÒÒTh¥'Dß¾µYÇ£úûÝ.´ëm$¡téA4ÿúKßÑÿ¤ºXý¥ÿ¤w\Ô½~~ßÖºK].¬zèR÷X!õx~?ÖëÿZ¾9dí!î¸\_ÿÎI^H\ºKúþëFîéD꿯Ò[èá4¿á'ÒúJ×Ѳ¾þöaÅwK§ÎZ¨Téº'Kó^üÁUúhØá{ô¾k@Úê¡ëðêM~'ºMt¿©XãJxtõuZüKû|]|&Òû~½²ô!±ut¯çðÁd¹¯Ó«~Òþ $ý¥ëé=lL>¿üÉiÿqÒÛ\Òýü0ªë;SȺ\ÆvTÙÙuÝúKÕ/þ¯Ø/t¿Ðdp¹Iè)¯Ru/ÌÏ$ºÔGfÚ\_ú¤õ9 +ÒýüWÿøÓ0h\,pÁT µ\-XÿúI-]%÷ÿëÒÿÙpTl\AQxxîÑî?®û3éR!jÒ}+W¯\_\_ý9xð©8\@¨'\mµHwï¿êWT®¯ßé}¿ ôk¤ÜÑæÚI;v¥3¬î«õÃRy[¤r]\_Vuõëÿ§·ã¥ JÒ÷ÕVî®?I ¨×IsImWÿÒzKÿôÿúé/iþõu±;¿¨\[A6÷ ŵVÒzô|Îþÿý[ë^¾çÿô RAK[Xa%µV%>¾ÒÔ$ô/íÿ¯ëî¯úüÇÖkb\BA' «PÕm'M+Kí}³ý>H/ÿû«ú¿ÿ ´L)îíot¦+ иÖè!ÿ×êÚW¿Å$#oÍê2_8ÊvBÌéÄhDDDDDCM²«A×Oià u»õëúPÒ×D%Lvxj A¢&A¤Áf""8a/ ×ÿ~lS¥j ªôѱ£s öØÕB.Ý-!í$×ýü&ÕÓÉyàlê'ÈIÒm(l$ÛFê0ôßOKïÿi!l?°àЮÉD¦Õ°e'¬:AÕé6Ú×_ÖÁÈñ§«ý7OÓ¯Nû«l׫Tô¿ÿ±0øN«õèõúë¡^Ú¡i~¿í{¤×ÿúÿ¿m}G»ii~¸ª ü¿ÿk\_ zF"ï\_ï¾Õk®ÿÇû§tªÒÿÿý~b?,Gù0?ü9ÿQt½ô¾«þÿý~¡öÿï×\_¯÷þÿýúúÒÿ^õoþÿÿëÿß«ÊÚÛ×ék~¿¿ýÚþýë;TÕbÚ÷þNº~ïÿû·¾v3³#ûV%Ý/T[ýoÛïÛºÿíºá38·ÈA´Ø"ÃH Þ ºpë|a¹; äÚR&vÿ½(84¥¸ÐÎüÅZ}+H;'¨ª0@é@àã´ö5 ´¿ø(hCDè2¸¥Úäîaß´JÐO Ä&h6)þ a²' ûGe^Ø?B""""#ñP¾vÛÿýÝ/ÿý×ÿý´§e@3ffÍFuG"^ÿúmjg&aha36iÐRT! Ðhó5EEÿºPÊ74-0TPS [#Ê ù@É ò,<ÐB@¹Ã53¨üöõAI°d°J¶Ì(ßv-6B 0^5 ᦠð-à 0ð«¥ÓAéØA´8d¹¤ÒAÒoA67CPíQvô:6©Ñº EÀtµÝ¯ztÚ~ -\s»=XayC ¾&ØþªÒ ÿëqÿý×^®»§§JÅÿI´ïÃ,áÒ}ÿZë^ô;õøôøÿ¯]?»:W^?Îÿÿõõÿû÷Çÿ¾¿
ÿÿUþ?\_ ÃúûÿëÉøDóõÿuÿ^Nò.·ÉAÿåqtAåtWßô÷þÚ\_ònU\"wüö×d(.5°RßW¿ý/×ýÿ¡ôú/¸{ü"Woôo28g;?ÿÿ¿û ÿô¿ÒðoýÃ}ò{X+eEil8<4Á]p¢Íím~µÿör§Oýx;ïùпÁBigj¹ß~%Áí[µû]tëU}WÃÿ¥ýý&
Ì£¸öKY\b¶ºþ¾Úï
¿»¯×ÿ9÷Wùäßÿ[ëï ìJ{ õ÷¿ #׶ÛK»ð¯Þ»ÿ^½]wéÿ»IgkÚb XìUÝò s´¤Kgÿw°¾i}¯þÿß_þ)a~à 8pmCj Áìl01A;bwwȦ#Õ=´PûJÈûðÌÖuD|È"Õû×þ¿¨0YP {8åfíbA Ô0Aæ!C9ØH áµ Z{ !Át¢ h%2:3#XÉÏ: «8Ês%eFF":ÿÔ|b"""""(Çd ¡Á3ÊÖH@ÓO±LBNà ئ plPD/Ã60 a4BÝ?N, a0O0)@Ê3l rà! ¢0ó#QyÐÎôSªøIÄDDDDDDö Ó(s°TbѱÅ
Pp wì!aEãD-ÓØÁBg¨ØÁPh· àÖD~#øL'ÐAaE ³ %r6DSÆ»õ.Áí82\ÙôݵPh èoIéÒá¤ØbgmªA½0 ê h!zaÔ& Lø§¡<3F"àÉãcÈLÙÔÊ9Òé[¦öéÁ/ºÝ.©½Ã,áJØ Ø1¥Ä®¤Ø1²TÒQ=´I·®¤^0TEÝØ(MB &g"-P@É3HÍá\ ë2P¿Üz׺ß÷]êÕéäî¯Aºä¢¢UB· + îh²4è&ÙCpqaU4a i\ai¦32Aæ?Ò]õ¤=8¾¸ôÿOCÝkÐî¿×¥»=6;֪ذAë¦Ã%ÎX%[FçµÃA/(&ÓÜÓPTl{Jí|½öÿýÿ÷Óïõ·KZþOOÈánôôߥ´àÌ 4êih é8d¸JʶÌÛ¤÷J£ð_}õ!¤¼lè3ï¯ÿþû·ßô7ªÿºûõ_Úumum¦'§>ûJä(þì=ºIùN¦B¤üÿǾü·H
ý×cõûÛÖÿ]}7Ýü0,G@»×oðz'LcþLr¡Êõþ";{ÿ·ô ðüwþû¥ëÖ¢·\_ãÓ~6!Vº\_ê ö/A}ÿÄG«ÿ]%ù{%ÝyðOaÞþÈ^ÿÅm'Ó\_ÿ\_ë¸#«}/÷"}{Kò×ÿï÷ûÿöä«á½ûÿÛDý¸ýïã%ÄïNBmN?þ¿I-ÿ·ýõÿ¯Ý÷¯
Koü6Ë÷ð;iu#ÿè'õ°k¶ÿïò}öÿº_5oÿÖÿßõý÷zI~jëÎ
ÿêÿàÿåØÿ _QZ÷íÿ÷¥éi~Æjÿ÷ÿß4ÁÛÿ ÿ®ézJ·ÿ÷ ü63%˦Ìÿµ¾ý/µo é_·ui7ôöö勵ûµóÉïÿý¿®ÿý}s¡NÉ Rn4¢Ø29'±)ëlIZbPøg8öEÆz½²?ö[ki6f»0+ûíi6«ûý~´½{÷Uý/x}&é ÈI4Å¡r1Õu ¨{ %vÇiEAÎã®DÎÙí6Ù pa(ûºau÷·ÿ÷±é[ëûµOW w<í8Ã0OA
ûW» °jÚj) PÄ(A ¯h4A¶ÔU¦¶¬0WI}+3_öø_¦á·Wýÿýò£ H¡¬þÛ×í% ßö°¡|>4"""8"5B" þ88 Å h8
ئ82;oa
bSúí(úû·\_2{ @Á å\4Áá¤ÓÃWH&bA ¤AÃM¯hîw¯¤""""""""9Èè Áì&iÃ8¯ø®ÝuÖÐ0¿þëÚK¨ÎÊ3qôt<÷ÿ¶õÔìPdqòlSP¥4DP¿«KÔu¯£Y !DM÷ &0@&3Á EFN RGDpÊsÈü=}|1AÒxÚ
@a»]Ô\ à§xE»320fl2Nøódjàª¤Ë AåZ:[¦ÐA¤pppÄ÷h0½ [µL"l\ ÃéAiÐ@ý?ôàôé:N»6ôpÛAH7(q3ðl0 ±a ²m\uzýmo×êµý×Û'K6NõÉfDF=·ôû¯Oõ½ÕÓÖ«tú·¨2ÎIºÿÿKý ïÿb®?ÿÿ]ÕS¯ ö¿ß¥þÚ[Áÿëÿýºÿã÷¯ÿéÿö2\_ýk×ëÿ+öMÅØKúûÿÿKX:\Gèîßüi&MðÅz÷÷ûåÿä(ý/×ܧ3±Ø/ïúÿëKÁ½ÿäÆïþMtü96[Ô2¸[ú0ÏÓï÷¯·ù¬ÿÿÿ÷K\_¬)-AÝL¿¥ÿ׿ú]{Ú\_~÷÷þÖ!<Éÿ¦Jÿû¾êÛÿÖ¼ÿ¿ýýM¥ßúy?Íh÷ÌßÐ3Ëïm'I»µ´\_½zÖÿuûû&yÙy!ºkz²(2%´¯³é³é°¶®×«µöÕ¿î£ëq¯Õ~0¶>IccÝÄ.>Òlõa°ÁlÍ~êÖgiHÈÅëõþÔgr?° PbvÒÓ $Å1VÄÎØeÍ?´¹û övSaO àÈb#HÀɤl¢bä|#4GèÍ3QÍ©§Êµçÿ^1FGQQ»;32X¿õá&ÄDDqª3ÁDC xam'l&4±.üN?ÓFSs hتªa6 Ó§
ºaìÔIfiÃ4e¡ Ñ#3^qÛÿ×éãB" ul2XEÇ °paàÄùI¤n´ JÊMºLOxsc £[ ]4mSMUÚ©²à¡ê|RäaEyNË!£ì¡f³:Ö¿ñdïí0~êÁ-÷´ÚXbw=?@Ä=zÐJæĬÝ&ÃI;º8a(iBzal"é¢Çg3H¶"¢d ÙDf3[ ò]zô!K§ÇëëúªunÓÝv´Ý7j S¯M6làËÚÒl+jÚJa®@ T" L"ì4\0äA°ÓSf0A°A?¿ëIÿÿÿºÖÛT¿ñéÅ·ªÇzö麧^ZׯzN ³ ôòN 'ʸ1îÓ£CA::6#smSÅÒê¿¿õÿÞÿçõýþÿõªïï}VÞïN'¦Ù(1é!Õ½Õ¸I6éÝ+n÷]GªÜ'þ[é ÿ¶«kÿÿÝv=×õjý6½?ôõ¯N%Û¨A´@ÚAÃÛKI%ä;®³1ûÿ·¶ÇþNë4 ×ñ]k^fôöÑǯïÿýÝS®úß[jÓjÝ:»ØaÒåê?ÿÚ\_¥ÿ&9UëÁÆ×Á¯ÿýúÈZxýºÿïÿýìiõvÒÿ®¾Ä+ª#Zÿ¼/ÿVKÿKèà·«Ð-ëù_ÿ¦ú§¯þ÷ýúÿ_î/×4(¥ûU_ýدm[þûô¿/×ÿÿŵ¹0Jº_ÿÿí¯Ç!F¿_kÿ2êMÐÒOÿþ,p]uÿõém¥u]¥ýþõçf~¯½õù1Ð">ÃþãÿÁ ÿëÿÚ6±±_÷ÿõ®ÿÿÿàßW¥ïÿÿZ·üòÿ!Èÿ¥oûÿõ õûúÅZ\_ÿýþý¿¶ÿÿ¥î ÿWòÄßÚþÿ}ºÿý\_\_öÿëÎÖw×b¿ÿ¯ßï×&ÿýÿãØg+î%ÿîѶװÌ×ý¥V°µ}î¿ÕÕñKý¯ÿ½u{ý£»ëÿå¹HRp¥ºR 0@áÓ Ái¦±Ò= ¯±)Ûà Dãù;g{®õ·Oþõýþºæ´ÿºWïÿõ-Ö %ö§p Øn§ÁÅ!u 8:¾!0¡» ÅE¶Ò¹jbqÛ öÿfß~ß±]?ÝvékÁ5Q:"fDDFà w )ëö®©®¶ÃI¡ ¨0jWVÇí¤È¹Øg½³ î¾ÚWíü®P¼82"""""""óÄD0B,§#¢V§a¥ wÇÁ Ýi]l3úWi6h¾ðFBéÒë¡´¤ ôínÄ&à #騦4 $ÃcV¶'ðí-|DDDDAaÐk8AÄ Ä °n @áµzß\_zñÄN! ¼± -Úïë\_ë¥"4"""?úô¿þv-:#Æj?ÿÕ-$¾ Î &nñ ÍSÈc"b.ÈLDuýÒ\_ÿOÐ@Ýѱª!GUá4hhDÐ&@d2$?8Ês"MäK$¾ÐcI°ÄøÒmÒ ^Áp¡Bg»a¡ ±E ÌØ&Rh##(¤ãþÐnJÛ ´18ìtÔi6?0ÄðÃ= à j ØFöt.Ø\. B 롧Ýnkz×WWa:NÉX4NÕ´U´½¨ ØýÐd?ïÓþ>²zëªi¿««Wúûc§I°ÆI6#Kþ¿ëº§Vׯuu¿¥m\_§ºN¯A ë÷5ÿýh¯·ÇÒêúmn
z±ÿï{K²]ÿø?õ÷ñ\·ZÎÈCÿëõ#õûì§ÿ²:.]nl¢Ü g~2ÜÎÿÿé|±HD¯þ¿ëÿнOÁu\_ÿ÷ÿh¯úþLSÿ#r´ÿV¥¸(/þë§÷Ký/ý×úõø!Ûú-ÆèÏѽ!2Fÿ¥þ¿ÿïûþÖEïÿõ¼Bï¤FD©!Oj[·î·Wíÿÿëÿº÷õö§FKÁ°» wVüQÒöï\_ûûõwÿ¾úüUUv÷\Am¥a&ÏýNÛzëï\_ßz¯·{ÿÿîú>É"m(ÇquݶzyX\_^Ò¾°·Wþݯ\_¯«
lB¦kÁǽnöÒ½NØg+Is°Ïý·GilÖÈÆFDMwÿûÖbÀS@@Ám4Ò¦ 0¡ àئ ÁðKAÛ¬H0 2!#6ÈA<ÙFHͼ$ 3¨ó ȶE)÷ÿ^8"%0IöÐa¢¡Xa1 @ØÃvB[»!0»h¸g"à9!0\à ùN «4ðÏpPDC¨© Bn4HÊ3^NgW¯ÿ a ! à ¢ñ &ÐAè Ø.70«èØÕ4la¼l 3TkF¶pi¢aËpÏ &r#Ï S60@ú!#¡Q)_ôDDDF ¡:Ém'5oí¤Ø1áM'¶ªA·Fêl4ÐI¸°¡B7èØñª5}BgÈ&ff d ñf!ÁæÌÔdY{þ}8Ö¶ÉÞg ׿zmêàË¢pÄ Øc§zI°dy ¦Á Út Ò¤xÕ 62 iO¼ ÁFB Adeä
Óô¼_ק¡[éöÚ^þéä®¿× Á¡ º&䢸2Ï÷OM%eêŪ40 -4la4k\xg|Ô0@ïÿßúßOýt?ÿ¯NÐÕô¾ß¯MWük[Ëh7N~àÉA½{i6 ~QqFÆxD.ý/ºÿ×û\_ÿ®ýªí¥õþøÿ÷]/Oü ÿL'§t¡ÒnKI°ààÆAÒmÚ\_ú$ø÷ÿøÿDA÷°}{Õÿÿÿ]/ëìwúÇñQ§Åú§\_}Õßdî¤ÞØdöHwõJúRëëþª¾:ä8Júñ¶±¿ÿÅ ?÷ÃÛ\_¿¦ý¶¿÷Õ×CZé;P6m¤ÿÿ#òc8n»ßáêAÿúKþ6B[rq~ wK®ýlWÿ\_ÎáÔ:blSÿÿÿÕá»·ë]òÆ[åÕ;0¿äQÌßà¿!Ì|%û¢x,=ÿ\_ïa.ÿ¿ÿû¬ê^û¥úÿ?ò˽äÆpçÃ}EÒê2+!{ÿÇý¯={KÞõÞðûú]ÿþõý~Uoéû}áÿre®ßýk¿KIoþ½^ßüöÿÿ¶¾ÿ÷ßáößÒü§ð[iuHÿù1o×M&Þ®×îÿºÿºÒÛ]{ÿþÍIûõ÷Áÿ\_þÒÈÿÿþ[pÈèÅa(gÓfj×zö¯]¿ÒÝm/i·ïzÛõÞß{êµÊÿúêú¾ÿÿÅQ2!ÄîVÐKiHM³Òýﾺ´¶ÿlÍ/]¥ß¯zGÞߥ¯¯ý+uÕÖI§úû~ºÞd46Úab tÐ pØ &ŸiFØAE[Lïc¶Ò¶ÊDÀ0ý³5ÿl"?iX\_ÿïKÿzµ¾ízMå\°an lBkÃI¬0Ò)¨@´ìP àð@Ø @تb¡Û %VØrµûKü+ViÕ¥û\ì^;\I7©\1Ø4nÄîè þ@apv¡Q²L#[ È"Çd¹ Á³ Á3F2xà ôN#
üKôì lt¤ï¹® 4 7ÒoAé£shØÛ °Í0(E» ¥ÿÝéúû§¦Òéߧ§I°û²mRmÒA>7t4#æBÃëÓÿNº^õ®·ºîÓiËV¶ÒVÁ' î"_þëÖÇÿܱúnï§ïVÛJ¹<¤ÓIÊ¡õãó ¿ðk¨§÷\_¯ãêÿºJøÿxü?û!HíG{"ѯëÁÿþÅkJ=ã#%Ñ1"V¿vþüø8þ=ÿè2Uý\2@C²Q,\Ë;Xêú \_êú]Åá'9F<l:÷ãxþ ÝJëC ó²d[»6ÉýÒÿÁ¿ý~Ýü±[wÿ&!ѽ¼LJÌòFK?ÿ¿ðîÒ¯:ÿׯ§,gÛkÃé¢1é¦d¨E½/Uû§'ÿÃÿy¨ïýïü(ùW¿ÜôX:©H}½m?ûÿ·¯SÞÿï\_ú;ÿÁý}ª!"NKÈhBïý¾K»[÷¿|÷ýþß{ú:ªîÉnvd´qáò&{Û°«ßý¤Ûx]z¿kúí[ÏoÿõãáBxH^ìaQNäKÚ\A0ÏpÎqpÁuf½³4ºÞÚí¤Ýíúÿ\_¯\_åqiÔ0Ãiª)cdXlLáÃIØ0Ú®ÝH
Ã=¥°Ïý¯Gig2ÿÿÿ¯UCB |Vå: º¸¦ h808m[b.î@Á}Äñf)H2laó È¿N¾ºýku鱿Òwéëÿÿ/¿ûú|9ÃhrÈõ!Éýúÿÿÿÿü{¥ðÁû§ÿO«Æ
ÿëýWÿý/ëÿ®Cpº&¼7í¥úÿÿÿï]í¥Æ¿ñ¯Kû
ÿ¿Ûý÷zú\_߯þIî½ù ÿ÷ÿþ¿,GKÑ!ËA Gþ½u¯3 ëïü3io¾m?ïÿû¥×a¿ÿð /ÿÿÿýúheëòdí\_þ(ÿ¿üG¿¯þõû^ÿ¼Ö½´¯óÉú?û§¿ÿ¿¯umÿ«Oûè \_ÿþ9PVÖ>Öë\_ÿÿ-ÿãn¿ûKÿ[zÿÿïºÿÿï¯ëá¯þÐÉJ;ZL=ºû]¼\ì5Ûí.ý½®ÿýÞ½k×otcÒ»ÿ÷ÞûÕÞë³¼Ú¨Îêl3ÚVì3Ü0±(|t-´?ûfzý×ßÝÞÝÕÿú}»Lÿ¿ß¥×ÿÿßÜemêÆ6\ ¨0@ئ'pê°ÒJÁ¦°¶Ô%ÚÚZÈ/ï1¥ òm/ýv~ÛýÝmþºÿjÿßáÔÈVa[ (4öÓV°Õ UÄ&!(<1@à v8m;a&6Ò[îØf)FL3ÔDÓfi³5÷öêÝÚÝkÿ_ïÿÑäS®" ͲP-e\0 )1Ö&½
í¥m4Ä Ó8 lmE1\_Ç ù;gû¦Ï¦Ï¦vV·Ûi/÷küÒÄDq]"" 0§rGMo°MNáMa¦0AÃpÃba&68ñQ)ëÈ¿3¶f£iÕ}""""""" éÚ 0^[ !;MAj pm6 M829ñ×kKfH0@ &S©c¦ÖÓh ì1]uqÄDDHLA"]¯q¢pÊwÿ}s±!A"'«Í³Ù jpò6ÿÿ¦pB¦|a ¤qvRÊHÀ§AHbC"ÉHGFAÝ¥ÁÃÛAÐA¥ôl!.D(3cD"tEÀh±ØAA 2pm/°dÔæ¸1¤ÔOuH6¶AÐÆ^AB/T ~z}+¾&Á\±+é:OZM¤ÐMÃL1 ÿõ÷T>ôú¸vVý]'ô:»8þ½ü\_ÿöúz½ÿ¿÷««+¬ ûûà×ÿÒÿ^ÇÆÅUéÅÿ¯ÙMw\_ÿ\_ü;ÿÇý¥ü?úÿïÞBP1È"ïË!ÎÖ"¸Õýµü7úOÿ×ìÁ°k¿p838ìXfC¤Èÿbàß°KÿßîØýO¶è5 MWöY¨ýÿÿòqAº ýuA6¡ôg&Ê#EþqMî»'Wïëðp{\È\_ûÐiÎÒ~PBê¿×óÛúÿý]ÃáÃý{ z\_ ¤Ç-Ä^µë÷ÒýëÒÔÌnæ´æÿµñ¥Åp¹nÓß¾Ï?I{þ×½ºëïÿÿW\_ }4î ÿ´¯Äü!×a÷é7ýmÓæoÿÝ~aÁt0í¥piZWÃ9Z_ºKå¸yNÈËÿëý¨¿ñVLB!1H:ba(0l4òÞ @! Ì!@¥i$gÈDPNE¹ÆSdBþõñ£¦#,sÄAS^Xè0
LTB6( àò¹ Ðal"î7 ì&lav|aäaA ï0 9hÌÞx!±J±?03¨ÎÖ]Rq!8ÄDhA@#VDÁC$bqè Ý:AÐ&ÐAÑ}Ð ÃÓ]0¼"ñCÝ0#[óc
tᢹa3²ÆRÁHè2B9eº=q ÿKAxÌÒ´'IéäàĬrxbV4I´¢U´H6¿ú° P[MA\xa0ò g"-a0@Ρû6GQã5È¢ÿí:¿ú·¿¿tv[¬gNpppyÇ.&äc¨Q+Nè·z zhØÕAð4©
D& E¨ NÁ&'þNÉóc3áÑôP¬:þûïu¯õõn°ßOT½:¶¯øöá¡d'I°Äl¤%s 6÷i!ÇhÜÚh mPaA] \g
84¡ÁAfýtôö65ØØé~KCþ½}\_ýÓÒWôëÓÉ[W ÚNÉDtÕ6> ©Ä¬iX1+)6( Ú/(ÜÛ}0ÁZ¦vBWF·´ýy�ÓàØ~þTË®
uÝ.µëïÿ¿úm¡î¾á=6»Óî S¤í6°« Jæôáá Ãë¥ÿÁÈ^!&Bþÿñÿ"Pô ~ÿm%ÿý'Þ¼~«IúuéÖêêÒëk§ºvdÓ§I»iZ\_øa°ä°ûþµüEG×ÿÿü{¥þú\_ëÿ¥ïN®´ý6ëÓÓ¯O Ba¯ü6Yvÿÿù1OɲÿÛKÿ×ÿøý¼×ÿþ¿üqúúÿë^ Ä We@¥ÿßû¯ëÿÿúù=wÿ¯ò /ÿñ¾¿ ·Kû¶¿Þ£ÜL¹ ÿÁ°|?þEûß÷õÿ÷÷ÿ\_üÎÚ@\_ÿõÿõ§ÿþºú÷0#¤ù´æ=ÍiÍ ýÿ¯ÿÿþÚ\_þþ¿\_ÿÿõÉú¾ÿûûõÿßu~÷zû×ÝuÛþïõï\_þuuþý{ýþ>¿^éuÝ~½-zÿ¯í|Ò×þÿ»muíÿûÿëo¯ÿýýÿßþ¿öKñÃIºý&ÿúû´¿m¿í¿×û\_ÿKu´¿é~ïë××Þ»ïÛKö!©Bm+JL0¶´½m-=yYþDÎÙþÏì05ì¿M×~ÝýÖöÇþÿ}ûÿºú\_û17Õº)L)n S¢¢¨¨0A°@Áñ@AÜln±±±)ìJv±»i,a¬ ¾Òoû Õ
¿oî¾þÝ/¯K¾prܬgk# (M&!Xb» pÃA
AAÕT0µ± m !'¥ $Ã9r%¶ÿû¯½7v¶ß«zÛ¢½ÎÒ^"""uB¶% ± ¥ A
éP©¯ \ 1 !1@Ó±;ïbT9û ÷÷\_i[v¥\_öÝm vè|DDDDGq}
ál\a+TÔ0AÃj í°{i0eÛ Å'ïG(ru·Jéb¹ßEuk&HqzCPýð×pØplSÆÔV0Aö í+u^¡tÿÿ¡g$jFFåZvìPb®¿þ¡ªúüDDDDG Ú_ëãúýi#½¯ú¿¯_ íNd4/!ýê»Ò.D(Á < Ê$ÈØ<ÙÍDt5Ï""õÒôµûjM¯¦!ÔB.ØA¢áB«.<0±Ã0AÞ@ì£6Ïdã 2°?ýk¥MÉA\Ð ß FÊ4¨ Þ¡ ¢Þ"î±¢(0 <)aÃ:ó¾ZOؤ°~2ÀêÝ6Î9fÄøÕ°:M
k®B´N+JÒêÚõûé{øÖð-&·õlg¦äp(0ÖFK¥®¿ÅõññÖûÛ®úøN }ñ&Å0Ò¾¿ëþ÷tõÖ¸ö+þ¾Ð½7b! ÅxÿÈØ»ø,/ºø÷ÿþª=èD/\_äÀ}PÍ2øò X; ýýkãc&ðËü"Yù!É [ÿé"0;!xÇù L·'en+äfwcÿï ^
pßýI°]´DÀ\_ÿÔRl²QõKý%:ýkAa¹1oËøDó«iL·$ÊéÙÙ¯ïþúÃ÷ý¼,èOþ¾è®iùó°+ßL¥ÿÿî»ÿþ¸oÿþ¼+Zm;i6×÷ûë¿~{÷¯{óÉïûý÷4§e>Ù¿ì\{v»WkÝÿ×ûÿÿþªôØ£ÛadE3X]wÛo {ß~«ÿ¥~¿ ð@ئ;á½°¬J}¥"¨i}>¬.gû/ûfjë°¿¯[×£× pjîÚA´6)¨8b¶'}¬q§p¬ÇÙÙVld0ñ²&>!#8yFn=æÈ ÑÆk"$EèW.ÿwø04 Á8öh1 pÓ
!8aÁ
ÎÅ&vÎ ¢ná#gøá3&<f É0@õ' d&h&df²::½m~""""FÐ!B@Á¥9¸aµ¬Z7¦y@LãÓUj»!xD.Ã04L mÈ·è4Xá#ÁO# ¤Ùtäc&ñd¹ºt1+'x8:!ÚN Ppbg£fPaÚè'½¦ Þ
¢ð±baLù äA~"²&D ÔFÔ&Gð§m àË¡(ôû@Ë¥Óí4Õ°AºlX ò?ïÞ%u&Ã#ÊN©EÛJ/ E¸2>vx &hÓ<4y@¥XÙFR N<êÈ?ëÖM×Óôß¿Zë¤ÿÓÓÓ\_î½6®í7P@Ë¢°Év%Îéѹ Å£``ªÍcP§.¸ ða0AØ f={÷ÿãÓúÿõôß\_ãÿOßNjí7\,eJÒzuh$Ødy ;¤Ïuú6 Å A°¸Eð~60Tþ¾ç/ÿ_ûÿê?¯_¿ßØú__×{{ûÿÔl&é°Ä¨x|18ðàúé6 4A©½&ÃÃý~¾ÿÿÿÿûðÈø"®H_tò"¿ÿ×_úݺºû¯M5^jôë¤é82èÿ×ÿÿþ¿õÿø/ü°tñÿ®ë_]ýþ#øý}]{®·ÕþþÇÿÿÿý´¿äܧ(;ÿ°ý|µÛ® 'ýDÿ×õõö¿¥¿ôÿý/ÿßÿþKõ¡Aàß^XËö¿1(ÿlÖ?CÍbÿ÷¯úÐ¥ÿÿþÿÿ^¿ü1_¿é~¹¬¼7_]\± _èÈ°%w©þCÿÇÿú ÷ú»¯ÿÿý Þ·õÿð~¯ðÅ×ý6v±XDûëò#ý/þ¿î¶Ý×ÿÒ×ý/¯õÿÍùÏÿù¶«¿KýsY%±_ßKïòf?Ë1þLeÒÿýõÿý¯úMºïúÿ¯ûé7úWÿÿ¸ /ý/×ÿþÿ!ûi{wýý¥¯úÝÕßÚýïë¯_émÿÿV}uÏ[÷ëÿûÿ{ûîMÉéÛi|´/Ã&.D¿È.]Na×°¾ÙtÒ«êÕ³D¾ÿ¯å½ö«ÿ¯ïÿýÿ¿Rn\fEh¸8lP h0ÆÅ@ÈG Á0Aö6NÅC¶'v·ÚKÚQ wÈ1²¦]C?ý
ýjúÕ&þÿôÖz÷·ÿ¾d\lã:pÝWµ|ã
Ú·ii AÃj@áµÅICñl29'±)éò&5ú°°Ï«3^ßa}Öû\_»ºïjøØ0A0B!(GTÉ)avÛLBCT0Aà »B ={ÛÃþì3ûuí»¥·ó%£""""#¢ç a2ÇGð×íiÜBaCm áÅ\w»M iw ÛaídLgþþI3¾ò°Ý DDDDDDDù©ÃØjPе ئ! 1Ll4Fõ
Z Óôb"x!DCA[4 Ða áÿU_ÛU b"""" ʬ¡cÿ÷þÚ_Xì¬Èwý~×B ºÍÒ$)!£9l§f·þý«iiW±Ú6pSM3äÁ&£ Á'3FN6عO YÔÊz×<tÝ6¯W $Û¾&xJªÌ&5"ãÓÁÑq ìx gAHÖld1s#ätØHE°'I°Ð81I°cA°bxa&M»Ð|0Õ$na;
ãc L .\ ñÄ& mk«½IDôlX¸AÒzl'ßÁ.:¤Ú70±8ôoL"áh3bÑ ´´õýôôôôþÿú@Ë9(¤ÞØ2è°vHák±Ðo"Ä.+kÒ×CßÓ}×¾ßÓ¯O«uAéÒnÓ¾\îÿ¯×úëý×Cãþ¿\_Óuô?ûøÿµßáú\_ÿõÿõÚÿÿëÿþBÒßñ ©´«þêôv¶ÈÝOÿüÏÿ þÃõõ>¿ÍÔøPP¡I°þµõ×.¿á½ºù\É\_IÿÒ|öéppøRlß~ÿÝÿø?ÿH/«ÿ&/ÿ]8påu ì£"JdeUÞéïûÿðKÞ¿ÿþ¡-¤?ço¦ùÐ{}×ßí\îëÃíÿ¥ýÿ量>-Þ;22]2RyKv¶ßÿ»©õ~ßïßï¯éJ ¤òðtÖÛÖwíÕ÷õ±ÿý¶ö´ßõþ¢ ¾IÇk?èë» ÅiDã¾DPÏþé{f{Õ
¿ý¼/wÿm¤ÚÇ÷ÿãú¯«[ $Ô0E±Úà iDãôº2ÙkeöÌÓa;÷4Jÿõú¡
Ã1
hC 6!b0A5úpdsbS¸ØeÍÞȹ¶X0yÙO#P¦y¸Ðd2: H¨¼ãóÑNgVS¢ wRÿÿÿ3L«(|±Ðan S[t 1MCi¤ÅD)à &\a0AÓ^3 à0@ÉÄ"P")�ÏpSÑ<0DTÉ$ æÌÔ?ÿÿ4!aGA ¯ SPÝãFΨØh¼n ý0ºal"q»dÀÓü¸i\lh9 w áge¢FÇ d hóÿÿãB""""" Ê-@CAÁæpbMÏÔ&¡´aÚ ¸AÒms[ aE¦n,!é£c ÑA òzòN¡ó ¤L ¨ÿø¸ý:O%}º×I¸A»ÁàÕ´ºl2>¤è,²#Ì}:ÄöÒtlqatnnza£!nnTB{Á§.å9fÈyFCÑ<®¿]tëtûÖVõÕ׺[Aºõ¦Wi°q\g Õºpd³:Na6y Áѱ¦G©xAáQð3 ÉãlÌ!DgIÒûZÇDZ[®×Ý{׬[nªº~.¸Mþ» °eÓ(rÏIÐAôEòH6v^/¨EóT!G
» àÓO¯þ5ýô¿·_ýôÿÿ]}þ-ïÿOON/«jàË¢t¤ÛFq@ò\ÁÁðtÄöÕ½bÓ¥¥ÿÆP,@dp¿í§á[Kõÿ÷õ\_Ø÷÷ÿ¥ëkÓÔz~õ¸AÁq«K®ÿõ¿Ý-t6Bð¯3 ÿÇü õ/ý¯ÿõÿ¾õÓô=7¾ûuòw«ªy7Z\_\_&0Ûÿ|!¢t¿¯Ø?µ;+ ÿ! ¯ý?úèUÿ¯õô¿èUéüy)×ïô ÐÿZ]l²íYÐ-¿ùb>úÙ!;ÿ°ÿ¯ã¯]ÿßõÿÿÿV
}{¥ýëø?Ò ×ÿðoAíËÿé?ÿÿÿß}þºÿ:û¿¬¿W\_ðoõî¿¿Á¿§J¿Á¿ýyu®·ÿÿýþõ~;\_ý®ywýuþkOj¾ÿÿáýt¿ó©ÿÿv¿ÿÿ¿ÿôëëý¿×ݵÒb¿»þ´´¿Ìÿîÿýîÿý¾¾ÿÿë]êíÿý^ÝÕÚÿ~ßM¶lßý¿¯¿óÉúÿÿßïÿ¯ýÿýåºÞVÛg½{°¾ÙÃ[KÌ/³5íÚXZý¿ýoKÿÛÿ÷µ¿ÿ½ïûÝäB í66'} ÈæÄîLw°ÚP͸0J%?\_aäH¥R&vÏöÿÿïû¿º·¯¿ëÿ¯¯ÿÒaF·m ¦ Ôx àبdrb¢PÜ ÁÅ6 Òôî8ôù¾Òaï²éolÑ6u»KÿÒÖö×ïºþ¢" 00A©PB 5ÕÝ°ê\8a¦¡7M kÇlV%>ºm¿ÿ´uëýþùÚõ!ã2DJ5D0A®îA±LP@àÜtPa&A9¥ò&xiD¡ò%¶Òÿ´¿
ðl³!0¥hM¶jÚ Bv! @ÁMAÐA¶NÕ[K\_~WÒ;)é/HDDDDDDA Áµ|§M[I±H8a pt½B¯Ö""""""" ÁÊÁðÿÿéjÿ\_êv/#æj"1¹ÐÉoÿ¥¤3±!¦a| 0A&ÁÅ5$ #èú<S\.è/×\ZhØ !ìØÁSM0¨±Ã4E!N\:B"$1Ѳ:Ó8üÕ!©)2<Ý&ÁMDXEÛFÈAzaFÆ F°Ó´BpgfdjGDu³º´l$ì Ø2æKùl ¥Äøöl1;´#u 0¨ÜдLv0XD/lakû§þOº½7Né$Û:M¤ÒOÈòÁßÐA½DGÓ×\_cýu¯Mô¯NºZ¶ í7OAÁLÿþ÷Å×úvúºöÕûë§ö¯uy;ÿCþDÖ¡ÿÿ¤¯ét5ëuïýãÿ÷õÊz·ÿë«ÿêôôMÊ¢#~LgVàßÿéÆ¿Åy@ëå¸<ÉkH3"Së×á÷Ð'ù2ÏÿòKÿà¿ÿ«þv¤Á"ÜTô?÷ù¨ ÿµÿTÿ#¢yÿI°ààøu_ÿàÿÿý=ÿÿ ¿ÿÌçeZMµûRnIÿÌÿô¿ÿïýï_Òý{õ WþI-ÑÍdÿïýÿûVôKþÿ{ÿÿýuoñ\·'ê¿öÿ×_Vÿãt½íé¥ÿô¿÷ öýþËýal+fÒjëîí+oí}þ¿Û¿×Ò§ 8a'qÃIobqìN=° n½³=·û3KëéöéÿÛ_»aQA ¦ ¨68°Û/ZLmoC×dP;~v¦)8¹Ðhã5²ó«¿ÿ¬V" ÁUì-
ãÓ8lBc B¦0AÃi8zma02:3ä0°@õ( å ó'dõÃæCeFl¤QçìæE¾³»Qà T:p & .ØI´6(atn !hØÕ0a·EÞÿ»§
Ðh· áN Ï àÏ &ãL|v2lgQ{6f³%Õ^:;GIIµ 6Ú6CÖ|07ú6t7úþæ·a±¨L ÏaáÈ fÁ$FÐS ¤ã! ÄDE'IÁDì' ÛqË~°p|$'ƹª¥xI6(Öа
´Ñ±¨D/©V·Óý65ÿßWúO êö©÷Òl}&ØAºhGXÚMºZMÚÅ6¸FÆÆy « 5B-Ã84A PAá£æ: Q.£¿þ=»ê«ÿ]:ÿ\_÷\_ÒôõÓi83 'I°Äï º&Á.¤Ú:AºztnzZza¨\£f}0Î 'Kÿÿÿ±M¾·ý~ÿ\_ëÇÛ\_ñzu»§øOJòw¦×K]'K§>N JÇÒ ,0¢ÐO§ÿÑ?ÁàýÝ%ÿûþÿßúÿÿ×Æêè]éµ÷¿ËtÉÞ é?#ëÉu&ÁXéi|~ ¿ÈR?¯ÿþÿÖÈCÿRø"ùà¡kÿþµ~¿§Jñ±èuÅzwoaW÷TêôMç®Ýò-\(ÿ¥ÿÿÿÃöÒÔì¬OKþË ¯þºê?þ꾬wÇÕvû®
kð^zû{{¥ÿÔOëðßü@Gÿøß(D¯¿þõðûªÒÿ[×ú\_ù¬mkÿÿÿ¯÷iW þúôLf¿úê éh~N'R!:0ÿïû ÿÿßÁצÒÿÿßÿóQûÿ ¿ÿÿýyuå öúüv ºÒøþ7~ÿ\_®NØýÿÿ\_ÿ÷ßWô¿û÷ÿßü6ý+'oI¿ëÿ]ÿêß×QÝWÿùïÕºÿ¯ÿïþï$È4¿ÿ¤øoúùb=·öþ×÷ÿk]UµÒõÿ]ïðE?¿¿ÿúÿðáúÛJÒü.ûÒGìÿöó ¿kn¬/am&Û\_ít¾Ç×úþý×÷{5nm=߯®µ¿ÿCdQ°ÁVwv{iC±öÄãág¸a{ì#¾EËÙÿì%í¯íýÿÝ·÷éíýõï'KïáI±öÓ8m Á\Rí0LTA° g;بpXN0U¸z DÏfk·vïþ.+ôµ_éßù6ÁCA ì&· =u±Mjá0]ÞÕÅ4pÅ á°ZÛ jÚ_Ùé¶ÿKõý»K_ÿûù¢¢èDDDHÌ¡Á8B|Äh0SCÚÓ°Å48a pÁ@á±Lp4(ïkaá¡
îÒj½¿
ÎÇ@û¤"8"""" eaU°ÓLTTSPË
ÁcLNôÕ4·¾Ëë
#q ¿üDDDDDJ0A[Ò0¡ qU 1@õUöÒÒqpËD¡ ÚößÿžºDDDDGÿ^ó±tqå;ÿöÒÿA ràÈbÅ:u#¡^L3SÿnI.©¦B!xä.0ì ÁN$ ÅÈdhdñ9Å%LÈ ë;\_%XA½%@h¸hØB Ø$¨Ø» tBcD'4Bwa¡d#4Á E%ÚaJÁGKPi4Æmº ÐNT6 °4l¦ §A%ú{W¸=An¬ìiöõovJätO" ¶Ntl'¿Å´ §îZ¯éºu¶¯Ãí® !§m$è6%ÌkÿO^ãÖ½¯¾?·ÖÚKõjý=ýõÿëý»ªñw¯±ºúÄìÿþÔ%Ôz8fPx]?Xªïßÿ2(ÌAc.CÿýèeÞµX-Ärè:èr!o×üðCµ¨UALFY×ûþÒ@»©jY4äóRì;õ#û ÿújøMNÖ¬¥ÄܵezýAlWè.åz ¹1(@¾Lcÿÿ¬8>uN(®ýè0¼0¿Kð}.·üAÿ²¹¨ßÌäUKûýÏRæ©ïþ×ê®I^¿øÿGb}dh® ¸:zEq'þë×ÒþÍ[ëßÖ;wÿÿÇÅT5«ALïPáµÿ¯¥ªÙîõ_ÿÛõI»5'KÞÿÿÇÇTû']/öë´í{]½i¿õÿÒÌßûkÿÿýt%V²E½°"Y0J ¿³Õv¶ÌÓ¥vlÑZajÛ½&ý¯õ¯þø®®d/%0@àØ D-Ô;b´ØwiÁͲ#ÇÈ+"Àaz¶Ã=év¾Ä¡ÿ³ÑOÿ뿪¬ì§SC: -¦Ã · P Ã
b¢L(t®)a± U0W¬0LÛ dÅ! $FÉI¸ÿ"FHYFCÑ"aEò$DdhD]ï\zõüDDDEñpT0g! ¡ i
´Å+V@á© ¡pEÃ T ØDÜ3Ã9á28ôÁìgÆB3ùñB Í
2"ò0@üÙ\_ÿV!4""""""" Í \ !<4#¯ØÜ"ÿ ^uEÚ@zá>èÞ C"ñ¨Fæ 60QÖ!.A£"ÌÙs#"?é×Äqd¹.jÚ8<ÀAÒl:0x8:Iñ½´l1¤Ø1\ÚAµuIá0ÆA¯.8JnP>0B>Æ\ÙVpÍfDѽkÖTîzÕðÒZ· Í!¿ özÛNú®'®Ðm+Y¤ÜuãI¶ $Á{TmèÞô.ÛIåÆL"d ! BJ3cÈÐÿ^Pëúõú¿Ö´>ßý=>oAë^;® &ëéúI°Æa t OmØaaátzla4lh×ÖÏr"wõªKñÿ\_èiýþýoÿJèUéêúoÝ[uÕWdîºZÛ Õ½<;$q+©D¬a&ѹ¤é 7 Ì"jZOÿßÿ¯ë\_ýWÖÚ8ô¿ïÕý~ºî»¤¡¦éééëúpôå%MÓtû¥ÓÉs¤M÷[Hô?õÿ÷ÿ¿öøû#
ÿêÔÀ_×Õ¿ýuÖývÿéúuý/tºÁ¤2öKi:_üSÿÿÿÿýú_\ÿümÖí+K×þ¿ÑÇi}~/cz½>·Ã LKÿ¯ÿýÿÿÿïÚRÄ)eþBwÿÿïò$_ÿõÿXv?þýLB×÷ýÿúׯÿ®Òÿÿ(×,½'KJAª¿øí'¯÷ô0VD¥këE×ÿþ·Wÿõýÿ¥ÿßÿû}íòÄeùbúµ_ÿÊ°²ÁÇö¿ïûýÿÿý~긦ûïÿïûÒÒëûÿý}ÿþLT[nCëþ¿Þ¿o_þÝëïÛÿ_ÿ·ûÿÿï¶úþÁÃrÄíûm¯Ý¯ûõ¯MÿÕÿºößïÓw÷_Ú_ÿþu"@ÿíy6(^ßÙ?Ûn¾Ì+[ ZÿWÓßî¯ÿ¸ø§Þ×_¸Þ¿ßßÃÿä"vr+I¦È£nÙ{}°eëK¸û 0a|m´½öËlô»iYt¬3ýûmSßz¿ïQ¯ë÷J{siïÿꧥSµ¦B6( á(5 0@ØØ @á´AÄ(¨6 4ÅÅá¶{~Ïûí¶Ý\_oÿÿß÷O÷ýR°áAHWà qÊpLï ×á[µ¬îNÕ± ÿ % $±M Ãb¦ &°ÎLJ{i{ ÿMw÷íáïû®v¤ªèÏñ|DDÄÌ@i; \§b¦ Øa AÛ÷"gd\ì3Ü5´½³?öÒãîI,ì$ áÄDDDDAÞ¬&¶!\6ÁÐ pðÃ%C´¡29'È
Ú]RyÚ¢ñ¡PUNí¤ì0Ä h6%ý ¿0@a)Çí1IÿëþDDDD¥Mzõõظʴ@ÑK_õ3Æ i0DTÈê"ÐMQ!#
çNd¡ ÿZ6BÁBa4Xá øAHÁED(ólÌdHjELÍÌüòúNuF·wuTÁùA] ¢yáOXá&8pdtAL&E>Í6)iiwppÆpÈú $Øpb{|OtÅ£[ ê¨ m ,a4lj!2}qÿÙ;Ö´[xAÁá6%ÔÉNÁ'H7AV&nPwacîÿB-î¿õôõÓÖÔÒµ°Ë0M¥b3ç¼(Xcÿ^5ÿúõuKÓ½4êÝ;Ó÷é~ÿþÿÿ_ëúúýp/Çï¿kõÿôÿþôW'CðD¢úþ½q]~Ô~¿¯ÿí©\íþWòiÓä8FÿÕwÿÿ÷ú21emó²¡þ¿º¯ëDŸDûÿ&)ÿÿý}S;&3"®ú×÷~¾ÿT¿ëÿßÿí×ѹìM?¯k¿ÿÿ¥ÿ»Óÿÿÿíxo_ÿ¥öÿ÷ÿ߯ê¿ÿÿéYËûdýoßïþ»ý/û¿ýÿÿoçuU¿µ°¯
öî¾ý¿ïµÛëþëÿ¿_;~ÒǬJvÃ<?§íOÙü-ëíÿú_þý~ëô)¨0µa6=n4ö^ágÈ'ì3Ûi·ýÿGk ÿ_ êÕ0@áÐ plP á±PÞ4¤Ý¶Ü{ðDW?sPÔ42:¿ÿÄDDDAp@gðeº 5¯ki«60 6 nÌ xgpj1à çÑñ0Agr¦¶í}ÐEõÿT6{Ì 2LO8f¢"2EÏøé48cI° t.EÃDLA0©
S CèÌy²5 f³ÒbqÊ An¨ØÍ#ã ®E¸g} ÊÃ362Èy>n>!N¢dSôÿÓí¥YI´pDt«a&Á'U ÓoL")]²0jpÎ ù M3ùñBhËñ4«B»Ók«4¶IëhîN¢|® Ob{h&õAô.êº/ PÐEã65FÀë½}eYtôõôÚMÐt±åõ é¸N È Õ°cK'çA°pb©êÚA6Æýñøþ·ß_Ðü;^^¾¹;«p·¶©. éºZýt«ýª¿ëëüã´¿ÿõz¦Åú裏Wÿòë¿ñÿñõ×ÿÿÿý&?_CW,ÀY-ÿÿþA ÿþ©´/Cßû/®îØm£¥½|V¿uÞ¿,ÿ&3ë÷ÿ§üGÇæ?!vÖ>Ì ºÿîÝÿúþ´¿äÆkù()"G Î ÅìõÄv ýkÿÿï_}´¾¿Ý7,G@¿ß¥,FLr>ÖÅû[îÔÓþî¿ý.Òÿ¿õzú
ø7¤ú¡0ºÿjýßÿÿþ½ÿÿ¿_Áü-+èØqþÃ=1)ÿ÷kûy×é(¦?ÿû·Ò×Ãûk}ÁKrHìö¡ lmCÛöú¿ðºïýýû÷õùBد}ù\Q²2JDØÓ![ í6jêEÏ äÚR.vÿÞª«Ûÿïí+Uÿ»àúNB=S#!:B!¦A)°Û JßØg/{¶f¶Íµÿ§ûúXu©kǸa!4b {ObNãV ¹Ä(îôÒ}Û3P¹\Kô·0B" ÓÓ[m]´8a @تj Á¼ %TÅ&ÇðµíjxB""%DDAeV 8L/Ø
¸a ¯ëXô½x"""" úKÖ¿Kÿÿ¯õÎÊ3 gLÙçSÈ?^úó±1¦1Ã3$ ?5é,è:¬öE_Çé Aµ Bó uanñáÐA óà!ÇA3ePFc(ê 5>ø1¤Ø1=µFç±<4la a1j&ôÁ FËJú ø¡< ÄtÔ&h3ÃÁ ~J+ ³§Ã=>d¸O\H8aa
è¼Å ¸4E¿ô4ßÿÓtÛÕBzn1þ%ÐiÔ_¤}-WCôãÞÓë×ÿÕ>ÓéÒOµ¨pï{ÿÝ/ÿק׫oU¿ÒÌYÐ~>è)×
ÿÿûí.î}kë2NHÇ×YÐ.>N¿øÿdEº0ÿ_þqßïëëùeëO?J:Këüu¥¿ûijºÿprÜIò_-ÕWÿA}hÿÉøoVÿ×ýºÿþH5ÚMÎÿÿz_ém¥ÿøo¯ú¿Ò÷ÿý ½%jvã&ÓýõëWIûþFí¥ÿïÚKÿÿùÚ¦vU«k
#wë¿ûKÿðöëÚOÿÿë hû<®%)0?kú^ÿ÷ù>ñéo¿ÿ¿¯~¾>ØëL(ÿjºZÅ×ý1éwñ±Wÿÿõí¨\®h¾Ã?íȽ¥a_öï¾»¦ý¯«Zmýo\ÉZ Õ MxÚÅD§þ åò%µ¯îÿ¦ÒÖíÍÔd8f¡QÆE¢ÍFtHãWôÿ¯°ÐAÃa0Ô: 8lPÒÝ=X3Ó¶ .Ä¡ú! .0ò2 Á Á3#duh\HG ÔÈ;S£×ÿ!ÐaL¡Ì>Xá·ý
vÔSÐAà ý¢gj¡=4nzhÜÔ&0ÀÑvÑ®a\4[Íà "Ç"PHAÈ0@ü;;SÊuþS¥ø#B"CÄA \x">íôÐbÐm\_#8(M'ºMCbgi6t è ߦ-toa4ljL\a¶pó²¨aC"ÐS êR¹©þ?ë g 8"ë 40vjÜ á ¤Þéi7N@àÄ®¥Aé6IÁí£s O UÎÅ·a0paf"Tò@BxÛó: ¤GÍègGÿÒþ´ºtººz×ÿýúu©ÖÁDÝ=l' ûÛ ^B6Ôn¦ " ¶padK¦%È Í A
<à§å#EùÆkÿoÿzzu|mÒ½ïz¿^×ûÓµ¯Ké6tµt®A©üL#sP±ª64^Q¬3c\hXDÜ3 ×0Aÿÿÿÿï»±Ý1Åé×\_ñêÿözéºÜ18îù\&ÁGA>5l1>0êtá'Ñ(S[ Ѹ4BVBÝ4k}/þ?JüðO]«µê=ÿÿý¾Úëÿí =JÙ;t¶baéitØ1ôy ul´Uáÿÿÿ\_w³!9 2%]ÿºõÿJ¼W¦Ôß}þµé¡\_÷ÿÿ¾P0mÁ°oÒ þ}øoúöfÒÿþ?Ó "ÑÿKë÷\_íü ëÿ¿ÿ s¡àçCü1 ÿKõÁ½yuQ1´Nûµÿ\_ÆÈ\y 'þ«ÿWÄ×ÿþN'6Íh9´ÿÌ5´¿õëÁ÷ÿ úZýo,O¼CCÝ D£ÿ¿ÿÇÿÛû¯Òõï׿í/ûú÷òtÿÞåYíû§ÿúË.åúÕÿ÷!ö³÷ÿºý«uý7ÿõ¶ÿÿÿ«ÿÞ:\_¥õÿÿe@ý´ûý+ÿåÈÕ>DË°ÏI÷\_h-t¾ÌÓ¯\_Ý[~é·ý{½Va?ÿúþúð÷ú\_ßßÿ\ìQ !Ab»m+ACí&NêÒ¾äÍ´¤L ÷í
û.Kû¾õÿcïÆ¿÷sR´¿ÿßþø)ØÊé\_´¶Ð pØ AÁb¢¢¤EA;ca4î'w|}ªûvßOázÞ¿oIoßïzëòw³±XËAjgeX è4'ª &¢½¶Úb 6 pûÓá=¶ûJ×ûZÿÒû«ã×í~ë§ DDDF^Ü¡¨ª70ðeèM8A$Dt}à aPAÃbØáIO¶¶é9û ÿ -³?ÛÚ°¿×n¿÷ü>}"(qUÃ4¨Á¼A@MVÁÚL3oqò$[Kî»/öÙ÷îô»¡/Ca(|ö°Åm]µ@±QLP m0¡´l4ÛI8ýÓ^Vé;""""80RÇ g8ðaì-« VÁ± mb¾>ÿÛK¡ 0HSøaü}zKÿ×Ý6¿Ïdbÿý[I/Hø3³!è28Ékda~û¤éL'¸E»¤A Ï d¹03hÍ#z0ÌÅ:5fÆh27HYȧ55ZlIê¯Ã!H ÚjÌ"ñªa ©&<)ÁN H¨EÀaÓM|Ó ð!M#b¡ÄZ|6ôq+M '¶1<4;²Æêm!Á¹
PÆÜ&a¢ ¶§áìA,ôÚ^«Ðu×a8mI qppÆ®ý+li6¯C TÃ0A±ï¸ÝÕ7WMíÓí=?²wdîi>ë®ÕÉ>ôtÖÓúî½®¿Ðãõ·ÖÛNý:àé_Ú®(?é/^«×µã_^¾ô5ãOMûûûª]^?Ûãøã¿úÿÿç^¼>×U¿é+ W3µ2¾ô«ëúõúÐëz~>Èè,kÿÚø|ì d1ØÙþTß,ÉÔLbõÿ_û[K¨øÿ÷¨ÃPá=H8ídeiïð¿þ«÷ÿùuþ¾LfHr(r=d:ÿ+ÛØ8tò Å4FKëéÿýûýúßKDC{Ëÿ·Räó#Ù2ë;%^ø:aÝ Mí¾ú»ýwõíö÷Jüÿ¿û®ðY º÷¹,d._ÒÒ}wÿûýÿK÷°ÿý/ikøøë¿õµþ¬wÓwöýë÷kêä ?ßkïÛÿÿ×æE^ ÿµ¦í+¿¯½½®Û¿öÿø§ÿÿññ îdaNÞÊA6zlÿzk·jûf°¾Þßÿ½{ÙÚÑ) f¢:ÿÿÿ ÐOµÅ1@·ÞÈ' NÓØaXújÉlÿi{ ÿeûKé°0VC ¤OFË!ÕgÖqÎ"&2!,§Yïýÿ»°½
8è4@á87j g±MA¾ÅØíSV+iZðÍ°74lapTÑ°0Q"à<àÏ à # @ÌÍK#&NÁÁrÁ'ì<Ùòúõ×] ä ÕezgD|250ýÃM{l(pØ
LP ìR&Òn±i6¡Iõ
Uà ¼ 0QêGÁ!8áá32ódHÉ¡gG¥úñfB! / WWJÒÃ%Ô94I¼tnla÷iI¶'v¬oáP Þ¢óÁDsü'j' áþt7(rÏ%ìGDÜJæO§Æº4B.A¢Ý%>Á2xÁ" Ð]¡:HD1Ñ#6D¨ãûþ4é^ÿ÷»m½ÕzéÅÙ;½?KÓé<&|aH Á ÌÌ ì:m!Ów>K©Æ»DùI:v->.ßá4L{4ªFÆ0{5ÆQ÷ëê[dïM÷TÛêÝaÃ'W0{Ø1ã@mPb{Ý&ÃÛI á^¶ÕµZþ
éý/úkëy+ÐzvöM¥m82æ°É³ßýw[¯ûjµ÷ö?¿cÐÓoõÓ«Óû¥çd¨geèïôÿþqÝ\_ÿõÚÓÿ[Þv²GjdvNH¶ú[Ç¡éuÿ"¾ßÿõzÿýlUNÁ®Q 'ÿ½¿ëÛõÿô¿(·4Òþ?äM÷Ùna ¿mt¿åÉÒoÿD ¡0ýþÆ]þ{ýkû¸i+;5Ø\v©/õí¥÷ÿ¥ÿBÿíÐ&ÿù$ÿo;\ó½Vr©ÙhÎÕÿí¥ÿï«ûÿú ~]¥ý~á;;Å%l\|·U=~ÚQõ¯íÖ¿ø7ëÿú¥¿ÿø7ëÒª^áÃM
ÿý1îßqLW÷dU?¿õÿú\_¿àþÿû¾°¿ÿùÿúúõú\_ëI¿ûüÏëô½s½oö¾Õ5íí¥×êÝþÿv¾Åwÿÿ×þÝzïSµ¿ýrØg½§ÿlÏÿþµá¥
ÿoÿ;) "Èhä|~pÍF@ßÈÈ
/JÿñíC6¼4 l ìC´+TïMCîJ-ÿlÿö\_é?0ñ
Aç dñDò@¥²!/$ Êm3VR×ÿÁ$-¦¶Ø%a$Õ8a¨ A 8A pãNã-íT#sÓFÆx!b¸ ØÔ&7 XáAÏE¸AäaAx!±JvBqOp[!¯ëÐB¯N"" ÁÿôÕº§apÚIäº Mñÿå:}ý}/à ¯ÿ»úÿÿÚÑ¢ ¶»ï]»ÿú^¶ýoôa}ÿþëëÿÛã)gej
?Ûýÿký®Ç×ú_ß·ÿÿøÿ¨TRH_µgûa᥶hº×ý+®&ÿ_ÛÕvïóÿÿê¸U«[ÃV-¤Ä§kñ(w %½¥gÿïûµ¶ùNå=ä¹3ffÍc&B8òÉO^¿JÚ @ب@±Qµ Á¦Q O¹ µØfwÙþ' t-Ãx~fdagÆ|AàfÍ ¥æÈê262|äG¢pÍd@oÿ}¡ ¯ 8aBjîðÕÎ8JôÅ pðA ¨aÔ5A þEãP¡S
OÁB.AaYr80&f8È4LKÿÜDFmÁ[ 4J ^íþkNÚjät#ÊMÈÇônÈìO Ø8;ÃPëéøѹö7aK¦d-eÑ°<g"\pe±\Å':ñ¿ë8¢éen @eªk
ÿ§¦tÜ0Ú\_#ÆaY áí¤á Óa:6 ¸a0 2v$ üè299ñf¯ºôq}¿ÿÚ¯¦ÃJ÷Óu ×h7M¾ì'VM:zn'¶ $A66Æ0(MDðÏSSPA8³Á'\ÈB:£½ªÿ¥ÿã¤ý\_×uÖ¿Ý]uõÓÖ¯7ÒM'Í& Òm$ JÍÇ(v"î&Ì$¤Bl "Ý"í\4BV\4DNBýSõÚÿûî¯^ô¿ýÓ«ÓcëþûÕûj¯NººN º&ØAéÁ!Ù!7#M> ÛeA6a0¨A\zÿñÿä ¿ÿÿýúþÝø¯OÕ{íÖ¶ëÓúM¬eÓ²QbTU°
Õ´F!= @Ø1>Bú\_×þ ÷÷ýÿÇü{ ¥ýÕn.ßÖëþ-þ/½tôéi_JØ2Æ^júJþùðßëë×÷ÐDu\_^ ßþ«ÿ¶ÿzÕ¿ý5÷êÝ ÿ¯c×^ïU²½ItïÿAÿï^¼åÔ"y×þ¿ÿÿã×Xduø/á/×û¯þõ¼tµnÚEaÿéß¾ÿý »¯Øa+ñéµÿ®ßB=Ê~Bø¯\_ý÷÷ê¡Lr¯ÿÿûëÿÖÿïýÿØ®ÿëþYòc]B'%ÿÿã´ý.·A/ßÿ¯ë«þ½êïþ«ÒýÿûýÒñð ½Ñ:¿H ?ðWú§÷ëëþë¯Ý®^þ»óÿ¶õßÿ·\_ð½\_ÿîL¿ýÊýßÛ4]¯ØUûoM&Êzþô¾+õ¶Ò¤¿¯ï÷Ò]}oÿéʱÿ KýNÌoPýXa~ÛXk÷ +iÙ\_Û´¬/ú·Õ¤Å:û÷÷õÿ×Ý}ÿÿWþàÚ¸ì±P plP Ú0Ní[J%;MdMDã¸{ -÷Ã9C 0¿òs m¯÷\_mõÿ¯ô¿ß·ïÿî¡NËì\Ì3²èî{^VÖ6a0bP i
¨6<0ÅD(äÁ;6 ½u ©×Û·]¾íµöÿoÿK¥÷X}HQ0Sºvã;NB"PDByàh0L§ô°°iS°¦¯8pÝ1 )îá=·PÈûfkjýëßíëjÞ¶ÿÿ}ÁÁé þ\""""""""""" ê ÀZ °Á àÃÇ $ÇiÄîï¦È¸ëÛ3Øamw]
~×ÿ¸88<;tcÄNͤöAT8l(i¨aØ LC¸¦'ħ,4¤MZ\0«°þê¿ e×A'B XR¸' ~Ó VH8 0¶¡)¦H8Ø
þðÃõ!!¡
ð ÐbÔ06 ÿu\_øqÿÄG "u÷\_÷ïIj±¥I~þúó²ä5ÙåÕuµW¥¥ô{>.¨Rm8fÌäGI%É{×|êÚ¥/"íÕúh· øÔ"Ü< ã10@ÉÍéÅfÈèG\á
k ^zNl0°aEj¨ ÞaEØgºf<4[¼ Ða¯ LB+¸ýz´®éò\ÒpcI¸ÛÔO úQ~pô.þr º·÷WõNl×~m[9.àé<aÁ\ÐAº/\_ïB\_U×'ÿMïN\_\¿í:Û¯ýÛZuñß÷§~µü{ÇLº0ÿ¤ÿ¿ëþ¿ì|UI¹n>àÐÞ´µ¯\_¿ÿÿÿ¬®±y13!&"ÃMß·×»uñÿòöÉÄÿú>o d|¯6wÙQë&ë&^?þCÿaôýzõI°péáNÇ3º3±/yÔný¥¿o¥òËÿÃ~'ÿÒèì®;)ÒlÉ {\_;0?öúÿ[ÿÿçQý/úê\,'ç«Ñöx;årDA«çÿý/ÿÿÁõKÿÒ¯Û¯ Ö¾o~÷M÷í×ßþß_÷õúWþ?äËôßíµÿ½¥¿ú¯Òÿô÷ÿ×çiìícVíý+3]nÙ¢ïJ¿þëI¿ÛRÒýëÿ× vS%v5µa
§Óa®Ã=ZN½¯í~Òý¨N¬2Fl¤tG.k÷ÿÿ7h àÃc ±Q ¯aÿCô¡¥ /±8ïPôBpg&3 Úäj03ÄtæÈê)!2,GC5AHÈOÏ/úÿ^- ÓM{ Ú§
&PpiEDî ùáEFæ©ÕK¢ïdÓ ò&BiA 30 f å"È<0ÿôá}á
§µÅJpa§m ô!BÒl76Pw
è?EÄA EãÓEä"ì0·¤o»©¢zǦpgv³aÉÐBógÿ¿ÄFDDACÈ2Ô5<'ÜXA¶J/Køè7ÒQ=´ O'A7 H0ÿ§àª¢yáS?3Íê)ͲhæLÒÕ[2\DDDD§'ÿM=;Ð2Îzvt&ÒôN%ÙãÕ&ú7B.Ã
HÊaì"ÇhhA
<)FlgAóeÑоÓbÞúîÛÖ>¯ëtÿêÞ^®¿Oá&ät¤ôW á4l ð?3cUÖ!8· á&3$2éBÇÿõ\_Øé?ßÓ÷¥þjýíSÓû'uÁ¯¦ú ÝHÎ'ƤüOl+laÔaPA.ò çàÿÿ½~ö=þtþ=\_ÐÓÿõôï@Ô'W¦Áp,é<M%Îl+¤ý."¼/ü
ù ]GÁ?àï^+ÿÿÿûúõÕÒת}¹;®Á+tûlt¤ïù?ì?ÜÐ;uüD~È^ÿÿþx/û÷ý·[ÿWú] ý{ï×Ý.XÎá¿Õ¿¯Èr;rêAÎýÿKÿé ×+ûÿþ¯ÿê+z_^¶<$ýàëî}/åí¢uXï÷÷ÿþ¿Éÿ¥×ÿkûÿëÿ}.ý-ÿT¶ÿî²(+û®¿×,GV%Ûÿÿî:ÿÁàýûûÿKÿÁ÷Ûÿÿÿ¿ðÚýiw¢cêAÿªÉÿ²o×Uõ'Z\_¥ßûù0þïKõÿëüÔiÿÿÿ,GþX<ÿ°þÞúþÝ×þ¬ßßÞ{ÿKûýºÿoëÿÿûÐ[þÙtÙö¿ÿûØZþÿ¦ßnÞ¿Oýmÿ>KKOïÿÿÿK\_Á¿ |J}ò%õ¯Vö\_÷4Wý¨fk»µûu´{ÿÿöý®úþû¿¿ÿܪ P áµ pÚà A(pàxm1ÞÁ5»Ia¨a°ÍÍ¥ò÷ak¿kÖ¾V¾ëÿ[×KþNAFÌÌfA(ð¦ãE.P»a}´ñJÚAô4 PbÛ @á´X¦'>êÛ´ Ú]ûw_·¶Ú÷ÿýßõ/³# AÜ¥dPdzÈ"$:(A-Qpka0° ]´ã¦lPN'á¬Jv°a&¤Kì3ÓiYuí+¾¿µÿú3³Ì¿àÈà>°¡ ä5íÇÄDDDDDHVÕ8 a2¶Jh Ú°Ä.pئÒ[`Ë°Îm¥"[=_a~úï×PD©æiðÁá®3P §: ¦Ã h8lP ll01Jwi/¯±ÿ]Òí¾DDDDq Á$? ,0 +¶¡ §¯¯·^3sr¨¯Õ´º¢f),¯A: [ ¨õ¶«Ðzµíé%qQÐÖ8÷÷ëk÷_~K^qÛi%O\§>gfeÑ#úÿw;ýÞÌ%]jPM3Ìñ0@ϳPï3²0ãñ¦iÆ!E!¥ëë!TÝp¡B
¡Bª¦¶5
0@x®Ý) =Yd4¥êÕRU ~
¯ª5¤T£¸éh,ÎIÐZJNý/Ó &èÐéë¥ÅE,RK¨ÿUÒZ^Iª\t«ut16"?Ò®¿ïZÿ^)b¢µñZ^ék[Ý=z¤ª«I+Z×_éªþéÝZ^éÊáwþ´©}Nt©Uê¾$¯I{¯ûßÌ©ØJÆ[Ögtz^¿ªÖUÿK^W²%:[¯èø:y]j WU½}%Køo^륥¯ÖÕ%î¯ÿÿ°tÃåp\ì×'GedböÿZI_×¥ýu¥® ÒU¶µèÈçkëÜÕAM0§uU®ôµtµÒõô¿ZHêôêþÿó$¯4¾o/áÒVS²ÿÖëúKZõú]júU±þ·ÿã
i~·¾ª0÷#´½zùO=ÿé/ªé£Õ%Zþ«¯ñÿúIwAu×KÖ¾.½U"J¾«ô½,÷ÑÑ$ëj·kÿõÿê½+õÖ/®¤{IW¬:Iá,¿ºéf®]uA5úvòúÒ'_ÿMÕcÓ[mÔWÅE1VU¥¦n½ |E!ëªHÝli7¥i òx Bl³d}ÎdS'3¢5DözÿÅ%]qÅ}R
QPP °KáW©«J+N¯Øð JAápÏ ' ãFfffx Í;4À
FñE(ÍKQùÆk2#ÏÿÄ0¡TUU UARlBKaRQ I( mÁTBè Þ¦ P°¦z 𧨷[³Ã &;
Dë ö³?kÿtÝ¥ëÿºW_õíþÎ+ügb£"3R%f[mF«éÁ0ª"LPA 8n½ÛÅ5ûmé6Úí¾Ün¿ê¯½ÞaB4 a«"£""AHD ÊLDÎ8[¿å@ ÓñA
wa¤îÐ@Ú Ô4ïa?¤Ü43PÖìÍ/m]}ö»wtðTÿ Fd ¥;:D B(ÉÐ5#^"4"""""$ìB" ÅdÝJ
àبÒn)bSÕØg=´¿ÝU¿ûÛº$;PTÿ$Ô"ñÚ 8A¨TÔ Ô(AÄðAï¨Ð® · AÆ66I4ÖS» 5O]\_³ûx uDß79¦¤n´»kMª½à©ª/@¨»k¶b" ëRÇ0Ú .Ã[0 ðíIÓ{ÓÐÎ.piKæ÷£uB&=I´PMýôt#,! ;GÚðÖ» á]ÔtýÒÛN¾X\'TDÞ®Ë:vÒ\hhDDDDDDFj1Á \ïÿô·F¥ôßOºÕ¤ÁuhBßIxA8Ò_ꮿFï®ëK¨êú^£é%Ç_õKúuñÿwK^¿ël®MR_þ¿ý/Öü~ñÿ«Ö¸¯I~Ö\yqæI/uo( ¯¤¦ïWjYÇýkõq!¥¥]WA4«õÿï§azOÕ/îáæïÖÔ#vô«½«#+2Ñ.ô¥ÛnMÞ"ÈUôô
_I/ÿUv!4)´©oÔ ðÁÄô!¯ëjWm~ôÐtà RIi!l^@sÛ¯ë¥xkðë´¾Ö¥ç
I}.ǧ×ýÒ¬_+zÚý{yÁ bLE Òm$ÖØ0]´¿5a/¨i0Ïü|2?´áoÿÙ áñA¨¦( Ú6+bÄ E %Ã#ûN>D ¥ì3Ö 0d²XÌ¢@ÌL)Ç8ö¼0[^íRá16 @á° Þ $6+Bq Ù(Q$,áaKp L{XkÚvVYÐÌñÄqDDq1 þpÁÿÕ 0õÁ(C¶KõþÁÿÒÁAÛ!ÿù¤N¥ êiyßkL¨M¸¬É,AXn"6 3Æêv)µ!ã;UìCçz; ÈJ%¹P¿L"~d\e8ÿþá8z}kûZMäSpáÁÞÿzæT¯úÿ¤Þº×ëÿÿÿÿþý¯ó: ÔÍèGDt%¯×ÿþSXÐ>AsìÀÊI±J#c: ?sdk2BÿÖ'£cFîEÃP¿eÅ0<4Ï &8a>Dº a¤ó:ÃýáÐpóA8::ìLô]ÛAé A°·á4l "ÞÀÔ<0¡ºÐapð¯úIá=;¤þ na]%}&á7ìO Òl8ßÿ_úm'¿Òÿدï(è/ÿÿoÿõÿÿÿÿ°¿ðmÿí/ÿΣûÿ¥¥/üÅ¿óiïªßÿü¾¿V½ì{¯þí×^«ÿïùËI@gÙ
ÞÌ-MÚÚÚþëÿÞßZ±_ßý}aXñÁ¶ý°
l-þîýÒþÒµÿ¿ïWH Crí ¨mE~~Áýõ×¥ú×ÿ¿ÿý´·A}|±ë.E¤]n¿ÿ¿ïÿõÕ/ê·Jÿüÿ÷½'ñÿÿÿ_Ú_ÿ¿¥ßýþF?¯{tû_ÿº_}o÷ýk¥îßù¢ÿ¿UÒõ÷ÿßÿ|ý¯Çûþ®N¾ýK¿vGÿöëþºÛû×Ò×þÿðï¾ú_iXÓíS´¾DÍï¶öË« ö
kûûuKêÿé/X th Á P pè;:AÇ|q;î LN?_lãOm×"fÛ3õ׫´äb N9C÷ÝØ^ÒÚna AÁíCS hpz %Z|%ÿDDDDDDDDDG#h0D4_ùc×^áÔR´pÿJ^DDDFGgX@ÁeD0½üõÐÿô]%×éÕ2ZIz ÌðJ«õýtÃPU¥¥T«zL<Kþ;¿¤ÆII¶ÉNùgýè1K§Þ äyChIa\î §ôDAFëñ &:DH D\_Õt!×ó\òж!vº°v°¿Ð\èò¨Sµ136wq(ÉjýÁ
@´ lP)"¸×ïÒ¬>êT<)Þ¦dj263´ïü H8®W #BêùÞ¯éU Ò bPýïCné$ÿMZ "¿yQ%¾ýB| ô×\¿ý[u ëþµý ¡$
ÿþûÿHTÙÿõõÿòl\QU\ì\E8gR" §"Cóÿ¯þþB\_¦áf0@õ(H<2u%Τ´jgK~¿Éµ/ÎÜtÒá}0¡BaQ±îÏ Q>XL!àfá3bÅ×03¨ÊDHWþA6¢]Wbhaôð ª.+^4k!ð¼b!4\0TÓ>4-ÃäL¦dAÏn³ .®ov ÒÒõ¦ÁF:Òl é6É é á=4^5[uôXéuÚéºzuoÕÙ( þjàÌtØIÒl>ä1+·¥ÛA6 Uþ ˶ªoNë\_õt(]éz½:½4ÛA´¼g:ÿ庫Ukÿø×êôÐÿ]?V½\_O]7õ×úÿ "céXß\'Òdt¿ÿúzú¾\ÿõþ¦ÅÓ×Áÿõ!øõþ¿ÿ×\_ïïþxdy¨">·ü"V½9 ¥ÿpDoÿÿ\_ÿãþãú<ÿÕ²>5imïÐ+þXÿ Oòë¿ÿÿò þ¿U%·¾Òé]߯Þÿý¿zÿÿË÷%úþȨc¸Å\ßúÿÿÿ¿ýw÷ý}ÿÿbÿ°Pt´ßÿÛýÿKûwûÿß\_ëZøDUÒø%ÿ\_ïô½oþÿýÿÿûð+\_ m¯¯·~ßö¿ÚÛkõþÿÿ{ø@ÔZÐVÎøa¶}6÷KÝ}³E ~µÿI×öõKß4 ¯@¬alV±lkpÐZq)Ý« +"ãJäÝ +3ûêíðQ;uÒi8l(a ÃàൠaÅ5m&-°ßÁ
¶EͰͶ%¿ðI?©ÙJ 0¥9VC¦ öÛ5m1A¦! h&(t0ÅTAúT3BÏADDDDDDDDDDA aJ§(p¥9¢#©c&LxkØMkúi1è:"""""4"""""?¢ K¨GC²&¤P{æCxPÊ\I!&JȧõLBUØ h'[ÿM#á6ЦôD7ë}ɪ¶DGM«,Ü'½ ëì4 Ký~®B0~6¿îý0wi°imïÿ2²ã¼ëÒº¸ ÿÓE½3Nw"C }÷þ±ábÑ 7}Çý_F£B£ðC# wÿÐhé%QA>q×MÜõôU¸ô £ 4»Y7UýUAÎõ´îÔ-?agapR²2aÆwsû§²Ü!µ5yÆ+IÜÎTÕAANÍGüiÕÎãG@|FÔK½¹3ubËG¸88wþXîí4ÜGI·ÛAB´wq_Bûý{cÚ@|6EB¯~V~ëu¿whNýÿ_× ÷Û´ðñJ½´½këÿëMxû«wÒÿÿ×ú´ÕÁÖçLRßoÿÿýzv¬PÞ×R¶ÞȹFO3æu,êD"gÆjaa Á'a2@Éãc:f/úJÉÑÝ%ý%sµcï²nA°÷w A¤]¸°º6^0Æo Æ EÛ paPzø[۶ШÁãÒÁ;éSjØv ÒÒpppò:jÉu& ¤A¢úa±0íè¾è'»^:Ö b\úÝ\ÐfIO§§éêÒkðwé´($®¬$è'W }ÐNÉcH'ö² 0G÷÷I._¯[ý]V·ü}Õ÷ÛVÒÒÒºu×קK¦×·H4éÈáûUZcmz_ÿÿÿ¶]W¥ÿ¯½oUûtõ®úÜ$ï HS±U´þßûÿGÁ>ÿõ}nýÒz|ÓtÔÞУ^é%§Òÿÿ#Ö¿t@ÿ_ëõû"Oß_ëëÃ\:ÒÒª»{iÿ%¯¿ïJ?¥¯ÿëüÇü·H=íAtCú_ÿü$úÿT ýí/ývÿ ü¢>¯è6kQhF+ÚÖëÿÿÿÛ¤¿íÿß_Ã,Cù12¤ØÈh×¥ôݵû¯ïÿþ¿µ +ï¿ÿ:ÿø¯ÃN¡¯8í$zõýß¿ôþ¦«û¿áþþÝþ!´?Ë?«¤µû^ÿ÷~¾Çþݤôõßúû_
ÉBm~×ékû¯ö×UÚ×íVÖ×Vú÷ÛÛ÷ýnÞXêLp÷t¯KßÛKöþ[^Ì.ûkôÝZí¯ýÿÛ~ðkI;±]tªÌáê86Ò ÄãS»}ñ)ò ûKIÇ°Ï}vM[f}÷fkøjÇÆ+³ÿéU¢:úbwm5µ 6¸PÁÔ;a¤ÃI
Ní&4Ø0UØ3 +ü4àã Á×í¥ZKdx0B"A
Ç´ÓðÕ»Å Ä ¸§6(866)~ ~~اWÓXR&?ýkõqÎuµ¿þ¿TÈOKÿÿÕáôl!2A,"ñª &N K±à!$©-^¼ÿtµúoõÃÐ",ÿ¯¦I{z¤ OtntZè4ÍÞþ·I~wÿÿ¯PÒ ¿ý-{¦é?«.8e.ØMîø<¡ËK¯?áí\þB7?x$'Û®«ÿÿ÷ Õ½þÞtÛuÒZØ3$/ÈÐ~·¤ûð«ÿz¿]\Kÿßú ßõÿõºÇ×ÿiúÛøký:I?z&9 "OK¸î¾ª ÿÕ½ýÿÿï½]z¤úTßÊÇ\_ÒÿÅ1ëê¶ãô¾ ÿþ¿£³]¾ÿ×ÿù@®¿úëOëþéSÿXöëýÿðºÿ¯þ;éÿÿÿ×þEBuOí×ðdp/þ~ªßUB"ëÿýïí¿îÚ\_ÿá»kúJú\_Ó¡"¿CKþÔ¸}Yÿ«¥×P ÿúXÚHÓÿ« %úÿ \_\_^îþ¿3ô½þA¤z /êµìºç¿a&»ÿ¯þ©þ·bkÿIwêºIjÿúBÚ\_[a$¶Q©uý>+ÕöO¯ïÿþ´~ºÌ3\_~¾éoê¥í®H6×t½}Ï2ã ·]}»ÿ¥PDX¯Wß×ÆK¿ÒïúëÿõGú\_ä\_¥iZÿóm¾®ÿï+oÒëõTD¾÷W¯muÕòGº\_é=u¯ÖôÿúîÝ\²cÁ Ú¾ý1×OÃkÿýo×PKÚÚßw}ZZý§ªJ\_<õ´\_4m×édøJ÷õ¯hiýïZ\_×k]^ÿ®¸kH¯nîAyöEÔ(4
®Ëp±]
Ñ/VV»iv ŵÿüW¸ÿÓúhíý}úá/¢¦ ° ¡ØAÃÝ´¢¢wºÇñ ¸{"m&ë´ú½ì0¿úÿÐ"o|kã¯é|BÕÓþx}êö)«l'´Óµ[ + &Z!C z±;þê7]%¡ ®4ëûï_»ÙjÒá %S²F\"AáB¦aÊx4ÓÔkiئ!+[ZV ÷Ñ¥ z $G!µwm¥·ýWü"Oý H \PN#3 !4QÍV¯é}$¯pGu¯ýïÿ|WýDKZO¤#ÿ Ð?]®p78k¤øíú\_þ+ÿÖJ$½/Ò¯¡mÿT!b·«\_Û^»ú'ô %3Uþªßôê¤ 7Õüã?þ°ëKt¤RLr/ ¡Wú¤«Ó¯Z¬±Ã ×ßï×÷]ÿqU\_Ñá I/ÚZÕiU±Ha¶ÓÒýªÿö¿Ð"?×"CB zét?öèqì>¯±ÿQÿÿ
ÿÚÚI%ÛòTûi´½ªL!v@ÿ|ÿ×z¯ý%¸ «a B¶ý´»Ð~ýøYµÿ¿ÿ]t¬w\z]ØKëG]?ïÿëþ¿ÿ× ÕÖU2Q\%ªÛK¥ ÿ×ûô¾¯ëõJ °°¼D~ö^ÓÔ{ðDuÿ}¿ ëÿÿlbÕé/z0äGåÒ«OxKzõ¿ÿ¯Øi¯ú õûK»¥¥ÿÿð@¯ã U(pÿ¤ìBIÐë÷ýoÿÿK÷üÚZöBØ>dc%)E}Wæ@H éüJíÿÿÿãZ0óÄ8|É@á)Ùí&3;Íq.@½Ý¸h"Üþþá,×øªõÿûT Åéè45zµP¥hR]ÈGú¢¬T^©]úNmcãúGÿuô¿ö7ÎÖÒzk®ppáøp0Ô.Êý5Õ3 Ï/ ã×ë]!Òc¦ îiææÚ{V½zÓ6>öºDÇxZO¸&»÷ÿÅpE×ýÖ Ðb«OCNµxü0I¿¯VÐÚÇ]¼\&ëÚ¯÷ÿÿpµß ÿé6´3ÕzNõmVóºÌ}}´Òõ¿ÿ@¯ø®ýdQþ¿ý/¤aÿ¶{ï{iIßïîöGZ ßþüºÿüuéþ?ºZëÕ¯]õÓk Õ-|Rî.¿×ýD/þü$÷t½s²ñþ¯×]VÝpíÒcµ¶M÷÷ÿ¯¿DïJy×D;¯£¨|Î(RFi#¨©gLu #Ýd ÔÿÔ¶ºô<
·½~ÿýÍÔ/ÞB:OÛii~
YD!fxT[!A× Ô ÐY!úºëº n¤ ïóiy#ÒÒþÿú ·ÿÄ$µö«¿£eBQ~Ë6 é£a4°¡=#ÞWP[ºJ~ðÒok¥¢>ë\&ü%ý~½¥
ÿàlWºôÜã'ºÒAº'ÊN®©?GÅÖë st4½ÜÛ^±UÛõÿýÖ]jøú +Ý´Õ{@MÍi/t¬gue.SÒÒéë^w¤ô´Þ lWáöýwM EÓãþ¼Wßø@O´ëÝ7J¿«ÿXé/¯¥¥bª½/QoûÚMÉòþïá/sÿôþ]lR[vÃKkþE].¾¿éýuÖ«\_ÒUéiHí·ÞþBæ:IßÐ\_õÿ¥o~: ë¶ékúßÿ¡ú]V´ëþ'Aþ >L
hàïK]R¸ðÿ\_Ä.äÇþ 7L
S±éÝýMNIuA%®:5¿Ù ¢ í¡xô¯°]¿þ¶<ºP²yá´¸"Wý/ÿûÐNBuYÏ\Ö½m ûðwn$q~ëmßÓú4Avüa»a4Ú]ÿKþÿÛI~L:ÿJõ]té?ì:éÁX{뻺×ÿø%§kåôMø µé¥ÿ¯þÅW ºÕ-/Eëý/Ûõ ôìéÈÜ
uv»ÒVø",zÿøG]ÿÄ µp¡¥õʵéwöõþ]uê´ªýuÒ}ZþìaÕ^µ0«ÞëþÿëVôÞ,4×êý-æWÿþºõêÿè>¶EÄ×ð©8kß}w}a cÇÅE]ØiÛWT¶þïÿKë¯KJºõé/ÿáê¬:ÿüt½×ýa¬ðPºìk°¿WHÖÕëýRüÍkúE ~ºò\_ïL:zê4ÒN»ÿõ\_ÛÒ°îÃUµ²ý¥i" °ÚÚß÷ê\_×/åûJôÒòýÔ¿N «jꯧèP:ý|oÿÿ&?o° £ÃIbqÜl4a%qìJ{iHùÔ.\Hq]iZº·Çj ÅêëõðmH»ù~¯ÿù7ûéh'Ã
¨¨®ÖÔ 0@áØLLT\a¥Té+\VÒtµæù ûïOÿþ¿mø
÷JÓ
%N[[ûm a0ÅB¤ p¢©ÅjÁa-ü:T² 8>¬=änßþïÿ¾´¤#ߨ¶82·¤E M}Jr0 Ó 'b§í þwVÈ7zý6¿ïÿ^ÿp¯}¯Ä DDDDDDG§õ k!ôCß.¿l»õÿýzÛÝ/q|¾©[ÕèEþ_"¸#IÞ 3æ{ÿõÿ¿ÖðßZ__é ÷¥ã±îuÿ¿áÿ{xWñé-ÿ_Ò¯GìWç°µ¿ÿøøÛ{¦Ûw²-$·éWÚúTÈ8ä·rÇixáÜ%_ÿ¬.¿½ÿô¯Þ?ÕÿHîý ´£ØJÿÿÿ¶ï¦ïíP«{¥ú_ÒPíÒ³ËÒp¿þµ º8=÷výÁ9êý¥úëô¬Òkë°¿þûaFÚvï«áP¨Ø0KÞ®«R lmñ_J ¨Duõ®Ý·íÝ5¢+ºb õ} ô ×ØøüwäÝ|§|m¾öTLp¢7õ~ðèoSk¡ûÿ×êê6ØÛÁxºL¾¾DÈl|%ØàÿúWÿ¿i=ïMá¤Öï[I?¿·Kô'ÿÿ;¶Ótý{ZLÎÿßxaz^0D~ ÁúSí¶ ½íÓ )ÿ[øÛPi°c_¥ð¿ÿçùÇîIîU;è) D5Çu¢ºúT}HwÄëóÊ5ÌA7þ¿±~Ûjú¦´»W$0
H0!U: H
DLdÉR5½n¿«Õ·¬¤ ÿ¿¡N»m´é^ ð@øTð¨¸h4<eVt\ô´ß\RDjôü>©>¿úç·Ö¶¶§ i£iÚáQp.à ýé6©W¦HÌá[ÿþ\qømØ·PëüÓH·ÍÎl)ÐDÞh¼i<Vò6Ø\74él:u¿ø´
P",'ý~+õa»·ØKzÐéèF [é6î|·¤ÚN1é6ŧÒI½.7I»¤øþÿ¾¿ç0@½»õA¥ÕñÒ|}÷úuß
úÜ'W §éaòÇj%éw®%¾wÇWÿîï±ï¦8îqÿÿ×\_÷8ô}Çî¿éX}SCé°×äøUIi×ÒýýD{K ëþ?÷êÿßÿ×»ßÐm-\k4Û¤ÿø"1úÌ;ÿð"QúÕYN·Yúã÷NõþêÿúJÃëÒ z\_©CükéÚÿÒÌfרD{úÉM×!}×-CýúÓú ¯ÝWC"Çÿÿõÿ¸I/¥ÞµÈédq;úÖ{ +ÿÿI ÚõÉÛ÷×àÌ\Ò2NÿíÿºÿômípØÔZº¤¿úöý¯úXa×ú ßsképþKÿÿ®º²"Im·½iO·ðÒÞ¿±ZÿÿJzê´ýý¯¥¯ÄÿþÝAxïµIôð^×KfÇl}ä(OZt}«÷ëõ Ã]zíÛ´½zIqúç{ïézïKûiw]ºM
ý-ҾõíWßVëÒ÷Ná×æA«<ºþï?ýzØkíò ÅÖ\V5Ò§ZôØaxjéPoü:÷!Zߤ.¿Þ(ÚñÞ¯ ÔWzëþÜ2 ñé- Ñ¡±8êªÿPÃKoS÷å»þ¿'ÿ;ëÿÿuåÇú¼'¿µ×}·Ø]úÇþ¡¶µd ô¿K |HAÒ{Q~¯ô ®¢7O±úéö»×ÿê4ïÒ¥ú8an-/ôÝ®ÞÕzñ]A4N?ÿFý¿ e ÐGh¼>ö¿ÿº÷IMÅ;KýÇÇKþZ÷¯ô §xÿãþ±ÿ
ÚÁ;×ÿÿúT é~ºZI~è4¾ï¥épÉ!ÿMÿüó˺Ò^>úÚ/Ò[KúzZ\_úkG~ä¡t¿¼%L0?ëÿ«ðCéwX û.i©+^Òô«Kÿ[TÚI\µ~íïÐd¦ '¯þ½[\_ô
iiúïºþÕÕ%t¯ïIÒß®öÖúô®C.ÎÌl0EÇOþ¿KK
ãÿSIýº4}ZD9ºzK´µýN¤þ¿D6}±®¤uúQeÁµõ£ëõÙC¾Í^+Gb¸DD>ÖÃTí/I[KH.·¢ §4i¶»¤\A%\_JÖõö¿ô:C=Ý°ýê»ëÿäXîáF·7O¶%?´´ªÒ ´%ïD¶½è%½^B×VM×ÿ[i~? ²ê¶Ï ¿¾õúÒû.éiA0àpAÔ=Á ì$P%PÒ´4ÕXí7I\ ?ýwÞýØ\_ªérËGÙ!Ú¿íoÇ|ºÇJ¿A÷I± ¢Ä&+0A¾Ô2Ų&WõKãÿ±Ûõä è554ÿv úÕz×ï~óÄDE! % ¦jg(ê{²ãR{^aÿ]zoJÿÿð«v{>Bû%È1cÿþ©~.øz ÄFRQeIQÿK¿õnëc¥«ú[PDy+6½ßõðE<¿ßt·xMÒÆIk=ïu\_#¯îþ.!¦×ý8ãê5Öy ½xMô«þ´î6ÖïZóÉ.µ\_Ô!ðoÿæüNÐúþìîÝ\\_éWcm,ñªÓ~5ÿ]Vm#ßÖµøûm×Kn²Çé$»µªÚmÒçô;×vÿãüOûõÐn5úªä(~í%ñÒþÿ<¿ÓÿýôAªû~Çú®aðaó;C»ÿþ.¶®ÿRuY´4mN\_ù(ïôÒA»CÕPZZO溺ÚÔ|Úÿÿ¸×ùßú×¼ÉÛ%;^¶½.¿ôÖm~ª\hÕýÕ-?öJÿ¤©+ûØ
A}ëßõ¿ëëHkÜyûøaÚKÓqÚ
ô©{ýñµUõ¯æÅÃ<'ÒÖXúgÿX~ÞµÒuÿÿºÐDÇ8¾¼7ê÷®íAÿIo͸Aýê×÷üö5¯ÿ7ê8õkÿý^Ý¿è$I¶gè8@©U}~?Ö½?п_òOlV¨>½×Ü=ÿJXÿoã¾þ½q±u¿®#FKÆ/û¯ý®vUÓ$";5[é%®ýC8}c¤·¥ÐbIÎ?ÇͪéM
àÈ~¿ê¶Ô¢:fÎiÅr»OI/I¶øl>~+Ä.dv¿]yµþq,GÑg ¸åv׿ú]&é[½Ûzf\ק %×}&Ô/ü\DAÿÿÔSkÍ?@¸°Ä7NXþ½ôºAPAºVÛOÁÿI}[û %#Rï¿K³ÁõÎ?ÿÿñ~!%iÿפÇt&=&ä§Í¶ó²ùÙWïê};Æ!= kXÿþ¿þ±#®qäHÿÿúÁjóé
ùÿ¤©.íï¾Å.AAÿÄÿÍqÍ¢\_Bm{dX3ÿöéWm·ø°»é%ôþGD2£ª^C>úüs½
Õm%Öëõö¹1ï«úýmß8ÿýì\U×8ª¦ÿá-
×ø©Ä6¯ûøI!÷Ø%è \Rû]QǯÕÿ¢8_ÿn¾?ÿûaROù ¯ö¾¿ðôúÿ¯¯W_¤íHüL.¹»U÷÷õ¿ÎfBÔ_õÿþøa-?¡ü~PþBc_þ5úÿI-&ÙMÉ?#ú¯A_¸þ¿ãöûðôïÿõô¡üuþ_?êCÁ¾Ãªÿõ×úðKÛ lS É:_é-ZýýÿÌ~EüAýÿûÖ¿TL¥Mþô ÿ¹b: ÷öù=ü_þ´K¿£¡õúÌ)n¿ñ½ï@¼úUëPãûÿêú×ûwýú uò@½®û÷÷¤Õµp|| T4´¯ê0½ÿ7ZWUý«£:ýk× ÃúJH_¥÷ý¿»ÿûÿ¿JºtE{ô
µ÷ùuÿë\#AÿǾXÿókëø&ÿÒ×îÿ]õõïÿïî¿Òé$Ú ê¥ÑµIMÿÖºûñ"Bfuä
æ?¶ßý~ßÒV×êíêÛÿïÏ}ïÿßþþ^Ø ¢"ªµ±ÿõ¿!ï±ô0 Òëú_H7ôê¨Í¯Úÿ¦ºßÖÝ®¿ß½~ZP
þ§¥¯ÿêµQøì]zKWãë× ÞøAvÞÛ ºÌ&Íë¿ÿmö«·kÚߤ[ ¡rÇ|7Téÿÿÿý^/>Èÿ\_úý7ÓIXi\ê »´¢Ø«±(È¥°Ïpà {göÁlºn«é"é!A:¸úªOþ¿ô(5×<~¿ÈjÕµÿ-쨯 ¦( àê 4I±LJ{iÆÄÎí/Ò ~Ó¯ÔUR·ã¿ëa{ïWǽ5Ò¿õçï»/L§ì¸L ì'ÛkMl+AÐb¯¤4J§ ´õëôIÿ¿ãÖºÃÖ«ãÛo¢"""""""" !:0aa%: aSL%áСàäv«I%ñÑ5Þý}~öb8³kûM^¯õûÛ¶"""?éxdÄ%@úªOú×áú·KYÜ?ýt,ñ¢Ý¥þþÄ6ÿÿ JdQ¸U×i-/ÿ×6ÿküw¯eCoR?õÿ$å½»i%÷UN>½.ÔÇÿã÷¡ÿÿà°CÑÛ×_ D}?nÛ½%ï¤Ú"ÞR±¾µ÷÷ÄD?^«Spí/ת 8î6öµþ¨-^§ô&ú§ÿ#è|ð/ÆÏ ö"@Vî¿úéÇ»wm%ýÐI%N!5X"%«ÿú÷±~+ÓCmõÿõXîºIÒ_²x]uZ§\ð¨qɿøMXaÚøÿ]¨=Þ«ü$ÈJëFÁZQæ×þcBA]ü!i¯\_¯.Ú[q\_Ô%MAÑ!ª?ÿ $uÿ ½¬<ÎþA¬¼mþºõìtÜ ÿªP¬DGô®?ÄëVǵ¦Ûkõúøh4ßé%Òþ´lþ:aÙü1£8O 0zøý°Ï´ZÁýGKJëköìz,úî!6ÿ¤È5ÁÃ6.4ì?G\_ÿýh±ØúüúÿÈ9[ÿIRH\&C@é|~+¾EZKüÃöu 2ô¿øÿ«ëGeKôù9}kén¹cþ¿× ÞÈ.'ì0iéö¯+3±AfC¿Ò H¯ï u çÿÿº÷¥¬0Çßþ\¯mrWú¯¯©ÜH3ÇÿI(ÿÿÛ¯®5¶é0m#þ¾ì~}Pêv©/aB Ñ(þ² u®í¯~¿ZÌ8ßÖEÕÿnÌRïntßÐIi%ÿõ¯uOßÿ^¿¿;× ¿×vÓ|
]j'c¾©VU{¯ÿÒ8éÿêÿÿÿ 6é&)#ÿõãnêêÖ¸kH+êx"-÷w« _ýPÃYÇÿÿõþzXõþ)¶Ò~?Øú ^¿þm8þª¡1¿Þ×õþùÜW]ë÷¯[áÿ¯Zÿñ®ÿúþ¿Ä.¾î6ÿÅî ÿô¿¢n .KãýC÷ÿýÕ\_ÞN¿¯Èì-éÿÿýA\UþAáê©7ïÿÿÒ¿óÙA^×\_ÿQ»®vP)8¤<ÙeÈÿ#(ò¹³:¢!ç Ôdº$2/d¢:\_æ B&àÿÿû# ÿÿÿéáówIÿÿßî· ù äGaKAá¢$Þf zaN C²& Ê3m2@Èy¦l±2{ö ô¿Ä¦CU×üD·L»ÿÁÿëmB(zXtÿÇ{zxEåBÁU}? ^=0¨¼sf & ¨¸ ÈL"c³ "s>3awø0i$v:IQþN84¿ÿþ½-k½éÚ¤Iu\_ôÞþtGÃÄ÷¤ÜOméÒx(Øõ' ØPʼ¸z¹º\_ ÿÿ!"9G¯ÿÿª¡\_õó7ngOïí·WWI¶d ¾ Ô\'¤ðtàÇMÓÈÀt6F:82<«m »l0J´µø@¤RÇ%BCB@¼?ÿ½¿ÿý7¶©:ÿ¯3¾Õõ«Óôý?ûtïtûÉ]-l4K²wméÒÛKØd¨$ºZý&ºKq8 ?ÿÒZëÿ÷Çi(m/ú×mþ£Ý}ÿuÿC¶ÿM¤8ý_·îþð@» r5;ò ½úÿýÿ\_º ºU ¿ï]Û6WÿÿïéáÃþýEwý§I·KRì) Z|Bfq":^Iù0 ƽûýþþÆJÿÖµ}ßïÿï÷ò ö½ý¯
×µb¸0KªIÝ P± Xݲ6 ¿úIëþÿ£X0ëÿǽÿÿúÿÙ>N~£òOã£@ü6ÕRÐý(+Ã!Ò Ç_¯ÿúwgq -ïSSßÿÿõüOÃa¿yåÿAÕßý?ä´¨ºT!k;"þ¿ÿ×éßN¾µRn;j¯¯ÿÿÝ~? ¾¹b\Ë ®O®Ð^± ÚõIÕÃUýÿ\_XþÒøcÐ"?ÿÒV'FÓ]ëÿUú¿Î¤ úKÿ¯è/þÇ ÅZI!@é8¦<0A3߯u~ú×®´ýªÚÿ÷¯ÿÃþÿÖÿÿö¾ :!6SSàÁ¨Í ûõ×¥7ô´õ'}}W^í~ÿþß¼÷ô½Úýú^þ;Üè"êKH \_ì28?ÿõëùÅö½XJûýê5ÛÿÿK^ýwÝßûuÿ\_I¾ÃA´\ÅÊ@¨>º!\_ÿûW¯µàÖZ\_Á]}´¿ý+^úôµ}5íºîßÖ½ºT Jb·ðéÿÿêýûõösw]wé.®Ã=½'Èá¥a|2>Ùÿÿëð»f]»»°·ö·°ÒK\Vù Kôø6ÿ×Ö¿úÿ\_Øv½ivnÇ $Ð"iÅ1W Kw +ImßpÁ[JÛ9°ÎK"@íÒ½Z#Ó°uN¾RÁtîßôÝúÄ×Ã\_þ\û´ ÒaMCiATP pÚ@Ába66(7¨6)+Ù «@\:}j ¡ê>þ¿ÿKûüö ÆØD¥F½·àÍ©2Ô"høa0
8åg$Áð§ØO V i
ÚØh+xdè'ìrÜ{øH?ûÐ×ü¿þùô÷ôºxBDNDDD0B82Z])R º·a¯ÿëýú _ÿ_Ãm."# $´µ vÊú¿ùïÿÿç¡Õ×ÿíß¾W¡Ø1K¯òR® l+øÿÿûê|AïÿL;s¾JµZW
¾:µ$69Ç)ëÿÿþ¾Ë_ÿ÷W~£Ò¶ q}zÉG¿8ÿÿúÁòiÿý7m¿.bÖèCø.Ýü¥¯¿Ã¶½ýÜ{úT¸0T2\5-ëÒwÿQï×ýùàeõ¿ÿWkþ 'ù4ÐJuçZW!°sGÿÿ®qw×zý|Meªý H& }eÃ(=µ[¥þ=y´m{×ýR×½×úôÛ¸a;ý±!äS@²á%Ú! ö¿Ìuwª®¿Û÷iúÏ××ôøÿ¸IH(0^±UuÿÁÿÿÇéÿk÷ØkýµÓßWÙdÈà6éw#}¡÷ûÄçø¯[_\ãé+ÒmÉÚ»ö4tLÁ® ¶ì ´ÿ×éuùåþ¿üÃÿzº{Cï´¨7;@h§ ºH·ëêý÷øçÿãÿ_´ÿÚÅ7×ò¾¡ÒZ;N®Ý ½×ëÿüõ×Ïj¿®»ë¯ïNÜ ³Hì²Ú +N¸mI;Õëúük!¯ïÿÿiéá7hµs cR¤d²¹Á©#M«aÚJ-ÿß&ÿü<¿ëÿíþ·_Ò±oZ
\Uý' h$¢oÿøÿß_¯úî3ÿô®7nxpý)Ì'M%[¥B_ }q¯ê5×¾º×ÞºWÿi!wôw×m þÿ{øþûþ¿ÿÿü+ÖK¿I¡aÒ¡ÿ·^w z¿Kú__ÿ_úXIÛúëZIW}NIÛÕkýß²¡ÿ÷ÿà¯_«Cÿë®í-_Ivjé%GúmtЯÿk×òxׯJ '¿¯ÿ¿ÿëºNuIUé~È°XGÖ·õ×ÿuïÿý_ҫȨeÉÉ$ÿ×ðÆN_Ï{×ÿÿý~á+¹Ø²A00@ÁgAHy¸= P!8¥@¾lÍgçNdB!ÝÁ´ÿé¶IÆó_]ÿ~¿ßÿÿúõ¯ëÿÓwA£cPFÆv'sPÔ"pEDøÏ&&; ,weòCÙñp? Í&DÍLÀÎ{´·ëä7e=´ªKÿÕí.°ÝUþ¿ÿÖ¿ú ·±=°a¡¤è&á7£A:Ñ±ê ¢úÐC¬5¦±BãM]pj ¸ Ó×KE»sXey%õÿ_¼1¶aÞycÿë÷GïJ®à[ ÷KIÁÒm'¦Ã{IºÕÒouj'kÃI¸Ò éF6 %ÿuöN>n>ª¿ßÿöñ7ýxÿ÷õ_ñPôÚ·Nõ%zÖ\îÉD$ôºO]8eÝB,áKIÁ.i\ù\¸}_»ÿ<«~¿ÿÿ½hî?Ôs§ÿÿ|T·þ·ÿ®ëïhVþºzî¿ÖýV/[½\éé°qÎÔtïþD©Óµªÿÿÿîÿþµý~¿ëR+¦ý:ÿcö+¿B½/û}ûþô5¯öí®ìÛKTøëëÿëÿµÿéÕÿ¥ 6ÿ²:.ÙDQµõÿø8i~;ÿÿ^»ÿUL1wïrGqK.µñþ?uïþª×Õ{\^ÿÁÆÁÔ~Cþ6BLþ´»Gÿ!_¯^¿é[ bËÝ}¼ÃºoBF3ùµÿÿþ§ÿׯ¯¾¶ÿ$°äDp®Æ]ÿÓ ýûýøý¯ÿÕmÌ·ì$-I|Ø.é5Öëÿ÷¯øÿÿ\_÷ÕSáxnX7²Æ4/þXûIÿG1\ä^ÿÿdZwmÝ.eéèÞÿïú~µÿÿµ 6þ¯Î°oؤÿa¿ÿKàø!õÿýdc¹ (! ¶Ý¾CM¸¤C3±Øÿ¦?Lg\_úÿëüÃÒ 7ü]ÿdÎÿIÿÿÿÿND ± í¶íÁ6 6óoñôÉ×óÛÿÿ¯ÿÿàý$oÒ=ù?lõ~ûÜ6þý¥ÿÿ×ÿÑ\ëØä´3vÝ·l)Çjx ¦Á¥b>¢Ë¯ý×ýc×ÿïÒ¤ßï÷n÷\_ëY«skþ8þÿ~ÿßìmÚh¬[ºEÕ1Áa6ÅAÚ!Ýûæ×^×Çý{+o´¯õþÛ]võoïºéÿþÿþÿDï´$È5gÛ Ùðö{mA+Ç¿\_üWèÍuõôøa&ÏÿÓ ÿÒVgݯa{Òé×ëÛþöëý-Jô[ ,í·^È7UÔÖ6 @W]sËÌ.xj¿ý¯ýW\_é&ï0U¸i0Ám+a¬%Øwm¯Ò´¶¿³ýý³ÿ¥ûuÐÔåÀÄí@SËd£é|/ÖÚbYìEþ·?ºÿÖ½ÿ ¾8qLTP lVÔ0@ÚlT0 Ý=5¶ 6ÒµÛJD80
z2éWùî BÐ5ÿÿñÿúÿÒý%nòÇ0¶qì ¬0aZmZµÄ&( uA ¦ @àØes¦
Õô÷ïÿ{Bÿùà?ÿýkø"?×A+éËøFG@ÁR¦ççaxaHÝìBkô;B/I}>¿ó8Ø+ïÿcUÿ¯øÕ²¬ß =Õ5¾Íÿö@ßÿÓç¯ÿ×ÿü%pwN?iéÿÿðC±ÿùÜjãÿ¯ÿºN_½lj]{¿ëÿp0ÿþ~f¿¯ÿ_àþÚ/_Ö×úþ§vmý_ÙÇÿÿÿG÷¬$Ý6ß×õKÿÿÿÔÿÐuZ¢÷Ö%\_ê$ÞÝûªú¯ÿÿÿm&ÿígqÒÅÿZÿþR½¾ºÿ´ÿýè&/ïëØÿþ¿ñ 'pÛöÒßS(CGý¯ýíßÿì^ÿþ
I»oØa/ý}WÿÿçpÈú@Åÿ+Ä5Qÿ¯\_{ lBûHWúÿÿbÒËÿ×4óýúIÝ·2l÷ØÔ}ÿûÿô uýÁõÿ#ëw|~ þê¿\_\_ýs§ÿÝÔÃÇ\_þ
$í·t³Ë¿ÿ}/æv¥¸B# +¶yP½/XïD0Û¿î\_öµøÒÁ1þm®Ò$8ׯ¤«Û~v)¤ûj=ÿýW¿üRA¿ÿ^±ÁqôDx( m¶ø)ØB&÷µî¿ÿ÷¯ÿÐJw!þ:õþ´ DCCa$f»ú©,#³GÕîúÿ×ë_¸@°ÁÿúÿÏcTqmøprßVA[÷ÿÿÿyE¸|Ìþ¢Çû>ËB¥»ÿª3ÁÿÒ÷¿ÿëývrCqoûý³Ù£i%MÛÿ¤8>®ÿýýÿ¿Ep¤]D| d,\_ûýûP@¿¯I$m¿ïD¢ÿ¤XÿOÿ¯ßÿR¸ØÃÿª¤¾
kqToþ\¿÷\_ÿúÿõñ® ß¿ÿTÇ× vØm¯ÿÿKÕÿÿÿë D Ç!÷aß\_ïõ,u¥oÿÿÿ÷ÿ÷ß×ú+äv Çwÿýפ ¯ qé6ÛuÿÿÒÿÿëÿ ßfaïöÿëét¢»ù²:JÍÔähGB!qÙN¿ÿI\_ÿý%½ÿAë °Æ?Kúÿ[m»ðì'¦adU 0I()8 gDpÉ«æÈÔD+'f¿ýuþµ÷ý,ãÿÿé Joúh ÜÙa0UFƶ ¸ " &60DÇ ø äA°ÈëL È« 8ôíÿ×ëþ>,êºÞéS ¶ïåÚOÛFæôh'ÓAè:nÍ ªh¼i/ ]°UEÃO¿Hëþÿ×ÿP7§(x÷ÿæë¶÷á6¯¼'ÜÒopbW´IéÒm'H8u ÅÚ\M &õ&.ßþ¯ÿþ:C@lñþï@¥¦í÷í}êäòºôêÿé>þé[6X¹téè?êßýzÿìaDÈ«þ¬jm¶Qzÿl_ën¯uûõúz}ß«ù;«uÿK§«ÿïÿ¢O&Á/×^¢ôg=»¥°×Øֽظë~û¤Þ>-ÿÖÒ¿øÿø¥¸1×zIÅjiì>ÿ¤1tßü×ÿö½Uÿékÿ-8éJõªúX^m÷¿éÜÐ?!E¿ñÈYïÿÿ!ÇéÔ}̹~ÿùýDÔ(+aôµkIíX«ü°·á¿&9c@ì6pÿ¯ø®#3 úë\_ÿûîÙJfA+ÿ®6Ø ×üô ñýxmåa¿ÿÿäÁrr8ÿéßßÿþ Ò&õéÿØ?¥ùV?òàß÷ÿýru¯ñÿ¥ýuò ìlDq^¤Ým0_ðÿþ_ÁòÈgýÿýðD~ÿüPA5´ Óÿr[¿¯¹¯~¯6Ú|ÖûKîµûô¯ý/^¿Z_&
dq$(ëÅ?ßþ½ï®Jíùëÿ]ÿßþµÿÿïü"b
D,ñMiÛ×õïîïþÿ·ý÷¿o½~Ò¯ÿÿÿ(AD:WWþîëÛ ¶f´¬%×é6hººû¯ûý}-_ßþû8AB ôº íI¬NûÈçk ÚVì0ºÚJÚIí¤lÿ¶gÚÿÒøÿækø ½µQü áPAÁµ ¸v\!E0Á(a(0Am$îÒ]m¶ %=[õÿÿ@@33ªø0@Î= Ü0
ÔTR 8lP@Ú6b?÷¬úÿúì ( ¥§""""#- #[:22:3¡AÊB
ÿ¯ÿÿújA I"9AkOÿÿ®ÚI4Ù}"é×Nÿׯÿú°»íª[èA& þ?ÿûÿÑÅÈ|¾Ptd Òz÷ïÿÕzéBéäTvÒJ³=PL&ÔȲÿû_ûÕ|&ØOe"mR¦Á4L+k&M²û×Ýÿ _ ÜÑ ª«ª áO KåyäfvÿÕWÿªõ¦Ö6S¦æzU¡H"¹)ËÍM¢4H) òæud»ÔYVΧd»wÇþQæ×ýý%A/ªj¨ÜÕxÁUO®G!.áL à ñ@ÏA§@°@@Za$SÝ|ÿìÚtB·ýݶÝ$º ÝÁXÁ&ÃÛ¦ô$+ ,-§¢òwFà ¦P!bÎEßEà A üË×ëýþstC¿¿uöúTúÈÇ«m:Ø1#Ãîñ=¶GwIÒjq,vv«Mº6iíEçñÿþ¿ m
ùim¶¤©í7=m6É\1û®àË Nl=ikÈçÁMEî÷M¤ôÈí¥úz¿ÿúÛDGþÿÁúUH$¿ã½?µOkwõNÓûÓ¤Ø=7O½80µýþ×ôÁvþ?\_ø@§t¾¿ÿL$xI}=-ïøï\_]ÿ÷¯¾½ÿÿïïNµÖPÆ¿þö ±l ª¿õðT´H A+ÿýÿþ¿÷þ8ÿþ=\_ãcÕÿï¢$¥Òøýix%ªíÿv× ÿßÐëm&çvÿð}ÿÿðõß];·ó î ¶"?ßýöÑH] \_ÿëÿ×ÿÙ"·ÇÙ@»û!H@ý½ÿÿ@
°ºú^ø ³¨uJ¨$ÿÿɧö½\_ÿûß\_>þ Ãÿïíe»; \_ÿ¾NJ:\_ÿÛ )Ñ I ?ÿýÕéµt¿\_ÛïäÓðgü7Jì?ýëöïÕ!6õ·öô»B ´ºJ¡Uë¿Ý{^¿÷ï$ éõðo
ÿðWU|DDZM¤¯ëîöOZ Ò\_®·Þµm/׬~ÿÊsÒΧßá]ìHÐ3a/ÿ¿R [.XJA ·[®¯¤ÅG×þs{ÿú\_óDýoÿ³ ö\_MÉ0=6[ý×û è$ êu¥û]úÿ÷÷ÿ׫ëýÒô~ÁÈ0\_7ã±þ¾¿ÞTH ð Ò´fk³ ©ÿ§þt¶þ×Óõ«^¿ÒõaÃæ \i¦T´"úá?û$A|Rl4¢ ãýdKÚþ¶}]C#öü.ö]=×auÛ¯Oö èJÄG°þõùôûºK¤+ÓTt ý§ &-´£aÄã´Ócí%bq÷W¤¼òÞ !ÛºývÌëéôA^[¦aá
A
°tÚAÃ!Å4Ä& ÃaC8}ÅT(qAÅþÁÌi¢#ßëöXä+ØîÁU% /ËHGj4N¦qÊtÉaÓ0M4[´ûµµµÒ°¯mü0Á´$E¨PÿûWý.A¿Òb !1û¨a!Uÿÿãk«¥Je@/ý E6"ëë^¹{àiRÐAýõò¸@m´D@Z#z¿\_üa;ÝÈaJ wKJravj J×ëìÞ Þêµú[ûzÖHqÛ XKUúÿµ :sX# IͤAªJ@_½Òü!A°@ Kÿö-á t&ôª JëîBYsL2È#
ÿÿ_² µ é%¨A «êÚéráæ¨æ2( ¤¡ß÷Óu%ÿTË$A.vãm&×þ2$,-}üpH3À ÒªA% '6¶ød Åõ²î$GWúÿÓ Ö1 DxAõåTªAô!#v4U $yÜÑ°ÖAÿõÈ6x6é>ÝR¨A 1÷ + ÞBC¡8 Ö^¿¶$ \_Ì/ÒTA.;¬(ð)dcí;úªÔ.0N8ÚIRA$pÁöR© ×0îkÿ×\_¢°E8 §õKTAs"¯b duöx ù½uÿúg$ BÙBç×ÒI,ÈTgcFdcÁ°R?6°1߯ÿ׿päÿúI$ WPä2îCH"/AøA¸D0?ÿ¿êÿõõ×ÛT´A,;~33¦iñtGÒÕ÷÷öñÿKç|A~côÐIw Ú@´ ¾v;CKÒßzÿëuÿ÷a;ô÷Ò¤¡É'ßIÒH'Ð;¦¯þ¼±ÿ¿ëúÿPMßéV]|ìÒÊåI cP¬BZÔøá°ØD<ÿ¦húÿõÇþ9ÑWÒZI$}tÕ"%>«T{cu]{þ¿ÿÿúÿN£¶µªI%÷õÑ)Û«úcl(é¿\_ßÿ¥ÿêE×ÛªTIÿAül¡·LU~¿ÿÿÿwÿ~ê´$¿\_õzD-13ÖºthnQ÷ÿUÿÿÿé}ôÚêé$ì¥÷ýöøÒ=7}6!&ºÿÿ®ºÄGõ_çûÖ S±Q y²5 3QÄcÛ<¿}"®q^à÷ÿ~ÿë\õ{Î>«IPEÛ
¸A&áL ð È3"ô C!FldÀÈy±1 "L2Fmºî¡ðçÇá(¦ ¿½ÿÿý.ÿ|}URH$ Å A½B/EÑy FÆ 2ÖEÛ80AA.Ã83ÃíLpfÆu³\Xf®)ÓÖ{lU¹{õþ¿ý|Mþ?ÕYWta$ O¸2ÎK«ÅSq>8@è''¶a¬O $Ú ¶'£ó ß Õ+ÁR àböà ±æ Ùÿÿ»ÿ¾¿bÄÍVµ¤I/®ôé<Rá:Mü é82Ô'VÃûÜ?i88:O#¢lø÷Q9ì-&Ð"4Cþ]õ»úü°ûþ»nh$J/×ô+uïV·Óu¥ÕÕÓÕÿÉ_ÐÛ$+L¸5%øB""ÿçp«×ÓwêA$«ÛKúþǤº§_éÅý_Ôzètw"WP@䬨jÉB$3ú_¾zÿ·U}tA §bÁ¯¥"òú® ûÿÿéׯ+¨ÀðØ·êïÒoáëé(J>ñÒù@I ¯ÿñ¡éÖìÙ×îRb¼¸\C6ãWÿúI$IhC×È@òMÞÃÿý~ÁþCÍDEz¡È'd¸ïõ¦ÇÛKÕRAPXoùc2b}Þ¿þN¢c^%¿ÛÑ?ÊptòB)9Ç3 ä4éèýÿßúZI$ø?ûþÁ¿õþÐ/ÿÃì±Éò8/¦:#qè1ø"íÿôûô´¨ ©a¿ÿÿÿúïÒÿü6¾ê~P r$Íßã[ý-$¯Oýïþÿ÷ÿûÍFßøDL0L:A.gÖoÒÇÒÒÒH ¿}ÿwõÞû§×ÿß½t õC3@@Ãit=~½Èk¥¤ Úïû}ßû¯ÕÞúýúßûÕØ¡HÊÃ;@ß@ëõêÞþXA$Ø]+Ûomýï»I»Óîþ½ý¼éµÉÃá$eRh>Ð}ÿ}n½% XÝ¥"grØg¯°¥íÖaZVf3[Ø\_m+ u·§}mBÐlZÿþÿTÒH Ch4 1@±@ÄîáªvØJ.)طħ®ÚÄãõ,3¯wé8o_ôØÔú$ ' · $à 8l(hB6 MB BbÔ0Á± ¨4) ÅëO;Ý'ÐcõöîT©$ """"uB Æ" ) 0
8ú /Ã0Ða~Üz äÓ»ÿÛÖ½$ """""""""""?À0ÞvÞýúMðõI$¨ µ±©Ã¸~Ûî«éAøhí¿ÿßÕ$I à Ýÿ^fkÿT ¾Í ¶Ûmÿ]þwxIZ/Ò^ÿý»þ½RÐ_¹É {·þ¿Ø5ëK/òþn÷õßKIa/û¿~m~ßúëV^ËÍ»ooÿoþ¤³µ ¿v!¸ ·ÛþÿUÕ'}4Dpþßôþ«\íL¾[{o¯ó¥TYÙX¯½¶ßûýµ¥Ò³º·òg#®ÿÿ®°TJçv ûñïƾ¼ tÚoý»·J«öA\³¿öÛmG ÿ¢ Täc
¯Ûîß\
ª¶B}%L6ÿ² :¶ÈàxD®"$0i÷Ö¯UAPaßóµ@'vCeDxKdR²r 8û©6kM¿\ " ®ÉÈ"ò !¬güê¤6(Óô6ßó´è"9-ÄIk \f£ò8h Ê)¿T¿¾M µÞ¨Bv@jѸø«¥M·Ö 'zKÎÉ#¢èpQÆjR# Ré\{Õ%ë ÄDJ@Ç!ѵò06àKn©xÙ1b´h·©[wüîJüz|Y¦4)&Û¿"ÉmmªJÃXD0Ë uT6ÿÊP¨.ÒøÒ%Ãbè1I»ET%|-<:IªL6é)G + ]b$;$ ¿¡ÈÈ@®$Ù2xáJCjoäT¤XW×t]Ð Í-$-¿±&àªÒ %© l8 ÌAÁ ¹jZa a8$½|]HÊ;a<à5¥¦\
[)ÃA:IWT'Èq!?B®0$T$éA.GA@F3!0OD_#ºKôCCJZ¥ ʵÍ\Èe^z4¢ Ýi6äAÊ"B8
ÖqÊ_LÈ3.Þl"ÒI>$PpÁ4ª ®Ì¦¦Ã?aW½$( ÒÈfoÌgúÃb © Í$0 0©Òmº8TD\4ÃR\ "%[ÈH» VM·+¡gk!¦ §iHTÃÄA5a°uYØ.GÊXeEÝHASaÒ'tmçe! FàÓ¢@iãn2¤Û|ìL3êÈᣲ.@ jd¹áÐI$Ûd¨ìÀ0'»¦TgBHA©ÉRT°a¼ÊÌ\¶$<î¬Ø.v¤VY¨I x¢7 ºH a¹Ø²6íT ï 59¥º Ø4úoUI²¤~x"ö$Ó¼©aÙFJQH)'Toáô K·ðrc$Ò!G)êëd\ eÑÙ.'cÃ!¦äAÒ!²u" ?¢+½R 6<íZ²?»aåfò&0ÄT© v/4ËÐ;>úP Gí±«u
pW¾LDì" iá|w
¨I& ÃûµúOÖ²D (¹+}PK~Aµáh$oãý4¡¯jpaÁI()Ã8.5¨"IRmY&3Á®d»m%A¿Ö©/ÆÁIQ�Ôì!Á\+¸ÁµUé¦ïÿ¯õ; Á:Ú G?pJ=%MøÚØVÒI7¶þ¿¯übäèàÃ@ûj")$õJÕ¸´Õ$Ý·ýìä"tkÃnÚÒJ
jþãªM»Öaù×$ÎDÄ21ò gÖDHöZ¤1B~pMø0DB$%°MãÒA½¿D< ü è0P@òC <SiÎ ©¬Ø,µeL olH$ÈETI"Þ9¬FÒm»ýDc{ÑÔjgAÈw
.&>!ñP2:0 A Ù¶jÙøbHI ¡(vV0:Ç3$Û þ#Ç¢;haWU ¨T-B"»¼ì0v Èà\)Ð[ �@l R©ñfHM ÅCh ÛëÒ{Õ&ôµcv/×Aó¸ðÃl0Î ç>Ç°Äør 6,AÏja+m×Ä~G2&A#ôD¶ è8<ppz¦Á/pÜ0Ã%8ج1:ÐAQ|I ÖC²r² F6êþ¯ éöþá= mðd] îÃaÃd\a°td\ fêHã °dÑI°Û°µoÿú}ú}ÿ¿Ã 4 0Á 6áA2 ¤yà ¬Ó®Ýÿÿzÿ¿_úO^à aD'v!e@7
TØaAÐIÁG(qÞ½ÿ^îÿ¯{ûýä{°Ãh 8°àà¡kIVÕ94ÕÙ)þî×wúïëÿð¤+'0à ¸PȤE¼8LxbBïJK[ã ìôIK~ß×ÿÿ÷ÿýbÎÃÃlvW¿IC)ÀÁA° ﶿý]ÿù6æ8lCh2sØÌØèÁÙ°zJêk ÅrÇû¨¯ÿ¾¿ÿ×ÿÉ°Ð!À]à ÄKt Ä2´¿V)¤ Ô¶ûá¯ÿýÿÿ&Ê28|0Øa83±0¸Ú¯¤ÔBH0Íæ?Xn·p¿ú¿¯ý/ÿU&ä6 a¶Qì ( ]SÒLeAbA»½K/ÿW¿ÿ_þ÷tBlAGhÈñØ0GDv\ÅÐÔF]´\m Ü7÷¿þúëþ¿^&Ú DReÃ7 ÄBbÈGÝ$"&]°möÿþé·ÿýS_uÔ!0ØFÀ86 s ¨Ó$²X/Cm÷þh] ˧VÖÒûÖêÕ?¯Y ð¡ìXldpÐQ7L$<}´áü[tÁ¶ª¿aÆÈ\ßáTmÁ"v{S:baФÈÀ©' vwTØIm$ßN¤m¥Ù65e,6vVÍÛ¸
Ði ðhÖ§ á´ÓÐ@ÁÔ6(8eºÐ®ÞDÄ;Ç4)¾Ý¡H BIÍØAðÝa¦ ö¾aÓ[ ÃWê£2Pa4íH) áDáÄFY" Ë-$ 0B" 0Á£lAc³phìHb mIÏ$ A! """#ÄÑkwla=,cLì6ÌÛwqA 2ß,wòÍ DûHquúÝ3hï¹ß InÞè2Z èîvÅAWÆVt;hCAÜÔ$[oiY}4A!A©Ûé3¸²ñãÛà Á¾A+Øa#±D'Á¸BC¿bTI¸ÆI%»,Ð ¤û´"©).+pÈlîßIxdV¦q ÓwÅ$Lqci_$´ÌäW#´Ûu¤°Ol;A.[ p© ¸Úb!UfqÁRK[J°$RI%$Èq´JdS¥#²kÒ L¨FѤlÉH&e®d¡é$,GguʦGuH^°êd<§f²é$ gò÷Gºé£BtAÕi!§èXIÄ ¿õÔì¢ò§Ò´Á «ÿÒ¨i,ªH$Ip!ÿIqñ4ÿýP$ëõ^\B KækÌ3L;6yEëz»äW\ÕÐIO¥ôE¸eXÓ00 Ð)C!ÆùL ¡ÊHq¦PÌ3@mQI Á:ñ8ì zXEá Á Fæ L·^0TD;)á¢è» øÐh
xPÀÁS:(aWI°cI©.to tÄñIÐwá\ lK«MÄð&x6^I ¯Mü¾ ©Á°Ë0+Ycð°ppe£V/McÒlLm$<ÛKÓ«Ðêõé?^Ý;ðßW«xAééýÑá¼":é?ýß½Fü}ÿzº¿zì2p>úÿµÿ]¯¥\_ý]þ¿ö©´Giî½÷ÿõ÷¿ÿ\_ÿ£ áïõî¿ÿÿ¾ýÿÿþVöá\_ùeë×ÿýÿú÷ÿoa«ÿÿÛ úúþ¿ëÿÿÿuQzÿÿÿ÷ÿýWÿþÿþ#+²84ÿ /úûÿÿÿ¿ÿÿs"G¿÷ó¯ÿïÿ×ëÿýHA-×ý¾þ×¾¿ÿÿëëµý¢ Y#)/ÿ¯ývÚ¯þ¾Úÿ¿ûVb=ÿa´Ð2è½ÕM¥·Ó÷úV·õÿi~ÊÓD~ífZ6(;ÆÚJÁm¤"~á¢Pùiz{iw°Y¥ppTà ÁØ àÚbÅ! £aApA±@Á {xA³7
ɹVÄA@Á0¤Ðiùà ppéäÊÓýP!qöÛmúO¸r4ÿ3\õ½PI O×[}vª}®\_Ó{µé\«ÅWþ ÿ Óû\_¸Åv§ôzZo\_¶ôÓÓðàUô]¾\ð½¢Tþ&×ð\¤h&EACñ^µUá$=MÚ^¿Tà¬à.´JÉõ?ªZMxDnäÐ\_ÑS@·ÚS.§w{äïáu{ µ°ZM¥2qúú~.¹;IÕzÒïÃêq\.%ò4¾aúj½/ëÕ\®ëJöø}ÈŧérÞ\]gkV©7jת-ê¨ÎëO×½?õªÛªGeß\_õaXëZ×ëëWôÓJ©6ÖÓ¢nûÿE½~¿E´¢²]|}qè:,tÅuéüj¾ª°Òi ýtÚãv] k«ÿú\l/þ·«M0wb ©í>½4°ÐÔì\üôSÄúÒvFÄ/iµ\C l È®aÁBÁ2&Kçê! 2c8f²: ¬êÎ/W%k±
¯Å°¼qX"à4B]0vxjkL&|a0 a4 @΢¡IÅS Ê#: /$ d$N%6áÏ�& ¨LAé i,Z75LhØ' "îpÑqìØÎ iá &fA h. nÅ!èCBÓpCI¹(Ô0Èñ¤Ø1=ØM $è Ú:ÓF¹qFæFǦ7hZè¼êÑvÐOB¸44ë²wiÕ¶aI´ºu´ºppbWBMtH$àÄùGZvÁéÅ " µ}]==:OONÛúO¿ôû«gênÓl©Vý7 Pu¼¿¦÷Ç¿~ºëIõêÇßúÿaWþ×ÿóèý.ØÿQíö¿±ûÿñ\_ïá /ò$Ç!(DGïÿýÉÿÈùÿò0ç\_ÿ\_ì¬ý{èz¯ìGd;ñB7ÿYÑÛÃ~ö¯!CáÏû~ßÿÿÿ¯(Wý[ Éè%õ oÁ»ÿÝ}÷ëààùHd3úb¾þýåçRõÿ¥ý>m9´Þïö½/ÿÃø}ªûýzßzµf¯6ôa½ýúZþiþ{ý7ÿí×õ¥ÿúÿKÿþõý´¤\ì3ÿú¯i7]{}?þíÖßë÷é/Õ´»M \Ûu|AiZÚPÏz¦ÝÒGìÿöäKßÿbpØ @ÁÅ)4¡¤Ç +U]r åãJ6ØvnöîPaL= ßa\_LTTSQQA6Øئ pê¨qQ(e¹Na@ &»aÓA
»Mm0
DDDDDQ)¯hG¤7ô·K®ßÿKôÚ]ÒnY Ú¶öÒÄ!áÿÿäÔeÙ"ÿÿÿ Cy6êDHdÊBÈ Ç)Ì9C9S\PP±^PeY\SãrìÃ9<˯8ç²)Ì9Ny°¯\e0jd[ÈæGËÌàÔGà
Å#äs#B9ÑÌf0£hµ äsB#æÆG¨+Á G28B;1Å#r9Ñ D|dpÈBTNÑD|FòèÃAp\eFX#Db'Dtb7ÙtGÈÑtG#áHè¡°d|¹Ì#¢vLtmDtGT""$qÈ<e�,DHãÖ!ÈQÊràÃã9C¡ÊÃrCÈAÈ+A#0åTÏÅS
(§(!RX)³Ò8"""""""""""""""""C$r$9+(rå9C2aü¬\Ê¡^RÉEÊЦp²¼¨!§±x4 qËG;(; m $ÁH|È{3D|°8#QDDL¹DtGD|Èù @´GP4DDDD®´ëÞ]eÑ#²98mÀðk#AàY-ÈèÑF.xùñ|£#²8d9àAtG2>G2>GDt]ù."ê""""""""""""""""""""?;[&?ÿÿÿÿÿÿÿÿ,ÅNñÿÿÿÿÿÿÿÿÿÿÿÿÿÿòÓQÿÿü�@� endstream endobj 17 0 obj <>/ProcSet 32 0 R/XObject 18 0 R>>/Rotate 0/Type/Page>> endobj 18 0 obj <> endobj 19 0 obj <>stream x]Ûrä6}wÿ¡c^v&¢!A}Û²w÷¬Z3Þð UEIôÔmXU{¿~�<(f]zRÔ!K"ód"øþþÝýðî¯?wù»ûÇwß~Ó\Õï2ûgïÒï\Ó\éw÷ËwþñóÝ?>þåþ÷o¿¹¹ÿæûéyvÕ'Uù®Êû?=¸iûaûÎ=çW9¿0»çn¹í/]�fW&�óÌ¿vÈv軡ZHå?8¬7Ý ¼ÈÔZ?üÞÍvÛÓ=Ê¡GeSÛ/ºGcïE¿ÏÖ«m?ïvׯWR ×cS¿\)P¾ËE|IlP^T�-IÈM;RFHYä÷u¿ÚãQWö÷õ×U7l¯þä?0>V±DÅ& <õa½xØ«ÐyÍãÊÝËÕttrñ5ï8ó=æ]&ºÊ�òq~uz4sÊÎEUéÚ
WQ:jø]WÕûI§ËÜ!r߶&mï¨3@\\_©ùB+
1Üî\_îÛQ�à
T�à·?çM¥~ûË4PÔ^3®ÞûçîRŪ4£�ÌÖûaßÛÃøÙíW²Pá4iñDª&Mágµ{ÊOõs· j¥ÐÙ=·;a ß%?·\_D¥±ZYb5Ñãdö+«'ýêiÚÊÊ�®µïÊÔmkÍÖ»nÕ®vÎçC÷ÔóïN7¸ë"NÜÓT%iãÅD#±(�aµ8õº]Í:AZOÐ °(Á!0þùØ[gªÞ¿A u5éK°^uÓ±¦ÎÀdFT¢|\kón3ô/Ý\9];}|ts«ô$åTgáåÔYM
9!s¢²±¥ÒT �pF¢fiJ|Åf³vûU¿û:ÕN«,ôw5wW;7øËöߧÔTÎlä ít¤ ã~/Ç Ùu´2ÑØ£QkFfÒ¯¡'ªô'Ng¡OÌ©²´÷õRES¾3ü¶)ÖNú¦ü9]íùC^ÁÏÛ~·OÝfu\sÀÑýé·à$ñbÐvM1êÍ^z¥x}îgÏ':Ú$<ÄDÙGRY@ðïÛ©äPÓ"Õͺí¶
ÞøWç$Ê@Ú|Êo úû8r7Jñwoï^a Ug4Éò%Zc³i}µSz'½p¨¤0ÖµæÛÙîðé<³MT�Æp®�±ºm7¼tX¶JÖ[cÈYçô
ÊsÞøîëûé$§òS¦ ÑP¤Ä¨|û5Z3v93öÿ£,Qü;Ê :DmAf¬Gµ×(�ôK;/¨¹îßåÇ}û¼Þ/æ» ºiÆáò.À:3:TõÐVÛF¤þK¢Lßþ¼ýí/SNG>ÌmE:.Ìæ/OÆÜòøõøn¶Ó¤é§ò\t-|ê^¬èZçЪ¬n µÈZ·2îÐM{k½L¤u¯1³mh{[ú÷k»#Oéq@ lQ\dÖº¶f4èVp¬_B2®~8&úYé %¾|°ZQ[W8¾×ü¥7xu:õí~ö<õrL6.¾Ò¿Ä/Û×á©[ÍúN°¹%ß:w7îBoÃw¹,Fÿ·_.»yÏüfß? ´5;vi¼v;Ï KUÚêK4àý²ÛJªÀX,¹(_ örb¨£õ=j(hYE»]Ö;Aº2à ÉÑRôb9Ú!ÎPB+È3Q²ÍnhHÇgé¥.ù°\_zÜÚxmÃl cÛ¾Fo,õzOàkeXdÚJÛ8JAOý˾ûéh\Ú=£Ê¼FÈÄÐE×)bÁ±pWVcs¢Ã±èÎcÔ$
5iÕ¡×V�@d:פ ëUÞã°^ÆÉdX8¸+ÿ&o ÏÑ[¤
µÕϤv¸à^ã殡¬v03rùy æ¢g^ûÝóaJ"{f7I$BÚíêãa ¦\_9ÛhÍ BÔÁ0¶�"ýµª^Aú^PX¹oJ(fa$Y¨ \ïwG?d5N¹ ÙÆÓC¥p¬"Gh·[´ý¼î�½AÚCªA°¬È¡ø±(éÁÊ ± NàÏHS";L&D8p°§<ÔZ@Ñàp ¼atóÒø\;l?"("Ú¡ÄîÜnÚ©Ú{?ØÛÝ°íöD@©7mºtr2³YT�YX/"ì7&þá@^EX}1 m�ñ¬6DK\o6gvqÿ£(Fú#l¾YÖVy÷�ã ÄøØ
6µFÖ 3ØnDð»ZOZó¥°çv{:Äö J<·7cSËÁü:ùyýJΤ߲ÆáNÈ%&y ä8¶§Øà꧶ææˬ~"ó^ØUÔ1ÃÖÃT"bÞowýêißo§HJ èHVw¯tgQ±ÄûFÀYﺪÒHïØ@«èS§®?4®J�ÎÏà÷¡{ÜoÛÅE¾ Áªq fí~+±B£lªg¬%ßg"ñ¬õK«øÃö 0vÜl¤8¤ìH 0gõÔÔb<íf³ègíÃBàÿÁI<Ió#öªª¯.àõQj~bìng¨ª0J-û ÎÈËz7DQì«à³È¤Ç2õ~»ë3;ÈsThÕ-Çø¶S¥Ö4ZÏPðíÒ&HØ9춸ÏÊüÜ}´lãÒÕkû©BÖÄ7¥)øk;kqÊ@U6Ú¡TêBuñÀj"ÙªÏÓý±Y´ýêJ¸qZ\qÏOåzíúûõriÍÙDbêÜÁü¿i/M ø£©G$R¸ý«\y ¥(8T¤f#Ç æ°Á9Í;P:,ÓIQhÎDæ¢É�Á,·Prí õø<|
ç[À%ä!&Êéز1;®M>Î9jÒøn«±ZOõu¯°¶Ç~SQ¸ÖúÐ$ô0ÌÇÙðhÑã×fëa.¨t»�µ[_ÎJìãu=P4jI-1²ü}Ü8ØZ 0&ñ3B-5é¼\=J{\d½ÍùD£ld>õ\X% ÖÖËæªqÀ#d¶áe-FËÛÙ ½Ä4bãÎv/=OC»ÔMå�s4¨¿^cðêb'¡¤ð[È\X )kÖ !ÒõîyÚR˺ûIr°¦<ºìÎé ¦ÐäEÈ{�÷E@Ì+ñ@ÞÖ4Ve~:òÜϧÝ׫)ÈzÀyR7?3w4(¤ü¹íi4Áo7Áº�H»XØå¸u3"1OÖçíLWX;Ãh²¿ÛîÄSöhö_ûÕ|ýíúäÖÎåÑhÏìÑ �^ªf%örw+¶Ý.¸¹£cÖ3|?ç&'àu9H°ä3öyÕ½J1v÷<¬×ËK×Kõ®QÁXý;Ô(C
íÃغ[a£Ð.øûk²KÈ~RÅçöEÜÈQ;ÐIGÊ©V®¥Õh~ߦ.T´È»~&¸o0BÈõ°Þ]üÝ-"Á{vFA5F¬Ö«ï¢krÈãÆåä£ÊFq·+ô±6é cöÉ3¬©^DH=R9Ã�SÝW¦)MÍr[N/ä¾æeTwOÜH+ýULº~ù°¶GòÜ 7ºBîh\D¿:
å ¥Gk}q ;Ç2¤6ôQÒα±TåYJ"SU í3í917Ógb¼$ ¦ ªÌë!ÃJ/n&墨°f³éÚÅI²w0Ç¥ xI³,ÑE52eÜÃGT£µ#ñEïxlûUpµ/ ¬Û�h7tPOM1Z"k(} v³±+sµ[[ÃÊò(FJvEòA4GçCÓ8TZYEÆr³VýÿI^ª©|$p$T¬Z3ñí}iý–42@4ÇÍg”Y|ZñÖæéîꤻEà)Ì2á¿p¢ê»ö,ñDͬ}BK¦µ{î@RQbÛÎf¤S¨2!HÏU$ñ>¯Øìíðº´ìööðv>!sAñÞTã¾æPeãjl X±EǸÚ[ xaçQQ©TØï\×h²i'²ÚÖ2þéY )Êúé ²ñUåËBôdFCoØå!ÅII;
Fõ¥Ïåì$²g¢Ùh(#D"ª(a8B9©SY&¯i'¹ö¸Ë¦l;?ÓáÄzXÈ))WÓaN©A¥3X5¸¸¢rÊÁDáfÝåªP$Ðï«é» ]�ð³0¤r¯]nýM³)äh ¼²Zy.Tå(J v¹ÞãÌÁÏö(4R& un½%$gkô'檮ȶìPÃ5(C ½Æ U/i! F|P0êÇQI%Ëé7nYD¿M´µ"E üe¡¡©®<d(hùDLef|ÅWóÞè3|'¾hd] mòS¤#è£URU['TÄ]·¶ Éöh5[ìçoqLHJ]´¯«åB0MB©©óv¹±°~D.R¥´Ã¹cíbÁ³
ÖÕÄZP©!'eb0Æ@Mgåa! êZ¯YðO$u=±m³û^p¤ËDæÿhÔÍ8lÑÄíAÑ' 'ÌèOK%F�)ÙE A^ðD÷=kçT6GmÔÏÚ·¬ï¢'¾åbyiþÌ"KµJÒ©)^d¦²òêx&m¼3b¾^vÛ]?IS;ظVC2».y¶¬þ©&¤,�Ò¯û$=5«pZ7ÞCBÖûÎRék°#Kæ%¤R êàÁ[ýê4%æ£0f¢ôa]s:{áyØ7ËK7l¿J;×Q?®§GaPTå8þ./EÀIÀìeHzÊß�-iËårj©ÊXÁ?ÿ¢£9ùB+KµÕä1drh
Òá&´Aw3@lZÉÛs ¼IöJî·Äõ,ïq½n
Q>Ö)[pÌ1Ù Ýv¿Øa´eVûï´ýcj©Ê þ={L¿è g9PD[
( tHm'&OLãɸ^Hïôa=Cªª|R §u [P¨ða½; 2©±\a»ïÁy¨D TýÖùK§¥1Qø\lð0¬÷¸ .- ½Î�à%<«±ày/>u01ú 9Es¢ cÇE÷Gï¤U u,(ÍöXphß0Òe°ä ªñübYÓÚ¸Ðiu^\up o¸@:62ÉÅzAîPÿ U³|±Ã[|.)ð%#JÀ±\àG< æ¨ÌfÑ~ãIÀÅØ$fÈäs�¯ÅÒ7h¸Ó0I®(ØSB¯¡&t«Á;ÚmÎ8£|´XY Ûü¼¦x ×òt©ÓYõ$ÿGµFÌôÐ6HÊ0J
Ì)· s5Íà(JöeÝÏ5½¡ÇÊ\vr úÓ½t¶<9\¹Ò!zbBÄB:9 98}|ÿ¢:PÕmn!&TJÆáÀß0¦¯T8ÿJ8åQÕI9hb 1§Cëf8"º?ºaÖoÏÊÚÃhûLõ!BN0EÎàT×çÿûiû¡£æá¨&)±�5MqÆyÿøØÅCýaÈj(r�Ä!bþq~[àÀøÔaÙzYC^[)|j#\eô\ئ¡¨óS
XÄßkñ$kÚ- "®
êõ¹½|
Ê}×ý«Ô_ö!C¾~æäoß·{wÍÐ~w
öK5ü»p.Kç"p°ð %¬£.¢©ðòèO
Ö ÞËñÚRa¤®5 î:IôLVfÙ~Ê)r¸»C<¸Íh¼Þãá|.ê.eûsY+ÅB§ªU7|TjC¡MV±Èa<>i½è¾[ÏfûM² ]q}m
yûZw]ªy+s'dmk!fA¥(OÏS1Õ¥¿èct'«èªÆß
Õ )ÝÏɶòÔU¢�Xï¤+¸÷ãìÙU2ñÜ\Ò9S£Ç+<ü×c:3Ê6@:«ãg»þ¥1.&éÜÛñ&ñ6Hìa´¯ññ m§ëp<ÿ2)
÷¼t&¨. }·d^ÃÕRA Üé«xòjÆãWmÜ Q/£ \_' Às .ÃHÚÖ4\_fÊreIXG¾=ûA:C7uZÍÅ+ðN¬%c¥éCNæËGÒ5Ïe.ãÁïÒ±NÃÀÝ4LS3SAE°ÄKJåPe!î&ë5¶ësY9M2E8ÒÊyÑx ;y+¸óCàxn·Þ¹\àºÍ$´D ñÜp¾¤ \~@ë.ßdº¼C::z àÊ ¶hLv@°·I&9"fíVØV4Z¦_]\¨êºyó|®$Á®ázëÈiáY£R.´®".-ñ áf¾.,Õ{pú¤#
z¯êÔç~óë-¼HÎa�nì-xàÁN°CVA˽tÀI;.Ý£¥¹
ë:(©\ØGÖ
¿©ãx^¥Ö5 dC0ÖvÑæ aMMGÛÇv±èäWÆÅ=éå¬ë¤ÇbÈ-¥ñÉbWú;;¦¶x~r o!9n %ÃóU$'×nJãùfÁLp\_¨!^ü!&=ê!PkÔ§ñ£)
#©}êýÒÍ êo\Rn»Ã³ÁÁe5nµà~¿TIDõ«í~Hê XÑÃÓº ϬáæÃó¬+ëq Hz$ hòJ¸Ä£2\�rY!Jñåi% �7lHÕfjUiL,±äÄØNù2 o~}îä¬V=VOn]#MÓªD¹å\_5+)Äh~Ï[<=¾:cÙ :'£Ñaè1Å8Üx¼õZ´¼d:o @D:¨Ê-õد.>b:¤cPaÑb¡9÷Çjdáçö¡ßÑØ ;,tç£æBÌç®Zç®°´ õ
¬ÃdP""ÅwÔ,iöó¥´Î5@\:¿¥-¹å«:àGC:$B°ó×SxL, JIîÕ abCyÕÑ¢oûå¥enDqÜý¢ðÈbеxÄTWX±éç{è1#ébÉ8vͱ<QR¡ÑxI{ñ¾bÑøk>Æ^ûp#±Î¼t~:Sm>ZrÙXÃÁó9Ìè¨R:PÃdr:° SË1ßÓÓéEø|\_¿tÉB>ãU BhU]m;6}-wÓIxÆtF+©dOäf%Ó.ò¶°Éºi¸K\¥¿0DÞDßÞ]O¢¢¤ãÛÓù¸¡E]ø4¨ä´OÑØ.ÛË×VuZ6ØìÙ3º²¼HÄ^A'½ÁÝÑÑoÒÛ´ÎÎ+;;tÜôW¦ !äëhºC!'ÄPíS4ßý¥÷ݺFRÜpJ0BÄ]�ÓëXaù Í!]¿{ºÓCpg| §Áá5@o>op Ä=§4àè¥ó>ßäÑÊ 3÷ø©xn³¥]-I,Ö[!Aô1'öLæï4 Þªt{QP0c¼yêº_z¦¤¾ËcþÒϤӴ+( ¶ü^ZÔáhËýMj\d2ÅÈX@ßus>,¥oX ´§Wqô\øV(Ö"¾�#ügɱo# òxXèw$Äè¨]Ñ_ ÉF(#´ÄìÌÌ¡g/íWåÉK/'ß�B<ÈC×þ%lÜ5Mg&ýRSþ8)ßßÜþôóýøþÜ%t}}P!M»ÞÞüzª ðrݸ|÷¾|øùóç_¦Oæ]ý6®ì")ôËõ/7_|ÓG%ôx,ýxöù§ø¥;, /·¿üóæ.ùXé}~èïw·~J«ÄO5Ê=5p}÷¯_®?ýp¢Õ¸Äjç(ÊKïn~ºýrw}ûùÓ8,Õ8xµ3õm»¾» ��.K8�>}þUînu5Ê�oéÓO'æ¿GHCEçÐø
rý·OòAÓË2ùF2öãç»Iß(K6î¾9Öô<´=+ÜãM#M]Ýx5]kFgÊäÏ>¬øa+SY^U¼ßz
ñÆýæ¯?~¾¼å4Õ¡öÿGÅ3NafÄ~àx 0ËnÙ¼l«üR?C5íÝÖ ãB [Ñ(wq̽ÿùóÇë/QïTTÑÁÊá>Î_®"¢ÊÕÔ!ok¸W�e©àìFOäÌR"|ú|wÿóRzä»nîn?üíE9WÁùþçͧÜl"/ø üýéúÖ ûÝïÿqô @«2ó¨¢hïSêTVÙ©ÙVJ¹ÙjOÿ÷?º9dÅtÎ@Eñt¯,ÿ®Ñõw¥±®ïYPn7¡ \A¢!jJC×ã%½_Ùwuabþó©«Ò[¿U^ÒgËoÿz»TÅ»ÖßþÏ·ÿÙ¢ endstream endobj 20 0 obj <>/Filter/CCITTFaxDecode/Height 3328/Length 103387/Name/Im23/Subtype/Image/Type/XObject/Width 2432>>stream ÿÿ $ Çÿÿÿÿÿÿÿÿÿÿÿÿú¨ÿÿÈ ³½TA 3
4 丼+aG 0ºNd36Aº´á ÞßZÎÊCn{û| m-7_ªÓAدßߢoI&úöqó£~ÿÿ´½7O«ÿ¤ÿ§û4Ué}Ú_Õ²nB¯½/ÿ4:þ÷m/ñþ\_ÿÿü/ßõÿ¿öÒí¿V×¥×Õ_ï_K{öÿ×¥ßûÿuÒJ
ý¿ÝzIvßñ]$¯®õßÒTÚ[ÝuÒKµÛmwé%mÖÝ¥ý$¶]°Áv®+$lP x ê@à ¶WD10f§È0AiÐüÿÿÿÿÿÿÿþ@Vhº6/EÑ.Ds#Ät\òÁ%òGftGË£è8 $p%$µMRÌ#ì#ÆÙ#Àæ Ìò:3EÑEÔ³ìáѹÑ#åÄ>)x²:#äx¼kDt""""2ÝB#yä|ÈR8!.Ðܺ. #£<ù¼""""""#;h2Å.ædx1AË;r0 ÿÿÿÿÿÿÿÿÿÿäRÓUÊãå¹J&Å ÈW2r¹dv³aÙn8ÌÆ aáÌó¼Ú¨ÁH®12Öî °zn÷°pÿßz\_;¢÷atåf2(ïµöÑSÜÒùVDò³(þ¡}üBã¨U~ïøÿëÿë×ÿÿzÿ¯ëÿþ¿þåry±!P) d1 2 8D[ÿëÿÿÿË0vna0A0@Í &ñçÅC0ÁH!FlyQ#ät"Go5ío¾Õuõëµ µí=7²wö¡:]5~÷^¾Ü'[¦è>ÛNðIÒýõ§Z÷þ
}ééõ¡§ªm+§ß¿uzÕÿ»¿nõÿ¿éU¯ßÒ«JÿWö?Cï½SV?úøCûЯMþ)7ÿ÷ïºÿ×ëßÓÐØ÷ÿפ¯µ~Mï´È"Q¬r)òz;´Ûßÿ~ÿߧ¯Oý·ëýë{ioë¯ï×ÿoõô¿öÿÿÿo^Ö«ÿÿÿ\_ÿá°qÁ²ð\_ò@\_öãý|&¿ÿÚ^ Òý~¤SÒÿV µþìý5®BlµÕÿÿÿ\ZÿÿdZ29-õdtG ý×üÿÌÿ 6Ñ!ÏSoúÁ»¤øD£þßûø"On¿ÿù Od"ÇûéG!Uëþ#ÿø¯aºõÿþ¼oÿWðqvÿ¯ÿöhCa¿ï@»í$ úöåÿX@»÷ýÞCìÁ¿°ÿä}$9Ç:7ɧa·É9@¿åÿòuvý×Kÿýÿú|0åüÏûïÉÿþC<"A½z^ßÿþ´ú\_ú¬½ú^þXp}\ ¨C××\_¿]^ ÚW]ÿKýý®ÿ^½ÿÒ~:^áðàýú®ûu\_ûþéoÿÿúJüÿð\ÇUmýÿþ¿ßçî¿\_¿íÿÿàßõÿ¯¿ý©½Í¯)ÓW½}qQëß·÷×¥÷ý]u¯ÿü?ÝëuýkþßÿîõWÿÿýuø÷þµß¾ý'¯ö×ë¿õÝ}×oÿ¥Uoþ¿úUûÿþ}=î¹?_¥þZ[Þh8®þý~ÿ½'NßßýõûÛÕ¥Wßÿ¯Ú½Z÷Ý·÷îëïݤþ¿R
ÿÿÅ__wÕÝuÝývÿÓÿº_ïÿÖ½×ýÿ·ÿ}ð ´®gý+ëì-na
]Û¿[í_í/JúÚÿß·ÿ·íÚávíwí&÷JÒÞïýº°¿úZúßíÿéZí¶éE64¡¤Ç (j¬4T=¨QÓn¶¼§íÕV ÿ¯u~Ù_ºlÍm6fº°½áÚPPÏýÕ¸göì¿i;\µNäDï_aaÿL3ÿö¾1ò/vÏì3ÊêQ lB6¡ÖÕ@ðà PAÁ¸ "@t¶QÄ&%NîídK« ÷i0ds^ v»¶8«a ¢¢¶)þ\Ò¬0Ð h Ò 4T4I0±A;cjAa)°²L ! &8h4îö°á»]á
nÄ&§1XL(aBh plPÃbáQL( (DáXª @Ã\ÃMTW»[) L'S¦ A40AÃa4¦Ã Åb""""""""""""""4" hA ,± ÁÔã8":3Ã0¥zehDC!h a0SÁJG(@Ìé×DFb"PDAieVmq""""""""""##õÿ×ÿ×ÿ\_ÿÿ/ëWëèhVÕWª¦a,(@Ðwb"9åV¬"Íá
;9£!fv3'ÊÃ"¸LGb#µ¶FònêÉÉNõ±©"e©$ÈÔ,áêv[)Ým|-©j7\¬^£ì.£ì"§z9C¢
iÿÒý/Kó$ùÙRKõèuËrϻ篩1mdSPï¤T=ÿZ¨ÿ®"¿ãõëÿô¿ëõÿúÿÿ\_¯ÿ¯ÿþÿþ¿É´fÙ§QF4!+"yO EÅA
F.²;DLFÍdxiA\¨E¸0cI´M ¤êÍÞm¾nþ5¸°¸aEÜ @Ã
ãc o !h×AT7T"\0
¨@ BaP@ØU×ôa¦ h"> Þ ê ªM'¶1;°¡B7Õ§ é£cU Ú0OPÅôõ@aA¹(«¤×òQ§ é6a-ú¾i:AÁÁùÐpd¹ » Têl1=ºOAH6 JÊNI°Ç:Ä®r:&Á ߤòJÎ%pµpt JìeeÑ; ºÁ °
Ñ6 q+zuxÎ1¥¤òBè%sI°pdøOìÕ¾zÒëéÖéÖú¦õoº×öÓté>N.¶öÃ,t¶þ®®ÉGl/zpi-½?«a jÕé7´àÓÓÕÓ¿Üá;¯ð¶Nûi6©ÿ¦y;¥¶W_ÿÐúôøø¯Ö×õcÖ»uëzôô¯ýºo}.®wÒºuVèGúÿ×ÿ§ÿñÿÕÚûññn
~þo¯ïo×ýÿ¯þïÓÿ¨ï_ýô
/ìúúßúêýoÿõÿúëëúëüo¯úÿzzöÒþ?ÿ×Á÷ÓÍ]ð¿º]wÿ\_ÿï][\_ÿ×÷ÿÿoÿ¿ÛÿÿÁÃõï\_û¿ÿÿÿ<´«ÿ\_$°Æ¿ÿæ>¯¥ô?<ÿ¿ÈCþ<Ö+ÿúù ÿôñþÿè~P\_\_ý
ÈZrÿÇÿý>]åÖÿù:ú¾ØÈzÿÁ¿×KÿÿÿþÃë¬Eßÿûàßÿö¨¿_ù1<ÿ»ÿ°Ø~hF-H_þ OÿÞß_þ׿ Xoùc_ÿÑ}ûo¿äÆt¿ëøoùb!×ÿÿEòÿÿûþµè%ÿ½ í¾Xÿ7ÿÿêXoÿ÷}¯Y0_÷ßý%ÿýÿXÿßðo¿Ð]ýêÿÿßýuïÒ÷_ßó Êôöÿ_ÿúýµÄ~ÿÿÿ\?]ûÿõûÚ_ý÷û¿Áÿúï¿÷ßÿ¯ÿëÿ¯ÿ×ðàþ¿ÿ¿ÿÿþýÿÿþD'µÝþÿ×ýýýÿÿþûÿúÿ]¿ÿïÿÿ_ß×ýßÿî½wÿëÕ´½»]}»[¯_zï«ÒúîúÿÝw¥çÞúþÿýïÿÞÿïßþ«ÝÜÖ¿êíÿûÿ׺ûîÖ×íë×ïûJï[K[Ûvïïÿíÿûk~¿kï¯éwïÚ_®ßºûïÿõÿ§½¥ÿÿÚ[ñMì3Jì3Ûµì-¤él3û{mØ[JýÕº[uÛKo¯þÕ¿_zõt ®Ý_÷½/Ýìÿöìí/×]/°µfý¿lÑÚ]¤ù¢ dHµaÆÚI±;í+I6 °GlÇq Ç°Ì=na¥"g»÷[gÓi/ilJ}ö #2Ú\_·\J}ö¼7vܲ°ÁSSûm+û´¯î%>Øam.EÎà i÷-´© kai ØOaAaAÅ( ØÄ&\\¡Ô(aÅ4ÅZM1áÛ &ÆØI;m&Å ¡´4 Øa µ Áà @Ú°A6¨1 pmC@±0@Ð0 h6¡ÅA!AÅ ÁAÅÃIØ¡ÄDA°BB0@ &qô×Xaa Ä&ApÅLPAÜPAµ 0Ú0Ä$ô ì1[ Ü0°Åía3¸":gtvqÎ<0D}¬5î¬BÝ
¨aIá
á
ûË,1 ,DDDDDDFm@ʬ4$ DD!}´´""""""""""""""""""""""4""?DDDGÿÿ×ýõÿ]~tó±Ö4êM9Cl©ÄyM.ÎÆ¢º¦B)l"½B¤Êæ£5[RÆd¥!#"¢-Ís#$v\2T0Ù:dµ{ùnìû $ ÔìLB0Ï¡Kªe]~ªÓ¿àá¯H0ph"WaÈÃDˤlbs$®Í/ÉÿßJ^F´Ð]ù\47é8Z\_i\_òIzIë'ùºp´íÖ§a/ÿõ¥¤ë×ðJÿQ»Õ¯ÿé%ôÒº^¾×ÿÿÿ/U\_®qó׿¯ÿÿëJ>ôé{º\_þe¡óe"ï8Êr:":rÈ^GDµ¤«ÿ/Òó#\2 f¯)ÓwÝ>H4B§ fdfDFØ!F DPÈD"Pß0<<)G"rª
#j"B9\ÉÖpÏ38Ï9Sóz4Ç3ÿoÈd§k g3H5( $F f$fÆt"OgÉ<ÀÏÐf°äæÆCtÑfódj.XT[¼"\ÍB] ¢Eå ÐhÖÑqE»6< ðpÑe&8E¸g?A\£Ã\XMTguAB¤T®\~¦FÆ D&)l ðÔ§[xa|£CÈ2M hxL"Ý¢'ahhÊ "Ü8AÁÌ$~¢U´ÄÐÜ+¤Õ Ú:? a0apÇAaÞÁBÏÚI¾IZOAõÑ ª
TqÄ>ý,J<4 L; éõÕÓØõTÜZ£[ÜA½0h&¨A6Ba ×M0\oc ÚM¿L'ʱºXbW4Ý'îÒM±H&é6Òt O+4é? Gz£½QÞuI×
ÊõG|ïðOt°va6°@àË¢}+0cI°Ä®tàæR|2\%cH4GI: W5y\&Á&ägåqn¯éûߨOû´JÒ¿é½l©°ÉEzO ÷d£ú^ÇÇ.õøxô:=!I!rñ÷é=S÷Mºú¼d®z~Nû»L'¨Ò}ºX}äï¸5 麮Åý^µë±§Wºß§ëééë¥z¥ZWWU´«ýIcý}i-V²Xá÷KÝß^¾ÅjèZ§èÅÕ¥ë[ÒèVýµ×éëµý½øuýú¾··Ýà÷¿ö=~%¥IU/Uú®©Åûô¿ÿÿߥ}¿ìáÕ\_íÿ·××çeB~Uÿÿþµïýÿ¯ÿÿ®¿øzý:õ«Ýj¿ÕÒTߺ\_û¯®¿õýªÿõûÿðD¡þ®½z]þÈQ¿·!|
§~?zÀZö¶¾¦ëUú®º\_¯KIRßïýþ?Y£ÿì§ò«Çñÿ¡ëÆÿE÷ßÿo뽿é¶ýÒ\_zÿÙ ©ô«Õ}{ÿªëÒ÷Ý-ÑGØ×ø"qøæbúúþ«õAWè%ÿßì%ýa¿ì6ÿW÷×ü^¾M=½ªªªÒúõUëJ¿¿õùb>ÿåáønÿ@¡×¢b}þLSò{RWúÿö·± ¾ðèõÿÖýV¿X;ÿ®½-~©WªúWû¥ûûÿÒ ðoýítõýoë\_×þýaëðo¯:?ÿÿï«ÿäAß]R¢ëhãªI\ÿÿÿÁÿÿëò]×ëýÿ·û÷ÿÿÿ \_ùäÿîß×}ýïzºûß×ý-/Õ¾½zÿK¤·Û\_ÿßn¿õýÿúm~¾ëõý~ÉÿÝ\_\_ý¾©½Íªÿ}ëúÆë狀ÒUÖ¾´½~©%ý//ý~¿îÿ}é/þ¿õº~ÿý[^ýýWýzºzÿô¿]½×ÝÞ÷\_Z\_\_Mõé\õY#Wúþòuûÿý¯¾¿ëý êí¾µï~áaÿ » ¶]}¥ \¾ýl%k¥i4ÝÝozj½þÍ},ÒÒRþªaZ;:z©uøAeõÔ¾bÚM®ÖØTofïKzû´µÿ×õÛ ]¥võ·k¶Ä+ Üu¡Üií¤ÄîäÿÈ60p¯Òárq¥Ã?Û]þÿÕacÒB¢«ÒØa.£ÒÖ\Ò´üJv¾¬0·ÈêÛKØg´JFÚU÷
±)Ûgøa&ï´¶îEÎÙý«6¨aÐ plSH6ÁAÐ@¢ @A±[ (lVÒªíØilÚ 0Rð¢¶) B\¯
«i0 t6¯ plP áµ %ØH#¤¨jh;b¨lTS6@b°Å M0Ó3 ¡ÐLãÞH{ímSÅD.S¬0 P Ú!x pÃ
BÅSUÕ0¨0
LUBª¤¤±Q <&׸avÓáQaS 0N!iáµm[R Ò![M'm 1""#2#V""HB! &IÌA 0B""B Öb" æÚDDê{¡êB"""PDDDDJ5 7DDDDDh[ª®ªI~é«éñ¿m.Ú¥¥ßI´»}0qÄ1 áʨÄG)µíaGj¬#º)iga]^YFÆB% vEHa3©97ɵÆAC+1Ñà@V!ÙY'M5+MUMÀ¦zÌ36kePl§d£"¤Aìtd4Jò59Ë£äO .\|Eà1'Ý ¤VH:hÈ18:Ø4Ö@!T ° 4AÀ( Ð ïpyxNàá£G¸EþÁðÛôîKèò½Ñ¹£[Ó|#cÒ Ï ·7ðü'3ðHPA÷V¤¨ª\_¥úUôÒmÚ¦ªoPT©7o[ÿÖGt½þ÷U tm,Ë¥
1¾«w?Ü
K\å ¹Ávÿÿéj©l/×è{ÇõUTÞõ§M¯«é·UNßÿ\_×\_ÿþ½ëNº¾®ú8þét½%ñÿt½qçÿõÿ¯K¯«íþ½j\_þÛxÿJ?é7C4vK'ÿ¿ú¿éÿtþ¿ë÷¥UÿõÿùÚÑA|!Á M¢'Ì ñaÞtÐLÈæ|¢:=fÌê~@¢1æ¿ÒCÿî9©z]ÿøã·ç²]
N< ~pÍCÈØÎ~ÈÐû!3j#PÌÑ õÍyã!£ÌÖgW¦Fàà á0SMn"aËp¢ñilDE'3gÏ>gÌ0Aä0
;TWT¿]-IÊAt¶ÒÒ\_KügA5z4ánXá#ã>a5.,pàì & M ùÁb{i6ÓI7ìXL a0AѲCHÜð´ÃD(ü(T.Bd'lðÖD 5Öÿ·A<&VÅûÕA£[H:5º ߦ ª
P0a #ó»-Ãüiѱ\£u¦¥}$ò].æd¼Ò©&÷¨Û´}ÃÅToZ60¾©º^ tlaRبúÕ_»&éÐMé_²·¤Ø9;:IºdV2> 7I°bW0ÄøÒnûÓ¶®úÝ=´t°Ð4¥1íª_Ô ßA~Y0"tV zyCu¥kûjÐpe¤[I¯éoºûÚ¿A´û}¸3$¶dà Ťé;UþÓ°&éô®´¾«÷õÕÿ¯ÛZvÕÿ§§}¹;éÅëT·ÿé'í×noûii/Nø¿_ZúOÕ:Ú]íé]7Mþÿÿÿ_ºÿøîÚ«ý_ÿú¾1½zT¡
ô´¼}.Mï I?_¶=}ëÿïtÝxãûÓî¸ßcúíÿÿÿ½kÖ»ÿÿõÿ]}-i&'êÝ/þ¡Ò±8õ¯Ýxkkw_ýýky/]kÿÿþÿþ£ÿøþ?´½´¿ÿÿKMWÖêAÿk¥ÿù_üq}ÿÿ^Èbè;úï¶ÿoDoÿÿ¯¶þÿÿ÷ÿÎÉ×ZP_ZÈØH6ûª_ñ°>¾ÝÿÿÝøüþ30¼
ÙWÿJYWÿ#Ë«¯É§¶¶ýú\_ðDÄD©ê'BEú$ülµ\_nZÒ®LSÿÿN¼Z¼èÈz&ÿÿ×ÿÿ»u\_ÿÿÖÚ\_ðïô|ä4¼ NOîº\_òö³AuúÕÿÿßÕv\_Oõºàõ÷¶÷ÿþÞºÿTÿ±]ý'Õ%KÕ]+Ò\_\_ÖKÿ;K5þý¿¯ÿÿ¿¯\_«®¯:ï^ÿÿÿû׶¾ßÿì/ýyt$õV×;ëÒT¿÷¯zý¿ÿ¿\×ÿ¯Ó^¾ÿدÿïïÞÖþ¿¯ÿªÿ×uIzº\_ÿæÞ´ý©oýßíßþõét÷þ¾ùïþÒïÿÛ÷¿ö;J?\_׺ÿz-õ¥kÿÿô©/ïÕÙ«í·]ûôëU«êöõ÷ý¾ëÿûnÕ«þÿÒõÿôNÕ.Í%oUöÒ´¼%KF÷®´Ú¶Ã:««\_´½+I»õÕít¯UÒ[ÛIØg.í{a¯aØgzÚ«þøgÖÝ{ ëèèM®öl-ý6é\ Dëû}CXûlþw¶l+¥awÃ#í×iXJÏíÝaÿp A6°¿ àØ
±@Ã@ÁÅÁc¶ ꩪÜAí¥SäC¤ Ò[ö%EOdS ´-4KÚ÷ Ý $Ä+ Åi±² &\Ó°r7QL0VA.Åv8
ÆÄï»JaïAÐawaii Aá¤B 8x lP v¡´ê¡ÃH Ò l$%²@It¨0h0Å1MCAA1
6 ± bAaCh]Ba0¡(DDDDDDHtæ,@Á §tGDtìãtÊÈ3»av´I RÇËRT´í;RObM²ò Ó(s -ª§ Dö\ a!)¸@ËDFk \ 0A""4-£"1DDDDDDDDDGDDDDDDGÆDURn¤ÿZ^®! ý/«¨¯¯Ö¯é/i?¯¥tËDa"fWí/´46Qz¦ ¡0A°¡!SñXÊã\°3±b &Á3Dv,&Bd+;«&ê¬
GuGg¿©Lå;GjAL¼å¸(PSdv62RE8ÉaDÜ,d´gd )&¢°{Pð²¹TkÿàîSÓÿGÙO¡èh88ppàîÑÀtëôÀü¡Ð }÷ܬ÷O J£±Ç ák4|¯Û¤ÿôD;$áp¿
¯ÅGúO¥÷ëkÿZ=þÿô¿þ¿ÿÿúýé ÿÿׯßïUýzÿÿÿÿ+ tH¢ÙdèD4HDH²QÞ¤ÅÒ#k¢4¯ýë;R"AQÆiùFF2!çVFùíòª³Vv´geY pax4ÊC eÍNÁ4ìEÆx42èäcÍÐs²QIÅ$d7dO'G ÎöGÞl£Èaï63 ÖÌ ø@³djÁ 0A0Á~ld1oá IYiÐ\\Fìµ]¼ j©øFæªÍ
éÏ B!^GAlðÓÑ#h0CA& 3Ã<3ÄÏA¨F¨MpMB.¢)hHp²¬aÃDMpì&aSL#sL"h¢(0
I °DÆ .CO60>I¤ðÒlÏ $Ø9;H6¡ìátn/ 5Ã0¢Ñ±/! ú¨OC¤u£Ä-BÅ 7Bá4lam0ÐaE ¨ é6bUìO +%[T p±èÙXÛL ¨Øõv'pT §¾ú_Oºï¥XC{.e¤Ø2\ùX%fwãIè Õ6 rCpd¹êë¤ò£½$TQã%Í&ô²h»N ¸Q+½'õmÙZa4a0«p1Ö,á7A°fo º'páI6].pbW¿¶úþöÝ®Nôÿôø5 ¦éô©Òn¿ÉDéúø¬iGN¤áè7O&Ûé´¬Òok¦º÷Õתéúzoúz¿¦ì_ëwõý~¬zõn¿éÅÞ©µÿÝmÓª^ÖO¿ÿûëêúCõÿï_WOø¿[øÿÖ¯«× mo_ûëýïþ«úÿï§þ)?×þ½WñªUëé}ïUÿ«~þë½ÿþëíÚKÿÿÿNÈj?ëúo\ Z÷µþÿÓt¿ëû_ÿë]-E/ëÿá}ïúÿÿþ"õ¦«þÿp¿ÿõù°"Qëû¿úß>õ¯úÿ( ÇÿUÿÿéz¯ãÿò^«ÿÿüF¿þP÷IHa?ý/r6Õ°w,_ýKïü"Wýïÿþ
ÿýûþ¿]zþCÿÁÞõïÿõÿ,GäêýçûÜ;ÿíÿNÿ\_ÿ÷úÒþ¿ÿïhÚ\_/d¿õÿëé}%×ËÿÐ-?µû×ÿúÿëûÐ\_×Z ÿÛIo ¯ýý~ýõÚþÿû\_\_öúú×\_ÿÿAu[þ¿}ßþ¿¥þÚª ÿØ
ý}=tßý}þº×þÿ×®éuÿïÿõKé?¥ÿ¿ÿ¾/ïÿÿ®ÿúÿëÿýý;TýþÛë\_K¿ÿ÷ÿ¦ÿÿõ¯÷þ¯\_ÿÿ¯ëÿ¯¿ßûo{·ÿþk¯ÿÁýukßýz·\_ZWý߯ý|VÛº»ÿ߯]-köû\_ÿºb¯ºïKÿÿ»¾ÿ\_®õý-´µµývÿöíoÛKO¿§KÍ/û¿Míý/þξU¥ú\_jÚÿ¯¥ÖþÛÿÚûÚõ¦¾¬k
ÿ׿]+J%>ØfßýÃ=CXí´°d}dLwúw ºþÓ«JíÚV½þÃJ×/ÚQPzÙöÌ×k ºû pM}¥l3ÊÒ\_\_lÏù?¶Ï²êöÃ.¡ ½µûôí²û¥am(0I
";bA b¡¶H8ĨiÆ«i'zqìJ 0ÍØfæÒbPïdT¶Á&ÐHU¥ «í ±)ñ§±)ìVÚR&ÊÚH4ÓM8lRwÆÅ\jJU\JziªwqÚQ;¸Ä&®0¿NÃ
lPAÓPÁÐ@è lB8tm pp¡µÇ S % tÃb¨QU z ÅÔ@ÁµP¨ plBiA° Ó ¶ ¨6 ¨mN8N 0@áÚ BaC""""" Á Å0B-i[ Ê +14ÊáðM ¨U ±Ó(pD|0V!¡aLDAv$b mB""2(ñ0hG"""#:"8ªZ ÿUýëZ^«ZýëKÖ«Zm}\E.£qq¨¦J´)!Ó¬&ågeÔ±;¨åvdòlXÙ\Dwo~©Ô ;3:ËH )^¡B+ 44Õ<,íNSº¯ù]fÝÑéGTxØpw0s \ÉdTyئvT½/~ÿ»úÒ
ÒÉlgj"£ÉoÒÌ´ó²Lì¦ò¿Ù$®kïÿLWBë×-'Qúú®¿ÿÿ÷þµÿÿt¿ëÿÿÿëÿª¿ßÿýýþóc:ç Ô2eHf¶AÞJßúÿÿ±;f£:Ù Ê´KWê@¢)Ð Ï@NvhÌÙÆà D÷æÌê\íS'ÍtèÇóffd4Q#¨È3duͨdm§A: IæÙÎCͲp§k0C0)DlPDR ÌA8!Áù±#Y±D( N!#6VQg¼Ù2.¸'i£s¸na¸^0©ÁS
¨ ¨×D%D#Î E½ ðÏD¬Ð8s°¢5¸ àÐh½hÈL"à2¬ÖÁS kgÈ Ñõ.XAáÈáDùõĬ¥h éÕ&Á'zT
]¼h toÁBziªi °T¢i\hØDké a<(@aU(&ô6©£^,UAD³\\£d:5´FÇ£cthz4Q±ê ±a\4 TÐA»M ÜZ6R Å
UôüON Á ÚM¤ëÌu°t l¤öäÒ. -&ÁÁéb{hÿXpbW4lÃ%VA°bW4Ä®i7%DøÒl4liÁ\è6 ÙA¤ÚUA°c ìB ¤ôý8bpbnÒpÉt2Ñz@ÔÍ&ÁøAI°dû$qñû{OétúOOûíryK té5ºÞàËY~Á°G§j&Ù8´Ó{N0EÓ¥ã ÷dîgwNÓÓ'tµ«dîOÝ7ÓtÜ'¦äóÝ;ú@é6·Nu~tØ;'iáÿÇÿ_þë¿ñºý¦Õ¿§ÿ§§ý?ô4××õïÓ¥t>ëÖ?ô;ø½õ®>=?c»Oïôÿ_îôý^ýÿ¯zÿïïV?ýv+þ?ÿ_ÿWî÷_ÿÔkÿ~¿^¿Kßÿú~«[ûK~ºuëÅÿoûÉÈWûöGKà¿þÃÿúßÿÿí׿ý.»ýÿ¯õ÷ø]iý{ÿÿûÿoÏíÿ_ÿX/üñäáþBÓþ»!¯Èbÿÿúÿò\_ÿ´>¿ÉÇÑÐ\_ÏøÍd'ù N?÷t?¯Ó\_ÿÿSÁ~´L üuõÛÿûÿÿÿÿõµIßÿõûéx"Qÿ
àÇ]/ ßJ?àßDíG\_¥ÿÿÿþþÿþÐKþøoÿðßýz%ÿÿþü ëÔ±]eèëËÒû\_P^è\å×ÿòÊøÿÿþó¿¯ýÿÿKßü§ÿÐ?ý×ÿÿÿþà ¯ÿÿmvýÿìR\_ãú^ý%}÷í¤¿ëÿÿÿÖ¢?¯ôÿ×ý?úáÿû³Qõýîÿúÿü1 ®¾¿ÿÕý½iuõé}{¯þÿ{õÿ^¿ÿßÿöÝû~«ÿå?þê÷ëÿ\_ÿÿÿæÕÿ÷¯ï\_þÏWîÿ\_ûýÿt÷ÿöï÷ÿû¯[»\_uÿýõÿû£ëuÿõÿýÿýußÚý?}ÿûi¥Õþ¿Z]ýïÿÚÿ¯ÿ¥ýúÿ·w]ú¯ÿÿÿµ×ÿÿÿt¿ûÒ×Þ¾ïnþõµÿµ×¿VõRw®ÖßZÅ~õöõÿõ¬È©Ùݳ4ÛÔ6ÍØ\_½}+õÛO[Û úÚÿûþÿ×\_¿÷½ëoð¿{®»a8a+]+ ka-.û½ÕvþëÛn¾þÒµ¤íø¥N lwa(6 =ßÜýò\}ûi60¶G¡ýÛ¿öékÿÈXîéHØgýy6ÿð3þDáäÀÁݱ)ða& +"?ØøaaaÞÃ=üs°Ïw¶«°Îv«È¼Ól#h8l(m8@á) S±@a Áà 80A8: &LLá±Li§»6Rva¤i¦©Ü[§l4;bci8lT0@á´6;b¬'W¦¡T8¨
¬ lP@ááÖÃba8ll0Á¸!a¦QDDFK"" Á!УR""8B34¨DDDDDDDDDDDDF(ä1ÿð¯××Òô½{\_¥õUë[úÿô®¿JÝ6yúì$ØKÓ) 0J Ç)¥¶WRå¯è©# hì+$ÈfE¦v4åt3³6W¬ÈÙRQge¤@È)\agko2¬íS $ËL ØU3ÒP¯àê´ÝT:v (:£Ä=z;s\::QÌ-÷ý!{"#°®²¯;:ÕJN°¾E?üHþ#ª¾´ª µÃXiT/]cë×ñÿÇÿþ¿Z÷ÿÿ\_ÿ÷þ«Þ¿ÿ;åÿÿûýÿÍè)#3q ΢Ìò'G#AçR&D»óÞjy®¯ÎÉQ."ë>BÙ¢ë: HÎDiÈê)8ÍC:£fuDLf¾qòÉÊÐüç²0ðEÀ¨×: &nC03ãilPNð@ÁÐA½ ý aªªÍá½¢Á
¢zéBávÑ ÆÀÎ Qp.ñÐa; \L¦aL#[ÿÿbÚ_ÿ·¯÷û×_÷_¿ÞÿÿÝoþ¿ÿÿÿ¯ ûÿúÿ«ý¾¾ï¤ºÃýþÌ3é^¾¾ÿkÿï÷O·úûwÿ}~ÿ_ßu î¿¿óÿêý¿ýöÿûÿûûëÞúKõ4O^ÿþ½ýÿýÿt÷þÚî½ýÿ¥®û¯{ÿûô¿½_õ¿úû½ß»K_'TO÷ý½é~í¯½Å·ûéß°ÌÒ궶¿v·tÛ¥ÿë¯}×Õ¥^ßõ¿{JïºÿÓöÞ뿽úµûî·nþÝêÞUí~¯ïÛ] Ú[g¦8?Øg¶%q¤ÙõÃ#úÃJÖtÙÛK÷ÿ¿÷ýÚMªÞ½{ }Ù¢z».[3VÃ3PV¿ÒL-lôÃ<®¤ÿ®ýOyíÚtqî¬9ÝÒìºöÏáb ÈaA¦cjÚLmØÓ¸Äã¸Ø29¦ÚIÚV}¶{éòDÎÃ9A ÓµûØîðÁØ2æÚ}°« ¹Ä(ïíza¤±La°0¦
1MCÖØ0^\Ó¸«¶+a¤iÛ Ã ¬0[Vaa5m1M!60¡´Å (21ÐAc áÛ@Á±QL :88ØÅÐ pØ @ÃL(M1A 0b8a¡¦¡´6!; °A 5 uLB @á±DDDêDA)Ð0AL®L§±94ÊÐal'a1 Û -ûl/ (L/ W½ÓAA
¥eC,p : ð¦ab Û OkÄB h4"!Q8aa B !ÄDDDDDDDDDDDDDD(£"#1¸ÒDB1Æþ¯Kúÿÿ¥Õt¿Iuýx¿]¿ôé}¤M«IvöÚHP ì0M6!sáAÁ0b"#!BdIéÔÉÑ6te®EØ2ÐÎÅñÎÅc%yÙ6· gcG;2n3\ íA Ã
Ï;3½QÞ T=ÕÎü"tL: ¢]¬¢pðptÁÁÃwÞ|>l)),É-6ÿtþýé\¨\ìGÎÕ,ÉJ;(º¢^Îô´¿ýªué×^¶ãúé¥6:ëÒþ¿µëëÿë÷ÿÿôÿ¤5õúúÿÿû\_ÿÒÿ2)'FÆt#8gAHdC¹ÆS¦Uäºÿÿÿ\_ç êdC5üyÕå$FDYçÑ%S¼ú"^u.\ 8g0ï0A e>¢ ! ! )DaNÏ°A <<sY{ °ÚZa ÚhÝFæÔnhBå4l¨3Tl pÎ QéôýìðÐr/B tBe0vá¦pÓ¸ <¸ à ÏA¢d#Xg Ot AwhmÕ È¯ O ñ';°¡Ò]é§^jûúMY3!IÃ8:MN l5ÝIe&Á\ê'¶Ù#êM¤Ø2\Ò´¤ò:&ÁçM'öHº%{0w¤H6 ³¤òH¾Aºv¥úM%Ev%EßvjÁ$ð@êáÞ¹Ø ë6s¶ {m¥MÓ×é{¯é6·MÉå'¦Öäîuu¨Aµzpe êðÔ&Õ÷û®ItôôᤶNþ©:MT'WI°z~äïÖ.¯OÒ]?Óõôõ¥ûO ÿ§§§é«zmíuë±[Æúë§ÇºÇº¿þëë¯ýi_¥i=~ôãÕékÓÐí×kw½_þ½¿Ý ýíÐØ´?ÿøÓÐïÿÿ´úøÚ±ÿü?ÿÿú½_ºiW_þ¯ÿ½ëüVõßÿúÕ¾ýj¿ý[êÿ¿ÿþíëõîµÿëßõúCký9ÿ¡ï_¿ûisý?ý¿õÝ/ë®þ«
ÿï+OëÓÿÿÿ$ ë²Oõý~Hÿÿ×_¯M^æÿÿÿÐÿí&ëãÿ¨ÿ¡ÿÇÿýx׺Zþ?×ü'ÿÿðoÿ÷þfm/ïÿz÷å¨ü±¹eÿÿ&'ÿÖEÄoKõ¾AÿÿÈqägù;ÿèGýd=6ßÉ9CGßDQÊÑ?_zå§ÿÿú_O?ÿ¥ÿßþ_àßëõÿü±otùb?¬½ÿ,Çùf?ËÐ_ÿ/^ß.½%ðöaý÷ÿø?ýû /Ð^ØKö_ÿßÚù¨ðÝû×ÿ¾õ¾Òûÿÿÿÿÿú ÿßûëïïÿï_¯ÿÁµÿö!]1 ö+ýwÿF~ûØvë×~¿ºn¿ÿÿÿýÿo×úÿ¿Ó®ýýÿÿ_mÿ~þ¼3 }{a~ɯß_Þ÷óÛ¹ïÿ·Uý|S[ÿ÷ý·¯{¿û¯ûý/ºnú÷n¿öÿúÛ÷¬Oÿ~¿þtÿ¿ÿ×´®Ò¿êÿý{ÒþҾاÿÿÿÿÿoíõúë_¯i_ßÿ{zêýÿÿO¯^¯úé~¿ë÷[i7i7ú·í¥ÛïuÚZMÿO÷ÿ·þí/oÒïð¿v¶·±WkõÚWíì3¢ÞÝûiw¯~·úëúUÿïì3· ¡¯´¬õi]r.xgû oPÒ³5Ý={fïï~Ûÿm·ÛoÚý°ÑÇÚiw¶g¶wíïÿfka
ûÏöϦÿî½.0ïþíö?{þ Ûÿý´¶(%qA8i1PÒl$¶8
l0Ba;= ò&vÏyÞÃ96ì3 ÝiH¸¹mN×l¹0±["æØ0¬4aìRi±PÐpØØ M°Ã lT4Õ;l[Ø«´ Õ;iݧ}ÚIðWqA
Å 0 @ت P mVàÁ±@Á´ b0A b¨1@èaÓPaDáB± ¨:bÐ pa A « áH4b q6Bm Gá8m! 80@ÃVF¦ 0RÜõ 8ATÓA§kà àè0LBÜ0L Ó #á×j3Î=§ a2¡ Ó)ÊS ÁN8">v´DDJADY,P!ÄG !ÄDDhDhDDZ!¨"""""""""""""""E D#"?¨ô´ÿôÕÿKú^¿ÿ¯ëúúõýëúmt¿K¶ºMÒöwöa¥a ¦µM# Åg@êÄ10B"#Ë�|°¬eº¶eNZc¶d³ê~W(ÈÌ#±"éL¤dÝ )P)[ÎÄ íAå ² "(¤B &ôe¥ +ACØ:aá ÁL3¤Ø>êú¾Ði0ÓTkzçf
çbÉ&ýü¡:nd;%O¥æL%&;NmS× oÎÁÞD¹R³"Eúák ½.Té¹ yþ©=ÍßþÅV;(pëãã¯p¥MÓ®Úoîµúq].¾?ÿ×ðDuéW¼ãÒí_ëÒÿKÕWõÿÔ{jºé}£¿ÿþÿÿÿÿÿëý=·^vX'3¨Ê|9gC"ª!?!»ÿÿ÷ÿ3©H"!ùÿ)2)ä£ÇJµÿÇÑnÑ p 0L804a¡iùÆÈÎ6GAI^l Rh<èSµ¦Cyª%q å'ùü§2LLwDäΤT#Ò5U» Á3FiF ÃÁi»3dá @ÁÜ«é-]/( à ¦Ì\Q¹ #p©¦ l~643c0r vK &; i È8a pAhÌ e ;Bà0'ÀQ;4fi;³".°AäaÍ a·M5ѹ¦!^0F¸A¡#WZ_ý¼OĬ¬¤tj'Ì1;JñzÅ h p±hl"îEä#sTÓ7±èØD/Ø(Aaµ!!rØ70Pa85
pi
p D?¦ð65îa&ÁVÑ GØ6qè ØbT4A5>B,zѲ $ét»¶ ÇMÓlÒ~ÁÓ°A°eϽÉt¸2]I¤¹ e&ÐAÐMÓ¤ØpÄ÷T tm l1=ѺڤUWI´aÑÖÒ 5'é5 M%XÜüôA6 Ç ´'I¸AÃ,w¼'wj}§UÒl1( (L(I/ûTôë´W¯éßþrWéÖ pjjété:îlsMéÖÁ¶tÛ&Ã,Ai:Nb AAÜðf=¶MÓð½éûõëz~¯t§¾«ÝÖ,ÿW¤ÿ÷ï«ÛÿÅ÷¡zÿýýw{öú}nºOý?ï¯ôýíÓÓi7Oý6þM>øI'ýtºo~§ÿ_ñ×êKÎçpÒþµÿÿëú÷êúé{ººñ¯ÿzOúþ/_W_®½tÿCý~ÿ·øÞwÿÿÿÿõß_¥N!±ûõn_ù~¿§ö ÿÛþÿÿÖÿÿ¿÷÷÷¾¿Ûõ¿ÿûK_ÿIi/ýëÿúoáêéU¾¥_þÿÇÑ8Où {þ"úDÈCëëûÿ¨_×ÿ¾¿ýªÿýÿëÁb}ãýÿÿÿþB üÖîýúÿòc?ò++×ì?ÿ_þø8ÿ×ÿǹ@Gÿÿú¿ÿû¯íòp·è?¯þýû¥#dnÒûÐGú_ÿ¿ÿ,g@»Uÿëí¤þLÅÐÜ×ûÿè
'ÿ¯Ñ4÷¿ÿÿ_õÁÍÉíOÿÿúýýþ Òè¥õÿ9 /¿ÿÿþ¿óXÿßv(úÒÁºÿúþËÐ_ÿµZõÿþÝ^æwWîßúþ¿ýyнÕÿÕ]+ßÿþ¿µÿÁþ¿ØD»þÛK:ÿý¿úëÿÿïÿ¯ÿÿ×ûúöúÿ¿ÿë¯ÿú^×çªÒþ¿½»öú¯û4Nºë«:Gûõ¿þÿëïKý¿ÿÿÿÿÿý¿}}»õ÷¯ÿÿÿßÑCÿKËù'ÿ«i}÷[öëÿ~¿ûî5ªÿõ9»ÿÿ½}þ¿ÿÒêÿÿöÒú×ý·¿nÿßô~×Î¥ëK®úÒµt¿¿º»û_ÿí¯÷ëÝß÷ÿu·^éõ×ÿÖí.µ~ýãï§ÿaûþÛm/[îµ÷¯z?.¹=kë¤Þfô·IlÎVÖÒ-íöÕMý
··t«_õû}¶ðºMÿºÞéÛ¾×~Ö¯vßµ´´»µÿjÝ+_ ¬[ÙvÌÕß÷í¥iÚÚ_ªøK#˯hý$®ýÓb¶-±¦È%°l·»Ód[§ðaÿ ¾RmðÏpÁáXû¦ÏýðÈýÖÞ®ÿ_·«õlÍZV¤Ý}§ü2ë²êëûG·f{ðÁXf&Òín OnÚWÝ_¢Ä{J\5M´õbSÃÐ@á°h1M1A:(0@Ápa- lB ,Rkq¶7ÚR.3±)ýȵ4¹Ãm(
K|JzzqÓi-ìWl6 a lBq ¨t6Á:4JªØPé[ ETB!6 ÝZ hD0h0M Úü4Û ¥8W Øa4ÕÅpÚb¬a Pb:b m b¡S|+U80ié0AÁ°¡ Ø 3ÈûRÇá§ÌÉá¬1 ûlSíBi«xb[ N,W-õµb""""""#""(ÔD ÁA"$": LB ±Ûµ¾aaªe0¦Lá)Ì<4DèÏ.#¡ a !B, ÄDDDDDDDDDDDDDDDDDRé""""#ÇUÿïI¤Kþ©Z]~©µ×ê/M¥út~¢-s d§C0¨äò'G2ɹ'ó 2&êDM´ÐTÁùÙ#;¶vôôþøtôË1J¼ì$[¤\_÷-Ô©ÿ&åF櫨O©ÊÌîM2jpB4ÿÿQQÿúoGwÿëú§×ÿ¯×ÿøåI}/ÿÿãô<'Í¿o6D1Ȭ2"ùȬvÿúãG¥êaá%D "ð030á' 2aP!;.3Äu0<HÍÅ ÎáÎÕF@Çd13lú6GæÈè;(Ê!æÈGäHy»: ¹²:UæÈù
!æÈê3'~aá ú¨]½4n£c C¨aS.W
»hÖh0làÐh&[³äAAaB4\x¹ÐgÈñ F°Ï ÉrÏL0a$ÑÍm0È "ì3Ã0ë¸s &ÁÁÁF¥|i:Mã©ùÛ? H;nÒm @Ú¤pôèÖ7B6A7¬X\_ £d.A¶/AªNA½ A°±0íÝ
A6x@ Ô Ý%¥÷ìÁªnÖpÓm'Ùt¥aá$Ød»#08N JÇH:\Óa>ã Jì&ä¦Ã%ðbWÁ=6Ù3¦Ù3KbW´jï ²òN/JûÐÿOzõûÓ»¥Ð~Õ鮺pj®Jûtº^àélªïN¸>µÓrWu¶^C¤ØzuzkÞÕÞ>n?þ·ïXÿB½u¿tþí½qtµ¿zwé±}צéÚ|êêët¯ÿïyXÖ½kÛø_ýÿÿúµúÿÿëÿÿÿÿªë^µ¿ÿý÷Kt¼Cv-ûÿ¡üAßQø%uúkþÿKõûõëÿ×ëïþ¿ßûÿ_ï¼ì(^Ýÿÿ¿×ðO¢4ô8¿¿ÛȨ;%xøÿÿ(÷ûßÿ}ÿøÝ#²ÑÞÿß\_ÎGÛðDñvjCÐ#ð\_þFÅ#¿$ÿKÿÿJ?ýÿÿò¯ÿõz/?ßöþô¯Ð\_ ´¹:ùc?ú'\_~LfXÿ \_ÿÿÿþ¼±?ÿË/¤õ\_Òþþt;þ¿
LBëÿýiPUô½ÿè/ÿ¸\_߯ïÿþ»jûÞ·úþ﮿®ïÿý÷«úú·ÿÿ¯÷ýÒÿýkß÷ïõù«0î¿êôoÓõý\/Þ¾ºÿ¿ÿÿëÿû\_ýïýÿ\_\_üÇWVþ¾HÒ~¾ÿ´ÿInß÷ߺÿÿþßÿÿöôý¯¿¿ÿÖö߯i:ûv¯i÷i=7kz\_i®Ýÿ¿ëû[Vûÿþ×ßý}ë]/ïû¯x«[êé´(fæÒbSîEê½[?ü
û YÙlÏý³5Ú]
¯
mnßý kiuikivÝZí/Ý/lÑ5a}ûJÒ´ S ¤ 7P)cNö%CNwqjÁÍO¶-ÜÒÄãïħ±)ì3Áí{×OrÚKpÁ»t"ãJEÎÃO¶÷ /Æà 0Á(iaNR|¨ L ÓPÁ
°@Ã Ø AÃØ @ØPÁÔ6 Ø Ã@áµAa$Ð plB0@àØ0Ål(mED(Т3$ ¸3ÙCà Ã0®îØXaa/ëì' aOË^PD|4Ä&è0CpéðÐkà ¡ 0AУ!DDDDDDDDDDhDDDDDDDDDDDGA~ÿý.¿¯].köÐ\LVÒ 6#È#±ÙØlídíO+¬ä8äÝV;-õFAyØÚ ©3Z#"]Pd¤d\R:R&DÝ@(Îç!ÈÎÒ;JåD¡yº Ôì:nWË)qßZ¶GßláÃU¹S.v°ê,tU=/ßûµNvVµÿÊ´I£µUïúëÿúýH¦vQoöÖ(WÈUOÿëxþ©i(ýPê¸ê#ÿ×ýÿÿñÇÿÿÿÿkÿÿÿëþv¦»ÿÿïÿ¿ùØ(HSP§LÝ#Xüá³óÙ¿ÿÿ¯s+o=ùª"qÕ~v(³ÈÃ3(DÇÈ,v 4ð açÅÁuÁÍÆðιFC³fk.ld1eÙDÆ|ÉÙèØÎfÈÖ3¨ódu hHb3dt8YÅðAÌ3x ü0@Ì3cRÃ=fÆOyFPQÁH# AJ6ô FöôѱBKZ ÐatÈÐÐi ðr)AL"Ü3xanÑE»®êÚÓ¯K´¥m¬ B!e¶G #Xíµ=¾Y;ò+Ìã-9nIy²¹¡ÎÔDÜ8µ$ì@åºÌQFJªÒ-ÌSÐ}Ó £<=Êþáðx{Ãè!|\5»;[ÿ廬£³5çf«+zT@A<®·× ôwK
é ½ «Auÿñÿø{þê>ÿþ¿úýSõ×þ¿ëëÿ׿êv¨¾ßÿïÿß®û@§52L#æ²DæÈøÑ:Gèd¼¥UùÑ\}ù\)å;%ï#"Wë"l"\_Ês"qWU¨§DqùÆk,e8ÿ=ìÅ FJddKÙ Dy«S±a±¨FÆÖlà:Ñ &d¹@ÉÚÙÀÅ !C3\ØΣ$fñNÉ# CPsFhÎX@Á N¡ÐC¤aÙ"'ºa2 ÊFa "Ü3ähAa4ÓÁUö('f øSÁ 3F 3æ50@Á ãEµH6@&ã Ð]4nzaðU
]´\£ÁBaÈDä.SÖÁÐsµ!ª5°pPPD,Íè¡°0TLÐj pa³60L#DQ°46¡ÓGæ f¶qNCå0v Aª?° #c ¦±
ABÙy [±'vªàÉsçÔD®i6Î6ÕªFçè&ÐM8bxi:ÅNêMAÁÒm é$ØA
H6ýuIÐ#<8:5´nªM£üØAu 6Ä®ÈÇGQ+VÐI6GU±Ô4b{aXöÕ$ØbV VÒl§FæFûO?oUtôºiÒÚzvJ ˦©ßVÒL ÚL(rÎK©82|%fé´»¤þ Ä'IÁBmaÞ ÁÕÝAºlOÛ ÒÒ¶§§j'Ià Ìp°fHØL'iÃ$ôé'iÚNÐlsONÕЫúôõÿ^ôýÓÓ¯\_WOÐÕ]^Óy<®¿÷úÚNëÓ~Kéÿ[§¦þ\_}¾owK}¾¶¸MЮßÝn½úOûõ¶ÿîî½8ô7¤/ëïúý:òUèw¯×ǯҫÜ]!üz±Ç'¡«^¿ëÇÇêÇî·ÞªôÚû¯±Òÿÿ×éïÿûúoÿÿýïþ=úÿõßëÚ½é{¯\_ÿýÿ¯ÿÿ×ïõÿ\_÷¯¿×éG÷õ\_ Þßÿýÿõ\_ïÓëð¯ÿá{Á}~ÔøMÿødysPÿ|ëýévêÿÓ\_v -ÿ·Èw ýµÿóáÿÊ@¿~üøëøÙF?\_òÿ༿ük_ÿà¿ù8_þÿõøþ¤O¯þý"c?õü±MO=7O?!Ï«öÿëûÈtD>¡òÿ¸Dïþþfg-¿ÿ$9Ýÿè¤<þ¸"Wÿÿöé\ ÿ&3¦Ø"yýzí]_ëuÿôúH/å§iuÚ_¢c?YcòÕùb:þ¡ÿUôÁ OÿþëAzÿ÷÷ÿë,Gëðmþÿïuý÷ÿý~ú_ìU^Å~ÿWÿÛ& ï¯úÿoé/mWÿúÿþ¿ý/¯¯ý¾ÿoóY¥ßú^®¿éiúÿiWø¿ /ÿºÿø}¿»ú\_ÿúþ¿×ÿþÒý\_\_ÿÒïÒÿÃö¿ÝGíÿ·¥ýþ¿hÃÎ&¿ÿÿ®D'ÿÞ¾¿ÿKýoÛµ\_ÿö÷\_ûÿíëÿÿ£EÒêý¥oÓþUÿuþź\_¥öÛ\_¿íµÿK\_ïë^ÿïíéïîýoÿ«¯ô¿b¯ûû¶¾ûì »IØfkí/ûµÿk÷õ¯[?wêßïkÚV¯íö¶¶Þì/Ý/ûö·¯á?þÒÛKo÷ÿ[V×X¥m\¸a&¤Lì3Õ0?¬.¨g®ôåe×Ùý{fi°¬0(eÔ34ÃGeÓu + Ã.¶ô ÙtÙl#]ïlÏm.; ÿvÞÜ ÑÇÙuvtÓgÿØgÿ#V°a}Ä' P h8
´4 b ßPÓ¶4!ªwŧ°¯ÖØaXö'w ,: ÛØJ!D¨wlbT8àÈç÷lWl$ÚIÃÓ¶N]ìW¦°§pvðQ\U± Ñn,p yNjðÕ5 ÁÓa8nÐ@àØ áC TSLBiLU5 6Ð h\Ðb 8mAÅ1Am6ÓM
AÁÐ lSA pÃ8uÔ0Ô6 ÜDDDDGBÐ' &#èNe9C¦ xaW &¶°Á0iÃA
- &; Щ38A!A¡ !@ ÐaCB"2Ü!0B"!a00NSºqF¡=0 £eÕp¡
ðÔpT6´© µI°J¶m¦±áÙC ´jÓA7¦¡PL;L Ô(aé
a 1t'§]-°éØA¶LX ]§öKM4H4è êätÓ¤Ú\WRlÎFq+)::MÀÆ¥éÐ?¤é7°cK¬àÄøÒpf:lgN}Sxqt&Ã,aCq+©a^Á'%{I°cÓ^ Jç# )7×éjÛÖ×MpðomêÙ(ÖnÐpd¤átÙ^º¾àÔ&þÝZé[òQ=úN¼éû Ò¿Þ¶ú¥¿Åè6®ÓrWIºoy(Oý85 ÷÷\_õzl÷ǾW§VÒ¯á5ZNé6>¶úÿõ¶§]/hkM/ÒÖèiþôé÷ÒÿI¿k«®ë[¡í/¯Ýoïu¯õÿÿÝ\êèkÿñÿׯ¨Øõúßÿë¥oÿúýtÿúÿþ»ÿôÿëý=+tõ×ÿ¿WÿÇíöð]þµ½uÿ\_ýuþÿ°ÿz×õßú´w×ëÿ~»ýþ¿Þ·@¿Ú¯ÿÿ÷ûi]éýük
×ü &¼VÿÔýþBx©+ø×þ>ÿûZÿzþ£ÿê=IÿÐÿÿCÿÿþ?åzåá½àEëþ«ÿÿßûAÁþ~üp¾÷K¯DAÇ\_ÿ×ÿËxDïü]ÿä%ÿùGéÿþÿýt ¿ýíW [ÿå×ÿþ]ÿ,Gþ-ÿ¯¶·¯²Æeÿ¯/j¿ÿÒ@¿ÑUÿ7\_þXÚÿÿ/j¿ÿÒë\_Òý×ÿÿý~¾þßz÷ÿý××ÿÞºïÿ¯ï¯ûÿÿÿÿÿiÿßúý/{ÿ[iuÓÿÿ×ðo&ÿÿÿþêõÿöûïÿ·¿ößë÷××Mûi?ë÷ýݯ}ýÿÿwÿÿÿßÝ·í÷¯ûý{Òµ¿µÿÿÿé~ýöëÿÿû¯ýÿmÞ·ÿ·ÿ]éþ×û»^×Þ{|Ö¿õ÷íïÿ:÷ïÿÿnïýدõ¿ÿô»ô¿tKÿþ»¿¿×\_ý{[íû\_výÿ¯Wænëo¿ÞÝ-}¦µí\_¿ÿïzû¿^÷µ×ïKzº¾ã!½/þäLì3Û ÒaöÌÓöïû Z ½»]~í{ªÚêÝév·÷öÝÕ¯ýûk{zßÿeÕ¤Ù
í§°«aa§®ßn[iXI¿·í¸A ØÙÀÒ]O[FéìN=Ð0 åÃ90aa,v» ôÃXöt»Ý0ÎZ\rù; å æÚL0þD/¯Øí9ÈËÿí{´£½¦%>S¶VÒÙ; å %ÚPa&ËQØa.ä\ì3 ; &bj@á´ pl(|TP abw± &Å1W4¡¨80¦ < plSà 1PµN0 ¨ai 0@Ð0@Á h1M Å;PÁÔb©Ð plTB 0a(Åõ! @á±B" 4 ©UjqÓ´Óì'iá°[VaÓPÚ Và ئBڶà p\01
ÛOÚì- +ÿO A &PÐ\ &TOa2aPjá
,p¶!K Ø[A
¸kÃDDDDDDDDGÄDAYJH0QDDDDDDDDDDDDDDDDDDDDDDDDDDFFDZ¬DGú¯¯ë¥ô½/\_×Ò\}WúE üïtÓª LDGØ©GbY2[ÊôEº&ãqÙxìB+ÄÜì dw¤¤¥{nXDÜÀÎëÞM0wOU<·²;êvfü5Fx~°{W%ü
÷º\_¿äØgúC«X]üù|ÈKÍ,íS)3²\_²ëùĹñüu\-EuP¬\QÿZÿÿÿ¯ÿõÿúÿÿÿÿ×þ¿ÿ\_ÿ×þeXÊ?=f¥ÿþÿük÷ÿ@Í ¤ìÁx Í£ÁE&ÆCÍAsdu#¦n4(çNÈ&FîvH2B8y!y hÙÔd4püè)S¢8yOY³!¢Gв-ê ѱ$fl!æÈÖªÓ0GÌÔ)(Êò@ó¨È#ffgS:¢tpó .Ò70¡0W¨.(\!8¸LÌq±ÁåA&±Ã@Ù
dlÁxd0
·,vxaìa8gÍE9@d|"Ü4-Ã>A"A', Ï ù"#°E¸fdFÆ Ñ¬3ðA gÆA\xgÆpa±Ãf[fAIAÁfGhá ¤Ø1=µt'v8Tn/AA á0¡ Í 0BðEã £g ²PL& 6
¦ ÞP
°7¯ÒP 5 abPå ¼.6bÑþ±Å¦ éi- j±i£[]074&áYÄãÃÜ82è¹.aéÒØ0M¤ßQ=´bUá¤ÛüÎàëI°Ép&ÁYJE$ÚAÒl1>R0ñ+ àèô'ÎØ2Î6JçM%ÐÓ%WRl%sÍ7GI8c ¬i6 JÍ$þ·[Óý+p×ÝûrQKIÃ,PÇ:Xi é6G´¤§cÁ]ªuè6 ²a ZZVpfA(½7æ¤ì'I·§îO4õm^M¥oÓP°ÁÛ%OOÐl¨F½Ò¡ëþ¿ÿéÇ¡þÿéé]¸5^¿úé®·ÕÒ÷Ûèzzý[¦ÿ|kúÆö½×¯ú]X¿^¿¿·ûþü1±×Wúë«¡ÿÿÿî½:Zøý}ÿë¯ïõ½×ÿëþþ>ÝVÿÿÒþ¿ÿ§ÿ}o_úߺÿý÷þÿý_ýoÿßþ·ëÿÿ¯_ÿwÿÿúÿÿÿ»þ¼²L~Hÿì(ûù\zÿ¢uü~¾Úïÿ׿ÿ[ÿä1É õü¯ïÿÃ#¬ØOÿ_÷äuûÁÿ¸¯ÿÿúüÿÿÿÿúþÿÿÈw_'ðDzô¿ì?ýrp¿ñßëÿ_Ëûaú/¼Wþ¤Ë?õù/þXËßëÿÿÿýnïÖaÿ_?îÿðßõ¾<ÿ¢u{÷ÿÿÿ! ~_Wÿ×ÿçTG^ýÿÿõÿÿÿýjÿéw·ôþÿÿàß÷®ß×ý}~¾ýAÉA¿_þ¿ýÿîµ_ÿþ×þ¿ÿþ×ÿ×ÿþ¿úýpÿÒ×÷úÿú×wkõÿÿ¿«ÿÞÿß÷õûÿõý>¾¿ÿ¯ÿþûíõÿÿ÷ÿï¥þ{r¿_¿ÿÿzÿÿïûý÷÷ýþ¿Þ¾ßýÿ_Úÿöºë'ú¿ëÿ¶ÛKýuÖïÿ·_öî¿·ÿ[»ÿßû¥nÿiZûuæ_¿ö¾ÞQkû×ïÿý¯ûÝÿh+]¿]-~û¥ÿw_º×½ÿÿï»ÿþµþ½6Ò¿¿ð¿ªëýö¿º_úÿÛ ÒÛ3ÁLJ|¿þ½°?aïØe׶gþÿÛ?Þȹ²>°ÏëÛ ÿò!pÈû¤Û¥ðÈÿpP0ødm/î; ÷ü0ïØú°?÷I°?ä[¥ÝvþÃ#í ÓA´
ÀqT´Å AC 'Å6aq§²(éo´DÙ+méÇ M=NøØi2 Á¤±VBb¡±¦Ã '{ A lP@ b¬ í¨jÄ+ íT4Ð|4m4î!]§q±0§ yPa;L+Ö( àÚ6Ab80Ä ÕpÚ6a¤AÁ 4ÓLB: ç5l bÕ¸an´6¦8lB (aÛQA L á¡ á´?¦""'H0L¢Ä"TôÒ¡X2À!0@ͨB DC""""""""8!F!± DA!¤""":®©Gþ@@ç3±HìI@§2V'¸Ë"QC v45+¥À¿ó#³´ñØ$v AÁ]Þø0J µPîVÿµûþû÷-̼ìB×6Jïýy¤Vú¦¿ õÔ{¯¨á®¿¡ÿý¯øÿZý~Ö¿ÿÿÿóÿëú²lPGS5¨Êɼ«O:¯þßÿ£§2¡u3»^{!>ðA ³daÉÑ ffO R(ÑÃ: h¢9²]æPeW">éu#PÉá2 bä£(,c" D:ÐDPq3Ï@¡d6a LÛEÑà5CÍ»£c5ëjëd$HÍò:ÆÆtͲJ<2£a »EÙ eàªl'\tBXhݢû±Ã 26Aá M0A È1J AõÓ±ÙáAÎóÃ40pÌÉ2vÁ%Ìù!üø¸AAá ,v 4Q°2PJD'ÐaÕAÉÉ9æ°ÁTdãLÐ"à4ð4AÐNØ7?5AIÁB.áPA°¨¸aÐAѺ¶65 PDÞ±Ï {~oMÚ7ÔÑ-B6¡4Xì ¨ÝÜ7U4à °P1+%c é7j,J¶z
µØ¤Ux°ª6 a ´\zeõ°ý¤î¤ì ÈæÌ í¤ÐpàĬ¤áÍcFæ OmhØØv¿ÒøÍ&Á\ÐMÖ"8: ´p%s@Á=:ONN¢|á D¬°v¤é818ú 0h:N N<¹ÓÉûdÈMpdøOðcI¿IýÝj¶NÒ^ú·àþ +Töë½6¾ÉF¤ÒON@é?ô©Õé÷WÁ=ni4Óï%WJÕÞPÕé½é×K^zø\_pðzv¹;½0dîwI\_êô\ãõãõxÛi-ÿOÕtÿ JÖöéÒ{¤ßO]cûõúM×WCþÿ×ÚUtõÓïWB?ºô>ã×jô+\_OCîõÿ×ë÷ÿûÕv?Øý¾øøí¿ûÿ{ïÿÜz«ìzÝ}ßûüþÿýþ¿Öý{Èbü}©@ôù]?V\ø}l?××õ×éZ^½kýÿûøܧýéWçQTÁ>ÿð¿×ÿ×[·^6¼ÿÉ8òJCd äCë:ü]ÿïü~RÈCÿ$ ßùH'l1²!?ëiÛö ÿY¬/ñÿd0ï\_]¿úÆiO6°LD®òpðD¯È 8"7½0ÿïÿã#ÿßÿü'°ÿÁRþþÿzÿ^xäÉ!ÏÒ&9BÐ"Wÿÿÿýÿ° m/k,»Ð/5Eöåú/¿¢3îëý&Òùb.'é×wôo¾Xý×ÿøAÖÐÿ _÷ÿõÿiKدëðuK_KðÃ¥ÿ ¯ûÿ¯XX7þï_Ò×ðÿ^»§ú¿ýWí¿ô¿ÿÿ¿ÿÅ^úýÿýRvIçBý/÷KÿÇþ¸6¿kû®ø>Rî×}·Óÿÿ×ÿëý%ôÝiVú÷~·½m¿áÿÿúÿôÿXúû×ë>ÿõÿúÿûþ¿ÿÿÿïýÿ뮪äÛÿ5/×ÖóVõç7þ¿½wþõÛï©·úþºý_<®Í[ÿ¯ÿÿoÝ¿÷ëý®¿Ò]«ukj¶ ݯÚ÷éSßFwÿ¯¯½úöºþí~ë»]~÷ý}§ëýþÿô¿p·¯íÿëûÕ¿÷ïQ·GEë öÙKlÍavþõ¶ïM+oU¿ê×µûoWvþþõØ__N×ÞM~Öõý¯Õ¿îk·f-°·ù¢ïí´µï´·×»l. NöÊLSä\ åPÁ=[lÿõþ¶}ûÌÕ
^ÿáï°>ë00>;äLm¥í0ÿé{û¿ÿlúôö%>í.Ø0_Q?°ò,÷mׯpÁ.ÒvíØPÛb¨6!5 lT0J¢
a&¡¦¸a&î'vÁÛG6%?MOÓH;bSvAÓÝ;I8qM Óc pé¤ôØÖÁ ¨Å!QP¨m 1@´60 4B Øða Ã
ØaD Ó l5A8q ¨µ pØ Å5 8mAß Øa&ØaC V0Å08 H4¾paÐ@áDA[(¸XaÃ;GA¯kÝùûÒ!4HÈb(Çb>txæ!Á0sp×>ËN ªÊÑÛ¡q%¤ô"DDF´-j¿ïK¯KI¥ÿQúÒÚßì -(0dráb1ùAܲÑfÙb- )QÎÕ"Så3³ì\ÉL¥Ê:Ø:i¾Ò%£ßûö¡ÿäýÖ¿-î¯{pÖñ +´¯×Xã×ï×ãýüÉÙdAL§ÿ¿;Y0A dbclØCby¶P)Ò63PÍYÑL2¬ÉsR ÉÑI)Ú¯, Rc6F¡'"¤ èEÄ8<è)Í:O"$f!£t#¢(G2dv+:P$F
!£Èf¦A^&¹[64k]°Vä"à32 9a !fNaÑ"hh-C8]
"¡ -ÃPp,vnÐfd.Dó@eYA"Ç É2 ó030P Ds'a0T ÙÁ¦K&;·fd!P£ ÉÙ¤ÎÌ è31 Ì[02¤13dEBäã! Om'A¤ÚAÐNCAº7h4\T\£u;M0P©¦ ¡Ø\Át¡a²&Ð@è pJ ÐP74= ØEÛUFæ©\5T0ÁRbÕ&Ý0¨ t0Õ \#``¨0STlè0ðH0vn
óI°Ä®r7TêàÆRpÄ÷Û4Äãѱ0MÚ h Ú¦²¨Ý «½:ZßþÜßdî¯MÖìþØIÝ'zI»Oº ºt'õt¹;ø0BÓ ÷Ð\_»'t¶Ù;¥Ä¨Ó K§ÒnJ,JM°~áéúI¸Í&ÒpbW0p}ѹõ¯×ãô6týÓÓ½StÜÕ»UþºÚýÿÚ^Ý>/¤ÐïMô;t>ÿ\_ÒÝ ´ôÛЧ§õzm]+§§tÿQ®¿ý¯Z\_×OCVõÿìkú±±ëô¿A{ëÓþ¿^¯¿=RéXý4ãm¥Ý=\_´Ú\_ïÿfÕðpaiÒ«÷^¡ÿþÿÖÿ\_ø^½Ã÷ÿÿÒ[\_ÿ궿×øÿ¿÷õõÖºëÿb/¯ñ°r\_~YÐWäA¯õëï®ÚIÙöad@.±ÈJ5öÿÅ(£òq?øÿÿÇåü ºK\_þúü5þLeG6¡°Á¥È8ëÉÆ^ö ¨ÿïÿÿëðþºa°>ÕúÁ}à?ü'ÿ¯òëöÿçÃôÁù KãÞ>¿dMyÀ¿W-[ O·ò=ÕÛDóëÿÿýW üºò˶/½¥¿Þ}w,TòÄpÿòÆ}¿ùc8D¯ì>Á·Ô?úÿØ0_ÿl#¨ÒÿúÐ_hþ¿ÿí¬±:þtßÚ\_ýýh/o¿è/ïÿÿ:õý \_~ÛYbþ]eõíO?ÿÁëÝ/µ]ò×ÿÿÿÄûÞL3¡¥í¥ÿzûz\_õ\_ÿÿªßõù~®½ïÁÐ^?~D'>ûöúpõûþ{i]}ÿvÝþÿ×®½»ÿõÿWþ÷pïÿÒþþ»k}^·ø6û·w^«º\_×ùÑú¯ÿ½{ϯûÍk7½{¥õÿ«n¿z\_÷Kûëÿ\_zÿ'þúþ¿\_÷mÿºñ÷®¯þ¿þìSÿï¿û^;¿¿]»×në\_ÿûßûýôµÿ4¿ßú{û\÷ëûfi»?ÿV»J¯Õ
ö÷Wi}~ÿm¶úÒõ¶ý
vv»ØK[Þ¿Û3\_}:\_wZ¯á.µÒ»JÛöÿïÓ¶ lL[ C Dï·VÊ P3õµ°ÿÏÿ°ßMù#Ò½{aÛ=ZÝD§{ÈKî%=[a=830öÚZ}°÷"ÛVÒöÏðV÷¥÷acnÒo{µú×8lBh1QHl$à´¨aA¢¬'lU¦'qJÚ Ø36á±LlR % $Õ´ÃH Ú6بP P¬M°a6(4 ØÚÃKø÷a Á&É7.Ä¡éÚVÏÄA Á0 §û jÓ6)
pÃP 8@±TÅv®!Cl0L&!E0i¦!.ì+V4é5pÁÅ ìµhá
¸ibA¦0A@ ØPéb @á°¡5b1fØ@B""""2Ü"#/DDDDèEÄDDDDDDfØ( Ò�DC g3¼4ÓA
a{Úq ¨aãÄkÄFDDDDDDDDO1_¯¤5úý~¿UúôëZòÌ;¹6-ý~)>=@ÄSÊãÆ"<È",Á¬ì]ILd 2¸±S²rú qI<%2 aàêaAY^ìö¯¼%ì4WOÙ2¥¹2ÎÄE^WQí9þÕȯîúá~+¦®-/ý.i××úÔ¼qV¿ÿÿÿú«ÿÿÿÿw8ýu2ÿ¯_úþéçfl¸§sÙ@¤À¤<áùÃ5 GEBdcçd£ìÙFHGò:'Dt0f¡|è2234"@ÉFdh¼áD3#{äLÄ>2a00>a|"ðh Á0@ÌÙöáÃí "Ç !í¥NÌØ v<gA#djÐR7çB$fÌÌÎ#H¡3Xò¡lèØÄîÐM±a4:ô70gðÆv±¨D+¦¡ ôÐA°¨PÖô 6qFÈCX@¼&Qù£uTNFæzhØ£s ¦ ±XAà CA ]ûðPTÑ ÆÆph05¸e!H³ ñ aU0|ÌÌ! 0A' ¹\[I°ÈóQ+)8bV0Äià ºJÐj¢: \Òlý êánnl ÈÇTañm'VÃI:Q+M¤Û@Äö×Û«¤AÝ.~n O \M±;ѳºmA0Î-4nª¦(PxFÇTßÓÖl Ø3 &Ò¶ jÚ¶Áúî S¥²wtÕë§VëðhAaº\_¤ì]VȤúMa82öUKo¥Òúpì àË¢p ÒtI°Ép' Ol ðpp¤Û¾·ÿ\_þûzuzn¶^}\·×½ õÞ·Ó¥úOVÐï¿t=Sôëz\_M¾ëÓ¥í¯¿Mëpö´\_
]¨Oï:\_é:Z×]}õõ¯OCWO¾;~¿uþÿõïú}|W¦õÿÿo.¾ëÿëëëñzmz¿îþ´áÒêß÷ßÁíÿÿ}+Ö½z\_¿ýÑÇ¥Ãÿo\_Vý/þýÿÿ·û£U×ÛK¯ÿÿèë÷Çÿ^:ÿô:ùÿñÓÿo¯¿W®ã®äOå@¿ÿ¼r$Ç¿ò#õÿ\_ï~?ÿ:þý×õýáÿÿ×ü¿a×ÿÿþ±ý×[ëín¶¿"»ÿÿØ5«Kþ?ÿÿÒ´¿í,ýR@±üÈ\_ÿ]?ùb:ïÿåþBiõ»ÿ,eë]¹1/ ¿ÿårißËú&7ÿë½÷å×÷ðD¯ÿ"kÁ¿õõûúÿñÿÿÿïþ¹b;ûþÿW¬-~½ÿënµÿÿþ»Jëþ×@¿ùb: þYO#íÿÿòe¿{òÿÿõþý{Kÿýö¸>½Uýy&múÿ¿ÿßëþõ×ÿ~ÿÿüÿÿåíWý{õÿõ\_Õoÿ\_¯ÒßëõþþÝýÿÿºOÿǯþþ¿·ýþj?{ÿïÿó=ïÿïöêûöë×íÛVyßߥ٩?êÛÿ½\_QÅ7~×úÿõþßÿ¿ïÿ¿íÿîþöúN¼{ÿúÿ½ÒöýÒ¿½µûÿûVú{mIÿôÿïÿïíÿ>×ý¿þﾯûûÒí-+}££º\_·í}7ß
ûo´}7ÿíí¥ÿé{Ú§µéoíý
þþÿÿíuÖôö Ã]a«i|Ý{¶êM¥³ íÓ"ÞᶻµlþȶҰ={m¥a$ØgÕ«gÿaûgþêþm¤ÕC=1(ýõì¿fkõöÛöÿëöûXjØÁ±APaØHpô±Ä%l%29ÜuØa Ô6('¸Mb¶Òb±µ¶PL1 ÒbiÐ;cÛ w@Ùëv)¨x Ú Ó½;ãÈæ·®Ã9°Î\Á»\_{>»V;ìþÊ¡ÓX0· 1APí AÃj lU1@´ýM!Øa ýÄ a1 « &APl&! &Xa
!!Ãk Bè0W ð
°¡86
866(nÅ pÐ0L á4Øí¶(G,¬Ð0BÚô"HDDf+=El2¶&0,§PDtÊrÐaÐaw± mPÚA!8ÐÄqÄDDDDDDDDDDIÐ2@BUÿ¯Þ¿®½zõúÿ¯Jìïi4 ÁÐvDDyM¨ì[.Y¥ òqKìH°%&ÓeºÛ+^lÉQ»úpUÊö\Ã;,¶°:ÁÔî¸84¦$÷[º<.²íÝ)\_%; ©ÚÊê¹ßyÙfvTôï¸;ëÒ®ì¢Z®¡i¯Ú÷û¯ÿZþº¨ÿÚþÿZÿÿ××ç×ÿÿÿÿßÒm,þD-ÿÿÿþÿà: æÆt{#¨¤¨eLf¶B,èùÙ&pÈ ñò yÆk",DΦDddA¼e9æ£"ùFDò\_ÉÈÌCèÏ#³èÙhDfhD© ò
¯úºý0¶ EÀg¨,paA ù dæfbçANÝ;R 4a0Aaf&á£AáfaC; F@dñ(fÆ Ê"x gA0A÷Af&; dñ±IàDS 2!äù¦pΣ4DìøüÐgh»Õºh&ÓAü'¦@ØEÜ#sÃs L(ABÙw %FÑÃ>IÕ ¨ÜT7 ¨
p"é£cFº70^60¡ ¦ &°3cTBaxÁ¨i ÖBlR¶% !apv ¨ØÑo0PgÁ¦pj±\43aj §áFÄ??ýdÿ¼ ý,ä'ÆtzX1þõt?Wÿ¾Nÿú×þþ»ëõþªßÿUðÿéoíÿ·½vþÿõÿÿú·ïõ¯®=/Ö¿ÿz÷Ö÷~¿ì^ñu¯oøõãïãÿ²> à¿Á
Óÿýkõdxÿí ¯¦GA|þ»ÿÿÿü%íþ¤¸ºà¿ÿþºÿöú_½z[Þ_ ý¯üGäá?d(¯Þû¯î#öÒþ+þ"¼íãÿÿý×J:]~ÿÔ/_¯ïÿ÷ý© _dR·Ñàí¯Ë~XòeäÆpD¯~ØDíp¿×ÖÿRÄúÿ&3 ¿ùc?ÿÿu [{¯á~X%'ÿëÿÿIëÍ^ø=/ícõ_×ëþ½ÿ ®¥ÿkúÿûÒý× Wÿÿõÿí¥¥é_ú ÿA?ÿÿÿûujÚHá·Ý Dòµ×þýïíý/õ×ÉWÿ~ÿþ×ý¿ýÒýzÿÿÿb_é}¥ïé/ý=ÿðÁ-/A~ ÚW ¾Çkÿß_ïßõÿ¿ý-ý/Þ¿ooýWÿÿëÿö»¤ÿ¿÷?¯ýõÿá^ôçWúI~×öëÿûßßÿæþ¾ë¿õ÷ö¿ÿ[wÿý~ßÿüûRüq_ÿÞ«ïÖ÷ÿÌ+J×ý¾ÿ¯öÿ¿ûÝ%ÿ¿û}×é]úOñ¿í¿úÿûVÿïÿÒWZþõìßÿõþÕëý(«óz÷¿û¿oý»¿¿ÿ~¿«§[¤Ý_´÷¶Ò»ü/¾ý¥ëþºjïí붶½öõúTö«ÞgG«ÓZl3trÙ; öÃt¡öÌ×Úße×V»}õÛ¤M®þý»«.gûí/Gá¤Ù÷¯ø{õì0E8a}ÙþhÃð¾¶4¶éÒïíýÓëþ6:6(< ídI;bSµíXïa¶ÒOÖ!0dsN¶;¶$È~iFÅ-±)ÚiÇ0[MUONícbáC§ÆÄ(i±«®%EWH »JÒÕ>C 5 #ÿJ¬/M>áim
l$á´Óa0ÁµPÁbÒNÐ aTÄ @Úz Aà SLSA ÃjA°ÚÄ @Áb paxkõjÁTÔPa£0¬0H0@!úv!XA4ºxÓz@Á Bj{L
á
± è²°Rì0k3 4!aM�DDD\DDDIñ ! !9¢épn!1 í¿jиªö¡ÄDDDDDDDDD ÇÒDQÑHDDGDDDf¤DDOÊÓA! zðûcB"7Õ/ÒÕ.«õ¥Öþ¿I/Òפ¿Ýÿ\_K×^ç|´¿\_Å[i{iA4i°Ágl&aáL â""""9MÄWÅ, dh3°Ì9©QFwñXìfÑ{;¼(S>É; þÿï]~?ÿ×þÿüìÎ'gÿëïÿóµSA<óa FùCͳٶ(¨9TqäÃ)ÙOµójG3A3t#¢5ähäRdèá³¢a@B¨!"0ÈÜQæÌêÄå2aáA 33>a0E¸dL Á04iXápa8gÌ Á0>A©0@ÌÈÌìÎÅH | "Ç ÍÁOA 2D!°Û44#du#¢6Ëy²$/'"O>¬Ol$:Xh0»jºè ØFæÕ 0²Í"ì3ÃL¤áN Í 0AÁm4mT#sÓ×Pª 0©.L Ba $BÓÜÕ0EãÓ¾j \tѯ¦ ØU ¨0²¡aôk ñA ùg gVOÛ»Û Éù.lÃ:M¤IÃÄ÷Ã'tH ÞñaU8Í' i8bW: ) Ø1°OpbWAõ°ÈìJÆÄ®iPxØhèpbW: %sI°Ætlaá ¤Ô#ÓF¶ª Öú7Qºý:kIØMTÞìÙ;ë¥~ðtáeþÒlðd¹\5N²wJÚNòVÒ¶Ný0tÛ'té'azsJªmü+T-§]°å¦Ø á&Á' Jí>OÒ}ÁÁÐ xoøî»ô4=6[ô÷¥¥ÿryz§dîw¯èw«hUèz®Ð¥më}:WZ¯Ný7V/MÓt>ëÓ÷ºO'§§dîü ÿú_¤ðõëºëÿë»þüUñúèî¾½=ýuï÷ï¥]þØþßÒ]?¿×^ãÓ¥ý ý6= ¥ÿï¾ëÓÿký~½o¨~ßÿÝuÿ¯_úÿúÿÿjo¿õëx~¿ßúÖ¿ÿïú·×ÿý_ö6:Ó´ÿÇî¾å:ú¯«©OÿÿÿÉÄÿÿÿý(ÿþDGÿúóA>?ýAW_ÿöýÿ{·ü±ÿPDqþ{ÿ÷òëýFl/úÿI×ïÿÿï¥ý~Áÿÿþ~«þ(_ÿÿò@µ¥ÿÈB¿ÿý~ÿòëá¿ÿôÿú_ååá¿ÿÿÿºùb:·ÿÕ/èòÄmB'ÿ¯ü'ÿÿ°à÷ÿÿÛ_Òÿü7ýý0ÒnÿÿWÿþþÿÿÿÖ¿ÿÿïÿKÿý´_ÿÿøE÷m/ý¶õÿÿúþ¿ÿ+Gý}±E8÷ÿ×ÿõ_ÿþÿ{tßÿä\ÿÿ菉¿ö!%þ¿k¯¥ÿ×àØ7ÿþߺÿÿÛ{ý^ô@î¿ûÿÿõ¯×ÿT·Û¯}ïý¿ÿÛÿe©/¯úoKÛKÿ:>ÿÿüWê½ï© OëºÚ:rý×ÿ{z_ýÿû×ÿ~Lo«×úZ_¥ßè%ý¾êÿÿÿÿÖÿû×ý×ûµ(õKÒù§uý}ÿº×ïÿô¿¿ïÿô»í-n¯ý"þïµ_ýºÒüòr.ßþÛ3ýÿµýÛT¿Òÿ_ úkûÿ¾ôµ÷Ií-´ªÿo_ª{J×mm+[I¿õ ¯ÚÚ]¯úÜ}kêiÝ}úiìÓºÓSïá¢S¾ÒῸg--´»[ûÈØgµa¡¡aÝ«uÿà -¤íþÛ?üí¥ #ø; HGpÁ&ÏZü3Ø_l#zÛGý¯Vþì-øA ÜaaA ÃPÒN(&|CA¥M¥±)þ ÷h Ð64
a4 $0 iS0ÅPI¦ÚLT0a\BjBbiø{®ÒØ®þÁÞð_Ú_M«~" ÁÑ «UûZ°µ8H086@µÃ6;Iáa
MXAh4:Ãá«NØI»LB8ajÐa¨p]0086ö¡Pa ¨0AC DîØ2;¼DDq¢RÄà k´ØQ$!:©¶6mXbÓ¬²X
Þ2ÇGL¨ÇL,5¯°¡§R¡&Å,S 8u£"""'N#""">!DDDDDDDDDEÄDDhCêNÿÒt¨qëÿ]T¾¿õ×¥®«ÿõJýú¯¡ÿÚ¤%¹híîM¾l4zÒTÁÊA Ø
bLIj å4·ê¦Ch¦ ÙÆwæJÙ_ Çf6æqYôùrªJ WÈ<ìÁJ©ùH88pé©+Uµl?æ£ÎôNîÖjd4ú>®Ðpá#á @ï¼µÿÒ;ó½|íO;I7À÷Øß×úL)ÛÿLíZ[ W]B¤úÿ5ÿª¸´¡6£®ºëzüoÿÓøøþ¾½¿øú_ÿÿÿz_ÿþÿÿÿú¯oÿü ³¡H± 9zÍYP¿ùìÔȼOùüd óùÈ+:Yª;ÅÑÕåQfELÔYÕçH\ü äYÁ0Aá$ ÑØ e jPÍÔd)çS#fu|íLÏ°L2Áa? ø ÁO0AÆÆC¢P@ò RD¨9fb ðAàÀæÈÔ)6GQÑêD èq'¬ò á Ü.è·£sÓ éBÛ4BaqDLÐg ©PG ÍäiÆÊ(A¢Á¦D&SF¶¸4BeaXð¢L& Ê6l&|h0¥9¸a4lBñ±álà0Ï Aðà;'Èz°È³M :MÄ«sóݺÏÎ0¨ á=0Ï T&Æ ¡5!BAB OuIÝ Ä¨hÐĨ i6%D ºH taRpÄðѱ±3°ªº66'ÕÐA´n¤êz ½=4mBhÜÑ°¶y ¦p°F¶"z
öÁ
ºNt§§ º'OÛ%EÉOm&ÁÃpb{i$ÛM¤t.ì&Ã,öÁÞAAÁpki6 ³ì1tc« ôá .~ 6pcJé´¯K%{]I°bWRm&¹ÐtnÃAÒo&/ªë£w[Wõ¸O½>ûÓÁáý º&äÝ<¡ÉÁîéÝ&á>×Té:ÛTé;/>¾ÓtÝ;Ò÷Oº['uqÿIõn+i×J ®aMé> 2Î0áÿ×J£O¸õøÿÿZMô?¿Éåuŧ§ÿ¯oÿtëÿ]_ÝcÿØãÓ¤ô5½ÿÓÓjôÝ7µNWOLk}~ì¼Oÿ®¯_ß·ÿÝ¿õÓëãö=;~¿ëÿu¯÷^±ßZþ¿¿×þÖõ¾¿ú×_±ñ("?ÿ«øé>Põü?ðZêõÿÿëïºõýÿÚ¯ý¯¯Ú~ ÿî½' vR+"¤Ù.ý¿ÿþÿõëÐõã%¿þzòOÙ î?ÿ¡ìG_ïõÿÿ¿Åy8_öB"ëüVø>#Ðÿ÷~+β ×ÿ¿¿ÿÐ_.¿»ø"yÿJÿ°þ«ÿ×àú_ßÿüOüÏü±"wÿaáÏÿ!èáÉÈÿðEÕ©92 dwëúè/ÿ_Ð_úýÿ3ÿÉ¢ckÿÿÿúÿëÿýýp}ÿòöf×,F¾þ¥
º@¶Ñ}ÿ×ÿ×ßݯ×ÿ_Ð:÷ÿë^÷ÿÿÿÿÿÿÿéäai{ÿ¥ßúë{ Aè/þºý}Ýwý/¯\_ðoëÿö÷\Çëÿÿÿ×oÿÛÿõÿÃôºëþ÷ÿ¬1 sQ¯ÿþ¿ÿ÷ÿÿ¯×ûþßÿþÿÚÿûßÿ~Ýmÿÿ§×ûöë=¾ßé~½ÜÕ-ëööþÿmu»µ¾¿^«×óMÿÿþü׿\_ÿëþýî¯ß¿ÿÿîúÿoýÖÿï÷KSÛëúGþÿþ×nÕ¶þít¾×ï£6ÿÿ÷»ÿ¿ßûÿ·\_ïnÒ»ý/þ·\_×»]+½ô½ë«#\_ÿ÷]ïo÷±(|ôħ·iuam+½Õ$ý¿ÿoo¤ïÿ«ikö]VÙº²ëÛ3ûn¬ºlÏû ý¾½
ÝnõeÓf ÍCA}¥Û}Ø^}´÷íµ÷5A
¢ Òî'|RÈ°%=>E¶È¹Øg/þDÆÛ3Z¾ýS»´´î-4Ø29¶ZÛ &ÁÛiFÁIìJzvȹQ\+lJ ÛbT8acÁm\½ºøgÇ
ø0A8fi[3VÃ.µjÖ ¡ ´ÅUa Ãv4ãNAà 10A ÃbAA(0Å1M(pl(a×´ Ô lBjBh1aPÃ#@á)aÅC D¨alB¶;cd°dw0@ÁIØL§S4ÚðMAÛ 0 @Á""!a Ðe 4B Ð °°YZPtc !ìAé~¸aSÄ&¡á¡ Ú 4 جDDDDDDR!@~"""""#B""""#ª1'Ø0B D0ÄDDFÝtb"""^êÝkõuéi+u¥ãm,\_ézQ´¢ %6!DrÄùbds%²]ëË=q²HÈÔí&ƹÎ髨C²¡ößR¼\i¦Þùi¬/Ü8pppYuðàÁ ¾E?}ïÓçcIò»ÎÄø´½.¦B?
TÞüw¿µüEFÚ\_ýküGKS\_ÿÿþþÝÿÿûÿ½ºë\ÙFAÄ2$Ñçùü³¯¯;JÈ+5¼äbR4ÈÕDjÛÍkÈäL fdxyÁ x!ù¶f2ò\_8DæN3¢<ÍfF|2èY�Á² 3A}(a3Hd1l¢$$FÇ·Ä<àÍ$F")äI ÓFæ¨ÜÕSCÝøMÞ¦²P5Lù ø33OmPj¡Ï £c "h#cE»63 á:M¤êèÐÑùº
p÷bUѱ« bx}áØa Þ¡bÒoAÒtpj+ÒmáZ.°i>pÄi7ÁY4 Ol$Û¸kWAÖ»¤ôàÌB¸bqàËCIÃnÜ0íêú¤àÆ ´æzWNÖ»Õn·Týé;'uè6ÉÆa§ ÁôÓ¤àáÁ+þ×éºnz¤¾Ý&é½¾ºîO<'KÕÿuß~÷ÿõôºÅö
zmnþõþÓOú×±éÇÇúöÇõëÛ×õñÿ¾±ÿúý´«ÿõ¿Kê÷®n»ý¯ÿ÷×רõÿÿ¿»¥õïß]ÿÿÿ_ÿý/¾¯ÿúZv=o_ýüúü/ïýúßÿëéðáÿû÷û¯ÇýE ý¦]Ð/þÁÁÿëøä)ûï¿ÉþßÔñ\rä#IkòÆ¿×÷ú ¯ÐD¿ýqÖÚ\
NBÓùà¿ù$ïÔ"wëëÁÏ÷û|Ѭ
ü# Ãûÿÿô¯ë0¾ëÿòc?ì0Ãÿ÷ü oÝÿïÐ/ÿöùôÿ,FX;\_}^»õ¿ß¥ÿÿ×·áÛÿ¯þRªÿÿJ¿×÷í«ÿþI?ÿÿû¯ûý/ÿú¿ÒÎ Çþ?ü?Õ}ÿõÿþÿ×ÿöý®\_ëþém×éwÿïààÿÿýÿ\_û®ÿý¿uÿïr;»ßwû}÷wÿûÕ¤ÿig¼êþßjÙ¯ûKÿ\_ûKö6¯ýÿÎ/ûW¯\_ëoÖ¯\_Kµû¾?ÿÚýõzX\_îí-}ÿ4{Û·ÿ}ýÑ£3°Ï}¥ï í}Û°¼4·«I¾Úÿÿí
°xiZ÷mö
³ [KýÛ3TÙïÛ3öÌþ½&cÛ &±§§°ÁUÝÛ åÃ$í(agþE=t/ûħ°Ám) Ä§Ü »ÚIúnÃáÜÛa¶ -¥im] 4AàÁ±Aj lSPb ´¢¤ú»KôõµµÅE5MC ±@ÓPÚbb1@´
P:¸D0h0©XMvlT Ð ÚpAÅE ÁÑ¢ a¥0è2c ÔÎPè0ÔÎ .Lýa8aÓ@Ìéà Êr3Ü5Gà LBÇÄDDDDFYX) Ákv¡Züq¡ÄDDDDDDDDDDDDDDE%¬q¡ÿºÿþ¯¯ékëý/úþ4¯õ;Iz\_ÒCó¾×Ýw4c0A¥L ÓU#¨Xå,2ºïÔÉÑh´Á^vªìqB!fwÌdn²S"lkX5gpó6F L ìhpåsH
3%r©EÑÄkÌFw©õÁü;Aß½øtÁ#;@B²) ýõû¸w;%Y\iÿñ¦>Öõÿü.¨ÔírVBôÇÿÿø4³±Þȯ+8ÿÿúþý~ª]/ÿÿë^ûéß
hÛÿÿ¾ÿ¯õ\_µêü ,WÿÿÖýúû¯·< Í >ÎÀ¤R4¨9A dìÑäLØÈc"¤SD&SÖPaòP°hÐÁ
9LA\4BgÉ"à0N3 ¡iA Nÿð}ØcK NíØ°Ü0 F¾¡w£w £` ¸L"Ç ,NíÞM£cÕØEä zl|\_ F·Ôh þ°kOEßO!¨d¾IÿþìvÃ8dºp88<zOÛ&ðìX@°q,áհŤÝ81¥ ¤àÆÉQ84 ÁÁý<pdþ¤Ý881+õ² O¹áPauèÍpçÍfͳ¿Î Ö.R#Ç4z(/ë§K§Úm'î'ßá>îú 佯8A¼'KdîÓÉßmîJàôßOÓ~]6¥¯Óºu ·§§ àgÄ
_A¡Ç!rAäùAP@PAç ø÷WÓ¥î¾þ-ûÓ¤ßé==7¶×Õè|zýzûþÇ/{»ýõë'¡jA¬+M'iüøDÑóhWëI?ÿÿZ_ý¯¿õ®=?ßõþµéþ¾¿ÿïßT«ñßVÃrݸD®íEh,XD±üX]R°<^H´ÿÿÛþ«K½¿ý~ÿ_þµá{êþ¿ø=öüã¿õþs2%´wJB[mÒ»# ¡Ô1¤û¿#¡ÿëú_wÄäD}¿ÿ_ÿ&ǦC ÿòó¨¿Ç!}Òÿõz}Úì'ðdEíjCFÌÌΣ$g 8f¡t?8ÍFLJÊ{ ÐTlk (EÀr 8AAá0Aå[0Î2q3cR@Èy²)£òb5y!hS cÍðÏóL"cxa4Xá GAE¸ò0GÅ ÐÍ8( Á ªt ØFú6MQ»^0%¨\5 ¢ xA$(!#3³38H]4ha0¸@hýü\ A¸´T}ÓAU5ô70L
ØÞt'I¹ê%d¦Òtn~®x°íBh×F
D(ðÑpõ± ¤Òl$ààáÁ&Ã#±+M²µÄ¬¤ààúA{i;]#svI´o½~¶îm÷Ò}éðcMIcdp«! p´tÁ>Ù¤ÉGtwd¢»4þpf.I¿Ò Öõªpö©:O¯úßO¯¥»òwÖ B®´êé;dIºz×\_N¥ÿiúÒ}õéûý.¾ÿÞ»Iɱn»z~¾Ç¯Çèß÷nIÕü>®¿Ü\_¡ÿÿ^·ú½±'¯^Wmh\_Ò]s"Ì® ºàïøz÷ÿÿõ×]}Õõý~õÿ¯ýïþþ¿þqÚ®ßõ®§iYÙBuÙ Ãÿÿd!5ÿûÁ¯ô½þ»c$ ñõÿñõȤ}ëÿáÕÒèë^¬4ûCÓÿü?è /ý~È@%ëýoß¿F?Òÿÿÿ¾Cϯÿïû© /¬}ä£ÿÒÃËOÿü?-;H"WëuýÿÿýG!v O¿.¿ÿ,GåøoßÿþXÛJðçoDêé®vkga^B½¿ûÿüöÒ@¾ûKÕ´\_¿½y9ÎÖ¿î¿ÿÿ÷é~ôý'ëõ ¿¿üÔ~Å%þÄó¨ÒÿÿþYÝWÿÿò÷ÿû¥j·ý£ÿ¡Å¾Ï'ýû Úÿà ü¯ïúûʱÿ¯ïõÿýýÿ_ÿûú¯ëoÛþþ+ûµëü÷î¯gÚ_Ý/þ¬þ¿·ÿ·ûÿ_ßÓqN¿Úú¥ÿ_ÿ·ßÿµvºéW÷ÿÝ÷û׿ÿ¿×ïÿþ¾ýU×Kã×b®èí©ª ×ý?Ófº_úӯ뵯ÚM®½Ú¿û_nÖ׿ûûÿz×µµ·«_nÓ ø f gA#dj¸Íó#Ç/8f³"¦DYE®ÕO´¿þáî×û®õ×Ûõׯò&{aì]{¶®Ã?þÃ?ȺÒöÒ?bPûA0Ïôöi³4ØFZ!8ØÁ4 Ï õT\aA@ìÌÌØL vxCq B)@#&)¨aAálTø%;þAy¾Dv¼J°keý³DéZMÅ0ôħÅ0Ði±@·AÛ $ÅCAµ)ø v lUûaRl1èÝbxhÝ kL0@
é£sGæ& <B60XA¢ apa\:r5lBA áPL4Û Ø pmE5 pØVÅ[Æ\ÐiZ6ØV ° j (a; $µ8lSMÚ¹Pa4Ðj@ÝA¨t º'aé°Nõ»JæM±¤ÒmZa< é:4""""L!ìA80N>®°Ó
®Ô6¡´6""""8Õ#^(ÄDDDö& ûn·_µï6ø5NÕÐpf=M1øuÒolDQÄDEñe(A
)Ê¡Ð\DDDE$¿ü^-êë¯Ð]OþëÖ¯ë¯\_ÿ\_×ß¡\_oñ§Ntª\_µ¿úºTí×\_ßÿÿÕC^ÇÿÔGÇå?¿õþ8ÿû#À»kÁû K®2ëà\_ý/ÿÿ?«ä'ÿ^4ü [ûÿi~ºÿD®´ÃG õ×þþµû¯ÿß¿øC·íË.åPéµmvïëÿ¯·]¾Ãýöý,êX6"6´½õý·V¾¿®¿¿ÁÃkúûëTÿ]#÷þë>Ýôû¯Û¬Wî÷^½ñþæÄb·vh¾×Û«K¿»#ö·íÛ]÷ôíp-ñ)ý°Î|0jEîÙbÏÿe׶f¬ºï¦ô¡:baC66¢¼ í;4Ø29Ç« åiDA\vµØ\¨ Òa á´bM!1 $W+)&ÆUDDDDhGg("" Á: ¤ù6Zp¥¸vd"èDFDDA¤Êå,7 ¹ÚÕK¿¹WÃáÕj¿ÿõÚÿµ¿Òøµ\_qZKZþëãë,Z´\_ÿC+
\6ÉÆuNdC,D¼KÎÖ ÈÞk2VdDOæD¢!ù/q®É\_¥ +Xá e &Í ÑÔ!t!"EÐ2(ddB:Å ÀÏ~3Æhaá ah 04f4:9¬Cùñsc$D3dIæÙ§QFqFOæãAÔ AÑ°Æ ¨Ñ 4k!BhÖÑp-ØEa3äcNa! È
se¸ÑQ²#G /!(M0¶ ¡|¨4XðMØFËíi]D B.fÆxgñ& 1 jÚMOðc áÃÝ't[´ ßßPÕL(L"à31"t&t xÚ?4¤Øb{ÅÛªNl1\Ø@Ú ÚAØVݾÄã½PA°ßÕBA¶Nôû¶É= îm&1IÒzpï¶aY1<4o£rAa0¿§}.ïé; IÃNAÖêÝj6è'%='H888pb©èVÒë/§Z§I¿^ºúm¦ÒxAN¤àÉs aëþôÝ7ºÝWÝÚî´¯®ë¾÷¦éÖÒ÷y+¥ez³Eÿÿôÿëi÷ZWNï]:NÓ1¡>ÇÒëß^×ß\_T¿KúÒõÚ+FÈ,äIu×õïëÃÇñÇÞ±éÖ¿zîßãü:ÿÿÿÔoîÿuÆ¿ÿýs ¹lï¢5ó0U
µÿýºnàëaÃ×ÿêÔVþ²)ð°wôÿÿýV°z;º÷ÿõ÷ëÔ5iþ¾ \_þÝ2ä%" GüïëÿKÁúùPCþ>ê?øàß××ý~?ÉiO>ýíÿÛ ÁÿäÀEÿßÿ>ßþØr¨#¬E~C >÷\_¨]ÿæB³ýdM´%õëÒëÛnL»oÿ¿\_ÿýßþÃrÄ~Le´¥÷Ë.ín½å×ÿåÈ.×I
ÈÜ1KÿûþAÃÿáÿÓý¿ùÚÙ}úX?ûñ_þUÖ¾ÿÿïþ>+áfþ_ѨÁ¶É:ÿUþÿ¥øÃöÿX;ø½ÿõzý¿ÿ×ÿëÿÐK÷ÿ8µ/pûÛý~¯ÿÿ©'ëõÏ'ýþzô¿½Õ¯ÿíÝÿÿ]ÚýºþêÔ÷û5îs¯¯·\_ÿßkþOßé7}íׯ{ÿ÷ÿ÷Ók
«\_ý^¶ºýÿþ׫«µ×ÿZßëá\_öõ¿o¾÷Wþí=}ÎÕ'Ò6Y>N^/2\ ÈÝçXGn¿M{Äà ±)ÿ§ &Ò×ÛÒZÛKõöÕ·ïúÚ8öþÝ$Ù¢ý³5ÃlÍnݶfµßø¶h[UÛ×ÖÍ ÐÁEÔ0L33á@Ú ÞMK¨ ò: ]I´l{¤Ø 8££/aLA "tNH0¨0½¤"""1"""""""""""#B"";ÓøÿÕz·×¿ÓÓÓ¥énÓ¤àÕ:]{Ó Òt°""""""":I¾×ÿý¼_¯ßÕêÆëÿ§ëÛ[º«µ´¾©VÁÿÿúKÿÿÿî¢Ú_i__?_úþ¿?ò@¾Þ ¯ÿëë_ê¿_ü?úþûC£ù¤Aûõü~N]ºþÿD×þºõÅü7,gðu¾}ÿ¿äÇ®÷õ"Üÿÿ°fd]þiZ×K2X¿ »í$ë× ]®ô~¿òÀ@¿ä䢥{KÿéÿÒöÿÿ]íkÿÿ"B ½za.ÚWßú¯½ýÿõ]¥®ßþ_û\®&ïÿÿk¯×ÿö½º^õÿá×ßlÈj AAÕÿë¬k÷õýõ~¯~¿ie_þ¾"1[û[_i°»×¯·úý±Zöéþïëýí&Ϲì0?ûGÚöðË«03_a}¸"?j¼5¿¾o#1A!CAµÞÚ U8ØàÁ{bSîØfî¢SýOs°Ïv¶±ÿÙþÄ&á§Ú(! ª P pÚ caAPð@á±H0¬'vÃ;#"v3ËbnfDDDFh0A Ða2c vËa
Mkµm-Õ Á¥-)Þ¤[Óµ\íLf7 ?DDDDAX@ è5U&L©LË## ~D]Qö^ ¦±÷ï_Òh(wþFÙ5[®ºÒæ@Iqki0¹W×ZXOø¸¤:¤µ¯¯ý}.¯ê¿ÿäÚõ¥û½ÿÅ)^½,Ùõ_ó)£¾DʵÿA$=ffgÄ 126F!VE2¿çjÎ ðÊ#dxº ÒED)0)Ô\£$HÍCÌ<£d|ÍC$ B2>ftGY²5 8P¡a0PTB<"ì3ÃL"áÈ¡PI°'fðÀ§AJ3o>!Ááñ9gVK¢<¸$ ¡45§D'à Ñð°ÏhÖáª!piF-ØBíôE088]ÜZ6IPA°è ÁUa #C80¨R a0 x"-&Ä8-ØØE¾&ªN¾èؤEä.n, z
X´:0JÓAjE» »A=&ÁætÒn='h Û¦ ê0J ¡!8ØÁT'§õA Ø0A\Òm'O8åÑ]:8dÏIÁí6 rG%Wjéß¨Ü ÝuµïN2QI½éJÿM#Y RN©7ª68¼0[¦ºa:ºOÓcJôémjÜ&׸? «V×Óá6ïþ]_N½7´òwÞK¶ =B ²QVÃ.ÁKíj·¯«ü~¬o¯¯Õ¶úuýú§^±Ìª3$GiY\5ÿû{ÿCãôô)]?´?×O^·'n×ÿéÿí×]}}v7t¿îáL¼gd£ßýWÿý¿WKmwq_üG§Û¯ÿÿúé꯯þ¿oü=z¿ÿ¾KÉÿÿÿüeÿ¯¾¿×i]}úëÿÐÿ( ÆU¥ÿÿúøÙOüu1á {Kÿþ¿ëÿÿÞºxô¸.ÿ_ýùÁ FÿÿýÿIkt¿y&Dwþ¹§ÿÿÿDÓεÿÈ#§íwþ#ÁoßúòëúòÅÿïëü²ï_3 üªÄòÿ¯ûH/ÿþXÏ׬|¬;ÿ¥¯ÿÿKôÿÿÿ®åXÝ^ýþ´!kKþÿÿº÷ëÿÕ×úáúßÿÿ¿×ýÿkþþö¿ú_ÿ¯ÿõo×ëÿ¥ýÿýéÚ_ÿÛ¿_×þëÿÿí»Iÿþ¿ÿî}ý[ÿýR¦ÿÿKKÿ¯ÿÿÿëþÝ[w¿_ÿÓ±Å?ëÿÿé}÷ÓëÝÿ÷Öîßýﯥ·_ÿîÿ^íoï¬oÿûúý¿Ôì/3Y ίÿí+ îÝ{x]´»^Ö(Õioßÿë¿ÿmµ¶ð~ézéwÚíÚ[_{Ú°q!IJ#bç8.tÎÑ0ÍFEO"¼TJw"ã´[Jû3VM½ZÙíÚ]íúéT}{¾Øg&%?bPù91)Ü3tm´¯¦Òä\þÃ=ÝWì3ßØPU \#phØgÉph=¦Ía¢!f@ìðÎ0A vF}ôgQJ3nPÜ Øa AÁ¶MC½8¥a¥" .ȧU %°Ïýû i[P@0A j áCi¨lSAà AÃØ4 0Ãb¡8liÛ¡0äïIÒ #cmhÐöA4am5@,0@©n D³m&åÆÖÎ )Ú¡5µaVìBAaC8Aa !1üBMV-±)ñ }öA0Ó´A
[¶bÒj¶LVáapê%¯ººpÄ àØ}êHé¹Ä®Ó XÒA´ Aݤޢ"" ûJ0ëb¨0¬0¨ mBø @êQÄDDDHhDDäXô»éºnü&ÁªoWcêiêÁ ^´Å"3oT¸(\8ç´O\ý×û==>Ú\_õÿ·Tþé;ßN²w\_ñª\_ê·×ÿëÿØúÕ¯ÿÚôýÐýWÿû#ë ¿îMÿÃÞ·õoVùÇizõêÒÿä8ÿïÿý�ö¿Gkÿ½uÿú]-Ë'zÿ뿯õñÿ«¯Ç¯|ðè/ÿu\_ÿ =µêX¯þêÿH\_øékü$ýw\_\_àÿ\_ÿî½|±1z\_ÿÃý¥û¯ÿæ£ßÿÿö×\_Îû ¥¿ìÖþÿÿßµëëêÒýý>+2µßÿKKÒïÿ<ë[oÛãt¾ð A×׿׫KcÝ/þIõõj?|E
uí~ÙÒl-´ºï¿í~êî»®¾ÔDvÅZ°¨4¢Sá >DÆÏú(¬$ÙzÛJÛàj lTUCQAÓMpÓ SÄ&6I6Ò¶·¬ä.a yc¨^ÐbA;8¦ªbÅt6( °¶ÈDv´¤ÝhÊÑÞb"#C1JgÚBòÜ !
Oá
êePÉtv3%qRc%Îdx¸ýDkF ó"ãÞ#¥êAÜÈ8áçòwýz[áÔ<Èk;\Ì«YùU]P?ê¾½¿z
¨OÿúÿÅ'ºþ¿ïú×ôëÿúÇ\_éþÇÿÿãùÕ/þºÿÿõiÅìº8D1çAùC ÿÎÖ yÐÃ5Ä¢\EN~r5².Î äXò}uA;0 f&a<2óC?æÆtΨdµ·GÑ\\_ ùÁ'a f2s@Ó0A(Cè£c:DZAé A´ £c,&zFÈñ&äQ²&B &F ÌÈÌÍ# R@BHÑó4fìÿæÎpæ Çæ¨0!Iá0©
n "$&±Xôl Ñ m<"l/ik F%u&ÐI6ME±w5+¥ß6Ô#DDDDGc ͨ öñ!æq+Dgî˨HX6 u¬888túúûWúîïëY7ûÛòcûúÿK¯¾Ò¿ÿýk¯ñQ×Í/øô¿ÿþ+ÿÕ/ÿÿ¯þdhFBäÚÙÿÿ¿øA¥a'aóc: Q~C#¡äè ²aýNÖE5¬d<ÜN<#lüQä1£1ÔR53ª9 Ô¬YFCG ,êE9åSEêMD&BaéAç y£ øºl\ÙD3Í¡.Ü©$432 ÔñááÝádPpÍ( Á¡#C0a ¤2x b""ÄðÒ
A×H#s é£d"áB~|&1ÚvxP8à¡4f®cA7Aµ.Ûë¥ aPÐEÃAª \ A ª
L"î¹\T'A¶:W(rÐO©X1á$ êèßFÇá Ó@%±alªJM¤I°d¹¤Ø1P0p}&ûjt%tÕ°bW4%e&Ð îßú¸½6¯'¸«tûÖ ³Òzê%sIߤjo\/Oìöù;梁¥OðºNÓ«{°IÕ»°zëéûÇ
Oÿ}itßëO¶éÒM·«ÄãÁåOaKïZÝýûßõN·M×®éuÓ½6[e$i:; >û}õÖÚÿô?·U¯~¿Ip×uþºýwN«ªzu½ëþ×ü®¬:Ù)ÙY×ë¿ß×áúÿôÞ\_¯ãÿúï×ÿþwÁÿÿ¾¾ºõõ¾¡NÓÿøý¸Êù}´·\_í××õøóXOþDuyÔI}ýÿãÿÚGÑ"y×ÿDöóã#¾?È¿ÿîQü
ýºA@áÿµèF&¯ÿÿýùF¿Ð» ½ÿùeûË\_}¿'Ò×ïmU´ÿ¶²þX8@¯åéÕÛö¿ÿ\_ùc?úתZGbÞKraiÿßÞá¿ÐUíÿý|&3ø¯ý/¿û[¤¬è/ûÿÿþ¿ú¯¯®¡z]ývô¿:Ú ÿ¿utKéa
ÞÚÒ¯ÿÖµô¶½ÿíÿÿõ¯ÿû·\_½mõø×ö×Kí+ö÷ïÎ׿¿ýêúï«oÿ\_ÿÿÿÿÿÿÿ÷÷¯ßïÿ¯ºÿô¿é?ßÿª»WFýzÒoÿ½õÿþ¿ÿíu´¿¼í¿ÿÿëº×¿Åm÷ýnÿÕî½?ÒïëK\_mnõó²!æÌÖ)~S®ÿÿì.ïß·§·v¿Õíëñý¥÷æ¥ôÛöô×Û ¾úõi;¥ +í~×mÌÉq$bÉ< !ñ ÑñOæÈÔ2B9:/{øýÚ EÎÃ9°Ïö½lº[°¿\_û´ÞÞ þAs³×°ÏjħüJÃ=µL0>ÖA-¥a&Òaaþħ°Ïì0EýÚ6C5¢A ÆÀýÞ¸Fº"unM4g 31Á0@Ó®gQXPà V86(<6 PÓXiGlvħÿîÚU¶vÌ×]Û8P ¦ Ø÷W %m(ÚQPPب
vl ê°¡$ð á<& ¤A°ªx!í&D (0l0¥Ç qì \0·MÚjAà¡Ià >ÇLýYe Ü0 i
iSW¡Ã©M6°Òa ðÁ8=á:»Ö µe7êÛSÉ OÒoI°cA6J¼ü¬i6""""""""NÁaCÐa¿Ò{©ØNÅ @ë9""""""""OK1B"" ̺5¥ßÓ¤ÿWÞ¸<&Ò¹+¥]²qÜЧZ²ÎKM¾" 0T&[ý¨IÇGß.ÿõwUÇßèw}餾wzt´¨DDFi]«êß×}_Zô¾¾?î?×¾¿¯&Õ ÿ_ÿ_£»Ø?ÿÿº_ÿ÷^©_$?ë]ÿïÿ©Qøùý5ÚþÿUJ¿á¶ë_ÿw§Vÿßå?ãªKuè/ýzÿ¯µºoùc2Äoäà¢ßNÿút¯¯÷ÿÿ¿ÝAÿÿë¡_@µ_È«¯¯ý¥ßÿîäaÿÿÿú_þYp´¾¿¿þë¿þÿÿÿ×ÿ¬BL õ÷b¿õc'Oÿß®ßÿÿ| N×ï:_ê»týæíïÝyÿ_êìDm}ºOöëúªÒ ý÷¿Ø_í{JpeÛë"v¶þ^ÙÛ3ÖÒÛ3Ò¿ÝÿaA 1U@´6ÒA %Ô0{ l0JÛI6Kd\ÜJw""gaá 2;ðw¶°Ä lSNÄhplP lP a ª Ô64 ·/QNÉd'"" !D,!0 e¡©!î[M'&edd" Íx)Ú2|(B"""#B"""" ÎÝB;KHÐé«ê[Ê«Ðj{¨tÔªRn'û£ÚGe« ÿº¥Kß]ÐA=i#.Êlí\ÿ¤êÖ1ge iz¯¨¨êãªá; «õÿëþ¿ú×ú_õÛKýô0Õ36k¢«ÿþ¬Ê2|¢óëK5#± úè³¥a3á A°@Î/6D1I¢(FËÈyÿú Ô& ð@ÈbðBHQÔRSA2¥²: ,ÉÅ)HÍQÙ EAh £h/0¸dAÁ,e@Á ~lC#BLíª=µ ª±ÐD.BaBÒí #[ £XhÅØga32A¢¨E»· ñü"Ç !#±èÝi¦&4 ¦ÕPÁ] ]´\4B!24Ï
°4ôl#T0ôµÑnѲbqé6y Ø\ Ø°·A j «6vAºo Á¤ÜÍ''ÈñÐ} ØbaÚ ´a
Ì#]D.l?¤ãZ('ª ´ ôÛ$ m&ÐAöÙ&Ã%ØM%Ù:N Jç"ºÕ°µÖßNTý¾pf?JØ m&ÒvH ),¥¤ØIØbwx¥ütXOO¸kéÒÒÒ°§ÞÖºWN !~'+ªñô¿úuúÛý'§õl4 &Òn']l ÷¯.¾¬}&éÿ¥}nÿ^¿ÖÒ̪ ï÷ý[þ¿¯¿^¿Ýtµ½z}/¬Wÿÿê¾?õï\_Øþßõîªdj ËSüÒ§þ¿ÿ\_õ¾ëõÿ¦ÿ¥Öµø\_÷õ¯ï×ýÿÿê¡<¥Ázrp¿ÿúÿë×ûÿþ?ÿÿÕ$µü 'ÿ½!ù@Où }ú<¢ýáËÕ¯ÿ_ÿÿÇÿñõÿçé/¯ëðD¯¯¬ þC'öwú[ÒGeK÷ýíkþêX·ÿÿÑ ýý+èþ©kµýÿÞXÖû# ÿ ˯å×õ ù ©Ifd
ú«×ÿÿþÿüëå;ÞLg_®½6éëþ¿¥ÿÿþ¿¢ê®ÖÿþÿÿûßÿÿUüª8ô¶!^ßÿ®ßÿø?ÿ¿úúÿê¿OuÒOÿýõ÷÷þÿýzÒõ /õÿîÿþ¿ÿÿïõÿ¿Ø_ï½÷«þº®ÿßþºUí~¿u»½×¯ùíïÿÛýõû_í/IïýkþÝ~þÿúÕÖ^¿¶¶ë÷oïúýý7÷íÚ#löQÙ@ÉI93¤U¯>ÿV×½]¥z¯}¶mÝw[mÝwÿJ:È÷
îµm»JïíõÿêÛS±rRä|ta 3PÌ(ï (Ô£$CG 6GRÎÔÎõLDtw½Ý¥" g¸;ù=׶f/¿¶×oé¶P!\ß"4¾¹nm&EÀÎ]ħ¯"Ý0g>äKvÙê%?UKFæ¸4B ôl G©xLàÏ .Ùò è04Ì"8CùñB ÁAA008l(t0ÅqLSPôm¸¶©ÈØg&ÏoPÈûgþà±MWPÚ86 ¦!¦)¦(8l(ap@äc Å lSPÃ9=¤é::A°Äñ é>
¤-Ól!f $la !pØÁPb"Bó:vjTL+NA8a B@Á±Ll4£5 Ô jPæ6]a{NAA }¯¦Ä&» Õ²w~Ò´ ±4é[Ø2Î0ñ+Ø9<É%Í XÒtM i8b{a$bwcB""""" !(AamxiÄ& áÄDD#[¡úÿÒÒz~O/Oá[§ ëË;úu¶9' uB"0@ÁXá 8áPMÿ®~?þ=õ×MÓ¿¤½=?Ýx'ÿÿ}[ú¯oý[ú×Ý>z½zz¤ÝªýwÿÿÿÿÚ8úÿÿý{ÿûÿÿÿÿõ¿íªßç×ÿåÿÿÿÿ½_ôº¯ô?××ÿïýëý}ß׬±ÝeÖXÉÚFI[_þõþºÿú¯¿þÒ¿ýkÅiú÷×ßÿÿ××ÿWþßUaëßïúé}÷_þôÚ¿ÿð¡_ïuî¿ïúû÷é7Ç·ÿãöõÝ×úékúUÿÓßýÿl3Ý®Û×ßÚV½·ßh&÷Û¿ÛúlTM´ÁI>DÏ $ħȴ¾JþD¦?°Ï¯²ý¯i
Å D.MA´6Hp¦©ØØÓØØ29öWw>[æI"¾#´DDÊÁaç SÊÚ°Ä+nw0öa Ä @Чp2ÜWäÑñ
+A1"0B! &SGËq¡C; éçf²¨QÈDD89^§
;+0{G\_èϼÓø8?;YFCªÊ¢ÒxBò~ ðfI^µêçod¯ÿ®¿º¿\_¯ÿÿ׿ÿ¯ÿ«õýzËL$AÙÒÿÿÿÿ M2¹@Èbä"8GR: cZ5¯×ç@L!8LØòaù"{!ä1²Èy¸ Sª8gÌ©2~pÊr%üÖa0T!»á0ï· ' Á füP!Fl}3©ùÆke[5å?Ñ 3jvÑêl´á"\"ì32AÈL"à32ahA¸ g0'a;$1;ÕìZ64 ÃFø@ÁSFæ(ÖFÆsÖÏND¸AEªBP@£>$\A°ômõaÛ¸OAª A¸´5»M=4nªO Ó°A°fnK4ôü&éÒ%sI´¤á é:AÝ Ã´Ñn4latkhÖð¡Á
¤é6 òD ÈÎ)[O¬t&äh±¥Æ6i½ÿA7N¾®¶ ÀÓ¤þÕÓ¤áÒpd³Q+6OA&Mè$>´µé°y<½7àÿÐpÓ'tºië ûÕIÙ8Z6Äï¾Å¦ñÖú齿õ÷Já6 º'§X/Òm&ô?¯ªºúý+«èU¾[/nIá?ý÷ÝSØõ¯Zc~¯¿N?Ú·ÿ^þ®£¿ÿÿûÿU×õíûÿý ^Ñ6SÙÚÎ üëkÃ÷î®Ýý¿Oø®¾ïõÿÿÿ_ÿÿýþµoL®S@ÁNÕHÈ\/òg'ÈAÙÿ§µûÿ×ÿÿ¤ëºÿþÿÿ×üÿüïi¾§ØDïøoÁ½üÿ!øT
ðo¿ûwÔX?ø¦êÿÿÿÿÿwüÿú_ù
áÁ }þߢúߨoûÃl@l?¯õ3ªñ!pqòaÿë÷ÿÿÿþ_ÿ\øN¿ô¿Î£ô°Áÿðáå·÷ÿ OüDdul?ßÿÿúÿþ¿è/ý}ÿ׿ëü?Öu?Á¿¯("å@ÉÆ¿ÿ ¿ðÜE§,vßë¿ÿªÿ¿ýµþ¿þ×ÿèÿZ}¼?ýÁÁ·ëÿKÿlÔ.ï·¯ÿ]_zÿÿ×þºúÿÿóûÿg7ú7¿Õ©´æÛ¿ÿÿ¸?í÷ßÿû¯_ÿúßÿoï¿oÿõýÿ×®ÿõz^÷:ù]Þü¤MþÛý+µÿõ×î¿ï××ô½^ýs°¸³£ý°¾¥öÖ.®Ò½þ¯úÛ¯þÿï¦ïlöûnzûÖÐVî½{i}é{}×uÝ@ì2¯'ÍÐB6GÌé335 è-24f¶KWEħöÈ®Òö'Ý[ik"TA:ZZ¶gº\_º\_þ}¿¶ößÈÒnm$öEÿ $ħȼÛIoWKö"ÇwÚLíØg¿m,&0[62 A
FfIL1Ã3 \Ì2ùöfdyÁ "6)Fl\Ó!f6¡¢8a AÃJJ¢¡¤J M+Jí$ù믳ý¤ÙlÍl5JÐbÄ @)¨:6! a@B8a c ´ ÄðÒI§a¡hZhi @Ú¦¨Ø5¼(OØM ðÓ:7} F˧iêyF¶"""""&xjc2°@ pí¯jØA°b6àÁ(/ÿýi:·M7L'Û Û«¤é1»Ó~ÚÐl0ÌÌïA5AÄG0ANb¢/OØôÚõuí>\_ëiAÕý{ôߤéx÷ûþ}¯þÿ±¯Åýñzn½ýßX×àÿ×ÿÝ^¾×\_ÿ´ºz}TFø/ò\_×ÿZÿöEVÿð\_ð¿ûÒ\<þÃÿÿßÿø8ëòÚD'ÿäôþÿÿÿÿaÉ9Da×%ðÇÿÔ'Iz×ÍCýÿÿÿðxCqúûÐ@ÿì"úýÁÿ¿ýÿ:ÿ\_Ú××륯÷ÿoýþêü=ÿ¯ïI׺V>¿/ûëÿ߯<øDuõþºéuÿWµÿ¯ûÿ[õÿ«m{K\_¿ÿ]ïÇa~Ì-mÒÿû¯ï{CÓµö\ÖÒ´©}lJw4]sm-?öÒÿIa¬5°¿ál%i^Ý
µ pÚB 8ll4 Ám$í8a(0J%=Oݪu»Q ¶Î%<È'- H0D4RǾLBL'iݱ p(1A
µ0v8T0cj ËQN;rÊAK"'Ø0@ÃA"N¨0¥Î¦uáò4ÎæÁJÞ¥¾-D tK¨QDqÇ÷MNÕs!0|?ýsÌ¿Ñö²ÚUþiï×ÿ¿®.b´Fï\_ƺêÁ/tÇ]}/¿ñUaz^ÿñÿÇ×ý}}cõ×ÿÐDtvÿþùÚ\SNdugS:çdÿ¯+²¹¨§PCü)8üã)Ì
y«ô¹Ú£@ a 0Á(f!±HÛ3Ò6) R6ÏBFùÖ2 CÑÈÎK>üÙR¨H& »#&~aB36!h'Æ\F £$ CG G ÔGR5¼èþ&[6º& ¬ 桪5°TA%»e8Ða0 naaÒ
D&Paá rÞH4ѱ£t0
£e 6 !<ÖÁ©qÓ4DxDÇ ùAaat:#t1<4té:67PÓ Pm´hhÙíabqÚa6PA´kZ¨!z
h ØEÅ¢"#Né¤'VRØ13µ -4ka= ªj60¶xF°¢'$ì éZN«ÓÓ'Ð1¤áAè6 k1+´á&áV2éG|¤ðIé´ëKJúm+2Î3 ] ðÒlØ1=Òm á'FèL jºm'Jÿé¸AÛäóÝ7üþnIÒõÜz¤ Bm+§ñÒÒ÷é÷ªwI¹;Ó]ÌzO<&Òmtzl$Ü6 zÿÿñÖ
zqúzÇýÅÒõ¯ßs²FK¿ÛIñ_{ÿèzl[é[®ýué hVä£þÔ~燐¿½ýÕzB\|AAKp&v^_õ_ÿýõýÿÿ^ºm/§h~ÿ"§÷]þáÿÖº[ë ÿ^8pSºÆM.õñÿaký¾½{ÿÿ[õÿ[ÿQ²Á}xë4 Ó"(ù Ot«ëøʱÁ
àáÌ¡%íþ;ÿêü '×õÙÒà ½ªßÿä¢"Ô ;íô"8"Wø'®¿¯ü"ó!qyYNúh¥ÿ,òCK\_®¼ý\_ïB#ãëÞ¿yØ:ú\Î~ü _ËK¯¯òÆÁmþ¥Gavwëô½Ý/ÿ
÷ßHíú×"zcùuÿÿöʲÒÿßÒü§é?륤þÈbú®ÎÉ'Uý×ÿþÿÿh-}wùu_ÿýø8:ûiUø?ô·úÿÿéþ¿þ) í«ýÿÿýuéoþ¾ßÿÿýï·ÿþÿ÷ýþÿÿÃûõÖ¿ý®÷ÿÿëÿÿõÿ{5,Úÿb»ÒײuÿöêÒ\_ÝýÿúÚÅväG^Ý~ëÿ\_þÿ½îßÿû·Úý~þµÿÿWýzúÿ½©õç¼êׯ¯°¬5}»Oûÿêÿµû¯ûÛÿíWí&ìÑZõanV¿uý¬0õ¥òÿïï\_ÁÁDΣ:#k" ó¨ó×gµìN>%;®Ï¦ëOÛ3WªÚUþ·º\_íÝÚïMÓØ2æÃ0Øam+ 1)û[ å %¯ìJ{£ öÚPÒB®\EîÃ=ÚÚZz5¹eÃ&[fAdæh0ñ x@ÁE ÐFfä<èGAaCjÝ66)SOîÛ«×±ìÿ½³4ÙØVØkß LS
5 <6\!0¡ àA8I´ plUUBÐvÆÈ°í(l¨ Ú¦q ¡ülªl Ò!:a\xh9çÉ0pÐ2:6Ak C ¦ b l0 Á;b¯NI;[# $ÄîØfÜJwm¤½ aOµ5ÝÄ$ÖÚ¶
Ta-48aÔ8¨A½]ByéÁXá6 JǤìOmÚ \ª a¼hÙzD=Á
ï±IéÃa[Ah0à SCbPÁ6Ò¸#6ÎSrg0 I°ulI]ôàÌtðÒbÒm+K$1I>IÄDDDDDDDDDD0Hê
a0°jÐlB±ÿZý뫦û龺u½éÃ['t¬¤Ý80C\_}V»×ñëñ§ÿJëo¡\_×Oû«ÿ\_ýûi\_î¯ÿþ5ÿ£ï:û"ÿÿ×úZÿ®õÿ!ÄßñÖ üí¥úü"4ÍWþY\_ùeø"yü0ÿÿA¢ÿéAÿþ¿þøuÞ¿RYú&\_ÉrÞ¼%ÿÿþùOÚÿÿûýЫÒÿ¯ÿü:¥ÿ·ýÿö è}+{úÿ§÷Ký¿¯ýÃÄuîõ¶ÿÿ¿Jÿÿïïò}¥¾ÝwúÿÕÔW÷zþÿ.îDæìÿ}0ÿ]jÚO~Ù¢oïÛÒ4±A SM¥±±Z|pÒ©û + åò'mry°Ïw¦vJ¥rH³ÁaÐbA8l(j)´Å1@Á´cÓcÎÔÈìDê¤\ÎÅY\¼dDAC" MBp /i4à !R$v GELíR;ê¡LfJÌDDGf¨!v)xS°¡#D8> ."ï¦S°íø:DÜ^Éoûø8:Ì$«ú4²ezázkÿ¾«
Çׯñó²oÿÿõ¯_ÿ[Uþ¾½/ÿÿÿ¥ÿÿÿÿÿ×Ë$ËÎÔò5öwËÿçd HG"þl£"£ój"!ShE ÌòA?U¥"d àC8ÊT>3ÿ2ÈdCGÔÉ|\_Ñ£@g äCÀhLÐa8pÈú Ò ,pdìÃDL6ðL&U#A²vH6MÚaì"»·@PþËA >A ! #\ÙFFÆR\NAð\6'70
¸L#sT0ÁU-BDä¡Ø] à§
ÆJÌ uFʦè Øa(0¾6Q¸5L"ñ
0DÇ]³ÃAÈ fùt+%~F4#D®i6r1ÔO&ÃNår ØItwTÚ8ú $ÝñÌÝX3A´ é:N÷@þÄöÒl1<4:7OL#s·¨%ÿzpi+¡§][û%d¢ª®z Ç¥¤ãJ ÀIÃIré÷ëIÛô¾õÿ§]¶t«ÉsÐJæh pÐmx÷«ý==~é= ·B)è-/ÿ}'«+¢ü|uô¿\_´¾ë¦ÕéÒzºpjµNØbS÷õÿýþßWÝVÒ]|=JÇ×ÜíH\(ìv"\kÿëQ÷ÿ×\_z¿]^¾zúÿ×þºÿ¯úõ×Uú]ý a¤Á%yߢ>TϬ. öý£½/ëúÖ¾¿ÿÒÖªÈûÿÇÿÈê¿UÖµô¾ÿ\á á5 EÄ: ,I"é%c59/Íä/ÿQýÿÁ(}\_ÿøý8ëï×ÿÿýÞ¿TµõÙú=t£SÖL:xDóO>ÿh´¿ÿþ#¿ÿý¸ûÍÛ#×ÿËÿËõ]/®¿Ø~¾NA´Û¦hØÿÐ^þÒöX¿t¾¾X'eh¿ÿúëùÚÐÿïuÿõýwþ¾ºÒ ß¿¥$ëH¾isû×m&뮽/úþÒ¿ûÿ¡õïõÓ$ÿ·ÿÿÿÿ·®/®´ªJ ÿÒéa6i?ðí³ÿúïm÷K\_þ¿ÿßüoÿj¿÷[ÿýkÒ\_¬¡ø7÷é©Rûÿߥ¥¥]ëúÿ÷ºWïÿ÷}ÿÿ\_óµ5ßÿþÛúíÞºõ¼-ÿÒKQúû§×ß}+ýoíXî8ÿµ½'ÿÿþïþÿo\_÷¯÷{§´/ë¥÷ZÝù þ¿×whÃúûÿõõm~Þ×}þþÖëþßÿWþß»uo¾ëêÒS 1Ò]ieüûVëR7"YÏù dùÌÐB gõ&äNdu"%¤>ÿ×¾ # +VÜ35ëðÖí'öÕÿô¿Ú°ÎWÈäÙ<ÊÒ{°Ïý6¸KJ8¥ÒB?µ¿ÁUp¡4Eã; « '@a 6!ÈãPA³qOÒÛ\èKøô Ãb®¢T6'w¬vÃv½6%a¡ ÿàÚ\_{Ýþ LP á´Ø$x@á±M¥qI«a+
ØJ¡U¢S¶½>H#vhUÐA¨FêZ¤¡t"añª£>µ¯¬!¥év°ÓPÃPè ab« aÅ! NØí¥Ürlö½ /h1 6áXa:biª®ªª)¨0Æ8=ëZ Áà´wª È>ªò¹a¤áUú÷èתÔDDD00
Íh° ÿjà ÐAÃj @Ãh4 c """"$J±(C,& !0ФLã¥TõIn ¸èª¿ Gu^ªàÍ=£ºTµ¼%ºXA }2@BNÎè5»^A¨DRô©Gk¤Ð¥ã¤+CUõ\§S1;ôê83Hq®ë¡UÿJ¿¯\PÖétè?¬]\ÿê©}.õÒé1(yÔûõåsõéuKZéuÿÿÕRõôÖû¥ø¿¥ëUׯé%I.©%\_êIC ï׿®=k¥×\_\_¤½ÿ¤+îïëþª´½}WýWIWKT«ïÿ½.Þª¿ú\_¥ôºÿþîÿ\_ëEÖ¯^¥®úÒTøÿý%¶ºú¯¯Hãõ×E×}$òµ"ëõΫÒùCÒú\êýu@ú\_U£¼êK\_ÿ§ë¶µ¥ûJÒõ\%eÖDÙÕz[ÕÖêºJ ©]uÚöºJDca\TS®ZZi[ /ýt¾»K/ª ¥¾Ú ú¾+µQLVÚ[ "g \´}¥ÅEV´¯S¾Òm%å¸v, aBa4¬Bê«b £¦¡R A´jö $Ú]W-ÉìHÄDEá, hTÔ&TXUàPÚLRJ¤lîRDE! P°àæZßêWý?I"¾©£\ìú^JKúíVKÒÿX¯T«ÿúHEëçkKýu¿ CÐBAÑÃóËÁ$& Ï(ʳ4Fç_ú " "Ç ù Ó!? ê2Fl¼ã)̪ Ìò>i 5 ÔSAJþqÌ"5ä|ëQù ¤ Aµ CXkA[x z$ãlÌRؤ2!åÜf2È£j.\ªhÙD&3ã Ñn-ØDÇDð0ACzt&Á\è0j0 éFÆK(F°Î>GÍ0 0A²:u8ÉA Oxi ¼m taB!îEØjEÄ"áë_ÕéÁ#êKNÚñ ©ÓF·
Tn¦ ¨ÖÈBA\F4O»àË¢pÉu'§IÒJ%[A6}FÆMÅïǺýV§cë4¬î4¡Ô A8EÜ ´l¯û_Õ7N½]{ËVYutÓÓ~Æëßî«^ý[º·];²QúmÇ«Éý&ÐAÒtëñ±éº×½ozzt¿]^ ?öÿ맮ºlo;ÖÐjWK[IËpoÿ{þwQ§ýþ^ú¹Ù(I 2Çþôµþþÿúwûzÿ´º;2f¸ BS1AJL3S"èèR$d+!ÿò(p^ýXkëÚøðÝï߿xúþÒ¯ÃZúëÅ{TÕÄà AN(Á.ÿÿR@^ª9ïðÍAX=ÃrÄWÒzÖA]þÅy O__O]aúv}64]ÛxoÿÃ%µaÿõF¼:X6P?Öû]åÿÉä9(ôÃ)ñü|l2eg÷ÉÆ´i66m7\íÖÐ/¾LCú_ªá»®÷ô¿ÿ_®X~y1_ Ø=U'N7ÎZN¨·össoûõïô¾·þþåÐ,<³a_pïûí¥ÿ_ÿÒ_h×˯,C©Çõûß¾ÚPZO]¿êÿì¯ü¤?¯ý,ô§·6¿ø¯ÿ¿ïþÁ×ÿü8¥£¿ÑÇÕúZøþ½[ÛÿT}=ûß÷¨?µëïýûë÷Ûë__ûweécíïýÖ·ûã_÷ý¿÷fú¯«zsÉïk¯Mÿ^ÿªý~ÿðéIc×ýýw®ßݯÿ
^Ûu¿·½mßÚV¬3ßpÈûûÿ}=ÝÖ¿î»ý»f¯JÒý¼éú]þ?íØ_ú» ÕßW_ÕݯÙOZI ìtÖ åìÛ´Ó ß÷ïÒKKÐ!ÿÿí¥ß3YvSdKäb#¦A8Ní>Ó "@þØg6îÕ´¸dÛ3Û°ºM+ÅD&x pØôö°fè§ÚM®ÙöÛ½$½Ö¼"Ýäq>õþêaUAN «ºm0¡ h8ÕC0i'Û 0dw»0U DZjPéà ;h á!ÅD(Pa ( 6 =raØg¾«KI[T5ÿ°a%ý¸5M(A 6pÃü¬Nh5øaS6±LS Pb""""""!dÝ i¸a8¦¡ÅÆÇ õ®{µµ®øªø°bV4aîèÎM 3Ôa B@Ê0a5´¨Û "6C paBK^ÚI<ÞÛÝ}Ô6·Ói7 ÒmØb7O¸ñ:#«ô§$½.pà ðÝ}?úôôêßñ½Ãöħ±;®xE¸i¯OMý:^?´ÐÒíÃT×ìAKëÿ_ÿ»èH®$Zú]aPà R ×ßýgþðe>º2Kãxû¥ÿ¿þü
ïþ/é~t¾¹bûýýü0þMÓ©Ò¶ë^¾ß¯ö£¯ÇµúÒoª^ë×ëþúÿÉOÝVëRiW¥è/ý÷ÿ®î·Jý$õõÿµ®¿ï87éVh ŵºH%z ´£O÷ïÿI×ëÃIoØ0E´Iz]" R÷Ói~×ËIh9 yt0AêÄ+aA:IWý{îêøi:V®,Î)(rqbÛ«ÁÅÊ - í°wiZU;N1(Å [.áÉÓ´O $ì1 áÓ
W&Ĭ!ÄDDhDDDAç.êT}ãê ¨#KMB·]NÃý!Ýk§Õ=/u~µiH§º_õmµÿ¿ÓuÎÉD$ êFÑtaJSÞ¿º]ØAT#DS BüÙã"üJª='Fº¨aH°3ãTDøAÁ0@Éh! ! Rn6dèêjF¢Moé=ÃÒ¤tza\gƨÖÑ® Ì0aÙ;$$0@ȸç h?ÖÉG ËB°Ç#@ J·¨A6p ªAì"(ÜÁ Eà <&1î/]-'dî PtÕ´´¬é¨ mC¤ØbaÚ6"é£X{¿îonNõï¼ÛKÁM°@é7KÃIð×ý^¿úz[h~ºm+ ±"È6vBkÛIÿ÷ÿþÆ´ïIÿ}éÕÌneqì3A¿ÿþ?õ¿µ~µõúÓ-ÖFS«\_S¶í:Kÿªÿ!&7èÖ=ÿöþ¿°áÜÈdv¶4»½ÿ÷+\_!öԵþ¿ôÓÝæK%ý~¿ Ë/ðÿÿÿ§ÿª'ý¿¿õo¯$Çî ÿþÿùs²}¯ÿz[×ÿþ ÛßUÿÿÿ-Ëãÿ]ý÷úötÕ¿úÿÿ÷X^»KþívÊuû×ëÿûXÿßTõoK½ïÒ·ÿ¿ýÎÉèD2!þ jMí´~íá¶ÌÕ«{ë«kï\_øAA Î'Íè!8GQHy! èGsyÖóèûV @Ø ¸am&ï¨a8duÝ}¥ÿhýÜѲñ&pB-Ù0pÏñAa£hf<2@k6ÈèðΣ0(FãANÕØjb LP@Ýßm¤ßÃ#þÚºN¤ÐN¦ÕQº¹ 0ÐEàaAp:hÖÉ2Q· !á#æ !H¾uDtGDèÅ#æGá¬^ÐùnB $ö6 ØpmA±§°f\~éü82\é¤1+Mºt}¤aÁðXA6¡PA°ß¨Fæq;(P5q¡B[)@Á±úN¿ð·iÒmé]}?w}^KÞàÅ$ÚAÁÁÃh yhg¬DDAl! cö8ÿ¿ÓÿNý$ßZON²v×KÝä¡®È-ä3q^ï{uþÇ»_ß׫^_^´¸["Ñ«ÿÃþþ×礷ýVûÒõïÖÚò ¦9 ÁÈ+9ìã§\²¨» £³!)ýYÙb/ßþÿ\_¯ýÕ|DDDDil\± ûtÁÿ¯û×ÿÿÇúëó ]p|6öÒoïðܲ·V¿ú×Èq¼ÿÿBSÉ2$ñD\þº½ÿÔ²º¾ÿ.¿«|8Ãë¯5¥kÿ÷ÿþ¿üAY ·³WOüå¾ýÿÒûuÿßûÿý÷ÿÒý+USën6\ÕÿO¯·ÿ\_ã¥z\_úýý?umoZ÷ZÞÿÿv¿aþÒ¿»í&û\_ºjû\_]&ûmuöÛðLT4%NÑ ´¬ JÁ?¦ÏþéX]ûI³4ÚÚûv¶)¨
@á¼0Bl$Ä(80í¥»V+PÁm71({§vzbPù6Xd] aÓ·AMqAÓLBj mAÐbb ¨0 aC"&d4DDEñÁÁ }Ð0B%HÓ )Aà ü0¸|).Ï2\Ñ2 S O""#2Èî÷;¥DCò[Õý ¦@×IÐ:Uÿ_ëEãî¿ÿëþ'5¯}ßôºoïÕµÿ¯ß_¥¯é|íPL£7Ê2ÿêzçÎ?F¶Ñ¢zòO3dìÎ0ÌÅ'?8Íw¥U¤ïýl$H8CU\BáM0l"á\x <ÐRFlAi,Um/Ûb Ó×#:ñtm´!m6=3äaDzßÈaÒÍhó%ñ\dí;¤àÌ{4ÚI°bVV'ºMán-)%éa|ffÿ¡]©<¥öÓÌpIøM'Ðíú6Rl ØÿÐÿø×OÓ÷ïM¥×iaÁíØ>¿þÓ[ÿ×ǧW5ßÒt®vrÝØ]½+þýµ×íI×K×kO¥~u26Îÿÿ¨8¦ 8ÞÃ7ZM¥ò.vÌ×kíª ´ßo4×ëJÁRvyDn=fÈyг¨È«"ιDMÑÿì Õ¬4ÓQ"ÐTtà @áëp¶Ä§®Ä§}¥õ¶:¶ÒÒ4!xÜ-Ú70T!¸ð83ÃG-ÃDI8ffÌP!;4 Rh! &P¨ódj(6C1LWmb :°ÒI7¸ý]jh ÒM&
ÕüÚa¨EÄ#sP[§L"ì4kBaqÖѲáq 8å ?¨¥j qQ ]h é6Éügy.àò:&Ð$ h t k?0pb{hÜA´
AH6+¦¹ÄDDDDD0B ¨OA¯MaBz]ÝäóÓî n M²QI×doc§ØNtKe]¤Üa®«qpΫ¦ý&é×]oúºv^ïZ\_î.ºV TõÜqÒþ¾ÿ¯î¿¼q¾]&ñïßëVëëëêÿ¾úú¿õÿÐýö+þßÿÿÿÿ×ým|ýx>éÿÿ\_ýÿÿò\_ÿd!ìøÿýþÿõ¯,KõÉà\_ÿØ{ÿëÿÿÿÿÿÿ^zéûjßÿÉ´¾Ûûþ¿ÿ¿ÿ^«òçQþ¿\_¯Þµ½ô¿þ»ß[\_{Áò1ÿýõ»«ÿõÿß¿{ÿÞ'ý/þiîvÿ¥Þëvëé\÷ÿïûÓAÙu}¿ÿÝZí÷]ßzïWkÿO\_÷´2ó /3ØJ0ùÿm-¶ÌþË«]vÛ«ï[U¿nÐbh0b¡µi+ w÷l3¥i2,:n¶EÏgùI®$¦a·jÓ5t ¦(6¡C pà ¤cªaD ÈAÛ0S ¤°S@Ît®E²¬b"""""F!3Í4sÄû1Q4Ðb¤á©
EÀh<"DB; ¤DDDDDDDDD[f¡Ò¨&Óa0hp3©ØAÏíôlt/JäÊ-]A<¨D4GdtGÎ",¶Fó¨À¤\Rà
8fdx#äh§!tÐqtA: fç$ é²Ó £ 40DJȶ""$@$Eb! b 8´i¦õ׿é+§h¸¢ß5EWÃôö8>ÿ}~ßôúI#§ñǨN½\_õá÷{¥ý×WT¶Õ¾¾©$±¿Aï÷wÿÿ;%>¿ê-ÿÿþý>û[¾ÿµáAÿ¤+[cÿúI®wwºûÚPD8××¥ú$& üW«ùÇÿýøMÒNÞY½-Uþ:36A»9RîCñXQCùÔ<"²¦B$"$þ
ká% íá! "ì\XE¸g&fx@Èb4ÉÄá?°¨7t§\_þ-é=©( Òé è&ê¸è Ú z &pÍ"c½ý&ýÿ￯o¯¥°ÃXý¤Ó0Ýí%IÁÁ=Áï½( ôßÒ]×÷ÿõ@ÿýÇBáÕô\_ÕþÓ777MîOÿ¾ÿûþÚ\_ñÎÖEÐJÖ\zÒÿöK^½§ûÿåïÿúõýÿÿïu;U"ï\0UuZÚKÿ~ý^ããY|gn3%þ£þ¿ÿÿܤ®»q¥©ý/ý÷÷½2Émó6Kÿôªô´·KúÖôºÒ /Iÿ½qô¿çAÁÛØ\_¨û ¯ÿfó5 ºÖëT©\é\_N Oþ¼GÿÉÈZ)¹Y®ÿÅHfÏüWë¬D? ¿[tº¯ÖÒô¿¥Ëÿðä0ðÃV½úÿê®ÉÁ,zûÿÒ ´·¥Õ:^»ÿÿ¿^Ã{t+\_ý+ÅGÚÿÿO]#ªÒ\_]+ë\_T¿×ý,êYÐ{ÿôC?þ¿ý--][éÒ¦ÒõÍ-ÿÿø|\_ëü5º«z÷ï±)ì1ézKzæk¯÷×é}gÓçÖvReBÿ¿îª¯ÿýCô\4·K¾Akat~¿UõpL ÍDÐRÛÏÆóDÈê2,2]2KÿÿÿÝ߯ûö¥ ¨a$TïXîÖ}Ö÷ý¯Ú¿øTÁT"í "ݪ £A; àC=HÙ Á 4Å!g@Ét#[ïïAÏÿ¸]rrÇ<úQÐ3¡IÅV+A=Z¸ M; %°Ï}öýt¯H:4=h7¦
P0 ¨Ù c ª>6²ãF°Ï ð¨·oè¡4ÌÁ9<b¼W4«ß" þEÿÐ2Ý5j ]&©¡Zq)÷ìB i6 JêM[HØÒpèÖõ 6@Ò ½ Ô0A Eä"¡Q±¢òvÖ÷ÿÃëwÚÿ\_¡´É¾ÛH0@áÐ pÚ v¦äîÓk²w9:TôìB ¾ é6/é àôè áÁÒtØFþÿÿö
Èä!¯Öú!Á 38[°U>;tõÐÿM®Órw«n´¯}öé'ÒtýëVÒ}ÿ"¿½ßøØ ""g ãú^\_ÛDZüWý\_ÿô·ô¿ýëWÿ·ÿý¿ï÷é>#ÿ«ÿéwO^éö>^£õÿÿþ÷Çõÿÿß²:\_}Xýý{zý¨ý߯þëÿûïµÿýýתî@I+øúöþDG¯Ùàÿÿï÷ÿÿýêÿõ¿!ÈúIì?ö×ëØ{¯ðÈY?×í}uOÿÿ}õ÷õ,G| [Ë/¯¿h÷×·ûpÿÿmëÿÚÿ·Ú_ÿö,/þºuá²?é_ÿ å×ð¿kþCùãÿýúÿ_ÿKëÛ÷üê[¸7ø6P?¤zà ï¿üVA¨
WïõÒ¿z÷ÿëÿ ý~Öÿ½ý+ÿÚÿém׿ڮy>ºÖºOëê÷{¿õþýoïÿï>¿Ò¿ïþ®ÿÚÛ«Üöïç·6ï¯}V»õÿÿp]¤ÃI¿k^-¿z×øÛUû×ú}kßú_ÿkÞ£°ÚPaOa0ýaÙkm(kim6Òn×«í¦½Pÿz§íïþ¹vKºBTSÔB¸a(¤Øõ´î×SûJ¹Cítg¶%ZðýioÕúä(ý঱H°Î",ñæ+pA
Óq A° aL(q<5Pbª!ALSPø/kÝà·w}ßj àýÄFNÈ(ñq'æ-8! ! !h0GÝ gÔB 5m0B"""""""#ÚlêZãÿo s 63_õÿÛÿL/®¿íó"2ÿ /®¿¿®¦Ð^ºÿµßÇKÒêvHf³Î²þëí ¯;íT,PNÞ¤1#tB³"±Ñ£eèØÑø Î? ÎpSæÚæÌÔDÁïI=PMpao£{ %ØÃA lØÖD¸A Ï
33Fg<óÞj2|4DèDDX.êé:Nñ¤Û\$¦6êL#sF?0PScTB60Òj4cÖ®¶ºéÒM¾¶ä¨%sI$êÒGÆH(Paõé}~N;úI;/zt/m§'x2ËÁæÞh& Z¿Õo餻ãÓî¥ü!®é%ÉRLØ°ÍB2VÎÕyÚ \_}½/wúK{úøíÿÿ¿Z\_·
; (Dóêµÿý¤õ|U®¿×ÿé,>°Ø0@àîMU«ï'WÿþÖºô¾5ÿþ·KúÕ/ðè:[ÿ\_ÿé|²õ3ü/ÿûëõT¾ó½}ÎÊë¾Õ»ÿ÷ÛKºÿåÿÿÿ]W÷Ká?þ¯ú¯õi~ëÿÿÿÿý-}Níÿúýÿwÿ|nuýÿÿéz¯Ú^¿Ö$Ø;[",=
mm{K?û«QÞ¯ÿouÿ×zTà mÿùfx!wÁ"aaaì0¶´Ý{lUçJý÷ÿÿõ;X
·ÿá ê!¨@4!xlBØ m±)ß~ém7¾¿ÝéZÿûúþ -¼!"6GQù³5C3¡ÏÏd¾aákaXx6¡± ¨~ \3sL3ÕÔ2:í³4Û}
ïÿ¥A)
\_·]xap"b0< ;Á:Ofb±¶f)Íó: #L=Ê« ÑA¦ ÊA´LlVáØ A««G6 ÅiDãÓkÞºQQª×½cA6 i6¡Fú7B!rFÆpjl Ï\2La03ähd¸C ÐÁ0M4ýDAì$é Ãb Ä& n ßiT.)wÒl2~Òm8Í&ôô(m&ÝB6±ên-4k¡ 0&;Pª¦HDDGÁa½¤!à ªá0öÿÝ'§ÝÒ´Á:M'Ý i6 ]K¯OÁ0öªoÒa0ª©=½[¯út¿ÿ~^t¿g÷W¦v®¶Ù(º!ûú üÑúêDD¡íÿºëýªö6?ýñC\_X½;i6´ôü Ú!æþ7é\ ¿ë×\_×þé¯ÿ\_÷ýC¤ ߤ&úÒMíÿO÷È ·ÿÔ-\þ¾é7ÿ]?¤Ý?J}¿ïGKÕ|¼|d'ä>ý#¨¾p-6ýi?¡ßo\_Yc?,fXÞú"ÕãdïÈ#¯Á]> \_}·Wé\_~ÿÿü6 ÿËòÊ~~ÚÚ@礶7éÇûßÿýòÈg®þ\_è/¿}øدû«ýv>¾ÿoÜ>ÿõÿëûi]W÷CõÕöÒÞýßýlÔÚõþõ¶ÿ¯®µ\_Ù ¿ì0Ýë[ë½î¼Í¯·¿ÿúû÷ÈgO÷ß± vÒݳ5w·Þ¾ý÷¯Úÿëìpú\_ó!ÐöÛDÍ°Î[ ÷uil³þÃ?à Mÿvº{^àÿ×çic#f¸bb Å V¶;b¢T5#¦ iß»OzbPý[ÿÿÁNÌCa· ¢
´ Ô0B 0AÁ°¡p¡l?¿ø=©Jó°³%F"""""CÒA0A ÓL§(tÊ´ÖÛúÿýÚgià¡Gqþÿÿ¦@ß"U×þÎõ¼/úÁÿ}×ãû&Ó^KRTY00ÿÿÿÆ|. ÿõ×ÿÿÿÿÿÿ^ïímyÚÀrtu5%ÿ¤MÕ}}éþñ2ÐL'DÈÖKÿ¨ýº®ÖÒè×PpPF°ÑxfÎ d<£'ÊÎ<ôkºo×ímxA¤1¤ £]Qy
-È£äxxAà>!üÌÆt"<ÉJmom&ÒêÛúMÈ£AÃÓ«ÄãÐATÕ¸OM"?L"Ça dPR'fbÓi}¯§éªm,BºXe m\Òxña V /TLQc» km&JA ØÕz\_ö×wá:L»ü&佫Èé]&Òl1=ѱª6Á¯b½w]kéucÿÿý==X5õ½; ºpÆfTz6¯ÿÿ[ýß×ëøïZþKktûlÕÌ°f±{iý¦úÿë×ÿÓü?Ûýþºüz°pço¥¢Ý#\³µí¿\_ÿÒû!iþõù룾Óëëö¬&vjÎÓ°¤ °^a~ëÿ{ú ~ì?øß«Xãð]<ÀOÿèv¬4׺¿ûZ×ÿÛKðßõ$êü¡G#ax«úηÊJ\_¸0 Ã0¿íÿ¿ø?~Å~ ×ð»í)u#^yõã
¦ëð !a=}[oÃÿ°µÁí}¿KÿÒùu9Cuþ"´;XÓ«þëüúÿç÷ÿÿzîô¿ÿü{ +m}þÿÿ×óÛïI{jÛ×\_ÿ\_ÕNøöØ_ÿÒ×ï¿cÿ_³,³ßÿöX
=´¤BbS»º¿î×ëôê®ÿ÷( Bc(G C"]å9^@ò'ÿÚ6¡6$ÓV¶r ¼Ä¡ëkðÒ³5o¶ßëØOOF¶²T0C±ÁA Á²x!&ÎÊ" yDH Rg,õt$ NÐj@@Ú $q È´´Ã96¾í£v'¤IûM( a
M±^B!G[´k¢çhðÏ &nÑ Çò.B ÐDDDDO°(NubAÃtªb OMr¦áô®Ð°JÊAºuAIîp Û XL(F1""" åÁ Ü0 m
6! ôúm'Iºa6¤àÌxIÒpàË<:6g¤Ød¹îëQ>0p¢VaØ0ÿ±ëô®½Þ¿z÷wûiëî¡>ÓÛÖÓý}ö¿±õì{¬W§±mVþÑdød|«"køßÿõ¯}×ê?\'ÿÁÿÿülRð_ßàùOûZÿ úåÿÿä8éõaÿ
õþBUÒ÷þéBÿÁÿÿõ!ûaµü"yÿöî¯ýi/þ ÿÿþø7þýÿ ë×ÿÛJ¾»ýÿþRàßýkß©Ðoºú·ö+]w5§ÿÛÿßðïúÿûÚWûÿ ô½ízÿ5¯ì-ÿ×ÿÒÿøØ^»ëÿ¿Ý»ú{µ¥þ¼öñLßì#Øg÷_Û ézVÞ¾¶¾ßý=zõئ A¤{Ú]ì3ßVGö]V¶¶avýký¿ãI
S j( Ô $@ØàQ¦Èp®-ONØöÒî[þÃ<¯äÈìÎû"> §.8ö¶!\x¢6¢j
@Ø @áÅ= l¾§nGR2-ÄDDDDDDHx3 h0B ! !2«A
N¾N:êdRB""""""""#+£Ñöx:×]ÁëÓæCKúª
ÿïÿÿçdFqFH yQÿÑ1Ælgnv2 f äÊn4J ÔµøFûº0Ѻ ¡1Ø(A
ÊrL"Ç ÉL » gQM ê3:ô8:N¥ JÈFú65T6P7BPE¸a0~gQJ5Rý+¬]¦ÒzpcIºIѳŠ°ÈEÃP¨ÝØÑ EÀa3ä:ÿ½-Ó¤ôÛ't°f=÷ ß%í+A8:Nabxh ØF&õú÷õøô;ÿõôé:^Ûþfº ÒtI°bW2ºZÿÚÁÿ¯ëÿb/´ßô={¯Óð[Ó3!¥ÿBÒUþµßºë¨ÿcë½jëuôhfDÆ[¨d³ÿÉI¶È0zò^þ¿÷ÿõï£ÁWÓ"åßèC¨@¿ðDóÈx[ì0]xò\Oéï§þàé^Õü«Wõ ¾Xá°çÞCÒý¿Æ©Gçéz0'ÿézKûá´ õ\ _·, ò!õê×Ugd?õmïÿëýsQ¯õè/ÈbÿRÀ©KèW]TÞÿÍk×ûÿ½ðõÿõôýìKñÿT¶¶·ÿûëïê}ýÿÛÿØbÿÿýµ°õá~×[½×Ò{_ϧ¿ÞzÿýX>'vÃ7CY1(}ì0FØ]¾úa{JÛÖ×ßý/ÎÒ
36DY÷ÿá
µ @ ¨PvÅ7¬3l0>Òl/i6fµµêþoPA #DxÁO>Í 2xÙ4ÎÔd\RHYÔ:#Ò5¿ SA ¤PÅ á¢xiÄ&XfÄ¡ö¬3ÜM½7$lŪ¦L]³0¡Ú!2· "Ü3Ã<;033 È Kóc: DDDDD0qÌ: -®(09ÇLPbL64 $ÆéÛ AyrôxM'ÐpbVWAÁá\_B6ÓA øD/phZa¼ L."""&L-«ba0 Û4ûÓô ³ÝÖà m8Í&ûDcAI°ÈìJÆÐl$- ß3Dt ûa]øôýoõ_¯Xj®{BÛ §[JÃ&$YÉcIø5"#Cÿ_þôÝ>ÿÓ¯úé>×ZNõ êøA:¯ÿþ¿·ïûÿOõÿõî¿ß Åïÿýëõëÿ¿uu¿ÿéx"1ýÿÒÇü5ÿ×õßÿÿáÇÿõðDÿøÕÿÿøÿëý «ÿÿ#|²Î²Ê¿ÿÿ"a¿þ¤ÚØ
8¤B.©éuÿÿú×û÷ÿÿòÊü°ÀÉÙH dàÜ eCÑ ´µÿÿïÿõÿÿÿÿÿ=P´ð|m+¯ÿï«öÿï]{ëÿ¯÷O´Ev®Ò½øaµÿúÓÛ÷~¿{ÿ»ÿôJÐ m£NO ¯ ÿý¥wmúݺ\_»¯ý¤Þ Ýô¾,3{þ·jÙí¶ý^Òµîÿîë¿n~ºþÞüÈ·c«8iÚħl0\_ÌPÍÒ&néºl#-M¤¾ÙÚWe×ÙKÑ1ôôý>2)#² Ó 0µÅ 1MÃØJ)ºcm$Ó¶)´4öïÕÿÿÿ [£ ! §x «¶ÐiAh67ïýrÝðáÌ£²¦D0@Á !0F´U:^6=·ÝÖØP¦VÄDDDDAmëí»ïÿø|8h(ú¼°ý¿ûîáÿä> ä6yïû_å)?Õ² d0®Ûÿ|)2¿W 0ÃÃVÿþ«¯V 0aàÃ÷çd¯ÿýÿ#I~ø@ÌÙ¶|\#ôHòÿÿ¡äX~ßØjÈ.3ÃÁFÏ|d1®kf¶uGò£ÿü§» ðÿôcFçªþ5´hÖÈÑ ( dì<Ò$f<ádÍH¾ïÝöý¾IõàÆ¤Û CHýH>¡0ÁB 'P$Ź²5énaù.ÿoÐz~O?¤ïtiaÁñÁÁí&º6B.Ý ôn "ÍAXᶼ²?ÿü¡\kÖíRÿxONÁ7 ðq+ºZ¤ØX°úÕï¿ÿÿ_×Þ/ÓïM´ÿé½&ÁDÜä²m¯W¯÷_ãÁÿu®õǧ¦ý'§×û¡7¹d(VºíÓ~Úûú:ÿãçz
ßýǯúW÷[-õÊõ °îØ[ eÕ¯½ßý+JèB×:ÿÕþ¾ºÓ+ÝäÈAÔÕÁL% à0Kw¾-ä¾ \0ÿ'";ÿü~;ÿÿ°ub£Ø¦7b¿ý_ÿOO [þ'ÿ_J;ýèínÈÇL8J׿ֺÿµºK:}ÿOáÿDÁjü":é£!;É.æ@Ó °Å) ªÿú·Óúé}zú }µãå²Ç ÐRn"uEF'\Dmß÷kìqZOÿÿÿZ¥ú¿ýx¨ü/ÿuúÿ×¥×·ÿû¯ÿý{áþíïjÿö¾é«¥ßÿ¶dkÑ/ýMÞħì3ÝU
´õ´KïTë{ºÿ;%3¨C¨A
: CF¡q(3]ÿ
&AÅ;Ĩi¥÷Ã+ {ÖÒ[Ý+ûM»h׸ "Çaìø,pÉ0A!³ÙÓ6Í,À§L£ÎÔSZ8eXÎ5£"ôDDA¤øa{P68¨lħl29ì3oÌ/l h'H: ´ýA èÜÁSD&TàôÌÆxr&B.' &8ey F ö 6)
Ab 0#Û åA¿Þ&ÒÒpt²AÒl[WÓ#ø. Ål @Ú !r¡ÄDDFe© !4<0b!Á±Té7ë«ïú¶t&à6qáW°cd¤Ø2YH:M² ¿0"'D|2_ã¥tëÿ ú :·6tØ3hJÿµ»¿±¿uºéÚo¡ëÿ¯Òþ§¦ÿûÁ°×_øÿÿ¼ýn¼úÿÿ!PBÐr uüÐ.º×ýÆÿ¿ª·_÷þ¼0ÛMaßôûÿõ¯ÿÿÿ¹à¿ÿ¡ýÿ÷ï¯ÿÿÿÿÿÿçR:& þøúõÿÿÿÿÿÒ_tºf·V«ß¯ÿÁÿÿÿÿñ§®Þûm=¿ïÿñÿÿÿÿÁ=-#Eß"]uûÿþþëï¯ÿ¡!;«ÿ«ëîÿ¾¿¿¨ÿÿvÿûX+KVïët½»®ïõÖ÷KïKïzÄ&V¤Ã6í+^ë¶Ï¦ÒüûAX_oöÛÿï©ôÄ$)PHHm¸ÛJø×bqò'»A'°Îv¾E¥´¨3_, aa1 V °Ðb @Á¦ Ø caM´¡0§fQØvdÄDDD·NZ5¡( Ó8÷à ØA»h Ø
U © ;[3·±Ð/rP1ãÒrx¯ß|³MÈt
Oþ ¡3S5³ÜØg ' ã:°R5 ÎùÌ») gFpÚ^;[þ á ûwÜ6ÒÖøwK¥\_t÷×m;¿ÐzöÚÿÃõ﮲\õMmÝ÷¢>w§"úÖù½©¨ãL£¥ßþ½Ô°dÃp@òÑ Ýè }yaÛJïáô ¡0LdCO$dCyHÿ×]ÞÃ{þ~¸oêî÷é#ã\(P\N"Dù4FȤ§Eå6D³ßÚ¿½jêÿ¿}=÷í¥¢cî°ÞÒIÑǤFAj¡0QááÐÝ3B<2Ff!á DÊw{ßý{÷Þü·ûuu÷÷ÒƳIíSg'§æ¡Òl1aPÒ\tÊ ððL#ðÁ +õ×{õúå÷Õ×[;¼k¥K Ü^1é6?ûQ>C AU· Nì0 xHØ. >A½}X¿ß·ÐÉå«öÒª}¿õÿ¥×cïï¶é¥ ÚW;?TÞôh èÙ½ï×½zöõu~¯÷þézQZéú÷ë´÷ôúU[ÓúWô¤ôO¬=]/Vëÿýuë®·ß¾º®¾?ô¾µôÒîñß´öÕm'úl®\ æ|
«ü9,ô¤-TµöÿoûÕ´º}¿ÿëU¬
§|/ÿ¼ãÿ¥ÿÿ¯\}ÜFFdW+6Cö±hl¡+ÛÿKWb¿ÿÿ®ªÝ7þh2÷«þÿußíõ{u\_õ':y^PÃ÷®6\«ú'ÿ«¯[Ûÿ¯¥^¡¿ô ÷þûKêú\_ÒÚõµú¦ÿè0꽿öÿöº÷þ¯Zý¤DÇë ¿Oý/ÿÿÞ¿õòý¢±þW\_´¶ÁêÿÿÒ׿Wïý\èãýAõô¿ßn¿ýþÖý´¤ëàGý §iVdGYQ~ü°d-Ê!RS~ßoÿ'K\^ßÿÒKI ¾¿ÿö¿ÿÿ÷û^?:êÖÿ¡pm ÏÛò2ïúÿÿ\_Öí¯×ÿõøßÿïöÒën®ô¿úñðßß»]¿ýwK]ývïêéid{Ì%ÿ×\_ºö¢¾ï¥ÿ××úÿÿó!eû¥ÿ·ÿýï÷·\_í(¤!T~m®·½ý?þÅ\_·]ÿçbÈÿðÿï¦þþÆþߢhÿËGJM,+lSáaò.[Kö¿¾ÿÞ¸ï]~D? ÒÁ6!ÁsL B6ÈÄPÔy È6F!ÿ÷ý®¯{ÿö¾Oþï÷aB:j PôÛ 'ml3˯ڴºmµµðÁ±¢ò xF¶|èGì¡&æÆtÐB¢45!3d|Î<Ùïûÿ¯û·ñëõï¥Ã°ÃMl+ Ø Ð pÁÆiÞÚ[°ÏlJa Ç Ý
鯨L#ED.Ba ±áN Q £L ÏAè>×ÿÛWÞ×õÖA±FðÐNÁpéÅ0¡Ä¨iíÕ;%wYzN aÁçO#
Mã p¬Oô¾¤aßP±iá%y 4½ûKÿÛSÿ]_÷W¾&ÄDDDDD\0¶M[PÁÖáõMÉÝéפôàÌ + $\¬¦¢|{av@dçl/þì0©þ¿|w¿ÅiÄDA:º_Ø¿÷Cuÿ½~¿Oß«üAØM;¥.83H\WñZßñ&>ö¿þ?z¤"#}}ÚRèëìq÷Hjõ×Iêÿ§÷¯ jÖ¶×þá¯þú½ë·¥ýïõÃÿÿküUþ¿ÿßkÿÃ_û°ýomÿØ^Ö w__ÞÈD×µýÿÿÿ¼00èÙt a´""""""?êLe½z¾ºØ<'_¡ûÇÿëÿÿ¾Òu¬ÿoë& ù_ÿÿö¿ûJé¼;Kÿð}ÿ^²À¿ÿ_Ú_¬µðý®·ýÛ¿ÿÿaéJúM×+m{¯÷×û¿÷ÿÿðÄ/ê1·õëþ{~»íý}ºÿÿ?Æ÷w´t_ý]¯öDZ½Þºÿú]zaþ¿n»¿_ íºî·ukÿév(C %XM°¯ÚR, NûaL3ýH¹{3VõkÝö·Ýè0¸§ÅIÚ°AÃcûM#¶Q)ÞȵoÑ?È A(¨D0O»v!4»vJÐ pÃ1LSPÁ 0@Û Ç(Á4" ÁÓ3Eðûwݤ""""""xÌ5Òn½%úK®´¿;P¼K:WJ
¡g.IÎ)è«Jð¨ÖÑ å»>JiR Aµ L0Jt¶JÁCT¼GÿÐÖþÖº,Ê×_aÿÈC´?,ü0uX9d /ûw,G¿¾t}^ïÊÿdo÷êûú
eÝ|úÿÿ_uk÷yþ¾ÚM
Ò¶ôË:ÿí¥»R&ròÏ BÃ4C!ì§d\Îÿ±
)¦(<#s<5 n 3C0AL)FlgAóeçQ4xά¨^} aaanz£t j¨ÜT6oFÀÑ Æ°Îa32 ¢@a8A ø5|0A!O='zpcH6 JÈI°H6 ñ8ì"íª (D>¦c¶h"!0a\¯áI´®J)m=téX2è¬ xM» l1+Õ,O Ú6ARn
kèUºm'WW¥nÞ{l Þý2¥¶úׯ¯´º¿üzý/êÛúmumº¯ÿõ®¾;õïïûtãéߺ\_þ?µ×Óéõÿ®ºí/ë×ïÅ/ñù OÿÜ [KûTö2tGe×÷,gþX·ù1ÉÌÈÝßÿÿèq´]úP¿ÿú@¾¿ü"yn¿²}k¿nþ¿ÿú\_ëõõþL$è½^¯êþÝýý/ÿÚI;×÷i\Q(ì-Ã2r;oï¾ëÛÞ·ÿýî½V×÷¤ÌÁ Rv|)àÎ0ufa Ä
¡dʶP VHg³FC3GDÒÛWÿîþï×ÿKÛÿÝÝÚz½á;Þµ«ôôö #a´¿nÒÿ×ï×õþí'Ië÷éßk÷ÿ½}§vÒMW .Ý.Ã?ÛueÓ ÿí
þÞ-{öôL¾: ûåû'9Ôêz'7èùgrÌy9¿åÍ˯è! ÃmA(6A;bèJ6$¿¹J ±íH½ìþÃ?ÕÓt)ôúÒ|&ïz½ÛÚÓצái ؤ0Ðih1Aa ÃéPýll0Uo~¦ý¿é»÷]¾õ{\_í'ÄA\0A Ñk@Ä(A§i Aà &!¯ô¯õôXï¥ÿæ\_ÓWýÕq¢" 1÷µØÿOôܱßßó;ëæ~Ì?Qô"?T=¿÷û|=ÿïÝÕý}[¿¯V» t¿ëþ^·¾¿]XJþëý
^ü
«kOÿ·²GR Õoûþ\A|ïþ¢¯3 °|RßþÞÈ ñ\_þû}þßFazþ¸oÿÿðÚßíÿ¥ÿô¹u\_!¶¿ï·Äû ßÿÔº9wÿ|º:ÎÞcú×óYEË;ÿûï§×nìèù(Tû¾Ãýß·]ûo×ß·þëzÅmÿßôÿé¼ÝÿÔoºZ÷»½ë~üºJãÞ]¿½ÔqU\_±¿êÝþ¿oÞÕ-{\_¿þ×õkºþý¥¿ïÿºï®»¶¾í¼5ûþ\_m\¾ýØi5X[}«\_È1á
Ø0é?°eÅÞ c[°dq~\AöÇ-Ì_÷ÇìWýû¸b©¿¿b·¯ÝGËT FIÈß >×½o°«a{õ» ø/Ú 5êqc4B" d¶DDD!:!îö®ú×XûY¥ïñÔBõÿñÿü´:Ù©õòÒSA 0A"²UzNêYB(GNÊÏßé0Ì(L"ò]²\3Aó|a ù0 Fá Ê3HÙEÈfNòÁ\ÒlpÐOÅ^uPMd!z]´lD ó0¦hóÆ=Õé´[Ü.h ø9;H8pbVBAÝPM¤îÊ
v- "l6Ô'¯º¯út½ôéºN=&Òw¹!êU#tmÖû|zúúKéþ÷ðkiÕÒÛ úë®þýÔuÿ^
{þ½öÒëþò%È¥ºZýÿûÿ~ÿ´¼×°qÁ÷Ý?ÿÿðÿþ¯ÕýòÄ<7$çÐ?ë^¿Ý.6BÿßYÔV¼ð]}a¸Bûß¿ûôBtÿþ¸"Cÿ' ¾®ó©duÿÿí¥ReÚÿûu__/µ÷¬>ÿÿýVåY?ê´Ä--?IG¥ùåÙ?ýÿþËX7ÿ¾ÒïÒ}Õýïäë¾ýoýykÿìÚ3ÿ]«Õø/[u{kú]þ¿øKúoëÃ&Ï÷ûfjÒÿü'kÿ·m6¿®ªWë¸a& vÞÃ=Ü0DDZ)é®G1fk[^ÿ»°ªßÚêÃ9 JècIáÛ´íÿ~ÝôÿN¯'öúmN ødocÒ×ýk¿ý^+ÓÐ÷N®Ó__ÿúa÷û¯[õWI=ut;ÿø¹]?ëkê·¿ûkþû¯ý0ÿÆN&þ¿È«ost\^#ä"#¿ß,»¿þGÈtþÿéN#и½¯ÿþyiÁt¾ûòĵ7èpmëÿÁÿÿô»zÓÖ¿n¡+©p¨>éÿ¿_½ýuûÿþ¼+Ý~þÙ«ý~×µ¯ô¯Ûÿ¯¯Ú_ß~}¶~ßýké{þÇþÖÒ]ooK¿ÿÛ×t¶ý+ö?¸ÿ8ø
iÜmmÝSir&vÌý³5atÒô&û££úüã´ºA b Èè$¤imX0V%>à n¡¬ÿr)ÿ!£Î%ÌdOþC¿ßÌa'L«Aáb:6ª8 @Øa(Åö¾hadæ(f @Ï6(A0 !;09OB óduÇHͳ1J(6_k è²8CÈÜ ÇGA 'álB 6¿Õ0TÏ ¡vî8a4Î Xü .Ã3Ä#[»hÖÎ Ñz øÕ!¥qq l »qv¢ûÄîÐMÑcý&ÐA °ÄÎöC0©ÅF¸dvE¸2;aûzIÃ"Á)8dWé?:NÄ6ÒoÒv82Äïé6 }\ÿ¦²7¤ö½´Ó ëëpeKþë®ÒÚy;·×תë=½.O¾ýþõïø¯OïãÐéW¯ûßýÿþýýt½ïÿiãõÿý{¨ßãþBþÿõ׿ ¶O\_ÿþú¯¯ì?ÐÿBøÞ¾ÒÉ~¿\_ÿÿV þ ûú¿D!Hþþéëó¾ÿÿöêM?Ù:þû[ùòƵÐWÚñ_ÿþ×ÿü÷ÿ$ËÿzÐ]¾0@ëÿÖüzýþõ}x}ÿZ«HDÿôÕ¿ÿßÿû5§ÿÚþÿÿ¥ßZýÿÿw¥¿ÚßÿìQj 3,ÆF4¨À ð0IÉlG ûKzvÿoïÿ{K¿oïký¦TµM; EÁÍ\Zï×ÇW#¿ºþÍ4¶ÌÕ¤Ùû.¿µëÖK[àà 8H6Té á± áðW{ Án0[ãM¤Øï{[ßßÃÚwÒlVÒiÛIÛQDP8 á!P lTS8m0@ÿîê»áÄDd!hBeZ0FUÃ,pA!ÐP5)ʽmÒµûÐÐZÝß]÷õ¿·JÓ½ÿÖÛZýÿ\_Wwß×UÛéo¿ÕÝÞýýzöë»ÿ¥ªúëü<®Z?¹·\_|úuAô»½"@h@ázÞÚëöÜG[ÝûÝ÷ÿöº¿×V´CÛ÷ë!ßÿÝw;Só-ï2GeïÓZ¶·\ìgf£+ ÈHgbez^¿]ÿÄ;ª|JF¯ïkÿ·¸:ê0tõ^)m¿ïýõ[ÈwÓÿO{ÿh¢v²e[ÿÈqýÓ_®¿ÅDkÿ®ÖÛÿÕÝ¥jðªtAõ¿ÿ~ªêµ¾ÿÿÿúöÿí}È}üWðÚÿoø e ÀÎ n62B@uÅ#dErq3c!P&EElíßÓõ¯ 4ka ¨vá Ö.Ú"HÙ0SÄ4Á§¾uí×T~a£}Ó Ò|+ l^pØm+]A´lÔ_ÿo÷ýØ ÚNún;&zM'ðzy/i:»(rÆÓmÐ1+ßN?iaK_}}íU^'¦ÒoÜ?Ú_ôé]bêî7ÓoO¶ôûèrÇw¡»zñýzuì}ÕÚo±¦÷±ªÛJõÿ¯îä"ÕkV¿úÿï×úï×»[mkþº×¯ÿ§]=ºZ
¯ëýkðÿáÿô¿ß]ÿZî÷õirpßï!ßwù5ãÆÖöPô?ëÞ¯ùÊwÁnyþ¿Ø~¾¾ÃÞC H8éõÁ¿ò OÿëzÚAßÞÿÿ ë,½e¨þÿÕýÿ½t»×¯Éÿÿúʱú¥ÿÿû÷[þÝUzÛ/ýïàÿú}uþ÷õÿÿ¯]ÿõ¿Púúþÿ¿s»ú_ûûÿÖõïz¯ýúWG?wé|öúVk^é>ÿýK{÷ý~×ßÿV÷ÿé¿»I»ë_íûýõÕô÷¯a}µ´¥¥ß¤¿¥i?[kvëûpÿ÷ F¿ùíWNÓØî$Ý]'"å]§u % ÷k"gô.%»Øg¶%?¿ª¦ÿéþ¡Ô0Ô8ØJ$t6$8aLhI¦)+@aC á±MC0AÃßIéýozá
øjéLVV!/±R¡ M0=m:öMZwÿþÿ """3Ý!Ù?ïý}áo-Ô"d}k÷õï àhÝt÷é÷ý½ä(ð©iË2@ðʨ8±PSËòͺå.·á'ÄDDDDDDDGZÿ þío_ {¯ Áw&ÁVüE!ùuÎ §!ä< !!
5g Í ©ÌÆgÆf>à0jà ¨ TFìsµ3#¢¼Ê2mmRx°è øAãÄA©°P§a¡M³8
ÉP¤ÁtBDÿÓÓa¸2V%? ÎÒÓIÕÝw]s$<Í>ÌÏ3O§@ðû§]ÿúéøa.ÖkÞõ½~ÿû[§ïô;cZß
}?ý?ðÿÿ^þôÝzÚÿJ+ò?ý[Rlq§¤¯ßßØ?üµAõþ¿óFßÕë÷ÿ
Q²èª#atUChøg:"1ÑH p ÿþ¿KºÿÿoÿDH B #´Ép"aó?#3óLÎ3'#L sóEP3DOäèd|Ê»®j÷ÿ¿Û\_øiÁïKáÜ&¡0¡Bi
(M4÷ Á0¡0¡0L!õÝÿä0¯ÿÚö¨:éR>0ªª>Qò´
Ññ¨P¡B
¸Aái-×F{ÿàëÿ½é?])úNðéÒ8éá):NR¹øUIÒGxGz£FBtp?^ë {ûïóØwëíom/ÐK'Ac ´µ¤±Ò]%?RØ: IÌä ´¿ßÒvôªµßoû»õï_úúJ¥¤.8¥ÿ×¥I%¥ µÒ^ÒÒs4Ûv¿ÿÿÞÝ'ª®¸¨|8uÒÿ×ÅCúºJ·õÆοÞ߶é}§~÷ëתªKéºëªéé.«I}%ôºÃfM¥ðfþ?[¯ÿo¿Zê×õÒÃUýuZ-/´ßûÚKlRlw +mS»\_m%¾úÿ-t¼Mj¿×U¥êȧ¥ö¸i76¡Å wõëú¯ëú]Xt½U%ékUÕ0é$·\_ÃðÖÖÕ°;ÿmWÑ?¤½}\J¿$¥ªÿÿÒ¨7¥÷\_tA"! È0zôÅ^´úýzëÒ@ëýéUkàê^¿ÓÜd)]ûôµÖ«ýWúÒõªõ¯WÒ]Û÷è=:ý®ëK=ýt¿_õéIVþ»û_ãI%ÿ]kYõ®¤½--z¥GÕ)Ñ$½wýïuÚÃéf¥ZFúYu¥FE¥úZ륤¦à¨ü·íM/þ?ØJ ]/Yj;It¤hß^Gú²?G©uöGzjÿRðèB«a«áUXZÑêá%Ðb
iD-RºMýd0ïY
J½d3@¨8a%¨Q ZTN+ëáB¤ð@Ül(ua%þC]BVïdWV" Á°a00A"5á "Aq! vT&ëÒÿøq®ìÚÿ+ÅnÞÃÕW]o!èÝ$û¯ÿ¦Õëµ¾F?Û÷Z×½¥¾ëªM_Ãt×¾ûÿ ïnµ¬V DëÞîø7¢]þßáû}ܯÿ&+¿úkõ°oô;ø½ðy6KE¬Uÿ¹éÿ½3´et¼®$YJÆf_ï¦YðaçiQØÐÈÔQhyü²eC;;¹£ùrs ·W®ÿïÿ0 ¡ ë3[:JQÄ>Ì-ð£Ð}oôÿïéáBj¹ºa¤ ûrøC D].ø\üùq¦÷ÓD ²íªM|¸i¤'¸TîDñõÄDA úáøAµFº ¿´Ñ¿H é[Ô"Þs;}ôGþÿ¦×}>=:¥ôßMðº
ÕÕmþ«Õ}.ö¿ýgV?®Pýï{GQÕÿq÷¡ÿZ¦ï÷þûû«ú;ë÷GtõþxëߥûKTô/¿ü×ýZëõïI×÷øéz^ºMÕ%þ·U·¾DųPιÃ)È×"0SñóQyòQ«¯ÿǧKnºÿQ¡«ü|Ö¿zZÿµÛ§®¿wûú\_¾» , ÿ¶·B½¯«×þ? ßÿïÿº¿ýöö°aW¯tñÒ]RºK f»û w×]þ¿úºPÿà×_àý}ñ@
Õ÷þ^ûÿ¥î¾È¯W÷ïýé¾?:ýò5ü+ñû!çP¤Yêñÿ×öúÿßíi~ºø"OÿI¿ØF¤£°×J>Þ4HÖ)ï þÒé´$SüúûþXN/÷ÃÿÐ/Ão ¿ÒÎûÒýn§"Ükôzòièý/è/û n¿ÒÿÁýÿ×ëÒþ½&ÒëKÿëvÿþ¿õ¿ÁÒÿÁ¿®Õ´µTú!¿ôæ T«þþýÿÿ¶÷ÿýûÿÓý\_íWÈþ½\_µý}o¿½ªß¯ï¤JSþ¤
×o©:}WÕ¿^H/uÝ\m-"\_õH¯¯öÃzßÿíõßí{µ´õøª·u©¦ºÝ×zæ[Û¯½¯YýîÙÑ^Ýý
öë«^ÿ ·kî×ÿØJÐI¢ ÖÒÞ©´Ñ-
Ò÷Ò«ISßo¶ ´¤Yz½¯ òí0û}¥ #?a¢S»J%;bPôö«pa"µî.Ò[½Ní$ïJÚ_k¶Øg°Å6[ÚH0b¡ê $ÅC jSPØP | Z8 QD0ì0 Õ¨q@$¢6LSA¶ì0¶(0L5µ Öam?µÔ¯ÌáK&3äã' )w Bgö¤xSÚ¶$îÛDc atØh ©ÆDDDRÄDE!¿¿¯ºº¯ÒN×ÖÖð¥¢Î¾.(ûmK ¶é¾õO¨P 1夰¤Øx²
Ùc-ÊÌìêW0ÌZAI¦eH)Ù_S"ø)H¡°é¦)w÷~îþT~PÈó²Ïéegò5ú´áW]ª°×ÔF×ñãúþGUÿÿÇëÿëÿÿÖ¿6d4r4ÑyZ$Fç/8ÍDtG9S}Ìd1lB4r',è24 #XÊ £[|ó5ÉÈÖd%òA®u, Á0¡|gÈxAäPÏ@ÁN'BüÙã#Çð9 Ð< Ð Èï ¢ÞI "à2&8håÍL Ïa0A|@È O=0ªÍBl!2D)P64k
H Bä
FÈAá0ª60¨Ù C ÚÓA×@aP@Ø.ÍS¸£cUL")£óTla~g¡ÐD/Ol1¤Ø9£IxjᡤØA+ bx ëÛIÃnº ÕÁYI°ÆéXbpbV4taÃÝbU°0tpl1¤Td¢o¬ Öt´÷YÃPe Ûë,áI¶Fõl?Ý4¶ë¦ÒÙ(¥aDìR¶¤énpe'[ ûIû xA4 º&éé×®«úÕ´¿ßª§ýÞ¿éïǾnw[iÿm/¾éééÒuÒxA÷ûúo¯Wúzÿÿÿü_ñ÷ºit:¯¨¯^?×[ýtÿB«ÿÿõÿÿ^û¥ÿß¿¿÷tqé+ÿW¯õ_öý}|Vÿëß_ýÿôõÿïÿäROß_Áï|è,ïÈAë!Ù±W¯ýZµ]ÿÿÿûÿþ?ÿt¿öB©~>>B´ô®ÈP/üz÷ÿÇÅKz¿ÿú_ÿ!éý{úýµù4ôFY ;Ö¡Öÿùc2b2Ä})ÿÜ9B:¿õÿÿì4¿þ]õëÿ ÿOöéÿÿÒÿ¿õÔ²ý7ëÿlRÿÿÿ¶¿ùÔ¿öó¡}ÒK5ßû×ÿÿÿÿÖßÓ ÿûÿÿÿÃÿ¯é}oÿÿëÿ÷ÿ[þ¿õó ÿïþû¥ßçÓÿ}çؼúþÿÞïÿö÷»û\_úÿT¿ÿoÿ¼Ö¯ÿÉ×ïÿë×ÿK~ûéÛÿþ÷u»û.Ò½~ÿ»]Wÿ\_ ý»÷µëÛ®ööÖîÿþû¾½wí-¾v¿ûv¶µn¿ÚÝ-^ÙÛ3]ÕêÚý®ÙìÍm¯nÌÕýöÙÛ3Ûn¶FÚQ[iD§È/ï½°ÎLJÄ¡÷"TCív5¶\púéØÚi'm¥Ã¶/ àÌL0ª"gXaa-¹Ta 6)
µ6¾ ªbb~8 Ãb a baÐ pÃ6(1°@µÓ $vXi«u¶½á:´Õiá(iÝ
A
Ä(´ !0¢ &h ÑCD}GAO´ð<01 \ !Ôb"""8¢""4"""#'úª\_I/}j½t]uõÖ×[iek [ 'Aµ¢êWÆ"#ËI.>¢ ÄÞ#¹¢+!² C2s;2ÜA̽¦väÙ¥NR^̵ W5n£Î êTÁÁÔnÙèûIÿ~¿îN]-üÒrþwߤòºÔºÏüv¶¶q^+t\_T}KK¯ÿÿ¯ÿïÿÿü· ªç|ÈÛþ¿ÿýSO_ÿ×ù Éü÷çp¼öKù×5">@²^%fEb'ÎE9ÿ¯;$!y²5 Tn4¨y#$Ðy2I¢@ÉÙÔüôF;Á ÆÏ Ìó7lS\i¸ ΢ñ¶q< @B ü3Fa0ó ïAôJ,ìXE¸gL -Ã>Jd Ó0DÊfA Ø dìÃ$ØL vhÁ0Aø ÊáD.Q±Ò X! Ð(&!2ØÍC ]°º5la
L(T.ÈM{64Fç°Du%g¥ _@ma ¨Pl£sFÄ"ñ£|þÁB¨Fæ <E»!pÞBã n°m¤±· h8bT5VÒm8}Ý'H6Ô$ Ä«àÄ«h&í¤Ü úQ«a¶HWÒ°c àÆprvtðA&Á¤HH< oI&ÃĬ¤á};itá÷]4ì Øe é82OwÒteÍ6ôlaáVg Òo¯þ«WÍ;'t'z¦× [»'opfGJ©ðf¥¶J ÌJýmÒé¿zmzwnþ·þ·zô5½uôëÿ¯÷ýº¼Fèwÿ_®
oöñ¾ë¡ëÿöÒéÿ¾¿÷þ¿ì~Këï÷õ Å(GFdäl#dC£0èöp|RÛ/#æCf¡Iãc¯õýzëýwïõþ£þÝ¿üSÿôß×÷õÿÞþßÿÿ×ÿA iì &;M{4 5°AÓ. íïÖµýðzý/û½/ûªúKÿÞÿ¡ÿÇ! Ò_J?ÿÿãq¯Ç¸ô/N8? ¡xëô<ãÿÿÇü
#ÿÿÓÿþÿÿÿêÁ¿íè¿ÿ_×D²µè)|µY=¤JÝ"¢VØI"Y£òU¿ä(pDïÈzDÿò¶ÿÿßÿí/×äÆ¿ÉvýIúûuû-u7"ð(@éภáAà¤xxAà ¦ä}ÿ"ùuÿÿ®XÐ_ÿü½ª_ÿÿÿ¯ü=¤¾¿«ü0ÞiáH¡ZO «J´5¥Ö×ÿý'ûûh«ÿþA}ÿëû§õý=ýÿ¹&a½ºÿü1 ëë¥K®]iz¿I~½uÿ¾ú_þ¬þÿóY××þÿèîÿõíýÿðÛ6¿W÷þ?úM×é~º]}të¥ïëûþ÷Û¯oðëý×ýÒÚ¿¥ÿÖqÿþkZ é¾ÿý/»ýkú®½jµê©%Ò¦ë¿ý¿ïÙ«ÿõ?ÿïÿ½úMÿïlxÿ÷v½úßÿÿôòêõ&\¥^½ä௥¾ïuû·[Wÿþÿÿ_û×öêûK×ÿý¿öîÒÛÿõþÒþþ¸¥ÒõÒÁ%ÿÿý½
ÛµÖú¶Ò÷ízý¥û߶IûKû«ûüÑv½¯»Ú³5úü3ÝûlÿÒëëÿõÿ_¯ÒIvë3°Ïu #Û?°aaaì3Þ²)7_ ÿØ]l/m¥Ã#[3I¦Æ;Zw"©Û"çÁvB=XeÎÒ^m v;Ni¬/Õ\_ª\_éz^¿ÿ PA´pØd~)S½Ç[a$ÐlRq;bwh4ÛI64í Ðl Ú : 5¨0\)¶)ØAØ a
úÖ¿¬zõë^%>ýM¤ì0XAal+ (¦ h68°A± è á°AC)æTUAl¡Îù ¢ (º4BDDFÃDUÿÖU]kOK× ¿^¢""""'T83ÔAa û0B""".""""""""£""ºúKÿ?¯ÿ_T¾µ"""#A'Kú×Òý´ë]m/õÿÒú¥¤ý¥U¯þÿë¥î½Zÿ¨Duþê¾ëk¯]¥úZêé_ø½¥L4º×mRë°·Ã uê5õb½ñ^W®Ä8{¯VÒ´¶ªôþ¯ðëÙ/µ\_,ë)/m& Úª¦NÓMVÕ4ÓÐaS] i ÄDDDDDDDDDDDMÃNÄ!Aг-;qh"|íB!3"¾W,ôÍr¸R20¤TfÆv0f± içgJ Nȳ²Hï%ÿáÓMßÔ4Ñî¿ 0@ù]$/ÿóO÷ª´Vgÿê¡ WæE7¨¨¯ýÜq¿×ÿÿÿû÷úÿäßW¯ú¯ÿð¿ëù}ÞudYeaEþ}d82Yñ´çdgGÑã:Êv¹B y¡ ²SdQ÷çUàxAfÙì gHÍ#dtQ$aED3Íøóá3ÄhÏ@ £60CDAÙ$«MLA0|ìñö f¶x ð£0!á!êPðCùñOp!Ái¼ØÈb
Ö!8Ö \a4ÏÁTB<"áLv!ááë\ô EÃ<0¢Swѱ· BhØÕ0¡ £cFÈ(TÚh#cª a«èØí\6"áMMÃFälN¼Z~6 MÚk~à ÁoOI´vÒÓA7º¥PI¡¤Øbxh îxº¤ôØvi ÝI8Fï¤Ø4 H}a&Ò ¥Á$Ø1GbVCæ¤àĬjÜÕ?¤Ûýñé6¶bW4K)I½uv¡°¨pá¥ÃV¸pÁé+§¤H4÷I\_JàÇQ?:Ò÷ y+°wa?1ût44þ|'×Ýé§^K§ú»uN·Z^×ÉÝ\_w§ýÿÝ&ý¶ÿvé+2è&öúv«jýºotºÕ¼\_öûéº~m´©½ºut¿Õº[NýPÕãÓ¶·¶W캩^K~½ Wë÷oÖ?÷ÿþ»jÿ¯]¥O\_×Ùñ ãc ðB\_b«¯÷þõïÿ¯ÿKKßúÝwתýzÿÿÕ÷ïëå?öÿö?U¯é < öµ÷þ¿ÿÃýëý/ÿ]Ñß\_u¯ÿöÇý~CþµÔÿ[£¨¿Æ×Çþx\qêô)ýÕÿê6Bëj¯·¡ûÿ}ªQ×¼}W\_úÑ8\_û\_ÁûôêPþ¿¯]iÆØ$O%Éÿ&òVt}ùûõÿR:\_ÿKätÚÈ!þµòc2Ä~÷°ãÿ¯|<ÿ¶·O?Ë-GåþÒ p¤ Wð
ÿþ¸! ¯¶¿¶²êëûí¥,¾¥û¢{RÝõú ÿ¿]úé ¾¿O¯üj¸]~ô=z×ÿÿ\Ëþ¿ëÿûKö×ÿ{¦ÿûþ¿ÿ¿K÷Ý&úþ¶åÿûõ~´¿hîþÿÿü?í«~ë¾ÿÿ\_Zÿë÷ÿ®¿ÝVûÿýúKé]á[û½þ/[¤ÞýÿýÿÖuw^Ý1M«n¿ÿ}ïÿ¿þýý-kÇÿ½Óé=v¿×
Qö¿þëúïþ¬ÆöÿÅwÞjW÷¦÷ÿë}¯ßÚ®õÖº]é7i]öºÿ?¤Ý¤ßÚÖ÷ÿ÷í]¤i}¥OqMà¿» +Û\_õõþ½Òõû¦»[Km³D·i7[kתA\«f}]¥fØkpÛ]»¯é´½¶Ó¿ÛvßþØí"[=7ú #q¶Ývû¥» .Í&àL3ìJ~kzIVºØ0H$ÜBËlJ°q)Úl3×ÛvÒMØfÛ·vNÃ9UÓ©êÃ95 óLP ÁÚiÛz¶Ã $b06KÄ A±±A+LP áµ\_ý{ö)[6av Áb¢!¦( éj]1NÁVÅ5l Ða'aaAÃa[AÄ 6®Ób¦®T+ ð Ûjÿÿþ ± #ªT# &T&aÊ Ó)ÉU§pÜ0A§¯bá
LM.B""""""" Á6ÌD¦d¥¯ÒIe""""""""""""""""""8ïõKþµÒí-%®¯ÖÚM¯ö¹×´¿p¤©úبúým5[m´ßÇ µ!"Õ/ÃB! a4=1AÓò̲ÜùFBè² ØÑ
U[àÈÑ ®W@R8ô¦õ+²&Øpê:áÁÁìvWôÎÕWÿíû¾v^r¸ö°geQØ%vvy ±_k]CIqÕuÕ
«_øÿÿñññþ¿ÿÿÿ_ÿúþ¿×ÿÿþd²B$f¼ÎÈ8ï<nqÈ2PgDpòdC"c¾Qg¢qr¬SA §ìJO?O¥¯¼Áª;ï ºN ÈM¥jé<'d QxM»Bëm?ïô4ÿ×tôý?é:Ýl!Vú®Õéµãÿ·ô?ÐíõïÓýk{}=ÓÿÓÓ{ÿ×ã&9XǪ|ëú~¶õz÷^ºÿUZÿÿ]zé}jê4õô5ÿëþ?úðèkþý´·ëï\_Kÿûõ¿]¯Uÿþ¿ÿÿõÿþÞ¿ÝpÿûPø=Á|õÓ÷«Vù??ïÿ$ ûýÅþ'ÿk¯ÿþûïðõ°~Nþ°VÒþ>¸¯ÿÿýø"Oÿÿ´°DÿÿÇïÿÿëþÁë\_!ÐJ=w%þC~F9´¥!Cÿ{ÿ×/½ÿÿáßäêÿɤ¾õÿ\_¿ûïå?õ ^ü±oÁ |±ÿÿþ÷ÿl0ÿ¯þU\_ïÿ\_ðßÿe¤µ¥·ÿý¿¯ÿúÿÿ¦!kÿÿþÛ½ýÿùпü?\_\_×úÿöëþ½?uþÿ×Ã8ÍÝÿþ«ÿÿýxþÍiõûÒ×m/ßK{«ÿÕË|ÚPÿ~·¥÷½w\_îýûý-~×Ï'ýmÿô¿ÕïÞloþÝ{®dT GSSÀæ ä
#±Á¹d|º#ävkEX! Dñ2ñtaßývÛõí/ÿ»ÿ}º\_þµÝÿÿ׿?v¿ëkAuÛ¾Þ¯ï½û4ð¦©ËÐDDq°f¡AlDDEïì/vwØ\_·l-ý·¥ÿv×ßpÈü5ØgûI? |0Ü+¥f¶°áöëä\º^Ú\_áö|;Á5üOØfæÒbSîD¨óħ¯"àg¦%;{°Ï¦äÝ|Maò.[KôNÕ´´TW¸úã»V)¦( ïM Ûi ýS×OºÐoöC0Å0a& Ái:8{Pð¶68i´N\)¦ÚApÒ6HUÒ0@á ¨7¦
6¶NAÄ Åº~©÷éªð×ì&!BÕ§L0m [A 0 Rpè h9NVÐA± 8qBA"tjlGAN¥AC©ÇË#R #E[FhиN A!Nû¾ý?"""""3N"""CÄDD0@ÁD0A "#¤""8ú¯ëþð""""#ÿüsÕd/¾ý¥}VMÊBî ý ýrëÖ©oª ¿Zou¢! üÔ&¾½¯ ÷ôþÿA¿ÿª]îÿþþÿ×o×ÿýÿéiuôÿ"÷ï~¯ý¥Rïþú{×V¿¶´ý+Ñzðý[è!£ß¯¸?I~Ê^[#ȯýÙ 1Èfαÿ¿Kn)«ë÷ýë÷§ÿÿÓþìþt |íg&H´ÆÚÿ¾-¾ÿ ü&V]R&æhÈ fI þ¯û ö¿ýp§kyÚ fHµ+:þý÷o¿ÿ£<6\¨zÜ¡Ït°oëÿ
§°ÿ»õ"h/ÛÒöA~D ÈD\þÈ'þÑ%¼ì³"W ~ºÿ®Iý¯ÿ.ÖÑ\.u]Fÿª¿íþÿÿãúýöëÛýþD ýÿÿ¾¢\_ÿÿëK×^ëþ|Wúûÿÿûÿ÷÷^©ß×úÿÎ# ½d['3¢Á0 Ê¸Í 2DAs|C<á(ÍÄ{( pa pgD¬ÐÁ ÌÉKO32 ¦pÏD0Be0pΣLù(E¸aí\_!\ uö×wôÚÿ ¥¦ 0ÁT#s íAT#t& Ño%Æ Lø á(Eàm6li5³ÃB!0×ÝÇþpPxµU¼Úqi .o5TmM¦¡HÖáÚý¯þ¹HíomÉQ6 q+M¬O¤1>8 l81ÓÏÌ1=áàJñ 4aÜO Øbw
H88pÁÁÁÒvHbl.Æj%c XÒl.rCDg%ÐįMîÆýÒl.þà {¢« ) Û' H4öÿòW¦Òpe'dí¥l'§vJôì'IðA¶!ÁWé,0:VÁÁ5ý:ÿ·§dï¤ô¡:ôáÁ¨M«ÓlÒëWdïuN ut÷ìÕà |Uì_úMÐÓµ]ôß|!ýé´¾©éá¦ôé.·®ëÞÖë¡þ¶§¯þºzì{¡uõ~ëëk]ÿcÿýWNúôã×b7_û[WB»õu¾ÿãÿý]tê?î½ëßÿuýôÿµþí}?ÿ×ÿëëÿûýõí%ÿ_øÿÃïÿ¯ÿÿþ:úU·Kõÿ¸jþ®·¯d ÿÿ¿ÿÇåôÈ£kë¯ÿÝû_ík¸?ÿä-(ÿÿ¿ñïÇ!xu© 'û÷íSë¾½¯ÿ¿ëÿþ|¿Ãý÷ÿÕwQ_ÇôÈ]ÿýÈ"?ÿü#¯¼:þA"Þ×X]{µð¼4ÿýþX/ÃaÏÿÿýë&3òc>ßÿáùb>·ÿËëÖqÊü²ùb:H÷é{¿«a|/kÿÿö÷ÿè/Á´ÿÿ×{}_ÿÿ:¿õ¯ÿÿã×ʱÿ¥þûb"B" ÊN2pDo#ð ¢i:k·úZÿýWx=ÿïýi}æ²ÿüïýïëÿÿëÁïmÿ_!ÿÿþÿ׬:ÿÿÿû}÷ÿa»ÿ÷ßÿß¾þÿ¿÷×ÿè]ýßýå ÿ¿ÿ«¿§þþëýç×ÿ÷¯ïýûGúÛ5¯Ø¯þ¿ßé7ÿ½¥õ{ÿ¿þ«º±ÿ½÷ÿþ¿ßÿÿßäý½nþï×ïþÝ}uÿzX[ï×ì.þÚ_]ÿÝX_ízÛÿoßõ÷nÒþêÕ=´µî $ÚR&ulÍCJ×ô¬/¶ºÞ[uM/Ûûø|0Eµ5gûïaZ¯pÏWÈ2ì ¶îÕrÞÁ=rÿtÛ ''O¶ !1)궬Jz|Jëq °°§Û åò$4w±PábµM;M4Ø 6ªÅEØÐa&({xA°'lP@å: SA ¦¡ 8ªpa¤ @\ @j $´ïa\SM0@ aa¤´¡ ¶ &!0i
m&Å Þ$!ÁH¤ÙÞl¢c>!á' (ÔC1!ahÊa ,pÍE&8h6" ÌÏ3Bh 2xÁx fÑ£sD&´Ñò a ¡�ÏA£C8BÈ©0PTDÈA\0Mð¶ &|i
EÃ,ãj Ä">- Þ5´>@´Ñ±£]B¨D,/MQê ¦ jôM£cé5T}0 Ça h#c8°ÐZ ´laH ÞÔlqhÙA64¬?ÉRn'Æm'¤ÚI6AÁÁë°´l1=Ò Pxl8:MXA«ÐnÁ\_4%}\ÙCzTã%ôar]Iº~l ôàÉsAI¾Þ5M¤ðïíý6 iÒ}ÜõmØ3^nýì¼ÕýSryV麹+BýmµioN¯¶ßÞONº]'«ëúÇÇÝ}m/ÿK®iúWÔW§Æ®vñéÒ~¬}%},n´ôûþÛßc·ïûßøõ^×÷]+×þÚ\_ÿë×íÿÿµõOÖ½ÿéjÿ׿ÿÞ¿ÿúü/ÿö ¥ÿ@¾¿÷ÿóA$-,ükÿ·ÿ\_ÿçÿÆfÿíãöÒ%A=0¿ú©@\_ãúïòCå ónaì?Òõÿ§ûÿÿÿå'ëDÆNÿÿÉP<·[ûçù6Ò¿Kì!@°Þß캻þÚßÿÿõoú ÿ÷VéÿKô(¯¤ÿ¯þýVtCÿÿÿÿÿ^:þëþ¯ú¥ÿé+Öÿ×÷ßÿúX>ÿëþë¿ÿÿ_µÿ_õ_·ÿý®»zÿþ¿ûÿ|ò{¯o×Ûîê¿^µëÿí{¯ýµÿ×öý¿ÿÿ{ÿß¾§×ûÿÿúM¿ÿé}þßú[ö¯«_ïµ®»]~ÞÚ_×}kÿémÚëìgM»Kÿý]/»_K´ºmöëö¿î»ýúûwë[.½³5 %k×Óm®&ÙÓïí+VëÛ3Võ $ÙXe×M
öFG^Ù«A[!Çví}Xi+q¦¬0[#J}¥ÃJÐVÃ9FÄ¡#êȧì[A^ȼøa&Á>I4Ø29ĨiÄ+a¢-±ÒĨ§± ã½%;Ãa%i¶'v² ñAÐ plTS P lPdXM1 tÖÁ@ @Á Ä&¡ÅPbié ¨m 4a´Å 5
a 4Ô!a BaPi ÐaaØR¶1 vå@A§c@B#6#X:ab" uqDDDDDDqGHDDGKõ^½WkýkôÚ_ý.þ¼¬ît0A¥N±ÝDDyi¤Ø$gaycVfËutʹXÌÈì´©ä& [äêú©Ù¬õM7 &²|8pt}Ý?Ügd¾¾Oÿ]Gõ|Z ÷ÿø¨ãããC%kÿÿåw hf¦k#C.x32s:s0ØÐ6 2q ÐÈ242s0 #FpÍ Ø1ÿúý0Nð?îw<\ýÿßÿïü&·§ÕÕéé]¯wÿ:£¨ÍháÆvæB#¡«Íl¥DäØXLdù²:t$"
c dDDäFÉá ÎÃ5O!èÔeè§ÿý¯§¤¿ô¿DL[NAdCèø¡C ¾C hú>( Á5uͲ"$Fx¹Á3l Bq ThDýôì ëuúý×i8}Ù;½4ôÛ'tnVþ½ºÿÞ¿®ý?ÓÓÓÓ´)kuÞëÙ÷,ý¯kÿ¿®Å"C»þ¯ÿO]k]ãÿMºª\_é]¿OCúÓ×ûÖ¿þ½î¸þÓt4õ}ê;þ+\Cô»éõ㥠Æm\_?ÕzÈ/ùäIÿ÷êºUëýõëûÿ\_ëo]ÿúõ¯þðô¿ÿÿuë¯áºöÚÿüWÈ/cÿúÝë½ÿëübî¡Nbúú;º¾µ^¿ë÷½5<ÿã÷ëãdA?××ø®öBçݪ\¿ÿëþ¾Öÿ{¯ÿý÷XøÁßuCÿÿÿ#¯t¢ÿZÿ¿àÊ@þÿýÿÿöúƽ·ý\_¯ÿ·\_ÿÿ Ý\_,X¡<ÿÝ|Wÿýþ?¾C£ÿ¼ºÿÿ,Cyb ¿ÿ^¥ßîßôËÆõìòS÷ûÿtÞ¾é¿×ÿõÒß@ÿ[JXÿúÿúRä;ÿÿ\_ü¨ ß
ÿ]¿ÿ~¤1w\_þúy=\^N§öòucù5Í~ÿtÿKÿtïÿÿëöÿÿûßÿéWKÿßÿßÜÖÿÞ ÷µÿ'?×%ûr$î6+Úÿïø{úµoý¿í+ÿÿÿúêÿý¿ÿ÷\_×÷¯]+ïÿÿ!ïïÿ¿¯½iÓwûëßëÍÆï¯ýE1½ûuÿõÞÚ÷\_ÿëûfßïýUúéÿOoÞW5´½í/ÿÿ®¾ëuþÿ®ßþýöþ·OVëÿÿkn±Mºÿݯ}ú®Þºé+}í·ïûj{Tö½õþø\_û]½¯{Kßi}ôÓíØ\_u»^Òÿô»¯º×Û4L5×ÿJÕî×xiv¤ßvé-¯V¥{ %ÿþb¤¿¶[÷ÈAÈÕ¬ÎȾۯëáöÏöýÖíØ0ýºþøim¤í¦Òï¶V8ïrÃ=^Ä¡êÙíCÓN t¡ ÛMnm-;«d,8½Øÿÿï¾øöÒ~?c ° áÁbqÅ6 °ÆÅ]¶iªÜBVÒ¦ +A dGµ ðb¡¡CC06QLP lS C @áÁAè6~ÿÿ×ÖÓôõýpé Ð Ã ¨aÕH0@Ø @à @اb ýÚ´ÚL0Ói Öb¡LD-Ø¥ o]8uÿûëïko·÷k(pB$ìD0@a °LîB &qÔÃGL±Ó;¦¥°°BèDC6 D3l²ÀRöP!×- ¯vOÿï^û[¯^íq¡DGD4L4A¡atÌ8)ÓTÐaS´ÓÕPaH£¯øÿÿÿúÿþýi0dØKòÕ4Z@<®+©)7,Deq¡Öjd±(H®\ÌGjÝ?%4ª }IÐäèNÈd fa3'3¹ )ÌeP #C0d¹¡\ÙÒ!æ":'d7°p{ÒÿM?\_´áëh={×Nøp@åÎÒæFHYðÿÚwoiúvÚ_öz ' µßB«J"O÷û¾w÷ÿ}¦©õqÿQóO3òsù?rCÝòò<ú{iißrC»Ñïÿÿý?»òu¡×¡\ÚA½y:þNt¢W$ è¿ÿÿ¤ÿKßÓÿïO÷Ûi}ûtõÖûÿó"FDÿNýucÿÒOv½~ÿýô5ÎÕ3f|£)Ë$ IHΣ;×%ìëTGD±Ù/øAÆØ"%@ÙÄ"BþlgQQIÅ!åFnÈÜO")4fbF £@>gQTdßÛü=ñù7öÅroÿõt/üì( 4Na; 0N!üø¦Ù Á<ò¦GLT0'Fñr\dÓ6.P!]xa7a¢!2)\· ùò È'&pýUE¼á þÿcÈT ¿_I}/û÷ô¢ñ¦¹¨L"ÍÔlBð ÃPª[AB¦¡jôk
ÖÁBhza<>3TnÕ] Û:6. °Ê]R]6AuPA®¨ÜDÿñÁõÇ_Ümkÿ÷ðÉ''Æ»¤Rm1\ÄðØIFêA°PÒAѺqat±è»~©8c§ÓMK©:M#ªXÒméX9< r(1+ªÛi+Æp}ÁÁ\Òª3¿úa×Ö|'Ó×ÓÉÿKÒ¿´Ó°sOX2è8a0·ÐpÉü +îÒt Ù#Ø8cöNõZVÓúV 2®®üO]:ïUÉE/KéÕ×únþ¾Lw÷Mkïúÿ¦º}ºûë¶^-¯Öéµ}ÛZÝ-v¬<öýöNët"ß«¤Þûý==jßЫ»tí¤¶Ð¯Þõýë{ýîÿþXõÿïÒô¤qÿ¿×v8Ú¾5¾ZWÿ«ÓÓô#ãô?ÿÿb¿úê?õéizT/ÿý®²tÉU!ûïuõ¶»]ýúoÿ¯õpÿÿª×ß¾Æ
oúÿÿ_¯á÷ûÿ_ÇÿÃ}ÿÿR~ÍC§®OݵÅqhßÇúÒ§ÿò(çû"ª×_Ú;ºþµÝéC]/õ×Êõu û¿!�þ±×þ¿ÆÔ²ýzpuÿ÷7=ÿ¾H«õ§¿¥þÿÿËþÿÿ¯ÿ÷W²kÖu btàëJ?þÿÿ÷ÿÿtµØë×þ÷^Úÿï¿H-ÿòÄ}\6ÿÿt¿ÿ×J÷Iø"OþXÄþÿùbïÿßå¨Éû·Z§ßûßÿ¾úéþ©t¯ßÿüê.ïø"?ý¯ÿïOùb? ð?zU÷ø7ÿÿúþ¾¤1}Øa&ÉYOÿöSa}³ëþ¯ÞU}_¥ÿüÿãí'¯ïû[£Xÿýº |Gê×ÿëÿÿýÿOÛpV!1ÐÝy ñØøþ/þLÿ¹ÿ ¼ýýÿûö·ÿýjý}?ö»ÿáÿëý×íÞ¾ÿÍ"ªÞ¯ÿT׿úßÖ·½uµÿÌÖASúÿÅ}·Lmî÷þ¾«÷ÿëçkþÝÝkïïOoj{ÒßÿjöA^ýï[î×ÿûkû¯ýþ×ø¢
ïÿÿöþ¾Ý½wöÛúúý¤Û{þ¾¡§ÿû Bk°Óý{÷ôõØJ aÿ¯ïÿ×û}Õ?»¿ý}~ì-÷_÷ÿ·v¯¶·ú^Þ !!á!t@ &CLB!&0^ìþ¼2?öM¥¯o߶×ïI¶×¾í¯°ìN>EËj½¤ÚW"c{=6{obPþD']rI¶ Ý°Îwi_È.n+¦Å]±PÐI¥i'wíiñ6Uil¾»äL±8þ%ö(8Pè Û Ãl$ASb¦ 0h PI¶(60H&4ôAà ÓAPa[à S pÚ0A°¡amm&Å-Å1tà &¦´I§¬RvîDC Ða4A
\÷iëÃ[\_¶0½¬4öÓê"""QÁJ """DÑ ¦"3Q¥DDDDDDDDDDD ºãªz\_]-|[KiX1+¬fBh+ÈÐeu1%2º·Gb^§ªàê\4?ÿý·|Óó²ëù¤ÿßa'´®#Ø\kýÿþºÿýÿóµ!ýäÄTemuåruL'ÉÄ$ gÐgDpᣠÌÞhÁ.(fHa hÈy²5&@³djñLÑìÀÈeæÈéfbtÍóD%ð§"o ÑcÁ|qÞ .(D(ð pi£YT3ä 4ïÐD&Pfd¡�Ápg<\:!pz74.JòÝC¤ J¶Ú ´l±8ì#|"íÅÁ(j-XµTm ŦJ8}81¤ÜÈóWI¶+AÁa;V aIáO¨'í©.ÓÈÎK Æi82X%wôÒ׺àÖ°¥ïMÓþ½?Z×Õ¤ÿMöÚ6PºO·M\_¨ÿ}m 7êûñ^´¿é¿éºÅéëé×ú§÷h{IºßÛ\_Çúë\_ÿëÿÿúÒýzýzÿµü?àÖµÿÿ¿¯ÿ×ýá×ïyÇ¥ëýn¿ÿ!/d'Ö+éü³ýÿÿöÿôK1=éýÍwÿÿûýÿÿ° gABtè3Á@ÈeYVBÆEGÿÿÿ×êðD¯ëeXË¿n«J=ÿþý½å÷ü? ¦¶"»#AH1B2Ð|3 ÿÿÿÝ$ ÿ@¿\_ëèÿÿþ ßä1\_þýþ B 0zÿÿÿÿûb\_ú ÿm'¯Kÿÿ¯ðÁÿÿûôï ==ÿõïÿï~ß\_·ö·×ÿ÷ßî÷µ¿ÿ¿M}4Õ4×íÚý«û׺èÁ\_ÿÿtô×Û§ÿó?µÏnîÿkÿ§¦Ó´ÿÛ×Ýÿëé \_µ×î8ªÿoÿ¯ýöÿï¿zóNý³ßIçÐ\Ò"E#ë»ýµn¿}Am}µ×ÞÒÖþíoÿ½+ é7·}¯§"ð~E)"õq¡È¼ÖtH¹!k6é|0=oµÿá^¿kk÷ö
ëÞKþELJzv æ äȹn÷ô?'Òu+ê)YàÒNØ®äKm¥Ý§Ä1)Ú{{nj¡N>Eç¤Ýiàµ#5pa&0±XA¶AÿÓ¥ïÿOMk°BA8i6 ÁÐ@Ø0A Ö0j
pØPÜ80I A:i {VI±Ko¿«ßïßMø¯ Õ8L㧠Pa>ïMÓU¸kjû /á5µêÄ&û3Dºª(Gׯÿ¯ù!þ¿§¬DDDDDDDQÄDDDDDDDDDDDDDD{»ðÁ~HÒôª½c¥ò ?{ñI_ÿµÌAt]0úÒ/yßý-±4ôµ×ùß¿úñPáÝÈÿÕÿß¿µ«jîí'²\èb¿ßë^ÓZïNé¸]û^÷É -2G]Þ¿m\ÿwïÜZfÁä%¹2/d^uÕ´ªDr(sý¾¯\_u«è=7OO¿bBíür)%ßÚ3Ö.6>¼0b¥ ÿ}µ×Öé¿}ßê#!ú[îêøx§ëÿü!ÿ×ðÈã¯ÿ¯ÿí%¿Ðÿ×ÿü¶ 2FvEï«÷úþ>lómÿ̬Aå&vUrh¢:#¢:.è®(gÌa0/í-ëî¿Ô'iûù¤,)Ù Íbc;$""" É.LGfPS9TKÿjýÿ©V=ý¿«ýÒéµµæE,Á jpáì#=ö:ûi+ݯjki{k¥Ü0@ààÐ}Yd?ÝßA tÿ°Èei:[iv°V +[ Ê:ïü7ÿÿêRYصü{ìqññLqÆÅ1ö½m}¶M}4ª)/þÖý5Öí5[\ßÄvÄGìõý¯Ýö¶©Ý¦ºëK^ÿªõÿµ¾ÓNûNÓ ¦¿¿ÿýûþOì&¶Ó ÓMÿÿÿýkpÖøi
´áA &5;$"SÞ§yÿîPf¡äHD2%Ù¨¼5ó¡Eù²5\è2ÙD0a 15°a,"%pÁ 0Íd ó'döC#¨ÌÅ>FÍÐgA3dFCÎEátHÊØ fÌ Ï Í ÁD?ÁøC>w Ís# "/Sa;GA £t(PÒÑ#´vÁO Á;P< "cxa322¬ ÏàÏAú60PÈTn P áPXxͪ´FÈEÅ}!^U:N1=ôL4 »èÝFÆ a ½4nhm EÃX¿TnúM §Äñ§jè?6Òläã 6¤Äöþì&ð ´l8>O%Ú}RxN%Îaä.ààé?¤ÚNé;¤ð¤ïïêîg.Ù(¸2Éá=:N.áïÓè/»ûÿxMí·I\ éZWN°éÿÿ]/ÿºoõ¯éÿ[¡ëÝm/o[¯ú\_ªJÿ¤þ?ýôؾýïÿc«×cwþÿ×í]x¯×øénßÿúû¶¾Úêïõÿ×cÕkÛ÷öêÿÿÿ맯÷ù dB?®î©ÿWð¿¯þ«¯þÿ ê/GEÅöDF©öEs¿\_ÿÐÿý§Ðüà{»·¿ëÿëå:ãòOý}& ýþ¯ðzWþ¿ýé×Ô7úzºMÿþþ[X/pD£ÿ¿ÃKJN®}þÃý'ÿþLcÿåoåíCËw÷é_ÿú½Kòë,G@¿ÿðØýx>ÿt×ÿúÿÿ¿ÚÑÿ·úOÿúÖkõþÿ¯¤ß® ߤÿ×÷þ»ÿÃíû÷_ÿßOÿÿ¥ÿïáÿ¼?ïÒo_nÿ÷ûýþv·¥Ûøu×¾¿zÿÿÿõ¼>çëç±Å?ëÿþ¿Ú÷ý¥·ÛýëíýÕm[Ò¶ÿÿ÷óëÒ¶û~ëï¯ûÿ¿¿í¾¿ö»ûfoêÖmu¾¿×ÿ] 7~ÿõºêØ_Ûþý»n¶ÿ¥¯·ö]>ÚL5Øakm_´¯êÿ¿kÕMÛ>¿VÒ_^¿±ÿ¶zþEÎÃ90Îm¬»ÃI4Ó¶^öJvJÓ ÷ löÃ=ħiþ¯Ú¡ 64á´®á¤Öi± i lUÁئ6A4ÅaÃX mU1 ¨aA ±Q1± ¨a p{ $A 1páa°M£PvNa -àÁa4áFÓ# ѧa= T&'à qûû¨iØ$&S|Ä0B $!f%®ZÁ bdEÆuk5:8AA 4¦UMziúi Ðz}4ÕSN×_x]uµOHD¶x0²%ÑÜEÏ"F=?n§pôú«Ó¯¾Jµ½i:[§ÿxÿþ#ú¥õþ½zýÓÿÅw!.wÇRåܶ×ÿÎì'Ú¬C^[Gcyn:,µ^±' ú\D? Cù¬e83³WK ª ÿ~ÑÔ=ý+ªÊIý.×z¿Ø88{wõô»ÿ¿ªÿ¾å¹Fd·ùÚ.äào?ê;5¼í=epd;}WÒïõK]4îEçõõ £ªëþÿÿOú!]
ÿÿÿ_ÿ-ÌÄÿK¦ßà§_ÿÿÿÿîÿÄþmÿÿÿUþ¿÷_ÿÿÉèDL¯:^}k¿2qI²ÈA ÑñLÈè)0!#(ÅÈbC±NÛ!3|¢ðÌÅ:&EBpS1:×ûÿ÷ßÙó x e NðCùñø gA ¸ÙD$fÆ@Ä4¢6YFm³2zÈbó!gh¸ ,vpÏðÑc¹Ë
¸ øÔ$A\áª5³Ã<4DO òÜ(o~?Þ>\Ô"z5/D/Oѱ°·
ÊalA]xi
]°PD#Ô¸ð)á¢Ï Q æ°Ñ³L0¬H¸h#ú0ü °PÑ® =bÕ A7R ¾ iÿÿ̾ á¦Õ&ѳé6V-ìZ ¸°ÍÚñ~Ô(A7M%ÎF®¼ éi8pzzI°IÃd°O&ßKAI°àò5zIýz÷ßýX3"¥.º úM÷L ý+ Rpd¹ ¤Ø2YGÁäºGDàÇ"ºÛÞú¯umúÁ]m½4ûô®ºWàÖªß{®EÖÒÒ¿Kïú½wþ®ë«Þnt®;ôÛàÒ·%a N=~íÿuÝOOMi[iv¯¾ÿUúßµöû[ÛZëÕêÿ]õî¶Ó¥×Z¾ýO\_WC¾û}c\_ãö¿uï¿ýkþÚ»öPÒ Ã C -øÿ\_Ò·§õÿ{~£õÿÿï^ëÿ8ÿµýjÿë¿ÿúýz¨Øþ66\6)×Iÿ¯ûÎô¿ÿ\_~¯ÿÿ×Ï.³²±Ò´¢ÈÿȦ¡ÿÿÿiF¾ßíûáÆÕ{Uµÿã×\Äÿº¿¾?ÿ<ßÿßÿÇó±qÞ»ïÿØ>"¿ÿÿÒú÷ô¿JÚvi«ü¤Nÿúé~ÿÿÿ_ÿþº/»öá»íòuÿÿöÒ«Ýÿ´kÚi¦jrÅ; _~ôÿ\§ýk¿ÿÿþ¶ÝúþàßKÿÿÿÿûi~,&´ &a5ÿÛT¿û[®÷ÿÜþÿ¿ûµ×÷të:ù¨ûoÿÿm'ÿ×þ°aDL¿·î¿ÿºÿýÿõýz×ßÓõÿu¤¸{ýÿî¿ïýµÿVÒ¿ýþµýÓÿ¿÷ÿßÕÿþºÿ[Õç·ÞÕÿ÷¦ýµ÷×ÿwu¯¿Å×½¯ÿ×ï_uïZÛVºýÛ©;Òÿ×{^µÞëöÛïÚý=ÖÿÛ×ßékÚ]þî½uzÕÒÛµVý}ºí¶ktÿ2Ü3ýÚV¾ôõ~é{w_ý¥koûiD§ü®ì0Eº´ 0ôùéO«KþÒirbS¶ê%äRjÒú²1¥êÚUÈöËÛK¸g6Õ>DÏ D¡ò.4¹)5 pQW $ÚLT; qMC¦Ò Á²:a¶)¨l0I¨{°a »b nÐAÃb Å6A @Áj H6i0®övÅ1M\Bh4ÕáJÐa¬1 §k W "0AA¦¿bµvK!'¯kå:aa¦!m1B""""fË0a0º 3¯X!0B ÏP!!¡Db"5]Rü¦jS%¦dY#¹ÄMrÓ+êdØ2©Äd$GD2n';%ðZ§ªØ:ôz0áÑá\_~©):ÚH\Í~ÑSÈõh)/R¢Ì¥kkª¼éuÇWªþ¿ÿëßþ¿×ïÿßÿýNÒò+ÿæOdùÃÈdDfk#.¹3$"sÌdj/:25@ÈdF.e9sµ6æ²æBOd#'ãbÄ$fÇæÈê2¬BÈdLójW"9 CÎÔhåóÙv(GÈñìÌl$\2v%Ì Á8Fg ùAÝ UA ÐÐa ±¢áºO ða ù&0Ñn±ÃÐ0з >AB-Ã>AgÈ,;BÕbaìW Ò7 ¡Q² i¨Fæo
Bñ¸0Dj(ØaBaxb{liÕÒwtXíqa=4h p·6A°'ÐAµMUT/\0@Ð88dy@¡Á>Õ$Øc¤8:ÛaªÏÌ á&ѹÛA°azMYØe Òpd¹ÉJêM¥{I´ª'ÊM¹A°bW4:zö\vJ/Ý$ì]'÷ØMºé4 ´
Ñ7O¿ÓÓÚn÷ïôÚMpzu}öõ}°Ð2m.ÉE-§§dî ÈuôÝï}ÕnÿúÓÕíw¿úßúúèUÞ¿§ÿéêü^«Ký÷vèUºmn×zèoþ©þǯ[QÓWññzn¾ÿþÿk÷Õ÷·þ½uéu÷Ö¿¯íÿ\8ÿýWü:×UÿÖ¾¾û×úðÖ×^½ÿ~·tþÿìè%~þÁÈûò(ê«\ÈO!íÿ{ßÿÿãdM<},ïñþþÿÿþ ¯øû¿\û(\_ýoÖ¿ÿÿ GP?¯¯õÿ^«þÿ·[øa·oüOÝ'nLP{ÿݪÿõÿü±8@¥ûþXÿäÆ{KÿAWúëáÃzÿ_V¨àÝ ½¯õ[ú¿ÿëù ëýkµê¿ÿô¿¦ûüêD¿J¿ÿüÔvßÿ×·_úÿðzÿÿýß¿ÒðÿÿzNþ?ýý~¾®ÿ÷ô¿ë×Kÿû}{ëÿÒõZý¿×ÿÿ×Ï,OÿÝÝý~}?_Oþî¿þÿ÷ k_vÿûëW¿KUï_Ø£Uuöÿ®ß÷VºýÚëk´·ýu×Kÿ¿½/÷T¶ý+í×ýo𮿥÷Ú^ß½kkíëö»Am¥Ûl5oûöï°»mýýÚ]¥i7ôÃÃ×0?Oj®¿í+´¶ýyHÈÖ¶Òlÿa{bPﶬ0Jwi'².vËJ%Mr´®Eç°Ïpa·Pa&˽PÓTúl%ià &»TÅ1Ó¶LJxQĨi»MLGPÌÐ lB m5AÅAL4à ÁÅEâ µ´ØNMX Ð0ðÄ @!E5 ,4ÐjÕ®(0pÚM @VaQ©Xá0· .&!5j B¶¼0Xb AµB""""-$!æ@ÌB83Vc B! OÈC8ã""#,ȽëñDDF?ïI%²¢±Ìåg8à»ê½b""#ýêþºÕÖ
ô¿ýJå©köB¦ñT BBj"!ÿÿÿÿÿÿÎÅ@åÁH2:0dtGD©£<'ÈñxÆGÈìeÅ.ÈèEÈÃ(¼\3GÌ Ær=D<ÉIËQFh_#¢9ѳ9FÑG.)³#Ù|D|!¶)s#ÅÈ×x!/ñ!p]G2è¼]Ù#¼\`VGDG®"""""""""#; æè2:#¢tvXÑ¢<¢:4DYTD#¾DþgE>c#æ2äs0ä|£¾aÌ">|B> endobj 22 0 obj <> endobj 23 0 obj <> endobj 24 0 obj <> endobj 25 0 obj <> endobj 26 0 obj <> endobj 27 0 obj <> endobj 28 0 obj <> endobj 29 0 obj <> endobj 30 0 obj <> endobj 31 0 obj <> endobj 32 0 obj [/PDF/Text/ImageB/ImageI/ImageC] endobj 33 0 obj <> endobj 34 0 obj <> endobj 35 0 obj <> endobj 36 0 obj <> endobj 37 0 obj <> endobj 38 0 obj <> endobj 39 0 obj <> endobj 40 0 obj <> endobj 41 0 obj <> endobj 42 0 obj <> endobj 43 0 obj <>stream 2016-10-03T15:20:03-07:00 2016-10-03T15:20:03-07:00 2016-10-03T15:20:03-07:00 Appligent pdfHarmony 2.0 uuid:02c05f25-a405-11b2-0a00-782dad000000 uuid:02c09d5c-a405-11b2-0a00-007f872bfc7f application/pdf Cotenants: Right to Possession v. Right to Contribution Colleen Clemente Prince 9.0 rev 5 (www.princexml.com) pdfHarmony 2.0 Linux Kernel 2.6 64bit Mar 13 2012 Library 9.0.1 endstream endobj 44 0 obj <> endobj xref 0 45 0000000000 65535 f 0000077383 00000 n 0000077687 00000 n 0000077719 00000 n 0000077786 00000 n 0000079120 00000 n 0000080391 00000 n 0000081705 00000 n 0000082928 00000 n 0000084193 00000 n 0000085496 00000 n 0000086723 00000 n 0000087983 00000 n 0000197355 00000 n 0000197662 00000 n 0000197695 00000 n 0000204650 00000 n 0000308779 00000 n 0000309086 00000 n 0000309119 00000 n 0000317505 00000 n 0000421114 00000 n 0000421215 00000 n 0000421319 00000 n 0000421427 00000 n 0000421529 00000 n 0000421634 00000 n 0000421743 00000 n 0000421842 00000 n 0000421940 00000 n 0000422040 00000 n 0000422144 00000 n 0000422239 00000 n 0000422288 00000 n 0000422385 00000 n 0000422439 00000 n 0000422585 00000 n 0000422681 00000 n 0000422733 00000 n 0000422913 00000 n 0000423031 00000 n 0000423119 00000 n 0000423191 00000 n 0000423264 00000 n 0000427271 00000 n trailer <<7E9AC00205A4B2110A00A084872BFC7F>]>> startxref 116 %%EOF~~~~~~ |
7741 | https://brainly.com/question/20366117 | [FREE] Simplify the expression: \frac{x^2 + ax - 2a^2}{3a^2 - 2ax - x^2} - brainly.com
5
Search
Learning Mode
Cancel
Log in / Join for free
Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions
Log in
Join for free
Tutoring Session
+89,6k
Smart guidance, rooted in what you’re studying
Get Guidance
Test Prep
+15,8k
Ace exams faster, with practice that adapts to you
Practice
Worksheets
+9k
Guided help for every grade, topic or textbook
Complete
See more
/
Mathematics
Textbook & Expert-Verified
Textbook & Expert-Verified
Simplify the expression:
3 a 2−2 a x−x 2 x 2+a x−2 a 2
1
See answer Explain with Learning Companion
NEW
Asked by apouche2020 • 12/30/2020
0:00
/
0:15
Read More
Community
by Students
Brainly
by Experts
ChatGPT
by OpenAI
Gemini
Google AI
Community Answer
This answer helped 286006 people
286K
5.0
1
Upload your school material for a more relevant answer
Answer:
−3 a+x x+2 a
Step-by-step explanation:
3 a 2−2 a x−x 2 x+a x−2 a 2
i) write ax as a difference
3 a 2−2 a x−x 2 x 2+2 a x−a x−2 a 2
ii) write -2ax as a difference
3 a 2+a x−3 a x−x 2 x 2+2 a x−a x−2 a 2
iii) factor out x from the expression
3 a 2+a x−3 a x−x 2 x(x+2 a)−a x−2 a 2
iv) factor out -a from the expression
3 a 2+a x−3 a x−x 2 x(x+2 a)−a(x+2 a)
v) factor out a from the expression
a(3 a+x)−3 a x−x 2 x(x+2 a)−a(x+2 a)
vi) factor out -x from the expression
a(3 a+x)−x(3 a+x)x(x+2 a)−a(x+2 a)
vii) factor out x+2a from the expression
a(3 a+x)−x(3 a+x)(x+2 a)(x−a)
viii) factor out 3a+x from the expression
(3 a+x)(a−x)(x+2 a)(x−a)
ix) factor out the negative sign from the expression and rearrange the term
(3 a+x)(a−x)(x+2 a)(−(−a−x))
x) reduce the fraction a-x
(3 a+x)(x+2 a)(−1)
−3 a+x x+2 a
Answered by rebeccarayshma •206 answers•286K people helped
Thanks 1
5.0
(2 votes)
3
Textbook &Expert-Verified⬈(opens in a new tab)
This answer helped 286006 people
286K
5.0
1
Physics for K-12 - Sunil Singh
Celestial Mechanics - Jeremy Tatum
Mechanics and Relativity - Timon Idema
Upload your school material for a more relevant answer
The expression 3 a 2−2 a x−x 2 x 2+a x−2 a 2 simplifies to −x+3 a x+2 a after factoring the numerator and denominator and cancelling the common factor. By identifying numbers that satisfy the required product and sum conditions for factoring, we can arrive at the simplified result. Always remember to check for any restrictions on the variables after simplification to avoid division by zero.
Explanation
To simplify the expression:
3 a 2−2 a x−x 2 x 2+a x−2 a 2 we will factor both the numerator and the denominator step by step.
Step 1: Factor the Numerator
The numerator is x 2+a x−2 a 2. We need to factor this quadratic expression. To do this, we look for two numbers that multiply to −2 a 2 and add to a. These numbers are 2 a and −a.
Thus, we can rewrite the numerator as:
(x+2 a)(x−a)
Step 2: Factor the Denominator
The denominator is 3 a 2−2 a x−x 2. To factor this, we rearrange it:
−x 2−2 a x+3 a 2
We can factor out −1:
−x 2−2 a x+3 a 2=−(x 2+2 a x−3 a 2)
Now, we need to factor x 2+2 a x−3 a 2. We look for two numbers that multiply to −3 a 2 and add to 2 a, which are 3 a and −a.
Thus, we rewrite the denominator as:
−(x+3 a)(x−a)
Step 3: Combine the Expression
Substituting back, we have:
−(x+3 a)(x−a)(x+2 a)(x−a)
Step 4: Cancel Out Common Factors
Since both the numerator and the denominator have a common factor of x−a, we can cancel it out (assuming x=a):
=−x+3 a x+2 a
Final Answer
The simplified form of the expression is:
−x+3 a x+2 a
Examples & Evidence
For example, if we put x=1 and a=2 in the simplified expression, we get −1+6 1+4=−7 5. This shows how substituting values into the simplified expression works.
The process of factoring quadratic expressions and cancelling common factors is a well-established method in algebra, ensuring accurate simplification of rational expressions.
Thanks 1
5.0
(2 votes)
Advertisement
apouche2020 has a question! Can you help?
Add your answer See Expert-Verified Answer
### Free Mathematics solutions and answers
Community Answer 5.0 1 simplify 3a(a^2-3a+4)-4(3a^3-2a^2)
Community Answer 1 Simplify (2 Points) 2a xa x 3a + b x 45
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25
New questions in Mathematics
If you work for an hourly wage, your gross pay is a function of the number of hours that you work. If you earn $8 per hour, complete the following table: | Number of Hours | 0 | 3 | 5 | 7 | 10 | 12 | :--- :--- :--- | Gross Pay | | | | | | | If you were to plot the ordered pairs on a set of axes, in which quadrant would the domain and range of the function lie?
The pressure, p, of water (in pounds per square foot) at a depth of d feet below the surface is given by the formula p=15+33 15d. If a diver is 100 feet below the surface, what is the pressure of the water on the diver? A. 15.0 pounds per square feet B. 38.63 pounds per square feet C. 1545.45 pounds per square feet D. 60.45 pounds per square feet
If PR=4 x−2 and RS=3 x−5, which expression represents PS? A. x−7 B. x−3 C. 7 x−7 D. $7 x+3
What can you conclude about a polynomial if its graph has 4 turning points? A. The polynomial has a degree of 4 B. The polynomial has a degree of 5 C. The polynomial cannot be factored D. The polynomial has only 3 zeros
Suppose that f(x)=l n(x 3+3)8. Find f′(3).
Previous questionNext question
Learn
Practice
Test
Open in Learning Companion
Company
Copyright Policy
Privacy Policy
Cookie Preferences
Insights: The Brainly Blog
Advertise with us
Careers
Homework Questions & Answers
Help
Terms of Use
Help Center
Safety Center
Responsible Disclosure Agreement
Connect with us
(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
Brainly.com
Dismiss
Materials from your teacher, like lecture notes or study guides,
help Brainly adjust this answer to fit your needs.
Dismiss |
7742 | https://www.omnicalculator.com/math/pentagon | Pentagon Calculator
With this pentagon calculator, you'll find essential properties of a regular pentagon: side, diagonal, height, perimeter, and area, as well as the circumcircle and incircle radius. Type any value, and the remaining parameters will be calculated on the spot.
If you are not sure what a pentagon is or how many sides a pentagon has, keep scrolling, and you'll find clarifying pictures with a short explanation.
What is a pentagon? How many sides does a pentagon have?
Pentagon is a 5-sided polygon. A pentagon can be simple or self-intersecting.
The sum of the internal angles in a simple pentagon is 540°, so every internal angle is equal to 108°. A regular simple pentagon has all five sides equal in length. (In this article, we use the term "regular pentagon" to describe a regular simple pentagon).
Area and perimeter of a regular pentagon
Area A of a regular pentagon can be calculated from the formula:
area = a² × √(25 + 10√5) / 4, where a is a side of a regular pentagon.
Also, you can find the area having the circumscribed circle radius:
area = 5R² × √[(5 + √5)/2] / 4, where R is the circumcircle radius.
Perimeter P of a regular pentagon is equal to the side length multiplied by the number of vertices. A pentagon is a five-sided polygon, so the perimeter is:
perimeter = 5 × a
How do I find the height and diagonal of a pentagon?
To calculate the height and diagonal of a regular pentagon, all you need to have given is the side length a:
diagonal = a × (1 + √5) / 2
height = a × √(5 + 2√5) / 2
A pentagon has five diagonals equal in length, which form a pentagram.
How to solve a regular pentagon using this pentagon calculator?
As we now know the pentagon definition, we can have a look at this step-by-step example:
Find out what is given. For a regular pentagon, one parameter is enough to find the remaining six.
Type the value into the pentagon calculator. Let's take the most famous, almost regular pentagon as an example – the Pentagon building, the headquarters of the US Department of Defense. From the Wikipedia page, we find out that it's 1414 ft wide – it's the height of the pentagram.
The Pentagon, 1,414 feet, 431m (Light blue)
RMS Queen Mary 2, 1,132 feet, 345m (Pink)
US Navy's nuclear-powered USS Enterprise, 1,123 feet, 342m (Yellow)
Airship LZ 129 Hindenburg, 804 feet, 245m (Green)
Imperial Japanese Navy's Yamato, 863 feet, 263m (Dark blue)
Empire State Building, 1,454 feet, 443m (Grey)
Knock Nevis supertanker, 1,503 feet, 458m (Red)
Apple Park main building, 1,522 feet, 458m (Green)
The pentagon parameters appear! They are:
Side – 918.9 ft;
Diagonal – 1486.8 ft;
Perimeter – 4594 ft (0.87 mi);
Area – 33.35 ac;
Circumcircle radius – 781.6 ft; and
Incircle radius – 632.4 ft.
Did you notice how enormous it is? Have a look at the perimeter – it's almost a mile! In reality, each side of the building is ~921 feet long – it looks like it's practically a regular pentagon!
Other regular shapes
If you are interested in other regular shapes, have a look at our great tools:
Triangle calculator
Square calculator
Hexagon calculator
Octagon calculator
FAQs
How do I calculate the area of a pentagon with side 2?
To compute the area of a regular pentagon from side length, you need to apply the formula:
area = a² × √(25 + 10√5) / 4.
Plugging in a = 2, we obtain area = 2² × √(25 + 10√5) / 4 = √(25 + 10√5) ≈ 6.882.
How do I calculate the apothem of a pentagon?
To compute the apothem of a pentagon with side length a, apply the formula:
apothem = 0.5 × a / tan(π/5)
By simplifying tan, we obtain the following:
apothem = 0.1 × a × √(25 + 10√5 ).
How do I calculate the pentagon internal angle?
To compute the internal angle of a pentagon:
Divide 360° by the number of sides: 360°/5 = 72°.
Subtract 72° from 180° to get the internal angle of a pentagon: 180° - 72° = 108°.
It follows that the sum of internal angles in a pentagon equals 5 × 108° = 540°.
© Omni Calculator
Did we solve your problem today?
Check out 24 similar 2d geometry calculators 📏
Area
Area of a rectangle
Area of crescent |
7743 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9218826/ | Interval laparoscopic appendectomy after laparotomy drainage for acute appendicitis with abscess: A case report - PMC
Skip to main content
An official website of the United States government
Here's how you know
Here's how you know
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.
Search
Log in
Dashboard
Publications
Account settings
Log out
Search… Search NCBI
Primary site navigation
Search
Logged in as:
Dashboard
Publications
Account settings
Log in
Search PMC Full-Text Archive
Search in PMC
Advanced Search
Journal List
User Guide
New Try this search in PMC Beta Search
View on publisher site
Download PDF
Add to Collections
Cite
Permalink PERMALINK
Copy
As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
Int J Surg Case Rep
. 2022 Jun 18;96:107319. doi: 10.1016/j.ijscr.2022.107319
Search in PMC
Search in PubMed
View in NLM Catalog
Add to search
Interval laparoscopic appendectomy after laparotomy drainage for acute appendicitis with abscess: A case report
Toshiyuki Suzuki
Toshiyuki Suzuki
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
Find articles by Toshiyuki Suzuki
1,⁎, Akiyo Matsumoto
Akiyo Matsumoto
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
Find articles by Akiyo Matsumoto
1, Takahiko Akao
Takahiko Akao
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
Find articles by Takahiko Akao
1, Seiji Kobayashi
Seiji Kobayashi
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
Find articles by Seiji Kobayashi
1, Hiroshi Matsumoto
Hiroshi Matsumoto
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
Find articles by Hiroshi Matsumoto
1
Author information
Article notes
Copyright and License information
1 Department of Surgery, Hanyu General Hospital, Hanyushi Saitama 348-8505, Japan
⁎
Corresponding author. t.suzuki@fureaihosp.or.jp
Received 2022 May 7; Revised 2022 Jun 13; Accepted 2022 Jun 15; Collection date 2022 Jul.
© 2022 The Authors
This is an open access article under the CC BY license (
PMC Copyright notice
PMCID: PMC9218826 PMID: 35738141
Abstract
Introduction
Immediate appendectomy for acute appendicitis with abscess has a high frequency of ileocecal resection and postoperative complications compared with interval appendectomy after conservative treatment. The optimal approach to acute appendicitis with abscess remains controversial.
Presentation of case
A 69-year-old woman was referred to our hospital for abdominal pain. A computed tomography scan revealed an enlarged abscess around the cecum. The diagnosis was perforated appendicitis with abscess, and conservative treatment was performed. Percutaneous drainage was difficult because the abscess was near the intestinal tract. Because of the persistence of symptoms on the fourth day of hospitalization, laparotomy drainage was performed, and the patient's condition improved afterwards. Colonoscopy was performed on an outpatient follow-up to rule out malignant tumors of the colon. Interval laparoscopic appendectomy was performed 3 months after discharge to prevent appendicitis. The postoperative course was uneventful.
Discussion
For this case of acute appendicitis with abscess, conservative treatment such as antibiotic therapy and laparotomy drainage was performed. Laparotomy drainage enabled us to approach the abscess directly and minimized the risk of its spread into the abdominal cavity compared to the laparoscopic approach. Interval laparoscopic appendectomy was more effective and easier for this case of appendectomy, wherein adhesions to the abdominal wall were expected compared to laparotomy.
Conclusion
Conservative treatment approaches, such as drainage and antibiotic therapy, can be first-line for appendicitis with abscesses. Interval laparoscopic appendectomy can be useful to resect the appendix and observe the abdominal cavity.
Abbreviations: CT, computed tomography; CRP, C-reactive protein; POD, postoperative day; WBC, white blood cell
Keywords: Appendicitis with abscess, Conservative treatment, Drainage, Interval laparoscopic appendectomy, Case report
Highlights
•
For appendicitis with abscess, drainage and antibiotics are the first-line approach.
•
Laparotomy drainage is an effective alternative to percutaneous drainage.
•
Colonoscopy should be done after conservative treatment to rule out tumors.
•
Interval laparoscopic appendectomy is effective for appendicitis with abscess.
1. Introduction
The optimal treatment of acute appendicitis with abscess is currently a matter of debate . A meta-analysis by Similis et al. revealed that conservative treatment of acute appendicitis with abscess was associated with significantly fewer overall complications compared to immediate appendectomy . However, another meta-analysis by Furgazzola et al. found that, among children with acute appendicitis with abscess, conservative treatment was associated with lower complication and readmission rates compared with immediate appendectomy . However, the treatment options for appendicitis with abscess in adults remain controversial.
Here, we report an adult case of perforated appendicitis with abscess treated with laparotomy drainage followed by interval laparoscopic appendectomy. This report aims to provide information regarding the treatment for appendicitis with abscess. This case report has been reported in line with the SCARE 2020 criteria .
2. Presentation of case
A 69-year-old woman sought consultation at our institution for decreased appetite and lower abdominal pain. The patient was noted to have a fever 10 days ago, which eventually resolved at the time of consultation (Body temperature: 36.1°C). Lower abdominal pain and tenderness were noted, but there was no rigidity or rebound tenderness. Laboratory tests revealed a white blood cell (WBC) count of 20,790/μl and C-reactive protein (CRP) of 37.6 mg/dL. Computed tomography (CT) scan revealed an abscess around the cecum and ascites in the pelvis (Fig. 1). The presence of appendicitis caused the appendix to collapse and form an abscess. Thus, the patient was diagnosed with perforated appendicitis with abscess. There were no signs of peritoneal irritation, and vital signs were stable. Therefore, conservative treatment with antibiotic therapy was initiated. Despite therapy with meropenem, the patient had a relapse of fever. Abdominal ultrasonography showed an enlarged abscess located behind the intestinal tract, making percutaneous drainage difficult. Therefore, laparotomy drainage was performed by a gastrointestinal surgeon at a district general hospital on the fourth day of hospitalization. The abdomen was opened a via gridiron incision. The area near the cecum was packed with purulent fluid. The purulent fluid was drained, and the peritoneal cavity was thoroughly irrigated with normal saline. A Penrose drain was placed in the abscess cavity. The symptoms improved after the surgery. The drain was removed on the 12th postoperative day (POD), and the patient was discharged on the 15th POD.
Fig. 1.
Open in a new tab
Computed tomography revealed an abscess around the cecum (blue arrow). (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
Colonoscopy was performed one month after discharge to rule out malignant tumors of the colon. An endoscopic mucosal resection was performed because of a polyp in the cecum. A pathologic examination revealed an inflammatory polyp (i.e., inflammatory granulation tissue), and there were no malignant findings.
On the follow-up, WBC and CRP values were within the normal range. On CT scan detected abscess around the cecum disappearing and appendix with gas (Fig. 2A–B). The risk of occult appendiceal neoplasm increases with increased age, with a 16% risk in patients aged ≥50 years. Given these findings, interval appendectomy is recommended for all patients with complicated appendicitis and aged ≥30 years. Thus, interval laparoscopic appendectomy was performed by a gastrointestinal surgeon at a district general hospital 3 months post-discharge. The abdominal wall and cecum at the site of the previous incision were adherent (Fig. 3A). The appendix was also mildly adherent to the surrounding tissue, and the appendix stump cutting surgery was easily performed (Fig. 3B). The status of the base of the appendix was apparently normal. The postoperative course was unremarkable, and the patient was discharged on the 2nd POD. The pathological finding was granulomatous appendicitis, and there were no malignant findings.
Fig. 2.
Open in a new tab
Computed tomography findings. (A) The abscess around the cecum was disappearing.
(B) An appendix with gas (red arrow). (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
Fig. 3.
Open in a new tab
Intraoperative findings. (A) The cecum (yellow arrow) had adhered to the abdominal wall (blue arrow).
(B) The adhesion between the appendix (red arrow) and cecum (yellow arrow) was mild. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
3. Discussion
The proper management of perforated appendicitis with abscess is a clinically important issue . It is technically demanding to perform immediate surgery for patients with acute appendicitis with abscess due to the distorted anatomy and difficulty in closing the appendix stump, with occasional ileocecal resection . Moreover, it is also associated with a higher morbidity compared to nonsurgical treatment (i.e., drainage and/or antibiotic therapy) . Some reports have noted the efficacy of appendectomy for appendicitis with abscess , , . However, because it was difficult to close the appendix stump, antibiotic therapy was done instead, in order to avoid ileocecal resection and to reduce postoperative complications.
In some cases of appendicitis with abscess, antibiotic therapy alone is not sufficient and case drainage may be effective . Abscess drainage can be done through percutaneous, laparotomy, and laparoscopic methods. In the present case, percutaneous or CT-guided drainage was difficult due to the possibility of intestinal injury at the abscess puncture site during the procedure on abdominal ultrasonography. Laparotomy drainage was performed in this case because it was an easier method. It provides direct access with the abscess cavity compared to laparoscopic drainage, and therefore dose not spread the abscess into the abdominal cavity. No complications occurred after laparotomy drainage.
Interval appendectomy after successful conservative treatment for appendicitis with abscess remains controversial. The rate of recurrence after conservative treatment for perforated acute appendicitis and phlegmon ranges from 12% to 24% , . To avoid this high chance of recurrence, some recommend regular selective interval appendectomy after initial conservative management . Therefore, interval laparoscopic appendectomy was performed in this case.
Laparoscopic surgery was chosen in order to avoid adhesions at the incision site of the first laparotomy drainage. In fact, the abdominal wall and cecum were highly adherent at the site of the previous incision (Fig. 3A). It is believed that if the laparotomy approach was chosen, it could have been difficult in this situation. The cecum might have been damaged, and it may have been difficult to find the appendix. Conversely, the laparoscopic approach was particularly effective in this case wherein the cecum and abdominal wall were adhered to. It was also easier to search for the appendix.
Low postoperative complications are one of the reasons for choosing interval surgery over immediate surgery. Immediate surgery in acute appendicitis with abscess is prone to postoperative ileus and residual abscess due to the presence of the abscess. Fortunately, no complications occurred after interval laparoscopic appendectomy in this case. Treatment of the abscess via antibiotic therapy or drainage followed by surgery may prevent complications related to the abscess and extended resection, such as ileocecal resection.
Interval surgery was also chosen because it can be performed alongside a colonoscopy while waiting. Renteria et al. reported that the rate of unexpected malignancy was 3% in of elderly (mean age: 66 years) and 1.5% of young (mean age: 39 years) patients who underwent appendectomy as primary treatment for acute appendicitis . Jonge et al. reported that adult patients undergoing interval appendectomy can be diagnosed with an appendix neoplasm in up to 11% of cases, in contrast to only 1.5% of patients undergoing immediate appendectomy . Recently, a randomized control trial by Mällinen et al. comparing interval appendectomy and follow-up with magnetic resonance imaging after initial successful conservative treatment of periappendicular abscess was prematurely terminated because of ethical concerns. During their interim analysis, there was an unexpected finding of a high rate of neoplasms (17%), with all neoplasms in patients older than 40 years . Therefore, tumors can actually be one of the causes of appendicitis with abscess. This means that for patients with appendicitis with abscess aged >40 years, conservative treatment should be done alongside colonoscopy to rule out malignant findings. The excision range should then be decided based on the presence or absence of malignant findings.
In this case, laparotomy drainage and interval laparoscopic appendectomy were performed, suggesting a risk of postoperative complications associated with two surgical procedures. However, no postoperative complications were observed. The proper choice of surgical approach can minimize postoperative complications. Thus, even two surgical procedures can be acceptable even for old age patients, such as this case.
4. Conclusion
Immediate appendectomy for acute appendicitis with abscess is associated with higher morbidity. Thus, conservative treatment by laparotomy drainage and antibiotic therapy should be performed in such cases. Subsequent colonoscopy should also be performed because of the possibility of malignant findings in patients aged >40 years. Interval laparoscopic appendectomy is effective to easily resect the appendix and observe the abdominal cavity.
Consent
Patient consent form was signed by the patient. Written informed consent was obtained from the patient for publication of this case report and accompanying images. A copy of the written consent is available for review by the Editor-in-Chief of this journal on request.
Declaration of competing interest
All authors do not have any conflicts of interest.
Acknowledgments
Acknowledgements
We would like to thank Enago for English language editing.
Sources of funding
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
Ethical approval
IRB/Ethics Committee of Hanyu General Hospital ruled that approval was not required for this study.
Provenance and peer review
Not commissioned, externally peer-reviewed.
Author contribution
Toshiyuki Suzuki: Conceptualization, Methodology, Software, Validation, Formal analysis, Investigation, Resources, Writing - Original Draft, Visualization, Project administration, Funding acquisition.
Akiyo Matsumoto: Data Curation, Supervision.
Takahiko Akao: Data Curation.
Seiji Kobayashi: Data Curation.
Hiroshi Matsumoto: Writing - Review & Editing.
Registration of research studies
There was no need to register this study because it is a case report.
Guarantor
Toshiyuki Suzuki.
References
1.Di Saverio S., Podda M., De Simone B., Ceresoli M., Augustin G., Gori A., et al. Diagnosis and treatment of acute appendicitis: 2020 update of the WSES Jerusalem guidelines. World J. Emerg. Surg. 2020;15:27. doi: 10.1186/s13017-020-00306-3. [DOI] [PMC free article] [PubMed] [Google Scholar]
2.Simillis C., Symeonides P., Shorthouse A.J., Tekkis P.P. A meta-analysis comparing conservative treatment versus acute appendectomy for complicated appendicitis (abscess or phlegmon) Surgery. 2010;147:818–829. doi: 10.1016/j.surg.2009.11.013. [DOI] [PubMed] [Google Scholar]
3.Fugazzola P., Coccolini F., Tomasoni M., Stella M., Ansaloni L. Early appendectomy vs. conservative management in complicated acute appendicitis in children: a meta-analysis. J. Pediatr. Surg. 2019;54:2234–2241. doi: 10.1016/j.jpedsurg.2019.01.065. [DOI] [PubMed] [Google Scholar]
4.Agha R.A., Franchi T., Sohrabi C., Mathew G., Kerwan A., Thoma A., et al. The SCARE 2020 guideline: updating consensus Surgical CAse REport (SCARE) guidelines. IJS. 2020;84:226–230. doi: 10.1016/j.ijsu.2020.10.034. [DOI] [PubMed] [Google Scholar]
5.Andersson R.E., Petzold M.G. Nonsurgical treatment of appendiceal abscess or phlegmon: a systematic review and meta-analysis. Ann. Surg. 2007;246:741–748. doi: 10.1097/SLA.0b013e31811f3f9f. [DOI] [PubMed] [Google Scholar]
6.Mentula P., Sammalkorpi H., Leppäniemi A. Laparoscopic surgery or conservative treatment for appendiceal abscess in adults?A randomized controlled trial. Ann. Surg. 2015;262:237–242. doi: 10.1097/SLA.0000000000001200. [DOI] [PubMed] [Google Scholar]
7.Gavriilidis P., de'Angelis N., Katsanos K., Di Saverio S. Acute appendicectomy or conservative treatment for complicated appendicitis (phlegmon or abscess)? A systematic review by updated traditional and cumulative meta-analysis. J. Clin. Med. Res. 2019;11:56–64. doi: 10.14740/jocmr3672. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Elkbuli A., Diaz B., Polcz V., Hai S., McKenney M., Boneva D. Operative versus non-operative therapy for acute phlegmon of the appendix: is it safer? A case report and review of the literature. Int. J. Surg. Case Rep. 2018;50:75–79. doi: 10.1016/j.ijscr.2018.07.031. [DOI] [PMC free article] [PubMed] [Google Scholar]
9.Mizuta N., Nakanishi T., Tsunemi K. Stump appendicitis occurred two and half years after first laparoscopic appendectomy for perforated appendicitis with abscess: a report of a case. Int. J. Surg. Case Rep. 2020;67:146–149. doi: 10.1016/j.ijscr.2020.01.033. [DOI] [PMC free article] [PubMed] [Google Scholar]
10.Rushing A., Bugaev N., Jones C., Como J.J., Fox N., Cripps M., et al. Management of acute appendicitis in adults: a practice management guideline from the Eastern Association for the Surgery of trauma. J. Trauma Acute Care Surg. 2019;87:214–224. doi: 10.1097/TA.0000000000002270. [DOI] [PubMed] [Google Scholar]
11.Darwazeh G., Cunningham S.C., Kowdley G.C. A systematic review of perforated appendicitis and phlegmon: interval appendectomy or wait-and-see? Am. Surg. 2016;82:11–15. doi: 10.1177/000313481608200107. [DOI] [PubMed] [Google Scholar]
12.Renteria O., Shahid Z., Huerta S. Outcomes of appendectomy in elderly veteran patients. Surgery. 2018;164:460–465. doi: 10.1016/j.surg.2018.04.027. [DOI] [PubMed] [Google Scholar]
13.De Jonge J., Bolmers M.D.M., Musters G.D., van Rossem C.C., Bemelman W.A., van Geloven A.A.W. Predictors for interval appendectomy in non-operatively treated complicated appendicitis. Int. J. Color. Dis. 2019;34:1325–1332. doi: 10.1007/s00384-019-03303-4. [DOI] [PubMed] [Google Scholar]
14.Mällinen J., Rautio T., Grönroos J., Rantanen T., Nordström P., Savolainen H., et al. Risk of appendiceal neoplasm in periappendicular abscess in patients treated with interval appendectomy vs follow-up with magnetic resonance imaging: 1-year outcomes of the peri-appendicitis acuta randomized clinical trial. JAMA Surg. 2019;154:200–207. doi: 10.1001/jamasurg.2018.4373. [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from International Journal of Surgery Case Reports are provided here courtesy of Elsevier
ACTIONS
View on publisher site
PDF (1.8 MB)
Cite
Collections
Permalink PERMALINK
Copy
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
On this page
Abstract
Highlights
1. Introduction
2. Presentation of case
3. Discussion
4. Conclusion
Consent
Declaration of competing interest
Acknowledgments
References
Cite
Copy
Download .nbib.nbib
Format:
Add to Collections
Create a new collection
Add to an existing collection
Name your collection
Choose a collection
Unable to load your collection due to an error
Please try again
Add Cancel
Follow NCBI
NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed
Connect with NLM
NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube
National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
Back to Top |
7744 | https://www.arxiv.org/pdf/2509.03147 | COLORED BASE-3 PARTITIONS, SEQUENCES OF POLYNOMIALS, AND PERFECT NUMBERS
KARL DILCHER AND LARRY ERICKSEN
Abstract. Motivated by the observation that the counting function of a cer-tain base-3 colored partition contains the even perfect numbers as a subse-quence, we begin by defining a sequence of polynomials in four variables and discuss their properties and combinatorial interpretations. We then concen-trate on certain subsequences that are related to the Chebyshev polynomials of both kinds. Finally, we consider several sequences of single-variable polyno-mials that have meaningful combinatorial interpretations as well as interesting zero distributions.
Introduction
There are countless different ways of writing a positive integer as a sum of other positive integers according to a fixed set of rules. The most famous among these are the classical partitions; they go back to Euler and are still an active area of research. Variants include compositions, overpartitions, colored partitions, partitions with restrictions of various kinds, and binary or base-b analogues of all of the above. Let us begin with the following base-3 partition rule and the corresponding counting function. Given an integer n ≥ 1, in how many ways can we write n as a sum of powers of 3 under the following conditions? (1.1)
At most one of each power may be overlined ,
at most one of each power may be marked by a tilde, and at most two of each power may remain unmarked. We also assume that at most one of an overline or a tilde are allowed, and that 30 = 1 is also an allowable power of 3. Such partitions are usually referred to as restricted colored (base-b) partitions, where in this case the three “colors” are: (1) “marked by a line”, (2) “marked by a tilde”, and (3) “unmarked”. Given the condition (1.1), we say that the restriction is of type (1 , 1, 2). We denote the number of partitions as given in (1.1) by S(n)and illustrate them with two examples.
Example 1. Let n = 3. Then the allowable partitions are 3, 3, ˜3, 1 + 1 + 1 , 1 + 1 + ˜ 1, 1 + 1 + ˜ 1,
and we have S(3) = 6.
2010 Mathematics Subject Classification. Primary 11P81; Secondary 11B37, 11B83.
Key words and phrases. Base-3 partitions, colored partitions, restricted partitions, recurrence relations, Chebyshev polynomials, zeros of polynomials. Research supported in part by the Natural Sciences and Engineering Research Council of Canada, Grant # 145628481.
1
arXiv:2509.03147v1 [math.CO] 3 Sep 2025 2KARL DILCHER AND LARRY ERICKSEN
Example 2. Let n = 12. To proceed systematically, we first consider the unmarked and unrestricted base-3 partitions, written more concisely as (1.2) (93) , (9111) , (3333) , (333111) , (331 . . . 1) , (31 . . . 1) , (1 . . . 1) .
The last three of these partitions have too many 1s to satisfy the conditions in (1.1). However, the first four in (1.2) lead to the following restricted colored partitions: (93) , (93) , (9˜ 3) , (93) , (93) , (9˜ 3) , (˜ 93) , (˜ 93) , (˜ 9˜ 3); (9111) , (911˜ 1) , (911˜ 1) , . . . , (˜ 911˜ 1); (333˜ 3); (333111) , (33311˜ 1) , . . . , (33311˜ 1) ,
where in the second and fourth subsets we only showed a few representative ones. The numbers of partitions in these four blocks are 9,9,1 and 9, respectively, for a total of S(12) = 28. We notice that the number of partitions, 6 and 28, in these two examples are perfect numbers. Of course, this could be just a coincidence, but it turns out that 496 and 8128 also occur in this way, for different n. One purpose of this paper is to explain this observation. However, apart from this rather unexpected connection, the even perfect numbers occur only as a subsequence of another well-known sequence, as we will see. In the process of explaining this, we will study several general and more specific sequences of polynomials and their combinatorial interpretations. In Section 2 we introduce a sequence of polynomials in four variables, which turns out to be a polynomial analogue of the sequence S(n) defined by (1.1). We also derive several properties of these polynomials, including combinatorial interpretations in terms of restricted base-3 partitions. In Section 3 we define two different but related subsequences of polynomials and derive several properties, including recurrences and connections with the Chebyshev polynomials of both kinds. We also explain the connection with perfect numbers. In Sections 4 and 5 we consider several sequences of single-variable polynomials that arise from specializing or equating certain of the four variables in the general case. These polynomial sequences have interesting zero distributions and specific combinatorial interpretations. We conclude this paper with some further remarks in Section 6. 2. Some generalities
The concept of colored (or multicolor) partitions is relatively recent; it was intro-duced by Keith as a generalization of overpartitions. Colored b-ary (or base-b)partitions were more recently studied by Ulas and ˙Zmija . In we developed a general theory of restricted multicolor b-ary partitions, along with polynomial analogues and characterizations of individual partitions. A more complete list of references can also be found in . Here, we are going to consider the following special case. We begin with a definition. COLORED BASE-3 PARTITIONS 3
Definition 2.1. Let Z := ( w, x, y, z ) be a quadruple of real or complex variables. We define the sequence S(n; Z) of polynomials by the generating function (2.1)
∞
X
n=0
S(n; Z)qn =
∞
Y
j=0
1 + wq 3j
1 + xq 3j
1 + yq 3j
zq 2·3j
.
The first few of these polynomials are shown in Table 1.
n S(n; Z)0 11 w + x + y
2 wx + wy + xy + z
3 wxy + wz + xz + w + x + y
4 w2 + 2 wx + 2 wy + wxz + x2 + 2 xy + y2
5 w2x + w2y + wx 2 + 3 wxy + wy 2 + wz + x2y + xy 2 + xz + yz
6 w2xy + w2z + wx 2y + wx + wxy 2 + 2 wxz + wy + wyz
+x2z + xyz + xy + z
Table 1 : S(n; Z) for 0 ≤ n ≤ 6. When Z = Z0 := (1 , 1, 1, 1), we set S(n) := S(n; Z0), and we compute (2.2) S(n)
n≥0
= (1 , 3, 4, 6, 10 , 12 , 13 , 15 , 16 , 18 , 22 , 24 , 28 , 36 , 40 , 42 , . . . ).
By expanding the right-hand side of (2.1) with w = x = y = z = 1, it is not difficult to see the following interpretation of the sequence S(n).
Proposition 2.2. For n ≥ 1, S(n) is exactly the number of base-3 partitions of n
under the conditions (1.1) .
We now see that the sequence (2.2) is consistent with Examples 1 and 2; the corresponding values have been highlighted. If we now expand the right-hand side of (2.1) with variables w, x, y, z in place, it is not difficult to see that we have the following combinatorial interpretation of the coefficients of the polynomials S(n; Z).
Proposition 2.3. If we write S(n; Z) in the form
(2.3) S(n; Z) = X
i,j,k,ℓ ≥0
c(i, j, k, ℓ ) wixj ykzℓ,
then among all colored base-3 partitions of n, restricted by (1.1) , the coefficient
c(i, j, k, ℓ ) counts those that have
i parts marked with an overline, and
j parts marked with a tilde, and
k single unmarked parts, and
ℓ pairs of unmarked parts.
It is easy to see with Table 1 for n = 3 that this is consistent with Example 1. The case n = 12 gives the following more meaningful example.
Example 3. Computations show that one of the monomials of S(12; Z) is 2 w2xyz .This corresponds to the two partitions (33˜ 3111) and (33311˜ 1) from Example 2, consistent with Proposition 2.3. 4 KARL DILCHER AND LARRY ERICKSEN
Another monomial of S(12; Z) is 3 wxz , which corresponds to the three partitions (911˜ 1), (˜ 9111), and (333˜ 3) from Example 2, again consistent with Proposition 2.3. Next we state a set of recurrence relations for the polynomials S(n; Z). They follow directly from Theorem 4.3 in .
Proposition 2.4. The polynomials S(n; Z) satisfy the recurrence S(0; Z) = 1 ,
S(1; Z) = w + x + y, S(2 , Z ) = wx + wy + xy + z, and for n ≥ 1,
S(3 n; Z) = S(n; Z) + ( wxy + wz + xz ) · S(n − 1; Z),(2.4)
S(3 n + 1; Z) = ( w + x + y) · S(n; Z) + wxz · S(n − 1; Z),(2.5)
S(3 n + 2; Z) = ( wx + wy + xy + z) · S(n; Z).(2.6) We finish this section with an easy consequence of Proposition 2.4. We note that the sequence sk(n) := k · 3n − 1, for a fixed k ≥ 1, satisfies the recurrence
sk(n + 1) = 3 · sk(n) + 2 with initial conditions sk(0) = k − 1. Therefore, upon iterating (2.6) and noting that the coefficient on the right of (2.6) is S(2; Z), we get the following explicit formulas.
Corollary 2.5. For any integers k ≥ 1 and n ≥ 0, we have
(2.7) S(k · 3n − 1, Z ) = S(k − 1; N ) · S(2; Z)n,
and in particular
(2.8) S(3 n − 1, Z ) = ( wx + wy + xy + z)n.
Upon setting w = x = y = z = 1 in (2.8), we get as an immediate consequence the fact that the number of base-3 partitions of 3 n − 1, restricted by (1.1), is 4 n.Analogous enumerations can be obtained from (2.7). Corollary 2.5 is related to a more interesting variant which we will now consider in Section 3. 3. Two different subsequences
In analogy to the subsequences considered in Corollary 2.5, we now define the two polynomial sequences (3.1) Qn(Z) := S( 3n−32 ; Z), Rn(Z) := S( 3n−12 ; Z),
where as before Z = ( w, x, y, z ). For greater clarity and ease of notation we now set
W1(Z) := wxy + wz + xz + w + x + y, (3.2)
W2(Z) := w2xy + w2z + wx 2y + wxy 2 + wxz + wyz + x2z + xyz. (3.3) Most of the properties of the polynomial sequences Qn(Z) and Rn(Z) depend on the following results.
Proposition 3.1. (a) The polynomials Qn(Z) satisfy Q0(Z) = 0 , Q1(Z) = 1 , and for n ≥ 2,
(3.4) Qn(Z) = W1(Z) · Qn−1(Z) − W2(Z) · Qn−2(Z).
(b) The polynomials Rn(Z) satisfy R0(Z) = 1 , R1(Z) = w + x + y, and for n ≥ 2,
(3.5) Rn(Z) = W1(Z) · Rn−1(Z) − W2(Z) · Rn−2(Z).COLORED BASE-3 PARTITIONS 5
Proof. We prove both identities jointly by induction on n. For greater ease of notation we suppress the multi-variable Z. We first verify the values for Q1, R0
and R1 with (3.1) and Table 1. We also note that Q2 = S(3), R2 = S(4), and by direct computation using Table 1 we can then verify that (3.4) and (3.5) holds for
n = 2. This is the induction beginning. We now assume that both (3.4) and (3.5) hold up to some n − 1. We wish to show that they then also hold for n, that is, as written in (3.4) and (3.5). Using the easy identities (3.6) 3n−j − 32 = 3 · 3n−j−1 − 12 , 3n−j−1 − 12 − 1 = 3n−j−1 − 32for j = 0 , 1, 2, we get with (2.4) that
Qn−W1Qn−1 + W2Qn−2
= S( 3n−32 ) − W1S( 3n−1−32 ) + W2S( 3n−2−32 )= S( 3n−1−12 ) − W1S( 3n−2−12 ) + W2S( 3n−3−12 )+ ( wxy + wz + xz )
S( 3n−1−32 ) − W1S( 3n−2−32 ) + W2S( 3n−3−32 )
.
Applying the two identities in (3.1) to the last two lines, we see by the induction hypothesis that both lines vanish, which proves (3.4). Similarly, using (3.6) again repeatedly, we get with (2.5) that
Rn−W1Rn−1 + W2Rn−2
= S( 3n−32 + 1) − W1S( 3n−1−32 + 1) + W2S( 3n−2−32 + 1) = ( w + x + y)
S( 3n−1−12 ) − W1S( 3n−2−12 ) + W2S( 3n−3−12 )
wxz
S( 3n−1−12 − 1) − W1S( 3n−2−12 − 1) + W2S( 3n−3−12 − 1)
.
Once again we apply the two identities in (3.1) to the last two lines and see that both expressions in large parentheses vanish by the induction hypothesis. This proves (3.5) and the proof by induction is complete. □
Before we continue with the general case, we specialize Z = Z0 := (1 , 1, 1, 1) and set qn := Qn(Z0) and rn := Rn(Z0). Then Proposition 3.1 simplifies to
q0 = 0 , q 1 = 1 , and qn = 6 qn−1 − 8qn−2 (n ≥ 2) ,(3.7)
r0 = 1 , r 1 = 3 , and rn = 6 rn−1 − 8rn−2 (n ≥ 2) .(3.8) The characteristic polynomial for the two recurrence relations is x2 − 6x + 8 = (x − 4)( x − 2), and it is not difficult to verify that the Binet-type formulas are, for
n ≥ 0,
qn = 12 (4 n − 2n) = 2 n−1 (2 n − 1) ,(3.9)
rn = 12 (4 n + 2 n) = 2 n−1 (2 n + 1) .(3.10) With (3.9) and using the definition of qn and (3.1), Proposition 2.2 now implies that our observation following Example 2 is indeed true in general. 6 KARL DILCHER AND LARRY ERICKSEN
Corollary 3.2. For any n ≥ 1, the number of base-3 partitions of 12 (3 n − 3) under the condition (1.1) is 2n−1(2 n − 1) . In particular, if n is a prime number such that 2n − 1 is also prime, then the number of such partitions is a perfect number. Conversely, all even perfect numbers can be expressed in this way.
The second and third statements come from the well-known characterization of even perfect numbers due to Euclid and Euler. If n is a prime, 2 n − 1 is called a Mersenne number, which may or may not itself be prime. It is not known whether there are infinitely many Mersenne primes. In analogy to Corollary 3.2, the numbers rn lead to the following consequence of Proposition 2.2.
Corollary 3.3. For any n ≥ 1, the number of base-3 partitions of 12 (3 n − 1) under the condition (1.1) is 2n−1(2 n + 1) .
We mention in passing that in the case n = 2 k, 0 ≤ k ≤ 4, the Fermat number 22k
1 is prime, but no other Fermat prime is known. It follows from a famous result of Gauss that a regular polygon with rn = 2 n−1(2 n + 1) sides is constructible with compass and straightedge if n = 2 k, 0 ≤ k ≤ 4. Returning to the general polynomial case, we use Proposition 3.1 to derive gen-erating functions for the polynomials Qn(Z) and Rn(Z).
Proposition 3.4. The polynomials Qn(Z) and Rn(Z) have the generating func-tions
∞
X
n=0
Qn(Z)qn = q
1 − W1(Z)q + W2(Z)q2 ,(3.11)
∞
X
n=0
Rn(z)qn = 1 − (wxy + wz + xz )q
1 − W1(Z)q + W2(Z)q2 .(3.12)
Proof. We multiply both sides of (3.11) by the denominator on the right and take the Cauchy product with the power series on the left. Then the constant coefficient is Q0(Z) = 0, the coefficient of q is Q1(Z) − W1(Z)Q0(z) = 1, and all other coefficients are 0 by (3.4). This shows that the numerator on the right is q, which proves (3.11). The identity (3.12) is proven analogously, using (3.5). □
Propositions 3.1 and 3.4 indicate possible connections with the Chebyshev poly-nomials of the first and second kind, Tn(v) and Un(v), which can be defined by their generating functions (3.13)
∞
X
n=0
Tn(v)tn = 1 − vt
1 − 2vt + t2 ,
∞
X
n=0
Un(v)tn = 11 − 2vt + t2 .
Numerous properties of these two polynomial sequences can be found, for instance, in . We can now state and prove the following identities, which are similar in nature to those in Proposition 3.3 of . COLORED BASE-3 PARTITIONS 7
Proposition 3.5. For all n ≥ 0 we have
Qn+1 (Z) = W2(Z)n/ 2 · Un
W1(Z)2W2(Z)1/2
,(3.14)
Rn(Z) = W2(Z)n/ 2 · Tn
W1(Z)2W2(Z)1/2
(3.15) + w + x + y − (wxy + wz + xz )2 Qn(Z),
with W1(Z) and W2(Z) as defined in (3.2) and (3.3) .Proof. Comparing (3.11) with the second identity in (3.13), we see that
q = tW2(Z)1/2 and v = W1(Z)2W2(Z)1/2 .
Equating coefficients of qn then gives (3.14). Next, with t and v as above, we rewrite the numerator on the right of (3.12) as (3.16) 1 − (wxy + wz + xz )q = 1 − vt + w + x + y − wxy − wz − xz
2W2(Z)1/2 t.
The term 1 −vt , together with the first identity in (3.13), leads to the first summand on the right of (3.15), while the remaining term on the right-hand side of (3.16) leads to the second summand on the right of (3.15). □
The connections in Proposition 3.5 between the polynomials Qn(Z), Rn(Z) and the Chebyshev polynomials will be particularly useful in the next section. 4. Single-variable polynomials: The case Z = (1 , 1, z, 1) There are numerous ways of specializing or equating the four variables in Z =(w, x, y, z ) so that we obtain single-variable polynomials. In this and the next sec-tion we concentrate on a few special cases which allow for reasonable combinatorial interpretations, while at the same time the polynomials in question have some interesting properties in their own right. This section will be devoted to what we consider the most interesting of these cases. For ease of notation we set Z1 := (1 , 1, z, 1) and
Q(1)
n
(z) := Qn(Z1), R(1)
n
(z) := Rn(Z1).
Then by (3.2) and (3.3) we have (4.1) W (1) 1 (z) := W1(Z1) = 2( z + 2) , W (1) 2 (z) := W2(Z1) = ( z + 1)( z + 2) ,
and Proposition 3.1, with R(1) 1 (z) = 2 + z, allows us to efficiently compute the polynomials Q(1)
n
(z) and R(1)
n
(z); see Table 2. Proposition 2.3 together with (3.1) provides the following combinatorial inter-pretations of the polynomials Q(1)
n
(z) and R(1)
n
(z).
Proposition 4.1. For n ≥ 1, let
(4.2) Q(1)
n
(z) =
n−1
X
k=0
cn(k)zk, R(1)
n
(z) =
n
X
k=0
dn(k)zk.8 KARL DILCHER AND LARRY ERICKSEN
Then cn(k) counts the number of colored base-3 partitions of 12 (3 n − 3) , restricted by (1.1) and having exactly k single unmarked powers of 3. Similarly, dn(k) counts the number of such partitions of 12 (3 n − 1) .
n Q(1)
n
(z) R(1)
n
(z)0 0 11 1 z + 2 2 2z + 4 z2 + 4 z + 5 3 3z2 + 12 z + 13 z3 + 6 z2 + 15 z + 14 4 4z3 + 24 z2 + 52 z + 40 z4 + 8 z3 + 30 z2 + 56 z + 41 5 5z4 + 40 z3 + 130 z2 + 200 z + 121 z5 + 10 z4 + 50 z3 + 140 z2 + 205 z + 122
Table 2 : Q(1)
n
(z) and R(1)
n
(z) for 0 ≤ n ≤ 5. Table 2 shows some obvious patterns in the leading and constant coefficients of the two polynomial sequences. Indeed, some straightforward inductions show that (4.3) cn(0) = 3n − 12 , cn(n − 1) = n, and dn(0) = 3n + 1 2 , dn(n) = 1 .
We will see below that these are special cases of a more general result. Next, we derive some special properties of the polynomials Q(1)
n
(z) and R(1)
n
(z). We first observe that with w = x = z = 1 in (3.15), the summand on the right of (3.15) vanishes, and with (4.1) we get
Q(1)
n+1
(z) = (z + 1)( z + 3) n/ 2 · Un
z + 2
p(z + 1)( z + 3)
!
,(4.4)
R(1)
n
(z) = (z + 1)( z + 3) n/ 2 · Tn
z + 2
p(z + 1)( z + 3)
!
.(4.5) All further properties are consequences of these two identities. We begin with the following rather surprising relations.
Proposition 4.2. For all n ≥ 0,
R(1)
n
(z) − Q(1)
n
(z) = ( z + 1) n,(4.6)
R(1)
n
(z)2 − Q(1)
n
(z)2 = (( z + 1)( z + 3)) n ,(4.7)
R(1)
n
(z) + Q(1)
n
(z) = ( z + 3) n.(4.8) The identity (4.6) is somewhat similar to (4.4) in , in the proof of which the identity (4.9) below was also derived and used.
Proof of Proposition 4.2. We use the defining identities sin θ · Un−1(cos θ) = sin( nθ ), Tn(cos θ) = cos( nθ ).
Multiplying both sides of the left identity by i, then adding both and using 2 i sin θ =
eiθ − e−iθ , 2 cos θ = eiθ + e−iθ , and u := eiθ , we get (4.9) u − u−1
2 Un−1
u + u−1
2
Tn
u + u−1
2
= un, n ≥ 1.COLORED BASE-3 PARTITIONS 9
We now set u = p(z + 1) /(z + 3) and verify that
u − u−1
2 = −1
p(z + 1)( z + 3) , u + u−1
2 = z + 2
p(z + 1)( z + 3) .
Upon substituting these two identities into (4.9) and multiplying both sides by (( z + 1)( z + 3)) n/ 2, we get
− (z + 1)( z + 3) (n−1) /2 · Un−1
z + 2
p(z + 1)( z + 3)
!
(4.10) + (z + 1)( z + 3) n/ 2 · Tn
z + 2
p(z + 1)( z + 3)
!
= ( z + 1) n.
Finally, with (4.4) and (4.5) we see that (4.10) is equivalent to (4.6). To prove (4.7), we use the well-known Pell-type equation
Tn(x)2 − (x2 − 1) Un−1(x)2 = 1 for the Chebyshev polynomials (see, e.g., [6, p. 9]), which in our case becomes (4.11) Tn
z + 2
p(z + 1)( z + 3)
!2
− 1(z + 1)( z + 3) Un−1
z + 2
p(z + 1)( z + 3)
!2
= 1 .
Upon multiplying both sides of (4.11) by (( z + 1)( z + 3)) n, we see that (4.7) imme-diately follows from (4.4) and (4.5). Finally, the identity (4.8) follows from dividing (4.7) by (4.6). □
By adding, respectively subtracting, (4.6) and (4.8), we obtain the following consequences.
Corollary 4.3. For all n ≥ 0 we have
Q(1)
n
(z) = (z + 3) n − (z + 1) n
2 =
n
X
j=0
nj
3n−j − 12 zj ,(4.12)
R(1)
n
(z) = (z + 3) n + ( z + 1) n
2 =
n
X
j=0
nj
3n−j + 1 2 zj .(4.13) Comparing Corollary 4.3 with the notation in (4.2), we see that the identities in (4.3) are special cases of (4.12) and (4.13). Proposition 4.1 combined with Corollary 4.3 now leads to the following combinatorial interpretations.
Corollary 4.4. Considering colored base-3 partitions restricted by (1.1) , for each
j = 0 , 1, . . . , n ,
(a) there are 12
nj
(3 n−j − 1) partitions of 12 (3 n − 3) that have exactly j single unmarked parts.
(b) there are 12
nj
(3 n−j + 1) partitions of 12 (3 n − 1) that have exactly j single unmarked parts.
Example 4. (a) For n = 2, the six partitions of 12 (3 2 − 3) = 3 are given in Example 1 (note the different use of n). Four partitions have no single unmarked part, two have exactly one, and there is no partition with two unmarked parts, consistent with Corollary 4.4(a). 10 KARL DILCHER AND LARRY ERICKSEN
(b) Taking again n = 2, the ten partitions of 12 (3 2 − 1) = 4 are, in the notation of Example 2: (31) , (31) , (3˜ 1) , (31) , (31) , (3˜ 1) , (˜ 31) , (˜ 31) , (˜ 3˜ 1) , (111˜ 1) .
We see that 12 (3 2 + 1) = 5 of these partitions have no single unmarked parts, while
12
21
(3 1 + 1) = 4 have one unmarked part and one partition, namely (31), has two unmarked parts. We conclude this section with a few more properties of the polynomials Q(1)
n
(z)and R(1)
n
(z). First, upon replacing z by z − 2 in the first equalities of (4.12) and (4.13), we get the following simple representations.
Corollary 4.5. For all n ≥ 0,
(4.14) Q(1)
n
(z − 2) =
⌊n−12⌋
X
j=0
n
2j + 1
zn−2j−1, R(1)
n
(z − 2) =
⌊n
2⌋
X
j=0
n
2j
zn−2j .
Concerning a related shift, it is a consequence of the identity (4.6) that R(1)
n
(z−1) and Q(1)
n
(z −1) differ by only the leading term zn of R(1)
n
(z −1). Also, as mentioned in entry A082137 of , the coefficients of this last polynomial appear in [4, Table 1] in a different combinatorial setting. We can also determine the zero distribution of the polynomials in question.
Proposition 4.6. All the zeros of the polynomials Q(1)
n
(z) and R(1)
n
(z) lie on the vertical line Re( z) = −2. The zeros are all nonreal with the exception of Q(1) 2n (−2) = 0 and R(1) 2n+1 (−2) = 0 for all n ≥ 0.Proof. We replace z by z − 2 in (4.4) and (4.5) and recall that the zeros of Un(v)and Tn(v) are all real and satisfy −1 < v < 1. This means that we need to solve the equation v = z/ √z2 − 1 for z, and we get (4.15) z = ±i
r v2
1 − v2 .
Since the expression in the square root is real and positive, this proves the first statement of the proposition. The second statement follows from Corollary 4.5. □
Single-variable polynomials: Some further cases
In this section we work out the details of two further cases, and in the end we identify six more cases that could be done by using the same methods. In all cases we only consider the polynomials Qn(Z). Analogous results could also be obtained with the polynomials Rn(Z) by making some necessary adjustments such as using (3.5) and the second part of (3.1). One main difference is that, in contrast to the situation in Section 4, the additional summand in (3.15) makes it difficult, if not impossible, to determine the distribution of zeros of Rn(Z). 5.1. The case Z = Z2 := ( z, z, z, z 2). In analogy to Section 4, we set
Q(2)
n
(z) := Qn(Z2).
By (3.2) and (3.3) we have (5.1) W (2) 1 (z) := W1(Z2) = 3 z(z2 + 1) , W (2) 2 (z) := W2(Z2) = 8 z4,COLORED BASE-3 PARTITIONS 11
and with Proposition 3.1 we can easily compute the entries in Table 2.
n Q(2)
n
(z)1 12 3z3 + 3 z
3 9z6 + 10 z4 + 9 z2
4 27 z9 + 33 z7 + 33 z5 + 27 z3
5 81 z12 + 108 z10 + 118 z8 + 108 z6 + 81 z4
6 243 z15 + 351 z13 + 414 z11 + 414 z9 + 351 z7 + 243 z5
7 729 z18 + 1134 z16 + 1431 z14 + 1540 z12 + 1431 z10 + 1134 z8 + 729 z6
Table 3 : Q(2)
n
(z) for 1 ≤ n ≤ 7. Before dealing with combinatorial interpretations of the polynomials Q(2)
n
(z), we prove some basic properties that are apparent from Table 3. As before, the connection with Chebyshev polynomials will be an important tool. Substituting the identities from (5.1) into (3.14), we get (5.2) Q(2)
n+1
(z) =
2√2z2n
· Un
34√2 (z + z−1)
.
We can now state and prove the following properties.
Proposition 5.1. For n ≥ 0, the polynomial Q(2)
n+1
(z) has degree 3n, its term with lowest degree has degree n, it is palindromic, and its leading coefficient is 3n.Furthermore, when n is even (resp. odd), Q(2)
n+1
(z) has only even (resp. odd) powers of z.Proof. With (3.4) and (5.1) we have the recurrence Q(2) 0 (z) = 0, Q(2) 1 (z) = 1, and for n ≥ 2, (5.3) Q(2)
n
(z) = 3 z(z2 + 1) Q(2)
n−1
(z) − 8z4Q(2)
n−2
(z).
Then the modified polynomials, which we define by eQ(2) 0 (z) = 0 and (5.4) eQ(2)
n+1
(z) := z−nQ(2)
n+1
(z) (n ≥ 0) ,
satisfy the recurrence given by eQ(2) 1 (z) = 1 and for n ≥ 2, (5.5) eQ(2)
n
(z) = 3( z2 + 1) eQ(2)
n−1
(z) − 8z2 eQ(2)
n−2
(z).
Now an easy induction using (5.5) and the initial terms shows that deg eQ(2)
n+1
(z) = 2n and that the leading coefficient of eQ(2)
n+1
(z) is 3 n.Next, by (5.2) and (5.4) we have (5.6) eQ(2)
n+1
(z) =
2√2z
n
· Un
34√2 (z + z−1)
,
which immediately shows that
z2n eQ(2)
n+1
( 1
z
) = eQ(2)
n+1
(z),
which means that eQ(2)
n+1
(z) is a reciprocal (or palindromic) polynomial. Finally, the statement concerning the parity of the powers follows from (5.3) by an easy induction. This completes the proof of the proposition. □12 KARL DILCHER AND LARRY ERICKSEN
With Proposition 2.3 and (3.1), along with Proposition 5.1, we can now state the following properties.
Proposition 5.2. For n ≥ 1, let
Q(2)
n
(z) =
3n−3
X
k=n−1
en(k)zk.
Then en(k) counts the number of colored base-3 partitions of 12 (3 n − 3) , restricted by (1.1) and having a total of k parts. Furthermore,
(a) the number of parts lies between n − 1 and 3n − 3,
(b) the number of parts cannot have the same parity as n,
(c) for 0 ≤ j ≤ ⌊ (n − 1) /2⌋, there are as many partitions with n − 1 + j parts as there are with 3n − 3 − j parts,
(d) there are 3n−1 partitions with n − 1 parts.
Example 5. (1) Let n = 2. Then Q(2) 2 (z) = 3 z3 + 3 z, consistent with Example 1 which shows 3 partitions with three parts and 3 partitions with one part. (2) Similarly, Q(2) 3 (z) = 9 z6 + 10 z4 + 9 z2 is consistent with Example 2, which shows 9 partitions each with six and with two parts, and 10 partitions with four parts. The polynomials Q(2)
n
(z) also have an interesting zero distribution, as we will now see.
Proposition 5.3. The zeros of all polynomials Q(2)
n
(z), n ≥ 2, lie on the two segments of the unit circle that satisfy |Im(z) | > 13 .
Proof. By (5.2), z is a zero of Q(2)
n+1
(z) if and only if (5.7) v = 34√2
z + 1
z
.
Solving (5.7) for z, we obtain (5.8) z = 2√23 v ± i
q
1 − 89 v2
and verify
|z|2 =
2√23 v
2
1 − 89
√2v2 = 1 ,
so the zeros of Q(2)
n+1
(z) lie on the unit circle. Finally, since the zeros of Un(v)satisfy −1 < v < 1, (5.8) shows that the zeros of Q(2)
n+1
(z) have imaginary parts
1/3 or < −1/3, which completes the proof. □
5.2. The case Z = Z3 := (1 , 1, z, z ). Here we set
Q(3)
n
(z) := Qn(Z3).
By again using (3.2) and (3.3) we have (5.9) W (3) 1 (z) := W1(Z3) = 4 z + 2 , W (3) 2 (z) := W2(Z3) = 3 z2 + 5 z.
Then Proposition 3.1 gives the recurrence Q(3) 0 (z) = 0, Q(3) 1 (z) = 1, and for n ≥ 2, (5.10) Q(3)
n
(z) = (4 z + 2) Q(3)
n−1
(z) − (3 z2 + 5 z)Q(3)
n−2
(z).COLORED BASE-3 PARTITIONS 13
With this, we can first compute the entries in Table 4.
n Q(3)
n
(z)1 12 4z + 2 3 13 z2 + 11 z + 4 4 40 z3 + 44 z2 + 28 z + 8 5 121 z4 + 158 z3 + 133 z2 + 68 z + 16 6 364 z5 + 542 z4 + 544 z3 + 374 z2 + 160 z + 32 7 1093 z6 + 1817 z5 + 2071 z4 + 1715 z3 + 1000 z2 + 368 z + 64
Table 4 : Q(3)
n
(z) for 1 ≤ n ≤ 7. We now prove some properties of the polynomials Q(3)
n
(z) that are apparent from Table 4.
Proposition 5.4. For any n ≥ 1, polynomial Q(3)
n
(z) has the following properties:
(a) The degree of Q(3)
n
(z) is n − 1.
(b) The leading coefficient is 12 (3 n − 1) .
(c) The constant coefficient is 2n−1.Proof. We prove parts (a) and (b) jointly, using (5.10). First, it is clear from the recurrence that the degree of Q(3)
n
(z) cannot exceed n − 1. Second, (5.10) gives the following recurrence for the leading coefficients ℓn: ℓ0 = 0, ℓ1 = 1, and
ℓn = 4 ℓn−1 − 3ℓn−2 (n ≥ 2). It is now easy to verify that the sequence 12 (3 n − 1) satisfies the same recurrence. This proves parts (a) and (b). Finally, (5.10) also gives Q(3)
n
(0) = 2 Q(3)
n−1
(0). This, together with Q(3) 1 (0) = 1, proves part (c). □
Once again, with (3.1) and Proposition 2.3 we obtain a combinatorial interpre-tation. Here we also use Proposition 5.4.
Proposition 5.5. For n ≥ 1, let
Q(3)
n
(z) =
n−1
X
k=0
fn(k)zk.
Then fn(k) counts the number of colored base-3 partitions of 12 (3 n − 3) , restricted by (1.1) and such that the number of single unmarked parts plus the number of pairs of unmarked parts equals k. Furthermore,
(a) there are 2n−1 partitions without unmarked parts,
(b) the maximal sum of the number of single unmarked parts and the number of pairs of unmarked parts in a partition is n − 1,
(c) this maximum is achieved by 12 (3 n − 1) partitions.
Example 6. Let n = 2. Then we see in Example 1 that two partitions have no un-marked parts, while the remaining four partitions have either one single unmarked part or one pair of unmarked parts. This is consistent with Q(3) 2 (z) = 4 z + 2. We will now show that the polynomials Q(3)
n
(z) have particularly interesting zero distributions. 14 KARL DILCHER AND LARRY ERICKSEN
Proposition 5.6. The zeros of all polynomials Q(3)
n
(z), n ≥ 2, lie on the segment of the circle with radius 78 and centered at z = 38 that satisfies Re (z) < 12 .Proof. Once again, we use the connection with Chebyshev polynomials given by (3.14). With (5.9) we have
Q(3)
n+1
(z) = 3z2 + 5 zn/ 2 · Un
2z+1 √3z2+5 z )
.
This means that z is a zero of Q(3)
n+1
(z) if and only if (5.11) v = 2z + 1
√3z2 + 5 z
is a zero of Un(v). We know that −1 < v < 1, and with the aim of solving (5.11) for z, we get the quadratic equation (4 − 3u)z2 + (4 − 5u)z + 1 = 0 , 0 ≤ u < 1,
where for simplicity we have set u = v2. The quadratic formula now gives the solutions (5.12) z = − 4 − 5u
8 − 6u ± i
s
28 u − 25 u2
(8 − 6u)2 ,
or equivalently,
z − 38 = 29 u − 28 4(8 − 6u) ± i
s
28 u − 25 u2
(8 − 6u)2 .
We note that the term inside the square root is always positive for 0 ≤ u ≤ 1. Then it is easy to verify that
29 u − 28 4(8 − 6u)
2
28 u − 25 u2
(8 − 6u)2 =
78
2
.
This, with (5.12), shows that the zeros lie on the circle in question. To prove the remaining statement, we let r(u) be the real part on the right of (5.12). Then r(0) = −1/2, r(1) = 1 /2, and since the derivative r′(u) = 4 /(4 − 3u)2
is always positive, r(1) = 1 /2 is in fact the supremum of the real parts of all the zeros of Q(3)
n+1
(z). This is due to the fact that the zeros of Un(v) get arbitrarily close to 1 and −1 as n grows, but ±1 are not zeros themselves. This completes the proof. □
5.3. Further cases. Without going into any detail, we now summarize a few ad-ditional cases that allow for some meaningful combinatorial interpretations. The polynomials in question are Qn(Z), as defined in (3.1). We write them generically as
Qn(Z) = X
k≥0
an(k)zk.
Furthermore, we refer to the colored base-3 partitions of 12 (3 n − 3), restricted by (1.1), simply as “partitions”. The methods used in Section 4 and in the previous two subsection can then be applied to the following cases. (1) Z = ( z, z, 1, 1): an(k) counts the number of partitions that have k marked parts. Qn(Z) is palindromic. (2) Z = ( z, z, z, z ): an(k) counts the number of partitions that have k parts, but with each pair of unmarked parts counted only once. COLORED BASE-3 PARTITIONS 15
(3) Z = (1 , 1, z, z 2): an(k) counts the number of partitions that have a total of
k unmarked parts. Qn(Z) is palindromic and its zeros lie on the unit circle. (4) Z = ( z, z, z, 1): an(k) counts the number of partitions whose total number of parts, excluding pairs of unmarked parts, is k.(5) Z = (1 , z, z, z 2): an(k) counts the number of partitions that have k parts that don’t have an overline. Qn(Z) is palindromic and its zeros lie on the unit circle or on a finite segment of the negative real axis. (6) Z = ( z, 1, z, z 2): This is the same as in (5), only with “overline” replaced by “tilde”. 6. Further Remarks
6.1. Some polynomial identities. We begin this section with a few more prop-erties of the polynomials introduced in (3.1). They all connect the two polynomial sequences with each other, but do not seem to have any obvious combinatorial interpretations.
Proposition 6.1. For all n ≥ 1 we have
wxz · Qn(Z) = Rn+1 (Z) − (w + x + y) · Rn(Z),(6.1)
Rn(Z) = Qn+1 (Z) − (wxy + wz + xz ) · Qn(Z).(6.2)
Proof. Upon replacing n by 12 (3 n − 1) in (2.5) we get (6.1) immediately from (3.1). With the same substitution in (2.4), we also get (6.2). □
If we consider (6.1) and (6.2) as modified difference equations, it becomes clear that in both cases we can use telescoping. In fact, without too much effort one obtains the following identities.
Corollary 6.2. For all N ≥ 1 we have
RN (Z) = ( w + x + y)N + wxz ·
N
X
n=1
(w + x + y)N −nQn−1(Z),(6.3)
QN (Z) =
N
X
n=1
(wxy + wz + xz )N −nRn−1(Z).(6.4) 6.2. Explicit zeros. The Chebyshev polynomials Un(v) and Tn(v) are known to have the zeros (6.5) vk = cos
k+1
n+1
π
, resp. vk = cos 2k+1 2n π , k = 0 , 1, . . . , n − 1; see, e.g., [6, pp. 6–7]. Substituting these values into (4.15), we get explicit formulas for all the zeros of Q(1)
n
(z) and R(1)
n
(z). Similarly, substituting the values of the first part of (6.5) into (5.8) and (5.12), recalling that u = v2 in this last case, we get explicit expressions also for the zeros of Q(2)
n
(z) and Q(3)
n
(z). 6.3. Divisibility results. Since the polynomials we have dealt with in this paper all have integer coefficients, it makes sense to ask about divisibility and irreducibility over the rationals. By (3.14), divisibility properties of the Chebyshev polynomials
Un(v) carry over to Qn(Z). In particular, since Un−1(v) is a divisibility sequence (see, e.g., [6, pp. 227 ff]), we have the following result. 16 KARL DILCHER AND LARRY ERICKSEN
Corollary 6.3. For j = 1 , 2, 3, the polynomial sequences Q(j)
n
(z) are divisibility sequences, that is, if m|n then Q(j)
m
(z)|Q(j)
n
(z).
Example 7. The polynomial Q(3) 6 (z) in Table 4 factors as 2(2 z + 1)(7 z2 + z +4)(13 z2 + 11 z + 4), and we see that Q(3) 2 (z) and Q(3) 3 (z) are among the factors. In general, the situation is less straightforward for the polynomials Rn(Z), but for R(1)
n
(z) the additional term in (3.15) disappears. The identity (4.5) then shows that divisibility properties of Tn(v) carry over to R(1)
n
(z). However, in contrast to
Un(v), the polynomials Tn(v) are only a partial divisibility sequence. For instance,
R(1) 6 (z) = ( z2 + 4 z + 5)( z4 + 8 z3 + 38 z2 + 88 z + 73), and we see from Table 2 that
R(1) 2 (z) divides R(1) 6 (z), but R(1) 1 (z) and R(1) 3 (z) do not. Finally, since compositions of functions are involved in (3.14), we cannot directly conclude that known irreducibility results for Chebyshev polynomials will carry over to our Q- and R-polynomials. For some related approaches and results, see Proposition 6.8 in .
References
K. Dilcher and L. Ericksen, Polynomial analogues of restricted multicolor b-ary partition functions, Int. J. Number Theory 17 (2021), no. 2, 371–391. K. Dilcher and L. Ericksen, Polynomials and algebraic curves related to certain binary and
b-ary overpartitions, Ramanujan J. 68 , 44 (2025).
. W. J. Keith, Restricted k-color partitions, Ramanujan J. 40 (2016), 71–92. A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, J. Algebra 278 (2004), no. 1, 342–359. OEIS Foundation Inc. (2011), The On-Line Encyclopedia of Integer Sequences ,
. T. J. Rivlin, Chebyshev Polynomials. From Approximation Theory to Algebra and Number Theory. Second Edition . Wiley, New York, 1990. M. Ulas and B. ˙Zmija, On p-adic valuations of colored p-ary partitions, Monatsh. Math. 188
(2019), no. 2, 351–368.
Department of Mathematics and Statistics, Dalhousie University, Halifax, Nova Sco-tia, B3H 4R2, Canada
Email address : dilcher@mathstat.dal.ca
1212 Forest Drive, Millville, NJ 08332-2512, USA
Email address : LE22@cornell.edu |
7745 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11589512/ | Volumetric segmentation analysis of the levator ani muscle using magnetic resonance imaging in pelvic floor function assessment - PMC
Skip to main content
An official website of the United States government
Here's how you know
Here's how you know
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.
Service Alert: Planned Maintenance beginning July 25th
Most services will be unavailable for 24+ hours starting 9 PM EDT. Learn more about the maintenance.
Search
Log in
Dashboard
Publications
Account settings
Log out
Search… Search NCBI
Primary site navigation
Search
Logged in as:
Dashboard
Publications
Account settings
Log in
Search PMC Full-Text Archive
Search in PMC
Advanced Search
Journal List
User Guide
New Try this search in PMC Beta Search
View on publisher site
Download PDF
Add to Collections
Cite
Permalink PERMALINK
Copy
As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
Diagn Interv Radiol
. 2024 Jul 8;30(4):220–227. doi: 10.4274/dir.2024.232586
Search in PMC
Search in PubMed
View in NLM Catalog
Add to search
Volumetric segmentation analysis of the levator ani muscle using magnetic resonance imaging in pelvic floor function assessment
Ayşenur Buz Yaşar
Ayşenur Buz Yaşar
1 Bolu Abant İzzet Baysal University, Training and Research Hospital, Department of Radiology, Bolu, Turkey
Find articles by Ayşenur Buz Yaşar
1,✉, Rüveyde Begüm Yüzok
Rüveyde Begüm Yüzok
1 Bolu Abant İzzet Baysal University, Training and Research Hospital, Department of Radiology, Bolu, Turkey
Find articles by Rüveyde Begüm Yüzok
1, Emine Dağıstan
Emine Dağıstan
1 Bolu Abant İzzet Baysal University, Training and Research Hospital, Department of Radiology, Bolu, Turkey
Find articles by Emine Dağıstan
1
Author information
Article notes
Copyright and License information
1 Bolu Abant İzzet Baysal University, Training and Research Hospital, Department of Radiology, Bolu, Turkey
✉
Corresponding author.
Received 2023 Nov 6; Accepted 2024 Jan 18; Collection date 2024 Jul.
Copyright© Author(s) - Available online at dirjournal.org.
Content of this journal is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.
PMC Copyright notice
PMCID: PMC11589512 PMID: 38375767
Abstract
PURPOSE
In this case-control study, we aimed to evaluate how muscle volume affects pelvic floor function by analyzing the levator ani muscle (LAM) using volumetric segmentation in addition to standard magnetic resonance (MR) defecography assessments.
METHODS
We enrolled 85 patients with varying degrees of pelvic floor dysfunction (PFD) and 85 age- and gender-matched controls in this retrospective study. All patients had MR defecography images, while all controls had pelvic MR images obtained for other reasons. Group comparisons were performed using independent samples t-tests and Mann–Whitney U tests. The receiver operating curve (ROC) was constructed to establish a cut-off value for a normal LAM volume. Interrater reliability was assessed by calculating the intraclass correlation coefficient. A P value of less than 0.05 was considered statistically significant.
RESULTS
Volumetric measurements revealed that the control group had higher LAM volumes, and the ROC curve analysis indicated a cut-off value of 38934.3 mm 3 with a sensitivity of 0.812 and specificity of 0.8 for PFD assessment using LAM volumetric measurement. Gender did not significantly affect LAM volume in the control group.
CONCLUSION
Alongside the useful structural and functional information acquired from MR defecography images, volumetric analysis, and three-dimensional reconstructions of LAM may help to improve the accuracy of the diagnosis.
Keywords: Segmentation, levator ani muscle, pelvic floor dysfunction, magnetic resonance defecography, volumetric measurement
Main points
• Pelvic floor dysfunction (PFD) encompasses various medical conditions affecting the supportive ligaments, fascial coverings, and the muscles in the pelvic region.
• The levator ani muscle (LAM) is a critical component, and damage or weakening of this muscle is a common cause of pelvic organ prolapse (POP) and related conditions like incontinence, dyspareunia, and pelvic pain.
• Our study indicates that individuals with PFD tend to have a lower LAM volume, with a specific cut-off value for muscle volume linked to a higher tendency for PFD.
• However, contrary to initial assumptions, no linear correlation between the severity of POP or pelvic floor relaxation and muscle volume was observed in this study.
Pelvic floor dysfunction (PFD) is a comprehensive term that refers to a broad group of medical conditions that can affect the suspensory ligaments, fascial coverings, and muscles supporting the pelvic organs.1,2 It is a common disorder with an estimated prevalence of 25 percent among women in the United States.3 The etiological factors of PFD include female gender, a history of vaginal childbirth, chronic constipation, pelvic surgery, obesity, genetic predisposition, menopause, and aging.1,3,4
The anatomical structures in the pelvic region include the bladder, prostate, uterus, vagina, and rectum, which are evaluated in three compartments: anterior (bladder and urethra), middle (uterus and vagina), and posterior (rectum, anal canal).2,4 These structures are attached by the endopelvic fascia, the pelvic diaphragm, and the urogenital diaphragm and function as a single unit.1 The levator ani muscle (LAM) is a complex funnel-shaped structure consisting of three main components: the pubococcygeus (pubovaginalis, puboprostaticus, puboperineal, puboanal), puborectalis, and iliococcygeus.4,5,6 Damage or weakening of the LAM is the most common cause of pelvic organ prolapse (POP), resulting in a distorted shape of muscle that tends to tilt more vertically and widen the levator hiatus. An insult to the pelvic floor muscles or ligaments can lead to urinary or fecal incontinence, dyspareunia, constipation, and pelvic pain.2,6
Radiologically, for the assessment of these structures, magnetic resonance (MR) defecography uses sagittal balanced steady-state gradient echo (GRE) (different vendors have similar sequences with different trade names, such as True-FISP, FIESTA, and balanced-FFE) images, and a reference line called the pubococcygeal line (PCL) is drawn from the lower border of the pubic symphysis to the last coccygeal joint.1,2,7 The distance perpendicular from the posterior wall of the anorectal component to the PCL is called the “M line”, which corresponds to a measure of the location of the anorectal junction. The “H line” is the distance from the inferior border of the pubic symphysis to the posterior of the anorectal component and represents the anteroposterior width of the levator hiatus (Figure 1).7
Figure 1.
Open in a new tab
Reference lines of magnetic resonance defecography measurements at rest (pubococcygeal line: yellow line; H-line: green line; M-line: red line). PCL, pubococcygeal line; FLP, foot left posterior; HRA, head right anterior.
MR defecography is a dynamic examination that allows evaluation at rest, during contraction, and defecation.1,2,4 In the anterior compartment, the position of the urethra and bladder can be assessed for diagnoses such as urethral hypermobility and cystocele. In the middle compartment, uterine or cervical prolapse can be observed. In the posterior compartment, pathologies like rectocele or rectal intussusception can be detected.4,8 The classic PCL, H, and M lines are utilized for the radiological grading of these pathologies. Additionally, measuring the anorectal angle (ARA) and its dynamic changes are important in diagnosing pelvic floor dyssynergia.9
The thickness and volume measurement of the LAM have been investigated through ultrasound (US), computed tomography (CT), and MR studies, and the factors influencing the muscle volume and architecture have been explored.10,11,12 In our study, we hypothesize that the LAM volume is lower in patients with PFD than in healthy individuals. Alongside routine evaluations in MR defecography, we aimed to assess the impact of muscle volumes on the POP by conducting a volumetric segmentation analysis of the LAM.
Methods
This retrospective study was approved by the Bolu Abant İzzet Baysal University Clinical Research Ethics Committee and written informed consent was waived by it (date of project: 22.08.2023; project decision number: 262). This research study was conducted in accordance with the Declaration of Helsinki. The STARD guidelines were followed for reporting and joint recommendations of the ESUR and ESGAR Pelvic Floor Working Group were followed for patient preparation, image acquisition, and interpretation.8,13,14,15
Patient selection
A total of 85 patients (M: 16, F: 69), aged between 19 and 92 years old, with varying levels of PFD, as well as 85 controls (M: 16, F: 69) with ages ranging from 20 to 85 years old, were included in this study. The control group consisted of individuals without pelvic floor pathology identified in MR imaging (MRI) assessments at rest but with pelvic MRI taken for other medical reasons. The participants’ MR defecography and pelvic MRI were recruited from picture archiving communication systems. All MRI were acquired at the radiology department between September 2020 and November 2023. Patients with poor quality imaging due to artifacts on MR scan, inadequate or incomplete imaging, or in whom volumetric measurements could not be performed were excluded from participation, as were control group participants with findings of PFD on pelvic MRI. There were no patients with a history of pelvic region radiotherapy or oncological surgery in the patient population and control group (Figure 2 shows a flowchart of the study).
Figure 2.
Open in a new tab
Flowchart of the study. MR, magnetic resonance.
Patient preparation
For MR defecography images, patients were asked to empty their bladders and bowels 1–2 hours before the MRI examination. The patient should be trained about the imaging steps (rest, squeeze, strain, Valsalva maneuver, and defecation) prior to the MRI examination. Immediately before imaging, 120–180 mL of rectal gel was injected gently through the anal canal in the decubitus position to fill the rectum. Preparation for a pelvic MRI involves wearing comfortable clothes, removing all metallic accessories, and discussing potential contrast allergies.
Magnetic resonance imaging protocol and image acquisition
The MRI scans of all patients participating in the study were performed on the General Electric Signa TM Explorer MR 1.5T closed system device (GE Healthcare, Chicago, Illinois, IL, United States) using a phased array body coil without contrast material administration. The patient was positioned lying horizontally with the face and torso facing up with knees elevated on a pillow. To protect the scanner during imaging, adult diapers and disposable sheets were used. MR defecography consists of both statical and dynamical sequences including sagittal, coronal, and axial static T2-weighted images at rest, sagittal and coronal cine balanced steady-state GRE while squeezing and straining, and coronal cine balanced steady-state GRE images during defecation at least three times until the rectum is emptied as much as possible (detailed information is summarized in Table 1). The entire defecography procedure duration varied between 15 to 30 minutes. Pelvic MR protocol includes sagittal and axial T2-weighted, coronal fat-saturated T2-weighted images, axial T1-weighted fast-spin echo images, axial diffusion-weighted images with a b value of 50 and 800, and axial liver acquisition with volume acceleration (LAVA) images. In cases where contrast media was required, dynamic sagittal LAVA images (for female patients only) and axial contrast-enhanced LAVA images were obtained. Pelvic MRI lasts approximately 20–30 minutes (Table 2).
Table 1. Magnetic resonance defecography protocol.
Plane MR sequence TR/TE (ms)Matrix size/NEX Slice thickness (mm)Spacing (mm)FOV (cm x cm)Information type Phase
Sagittal Static, T2 PROPELLER 7570/152.32 288 x 288/4 4 1 25 x 25 Structural Rest
Coronal Static, T2 PROPELLER 3859/148.99 288 x 288/4 4 1 34 x 34 Structural Rest
Axial Static, T2 PROPELLER 5195/121.3 320 x 320/4 4 1 35 x 35 Structural Rest
Sagittal midline Dynamic, Cine FIESTA 4/1.4 256 x 288/4 6 1 25 x 25 Functional Squeeze (Kegel)
Coronal Dynamic, Cine FIESTA 5/2.02 256 x 288/4 6 1 25 x 25 Functional Squeeze (Kegel)
Sagittal midline Dynamic, Cine FIESTA 4/1.91 256 x 288/4 6 1 25 x 25 Functional Strain (Valsalva)
Coronal Dynamic, Cine FIESTA 5/2.01 256 x 288/4 6 1 25 x 25 Functional Strain (Valsalva)
Sagittal midline Dynamic, Cine FIESTA 5/1.9 256 x 288/4 6 1 25 x 25 Functional Defecation (at least 3 times)
Open in a new tab
MR, magnetic resonance; TR, time of repetition; TE, time of echo; NEX, number of excitations; FOV, field of view.
Table 2. Pelvic magnetic resonance protocol.
Plane MR sequence TR/TE (ms)Matrix size/NEX Slice thickness (mm)Spacing (mm)FOV (cm x cm)
Sagittal T2 PROPELLER 6.121/113.74 288 x 288/4 5 1 25 x 25
Axial T2 PROPELLER 4.146/96.62 300 x 300/4 5 1 32 x 32
Coronal T2 FAT-SAT PROPELLER 6.446/96.82 300 x 300/4 5 1 32 x 32
Axial T1 fast spin echo 552/10.28 320 x 224/4 5 1 32 x 32
Axial Diffusion-weighted imaging b value: 50–800 5.835/78.90 256 x 288/4 5 1 32 x 32
Axial Apparent diffusion coefficient 5.835/78.90 128 x 128/4 5 1 32 x 32
Axial LAVA 6/3.15 280 x 192/4 6 1 32 x 32
Sagittalbr>(female protocol)Dynamic, contrast-enhanced LAVA 3/1.84 320 x 192/4 4 1 32 x 32
AxialContrast-enhanced LAVA 6/3.15 280 x 192/4 4 1 32 x 32
Open in a new tab
Only obtained for contrast-enhanced studies. MR, magnetic resonance; TR, time of repetition; TE, time of echo; NEX, number of excitations; FOV, field of view; LAVA, liver acquisition with volume acquisition.
Routine magnetic resonance defecography and pelvic magnetic resonance imaging interpretation
Two radiologists (E.D. and A.B.Y.) evaluated the MRI in consensus. An independent radiologist (R.B.Y.), who was blinded to the outcomes, concurrently interpreted a randomly selected subset of 35 cases. All metric measurements including PCL, H, and M-lines at rest and defecation, ARA at rest, squeezing and defecation, levator plate angle (LPA, the angle between the levator plate and PCL) at maximal straining, and urethral angle (the angle between the urethra and PCL) were completed on midline sagittal MRI. The severity of cystocele and uterine prolapse was graded according to the depth of the herniation under the PCL as mild (less than 3 cm), moderate (3 to 6 cm), and severe (greater than 6 cm).8,16,17 A rectocele is characterized by the rectal wall extending beyond the anticipated typical shape, with grading based on the extent of protrusion: small (<2 cm), moderate (2-4 cm), and large (>4 cm), determined by the depth of the bulge.18 The assessment of pelvic diaphragm relaxation based on M-line lengths was evaluated using the H-line, M-line, and organ prolapse (HMO) classification system. A normal hiatal position was defined as an M-line measurement between 0 and 2 cm (grade 0), while mild descent was categorized as an M-line measurement ranging from 2 to 4 cm (grade 1). Moderate descent was characterized by an M-line measurement between 4 and 6 cm (grade 2), and severe descent was indicated when the M-line measurement exceeded 6 cm (grade 3).19 On the other hand, a normal hiatal width was defined as an H-line measurement between 0 and 6 cm (grade 0), while mild hiatal enlargement was categorized as an H-line measurement ranging from 6 to 8 cm (grade 1). Moderate enlargement was characterized by an H-line measurement between 8 and 10 cm (grade 2), and severe enlargement was indicated when the H-line measurement exceeded 10 cm (grade 3).19 Urethral hypermobility is a condition of excessive horizontal translation (more than 30°) of the urethra due to a weak pelvic floor.1
Levator ani muscle manual segmentation and volumetric measurements
For analyzing the medical image data, a free and open-source imaging package software [three-dimensional (3D) Slicer version 5.2.2 for Mac OS X] was utilized. A radiologist with eight years’ experience (A.B.Y.), and a radiology resident with five years’ experience (R.B.Y.) manually segmented the LAM from the contiguous axial T2-weighted MRI slices using the “Segment Editor” module in the 3D Slicer software. The anterior boundary of the LAM is defined as the pubic symphysis, whereas the posterior boundary of the LAM is defined as the coccyx. The muscles surrounding the anal canal and rectum were delineated. Quantitative information, including the number of voxels, the volume of the muscle, minimum, maximum, mean, and median values, standard deviation, and surface area, derived using the “Segment Statistics” module, was noted. For each patient and control, 3D reconstruction models of the LAM were also created (Figures 3, 4). The average time to segment the LAM required 10 minutes per patient.
Figure 3.
Open in a new tab
An example of levator ani muscle (LAM) segmentation and three-dimensional (3D) image reconstruction in a healthy individual. Image (a) represents a completed volume rendered segmentation model of the LAM. On the right is a screenshot of the 3D-Slicer software “Segment Editor” module. Image (b) shows an axial T2-weighted series, which are the source images we loaded for segmentation analysis. Images (d, e) show a coronal and sagittal view of the LAM, reconstructed by 3D-Slicer to edit the segment, and image C is a real-time 3D model of LAM.
Figure 4.
Open in a new tab
Pipeline of the study. MR, magnetic resonance.
Statistical analysis
All statistical analyses were conducted using the SPSS 24.0 software (IBM Corp., Armonk, NY, USA). Metric measurements and quantitative segmentation results were reported with means and standard deviations. The normality of distribution was assessed using the Shapiro–Wilk test. Normally distributed data were compared using the independent samples t-test, while non-normally distributed data were evaluated using the Mann–Whitney U test. A chi-square test was used to compare the observed frequencies of categorical data. Subgroup comparisons within the patient group were done using the Kruskal–Wallis test. The receiver operator characteristic (ROC) curve was drawn to assess the sensitivity and specificity of the volumetric measurement, and the optimal cut-off value was selected. The intraclass correlation coefficient was used to estimate the interrater reliability of the MR defecography measurements. A P value of less than 0.05 was considered statistically significant.
Results
Demographic characteristics
The mean age was calculated as 53.54 ± 15.7 years for the patient group and 51.99 ± 13.4 years for the control group. There was no significant difference between the groups in terms of age. In both groups, the distribution of men and women was equal. Among 69 female participants, 14 in the case and 13 in the control groups had undergone a hysterectomy (P value, 0.632).
Routine magnetic resonance imaging of pelvic floor and magnetic resonance defecography findings
Pelvic floor measurements were performed in both the patient and control groups where applicable. The mean PCL length was calculated as 102.24 ± 9.9 mm, the H-line at rest was 49.35 ± 9.8 mm, and the M-line at rest was 15.97 ± 11.7 mm for the patient group. Conversely, the mean PCL length was calculated as 103.06 ± 10.9 mm, the H-line at rest as 31.06 ± 5.4 mm, and the M-line at rest as 6.12 ± 3.2 mm for the control group. The remaining measurements were only performed in the patient group. The H-line at defecation was calculated as 69.07 ± 17.1 mm, and the M-line at defecation was 43.21 ± 21.3 mm. The average ARA angle at rest was 96.11 ± 17.04°, 82.71 ± 17.8° at strain, and 113.01 ± 22.4° at defecation. The mean LPA at maximal straining was 37.86 ± 20.1°; 61.2 percent of the patients (n = 52) had urethral hypermobility. Only 4 patients had peritoneocele (Table 3). The most common pathologies were grade 1 cystocele (n = 36, 42.4%) and mild hiatal enlargement (n = 35, 41.2%) followed by grade 1 anterior rectocele (n = 32, 37.6%), grade 2 anterior rectocele (n = 26, 30.6%), and mild pelvic floor descent (n = 25, 29.4%). Data regarding POP and pelvic floor relaxation are outlined in Table 4.
Table 3.Magnetic resonance imaging measurements of patient and control groups.
Patient group Control group P value
Age (year, mean ± SD)53.54 ± 15.7 51.99 ± 13.4 0.489 a
Gender distribution (M/F, n)16/69 16/69 1 b
Hysterectomy rate of women (n, and percent)14 (16.5%)13 (15.3%)0.632 b
PCL rest (mm, mean ± SD)102.24 ± 9.9 103.06 ± 10.9 0.609 a
H-line rest (mm, mean ± SD)49.35 ± 9.8 31.06 ± 5.4 P 0.001 c
M-line rest (mm, mean ± SD)15.97 ± 11.7 6.12 ± 3.2 P 0.001 c
H-line defecation (mm, mean ± SD)69.07 ± 17.1 N/A N/A
M-line defecation (mm, mean ± SD)43.21 ± 21.3 N/A N/A
ARA rest (°, mean ± SD)96.11 ± 17.04 N/A N/A
ARA strain (°, mean ± SD)82.71 ± 17.8 N/A N/A
ARA defecation (°, mean ± SD)113.01 ± 22.4 N/A N/A
LPA maximal straining (°, mean ± SD)37.86 ± 20.1 N/A N/A
LAM volume (mm 3, mean ± SD)33,214.6 ± 11884.6 48,107.9 ± 12274.2 P 0.001 a
LAM surface area (mm 2, mean ± SD)15425.8 ± 4022.2 19,458.4 ± 4467.9 P 0.001 a
LAM number of voxels (mean ± SD)12896.5 ± 5202.9 18,778.1 ± 6784.1 P 0.001 a
n, percentage
Urethral hypermobility52 (61.2%)N/A N/A
Peritoneocele4 (4.7%)N/A N/A
Open in a new tab
a Independent samples t-test results; b Fisher’s exact test results; c Mann–Whitney U test results. SD, standard deviation; M, male; F, female; n, number; PCL, pubococcygeal line; ARA, anorectal angle; LPA, levator plate angle; LAM, levator ani muscle.
Table 4. Magnetic resonance defecography assessment of patient group.
Absent Grade 1 Grade 2 Grade 3
(n, percent)(n, percent)(n, percent)(n, percent)
Cystocele41 (48.2%)36 (42.4%)7 (8.2%)1 (1.2%)
Uterine prolapse40 (72.7%)11 (20%)4 (7.3%)0 (0%)
Anterior rectocele21 (24.7%)32 (37.6%)26 (30.6%)6 (7.1%)
Healthy Mild Moderate Severe
(n, percent)(n, percent)(n, percent)(n, percent)
Hiatal enlargement (n, percent)26 (30.6%)35 (41.2%)20 (23.5%)4 (4.7%)
Pelvic floor descent (n, percent)15 (17.6%)25 (29.4%)23 (27.1%)22 (25.9%)
Open in a new tab
n, number.
Interrater reliability of magnetic resonance defecography assessment
To determine interrater reliability, 35 patients were selected randomly and two reviewers (blinded to each other) interpreted the PCL line at rest, H-line at rest and defecation, M-line at rest and defecation, ARA at rest, maximal strain and defecation, and LPA. The intraclass correlation analysis revealed excellent agreement (Table 5).20
Table 5. Interobserver correlations of routine magnetic resonance defecography measurements.
Observer 1 Observer 2 ICC P value
Measurements (n = 35)(Mean ± SD)(Mean ± SD)(95% CI)
PCL rest (mm)100.49 ± 10.7 102.09 ± 10.1 0.985 (0.970–0.992)0.001
H-line rest (mm)50.9 ± 9.9 51.2 ± 10.1 0.992 (0.985–0.996)0.001
M-line rest (mm)19.5 ± 12.2 18.6 ± 12.1 0.976 (0.952–0.988)0.001
H-line defecation (mm)71.9 ± 18.5 74.2 ± 18.5 0.955 (0.914–0.977)0.001
M-line defecation (mm)47.5 ± 21.5 48.8 ± 21.2 0.967 (0.936–0.983)0.001
ARA rest (°)99.8 ± 15.7 98.7 ± 14.6 0.948 (0.9–0.973)0.001
ARA strain (°)84.9 ± 18.9 83.4 ± 17.2 0.943 (0.891–0.971)0.001
ARA defecation (°)112.3 ± 22.8 113.7 ± 22.4 0.991 (0.982–0.995)0.001
LPA maximal straining (°)35.6 ± 19.3 37.3 ± 19.6 0.967 (0.936–0.983)0.001
Open in a new tab
ICC, intraclass correlation coefficient; CI, confidence interval; n, number; SD, standard deviation; PCL, pubococcygeal line; ARA, anorectal angle; LPA, levator plate angle.
Levator ani muscle volumetric measurement
The mean number of voxels was calculated as 12,896.5 ± 5202.9 for the patient group and 18,778.1 ± 6784.1 for the control group. The mean volume of LAM was quantified as 33,214.6 ± 11,884.6 mm3 for the patient group and 48,107.9 ± 12,274.2 mm3 for the control group. The mean surface area of the patient group was 15,425.8 ± 4,022.2 mm 2 and 19.458.4 ± 4,467.9 mm 2 for the control group (Table 3).
Association of levator ani muscle volume and pelvic floor dysfunction
Voxel numbers, LAM volumes, and surface area were higher in the control group. Since the data were not normally distributed, the number of voxels, segment volume, and surface area of the patients and controls’ LAM were compared using the Mann–Whitney U test. For each variable, a statistically significant difference was observed (P values were<0.001). PFD is defined as the presence of conditions that may affect any compartments, including hiatal enlargement, pelvic floor descent, cystocele, uterine prolapse, rectocele, and peritoneocele. The ROC curve analyses were performed to evaluate the sensitivity and specificity of the volume and surface area measurement of the LAM on PFD. A cut-off value of 38,934.3 mm 3 was set with a 0.812 sensitivity and 0.8 specificity for the LAM volume. The area under the curve (AUC) was computed as 0.834. For surface area measurement, the AUC was calculated as 0.753, and the cut-off value was set as 16,639.4 mm 2 with a sensitivity of 0.753 and specificity of 0.706 (Figure 5). We also compared the mean volume of the LAM in the patient group, depending on the severity of hiatal enlargement, pelvic floor descent, and POP. When the disease worsened, no statistically significant change in the muscle volume was observed (P values were 0.440, 0.929, and 0.732, respectively).
Figure 5.
Open in a new tab
The receiver operating curve analyses of the LAM volume measurement and surface area in the presence of pelvic floor dysfunction. Volumetric measurement: AUC: 0.834, cut-off value: 38,934.3 mm 3 with a sensitivity of 0.812 and specificity of 0.8 (a). Surface area measurement: AUC: 0.753, cut-off value 16,639.4 mm 2 with a sensitivity of 0.753 and specificity of 0.706 (b). LAM, levator ani muscle; AUC, area under the curve.
Effect of gender on levator ani muscle volume
No statistically significant difference was observed between female (n = 69) and male (n = 16) controls concerning the number of voxels, LAM volumes, and surface areas (P values were 0.419, 0.449, and 0.449, respectively).
Effect of aging on levator ani muscle volume
A weak negative correlation was observed between the age and LAM volume in only the patient group (r: −0.227, P value, 0.037). All participants (n = 170) were divided into two groups according to their age; individuals older than 50 years comprised the first group (n = 93) and the remaining individuals represented the second group (n = 77). The average volume of the LAM was significantly lower in the first group (P value, 0.019).
Effect of history of hysterectomy on levator ani muscle volume
No statistically significant difference was observed among the women controls based on their history of hysterectomy (n = 69, and P value is 0.671).
Discussion
In the diagnosis of PFD, clinical examination is generally indefinite in isolation and may lead to the underestimation of pathologies and involved compartments.2, 15 Various imaging modalities are employed to assess the pelvic floor, particularly the LAM, which represents the active component, including translabial–endovaginal US, CT, fluoroscopy, and MRI.15, 21 MR defecography imaging offers exceptional spatial and contrast resolution, enabling the delineation of even small tears or injuries, and providing detailed anatomical and functional information.3 In this regard, it plays a crucial role in the concurrent assessment of pelvic organs and pelvic floor muscles without radiation exposure and contrast media administration, unlike dynamic fluoroscopic defecography.22
In this retrospective case-control study, our primary objective was to investigate the relationship between the volume of the LAM and PFD, and its potential contribution to routine MR defecography measurements. As we assumed, our results confirmed the presence of a correlation between a decreased LAM volume and PFD. Patients with an LAM volume calculated below 38,934.3 mm3 have a higher tendency toward PFD. However, contrary to our initial hypothesis, we did not observe a linear correlation between the severity of POP or pelvic floor relaxation and muscle volume. The small number of subgroups in the patient group may have affected the reliability of the subgroup comparison results. Further work on larger populations is thus needed to validate our results.
Previous studies focused on LAM segmentation based on transperineal or endovaginal US and MRI.10, 11, 12 Rabbat et al.23 proposed using deep learning algorithms to automate LAM segmentation as a means to improve the diagnostic ability of the US. Another study utilizing MRI suggests a modified Chan–Vese segmentation model, which uses intensity information and the influence of shape to segment the LAM in axial slices.24 Compared to manual segmentation, automated segmentation models may shorten the time taken to complete the procedure, which can assist physicians in executing muscle identification, segmentation, 3D reconstruction, and automatic volume measurement.22
In the current study, we chose manual segmentation, despite the extended time of the process, because it is the reference standard. Despite employing manual segmentation, the fact that one segmentation could be completed in approximately 10 minutes demonstrates its feasibility and appropriateness for clinical work.
As predicted in prior studies, our study participants mostly included women. One of our secondary objectives was to assess the influence of gender on LAM volume in healthy participants. A publication by Cheung et al.25 reports that the LAM has extraordinary androgen sensitivity in rodents and humans. We hypothesized that the volume of the LAM in women may be lower than in men, potentially giving rise to vulnerability to pelvic floor disorders. However, we found that LAM volume, LAM surface area, and the number of voxels were similar for both genders. This suggests that it is the difference in processes, such as pregnancy and childbirth, rather than gender, which may be at play. Cheung et al.25 investigated the LA and walking muscle volumes in patients with prostate cancer receiving androgen deprivation therapy and found that the therapy process caused muscle volume loss. The limitation of their study is that patients with prostate cancer often concurrently receive radiation therapy (RT); it should thus be kept in mind that the outcomes may have been influenced by RT rather than the androgen sensitivity of the muscle. There is therefore a need to elucidate the molecular mechanism of androgen sensitivity of the LAM.
Wyman et al.26 conducted a study to evaluate the relationship between LAM volume, age, and body mass index (BMI). Interestingly, the results showed that an increased age in female participants correlated with an elevated LAM volume; however, there was no correlation between BMI and muscle volume. The study authors assumed these results to be related to a reduction in the strength and integrity of the LAM resulting from sarcopenia.26 In their previous paper, the authors evaluated whether the estimated LA subtended volume (eLASV) could predict the success of POP surgical treatment. Their results indicated that patients with a higher eLASV had an increased risk of surgical failure.12 The main question concerning this study is the absence of a clear definition of the volumetric measurement process described in their paper; accordingly, these findings must be interpreted with caution. In contrast to earlier findings presented by Wyman et al.26, our study reveals that older individuals have a lower LAM volume.
A similar study to the current research was carried out by Nandikanti et al.27 to evaluate the LA bowel volume variation between resting and straining states in patients with POP and healthy controls. The results indicate that hiatus size and bowel volume change during straining.27 To the best of our knowledge, this is the first study comparing LAM volume and routine MR defecography measurements.
This study has some limitations. Due to the retrospective design of the research, we were unable to gather selected information, including BMI, history of pregnancy/vaginal birth, and abortion, which indicate a constant relationship with pelvic floor insufficiency. Additionally, we only assessed LAM volume; however, anal and urethral sphincters, internal obturator muscle, coccygeus muscle, and perineal muscles also play roles in PFD. Finally, the technique for the axial T2 pelvis MRI for the controls was not exactly matched to what was used for MR defecography.
In conclusion, a lower LAM volume appears to show a direct correlation with an increased probability of PFD. Our results did not reveal a linear correlation between the severity of POP or pelvic floor relaxation and muscle volume. Future research with a larger participant pool is warranted to further investigate this matter.
Acknowledgments
We would like to thank our MRI team.
Footnotes
Conflict of interest disclosure
The authors declared no conflicts of interest.
References
1.Salvador JC, Coutinho MP, Venâncio JM, Viamonte B. Dynamic magnetic resonance imaging of the female pelvic floor-a pictorial review. Insights Imaging. Insights Imaging. 2019;10(1):4. doi: 10.1186/s13244-019-0687-9. [DOI] [PMC free article] [PubMed] [Google Scholar]
2.Kanmaniraja D, Arif-Tiwari H, Palmer SL. MR defecography review. Abdom Radiol (NY) 2021;46(4):1334–1350. doi: 10.1007/s00261-019-02228-4. [DOI] [PubMed] [Google Scholar]
3.Wu JM, Vaughan CP, Goode PS. Prevalence and trends of symptomatic pelvic floor disorders in U.S. women. Obstet Gynecol. 2014;123(1):141–148. doi: 10.1097/AOG.0000000000000057. [DOI] [PMC free article] [PubMed] [Google Scholar]
4.Maccioni F, Alt CD. MRI of the pelvic floor and MR defecography. 2018 Mar 21. In: Hodler J, Kubik-Huch RA, von Schulthess GK, editors. Diseases of the Abdomen and Pelvis 2018-2021: Diagnostic Imaging - IDKD Book [Internet]. Cham (CH): Springer; 2018. Chapter 2. [Google Scholar]
5.Margulies RU, Hsu Y, Kearney R, Stein T, Umek WH, DeLancey JO. Appearance of the levator ani muscle subdivisions in magnetic resonance images. Appearance of the levator ani muscle subdivisions in magnetic resonance images. Obstet Gynecol. 2006;107(5):1064–1069. doi: 10.1097/01.AOG.0000214952.28605.e8. [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Gowda SN, Bordoni B. Anatomy, abdomen and pelvis: levator ani muscle. [Updated 2022 Oct 26]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2023 Jan-. [PubMed] [Google Scholar]
7.Li M, Wang B, Liu X, Qiao P, Jiao W, Jiang T. MR defecography in the assessment of anatomic and functional abnormalities in stress urinary incontinence before and after pelvic reconstruction. Eur J Radiol. 2020;126:108935. doi: 10.1016/j.ejrad.2020.108935. [DOI] [PubMed] [Google Scholar]
8.El Sayed RF, Alt CD, Maccioni F. Magnetic resonance imaging of pelvic floor dysfunction - joint recommendations of the ESUR and ESGAR Pelvic Floor Working Group. Eur Radiol. 2017;27(5):2067–2085. doi: 10.1007/s00330-016-4471-7. [DOI] [PMC free article] [PubMed] [Google Scholar]
9.Haliloglu N, Erden A. Magnetic resonance defecography findings of dyssynergic defecation. Pol J Radiol. 2022;87(1):181–185. doi: 10.5114/pjr.2022.114866. [DOI] [PMC free article] [PubMed] [Google Scholar]
10.Zhang X, Xiang Y, Yao J, et Automatic segmentation of the female pelvic floor muscles on MRI for pelvic floor function assessment. Quant Imaging Med Surg. 2023;13(7):4181–4195. doi: 10.21037/qims-22-1198. [DOI] [PMC free article] [PubMed] [Google Scholar]
11.Asif Z, Tomashev R, Peterkin V. Levator ani muscle volume and architecture in normal vs. muscle damage patients using 3D endovaginal ultrasound: a pilot study. Int Urogynecol J. 2023;34(2):581–587. doi: 10.1007/s00192-022-05366-4. [DOI] [PubMed] [Google Scholar]
12.Wyman AM, Tanner JP, Bassaly R, Hoyte L, Greene K. Levator ani muscle (LAM) anatomical volume compared to age and body mass index (BMI) of patients. Int Urogynecol J. 2020;31(7):1457–1461. doi: 10.1007/s00192-019-04141-2. [DOI] [PubMed] [Google Scholar]
13.Bossuyt PM, Reitsma JB, Bruns DE. STARD 2015: an updated list of essential items for reporting diagnostic accuracy studies. Radiology. 2015;277(3):826–832. doi: 10.1148/radiol.2015151516. [DOI] [PubMed] [Google Scholar]
14. [Google Scholar]
15.Lalwani N, Khatri G, El Sayed RF. MR defecography technique: recommendations of the society of abdominal radiology’s disease-focused panel on pelvic floor imaging. Abdom Radiol (NY) 2021;46(4):1351–1361. doi: 10.1007/s00261-019-02160-7. [DOI] [PubMed] [Google Scholar]
16.Chamié LP, Ribeiro DMFR, Caiado AHM, Warmbrand G, Serafini PC. Translabial US and dynamic mr imaging of the pelvic floor: normal anatomy and dysfunction. Radiographics. 2018;38(1):287–308. doi: 10.1148/rg.2018170055. [DOI] [PubMed] [Google Scholar]
17.Yoon I, Gupta N. Pelvic prolapse imaging. [Updated 2023 May 1]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2023 Jan-. [PubMed] [Google Scholar]
18.Thapar RB, Patankar RV, Kamat RD, Thapar RR, Chemburkar V. MR defecography for obstructed defecation syndrome. Indian J Radiol Imaging. 2015;25(1):25–30. doi: 10.4103/0971-3026.150134. [DOI] [PMC free article] [PubMed] [Google Scholar]
19.Picchia S, Rengo M, Bellini D. Dynamic MR of the pelvic floor: influence of alternative methods to draw the pubococcygeal line (PCL) on the grading of pelvic floor descent. Eur J Radiol Open. 2019;6:187–191. doi: 10.1016/j.ejro.2019.05.002. [DOI] [PMC free article] [PubMed] [Google Scholar]
20.Landis JR, Koch GG. The measurement of observer agreement for categorical data. Biometrics. 1977;33(1):159–174. doi: 10.2307/2529310. [DOI] [PubMed] [Google Scholar]
21.Guo M, Li D. Pelvic floor images: anatomy of the levator ani muscle. Dis Colon Rectum. 2007;50(10):1647–1655. doi: 10.1007/s10350-007-0262-1. [DOI] [PubMed] [Google Scholar]
22.Boyadzhyan L, Raman SS, Raz S. Role of static and dynamic MR imaging in surgical pelvic floor dysfunction. Radiographics. 2008;28(4):949–967. doi: 10.1148/rg.284075139. [DOI] [PubMed] [Google Scholar]
23.Rabbat N, Qureshi A, Hsu KT. Bioengineering (Basel). 2023;10(8):894. doi: 10.3390/bioengineering10080894. [DOI] [PMC free article] [PubMed] [Google Scholar]
24.Ma Z, Jorge RN, Tavares JM. A shape guided C-V model to segment the levator ani muscle in axial magnetic resonance images. Med Eng Phys. 2010;32(7):766–774. doi: 10.1016/j.medengphy.2010.05.002. [DOI] [PubMed] [Google Scholar]
25.Cheung AS, Cunningham C, Ko DD. Selective loss of levator ani and leg muscle volumes in men undergoing androgen deprivation therapy. J Clin Endocrinol Metab. 2019;104:2229–2238. doi: 10.1210/jc.2018-01954. [DOI] [PubMed] [Google Scholar]
26.Wyman AM, Rodrigues Jr AA, Hahn L. Estimated levator ani subtended volume: a novel assay for predicting surgical failure after uterosacral ligament suspension. Am J Obstet Gynecol. 2016;214:611. doi: 10.1016/j.ajog.2015.11.005. [DOI] [PubMed] [Google Scholar]
27.Nandikanti L, Sammarco AG, Chen L, Ashton-Miller JA, DeLancey JO. Levator bowl volume during straining and its relationship to other levator measures. Int Urogynecol. 2019;30(9):1457–1463. doi: 10.1007/s00192-019-04006-8. [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from Diagnostic and Interventional Radiology are provided here courtesy of Turkish Society of Radiology
ACTIONS
View on publisher site
PDF (767.7 KB)
Cite
Collections
Permalink PERMALINK
Copy
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
On this page
Abstract
Main points
Methods
Results
Discussion
Acknowledgments
Footnotes
References
Cite
Copy
Download .nbib.nbib
Format:
Add to Collections
Create a new collection
Add to an existing collection
Name your collection
Choose a collection
Unable to load your collection due to an error
Please try again
Add Cancel
Follow NCBI
NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed
Connect with NLM
NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube
National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
Back to Top |
7746 | https://www.ncbi.nlm.nih.gov/books/NBK10806/ | Phototransduction - Neuroscience - NCBI Bookshelf
An official website of the United States government
Here's how you know
The .gov means it's official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Bookshelf
Search database
Search term
Search
Browse Titles
Advanced
Help
Disclaimer
NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health.
Purves D, Augustine GJ, Fitzpatrick D, et al., editors. Neuroscience. 2nd edition. Sunderland (MA): Sinauer Associates; 2001.
By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.
Neuroscience. 2nd edition.
Show details
Purves D, Augustine GJ, Fitzpatrick D, et al., editors.
Sunderland (MA): Sinauer Associates; 2001.
Search term
Phototransduction
In most sensory systems, activation of a receptor by the appropriate stimulus causes the cell membrane to depolarize, ultimately stimulating an action potential and transmitter release onto the neurons it contacts. In the retina, however, photoreceptors do not exhibit action potentials; rather, light activation causes a graded change in membrane potential and a corresponding change in the rate of transmitter release onto postsynaptic neurons. Indeed, much of the processing within the retina is mediated by graded potentials, largely because action potentials are not required to transmit information over the relatively short distances involved.
Perhaps even more surprising is that shining light on a photoreceptor, either a rod or a cone, leads to membrane hyperpolarization rather than depolarization (Figure 11.5). In the dark, the receptor is in a depolarized state, with a membrane potential of roughly -40 mV (including those portions of the cell that release transmitters). Progressive increases in the intensity of illumination cause the potential across the receptor membrane to become more negative, a response that saturates when the membrane potential reaches about -65 mV. Although the sign of the potential change may seem odd, the only logical requirement for subsequent visual processing is a consistent relationship between luminance changes and the rate of transmitter release from the photoreceptor terminals. As in other nerve cells, transmitter release from the synaptic terminals of the photoreceptor is dependent on voltage-sensitive Ca 2+ channels in the terminal membrane. Thus, in the dark, when photoreceptors are relatively depolarized, the number of open Ca 2+ channels in the synaptic terminal is high, and the rate of transmitter release is correspondingly great; in the light, when receptors are hyperpolarized, the number of open Ca 2+ channels is reduced, and the rate of transmitter release is also reduced. The reason for this unusual arrangement compared to other sensory receptor cells is not known.
Figure 11.5
An intracellular recording from a single cone stimulated with different amounts of light (the cone has been taken from the turtle retina, which accounts for the relatively long time course of the response). Each trace represents the response to a brief (more...)
The relatively depolarized state of photoreceptors in the dark depends on the presence of ion channels in the outer segment membrane that permit Na+ and Ca 2+ ions to flow into the cell, thus reducing the degree of inside negativity (Figure 11.6). The probability of these channels in the outer segment being open or closed is regulated in turn by the levels of the nucleotide cyclic guanosine monophosphate (cGMP) (as in many other second messenger systems; see Chapter 8). In darkness, high levels of cGMP in the outer segment keep the channels open. In the light, however, cGMP levels drop and some of the channels close, leading to hyperpolarization of the outer segment membrane, and ultimately the reduction of transmitter release at the photoreceptor synapse.
Figure 11.6
Cyclic GMP-gated channels in the outer segment membrane are responsible for the light-induced changes in the electrical activity of photoreceptors (a rod is shown here, but the same scheme applies to cones). In the dark, cGMP levels in the outer segment (more...)
The series of biochemical changes that ultimately leads to a reduction in cGMP levels begins when a photon is absorbed by the photopigment in the receptor disks. The photopigment contains a light-absorbing chromophore (retinal, an aldehyde of vitamin A) coupled to one of several possible proteins called opsins that tune the molecule's absorption of light to a particular region of the spectrum. Indeed, it is the different protein component of the photopigment in rods and cones that contributes to the functional specialization of these two receptor types. Most of what is known about the molecular events of phototransduction has been gleaned from experiments in rods, in which the photopigment is rhodopsin (Figure 11.7A). When the retinal moiety in the rhodopsin molecule absorbs a photon, its configuration changes from the 11-cis isomer to all-trans retinal; this change then triggers a series of alterations in the protein component of the molecule (Figure 11.7B). The changes lead, in turn, to the activation of an intracellular messenger called transducin, which activates a phosphodiesterase that hydrolyzes cGMP. All of these events take place within the disk membrane. The hydrolysis by phosphodiesterase at the disk membrane lowers the concentration of cGMP throughout the outer segment, and thus reduces the number of cGMP molecules that are available for binding to the channels in the surface of the outer segment membrane, leading to channel closure.
Figure 11.7
Details of phototransduction in rod photoreceptors. (A) The molecular structure of rhodopsin, the pigment in rods. (B) The second messenger cascade of phototransduction. Light stimulation of rhodopsin in the receptor disks leads to the activation of a (more...)
One of the important features of this complex biochemical cascade initiated by photon capture is that it provides enormous signal amplification. It has been estimated that a single light-activated rhodopsin molecule can activate 800 transducin molecules, roughly eight percent of the molecules on the disk surface. Although each transducin molecule activates only one phosphodiesterase molecule, each of these is in turn capable of catalyzing the breakdown of as many as six cGMP molecules. As a result, the absorption of a single photon by a rhodopsin molecule results in the closure of approximately 200 ion channels, or about 2% of the number of channels in each rod that are open in the dark. This number of channel closures causes a net change in the membrane potential of about 1 mV.
Equally important is the fact that the magnitude of this amplification varies with the prevailing levels of illumination, a phenomenon known as light adaptation. At low levels of illumination, photoreceptors are the most sensitive to light. As levels of illumination increase, sensitivity decreases, preventing the receptors from saturating and thereby greatly extending the range of light intensities over which they operate. The concentration of Ca 2+ in the outer segment appears to play a key role in the light-induced modulation of photoreceptor sensitivity. The cGMP-gated channels in the outer segment are permeable to both Na+ and Ca 2+; thus, light-induced closure of these channels leads to a net decrease in the internal Ca 2+ concentration. This decrease triggers a number of changes in the phototransduction cascade, all of which tend to reduce the sensitivity of the receptor to light. For example, the decrease in Ca 2+ increases the activity of guanylate cyclase, the cGMP synthesizing enzyme, leading to an increase in cGMP levels. Likewise, the decrease in Ca 2+ increases the affinity of the cGMP-gated channels for cGMP, reducing the impact of the light-induced reduction of cGMP levels. The regulatory effects of Ca 2+ on the phototransduction cascade are only one part of the mechanism that adapts retinal sensitivity to background levels of illumination; another important contribution comes from neural interactions between horizontal cells and photoreceptor terminals.
Once initiated, additional mechanisms limit the duration of this amplifying cascade and restore the various molecules to their inactivated states. The protein arrestin, for instance, blocks the ability of activated rhodopsin to activate transducin, and facilitates the breakdown of activated rhodopsin. The all-trans retinal then dissociates from the opsin, diffuses into the cytosol of the outer segment, and is transported out of the outer segment and into the pigment epithelium, where appropriate enzymes ultimately convert it to 11-cis retinal. After it is transported back into the outer segment, the 11-cis retinal recombines with opsin in the receptor disks. The recycling of rhodopsin is critically important for maintaining the light sensitivity of photoreceptors. Even under intense levels of illumination, the rate of regeneration is sufficient to maintain a significant number of active photopigment molecules.
By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.
Copyright © 2001, Sinauer Associates, Inc.
Bookshelf ID: NBK10806
Views
Cite this Page
Disable Glossary Links
Related Items in Bookshelf
All Textbooks
Recent Activity
Clear)Turn Off)Turn On)
Phototransduction - NeurosciencePhototransduction - Neuroscience
Your browsing activity is empty.
Activity recording is turned off.
Turn recording back on)
See more...
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
PreferencesTurn off
External link. Please review our privacy policy.
Cite this Page Close
Purves D, Augustine GJ, Fitzpatrick D, et al., editors. Neuroscience. 2nd edition. Sunderland (MA): Sinauer Associates; 2001. Phototransduction. Available from:
Making content easier to read in Bookshelf Close
We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in.
The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders.
Cancel Download
Share
Share on Facebook
Share on Twitter
URL |
7747 | https://pubmed.ncbi.nlm.nih.gov/20075580/ | Perplexing epigastric pain-coincident myocardial infarction and acute pancreatitis - PubMed
Clipboard, Search History, and several other advanced features are temporarily unavailable.
Skip to main page content
An official website of the United States government
Here's how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Search: Search
AdvancedClipboard
User Guide
Save Email
Send to
Clipboard
My Bibliography
Collections
Citation manager
Display options
Display options
Format
Save citation to file
Format:
Create file Cancel
Email citation
Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page.
To:
Subject:
Body:
Format:
[x] MeSH and other data
Send email Cancel
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Add to My Bibliography
My Bibliography
Unable to load your delegates due to an error
Please try again
Add Cancel
Your saved search
Name of saved search:
Search terms:
Test search terms
Would you like email updates of new search results? Saved Search Alert Radio Buttons
Yes
No
Email: (change)
Frequency:
Which day?
Which day?
Report format:
Send at most:
[x] Send even when there aren't any new results
Optional text in email:
Save Cancel
Create a file for external citation management software
Create file Cancel
Your RSS Feed
Name of RSS Feed:
Number of items displayed:
Create RSS Cancel
RSS Link Copy
Actions
Cite
Collections
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Permalink
Permalink
Copy
Display options
Display options
Format
Page navigation
Title & authors
Abstract
Similar articles
Cited by
Publication types
MeSH terms
Related information
Case Reports
Intern Med
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2010;49(2):149-53.
doi: 10.2169/internalmedicine.49.2367. Epub 2010 Jan 15.
Perplexing epigastric pain-coincident myocardial infarction and acute pancreatitis
Cheng-Hsueh Wu1,Kang-Ling Wang,Tse-Min Lu
Affiliations Expand
Affiliation
1 Division of Cardiology, Department of Medicine, Taipei Medical University-Shuang Ho Hospital, Taipei, Taiwan.
PMID: 20075580
DOI: 10.2169/internalmedicine.49.2367
Item in Clipboard
Case Reports
Perplexing epigastric pain-coincident myocardial infarction and acute pancreatitis
Cheng-Hsueh Wu et al. Intern Med.2010.
Show details
Display options
Display options
Format
Intern Med
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2010;49(2):149-53.
doi: 10.2169/internalmedicine.49.2367. Epub 2010 Jan 15.
Authors
Cheng-Hsueh Wu1,Kang-Ling Wang,Tse-Min Lu
Affiliation
1 Division of Cardiology, Department of Medicine, Taipei Medical University-Shuang Ho Hospital, Taipei, Taiwan.
PMID: 20075580
DOI: 10.2169/internalmedicine.49.2367
Item in Clipboard
Cite
Display options
Display options
Format
Abstract
Acute myocardial infarction (MI) complicated with acute pancreatitis has been rarely reported. A 68-year-old man presented to our department 15 hours after development of epigastric pain. In addition to his symptoms, the elevated serum pancreatic enzymes and the image study on abdominal computerized tomography all led to the diagnosis of acute pancreatitis. Elevated cardiac biomarkers and a standard 12-lead electrocardiogram (ECG) demonstrating ST-segment elevation in 5 of the 6 precordial leads suggested an attack of MI. Oral intake was resumed after medical management for his acute pancreatitis and acute MI. Coronary angiogram on day 11 revealed total occlusion of the middle segment of the left anterior-descending coronary artery. Subsequently, angioplasty with stenting was done. The patient was discharged without significant complications. It is critical to make a rapid but detailed differential diagnosis of abdominal pain. Even though acute pancreatitis-associated ECG abnormalities have been reported previously, any ECG abnormalities in a patient presenting abdominal pain should be evaluated and treated cautiously. Thorough clinical evidence, including history, physical findings, ECG, image studies and serum biomarkers, are informative in seeking and analyzing possible etiologies.
PubMed Disclaimer
Similar articles
Pseudo-Wellens syndrome, acute pancreatitis, and an anomalous coronary artery: a case report.Effoe VS, O'Neal W, Santos R, Rubinsztain L, Zafari AM.Effoe VS, et al.J Med Case Rep. 2019 Dec 30;13(1):387. doi: 10.1186/s13256-019-2315-1.J Med Case Rep. 2019.PMID: 31884973 Free PMC article.
Acute pancreatitis presenting as acute inferior wall ST-segment elevations on electrocardiography.Makaryus AN, Adedeji O, Ali SK.Makaryus AN, et al.Am J Emerg Med. 2008 Jul;26(6):734.e1-4. doi: 10.1016/j.ajem.2007.11.008.Am J Emerg Med. 2008.PMID: 18606342
Abdominal pain and ECG alteration: a simple diagnosis?Pezzilli R, Bellacosa L, Barakat B.Pezzilli R, et al.Adv Med Sci. 2010;55(2):333-6. doi: 10.2478/v10039-010-0016-5.Adv Med Sci. 2010.PMID: 20494872
Acute pancreatitis presented with diffuse ST-segment elevation: A case report and literature review.Hsieh YL, Wu SH, Liu CY, Lin WC, Chen MJ, Chang CW.Hsieh YL, et al.Medicine (Baltimore). 2024 Feb 16;103(7):e37245. doi: 10.1097/MD.0000000000037245.Medicine (Baltimore). 2024.PMID: 38363907 Free PMC article.Review.
Pancreatitis with electrocardiographic changes mimicking acute myocardial infarction.Khairy P, Marsolais P.Khairy P, et al.Can J Gastroenterol. 2001 Aug;15(8):522-6. doi: 10.1155/2001/604386.Can J Gastroenterol. 2001.PMID: 11544536 Review.
See all similar articles
Cited by
Coronary thrombosis in acute pancreatitis.Sanghvi S, Waqar F, Effat M.Sanghvi S, et al.J Thromb Thrombolysis. 2019 Jan;47(1):157-161. doi: 10.1007/s11239-018-1741-z.J Thromb Thrombolysis. 2019.PMID: 30269287
Triggering acute pancreatitis complicated with acute myocardial infarction by marijuana: a rare case report.Hajimoradi B, Safi M, Pishgahi M, Alirezaei T, Jebreil Mosavi M.Hajimoradi B, et al.Acta Biomed. 2021 Apr 30;92(S1):e2021035. doi: 10.23750/abm.v92iS1.8269.Acta Biomed. 2021.PMID: 33944859 Free PMC article.
Publication types
Case Reports
Actions
Search in PubMed
Search in MeSH
Add to Search
MeSH terms
Abdominal Pain / complications
Actions
Search in PubMed
Search in MeSH
Add to Search
Abdominal Pain / diagnosis
Actions
Search in PubMed
Search in MeSH
Add to Search
Aged
Actions
Search in PubMed
Search in MeSH
Add to Search
Diagnosis, Differential
Actions
Search in PubMed
Search in MeSH
Add to Search
Electrocardiography / methods
Actions
Search in PubMed
Search in MeSH
Add to Search
Humans
Actions
Search in PubMed
Search in MeSH
Add to Search
Male
Actions
Search in PubMed
Search in MeSH
Add to Search
Myocardial Infarction / complications
Actions
Search in PubMed
Search in MeSH
Add to Search
Myocardial Infarction / diagnosis
Actions
Search in PubMed
Search in MeSH
Add to Search
Pancreatitis / complications
Actions
Search in PubMed
Search in MeSH
Add to Search
Pancreatitis / diagnosis
Actions
Search in PubMed
Search in MeSH
Add to Search
Related information
MedGen
[x]
Cite
Copy Download .nbib.nbib
Format:
Send To
Clipboard
Email
Save
My Bibliography
Collections
Citation Manager
[x]
NCBI Literature Resources
MeSHPMCBookshelfDisclaimer
The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited.
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
7748 | https://math.stackexchange.com/questions/630343/is-there-any-difference-between-linear-dependence-collinearity-and-coplanarity | Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Is there any difference between linear dependence, collinearity and coplanarity?
Ask Question
Asked
Modified 11 years, 8 months ago
Viewed 12k times
8
$\begingroup$
The way it seems to me, linearly dependent vectors have to be collinear, and collinear vectors have to be coplanar. However, since a plane doesn't really have a direction, I'm assuming coplanar vectors can point in different directions as long as their lines exist on the same plane. Or do coplanar vectors/points also have to point in the same direction? If so, what's the practical difference between these concepts? I'm wondering this in terms of orientation and position in three-space, not in terms of whether the math is done differently or not.
For example, what would be the difference between 3 linearly dependent vectors, 3 collinear vectors and 3 coplanar vectors?
EDIT: So far I still can't visualize the difference in 3D space. It's not that I don't understand that the math is different, I just want to be able to clearly visualize what the similarities and differences are, because if I don't then the math won't make sense to me. I need to understand what the math does in order to make it stick.
linear-algebra
vectors
Share
edited Jan 7, 2014 at 18:04
ThreethumbThreethumb
asked Jan 7, 2014 at 16:13
ThreethumbThreethumb
96966 gold badges2525 silver badges3737 bronze badges
$\endgroup$
2
$\begingroup$ 3 linearly dependant vectors are coplanar, it is necessary and sufficient (and easy to see, from a relation $\lambda_1v_1 + \lambda_2v_2 + \lambda_3v_3=0$). However, 3 colinear vectors span a dimension 1 subspace, whereas coplanar vectors can span a dimension 2 subspace (it can be 1 if the vectors are actually colinear). This is the only difference. $\endgroup$
zarathustra
– zarathustra
2014-01-07 16:24:46 +00:00
Commented Jan 7, 2014 at 16:24
$\begingroup$ Linearly dependent vectors need not be co-linear. For example you can take the coordinate axis in $\mathbb{R}^2$ and the vector $\mathbf{v}=\begin{pmatrix} 1 \ 1 \end{pmatrix}$. Clearly ${\mathbf{v},\mathbf{e}_1,\mathbf{e}_2}$ is a set of linearly dependent vectors but none of them are co-linear. $\endgroup$
Ben
– Ben
2014-01-07 16:27:27 +00:00
Commented Jan 7, 2014 at 16:27
Add a comment |
3 Answers 3
Reset to default
3
$\begingroup$
The way it seems to me, linearly dependent vectors have to be collinear, and collinear vectors have to be coplanar. However, since a plane doesn't really have a direction, I'm assuming coplanar vectors can point in different directions as long as their lines exist on the same plane.
Planes can certainly have a direction. In fact, planes have two directions, which can be specified by two vectors that span them. However, you are correct that coplanar vectors can point in different directions, so long as their lines are in the same plane. This is where the word coplanar comes from:
co-
together; mutually; jointly
planar
Of or pertaining to a plane.
So it means "together in a plane". Any group of vectors that are in the same plane are coplanar.
Or do coplanar vectors/points also have to point in the same direction?
No, as above, it only means in the same plane. If vectors are in the same plane and point in the same direction as you suggest, they are on the same line. Colinear of course means "together on a line". All colinear vectors happen to also be coplanar, but this is not embedded in the definitions of the words.
If so, what's the practical difference between these concepts? I'm wondering this in terms of orientation and position in three-space, not in terms of whether the math is done differently or not.
To address linear independence, I'll say the following.
All colinear vectors are linearly dependent, almost trivially by the definitions of colinearity and linear dependence.
For a set of non-zero coplanar vectors, none of which are colinear (i.e., they point in different directions), any two of the set can be considered linearly independent. This is because there are fundamentally only two orthogonal directions to "go" on a plane. As long as you have two vectors that are not colinear but lie in a plane, you can write any other member of the planar subspace as a linear combination of those two vectors.
Share
edited Jan 12, 2014 at 16:17
answered Jan 7, 2014 at 17:11
rajb245rajb245
4,8751818 silver badges3636 bronze badges
$\endgroup$
Add a comment |
1
$\begingroup$
A set $S$ of vectors is called collinear iff $\dim\bigl(\mathrm{span}(S)\bigr)=1$ and iff that dimension is $2$ the vectors are called complanar.
Share
answered Jan 7, 2014 at 17:41
Michael HoppeMichael Hoppe
18.7k33 gold badges3535 silver badges5252 bronze badges
$\endgroup$
Add a comment |
0
$\begingroup$
Any two vectors (which are NOT linearly dependent, i.e. not a scaling of each other) establish a plane. Given another vector which is in the span of these vectors, it is "coplanar" with them (in the same plane). So being coplanar does mean linear dependence (to the basis of a given plane). Colinear is the same idea but more general, the dependence doesn't have to be in a plane, it can be a hyperplane etc.
As a motivation, if we take the standard basis $E={e_1,e_2,e_3}$ and the vector $v=(1,1,1)$, $v$ is linearly dependent with the elements of $E$, but not coplanar with any pair of ${e_i , e_j }$. However, $w=(2,4,0)$ is coplanar with ${e_1,e_2}$ (these examples are easier to see because of use of standard basis). By the way, Arnold's Classical Mechanics does make appeal to these ideas quite frequently.
Share
answered Jan 7, 2014 at 16:28
SquirtleSquirtle
6,9833939 silver badges6666 bronze badges
$\endgroup$
Add a comment |
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
linear-algebra
vectors
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
linear dependence and coplanar vectors
What exactly does linear dependence and linear independence imply?
1 What is the term for 2 vectors at 45 degree angle to each other
0 Linear algebra linear dependence, independence and spanning sets?
What is the difference, geometrically, between row vectors and column vectors?
0 What is the relationship between linear dependence of vectors and being coplanar?
Hot Network Questions
Switch between math versions but without math versions
How to home-make rubber feet stoppers for table legs?
What is this chess h4 sac known as?
What "real mistakes" exist in the Messier catalog?
Why is the fiber product in the definition of a Segal spaces a homotopy fiber product?
How can blood fuel space travel?
How to use cursed items without upsetting the player?
Analog story - nuclear bombs used to neutralize global warming
The altitudes of the Regular Pentagon
What meal can come next?
Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth?
Does the curvature engine's wake really last forever?
Who is the target audience of Netanyahu's speech at the United Nations?
Is there a specific term to describe someone who is religious but does not necessarily believe everything that their religion teaches, and uses logic?
How long would it take for me to get all the items in Bongo Cat?
Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters?
What happens if you miss cruise ship deadline at private island?
Exchange a file in a zip file quickly
Why do universities push for high impact journal publications?
how do I remove a item from the applications menu
What’s the usual way to apply for a Saudi business visa from the UAE?
Change default Firefox open file directory
Two calendar months on the same page
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
more hot questions
Question feed |
7749 | https://math.stackexchange.com/questions/2560913/im-stuck-finding-the-maclaurin-series-for-fx-ln15x | calculus - I'm stuck finding the maclaurin series for $f(x) = \ln(1+5x)$ - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
I'm stuck finding the maclaurin series for f(x)=ln(1+5 x)f(x)=ln(1+5 x)
Ask Question
Asked 7 years, 9 months ago
Modified7 years, 9 months ago
Viewed 4k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I've calculated the first 4 4 derivatives of f(x)f(x) and evaluated each at x=0 x=0. I now have the following maclaurin series:
0+5 x−25 2!x 2+250 3!x 3−3750 4!x 4 0+5 x−25 2!x 2+250 3!x 3−3750 4!x 4
I am awful at recognizing patterns, and need to come up with a series for this from n=1 n=1 to infinity. Additionally, if anyone has any tips for recognizing patterns like this, let me know, because it's something I need to be better at.
calculus
sequences-and-series
taylor-expansion
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Dec 11, 2017 at 1:55
Pizza or Pasta
17 4 4 bronze badges
asked Dec 11, 2017 at 1:18
user482496user482496
11 1 1 silver badge 2 2 bronze badges
1
Well, 1/(1+x)1/(1+x) is 1−x+x 2−x 3+x 4−x 5+....1−x+x 2−x 3+x 4−x 5+.... You can see the pattern there, right? When you integrate it, the left hand side is ln(1+x)ln(1+x). Can you see the pattern in the right hand side? Now substitute 5 x 5 x in place of x x. What's the pattern now?Michael Hartley –Michael Hartley 2017-12-11 01:21:41 +00:00 Commented Dec 11, 2017 at 1:21
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
If you know that ln(1+x)=∑∞k=1(−1)k−1 x k k ln(1+x)=∑k=1∞(−1)k−1 x k k, just substitute 5 x 5 x for x x to get ln(1+5 x)=∑∞k=1(−1)k−1(5 x)k k=∑∞k=1 x k(−1)k−1 5 k k ln(1+5 x)=∑k=1∞(−1)k−1(5 x)k k=∑k=1∞x k(−1)k−1 5 k k.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Dec 11, 2017 at 4:05
marty cohenmarty cohen
111k 10 10 gold badges 88 88 silver badges 186 186 bronze badges
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
5=5 1∗0!5=5 1∗0!
25=5 2∗1!25=5 2∗1!
250=5 3∗2!250=5 3∗2!
3750=5 4∗3!3750=5 4∗3!
Perhaps this may help
f(x)=−∑i=1∞(−5)i i x i f(x)=−∑i=1∞(−5)i i x i
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Dec 11, 2017 at 1:23
Harry AlliHarry Alli
2,121 1 1 gold badge 11 11 silver badges 17 17 bronze badges
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
When looking for the general term of a Taylor series, it's best to not evaluate the products. E.g.,
f(0)f′(0)f′′(0)f(3)(0)f(4)(0)=−5(1+5 x)1∣∣∣x=0=−5 2(1+5 x)2∣∣∣x=0=−2⋅5 3(1+5 x)3∣∣∣x=0=−2⋅3⋅5 4(1+5 x)4∣∣∣x=0=−0=−5=−5 2=−2⋅5 3=−3⋅2⋅5 4 f(0)=−0 f′(0)=−5(1+5 x)1|x=0=−5 f″(0)=−5 2(1+5 x)2|x=0=−5 2 f(3)(0)=−2⋅5 3(1+5 x)3|x=0=−2⋅5 3 f(4)(0)=−2⋅3⋅5 4(1+5 x)4|x=0=−3⋅2⋅5 4
The general term should be apparent after this: f(n)(0)=(−5)n−1(n−1)!f(n)(0)=(−5)n−1(n−1)! for n≥1 n≥1.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Dec 11, 2017 at 1:31
youngsmasheryoungsmasher
457 4 4 silver badges 9 9 bronze badges
1
So f(n)(0)/n!=...f(n)(0)/n!=....marty cohen –marty cohen 2017-12-11 04:06:37 +00:00 Commented Dec 11, 2017 at 4:06
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
calculus
sequences-and-series
taylor-expansion
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
1maclaurin series question!?
0Maclaurin series stuck at finding L n L n
1Maclaurin Series - finding the co-efficients for functions that require the product rule
0Maclaurin series for x 6⋅e−x x 6⋅e−x
0Finding a Taylor series for f(x)=(x 2+2 x)e x f(x)=(x 2+2 x)e x?
1Maclaurin series for f(x)=1 1+x+x 2 f(x)=1 1+x+x 2
2Finding the Maclaurin series for f(x)=x ln(x+1)f(x)=x ln(x+1)
0Finding Maclaurin series for antiderivative, such that series satisfies some initial condition
1Finding a function from its Maclaurin series
1Maclaurin series of sin(5 x 2)sin(5 x 2)
Hot Network Questions
Is encrypting the login keyring necessary if you have full disk encryption?
How to start explorer with C: drive selected and shown in folder list?
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
Spectral Leakage & Phase Discontinuites
Why do universities push for high impact journal publications?
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
The rule of necessitation seems utterly unreasonable
how do I remove a item from the applications menu
Is it safe to route top layer traces under header pins, SMD IC?
Clinical-tone story about Earth making people violent
Why include unadjusted estimates in a study when reporting adjusted estimates?
Why are LDS temple garments secret?
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
Non-degeneracy of wedge product in cohomology
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
Is existence always locational?
Calculating the node voltage
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Checking model assumptions at cluster level vs global level?
Discussing strategy reduces winning chances of everyone!
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
My dissertation is wrong, but I already defended. How to remedy?
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7750 | https://gmatclub.com/forum/an-escalator-moves-downward-from-street-level-to-a-subway-platform-at-236880.html | An escalator moves downward from street level to a subway platform at : Problem Solving (PS)
gmatclub
FORUMS
GMAT
MBA
RESOURCES
DEALS
REVIEWS
CHAT 1
All Forums Index Start a New Discussion General GMAT Questions Quantitative PS Verbal CR RC Data Insights DS G&T MSR TPA Ask GMAT Experts Share GMAT Experience
All Business School Discussions BSchool Application Questions Admitted - Which School to Choose? Profile Evaluations by Experts Student Loans & Financing Master's Programs (MiM, MFin, MSc) PhD in Business
GRE Forums General GRE Discussions Tech Support Downloads Blog
Diagnostic Quiz Question Banks GMAT Focus Tests Forum Quiz Error Log Question of the Day Study Plan
Business School Hub Decision Tracker What are my Chances Tool Request a Profile Evaluation Interview Debriefs Sample Essays Ask Admissions Consultants
GMATScore Calculator Study Plan Math Book Flashcards Practice Tests Overview Timing Strategies
MBAMBA Rankings 2025 - 2026 Deadlines All Application Fee Waivers Interview Release Dates Average GMAT Focus Scores School Stats Waitlists Scholarships & Negotiation Best MBA for Consulting, Finance, Tech MBA ROI Calculator
MBA ProgramsTuck (Dartmouth) Yale SOM Kenan-Flagler (UNC) Kelley (IU) Harvard Kellogg (Northwestern) Wharton (Upenn) Booth (Chicago) Sloan (MIT) Stanford Fuqua (Duke) ISB INSEAD
Goizueta (Emory) UC Riverside NTU CBS (Columbia) Haas (Berkeley) Ross (Michigan) Anderson (UCLA) Darden (UVA) Stern (NYU) LBS Johnson (Cornell) See All
GMATMarketplace - Compare all GMAT Deals e-GMAT - Save up to $525 Magoosh - Save up to $215 Manhattan Prep - Save up to $225 Target Test Prep - Save up to $579 GMATWhiz - Save up to $725 Experts' Global - Save up to $418 GMAT Tutoring Deals Compare All $500 Prodigy Loan Cashback
ADMISSION CONSULTINGAccepted.com Admissionado ApplicantLab Gatehouse Admissions mbaMission Square One Prep Stratus Admissions Counseling The Red Pen Admit Expert Amerasia ARINGO MBA Admissions Consulting ARLee Consulting Avanti Prep Embark MBA Experts' Global Compare All
Fortuna Admissions Gurufi Ivy Groupe Leland MBA Admit MBA and Beyond MBA Prep School Menlo Coaching OnCourse Vantage Personal MBA Coach Prep MBA Sia Admissions Stacy Blackman Consulting UniAdvise Vantage Point MBA Vikram Shah Consulting Compare All
STUDENT LOANSProdigy Finance Juno Financing Earnest Financing
See All Verified & Authentic Reviews GMAT COURSE REVIEWSe-GMAT Reviews Magoosh Reviews Manhattan Prep Reviews Target Test Prep Reviews Experts' Global Reviews GMATWhiz Reviews GMAT Tutors' Reviews See All Student Loan Reviews
Admissions Consulting ReviewsAccepted.com Reviews Admissionado Reviews ApplicantLab Reviews Gatehouse Reviews mbaMission Reviews Square One Prep Reviews Stratus Admissions Reviews The Red Pen Reviews Admit Expert Reviews Amerasia Reviews ARINGO MBA Admissions Consulting Reviews ARLee Consulting Reviews
Avanti Prep Reviews Embark MBA Reviews Experts' Global Reviews Fortuna Admissions Reviews Gurufi Reviews Ivy Groupe Reviews Leland Reviews MBA Admit Reviews MBA and Beyond Reviews MBA Prep School Reviews Menlo Coaching Reviews OnCourse Vantage Reviews Personal MBA Coach Reviews
Prep MBA Reviews Sia Admissions Reviews Stacy Blackman Consulting Reviews UniAdvise Reviews Vantage Point MBA Reviews Vikram Shah Consulting Reviews See All
Business School ReviewsHBS Reviews Wharton Reviews Stanford GSB Reviews Kellogg Reviews Booth Reviews Sloan Reviews ISB Reviews INSEAD Reviews Columbia Reviews Yale Reviews LBS Reviews Fuqua Reviews Ross Reviews Anderson Reviews See All
MAIN CHATROOM 1
- General GMAT Chat
- GMAT Questions Chat
- Manage WhatsApp Connection
- MBA Applicant WhatsApp Group
- GMAT Prep WhatsApp Group
- More GMAT & MBA Chatrooms
Sign InJoin now
Forum HomeGMATQuantitative QuestionsProblem Solving (PS)
GMAT Problem Solving (PS) Questions
It is currently 28 Sep 2025, 14:51
Watch on YouTube!GRE Quiz (20 Questions) @9:30am ET / 7pm ISTGMAT Quiz (30 Questions) @10:30am ET
Toolkit
Forum QuizPractice TestsQuestions BankDecision TrackerError LogMy PostsHot TopicsChat
Forum Home
GMAT
Quantitative Questions
Problem Solving (PS)
HomeGMATMessagesTestsSchoolsEventsRewardsChatMy Profile
Ask QuestionGMAT ForumsMBA ForumsExecutive MBAMasters ProgramsGRESupportHome
Make PostGeneral QuestionsQuantitativeVerbalData InsightsAsk GMAT ExpertsGMAT Experience
Start TestGMAT Focus TestsFree TestForum Quiz 3.0EA TestsGRE TestsError Log
Update StatusDecision Tracker
AndersonBoothCarlsonColumbiaCox (SMU)DardenFosterFuquaGoizuetaHaasHarvardHEC ParisIESEIIM AhmedabadIIM BangaloreIIM CalcuttaINSEADISBJohnson (Cornell)Jones RiceJudge CambridgeKelleyKelloggKenan-FlaglerLBSMarshall (USC)McCombsMcDonoughMendozaMerageNUS SingaporeOlin St. LouisOwen VanderbiltRoss (Michigan)RotmanSaidSchellerSchulichSchulichSimonSloan MITStanford GSBSternTepperTerry GeorgiaTuckUCR BusinessW.P. CareyWhartonYale
JoinPrize Fund ($30,000)RulesMy pointsQuestions
MBA Spotlight Fair Nov 13-14, 2024Sign upBenefitsApp Fee waiversActivate Tests
Latest VideoLatest PostLatest PostLatest TweetLatest PostLatest PostLatest PostLatest Post
Special Offer: Get 25% Off Target Test Prep GMAT Plans
Tier StatusPointsRedeemActivityTop MembersHow to Use?
New ChatMain ChatroomGeneral GMAT ChatPractice Questions ChatStudy Buddy ChatWhatsApp for GMATWhatsApp for MBA Applications
EditMy BookmarksMy PostsFollow FeedMy GMAT ScoreMy SchoolsNotificationsSettingsLog-In
Send PMInbox:SentDrafts
Ask QuestionGMAT ForumsMBA ForumsExecutive MBAMasters ProgramsGRESupportHome
Make PostGeneral QuestionsQuantitativeVerbalData InsightsAsk GMAT ExpertsGMAT Experience
Start TestGMAT Focus TestsFree TestForum Quiz 3.0EA TestsGRE TestsError Log
Update StatusDecision Tracker
AndersonBoothCarlsonColumbiaCox (SMU)DardenFosterFuquaGoizuetaHaasHarvardHEC ParisIESEIIM AhmedabadIIM BangaloreIIM CalcuttaINSEADISBJohnson (Cornell)Jones RiceJudge CambridgeKelleyKelloggKenan-FlaglerLBSMarshall (USC)McCombsMcDonoughMendozaMerageNUS SingaporeOlin St. LouisOwen VanderbiltRoss (Michigan)RotmanSaidSchellerSchulichSchulichSimonSloan MITStanford GSBSternTepperTerry GeorgiaTuckUCR BusinessW.P. CareyWhartonYale
JoinPrize Fund ($30,000)RulesMy pointsQuestions
MBA Spotlight Fair Nov 13-14, 2024Sign upBenefitsApp Fee waiversActivate Tests
Latest VideoLatest PostLatest PostLatest TweetLatest PostLatest PostLatest PostLatest Post
Special Offer: Get 25% Off Target Test Prep GMAT Plans
Tier StatusPointsRedeemActivityTop MembersHow to Use?
New ChatMain ChatroomGeneral GMAT ChatPractice Questions ChatStudy Buddy ChatWhatsApp for GMATWhatsApp for MBA Applications
EditMy BookmarksMy PostsFollow FeedMy GMAT ScoreMy SchoolsNotificationsSettingsLog-In
Send PMInbox:SentDrafts
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error LogLearn more
JOIN GMAT CLUB
Become part of our 1M+ member community!
×
GMAT Registration Benefits:
Vast Question Bank:
Practice thousands of GMAT questions with top expert solutions.
Error Tracking:
Identify and improve upon mistakes efficiently using our Error Log.
Expert Support:
Get the latest tips and news from our top GMAT professionals.
Sing up now
it’s free and easy!
Join now
Sign In
?
GMAT Club Timer Informer
Hi GMATClubber!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
We’ll give you
an estimate of
your score
learn more
We’ll provide
personalized question
recommendations
learn more
Your score will improve
and your results
will be more realistic
learn more
Is there something wrong with our timer?Let us know!
Thanks! I'll try it now
Add Mistakes & Note
[x] Calculation Error
[x] Careless Mistake
[x] Anxiety
[x] Conceptual Gap
[x] Time Mismanagement
[x] Vocabulary Void
Add Note
Cancel Add
Request Expert Reply
Please wait...
ConfirmCancel
Events & Promotions
Sep 22 Special Offer: Get 25% Off Target Test Prep GMAT Plans
Sep 28 GRE Practice Test #41 - Free GRE Practice Covering Quant, Text Completion, Sentence Equivalence
Sep 28 GMAT Quiz #44 - Live GMAT Practice Covering Quant, Verbal, RC, & Data Insights
Sep 30 5 Quant & DI GMAT Hacks That Can Save You 10+ Minutes and Add 50+ Points
Sep 30 $325 Off TTP OnDemand Masterclass
Sep 30 Free Master’s in Europe? Scholarships & Hidden Hacks Explained | Masters (Ep4)
Oct 01 MBA INTERVIEW PREP SERIES
Oct 04 Free Webinar – Achieve 90th%ile score on GMAT CR and TPA
Oct 05 Ace Arithmetic on the GMAT Focus Edition
September 2025
SunMonTueWedThuFriSat
31
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Special Offer: Get 25% Off Target Test Prep GMAT Plans
23
24
25
26
27
28
GRE Practice Test #41 - Free GRE Practice Covering Quant, Text Completion, Sentence Equivalence GMAT Quiz #44 - Live GMAT Practice Covering Quant, Verbal, RC, & Data Insights
29
30
5 Quant & DI GMAT Hacks That Can Save You 10+ Minutes and Add 50+ Points$325 Off TTP OnDemand Masterclass Free Master’s in Europe? Scholarships & Hidden Hacks Explained | Masters (Ep4)
1
MBA INTERVIEW PREP SERIES
2
3
4
Free Webinar – Achieve 90th%ile score on GMAT CR and TPA
Register Oct 05 Ace Arithmetic on the GMAT Focus Edition 11:00 AM IST
01:00 PM IST Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions.
Save Now! Sep 22 Special Offer: Get 25% Off Target Test Prep GMAT Plans 12:00 PM EDT
11:59 PM EDT The Target Test Prep GMAT Flash Sale is LIVE! Get 25% off our game-changing course and save up to $450 today! Use code FLASH25 at checkout. This special offer expires on September 30, so grab your discount now!
Set a Reminder Sep 28 GRE Practice Test #41 - Free GRE Practice Covering Quant, Text Completion, Sentence Equivalence 06:30 AM PDT
07:15 AM PDT Solve GRE practice problems covering Quantitative reasoning, Verbal Reasoning, Text Completion, Sentence Equivalence, and Reading Comprehension Problems. Take this GRE practice quiz live with peers, analyze your GRE study progress, and more.
Set a Reminder Sep 28 GMAT Quiz #44 - Live GMAT Practice Covering Quant, Verbal, RC, & Data Insights 07:30 AM PDT
07:15 AM PDT Join us in a live GMAT practice session and solve 30 challenging GMAT questions with other test takers in timed conditions, covering GMAT Quant, Data Sufficiency, Data Insights, Reading Comprehension, and Critical Reasoning questions.
Set a Reminder Sep 30 5 Quant & DI GMAT Hacks That Can Save You 10+ Minutes and Add 50+ Points 08:30 AM PDT
09:30 AM PDT Do you constantly run out of time in the GMAT Quant and Data Insights sections? Losing 10+ minutes can cost you your target GMAT score. In this session, we reveal 5 powerful GMAT hacks that top scorers use to boost efficiency, accuracy, and confidence.
Save Now! Sep 30 $325 Off TTP OnDemand Masterclass 10:00 AM EDT
11:59 PM EDT Get a massive $325 off the TTP OnDemand GMAT masterclass by using the coupon code FLASH25 at checkout. If you prefer learning through engaging video lessons, TTP OnDemand GMAT is exactly what you need.
Watch Now! Sep 30 Free Master’s in Europe? Scholarships & Hidden Hacks Explained | Masters (Ep4) 10:00 AM PDT
11:00 AM PDT What if studying in Europe didn’t mean spending €40,000–50,000? What if you could study at HEC Paris, Bocconi, or ESSEC—and pay close to zero?
Register Now! Oct 01 MBA INTERVIEW PREP SERIES 09:00 AM PDT
11:00 AM PDT If you’re aspiring to join one of the top 20 MBA programs, we’ve got an exciting opportunity just for you! Welcome to the GMAT Club's Free Interview Prep Event
Register Today! Oct 04 Free Webinar – Achieve 90th%ile score on GMAT CR and TPA 11:00 AM IST
01:00 PM IST Attend this session to learn how to Deconstruct arguments,Pre-Think Author’s Assumptions, and apply Negation Test.
Register Oct 05 Ace Arithmetic on the GMAT Focus Edition 11:00 AM IST
01:00 PM IST Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions.
Save Now! Sep 22 Special Offer: Get 25% Off Target Test Prep GMAT Plans 12:00 PM EDT
11:59 PM EDT The Target Test Prep GMAT Flash Sale is LIVE! Get 25% off our game-changing course and save up to $450 today! Use code FLASH25 at checkout. This special offer expires on September 30, so grab your discount now!
Back to Forum
Create Topic
Reply
An escalator moves downward from street level to a subway platform at
TheUltimateWinner
TheUltimateWinner
Joined: 23 Feb 2015
Last visit: 27 Sep 2025
Posts: 2,489
Concentration: Finance, Technology
Schools:Harvard '24Judge '23LBS '24McCombs '24MendozaTepper '24Ross '24CoxMarshall '24FisherKelley '24JonesMcDonough '24Owen '24TerryBUSimon '24KatzGWUMasonBabsonGabelliFoster '24Goizueta '24SchellerStern '23CBS '23Booth '23Haas '25Wharton '25Stanford '26Johnson'23Fuqua '23Kellogg '23Kenan-Flagler'23INSEAD Aug 22 intakeOlin '23Said '23Sloan '23Carey Arizona "22Yale '23UC DavisMadison '23Anderson '23CarrollCarlson '23Darden '23
Schools:Harvard '24Judge '23LBS '24McCombs '24MendozaTepper '24Ross '24CoxMarshall '24FisherKelley '24JonesMcDonough '24Owen '24TerryBUSimon '24KatzGWUMasonBabsonGabelliFoster '24Goizueta '24SchellerStern '23CBS '23Booth '23Haas '25Wharton '25Stanford '26Johnson'23Fuqua '23Kellogg '23Kenan-Flagler'23INSEAD Aug 22 intakeOlin '23Said '23Sloan '23Carey Arizona "22Yale '23UC DavisMadison '23Anderson '23CarrollCarlson '23Darden '23
Posts: 2,489
Post URL26 Mar 2017, 04:05
Show timer
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
a 10%
b 46%
c 11%
d 4%
e 30%
A
B
C
D
E
Hide
Show
History
My Mistake
Official Answer and Stats are available only to registered users.Register/Login.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95% (hard)
Question Stats:
30% (02:41) correct 70%(02:15) wrong based on 630 sessions
History
Date
Time
Result
Not Attempted Yet
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Source: Manhattan GMAT Challenge question
Show Hide Answer
Official Answer
Official Answer and Stats are available only to registered users.Register/Login.
The best tool for practice:
Signature Read More
9
Kudos
Add Kudos
104
Bookmarks
Bookmark this Post
Most Helpful Reply
devarshi9283
devarshi9283
Joined: 25 Jul 2011
Last visit: 22 Sep 2025
Posts: 54
Location: India
Concentration: Strategy, Operations
Schools:NUS'20(S)Jones'20(A$)Mays'20(A$)Jindal'20(A$)Cox'20(A$)Kenan-Flagler'20(WL)
GMAT 1:740 Q49 V41
GPA: 3.5
WE:Engineering (Energy)
Schools:NUS'20(S)Jones'20(A$)Mays'20(A$)Jindal'20(A$)Cox'20(A$)Kenan-Flagler'20(WL)
GMAT 1:740 Q49 V41
Posts: 54
Post URL17 Aug 2017, 10:15
sarugiri
Is there a easy way to solve this question.?
Show more
When the person is not aided by Escalator movement..he takes 70 steps...so total distance in terms of steps is 70
When his movement is aided by escalator ..he has to take only 36 steps, i.e., rest of the 34 steps are taken care by escalator movement.
Or we can say that for every 36 steps taken by person..escalator is helping him with 34 steps..
Now..when he tries to move up in a downward escalator..his same movement will now be opposed by escalator,
So...for each of his 36 steps, escalator will oppose him with 34 steps..so resultant will be 2 steps only..
As the total distance in terms of steps is 70...and for each of his 36 steps..person is able to cover only 2 steps
due to downward motion of escalator..so total no. of steps required by person will be (3670/2)=1260 steps...Hence option E
Give kudos plz if you find above solution useful..
101
Kudos
Add Kudos
14
Bookmarks
Bookmark this Post
sarathgopinath
sarathgopinath
Joined: 22 Aug 2016
Last visit: 04 Feb 2025
Posts: 68
Location: India
Concentration: Operations, Finance
GMAT 1:650 Q49 V29
GMAT 2:710 Q49 V37
GPA: 3.5
WE:Other (Education)
Products:
GMAT 2:710 Q49 V37
Posts: 68
Post URL26 Aug 2017, 05:55
iMyself
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Source: Manhattan GMAT Challenge question
Show more
This is how I usually go with questions that involve rate. Whether it is Time, Speed, and distance or time and work.
I don't know my total distance. So I assume it to be the LCM(70,36)=1260.
Let us just give a unit for distance, 1260 miles.
Wesley covers 1260 miles in 70 steps. His speed is 1260 70 1260 70=18miles/step.
Wesley and the escalator together cover 1260 miles in 36 steps. So the combined speed is 1260 36 1260 36 = 35miles/step
This means the speed of escalator is 35-18 = 17 miles/step.
Now when he is going up, the escalator's motion is against his motion. So the net effect of rates will be Wesley's speed - Escalator speed = 18-17=1mile/step.
Distance=1260 miles.
Speed = 1 mile/step.
So number of steps = 1260 1 1260 1 = 1260.
24
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
General Discussion
Shubhranil88
Shubhranil88
Joined: 05 Dec 2016
Last visit: 21 Oct 2024
Posts: 7
Posts: 7
Post URL26 Mar 2017, 05:10
iMyself
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Source: Manhattan GMAT Challenge question
Show more
SOLUTION:
Let the level distance between street to Subway = D
Let the speed at which Wesley moves when escalator is still= u
Let the speed at which escalator moves up or down = v
When escalator is stand still he needs 70 steps
Therefore D/u = 70 => u = D/70 ............. (I)
Also when both escalator and Wesley moves in same direction he needs 36 steps
Therefore D/(u +v) = 36 => u + v = D/36 ........... (Ii)
Solving both (i) & (ii) we get----
v= 17D/(35x36) ..........(iii)
Hence u - v = D/(35x36)
Therefore Number of steps required when escalator and Wesley are moving in opposite directions-----
= D/(u-v)
= D/ (D/(35x36))
= 35x36
= 1,260 (which is greater than 1000 steps). Hence E
7
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
warrenlee206
warrenlee206
Joined: 28 Jul 2016
Last visit: 20 Jul 2018
Posts: 1
Location: United States (CA)
Schools:LBS'19(D)INSEAD Jan '17Anderson '20Anderson FEMBA'21(A)
GMAT 1:700 Q42 V44
GPA: 3.01
Schools:LBS'19(D)INSEAD Jan '17Anderson '20Anderson FEMBA'21(A)
GMAT 1:700 Q42 V44
Posts: 1
Post URL27 Mar 2017, 23:33
Shubhranil88
iMyself
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Source: Manhattan GMAT Challenge question
SOLUTION:
Let the level distance between street to Subway = D
Let the speed at which Wesley moves when escalator is still= u
Let the speed at which escalator moves up or down = v
When escalator is stand still he needs 70 steps
Therefore D/u = 70 => u = D/70 ............. (I)
Also when both escalator and Wesley moves in same direction he needs 36 steps
Therefore D/(u +v) = 36 => u + v = D/36 ........... (Ii)
Solving both (i) & (ii) we get----
v= 17D/(35x36) ..........(iii)
Hence u - v = D/(35x36)
Therefore Number of steps required when escalator and Wesley are moving in opposite directions-----
= D/(u-v)
= D/ (D/(35x36))
= 35x36
= 1,260 (which is greater than 1000 steps). Hence E
Show more
Can you elaborate more on step iii?
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sarugiri
sarugiri
Joined: 22 Oct 2016
Last visit: 30 Oct 2017
Posts: 17
Posts: 17
Post URL04 Apr 2017, 06:56
Is there a easy way to solve this question.?
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NamVu1990
NamVu1990
Joined: 18 Aug 2017
Last visit: 10 Sep 2017
Posts: 24
GMAT 1:670 Q49 V33
GMAT 1:670 Q49 V33
Posts: 24
Post URL26 Aug 2017, 03:40
devarshi9283
sarugiri
Is there a easy way to solve this question.?
When the person is not aided by Escalator movement..he takes 70 steps...so total distance in terms of steps is 70
When his movement is aided by escalator ..he has to take only 36 steps, i.e., rest of the 34 steps are taken care by escalator movement.
Or we can say that for every 36 steps taken by person..escalator is helping him with 34 steps..
Now..when he tries to move up in a downward escalator..his same movement will now be opposed by escalator,
So...for each of his 36 steps, escalator will oppose him with 34 steps..so resultant will be 2 steps only..
As the total distance in terms of steps is 70...and for each of his 36 steps..person is able to cover only 2 steps
due to downward motion of escalator..so total no. of steps required by person will be (3670/2)=1260 steps...Hence option E
Give kudos plz if you find above solution useful..
Show more
Amzing solution. How did you come up with this?
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sebastian Shoaib
Sebastian Shoaib
Joined: 17 Oct 2016
Last visit: 27 Nov 2017
Posts: 7
Posts: 7
Post URL16 Sep 2017, 06:47
Intuitively speaking, it seems that the speed of the escalator and the speed of Wesley are equal (while descending he covers 2 steps when the escalator is on). So, if the speed of the escalator and Wesley are unchanged when he ascends, every upward step Wesley takes will be cancelled by the downward movement of the escalator. In that case (E) looks like the best answer.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bkpolymers1617
bkpolymers1617
Joined: 01 Sep 2016
Last visit: 11 Dec 2017
Posts: 125
GMAT 1:690 Q49 V35
Products:
GMAT 1:690 Q49 V35
Posts: 125
Post URL19 Sep 2017, 04:07
Answer
Assume that the escalator is 70 steps long, since it takes Wesley 70 steps to descend when the escalator isn’t moving.
When the escalator is turned on, Wesley takes only 36 steps to descend, which means the escalator is doing the work of 70 – 36 = 34 steps in the time that it takes Wesley to descend.
For every 36 steps Wesley takes, then, the escalator “takes” 34 steps. If Wesley reverses his trip and walks upward, against the escalator, then Wesley “gains” 2 steps on the escalator in the period of time it would normally take to descend to the platform. That is, for every 36 steps that Wesley takes, he actually moves 2 steps up the escalator.
Since he has to “gain” a total of 70 steps in order to make it to the top of the escalator, he must gain a total of 2 steps 35 times. In total, then, he takes (36)(35) steps. That number is greater than 1,000. (The exact number is 1,260 steps, but don’t do math that you don’t have to do!)
The correct answer is (E).
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
ScottTargetTestPrep
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 27 Sep 2025
Posts: 21,497
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,497
Post URL14 May 2018, 16:55
iMyself
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Show more
Since Westley takes 70 steps to descend on the escalator when it’s turned off, we see that the escalators has 70 steps.
We also see that the escalator provides an extra 70 - 36 = 34 steps when it’s turned on. However, this really means for every 36 steps Wesley is moving (upward or downward), the escalator is moving downward 34 steps. Therefore, when he is moving upward 36 steps, the escalator is working against him 34 steps. So he only has a net rate of 36 - 44 = +2 steps for every 36 steps he is moving upward on the escalator.
Since the escalator has 70 steps and 70/2 = 35, he needs to walk 36 x 35 = 1,260 steps upward on the escalator in order to reach street level from the platform.
Answer: E
Scott Woodbury-Stewart | Founder and CEO | scottwoodburystewart@targettestprep.com
Flash Sale: Get 25% Off Our GMAT Plans | Use Code:FLASH25
Try OnDemand GMAT, the Most Comprehensive Video Course
Perfect 5-Star Ratingon GMAT Club
See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews
Signature Read More
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
naturalimproviser
naturalimproviser
Joined: 16 May 2018
Last visit: 02 Feb 2021
Posts: 26
Location: India
Schools:Queen's MBA'20
GMAT 1:610 Q44 V30
GMAT 2:650 Q44 V35
GRE 1:Q160 V150
GPA: 3
WE:Advertising (Advertising and PR)
Schools:Queen's MBA'20
GMAT 2:650 Q44 V35
GRE 1:Q160 V150
Posts: 26
Post URL06 Jun 2018, 06:33
AsadAbu
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Source: Manhattan GMAT Challenge question
Show more
I did it this way.
Let us assume that Wesley takes 1 step per second.
That means it takes Wesley 70 seconds to reach the subway level, and also the fact that there are 70 steps between the street level and the subway.
When the escalator is on, It takes Wesley 36 seconds to reach the subway level, that means he covered 36 steps at his usual pace, and the speed of the escalator helped him reach the cover the rest of the distance which would be 34 steps.
That means in 36 seconds Wesley covers 36 steps of distance and the escalator covers 34 steps of distance.
Now we can conclude that the escalator is a bit slower than Wesley.
When Wesley tries to reach up from the subway level to the street level while the escalator is on, the escalator would work in the reverse direction hence impeding his pace. So for every 36 steps the escalator would effectively cancel out 34 steps of distance as it is moving in the opposite direction.
Progress made by Wesley in 36 seconds/steps would be - 36-34 = 2
if he covers a distance of just 2 steps in 36 street level.seconds it would take him
(70/2)36 seconds/steps to reach the street level.
which is 1260 steps.
Hence, E.
Kudos Please!
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATGuruNY
GMATGuruNY
Tutor
Joined: 04 Aug 2010
Last visit: 29 Aug 2025
Posts: 1,344
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,344
Post URL31 May 2020, 03:59
Asad
An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform. If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
(A) Between 50 and 100 steps
(B) Between 100 and 200 steps
(C) Between 200 and 500 steps
(D) Between 500 and 1000 steps
(E) Over 1000 steps
Show more
DOWNWARD:
When the escalator is turned OFF, the number of steps taken by Wesley to travel downward = 70.
When the escalator is turned ON, the number of steps taken by Wesley to travel downward = 36, implying that the number of downward steps attributed to the escalator = 70-36 = 34.
Implication:
For every 36 steps taken by Wesley, the escalator moves downward 34 steps.
UPWARD:
For every 36 steps Wesley takes UPWARD, the escalator will move him DOWNWARD 34 steps, with the result that the net movement upward = 36-34 = 2 steps.
In other words, 36 upward steps taken by Wesley = a net upward movement of 2 steps.
Since Wesley must travel upward a total of 70 steps, we can set up the following proportion:
36−u p w a r d−s t e p s−t a k e n−b y−W e s l e y n e t−g a i n−o f−2−s t e p s=x−u p w a r d−s t e p s−t a k e n−b y−W e s l e y 70−s t e p s 36−u p w a r d−s t e p s−t a k e n−b y−W e s l e y n e t−g a i n−o f−2−s t e p s=x−u p w a r d−s t e p s−t a k e n−b y−W e s l e y 70−s t e p s
36 2=x 70 36 2=x 70
18=x 70 18=x 70
x=18∗70=1260 x=18∗70=1260
Show Spoiler
E
GMAT and GRE Tutor for over 20 years
Recent success stories for my students:admissions into Booth, Kellogg, HBS, Wharton, Tuck, Fuqua, Emory and others.
Available for live sessions in NYC and remotely all over the world
For more information, please email me at GMATGuruNY at gmail
Signature Read More
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kinshook
Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 28 Sep 2025
Posts: 5,760
Location: India
Schools:HBS '26CBS '26Wharton '26
GMAT 1:690 Q50 V34
WE:Engineering (Transportation)
Products:
Schools:HBS '26CBS '26Wharton '26
GMAT 1:690 Q50 V34
Posts: 5,760
Post URL14 Jun 2022, 06:43
Given: An escalator moves downward from street level to a subway platform at a constant rate. When the escalator is turned off, Wesley takes 70 steps to descend from street level to the platform. When the escalator is turned on, Wesley needs only 36 steps to descend from street level to the platform.
Asked: If Wesley begins at the platform and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach street level? (Assume that Wesley walks at a constant rate in all scenarios.)
For every 36 steps taken by Wesley, escalator moves 70-36=34 steps.
For every 36 steps taken by Wesley, escalator opposes 34 steps, resulting in 36-34= 2 effective steps
For climbing 70 steps, steps required by Wesley = 7036/2 = 1260 > 1000 steps
IMO E
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kungfupanda2393
kungfupanda2393
Joined: 07 Jun 2022
Last visit: 29 Sep 2022
Posts: 2
Posts: 2
Post URL26 Jun 2022, 01:12
This is an excellent problem; can someone share the source please?
Excellent because it tricked me into thinking if steps reduce by ~1/2x, they should increase by 2x but a trained mind would know % increase/decreases or upstream/downstream logic doesn't work that way
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bumpbot
bumpbot
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,132
Posts: 38,132
Post URL26 Jul 2025, 22:25
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources
Signature Read More
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NEW TOPIC
POST REPLY
Question banks
Downloads
My Bookmarks
Important topics
Reviews
Similar topics
Similar Topic
Author
Kudos
Replies
Last Post
In a subway station, there is a long escalator.
sherxon
by: sherxon
19 Oct 2023, 12:15
4
31
There is an escalator that moves at a constant rate from one floor up
Bunuel
by: Bunuel
01 Nov 2024, 05:53
2
19
Mike and Bryan walk up a moving escalator. The escalator moves at
Bunuel
by: Bunuel
06 Aug 2024, 14:27
1
17
Streets L, M and N are straight and level, and they intersect to form
Bunuel
by: Bunuel
11 Nov 2018, 00:51
7
3
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten
Bunuel
by: Bunuel
07 Nov 2024, 05:12
3
5
Moderators:
Bunuel
Math Expert
104347 posts
Krunaal
Tuck School Moderator
795 posts
Prep Toolkit
Announcements
How to Analyze your GMAT Score Report
Monday, Sep 29, 2025
11:30am NY / 3:30pm London / 9pm Mumbai
CLOSE SAVE
SUNDAY Quizzes! GRE Quiz @9:30am ET & GMAT Quiz @10:30am ET
Sunday, Sep 28, 2025
9:30am NY / 1:30pm London / 7pm Mumbai
Sunday, Sep 28, 2025
10:30am NY / 2:30pm London / 8pm Mumbai
CLOSE SAVE
What Makes a Great MBA Interview? Find Out from MBA Consultants and Students
CLOSE SAVE
An Emory MBA Powers Growth
Emory powers growth—professionally and personally. Built on a foundation of interdisciplinary learning, leadership development, and career readiness, our top-20 MBA program delivers a high return on investment with top-5 career outcomes.
CLOSE SAVE
Tuck at Dartmouth
Step Inside the Tuck Experience: Discover Tuck Events This Fall Current Tuck MBA students and Discover Tuck co-chairs share advice for prospective students and reflect on why they chose Tuck for their MBA.
Read More
CLOSE SAVE
AGSM at UNIVERSITY OF CALIFORNIA RIVERSIDE
CLOSE SAVE
Latest Posts
More Latest Posts
Calling all ISB PGP Applicants: 2026 Intake Class of 2027
by:PivotKing
19 mins
Seven children are playing musical chairs. When the music stops, in ho
by:Vaishbab
46 mins
Applications
by:Scazzu16
48 mins
In a group of 11 members, 5 have 4 siblings each and 6 have one siblin
by:gmatophobia
49 mins
Ida had five cards with matching envelopes
by:gmatophobia
1 hr
Copyright © 2025 GMAT Club
Terms & ConditionsForum Rules
Get our application on:
App StorePlay Store
Follow us on:
Contact us:
SupportAdvertise
GMAT® is a registered trademark of the Graduate Management Admission Council ™
GMAT Club's website has not been reviewed or endorsed by GMAC®
Terms & ConditionsForum Rules
The post is bookmarked successfully
view my bookmarksreturn to the post
[x] don't show this message again
by Problem Solving (PS)
JOIN NOW
Check Answer
×
You must be logged in to check and save answers
LoginRegister
Or continue with
GoogleApple |
7751 | https://math.stackexchange.com/questions/769136/how-to-create-a-two-circle-venn-diagram-with-3-equal-sections | geometry - How to create a two circle Venn diagram with 3 equal sections? - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How to create a two circle Venn diagram with 3 equal sections?
Ask Question
Asked 11 years, 5 months ago
Modified9 years, 2 months ago
Viewed 2k times
This question shows research effort; it is useful and clear
5
Save this question.
Show activity on this post.
I had a student ask if I could draw a Venn diagram in which each region was of equal area. I have played around with this a little but have not landed on an answer I'm satisfied with.
I was able to find the area of the overlapping region by subtracting the area of the triangle from the sector. I set the area of the overlapped area equal to the area of half of my circle. I am trying to get this down to two variables so that I can plug in a set radius and find an angle or something to that effect.
So far I have been able to simplify this down to radius, triangle height, and angle. Any help would be greatly appreciated.
geometry
circles
area
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Apr 25, 2014 at 17:56
Matthew Conroy
11.4k 4 4 gold badges 34 34 silver badges 36 36 bronze badges
asked Apr 25, 2014 at 17:49
MathZombieMathZombie
63 6 6 bronze badges
1
2 See equation 17 here: mathworld.wolfram.com/Circle-CircleIntersection.htmlMatthew Conroy –Matthew Conroy 2014-04-25 18:04:05 +00:00 Commented Apr 25, 2014 at 18:04
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
6
Save this answer.
Show activity on this post.
Start with the unit circle. It has area π π. So you want your areas to be 1 2 π 1 2 π each. Which means that half the lens-shaped intersection will have area 1 4 π 1 4 π. You could integrate from a given position x x to the right to get that:
∫1 x 2 1−t 2−−−−−√d t=1 2 π−x 1−x 2−−−−−√−arcsin x=1 4 π x 1−x 2−−−−−√+arcsin x=1 4 π x≈0.403972753299517209318961740066∫x 1 2 1−t 2 d t=1 2 π−x 1−x 2−arcsinx=1 4 π x 1−x 2+arcsinx=1 4 π x≈0.403972753299517209318961740066
I doubt that equation can be solved except numerically, it seems pretty transcendent to me. In any case, now you can compute
θ=2 arccos x≈2.30988146001005726088663377931≈132.346458834092919049353117798°θ=2 arccosx≈2.30988146001005726088663377931≈132.346458834092919049353117798°
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Apr 25, 2014 at 22:32
MvGMvG
44.2k 9 9 gold badges 94 94 silver badges 186 186 bronze badges
1
Reminds me of a question I asked myself in the past.MvG –MvG 2014-04-25 22:35:54 +00:00 Commented Apr 25, 2014 at 22:35
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
I was just doing this as well. Came out with a numeric approximation like MvG, not very satisfying. My working involves just trig, no calculus.
A c i r A c i r: Area of a circle = π r 2 π r 2
A m i d A m i d: Area of the central region of the Venn diagram
A s e g A s e g: Area of a segment of a circle =1 2 r 2(θ−s i n θ)=1 2 r 2(θ−s i n θ)
A m i d=2∗A s e g A m i d=2∗A s e g
A m i d=r 2(θ−s i n θ)A m i d=r 2(θ−s i n θ)
In order that the areas of each part of the Venn diagram be equal,
A m i d=A c i r−A m i d A m i d=A c i r−A m i d
2∗A m i d=A c i r 2∗A m i d=A c i r
2 r 2(θ−s i n θ)=π r 2 2 r 2(θ−s i n θ)=π r 2
2(θ−s i n θ)=π 2(θ−s i n θ)=π
θ−s i n θ=π 2 θ−s i n θ=π 2
I can't think of a way to solve for θ θ here. Approximation:
θ≈2.31 θ≈2.31
Now, to find the correct distance between the centres, a right triangle with a corner on the centre of a circle and on one of the intersection points:
diagram
θ 2≈1.155 θ 2≈1.155
c o s(1.155)=d r c o s(1.155)=d r
r c o s(1.155)=d r c o s(1.155)=d
So, to get 3 equal areas, the separation distance between the centres of the circles should be:
2 d=2 r c o s(1.155)2 d=2 r c o s(1.155)
Here's a demonstration.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Aug 2, 2016 at 3:47
MichaelMichael
1 1 1 bronze badge
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
geometry
circles
area
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
28Divide circle into 9 pieces of equal area
Related
1Circle geometry problem to find a relation between angles
2How to find area of remaining square with shapes inside
0Calculating area of a shape where there is a circular edge
2Random position within overlapping circles (Venn Diagram)
3Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is π r 2 2 π r 2 2
1Circle geometry, finding area of the minor segment
1Calculating the area between three tangent circles of unit radius, and using the result to show π 2<12 π 2<12
Hot Network Questions
ICC in Hague not prosecuting an individual brought before them in a questionable manner?
In the U.S., can patients receive treatment at a hospital without being logged?
How do you create a no-attack area?
Drawing the structure of a matrix
в ответе meaning in context
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
Direct train from Rotterdam to Lille Europe
alignment in a table with custom separator
How different is Roman Latin?
Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road?
How to locate a leak in an irrigation system?
Bypassing C64's PETSCII to screen code mapping
Proof of every Highly Abundant Number greater than 3 is Even
Xubuntu 24.04 - Libreoffice
How to solve generalization of inequality problem using substitution?
Riffle a list of binary functions into list of arguments to produce a result
how do I remove a item from the applications menu
The geologic realities of a massive well out at Sea
Vampires defend Earth from Aliens
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Alternatives to Test-Driven Grading in an LLM world
Does "An Annotated Asimov Biography" exist?
I have a lot of PTO to take, which will make the deadline impossible
Lingering odor presumably from bad chicken
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7752 | https://www.quora.com/If-the-empirical-formula-of-a-compound-is-CH2-and-its-molecular-formula-is-56-amu-what-is-the-molecular-formula | If the empirical formula of a compound is CH2 and its molecular formula is 56 amu, what is the molecular formula? - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Chemistry
The Amu
Average Molar Mass
Empirical Formula
Molecular Ratios
Unified Atomic Mass
Organic Compounds
Chemical Formulas
Molecular Equation
5
If the empirical formula of a compound is CH2 and its molecular formula is 56 amu, what is the molecular formula?
All related (41)
Sort
Recommended
Ted Krapkat
46 years as a chemist in the cement & aluminum industries. · Author has 996 answers and 4.9M answer views
·Updated 1y
Related
What is a compound with a molecular mass of 70.0g/mol and an empirical formula of CH2?
This problem can be expressed in terms of two simultaneous algebraic equations.
If we let the molecular formula of the compound in question be C x x H y y then, from the empirical formula of CH 2 2, we can write the following equation;-
y = 2x
Since the molar mass of the compound is 70 g/mol and the atomic masses of C and H are 12 and 1 g/mol respectively, we can write the following equation;-
12x + y = 70
Now, substituting 2x for y in this equation, and solving for x we have;-
12x + 2x = 70
14x = 70
x = 70/14
x = 5
(Despite unfounded claims to the contrary by Ernest Leung in his answer, 70 is indeed divisibl
Continue Reading
This problem can be expressed in terms of two simultaneous algebraic equations.
If we let the molecular formula of the compound in question be C x x H y y then, from the empirical formula of CH 2 2, we can write the following equation;-
y = 2x
Since the molar mass of the compound is 70 g/mol and the atomic masses of C and H are 12 and 1 g/mol respectively, we can write the following equation;-
12x + y = 70
Now, substituting 2x for y in this equation, and solving for x we have;-
12x + 2x = 70
14x = 70
x = 70/14
x = 5
(Despite unfounded claims to the contrary by Ernest Leung in his answer, 70 is indeed divisible by 14)
Therefore, the molecular formula of the compound is C 5 5 H 10 10.
There are ten structural isomers which match this formula… five alkenes and five cyclic alkanes;-
Question Originally Answered:“What is a compound with a molecular mass of 70.0g/mol and an empirical formula of CH2?”
Your response is private
Was this worth your time?
This helps us sort answers on the page.
Absolutely not
Definitely yes
Upvote ·
9 5
Sponsored by Grammarly
92% of professionals who use Grammarly say it has saved them time
Work faster with AI, while ensuring your writing always makes the right impression.
Download
999 210
Related questions
More answers below
If the empirical formula of a compound is CH2 and its molecular formula is 56 amu ten, what is the molecular formula?
A compound had an empirical formula of CH2 and a molecular mass of 28 amu. What is its molecular formula?
How is the molecular formula of a compound related to its empirical formula?
What are the 30 examples of empirical formulas and molecular formulas?
Is empirical related to the molecular formula?
Brad Yundt
Mechanical engineer and Excel aficionado · Author has 3.6K answers and 18.4M answer views
·7y
Originally Answered: What is the empirical formula of a substance whose molecular formula is CHO2? ·
As a mechanical engineer, please accept that my knowledge of chemistry is both limited and distant in the past. Nevertheless, I cannot think of any uncharged molecule whose molecular formula is CH02. And even if I could think of such a molecule, its empirical formula would also be CHO2.
If you let formic acid disassociate into H+ and a negatively charged ion, you get the formate ion which has the structure shown below and satisfies your molecular formula:
Now if your question were instead to give the molecular formula of a substance whose empirical formula is CHO2, that is something within my ke
Continue Reading
As a mechanical engineer, please accept that my knowledge of chemistry is both limited and distant in the past. Nevertheless, I cannot think of any uncharged molecule whose molecular formula is CH02. And even if I could think of such a molecule, its empirical formula would also be CHO2.
If you let formic acid disassociate into H+ and a negatively charged ion, you get the formate ion which has the structure shown below and satisfies your molecular formula:
Now if your question were instead to give the molecular formula of a substance whose empirical formula is CHO2, that is something within my ken. Consider oxalic acid, which has a formula C2H2O4, twice that of the given empirical formula.
Upvote ·
9 2
CheeHock Chew
Author has 3.9K answers and 2.6M answer views
·1y
If the empirical formula of a compound is CH2 and its molecular formula is 56 amu, what is the molecular formula?
Note:
Molar mass of :
C = 12.011 g/mol
H = 1.008 g/mol
If the empirical formula of a compound is CH2, its molecular formula is (CH2)n, where n is the number of empirical molecules making up the actual molecule.
Given that its molar mass is 56 g/mol, then (CH2)n = 56 or (1 x 12.011 + 2 x 1.008)n = 56 or (14.027)n = 56. On solving for n = (56/14.027) = 4.
Hence, the molecular formula of the compound is (CH2) x 4 = C4H8.
Your response is private
Was this worth your time?
This helps us sort answers on the page.
Absolutely not
Definitely yes
Upvote ·
Daniel Iyamuremye
Former Senior Lecturer (Retired) (2000–2018) · Author has 12.1K answers and 2M answer views
·Jun 12
CH2= 12 + 2 = 14 amu
56/ 14 = 4
Number of carbon atoms = 1×4= 4
Number of hydrogen atoms = 2x4 = 8
The molecular formula is: C4H8
Upvote ·
9 1
Related questions
More answers below
Why do some compounds have the same empirical and molecular formula?
How do you find the molecular formula from empirical?
A compound has a composition of 63.6% N and 36.4% O. If its density is 1.96 g/L, what is its empirical formula and molecular formula?
What is the empirical formula and the molecular formula?
Is CO an empirical formula or a molecular formula?
Meave Gilchrist
B.S. in Bachelor of Science in Biology&Chemistry, University of Texas Pan American (Graduated 1985) · Author has 6.6K answers and 4.7M answer views
·9mo
Calculate the empirical formula mass.
12.011 amu + (2 × 1.008 amu) = 14.027 amu = 14 u rounded to two significant figures
Divide the molecular formula mass by the empirical formula mass.
56 amu/14 amu = 4
To determine the molecular formula, multiply the subscripts of the empirical formula by 4.
C₍₁ₓ₄₎H₍₂ₓ₄₎ = C₄H₈
Upvote ·
Sponsored by EGNITION
Automatically sort and organize large inventory on Shopify.
This Shopify app will organize products in collections automatically based on flexible sorting rules.
Get the App
2.5K 2.5K
Ron Morel
B.S. in Chemistry, United States Air Force Academy (Graduated 1979) · Author has 13.9K answers and 2.7M answer views
·1y
Originally Answered: If the empirical formula of a compound is CH2 and its molecular formula is 56 amu ten, what is the molecular formula? ·
CH 2 2 adds up to 14
56/14 = 4
CH 2 2 x 4 = C 4 4 H 8 8.
No way to know the exact formula including branching and double bonds.
Upvote ·
9 1
9 1
Trevor Hodgson
Knows English · Author has 11.8K answers and 12.3M answer views
·3y
Originally Answered: 2. The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is its molecular formula? ·
I will answer the question on the basis that the molar mass is 70 g/mol
It is impossible for the molecule to have a mass of 70 g
Formula mass CH2 = 12+12 = 14g/formula
Number of formulas in 70 g : 70 g / 14 g/formula = 5
molecular formula = (CH2)5 = C5H10
Upvote ·
Sponsored by JetBrains
DataGrip, a powerful GUI tool for SQL.
Smart code completion, on-the-fly analysis, quick-fixes, refactorings that work in SQL files, and more.
Download
999 740
Tolulope Oyinlola
Former Production Manager (2016–2021) · Author has 198 answers and 157K answer views
·3y
Originally Answered: 2. The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is its molecular formula? ·
Empirical formular = CH2
Molecular formular = (CH2)n
(C{12}+H{12})n=70
(14)n=70
n= 70/14
n=5
(CH2)5
Molecular formular= C5H10
Thank you I believe this answers your question
Upvote ·
9 1
Gloria Dumaliang
College Professor/ Physics Teacher (2016–present) · Author has 1.2K answers and 735.5K answer views
·3y
Originally Answered: A compound with the empirical formula CH2 has a molar mass of 70 g/mol. What is the molecular formula for this compound? ·
Get the empirical formula mass of CH2. The Empirical formula CH2= 14 g/mole.
Molecular Formula = ( Empirical Formula)x
Get x = divide the molar mass by the empirical formula mass
x = 70g/mole/ 14 g/mole = 5
Answer: Molecular Formula = (CH2)5 = C5H10 ( pentene)
Upvote ·
Sponsored by RedHat
Know what your AI knows, with open source models.
Your AI should keep records, not secrets.
Learn More
99 36
John Maiko
High School Teacher at Teachers Service Commission (2007–present) · Author has 5.9K answers and 3.5M answer views
·Updated 5y
Originally Answered: A compound with the empirical formula CH2 has a molar mass of 70 g/mol. What is the molecular formula for this compound? ·
1 mole=70g/RMM
RMM=70
(CH2)n=70
(12+2)n=70
14n=70
n=5
Molecular formula C5H10(pentene)
Upvote ·
9 1
Kumaraswamy Sathiavasan
MSc in Chemistry & IAS officer(retd.) · Author has 9.2K answers and 14.4M answer views
·1y
Originally Answered: If the empirical formula of a compound is CH2 and its molecular formula is 56 amu ten, what is the molecular formula? ·
Empirical formula = CH2
Empirical formula mass = 121 + 12 = 16
Molecular mass = 56
Molecular mass/empirical mass = 56/14 = 4
Molecular formula = (CH2)4
= C4H8 (answer)
Upvote ·
9 2
Gerry St-Aubin
Retired High School Chemistry, Biology and Science Teacher · Author has 1.2K answers and 2.5M answer views
·2y
Related
A compound had an empirical formula of CH2 and a molecular mass of 28 amu. What is its molecular formula?
The molecular formula will be a whole number multiple of the simplest (empirical) formula. As a result, the molecular mass will also be the same whole number multiple of the empirical formula mass.
So, determining the molecular formula from the empirical formula is as easy as comparing their mass.
Empirical formula mass (EFM) = mC + 2(mH) = 12 u + 2(1 u) = 14 u
You are given the molecular mass (MM) = 28 u
Multiple = MM/EFM = 28/14 = 2
This means that the molecular formula is 2X the empirical formula:
2(CH2) = C2H4 which is CH2=CH2 or ethene
Upvote ·
9 1
Aidan Cooney
Consultant Formulation Scientist/Research Chemist at Self-Employment (1996–present) · Author has 584 answers and 756.8K answer views
·5y
Related
What is the molecular formula if the molar mass is 350.75 g/mole and the empirical formula is CH2?
350.75/14.027 = 25.005
Therefore Molecular formula = C25H50
Right, how did I get this answer? what I did was I divided the molar mass by the empirical formula mass. I got the empirical mass by adding relative atomic mass of carbon (12.011) to 2 x the relative atomic mass of hydrogen (2 x 1.008). Relative atomic mass does not have any units because it is a weighted average of the mass numbers of all the isotopes of the particular element however we will give the empirical mass the same units as molar mass which is grams per mole (so the units will cancel out). So we end up with 25.005 which we c
Continue Reading
350.75/14.027 = 25.005
Therefore Molecular formula = C25H50
Right, how did I get this answer? what I did was I divided the molar mass by the empirical formula mass. I got the empirical mass by adding relative atomic mass of carbon (12.011) to 2 x the relative atomic mass of hydrogen (2 x 1.008). Relative atomic mass does not have any units because it is a weighted average of the mass numbers of all the isotopes of the particular element however we will give the empirical mass the same units as molar mass which is grams per mole (so the units will cancel out). So we end up with 25.005 which we can round off to 25. Since the empirical formula is CH2, all we do is multiply the number of carbons by 25 and the number of hydrogens by 25, which gives us C25H50.
Upvote ·
9 3
Related questions
If the empirical formula of a compound is CH2 and its molecular formula is 56 amu ten, what is the molecular formula?
A compound had an empirical formula of CH2 and a molecular mass of 28 amu. What is its molecular formula?
How is the molecular formula of a compound related to its empirical formula?
What are the 30 examples of empirical formulas and molecular formulas?
Is empirical related to the molecular formula?
Why do some compounds have the same empirical and molecular formula?
How do you find the molecular formula from empirical?
A compound has a composition of 63.6% N and 36.4% O. If its density is 1.96 g/L, what is its empirical formula and molecular formula?
What is the empirical formula and the molecular formula?
Is CO an empirical formula or a molecular formula?
What information do we need to determine the molecular formula of a compound from the empirical formula?
What is the molecular formula for a compound with the empirical formula: CaCl2 and a molecular mass of 330g?
What is the difference between an empirical formula and a molecular formula?
How does the molecular formula of a compound compare with the empirical formula?
What is the relationship between an empirical formula and a molecular formula?
Related questions
If the empirical formula of a compound is CH2 and its molecular formula is 56 amu ten, what is the molecular formula?
A compound had an empirical formula of CH2 and a molecular mass of 28 amu. What is its molecular formula?
How is the molecular formula of a compound related to its empirical formula?
What are the 30 examples of empirical formulas and molecular formulas?
Is empirical related to the molecular formula?
Why do some compounds have the same empirical and molecular formula?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025 |
7753 | https://www.chegg.com/homework-help/questions-and-answers/1-compute-volume-tetrahedron-illustrate-tetrahedron-vertices-0-0-0-quad-2-0-0-quad-0-3-0-q-q105849579 | Solved 1) Compute the volume of a tetrahedron. (a) | Chegg.com
Skip to main content
Books
Rent/Buy
Read
Return
Sell
Study
Tasks
Homework help
Understand a topic
Writing & citations
Tools
Expert Q&A
Math Solver
Citations
Plagiarism checker
Grammar checker
Expert proofreading
Career
For educators
Help
Sign in
Paste
Copy
Cut
Options
Upload Image
Math Mode
÷
≤
≥
o
π
∞
∩
∪
√
∫
Math
Math
Geometry
Physics
Greek Alphabet
Math
Advanced Math
Advanced Math questions and answers
1) Compute the volume of a tetrahedron. (a) Illustrate the tetrahedron that has vertices at (0,0,0),(2,0,0),(0,3,0),(0,0,6), in Cartesian coordinates. The top face of the tetrahedron is part of the plane 6x+ 4y+2z=12, or equivalently, z=6−3x−2y. This tetrahedron sits inside a box with side lengths 2,3 and 6 . The volume of this box is V=2×3×6=36 cubic units.
Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
See Answer See Answer See Answer done loading
Question: 1) Compute the volume of a tetrahedron. (a) Illustrate the tetrahedron that has vertices at (0,0,0),(2,0,0),(0,3,0),(0,0,6), in Cartesian coordinates. The top face of the tetrahedron is part of the plane 6x+ 4y+2z=12, or equivalently, z=6−3x−2y. This tetrahedron sits inside a box with side lengths 2,3 and 6 . The volume of this box is V=2×3×6=36 cubic units.
Show transcribed image text
There are 3 steps to solve this one.Solution 100%(2 ratings) Share Share Share done loading Copy link Step 1 (a) The tetrahedron with vertices (0,0,0),(2,0,0),(0,3,0), and (0,0,6) can be visualized as follows: (b) The fraction of the volume ... View the full answer Step 2 UnlockStep 3 UnlockAnswer Unlock Previous questionNext question
Transcribed image text:
1) Compute the volume of a tetrahedron. (a) Illustrate the tetrahedron that has vertices at (0,0,0),(2,0,0),(0,3,0),(0,0,6), in Cartesian coordinates. The top face of the tetrahedron is part of the plane 6 x+4 y+2 z=12, or equivalently, z=6−3 x−2 y. This tetrahedron sits inside a box with side lengths 2,3 and 6 . The volume of this box is V=2×3×6=36 cubic units. The volume of the tetrahedron must be some fraction of this. (b) What do you think this fraction is? (c) Set up a double integral that will compute the volume of this tetrahedron. (d) Evaluate the double integral. Suggestion: work slowly and at each step check your algebra and arithmetic before proceeding. (e) Set up a triple integral that will compute the volume of this tetrahedron. (f) Evaluate this triple integral as an iterated integral so that you integrate first with respect to z. After integrating with respect to z, your computation should connect with your computations in part (d). After this point it is not necessary to repeat these computations.
Not the question you’re looking for?
Post any question and get expert help quickly.
Start learning
Chegg Products & Services
Chegg Study Help
Citation Generator
Grammar Checker
Math Solver
Mobile Apps
Plagiarism Checker
Chegg Perks
Company
Company
About Chegg
Chegg For Good
Advertise with us
Investor Relations
Jobs
Join Our Affiliate Program
Media Center
Chegg Network
Chegg Network
Busuu
Citation Machine
EasyBib
Mathway
Customer Service
Customer Service
Give Us Feedback
Customer Service
Manage Subscription
Educators
Educators
Academic Integrity
Honor Shield
Institute of Digital Learning
© 2003-2025 Chegg Inc. All rights reserved.
Cookie NoticeYour Privacy ChoicesDo Not Sell My Personal InformationGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
Do Not Sell My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Sale of Personal Data
[x] Sale of Personal Data
Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link.
If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences.
Targeting Cookies
[x] Switch Label label
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices
%20Compute%20the%20volume%20of%20a%20tetrahedron.%20(a)%20%7C%20Chegg.com&p=https%3A%2F%2Fwww.chegg.com%2Fhomework-help%2Fquestions-and-answers%2F1-compute-volume-tetrahedron-illustrate-tetrahedron-vertices-0-0-0-quad-2-0-0-quad-0-3-0-q-q105849579&r=<=2820&evt=pageLoad&sv=2&cdb=AQET&rn=515671)
mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
7754 | https://npflueger.github.io/teaching/math1a/lecture24.pdf | Lecture 24: Indeterminate forms Nathan Pflueger 6 November 2013 1 Introduction Last time, we saw one method for dealing with limits of functions f(x) g(x) , where both f(x) and g(x) have limits of ∞or 0; such limits are said to be in the indeterminate forms ∞ ∞or 0 0. Today we will examine several other similar situation, where a function is built up from two other functions, but the limits of these two functions is not sufficient information to determine the overall limit. In most such cases, the necessary idea is to somehow reformulate the limit in a form that previous techniques can approach.
The reference for today is Stewart §4.5.
2 List of indeterminate forms The following expressions are all called indeterminate forms.
0 0 ∞ ∞ 0 · ∞ ∞0 00 1∞ ∞−∞ Remember that ∞is not a real, honest number, but a shorthand for a limiting process. Here are more precise statements of what it means that each of these forms are indeterminate: • 0 0 is indeterminate, because knowing that lim x→c f(x) = 0 and lim x→c g(x) = 0 is not enough information to determine lim x→c f(x) g(x) .
• ∞ ∞is indeterminate, because knowing that lim x→c f(x) = ∞and lim x→c g(x) = ∞is not enough information to determine lim x→c f(x) g(x) .
• 0·∞is indeterminate, because knowing that lim x→c f(x) = 0 and lim x→c g(x) = ∞is not enough information to determine lim x→c f(x) · g(x).
• ∞0 is indeterminate, because knowing that lim x→c f(x) = ∞and lim x→c g(x) = 0 is not enough information to determine lim x→c f(x)g(x).
1 • 00 is indeterminate, because knowing that lim x→c f(x) = 0 and lim x→c g(x) = 0 is not enough information to determine lim x→c f(x)g(x).
• 1∞is indeterminate, because knowing that lim x→c f(x) = 1 and lim x→c g(x) = ∞is not enough information to determine lim x→c f(x)g(x).
• ∞−∞is indeterminate, because knowing that lim x→c f(x) = ∞and lim x→c g(x) = ∞is not enough information to determine lim x→c(f(x) −g(x)).
You’ve found some examples in your homework to show that these forms are all indeterminate.
In contast, forms like 0 ∞, ∞+ ∞and 0∞are determinate, because if they arise in a limit, the answer is unambiguous (in these cases, these forms resolve to 0, ∞, 0, respectively).
Keeping straight which forms resolve unambiguously, and which are indeterminate, can be confusing.
Here is a good rule of thumb: a form is indeterminate if it is sensitive to small errors. Here’s what I mean: • Consider 0 0. Think of both of these zeros as being imprecise measurements. So it could be that one zero is actually 0.0001, while the other is actually 0.0000001, but your measuring device is only accurate enough to measure them both as zero. Then the ratio could be 0.0001 0.0000001 = 1000, but it could also be 0.0000001 0.0001 = 0.001. So this arithmetic is sensitive to errors in the measurements, so it is indeterminate.
• Consider ∞ ∞. Think of ∞as just meaning “offthe chart;” it is a number too large for you to measure.
So one ∞could stand for 1000, while the other stands for 1000000. The ratio of these could be 1000, but it could also be 0.001. So the arithmetic is sensitive to error about just how big these two “offthe charts” numbers are.
• Consider 0 ∞. Here 0 is just some small number (possibly with a bit of error), and ∞is some huge number (offthe charts of your measuring device). But you can see that even if you don’t know just how far offthe charts the ∞is, or just how close to 0 the 0 is, the quotient will nonetheless be an immeasurably tiny number. So this form is perfectly well-defined: it is definitely 0.
If you apply this criterion carefully, you will be able to tell whether a particular form determines the limit uniquely or not. For example, consider the most confusing indeterminate form of the bunch: 1∞. Then all you really know about the ∞is that it’s too large to measure; lets say it is at least 1000000. And all you know about the 1 is that it is within 0.0001 of 1. This isn’t good enough to have even an approximate sense for the result of exponentiation: it might actually be 1.00011000000 ≈1043, but it could also be 0.9991000000 ≈10−43.
This arithmetic is extremely sensitive to even small errors in the measurement of “1” and “∞,” so it is indeterminate.
The point is that if you obtain an indeterminate form when you evaluate the limits of two parts of a function separately, you will need to do some more analysis to find the limit. We saw one tool last time that often resolves 0 0 or ∞ ∞, called L’Hˆ opital’s rule. We will now examine some examples of how to resolve the other types. This will often involve converting these other types into a from where L’Hˆ optial’s rule applies, but not always; in some cases we will want to draw on more elementary techniques.
3 Resolving ∞· 0 The typical way to resolve ∞· 0 is to convert it to 0 0 or ∞ ∞by either taking the reciprocal of the ∞and putting it in the denominator, or putting the reciprocal of the 0 and putting it in the denominator. Here is an example of both of these.
2 Example 3.1. Consider lim x→∞ x · sin 2 x . This has the form ∞· 0 if you try to evaluate directly. But if you flip the x to the denominator lim x→∞ sin 2 x 1/x !
then this limit has the form 0 0.
We have simply taken the ∞, and transformed it into a 0 in the denominator. This limit can be done with L’Hˆ opital’s rule.
lim x→∞ sin 2 x 1/x !
= lim x→∞ cos 2 x −2 x2 −1/x2 = lim x→∞2 cos 2 x = 2 cos(2/∞) = 2 cos(0) = 2 So in this case, we could evaluate the limit by flipping it to 0 0 and using L’Hˆ opital’s rule.
Example 3.2. In this example we’ll change ∞·0 to ∞ ∞. Consider lim x→∞(xe−x). This has the form ∞·0. Move the 0 to an ∞in the denominator.
lim x→∞xe−x = lim x→∞ x ex = lim x→∞ 1 ex = 1 ∞ = 0 It is not always easy to tell whether you should convert ∞· 0 to 0 0 or ∞ ∞. You may need to try both options, and see which one makes the problem simpler to solve. As always, there are no hard fast rules; you should just experiment with many examples and try to get a sense for what is most effective in various situations.
4 Resolving 1∞ In the case of 1∞, the usual tack is to take the logarithm of the expression in question. This results in a new expression, with will have a limit of the form ∞· 0. Resolve this indeterminacy using ideas from the previous section, and then raise e to the result. As a first example, consider the following.
lim h→0(1 + 2h)1/h This can be rewritten as following, by taking the logarithm of the expression.
lim h→0(1 + 2h)1/h = lim h→0 eln((1+2h)1/h) = elimh→0 ln((1+2h)1/h) 3 Now evaluate the limit in the exponential separately.
lim h→0 ln (1 + 2h)1/h = lim h→0 1 h ln(1 + 2h) (has form ∞· 0) = lim h→0 ln(1 + 2h) h (Convert to 0 0) = lim h→0 2/(1 + 2h) 1 (L’Hˆ opital) = 2 Therefore the original limit is: lim h→0(1 + 2h)1/h = e2.
The reason the logarithm is a useful tool here is precisely that it transforms exponentiation into mul-tiplication. This is one of the main reasons that logarithms became centrally important in mathematics: they convert complicated operations into simpler operations, and therefore make many computations much easier. We’ve already seen this feature once in this course, in the context of logarithmic differentiation.
5 Resolving 00 and ∞0 For both 00 and ∞0, the usual technique is the same as for 1∞: evaluate instead the limit of the logarithm of the expression, and then exponentiate the result. Here are a couple examples.
Example 5.1.
lim x→∞x1/x = elimx→∞1 x ln x lim x→∞ 1 x ln x = lim x→∞ ln x x (form ∞ ∞) = lim x→∞ 1/x 1 (L’Hˆ opital) = 0 ⇒lim x→∞x1/x = e0 = 1 Example 5.2.
lim x→0+ xx = elimx→0+ x ln x lim x→0+ x ln x = lim x→0+ ln x 1/x (form ∞ ∞) = lim x→0+ 1/x −1/x2 (L’Hˆ opital) = lim x→0+(−x) = 0 ⇒lim x→0+ xx = e0 = 1 4 6 Resolving ∞−∞ The typical approach to resolving ∞−∞is to re-express the expression as a fraction, by finding some common denominator. Typically this fraction will also be in an indeterminate form, which will need to be dealt with. Here are two examples.
Example 6.1. Consider lim x→∞ p x2 + x −x . This can be converted to a fraction by the algebraic trick of “multiplying by the conjugate:” p x2 + x −x = p x2 + x −x · √ x2 + x + x √ x2 + x + x = (x2 + x) −x2 √ x2 + x + x = x √ x2 + x + x From here, if you try to evaluate the limit lim x→∞ x √ x2 + x + x , you will see that it is in the form ∞ ∞. So you could apply L’Hˆ opital’s rule at this stage (and this is one valid way to compute the limit). In this case, it is somewhat easier to evaluate this limit by elementary means, however: divide both the numerator and the denominator by x and substitute directly.
x √ x2 + x + x = x √ x2 + x + x · 1/x 1/x = 1 1 x √ x2 + x + 1 = 1 p 1 + 1/x + 1 ⇒lim x→∞ x √ x2 + x + x = 1 √1 + 0 + 1 = 1 2 Therefore lim x→∞ p x2 + x −x = 1 2 as well.
Note. This computation did not require L’Hˆ opital’s rule at all; we used only methods from earlier in the course. In fact, this problem was originally proposed as a problem on last week’s midterm exam, but was removed for length reasons.
Here is a second example, where the conversion to a fraction is more transparent, but the ensuing limit computation requires a bit more work.
Example 6.2. Consider the limit lim x→0 1 sin x −1 x . This is in the form ∞−∞if x comes from the right, or −∞+ ∞if x comes from the left. Either way, it must be reformulated. In this case, just find a common denominator and do the fraction arithmetic.
1 sin x −1 x = x −sin x x · sin x Now this is in the form 0 0. We can attack it with l’Hˆ opital’s rule.
5 lim x→0 1 sin x −1 x = lim x→0 x −sin x x · sin x (form 0 0) = lim x→0 1 −cos x x cos x + sin x (L’Hˆ opital; still has form 0 0) = lim x→0 sin x −x sin x + 2 cos x (L’Hˆ opital again) = 0 −0 + 2 (direct substitution) = 0 So this limit is equal to 0.
6 |
7755 | https://www.mathworks.com/matlabcentral/answers/316950-1d-heat-conduction-using-explicit-finite-difference-method | 1D Heat Conduction using explicit Finite Difference Method - MATLAB Answers - MATLAB Central
Skip to content
MATLAB Answers
MATLAB Help Center
Community
Learning
Get MATLABMATLAB
Sign In
My Account
My Community Profile
Link License
Sign Out
Contact MathWorks Support
Visit mathworks.com
Search Answers Answers
Help Center
Answers
MathWorks
MATLAB
Help Center
MathWorks
MATLAB Answers
File Exchange
Videos
Online Training
Blogs
Cody
MATLAB Drive
ThingSpeak
Bug Reports
Community
MATLAB Answers
File Exchange
Cody
AI Chat Playground
Discussions
Contests
Blogs
More
Communities
Treasure Hunt
People
Community Advisors
Virtual Badges
About
Home
Ask
Answer
Browse
MATLAB FAQs
More
Contributors
Recent Activity
Flagged Content
Manage Spam
Help
You are now following this question
You will see updates in your followed content feed.
You may receive emails, depending on your communication preferences.
1D Heat Conduction using explicit Finite Difference Method
Follow
103 views (last 30 days)
Show older comments
Derek Shawon 15 Dec 2016
Vote 1
Link ×
Direct link to this question Cancel Copy to Clipboard
⋮
Vote 1
Link ×
Direct link to this question Cancel Copy to Clipboard
Answered:Alugunuri on 8 Feb 2023
Accepted Answer:michio
Open in MATLAB Online
Hello I am trying to write a program to plot the temperature distribution in a insulated rod using the explicit Finite Central Difference Method and 1D Heat equation. The rod is heated on one end at 400k and exposed to ambient temperature on the right end at 300k. I am using a time of 1s, 11 grid points and a .002s time step. When I plot it gives me a crazy curve which isn't right. I think I am messing up my initial and boundary conditions. Here is my code.
Theme Copy
L=1;
t=1;
k=.001;
n=11;
nt=500;
dx=L/n;
dt=.002;
alpha=kdt/dx^2;
T0(1)=400;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha(T0(i+1)-2T0(i)+T0(i-1));
end
T0=T1;
end
plot(x,T1)
2 Comments
Show None Hide None
KSSVon 15 Dec 2016
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
First place, it is not giving any curve..there is a error in your code. Please recheck your code once.
Derek Shawon 15 Dec 2016
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
I am not sure I understand correctly. In the above I wrote this equation to be iterated
With Boundary conditions
and Initial Conditions
with T0=400k and TL=Ti=300k
I am not sure how to set these boundary conditions in the code. Or if there is a curve I need to derive before doing the iterations.
Sign in to comment.
Sign in to answer this question.
Accepted Answer
michioon 15 Dec 2016
Vote 4
Link ×
Direct link to this answer Cancel Copy to Clipboard
⋮
Vote 4
Link ×
Direct link to this answer Cancel Copy to Clipboard
Edited: michioon 15 Dec 2016
Open in MATLAB Online
It seems your initial condition and boundary conditon (x = L) are missing in the code. Try
Theme Copy
L=1;
t=1;
k=.001;
n=11;
nt=500;
dx=L/n;
dt=.002;
alpha=kdt/dx^2;
T0=400ones(1,n);
T1=300ones(1,n);
T0(1) = 300;
T0(end) = 300;
for j=1:nt
for i=2:n-1
T1(i)=T0(i)+alpha(T0(i+1)-2T0(i)+T0(i-1));
end
T0=T1;
end
plot(T1)
could make a good example for you.
9 Comments
Show 7 older comments Hide 7 older comments
Derek Shawon 15 Dec 2016
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Although your code didn't plot the graph properly for me it helped me figure out how to define the boundaries. I was able to plot it properly by switching around a few things. Thanks for your help!
michioon 15 Dec 2016
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Open in MATLAB Online
Glad to know that you figure things out. One additional tip is vectorization instead of for-loop, ie.
Theme Copy
for i=2:n-1
T1(i) = T0(i) + alpha(T0(i+1)-2T0(i)+T0(i-1));
end
is equivalent to
Theme Copy
T1(2:n-1) = T0(2:n-1) + alpha(T0(3:n)-2T0(2:n-1)+T0(1:n-2));
The later runs much faster.
Kunpeng Liaoon 13 Oct 2017
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Open in MATLAB Online
Are the time step and grid spacing missing? I guess it would be:
Theme Copy
if true
T1(2:n-1) = T0(2:n-1) + alphadt/dx^2(T0(3:n)-2T0(2:n-1)+T0(1:n-2));
end
Torstenon 16 Oct 2017
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Look at the definition of "alpha" in the code ...
Best wishes
Torsten.
Siddarth Banerjeeon 1 Nov 2022
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
@Derek Shaw Can you please share the code. I am also struck in the same and cannot figure out the correct code for plotting..
Thanks
Torstenon 1 Nov 2022
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Open in MATLAB Online
Ran in:
Theme Copy
L=1;
t=1;
k=.001;
n=11;
nt=10000;
dx=L/n;
dt=.002;
alpha=kdt/dx^2;
T0=400ones(1,n+1);
T1=300ones(1,n+1);
T0(1) = 300;
T0(end) = 300;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha(T0(i+1)-2T0(i)+T0(i-1));
end
T0=T1;
end
plot((0:n)L/n,T1)
SYML2ndon 4 Dec 2022
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Edited: SYML2ndon 4 Dec 2022
Hi @Torsten I have a question that regards this code.
If one end should be at 400K and the other end is at 300K.
Why, before the for cycle, is T0 = [300,400,..,400,300]?
For me it should have been T0=[400,300,...,300,300]. If T(0,t)=T0=400 K, then this boundary condition should be fixed during the whole calculation. Can you explain?
Torstenon 4 Dec 2022
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Open in MATLAB Online
Ran in:
Yes, code should be
Theme Copy
L=1;
t=1;
k=.001;
n=11;
nt=10000;
dx=L/n;
dt=.002;
alpha=kdt/dx^2;
T0=300ones(1,n+1);
T0(1) = 400;
T1 = T0;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha(T0(i+1)-2T0(i)+T0(i-1));
end
T0=T1;
end
plot((0:n)L/n,T1)
SYML2ndon 4 Dec 2022
×
Direct link to this comment
Cancel Copy to Clipboard
⋮
Link×
Direct link to this comment Cancel Copy to Clipboard
Thank you @Torsten
Sign in to comment.
More Answers (3)
youssef aideron 12 Feb 2019
Vote 4
Link ×
Direct link to this answer Cancel Copy to Clipboard
⋮
Vote 4
Link ×
Direct link to this answer Cancel Copy to Clipboard
Open in MATLAB Online
here is one, you can just change the boundaries
Theme Copy
clear
clc
clf
% domain descritization
alpha = 0.05;
xmin = 0;
xmax = 0.2;
N = 100;
dx = (xmax-xmin)/(N-1);
x = xmin:dx:xmax;
dt = 4.0812E-5;
tmax = 1;
t = 0:dt:tmax;
% problem initialization
phi0 = ones(1,N)300;
phiL = 230;
phiR = phiL;
% solving the problem
r = alphadt/(dx^2) % for stability, must be 0.5 or less
for j = 2:length(t) % for time steps
phi = phi0;
for i = 1:N % for space steps
if i == 1 || i == N
phi(i) = phiL;
else
phi(i) = phi(i)+r(phi(i+1)-2phi(i)+phi(i-1));
end
end
phi0 = phi;
plot(x,phi0)
shg
pause(0.05)
end
0 Comments
Show -2 older comments Hide -2 older comments
Sign in to comment.
okncylnon 22 Dec 2017
Vote 0
Link ×
Direct link to this answer Cancel Copy to Clipboard
⋮
Vote 0
Link ×
Direct link to this answer Cancel Copy to Clipboard
implicit?
0 Comments
Show -2 older comments Hide -2 older comments
Sign in to comment.
Alugunurion 8 Feb 2023
Vote 0
Link ×
Direct link to this answer Cancel Copy to Clipboard
⋮
Vote 0
Link ×
Direct link to this answer Cancel Copy to Clipboard
how to write code with neumqn BCs ?
0 Comments
Show -2 older comments Hide -2 older comments
Sign in to comment.
Sign in to answer this question.
FEATURED DISCUSSION ###### MATLAB EXPO 2025 Registration is Now Open! November 12 – 13, 2025
Registration is now open for MathWorks annual virtual event MATLAB EXPO 2025...
Devin in MATLAB EXPO
0
View Post
See Also
MATLAB Answers
plotting with for loop
1 Answer
how can I get my results in a csv file rather than a figure
1 Answer
How can I turn this Explicit method code to Implicit method code ?
1 Answer
Entire Website
throw solved by ode45
File Exchange
Finite Difference Explicit Method for Fick's 2nd Law
File Exchange
SEIRS epidemic model
File Exchange
Categories
MATLABMathematics
Find more on Mathematics in Help Center and File Exchange
Tags
1d heat conduction
Products
MATLAB
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
7 Ways to Make Embedded Software Safe and Secure
Read white paper
An Error Occurred
Unable to complete the action because of changes made to the page. Reload the page to see its updated state.
Close
×
Select a Web Site
Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: United States.
United States
United States (English)
United States (Deutsch)
United States (Français)
United States（简体中文）
United States (English)
You can also select a web site from the following list
How to Get Best Site Performance
Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.
Americas
América Latina (Español)
Canada (English)
United States (English)
Europe
Belgium (English)
Denmark (English)
Deutschland (Deutsch)
España (Español)
Finland (English)
France (Français)
Ireland (English)
Italia (Italiano)
Luxembourg (English)
Netherlands (English)
Norway (English)
Österreich (Deutsch)
Portugal (English)
Sweden (English)
Switzerland
Deutsch
English
Français
United Kingdom(English)
Asia Pacific
Australia (English)
India (English)
New Zealand (English)
中国
简体中文 Chinese
English
日本 Japanese (日本語)
한국 Korean (한국어)
Contact your local office
Trust Center
Trademarks
Privacy Policy
Preventing Piracy
Application Status
Terms of Use
Contact Us
Your Privacy Choices
© 1994-2025 The MathWorks, Inc.
Select a Web Site United States |
7756 | https://www.hanyuace.com/blog/hsk-2-grammar-using-zhe-and-na-to-indicate-location | Services
Helpful Links
Downloads
© 2024. HanYuAce. All rights reserved.
HanyuAce offers free HSK exam practice tests, online mock exams, and Chinese language resources to help learners at all levels. Whether you’re preparing for the HSK 1, HSK 4, or HSK 6, our platform provides everything you need to master your Chinese proficiency test.
HSK 2 Grammar: Using 这 (zhè) and 那 (nà) to Indicate Location
For HSK 2 learners, mastering the use of demonstrative pronouns 这 (zhè) and 那 (nà) is crucial for describing locations accurately in Chinese. These versatile words not only help you point out objects but also play a significant role in expressing spatial relationships. Let's dive into how you can use these essential grammar elements to boost your Chinese language skills.
Understanding 这 (zhè) and 那 (nà)
Before we delve into their use for indicating location, let's refresh our understanding of these demonstrative pronouns:
In their basic form, these words are used to point out specific objects or people:
Using 这 (zhè) and 那 (nà) with Location Words
When it comes to indicating location, 这 and 那 are often combined with specific location words to create more precise spatial descriptions. Here are some common location words you'll encounter at the HSK 2 level:
To use these with 这 and 那, we follow this structure:
这/那 + 个 (measure word) + noun + location word
For example:
Practical Examples in Sentences
Let's look at some complete sentences to see how these structures work in practice:
我的钱包在这个包里。(Wǒ de qiánbāo zài zhè ge bāo lǐ.)
My wallet is in this bag.
他的车停在那个商店前。(Tā de chē tíng zài nà ge shāngdiàn qián.)
His car is parked in front of that shop.
猫睡在这张沙发上。(Māo shuì zài zhè zhāng shāfā shàng.)
The cat is sleeping on this sofa.
我们的教室在那栋楼里。(Wǒmen de jiàoshì zài nà dòng lóu lǐ.)
Our classroom is in that building.
Expanding Your Usage
As you progress in your Chinese studies, you'll find that 这 and 那 can be used in more complex ways to describe location:
Combining with Directional Complements
You can add directional complements to create more specific location descriptions:
Example:
书在这里面。(Shū zài zhè lǐmiàn.) - The book is in here.
Using with Time Expressions
这 and 那 can also be used with time expressions to indicate proximity:
Example:
这个周末我们去北京。(Zhè ge zhōumò wǒmen qù Běijīng.) - We're going to Beijing this weekend.
Common Mistakes to Avoid
Confusing 这里 (zhèlǐ) and 那里 (nàlǐ):
Forgetting the measure word:
Incorrect: 这书桌上 (zhè shūzhuō shàng)
Correct: 这张书桌上 (zhè zhāng shūzhuō shàng)
Using the wrong location word:
Remember that some location words have specific uses. For example, 上 (shàng) for "on top of" and 里 (lǐ) for "inside".
Practice Exercises
To reinforce your understanding, try these exercises:
Fill in the blanks with the correct form of 这 or 那:
Translate these sentences into Chinese:
Create two sentences using 这 and 那 with different location words.
Cultural Note
In Chinese culture, the concept of space and location is often viewed differently from Western perspectives. The use of 这 and 那 reflects not just physical distance, but can also indicate psychological or social distance. For example, using 这 might imply familiarity or closeness, while 那 could suggest distance or unfamiliarity.
Conclusion
Mastering the use of 这 (zhè) and 那 (nà) for indicating location is a key skill for HSK 2 learners. It allows you to describe your surroundings more accurately and navigate conversations about place and space with greater confidence. Remember to practice these structures regularly in your speaking and writing to fully integrate them into your Chinese language toolkit.
As you continue your HSK journey, you'll find that these demonstrative pronouns become increasingly useful in more complex grammatical structures. Keep practicing, and soon you'll be using 这 and 那 as naturally as a native speaker!
Blogs you may also like
Rare Chinese Characters You May Want to Know for Advanced HSK Levels
Explore uncommon Chinese characters that appear in higher HSK levels. Learn their meanings, usage, and importance for advanced Mandarin proficiency.
Basic Chinese Sentence Structure: How to Form Simple Sentences in Mandarin
This comprehensive guide explores the fundamental structure of Chinese sentences, focusing on simple constructions essential for HSK preparation. It covers key grammar points, vocabulary usage, and provides practical examples to help learners master basic Mandarin sentence formation.
Using HSK Study Apps to Enhance Your Learning
Dive into the benefits of using HSK study apps to master Chinese. From vocabulary to grammar and cultural insights, these apps can revolutionise your HSK preparation.
Ultimate Guide to Drinking in China for HSK Students
This guide explores the cultural nuances of drinking in China, incorporating essential HSK vocabulary and grammar to help students better understand and engage with local customs.
Stories you may also like
Three Men Make a Tiger
The Map is Exhausted and the Dagger is Revealed
Breaking the Cauldrons and Sinking the Boats
The Humiliation of Passing Under the Thigh |
7757 | https://kekenet.com/gaokao/201409/331507.shtml | 2011年高考数学真题附解析(重庆卷+文科)_高考数学_可可英语
首页 - 手机APP下载 - 资料 - 查词 - 公众号
关注「可可英语」微信公众号，英语学习有趣又有
专项提升
听力训练
背单词
名著精读
情景会话
英语听力
英语听力
经济学人
分级读物
美文诗歌
广播
周刊
时报
外刊
英语口语
英语口语
精选播客
情景会话
背句子
英语考试
四级
六级
翻译
考研
BEC
专四
专八
职称
SAT
托福
雅思
GRE
基础
少儿
小学
中考
高考
综合
阅读
影视
歌曲
单词
网址
查词
翻译
下载
手机APP下载
iPhoneAndroid桌面版
高考
导航
基础英语
少儿英语：育儿经验儿歌视频学知识少儿游戏少儿口语少儿单词英文读物中文故事儿歌MP3品牌课程单词语法：单词教程语法教程在线背单词
中考英语：初中英语听力教材中考英语真题中考作文中考英语阅读中考英语语法中考数学中考语文中考物理中考化学中考词汇背诵中考作文批改
高考英语：高考资讯高考经验高考语文高考数学高考理科综合高考文科综合高考英语题库高考听力真题高中教材听力
考研频道：考研动态考研经验考研英语阅读考研写作考研翻译考研英语真题考研政治考研数学考研专业课真题考研作文批改考研词汇背诵
能力提升
英语听力：英语语音听力入门视频听力实战听力媒体资讯娱乐听力有声读物英语美文英文演讲可可之声品牌听力经济学人听写训练
英语口语：入门口语经验商务口语实战口语品牌口语行业口语俚语可可口语背句子口语秀情景会话每日英语
在线广播：VOA慢速VOA常速BBC广播CNN广播CRI广播NPR广播科学美国人ABC 新闻AP newsPBS高端访谈NBC晚间新闻
双语阅读：图文阅读时尚双语双语资讯故事散文名著小说海外文化原著下载
影视英语：影视资讯欧美电影美剧学习英剧学习影视学习新片佳片影视音乐
英文歌曲：听歌学英语音乐咖啡厅专题地带摇滚沸腾嘻哈态度动感节拍优雅情调乡土气息欧美MV翻唱歌曲Flash歌曲
英语考试
英语四级：四级资讯四级阅读四级词汇四级语法四级写作四级翻译完型填空四级听力四级题库资料下载四级听力训练单词背诵
英语六级：六级资讯六级词汇六级阅读六级写作六级试题六级翻译六级听力六级改错六级下载六级听力训练六级单词背诵
口译笔译：学习课程学习经验上海中级口译上海高级口译CattiMTI翻译硕士精选翻译素材口译资料下载在线翻译
雅思考试：雅思课程雅思经验雅思资讯雅思词汇雅思口语雅思听力雅思阅读雅思写作雅思机经雅思预测雅思词汇背诵雅思作文批改
托福考试：托福动态托福词汇托福语法托福备考经验托福机经托福预测托福听力托福口语托福写作托福阅读托福词汇背诵托福作文批改
其他综合
新概念英语：新概念第一册新概念第二册新概念第三册新概念第四册新概念资料下载
品牌课程：走遍美国老友记BBC新闻精讲经济学人双语版可可茶话会BBC纪录片精讲有声读物社交美语听歌学英语
高考资讯
志愿报考
高考英语
高考语文
高考数学
理科综合
文科综合
高考单词背诵
高考作文批改
资讯
高考录取分数线 | 状元访谈 | 高考录取查询 | 高考成绩查询
指南
专业排名 | 专业就业前景
英语
词汇 | 语法 | 高考听力 | 高考英语作文
语文
高考满分作文 | 名师视频
您现在的位置： 首页>高中英语>高考数学>高考数学真题>重庆高考数学真题> 正文
2011年高考数学真题附解析(重庆卷+文科)
时间:2014-09-25 14:06:32 来源:可可英语 编辑:Ookamie 可可英语APP下载|可可官方微信:ikekenet
字号：大 | 中 | 小
评论 打印收藏本文
一、选择题：本大题共10小题，每小题5分，共50分。在每小题给出的四个备选项中，只有一项是符合题目要求的。
3.曲线在点（1，2）处的切线方程为
(A) (B) (C) (D)
【命题意图】本题考查利用导数求函数的切线，是容易题.
【解析】∵=,∴切线斜率为3，则过（1,2）的切线方程为，即，故选A.
【答案】A
4.从一堆苹果中任取10只称得它的质量如下（单位：克）
125 120 122 105 130 114 116 95 120 134
则样本数据落在[114.5,124.5）内的频率为
(A)0.2 (B)0.3 (C)0.4 (D)0.5
6.设=,=,=,则,,的大小关系是
(A) ＜＜ (B) ＜＜ (C) ＜＜ (D) ＜＜
【命题意图】本题考查对数函数的图像与性质，是简单题.
【解析】∵与在（0，+∞）都是减函数，且0＜＜1,0＜＜1，
∴=＞0，=＞0，
又∵在（0，+∞）上是增函数，且0＜＜1，∴=＜0，即最小，只有B符合，故选B.
【答案】B
7.若函数=（＞2）在=处有最小值，则=
（A） （B） （C）3 （D）4
9.设双曲线的左准线与两条渐近线交于 两点，左焦点为在以才为之直径的圆内，则该双曲线的离心率的取值范围为
（A） （B） （C） （D）
【命题意图】本题考查双曲线的性质、点与圆的位置关系，考查学生转化与化归能力、解不等式能力，难度较大.
【解析】双曲线的左准线为=，渐近线方程为，联立解得（，），
∴=，根据题意得，＜，即，即，即，即，即，又＞1，,1＜＜，故选B.
【答案】B
10.高为的四棱锥的底面是边长为1的正方形，点、、、、均在半径为1的同一球面上，则底面的中心与顶点之间的距离为
（A） （B） （C） （D）
【命题意图】本题考查四棱锥与其外接球的相关知识，考查空间想象能力、转化化归能力以及运算求解能力，是难题.
【解析】如图，设四棱锥的外接球球心为，则⊥面，在中，=1，,∴=，
∵设四棱锥的高=，∴∥且=，
取的中点，连结，则四边形为矩形，∴⊥，=，在中，=1，则=，∴=，
在中，== ,故选A.
【答案】A
点击此处下载文档(rar格式，335.29KB)
来源:可可英语
上一页下一页下载《重庆高考数学真题》更多资料>>
保存到QQ日志登录QQ空间
下载该资料的人还下载了：
2014年高考数学真题附解析(重庆卷+理科)2014-08-10
2009年高考数学真题附答案(重庆卷+理科)2014-09-07
2013年高考数学真题附解析(重庆卷+文科)2014-10-04
2013年高考数学真题附解析(重庆卷+理科)2014-10-06
2012年高考数学真题附解析(重庆卷+文科)2014-10-18
文章关键字： 高考数学文科高考数学真题重庆高考
上一篇：2009年高考数学真题附答案(重庆卷+理科
下一篇：2013年高考数学真题附解析(重庆卷+文
发布评论我来说2句
特别推荐
· 小学英语课文点读免费下载
· 中高考英语智能提分神器
· 听说训练神器 省时更省力
最新文章
可可英语官方微信(微信号:ikekenet)
每天向大家推送短小精悍的英语学习资料.
添加方式1.扫描上方可可官方微信二维码。
添加方式2.搜索微信号 ikekenet 添加即可。
我们也在这里：
可可节目精选 Xue.kekenet.com
人群 --
小学生
中学生
大学生
留学生
上班族
专业人士
零基础
格式 --
视频
资讯
Flash
字幕
MP3
双语
主题 --
单词(37)
发音(10)
语法(7)
名著(15)
故事(12)
旅游(6)
求职(12)
公开课(17)
俚语(9)
美文
演讲
翻译
趣味娱乐(38)
有声读物(14)
媒体报道(88)
电影(29)
美剧(30)
音乐
听力 --
高级
初级
中级
口语 --
情景会话
入门口语
职场口语
商务口语
实战口语
面试口语
品牌 --
剑桥雅思
新概念
PBS
CRI
科学美国人
走遍美国
新东方
赖世雄
可可出品
BBC
VOA
NPR
APnews
CNN
李阳
经典
教材 --
少儿(20)
小学(31)
初中(49)
高中(11)
大学(29)
研究生(0)
考试 --
专四(16)
六级(20)
四级(43)
高考(23)
中考(30)
专八(1)
GRE(8)
雅思(7)
口译笔译(5)
BEC(9)
托福(15)
应用
查词
背单词
听写训练
口语秀
写作
每日英语
iphone版
安卓版
关于本站 - 联系我们 - 人才招聘 - 可可YY课堂 - 免责条款 - 意见反馈 - 网站导航 - 设为首页 - 加入收藏;)
Copyright @ 2005-2023 www.kekenet.com online services.
All rights reserved.
京ICP备11028623号-4京公网安备11010802011079号
APP建议订阅
可可小助手微信 微信扫一扫 获取精彩英语节目
可可英语APP 扫一扫 下载我 |
7758 | https://emedicine.medscape.com/article/384024-overview | For You
News & Perspective
Tools & Reference
Edition English
Medscape Editions
About You
Professional Information
Newsletters & Alerts
Your Watch List
Formulary Plan Manager
Log Out
Log In
Medscape Editions
English
Deutsch
Español
Français
Português
UK
Univadis from Medscape
About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out
Register Log In
close
Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel
processing....
Tools & Reference>Radiology
Parathyroid Adenoma Imaging
Updated: May 18, 2021
Author: J Levi Chazen, MD; Chief Editor: Eugene C Lin, MD more...;)
Share
Print
Feedback
Close
Facebook
Twitter
LinkedIn
WhatsApp
Email
Sections
Parathyroid Adenoma Imaging
Sections Parathyroid Adenoma Imaging
Practice Essentials
Computed Tomography
Magnetic Resonance Imaging
Ultrasonography
Nuclear Imaging
Angiography
Show All
Media Gallery;)
References;)
Practice Essentials
A parathyroid adenoma is a benign tumor on a parathyroid gland. In primary hyperparathyroidism due to adenomas, the normal feedback on parathyroid hormone production by extracellular calcium seems to be lost, resulting in a change in the set point. However, this is not the case in primary hyperparathyroidism from parathyroid hyperplasia; an increase in the cell numbers is probably the cause. In approximately 85-90% of cases, primary hyperparathyroidism is caused by a single adenoma. In 10-15% of cases, multiple glands are involved (ie, either multiple adenomas or hyperplasia). [1, 2, 3, 4]
Primary hyperparathyroidism (HPT) is a condition characterized by an inappropriate excess of parathyroid hormone (PTH) secretion. The elevated PTH levels result in hypercalcemia and hypophosphatemia, with associated medical comorbidities including calculus formation, bone and abdominal pain, polyuria, and depression. An estimated 100,000 new cases occur per year in the United States, with a 2.5:1 female-to-male predominance. Primary HPT is caused by a single parathyroid adenoma in 90% of patients and multigland disease in approximately 10%. Rarely, patients develop hyperparathyroidism secondary to a parathyroid carcinoma.
Imaging studies to detect parathyroid adenomas should be performed only after the diagnosis of primary HPT is established on the basis of biochemical findings. The diagnosis of primary hyperparathyroidism is made when hypercalcemia and elevated PTH levels are present. National Institutes of Health (NIH) guidelines recommend parathyroidectomy in all symptomatic patients and asymptomatic patients younger than 50 years, although some authors recommend more expansive criteria for surgical therapy. Preoperative localization studies are increasingly becoming the standard of care, as they enable minimally invasive surgical techniques. Many surgeons advocate for 2 concurrent examinations to definitively identify the site of disease, and almost all authors advocate for intraoperative parathyroid hormone level monitoring.
NICE (National Institute for Health and Care Excellence) recommends cervical US and suggests a second modality, such as MS, if additional guidance is needed for determining surgical approach. The American Association of Endocrine Surgeons (AAES) also recommend US, along with another imaging modality with high resolution for surgical planning. [1, 8, 9, 2]
Preoperative localization studies are also critical in patients with prior neck surgery or recurrent postoperative hypercalcemia. These patients can be particularly challenging for the imager because of distortion of normal anatomic landmarks and an increasing incidence of ectopic adenomas.
Noninvasive parathyroid imaging studies include technetium (Tc)-99m sestamibi scintigraphy, ultrasound, computed tomography (CT) scanning, magnetic resonance imaging (MRI), and positron emission tomography (PET) scanning. Tc-99m sestamibi is considered the most sensitive and specific imaging modality, particularly when used with single-photon emission CT (SPECT). PET with or without simultaneous CT scan (PET/CT) can be used but is an expensive imaging modality. [1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
Although more common in the past, invasive procedures such as parathyroid selective arteriography and/or selective parathyroid venous sampling are now rarely performed. Level III evidence suggests a role for preoperative imaging before minimally invasive parathyroidectomy. Ultrasonography and nuclear scintigraphy both have level II evidence as a first-line examination. The authors advocate parathyroid ultrasonography as a first-line test given the lack of ionizing radiation, low cost, and ease of use. Confirmatory testing with nuclear scintigraphy or, increasingly, multiphase CT, is often obtained if surgery is considered.
Next:
Computed Tomography
Multiphase CT has become the standard of care for diagnostically confounding cases or, at some institutions, for first-line preoperative localization given the high accuracy. The 4D-CT scan was originally described using 4 separate images phases, 1 noncontrast and 3 post contrast. [11, 18, 19] The authors recommend using a 3-phase technique to reduce the radiation dose. The first noncontrast images can be helpful to distinguish adenomas from the intrinsically dense iodine-rich thyroid gland, while the early and delayed postcontrast images highlight the hypervascular nature of adenomas and their characteristic early washout.
CT acquisition involves intravenous administration of nonionic iodinated contrast at a dose of 2 mL per kg of patient weight to a maximum of 120 mL. An intravenous infusion rate of 4 mL per second via an 18-gauge catheter works well for enhancement conspicuity. The authors recommend a 3-phase 4D-CT protocol with helical acquisition at 1.25-mm slice thickness from the bottom of the mandibular teeth down to the carina with noncontrast, early postcontrast, and delayed postcontrast imaging. A helical postcontrast sequence can be obtained 25 seconds after the start of intravenous contrast administration. Following a subsequent 30-second delay, a delayed-phase sequence can be acquired.
Prior to the intravenous administration of contrast material, parathyroid adenomas demonstrate attenuation similar to that of muscle. Parathyroid adenomas tend to be hypervascular structures with variable contrast enhancement and early washout. A hypervascular soft-tissue mass near the expected location of the parathyroid glands is considered to represent a parathyroid adenoma. Ectopic glands may be seen, most commonly within the mediastinum. A 4D-CT study is illustrated below with the characteristic hypodensity relative to normal thyroid, early avid enhancement, and washout.
Axial CT images in noncontrast (A) early post-contrast (B) and delayed post-contrast (C) phases demonstrate a subtle left tracheoesophageal groove lesion with characteristic early enhancement and washout. This case illustrates the sensitivity of 4D-CT for small adenomas.
View Media Gallery)
Intrathyroidal lesions may be subtle; detection is greatly improved by a multiphase technique, as in the image below. 4D-CT is sensitive for small adenomas that may elude other imaging techniques.
Axial CT images in noncontrast (A) early post-contrast (B) and delayed post-contrast (C) phases demonstrate an intrathyroidal lesion with subtle hypodensity on precontrast imaging and delayed enhancement. This enhancement pattern is seen less commonly than early enhancement and washout.
View Media Gallery)
A prominent feeding artery at the margin of a parathyroid adenoma (polar vessel), first described on ultrasound, may be seen in up to two thirds of parathyroid adenomas on high-quality 4D-CT studies. This finding can increase diagnostic confidence that a suspicious lesion is a parathyroid adenoma instead of a lymph node or thyroid nodule.
4D-DCT is increasingly being used for preoperative parathyroid localization and, in some studies, has been shown to be associated with a reduced number of glands explored and increased use of minimally invasive parathyroid surgery. [11, 18, 19] In a study by Jategaonker et al, the sensitivity and specificity of 4D-CT was 90.4% and 100% respectively, versus 63% sensitivity and 93.7% specificity for nuclear imaging studies and 41% sensitivity and 95% specificity for ultrasound.
CT is limited by the ionizing radiation required for the multiphase technique. However, newer dose-reduction techniques, including automatic tube current modulation, appear to be effective in reducing the radiation dose.
Most modern multiphase 4D-CT techniques report a sensitivity and specificity exceeding 90%. The 4D-CT scan appears superior to both ultrasound and nuclear studies in side-by-side comparisons, particularly when attempting to diagnose the precise site of disease before minimally invasive surgery.
In a study by Mekel et al of 27 patients who underwent parathyroidectomy for primary hyperthyroidism, 4D-CT identified a single adenoma in 26 patients, in whom a parathyroid adenoma was confirmed in surgery. Sensitivity was 81.4% and positive predictive value was 100% for 4D-CT in this study.
False-negative findings result from small or ectopic parathyroid glands, poor visualization of neck structures as a result of streak artifact or distorted neck anatomy due to prior surgery, and misinterpretation of a parathyroid adenoma for a thyroid nodule. Multigland disease, including 4-gland hyperplasia, can be particularly challenging to diagnose, even with modern CT technique.
Previous
Next:
Magnetic Resonance Imaging
A typical MRI protocol for detecting parathyroid adenoma involves the acquisition of images through the neck and mediastinum. Axial, coronal, and sagittal views are typically acquired. A surface neck coil is used to image the neck and a chest coil for the mediastinum.
Images are obtained from the hyoid bone to the lung apices by using T1- and T2-weighted spin-echo sequences. T2 acquisitions are typically fat-suppressed to increase conspicuity of the glands, with short tau inversion recovery (STIR), frequency-selective fat saturation, or DIXON-based fat-suppression techniques. A section thickness of 3 mm is typically acquired, with minimal intersection gap. Postcontrast T1 imaging is typically fat suppressed to increase enhancement conspicuity.
MRI findings
Normal parathyroid glands usually are not seen on MRI. Parathyroid adenomas are identified as soft-tissue masses in the expected location of the parathyroid glands. They characteristically demonstrate low T1 signal, avid enhancement, and high T2 signal. An axial STIR image of a parathyroid adenoma showing the characteristic T2 hyperintensity is shown below.
Axial STIR image showing a T2 hyperintense lesion in the left tracheoesophageal groove representing a parathyroid adenoma at surgery.
View Media Gallery)
As many as 30% of abnormal parathyroid glands do not have typical MRI signal intensity characteristics. Atypical patterns include high signal intensity on T1-weighted images and low-to-medium signal intensity on T2-weighted images, low signal intensity on both T1- and T2-weighted images, and high signal intensity on both T1- and T2-weighted images. Low signal intensity on both T1- and T2-weighted images reflects cellular degenerative changes, old hemorrhage with hemosiderin-laden macrophages, and fibrosis in the abnormal gland. High signal intensity on both T1- and T2-weighted images has been associated with hemorrhage without significant degenerative or fibrotic changes.
Fat suppression can be problematic in the neck, requiring special attention to technique. There are little data supporting advanced MR techniques such as MR perfusion or dynamic contrast enhanced imaging.
Sensitivity for MRI ranges appears significantly lower than that for CTin the 70% range. However, MRI appears especially useful in detecting ectopic mediastinal glands, with sensitivities exceeding 80%.
False-positive findings are reported to result from the misidentification of the following as parathyroid adenomas: enlarged lymph nodes, thyroid nodules (adenomas and/or exophytic colloid cysts), enlarged cervical ganglia, and other neck masses such as sarcoid nodules and neurofibromas. Enlarged lymph nodes have signal intensity characteristics similar to those of abnormal parathyroid glands. Abnormal parathyroid glands are expected to be medial to the carotid sheath, whereas lymph nodes are most frequently situated around or lateral to the sheath.
False-negative findings most commonly result from small parathyroid glands. The mean volume of detected abnormal glands has been reported as 3.5 cm3, while the mean volume of missed glands is 1.4 cm3. Other reported false-negative findings result from concomitant thyroid disease, anatomic distortion due to prior surgery, ectopic glands (especially intrathyroidal glands), and atypical signal intensity characteristics.
In a study of 11 patients with parathyroid adenoma in whom 4D-MRI was used, parathyroid adenomas were identified in 10 patients. In 9 patients, there was an exact match as compared to ultrasound and sestamibi scan (MIBI). Sensitivity of 4D-MRI was 90% (100% after optimization), and specificity was 100%.
Previous
Next:
Ultrasonography
The use of high-frequency 12- or 15-MHz transducers has become the standard for interrogating the parathyroid glands. The patient should be supine with his or her neck hyperextended. The examination should proceed from the carotid bifurcation superiorly to the sternal notch inferiorly and the carotid artery/internal jugular vein laterally. [30, 31]
NICE (National Institute for Health and Care Excellence) recommends cervical US and suggests a second modality, such as MS, if additional guidance is needed for determining surgical approach. The American Association of Endocrine Surgeons (AAES) also recommend US, along with another imaging modality with high resolution for surgical planning. [1, 8, 9, 2]
Normal-sized parathyroid glands are usually not visualized with ultrasound. On gray-scale images, parathyroid adenomas appear as a discrete, oval, anechoic or hypoechoic masses located posterior to the thyroid gland, anterior to the longus colli muscles, and, frequently, medial to the common carotid artery.
An echogenic line that separates the thyroid gland from the enlarged parathyroid gland can usually be seen. Larger adenomas are more likely to have cystic change, lobulations, increased echogenicity due to fatty deposition, and occasional calcifications.
Color Doppler ultrasound has been used to localize enlarged parathyroid glands. Parathyroid adenomas tend to be hypervascular lesions. An extrathyroidal artery may lead to a parathyroid adenoma in up to 83% of patients. An extrathyroidal feeding vessel may provide a roadmap to an otherwise inconspicuous gland. Characteristically, the extrathyroidal artery enters at one pole of the gland (polar arterya finding that can occasionally be appreciated on 4D-CT as well). (See the image below.) However, color Doppler sonograms of parathyroid adenomas may not show increased vascularity until the lesions are 1 cm in size.
Ultrasound imaging with color Doppler interrogation showing a hypoechoic lesion posterior to the thyroid gland with a polar vessel sign (arrow).
View Media Gallery)
The sensitivity of ultrasound in detecting parathyroid adenomas has been reported from 55 to 83%. Ultrasound is especially limited in the mediastinum because of a poor or absent acoustic window. The specificity of ultrasound in detecting parathyroid adenomas ranges from 40 to 98%.
In a study of 146 patients with parathyroid adenoma (88% single gland, 7% multigland, and 5% negative explorations) by Touska et al, the sensitivity and specificity of SPECT/CT-guided ultrasound were 83% and 96%, respectively, with sensitivity being higher for single gland (87%) than multigland disease (70%).
False-positive findings result when thyroid nodules, enlarged lymph nodes, the esophagus, longus colli muscles, and perithyroid veins are mistaken for enlarged parathyroid glands.
False-negative studies result from small parathyroid glands; ectopic locations lacking an adequate acoustic window; and poor visualization of neck structures due to previous surgery, thyromegaly, or body habitus.
Localization of adenomas in the mediastinum is limited because of the lack of an acoustic window and the difficulty in visualizing structures posterior to the air-filled trachea and esophagus. If an intrathyroidal lesion is detected, the lesion cannot confidently be differentiated as a parathyroid adenoma or thyroid nodule. Aspiration biopsy is required.
Previous
Next:
Nuclear Imaging
Radiopharmaceutical agents
Nuclear medicine imaging of the parathyroid glands was introduced in the late 1970s using thallium-201 (201Tl) as an imaging agent. Subsequently, the 201Tltechnetium-99m (99mTc)-pertechnetate (thallium-pertechnetate) subtraction method was described, which became the first widely accepted method for radionuclide imaging of the parathyroid glands.
Thyroid tissue takes up both 201Tl and 99mTc-pertechnetate. Abnormal parathyroid tissue, such as parathyroid adenomas, hyperplastic parathyroid glands, and parathyroid carcinoma, takes up 201Tl but not 99mTc-pertechnetate. Separate 201Tl and 99mTc-pertechnetate images are obtained in a single session without moving the patient, and a subtraction image is generated. An example of a dual-tracer acquisition with 99mTc-pertechnetate and 99mTc sestamibi is shown below.
Planar acquisition from dual-tracer 99mTc pertechnetate (upper left) and 99mTc sestamibi (upper right) imaging protocol showing subtracted image (lower right) with a right lower pole parathyroid adenoma.
View Media Gallery)
In the 1990s, 99mTc sestamibi was introduced for parathyroid imaging for both subtraction and single-agent, dual-phase technique. 99mTc tetrofosmin was subsequently discovered to have imaging characteristics similar to 99mTc sestamibi. The discussion focusd on 99mTc sestamibi and 99mTc tetrofosmin, which have largely supplanted thallium in modern imaging protocols. [16, 17]
Technetium-99msestamibiimaging and technetium-99mtetrofosmin imaging
Technetium-99m sestamibi is taken up by both thyroid tissue and abnormal parathyroid tissue. [16, 17] A dual-phase single-agent technique is based on the differential washout of 99mTc sestamibi from thyroid tissue compared to abnormal parathyroid tissue. The rate of washout from abnormal parathyroid tissue, such as parathyroid adenoma, is much slower than that of normal thyroid tissue.
One significant difference between 99mTc tetrofosmin and 99mTc sestamibi is the differential washout of the radiotracer from the thyroid gland. The thyroid washout rates are slower for 99mTc tetrofosmin than for 99mTc sestamibi.
A typical protocol involves the intravenous injection of 20-25 mCi of 99mTc sestamibi or 99mTc tetrofosmin, followed by the acquisition of early and delayed images of the neck and upper thorax and/or mediastinum. The initial early image is obtained 10-15 minutes after the injection, and delayed imaging is obtained 1.5-3 hours after the injection. Single-photon emission CT (SPECT) imaging can be performed for improved anatomic localization. [38, 13, 14] A typical protocol for dual-phase scintigraphy with 99mTc sestamibi or 99mTc tetrofosmin follows :
Examine the patient's neck for palpable thyroid or other neck masses.
Place the patient in a supine position with his or her neck hyperextended under the gamma camera.
Set energy windows to 140 keV ± 28.
Intravenously inject 20-25 mCi of 99mTc sestamibi or 99mTc tetrofosmin.
Use a camera with a large field of view and a high-resolution collimator.
Obtain anterior images of the neck and upper thorax and/or mediastinum. Image acquisition time is 10 minutes.
Obtain early thyroid-phase images 10-15 minutes after the injection.
Obtain delayed parathyroid-phase images 1.5-3 hours after the injection.
Optional imaging procedures include the following:
Right and left anterior oblique views can be obtained.
If nodules are palpable, delayed images can be obtained after 5-6 hours.
99mTc pertechnetate can be injected after delayed imaging.
Sodium iodide I-123 images may be obtained in a second examination on the next day.
SPECT can be performed for improved anatomic localization.
The agent may also be injected prior to minimally invasive parathyroid surgery on an outpatient basis, and a hand-held gamma probe can be used intraoperatively to guide the incision and localize the abnormal gland.
The sensitivity of the 99mTc sestamibi or 99mTc tetrofosmin techniques in detecting parathyroid adenomas has been reported ranging from 70 to 100%. Early SPECT imaging has a reported sensitivity of 91-96%.
In a study of 146 patients with parathyroid adenoma (88% single gland, 7% multigland, and 5% negative explorations) by Touska et al, the sensitivity and specificity of SPECT/CT-guided ultrasound were 83% and 96%, respectively, with sensitivity being higher for single gland (87%) than multigland disease (70%).
False-positive findings in 99mTc sestamibi and 99mTc tetrofosmin dual-phase imaging include uptake in coexisting thyroid nodules, which are interpreted as parathyroid adenomas. False-negative findings include parathyroid lesions, which are too small to be detected, and unusually rapid washout from a parathyroid adenoma.
(11)C-methionine positron emission tomography/computed tomography (Met-PET/CT), in a study by Lenschow et al, raised the rate of correctly localized single parathyroid adenomas in patients with negative cervical ultrasonography and MIBI-SPECT/CT and increased the number of focused surgical approaches. Cervical US localized a single parathyroid adenoma in 10 of 17 patients (59%), while MIBI-SPECT/CT identified 11 of 17 single adenomas (65%). In the remaining 6 patients, Met-PET/CT identified 5 single adenomas. This step-up approach correctly identified single adenomas in 16 od 17 patients (94%).
Previous
Next:
Angiography
Invasive examinations, such as parathyroid arteriography and parathyroid venous sampling, can be considered when the findings of noninvasive imaging modalities are nondiagnostic.
Angiographic and perfusion characteristics of parathyroid adenomas were first described in the 1970s and were used to identify abnormal hypervascular glands with preoperative digital subtraction angiography. A characteristic blush has been described following the injection of intraarterial contrast, indicating the presence of an abnormal gland that is hypervascular compared with normal thyroid and parathyroid tissue. Similar hypervascular and washout principles are applied to modern cross-sectional imaging interpretation.
Typically, the paired inferior thyroid arteries supply the parathyroid glands. However, parathyroid gland blood supply can also be derived from the superior thyroid arteries, small branches of the laryngeal and tracheoesophageal arteries, and, occasionally, a thyroidea ima artery. The superior thyroid arteries are branches of the external carotid arteries. The inferior thyroid arteries are branches of the thyrocervical trunk.
Digital subtraction angiography (DSA) and/or conventional arteriography can be used to localize a parathyroid adenoma. Selective arteriography of both thyrocervical trunks, both internal mammary arteries, and both common carotid arteries is recommended. The thyrocervical trunks are examined to assess intrathyroid glands, juxtathyroid glands, and glands that have descended into the superior mediastinum in the tracheoesophageal groove.
For ectopic mediastinal or thymic glands, the internal mammary arteries can be studied. The common carotid arteries are injected to detect parathymic or juxtathyroid glands. Occasionally, arteriography of the aortic arch, and possibly the innominate artery, is performed to search for a thyroidea ima artery if findings from the aforementioned selective angiograms are negative.
The risks of parathyroid arteriography are stroke and spinal cord injury. Arteriography of the common carotid arteries and, especially, superselective catheterization of the superior thyroid arteries, often require extensive manipulation of the guidewires and catheters in the region of the carotid bifurcation.
Selective venous sampling and parathyroid hormone measurements are performed to determine the general location of a parathyroid adenoma. A parathyroid arteriogram should be performed first because this serves as a guide or road map to the more variable parathyroid venous pathway. An end-hole catheter without side holes should be used to prevent the mixing of blood from adjacent veins. Sampling of small veins is the goal. Attempts should be made to sample the right and left thymic veins, inferior thyroid veins, and vertebral veins (if the middle and inferior thyroid veins were ligated in a previous operation). A peripheral vein sample is obtained, and a 2-fold gradient between the parathyroid hormone concentration in the sampled vein and that of the peripheral vein must be observed.
The sensitivity of DSA has been reported as 49% and that of parathyroid venous sampling ranges from 70-80%.
Previous
References
Pokhrel B, Levine SN. Primary Hyperparathyroidism. 2021 Jan. [QxMD MEDLINE Link]. [Full Text].
Wolfe SA, Sharma S. Parathyroid Adenoma. 2021 Jan. [QxMD MEDLINE Link]. [Full Text].
Fraker DL, Harsono H, Lewis R. Minimally invasive parathyroidectomy: benefits and requirements of localization, diagnosis, and intraoperative PTH monitoring. long-term results. World J Surg. 2009 Nov. 33 (11):2256-65. [QxMD MEDLINE Link].
Elaraj DM, Clark OH. Current status and treatment of primary hyperparathyroidism. Perm J. 2008 Winter. 12(1):32-7. [QxMD MEDLINE Link]. [Full Text].
National Institutes of Health. NIH conference. Diagnosis and management of asymptomatic primary hyperparathyroidism: consensus development conference statement. Ann Intern Med. 1991 Apr 1. 114(7):593-7. [QxMD MEDLINE Link].
Eigelberger MS, Cheah WK, Ituarte PH, Streja L, Duh QY, Clark OH. The NIH criteria for parathyroidectomy in asymptomatic primary hyperparathyroidism: are they too limited?. Ann Surg. 2004 Apr. 239(4):528-35. [QxMD MEDLINE Link]. [Full Text].
Grant CS, Thompson G, Farley D, van Heerden J. Primary hyperparathyroidism surgical management since the introduction of minimally invasive parathyroidectomy: Mayo Clinic experience. Arch Surg. 2005 May. 140(5):472-8; discussion 478-9. [QxMD MEDLINE Link].
Wilhelm SM, Wang TS, Ruan DT, et al. The American Association of Endocrine Surgeons Guidelines for Definitive Management of Primary Hyperparathyroidism. JAMA Surg. 2016 Oct 1. 151 (10):959-968. [QxMD MEDLINE Link].
Jawaid I, Rajesh S. Hyperparathyroidism (primary) NICE guideline: diagnosis, assessment, and initial management. Br J Gen Pract. 2020 Jul. 70 (696):362-363. [QxMD MEDLINE Link].
Smaxwil C, Aschoff P, Reischl G, et al. [18F]fluoro-ethylcholine-PET Plus 4D-CT (FEC-PET-CT): A Break-Through Tool to Localize the "Negative" Parathyroid Adenoma. One Year Follow Up Results Involving 170 Patients. J Clin Med. 2021 Apr 13. 10 (8):[QxMD MEDLINE Link].
Jategaonkar AA, Lerner DK, Cooke P, et al. Implementation of a 4-dimensional computed tomography protocol for parathyroid adenoma localization. Am J Otolaryngol. 2021 May-Jun. 42 (3):102907. [QxMD MEDLINE Link].
Jackson R, Chew D, McClean S, England J. Factors related to a non-localising technetium 99m sestamibi scan result during parathyroid adenoma imaging in primary hyperparathyroidism. Clin Otolaryngol. 2021 Mar. 46 (2):357-362. [QxMD MEDLINE Link].
Raruenrom Y, Theerakulpisut D, Wongsurawat N, Somboonporn C. Diagnostic accuracy of planar, SPECT, and SPECT/CT parathyroid scintigraphy protocols in patients with hyperparathyroidism. Nucl Med Rev Cent East Eur. 2018. 21 (1):20-25. [QxMD MEDLINE Link].
Touska P, Elstob A, Rao N, Parthipun A. SPECT/CT-Guided Ultrasound for Parathyroid Adenoma Localization: A 1-Stop Approach. J Nucl Med Technol. 2019 Mar. 47 (1):64-69. [QxMD MEDLINE Link].
Czarnecki CA, Einsiedel PF, Phal PM, et al. Dynamic CT for Parathyroid Adenoma Detection: How Does Radiation Dose Compare With Nuclear Medicine?. AJR Am J Roentgenol. 2018 May. 210 (5):1118-1122. [QxMD MEDLINE Link].
Nafisi Moghadam R, Amlelshahbaz AP, Namiranian N, et al. Comparative Diagnostic Performance of Ultrasonography and 99mTc-Sestamibi Scintigraphy for Parathyroid Adenoma in Primary Hyperparathyroidism; Systematic Review and Meta- Analysis. Asian Pac J Cancer Prev. 2017 Dec 28. 18 (12):3195-3200. [QxMD MEDLINE Link].
Wong KK, Fig LM, Gross MD, Dwamena BA. Parathyroid adenoma localization with 99mTc-sestamibi SPECT/CT: a meta-analysis. Nucl Med Commun. 2015 Apr. 36 (4):363-75. [QxMD MEDLINE Link].
Matthews C, Matthews A, Safavi K. Clinical Images: Four-Dimensional Computed Tomography-Future of Preoperative Parathyroid Adenoma Imaging. Ochsner J. 2017 Fall. 17 (3):220-222. [QxMD MEDLINE Link].
Leong D, Ng K, Boeddinghaus R, Lisewski D. Three-phase four-dimensional computed tomography as a first-line investigation in primary hyperparathyroidism. ANZ J Surg. 2021 May 12. [QxMD MEDLINE Link].
Phillips CD, Shatzkes DR. Imaging of the parathyroid glands. Semin Ultrasound CT MR. 2012 Apr. 33(2):123-9. [QxMD MEDLINE Link].
Chazen JL, Gupta A, Dunning A, Phillips CD. Diagnostic accuracy of 4D-CT for parathyroid adenomas and hyperplasia. AJNR Am J Neuroradiol. 2012 Mar. 33(3):429-33. [QxMD MEDLINE Link].
Beland MD, Mayo-Smith WW, Grand DJ, Machan JT, Monchik JM. Dynamic MDCT for localization of occult parathyroid adenomas in 26 patients with primary hyperparathyroidism. AJR Am J Roentgenol. 2011 Jan. 196(1):61-5. [QxMD MEDLINE Link].
Bahl M, Muzaffar M, Vij G, Sosa JA, Choudhury KR, Hoang JK. Prevalence of the Polar Vessel Sign in Parathyroid Adenomas on the Arterial Phase of 4D CT. AJNR Am J Neuroradiol. 2013 Aug 14. [QxMD MEDLINE Link].
Rodgers SE, Hunter GJ, Hamberg LM, et al. Improved preoperative planning for directed parathyroidectomy with 4-dimensional computed tomography. Surgery. 2006 Dec. 140(6):932-40; discussion 940-1. [QxMD MEDLINE Link].
Mekel M, Linder R, Bishara B, Kluger Y, Bar-On O, Fischer D. [4-dimensional computed tomography for localization of parathyroid adenoma]. Harefuah. 2013 Dec. 152 (12):710-2, 753. [QxMD MEDLINE Link].
Johnson NA, Tublin ME, Ogilvie JB. Parathyroid imaging: technique and role in the preoperative evaluation of primary hyperparathyroidism. AJR Am J Roentgenol. 2007 Jun. 188(6):1706-15. [QxMD MEDLINE Link].
Auffermann W, Guis M, Tavares NJ, Clark OH, Higgins CB. MR signal intensity of parathyroid adenomas: correlation with histopathology. AJR Am J Roentgenol. 1989 Oct. 153(4):873-6. [QxMD MEDLINE Link].
Stevens SK, Chang JM, Clark OH, Chang PJ, Higgins CB. Detection of abnormal parathyroid glands in postoperative patients with recurrent hyperparathyroidism: sensitivity of MR imaging. AJR Am J Roentgenol. 1993 Mar. 160(3):607-12. [QxMD MEDLINE Link].
Merchavy S, Luckman J, Guindy M, Segev Y, Khafif A. 4D MRI for the Localization of Parathyroid Adenoma: A Novel Method in Evolution. Otolaryngol Head Neck Surg. 2015 Nov 23. [QxMD MEDLINE Link].
Leupe PK, Delaere PR, Vander Poorten VL, Debruyne F. Pre-operative imaging in primary hyperparathyroidism with ultrasonography and sestamibi scintigraphy. B-ENT. 2011. 7(3):173-80. [QxMD MEDLINE Link].
Gooding GA. Sonography of the thyroid and parathyroid. Radiol Clin North Am. 1993 Sep. 31(5):967-89. [QxMD MEDLINE Link].
Hopkins CR, Reading CC. Thyroid and parathyroid imaging. Semin Ultrasound CT MR. 1995 Aug. 16(4):279-95. [QxMD MEDLINE Link].
Johnson NA, Yip L, Tublin ME. Cystic parathyroid adenoma: sonographic features and correlation with 99mTc-sestamibi SPECT findings. AJR Am J Roentgenol. 2010 Dec. 195(6):1385-90. [QxMD MEDLINE Link].
Lane MJ, Desser TS, Weigel RJ, Jeffrey RB Jr. Use of color and power Doppler sonography to identify feeding arteries associated with parathyroid adenomas. AJR Am J Roentgenol. 1998 Sep. 171(3):819-23. [QxMD MEDLINE Link].
Young AE, Gaunt JI, Croft DN, Collins RE, Wells CP, Coakley AJ. Location of parathyroid adenomas by thallium-201 and technetium-99m subtraction scanning. Br Med J (Clin Res Ed). 1983 Apr 30. 286(6375):1384-6. [QxMD MEDLINE Link]. [Full Text].
Coakley AJ, Kettle AG, Wells CP, O'Doherty MJ, Collins RE. 99Tcm sestamibi--a new agent for parathyroid imaging. Nucl Med Commun. 1989 Nov. 10(11):791-4. [QxMD MEDLINE Link].
Lind P. Parathyroid imaging with technetium-99m labelled cationic complexes: which tracer and which technique should be used?. Eur J Nucl Med. 1997 Mar. 24(3):243-5. [QxMD MEDLINE Link].
Wong KK, Fig LM, Gross MD, Dwamena BA. Parathyroid adenoma localization with 99mTc-sestamibi SPECT/CT: a meta-analysis. Nucl Med Commun. 2015 Apr. 36 (4):363-75. [QxMD MEDLINE Link].
Taillefer R. 99m Tc-sestamibi parathyroid scintigraphy. Nuclear Medicine Annual. 1995. 51-75.
Lenschow C, Gassmann P, Wenning C, Senninger N, Colombo-Benkmann M. Preoperative ¹¹C-methionine PET/CT enables focused parathyroidectomy in MIBI-SPECT negative parathyroid adenoma. World J Surg. 2015 Jul. 39 (7):1750-7. [QxMD MEDLINE Link].
Miller DL, Doppman JL, Krudy AG, et al. Localization of parathyroid adenomas in patients who have undergone surgery. Part II. Invasive procedures. Radiology. 1987 Jan. 162(1 Pt 1):138-41. [QxMD MEDLINE Link].
Miller DL. Endocrine angiography and venous sampling. Radiol Clin North Am. 1993 Sep. 31(5):1051-67. [QxMD MEDLINE Link].
Dillenberger S, Bartsch DK, Maurer E, Kann PH. Single Centre Experience in Patients with Primary Hyperparathyroidism: Sporadic, Lithium-associated and in Multiple Endocrine Neoplasia. Exp Clin Endocrinol Diabetes. 2020 Oct. 128 (10):693-698. [QxMD MEDLINE Link].
Media Gallery
Axial CT images in noncontrast (A) early post-contrast (B) and delayed post-contrast (C) phases demonstrate a subtle left tracheoesophageal groove lesion with characteristic early enhancement and washout. This case illustrates the sensitivity of 4D-CT for small adenomas.
Axial CT images in noncontrast (A) early post-contrast (B) and delayed post-contrast (C) phases demonstrate an intrathyroidal lesion with subtle hypodensity on precontrast imaging and delayed enhancement. This enhancement pattern is seen less commonly than early enhancement and washout.
Axial STIR image showing a T2 hyperintense lesion in the left tracheoesophageal groove representing a parathyroid adenoma at surgery.
Ultrasound imaging with color Doppler interrogation showing a hypoechoic lesion posterior to the thyroid gland with a polar vessel sign (arrow).
Planar acquisition from dual-tracer 99mTc pertechnetate (upper left) and 99mTc sestamibi (upper right) imaging protocol showing subtracted image (lower right) with a right lower pole parathyroid adenoma.
of 5
Tables
Back to List
Contributor Information and Disclosures
Author
J Levi Chazen, MD Assistant Attending Physician, Department of Radiology, Division of Neuroradiology, New York-Presbyterian Hospital; Assistant Professor of Radiology, Weill Cornell Medical CollegeDisclosure: Nothing to disclose.
Coauthor(s)
C Douglas Phillips, MD, FACR Director of Head and Neck Imaging, Division of Neuroradiology, New York-Presbyterian Hospital; Professor of Radiology, Weill Cornell Medical CollegeC Douglas Phillips, MD, FACR is a member of the following medical societies: American College of Radiology, American Medical Association, American Society of Head and Neck Radiology, American Society of Neuroradiology, Association of University Radiologists, Radiological Society of North AmericaDisclosure: Nothing to disclose.
Specialty Editor Board
Bernard D Coombs, MB, ChB, PhD Consulting Staff, Department of Specialist Rehabilitation Services, Hutt Valley District Health Board, New ZealandDisclosure: Nothing to disclose.
C Douglas Phillips, MD, FACR Director of Head and Neck Imaging, Division of Neuroradiology, New York-Presbyterian Hospital; Professor of Radiology, Weill Cornell Medical CollegeC Douglas Phillips, MD, FACR is a member of the following medical societies: American College of Radiology, American Medical Association, American Society of Head and Neck Radiology, American Society of Neuroradiology, Association of University Radiologists, Radiological Society of North AmericaDisclosure: Nothing to disclose.
Chief Editor
Eugene C Lin, MD Attending Radiologist, Teaching Coordinator for Cardiac Imaging, Radiology Residency Program, Virginia Mason Medical Center; Clinical Assistant Professor of Radiology, University of Washington School of MedicineEugene C Lin, MD is a member of the following medical societies: American College of Nuclear Medicine, American College of Radiology, Radiological Society of North America, Society of Nuclear Medicine and Molecular ImagingDisclosure: Nothing to disclose.
Acknowledgements
Hussein M Abdel-Dayem, MD Professor of Radiology, New York Medical College
Disclosure: Nothing to disclose.
Fred S Mishkin, MD Professor of Radiology in Residence, University of California at Los Angeles School of Medicine; Director, Division of Nuclear Medicine, UCLA-Harbor Medical Center
Fred S Mishkin, MD is a member of the following medical societies: American College of Angiology, American College of Radiology, American Heart Association, American Medical Association, California Medical Association, and Society of Nuclear Medicine
Disclosure: Nothing to disclose.
Glenda L Romero-Urquhart, MD Diagnostic Radiologist, West Coast Radiology Center
Glenda L Romero-Urquhart, MD is a member of the following medical societies: American College of Radiology
Disclosure: Nothing to disclose.
Panukorn Vasinrapee, MD Assistant Clinical Professor of Radiology, University of California at Los Angeles School of Medicine; Associate Chief, Nuclear Medicine, Department of Radiology, Harbor-UCLA Medical Center
Panukorn Vasinrapee, MD is a member of the following medical societies: American Roentgen Ray Society, Radiological Society of North America, and Society of Nuclear Medicine
Disclosure: Nothing to disclose.
Close;)
What would you like to print?
What would you like to print?
Print this section
Print the entire contents of
Print the entire contents of article
Sections Parathyroid Adenoma Imaging
Practice Essentials
Computed Tomography
Magnetic Resonance Imaging
Ultrasonography
Nuclear Imaging
Angiography
Show All
Media Gallery;)
References;)
encoded search term (Parathyroid Adenoma Imaging) and Parathyroid Adenoma Imaging
What to Read Next on Medscape
Related Conditions and Diseases
Neuroendocrine Tumors Guidelines
Carcinoid Tumor
Fast Five Quiz: Neuroendocrine Tumors
Fast Five Quiz: Managing Neuroendocrine Tumors
Carcinoid Lung Tumors
Systemic Treatment of Metastatic Gastroenteropancreatic Neuroendocrine Tumors
Thymic Tumors
News & Perspective
Drug Combo for Common Neuroendocrine Tumor: A New Standard?
Cabozantinib Improves PFS in Advanced Neuroendocrine Tumors After Prior Treatment
FDA OKs Cabozantinib for Pancreatic Neuroendocrine Tumors
Molecular Marker Testing for Evaluation of Thyroid Nodules
Breast Cancer e-Tumor Boards: Case 5: Metastatic High-Grade Neuroendocrine Carcinoma
A Practical Approach to Managing Neuroendocrine Tumors
Drug Interaction Checker
Pill Identifier
Calculators
Formulary
Pancreatic Cancer: Difficult Diagnosis, Ominous Outlook
2002 2500010-overview Diseases & Conditions
Diseases & Conditions Neuroendocrine Tumors Guidelines
2003 /viewarticle/1002665 Clinical Review
Advanced Neuroendocrine Tumors: Integrating Emerging Targeted Therapies Into Patient Care 0.75 CME Credits Clinical Review
You are being redirected to Medscape Education
Yes, take me there
0.75 CME
Advanced Neuroendocrine Tumors: Integrating Emerging Targeted Therapies Into Patient Care
2010 surufatinib-4000239 Drugs
Drugs surufatinib
2010 netspot-gallium-ga-68-dotatate-1000115 Drugs
Drugs gallium Ga 68 dotatate |
7759 | https://math.stackexchange.com/questions/2383729/arithmetic-sequences-in-finite-set | number theory - Arithmetic Sequences in Finite Set - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Arithmetic Sequences in Finite Set
Ask Question
Asked 8 years, 1 month ago
Modified8 years, 1 month ago
Viewed 329 times
This question shows research effort; it is useful and clear
3
Save this question.
Show activity on this post.
Consider a finite set S⊂N≥1 which contains no arithmetic sequences of length greater than l. For each arithmetic sequence of length l contained in S, place the greatest element of this arithmetic sequence in a new set, which we call ¯S. For what values of l is it possible for |S∖¯S|<|¯S|
For l=3, the set A={1,17,41,52,69,79,81,86,87,89,92,93,94} (which has no arithmetic sequences of length 4) has ¯A={81,86,87,89,92,93,94}, so the inequality can be satisfied. However, it seems like finding such a set should not be possible for l≥4, but I can't find a nice proof.
number-theory
elementary-set-theory
arithmetic-progressions
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Aug 7, 2017 at 17:28
RSpecielRSpeciel
asked Aug 5, 2017 at 18:15
RSpecielRSpeciel
2,559 20 20 silver badges 39 39 bronze badges
0
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
There exists a set S with the properties you want for every value of l. Here are a few examples for the smaller values of l:
For l=2: |S∖ˉ S|=1<|ˉ S|=2 S={1,2,4} ˉ S={2,4}
For l=3: |S∖ˉ S|=3<|ˉ S|=4 S={1,2,3,5,6,8,9} ˉ S={3,5,8,9}
For l=4: |S∖ˉ S|=6<|ˉ S|=7 S={1,2,3,4,6,7,8,9,11,12,13,14,16} ˉ S={4,8,9,12,13,14,16}
For l=5: |S∖ˉ S|=16<|ˉ S|=17 S={1,2,3,4,5,7,8,9,10,12,13,14,15,17,18,19,20,22,23,24,25,26,33,34,35,36,37,39,43,44,45,46,47} ˉ S={5,9,13,17,22,23,24,25,26,33,34,37,43,44,45,46,47}
For l=6: |S∖ˉ S|=15<|ˉ S|=16 S={1,2,3,4,5,6,8,9,10,11,12,13,15,16,17,18,19,20,22,23,24,25,26,27,29,30,31,32,33,34,36} ˉ S={6,12,13,18,19,20,24,25,26,27,30,31,32,33,34,36}
For l=7: |S∖ˉ S|=78<|ˉ S|=79 , max(S)=239
For l=8: |S∖ˉ S|=109<|ˉ S|=110 , max(S)=335
For l=9: |S∖ˉ S|=213<|ˉ S|=214 , max(S)=633
For l=10: |S∖ˉ S|=45<|ˉ S|=46 , max(S)=100
As the sets become rather large I left out the lasts few examples. These are the first sets I encountered in the following procedure.
We start for the chosen value of l with the set S={1,2,…,l−1} which does not contain an arithmetic sequence of length l. (Optional ˉ S={l−1}). We also use an index n=l. Now we can use the following algorithm:
Add n to the set S
Check wether there is any arithmetic sequence of length l in S that ends in n. If that is the case we remove n from S, if not we have a new set that does not contain any arithmetic sequences of length l or longer.
Optional: If n was accepted we can check whether there is any arithmetic sequence of length l−1 in S that ends in n. If that is so, we add n to ˉ S.
Increase n by one and go back to step 1.
After that it is only repeating until the number of elements in ˉ S exceeds half the size of S.
I have not tried to prove the existence of such a set for any value of l, but I see no obvious reason why they would seize to exist for larger values of l. It should also be noted that there might very well be sets with the required properties that are smaller in size and of which the largest element in S is less than the algorithm above gives.
It is, however, possible to construct a set for any value of l in a systematic fashion. Hereto we start with the following sets: S 0=1
ˉ S 0=∅ and apply the following iterative scheme: S n+1=∪l−1 k=0[S n+k Δ n]S n+1=∪l−2 k=0[ˉ S n+k Δ n]∪[S n+(l−1)Δ n] where Δ n is not unique but should be large enough to avoid any problems with generating additional sequences. It is sufficient to choose Δ n≥2 max(S n−1)−1.
We find that |S n|=l n
|ˉ S n|=l n−(l−1)n and if we choose Δ n=(2 l−1)n−1 the largest element in S n is given by max(S n)=(2 l−1)n+1 2.
From this it follows that for |S n∖ˉ S n|<|ˉ S n| we require (l−1)n−ln 2 ln(1−1 l)
.
For the case l=3 we would obtain the set S={1,2,3,6,7,8,11,12,13}
and for l=4 we get S={1,2,3,4,8,9,10,11,15,16,17,18,22,23,24,25,50,51,52,53,57,58,59,60,64,65,66,67,71,72,73,74,99,100,101,102,106,107,108,109,113,114,115,116,120,121,122,123,148,149,150,151,155,156,157,158,162,163,164,165,169,170,171,172}.
And finally, we get that |ˉ S n||S n|=1−(1−1 l)n
and hence lim n→∞|ˉ S n||S n|=1.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Aug 6, 2017 at 8:23
answered Aug 5, 2017 at 23:54
Ronald BlaakRonald Blaak
3,317 12 12 silver badges 14 14 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
number-theory
elementary-set-theory
arithmetic-progressions
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
2Is this set of integer sequences countable?
1A Question Regarding Diagonalization
7Sequences avoiding three-term arithmetic progressions
6Showing cardinality of all infinite sequences of natural numbers is the same as the continuum.
2Erdos conjecture about Fat sequences?
0cardinality of the set of all rational sequences taking exactly one value between two integers
1Intersections and Finite Unions of Closed Sets (Proof Verification)
3If the set of specific sequences of 0's and 1's is countable then it can be used to count the set of all infinite sequences of 0's and 1's. Right?
Hot Network Questions
Are there any world leaders who are/were good at chess?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
How different is Roman Latin?
Program that allocates time to tasks based on priority
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
The geologic realities of a massive well out at Sea
How to locate a leak in an irrigation system?
How to start explorer with C: drive selected and shown in folder list?
Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-?
Why are LDS temple garments secret?
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
How do you emphasize the verb "to be" with do/does?
What were "milk bars" in 1920s Japan?
How to home-make rubber feet stoppers for table legs?
Discussing strategy reduces winning chances of everyone!
I have a lot of PTO to take, which will make the deadline impossible
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
Can a cleric gain the intended benefit from the Extra Spell feat?
ICC in Hague not prosecuting an individual brought before them in a questionable manner?
Languages in the former Yugoslavia
Implications of using a stream cipher as KDF
Why multiply energies when calculating the formation energy of butadiene's π-electron system?
Do we need the author's permission for reference
Why include unadjusted estimates in a study when reporting adjusted estimates?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7760 | https://www.cuemath.com/numbers/composite-numbers/ | LearnPracticeDownload
Composite Numbers
Composite numbers are numbers with more than two factors. Numbers can be classified on the basis of the number of factors that they have. If a number has just two factors - 1 and the number itself, then it is a prime number. However, most numbers have more than two factors, and they are called composite numbers. On this page, we will learn the difference between prime and composite numbers, the smallest composite number, and odd composite numbers. The last one is interesting because there are several odd composite numbers, unlike 2, which is the only even prime number.
| | |
--- |
| 1. | What are Composite Numbers? |
| 2. | How to Find Composite Numbers? |
| 3. | Properties of Composite Numbers |
| 4. | Types of Composite Numbers |
| 5. | Smallest Composite Number |
| 6. | FAQs on Composite Numbers |
What are Composite Numbers?
A number that is divisible by a number apart from 1 and the number itself, is called a composite number. For example, 4 is a number that is divisible by 1, 2, and 4, therefore, it is a composite number.
Composite Number Definition
Composite numbers can be defined as natural numbers that have more than two factors.
Meaning of Composite Number
Composite number means a number that is divisible by smaller positive numbers apart from 1 and itself. For example, 6 is a composite number because it is divisible by 1, 2, 3, and 6. However, 19 is not a composite number because it is divisible by only two numbers 1 and 19 itself. There is no other number that divides 19, so, 19 is a prime number. Let us learn more about composite numbers with examples.
Examples of Composite Numbers
4, 6, 8, 9, and 10 are the first few composite numbers. Let us take 6 and 8. In the above example, 6 and 8 are called composite numbers because they have more than 2 factors. Let us proceed to understand the important properties of composite numbers.
How to Find Composite Numbers?
In order to find a composite number, we find the factors of the given number. If the number has more than two factors, then it is composite. The best way to figure out a composite number is to perform the divisibility test. The divisibility test helps us to determine whether a number is a prime or a composite number. Divisibility means that a number is divided completely (with no remainder) by another number, other than 1 and itself.
To do this, we need to check to see if the number can be divided by these common factors: 2, 3, 5, 7, 11, and 13. If the given number is even, then start checking with the number 2. If the number ends with a 0 or 5, then check it by 5. If the number cannot be divided by any of these given numbers, then the number is a prime number. For example, 68 is divisible by 2, which means it has factors other than 1 and 68, so, we can say 68 is a composite number.
Properties of Composite Numbers
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Note the properties of a composite number listed below:
All composite numbers are completely divisible by smaller numbers that can be prime or composite.
Every composite number is made up of two or more prime numbers.
Let us have a look at the properties of the composite number 72 in order to understand the concept in a better way.
This figure shows that by multiplying these positive integers we get a composite number.
First 10 Composite Numbers
The first 10 composite numbers can be listed as 4, 6, 8, 9, 10, 12, 14, 15, 16, and 18. All these numbers have factors other than 1 and the number itself. If we start counting from 1, we know that 1 is neither a prime number nor a composite number. Then 2 and 3 are prime numbers, so we start with 4 as the first composite number. In the same way, if we asked to list the first five composite numbers, we can list them as follows:
First Five Composite Numbers
The first five composite numbers are 4, 6, 8, 9, and 10.
Types of Composite Numbers
The two main types of composite numbers in math are odd composite numbers and even composite numbers. Let us have a look at the two of them individually:
Odd Composite Numbers
All the odd numbers which are not prime are odd composite numbers. For example, 9, 15, 21, 25, 27 are odd composite numbers. Consider the numbers 1, 2, 3, 4, 9, 10, 11, 12 and 15. Here 9 and 15 are the odd composite numbers because these two odd numbers have more than 2 factors and they fulfill the condition of composite numbers.
Even Composite Numbers
All the even numbers which are not prime are even composite numbers. For example, 4, 6, 8, 10, 12, 14, and 16, are even composite numbers. Consider the numbers 1, 2, 3, 4, 9, 10, 11, 12, and 15 again. Here 4, 10, and 12 are the even composite numbers because these even numbers have more than 2 factors and they fulfill the condition of composite numbers.
Smallest Composite Number
The smallest composite number is 4. A composite number is defined as a number that has divisors other than 1 and the number itself. Now, as we start counting: 1, 2, 3, 4, 5, 6, .... so on, we see that 1 is not a composite number because its only divisor is 1. And 2 is not a composite number because it has only two divisors, i.e., 1 and the number 2 itself. 3 is not a composite number because it has only two divisors, i.e., 1 and the number 3 itself. However, when we come at number 4, we know that its divisors are 1, 2, and 4. So, the number 4 satisfies the criteria of a composite number. Therefore, 4 is the smallest composite number.
Difference Between Prime and Composite Numbers
A number that is divisible by a number other than 1 and the number itself, is called a composite number. This means that composite numbers have more than 2 factors. For example, 4 and 6 are composite numbers. Whereas, a number that is divisible only by 1 and itself is called a prime number, like, 2, 3, and 5. This means that a prime number has only 2 factors, 1 and the number itself.
☛ Related Articles
Real Numbers
Whole Numbers
Rational Numbers
Irrational Numbers
Prime and Composite Numbers Worksheets
Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12. Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs.
Composite Numbers Examples
Example 1: Which of the following is a composite number?
a.) 34
b.) 31
c.) 39
Solution:
a.) 34 has factors more than 1 and itself, factors of 34 = 1, 2, 17, 34. So, it is a composite number.
b.) 31 has only two factors, factors of 31 = 1, 31. So, it is a prime number.
c.) 39 has factors more than 1 and itself, factors of 39 = 1, 3, 13, 39. So, it is a composite number.
Therefore, 34 and 39 are composite numbers.
2. Example 2: Fill in the blanks:
a.) The smallest composite number is __.
b.) The smallest odd composite number is __.
Solution:
a.) The smallest composite number is 4.
b.) The smallest odd composite number is 9.
3. Example 3: State true or false with respect to composite numbers.
a.) All even numbers are composite numbers.
b.) 1 is a composite number.
Solution:
a.) False, 2 is an even number but it is a prime number. Therefore, all even numbers are not composite numbers.
b.) False, 1 is neither a prime number nor a composite number.
Show Solution >
go to slidego to slidego to slide
Have your child solve real-life challenges using math
Have your child apply concepts learned in school in the real world with the help of our experts.
Book a Free Trial Class
Practice Questions on Composite Numbers
Try these >
go to slidego to slide
FAQs on Composite Numbers
What are Composite Numbers in Math?
Composite numbers are those numbers that have more than two factors. In other words, composite numbers have factors other than 1 and itself. For example, the number 6 is a composite number because it has 1, 2, 3, and 6 as its factors.
How to Find Composite Numbers?
To find whether a given number is composite, we just find the number of factors it has. If it has more than 2 factors, then the number is composite. For example, 4 has three factors, 1, 2, and 4. So, it is a composite number.
Is 2 a Composite Number?
No, 2 is not a composite number, it is the only even prime number. A composite number has more than two factors, other than 1 and the number itself. Here, 2 has only two factors, 1 and 2 itself. Therefore, 2 is a prime number.
What are the Composite Numbers Between 1 and 100?
The composite numbers between 1 and 100 are,
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100.
What are Prime and Composite Numbers?
A number that is divisible by a number other than 1 and the number itself, is called a composite number. For example, 4 and 6 are composite numbers. Whereas, a number that is divisible only by 1 and itself is called a prime number, like, 2, 3, and 5.
What is an Odd Composite Number?
All the odd numbers (excluding 1) which are not prime are odd composite numbers. For example, 9 and 15 are the odd composite numbers.
What is the Smallest Composite Number?
A composite number has more than two factors, other than 1 and the number itself. Here, 4 satisfies this condition, and there is no other number smaller than 4 which is composite. Therefore, 4 is the smallest composite number.
What is the Difference Between a Prime and a Composite Number?
A prime number has only two factors, 1 and the number itself, whereas, a composite number has more than two factors. For example, 5 has only two factors, 1 and 5, so it is a prime number. Whereas 6 has four factors, 1, 2, 3, and 6, so it is a composite number.
What are Consecutive Composite Numbers?
Consecutive composite numbers are those composite numbers that continuously follow each other without any prime number in between. For example, the first few consecutive composite numbers can be listed as 4, 6, 8, 9 10, and so on.
How many Composite Numbers are there from 1 to 100?
There are 74 composite numbers from 1 to 100. They can be listed as follows, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100.
What Composite Numbers are Between 6 and 11?
The composite numbers between 6 and 11 are 8, 9 and 10 because these numbers have more than 2 factors. The rest of the numbers between 6 and 11 are prime numbers, i.e., 7 is a prime number.
Which of the Factors of 120 are Composite Numbers?
The factors of 120 can be listed as, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120. Among these, the composite numbers are 4, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
What are the First Four Composite Numbers?
The first four composite numbers can be listed as 4, 6, 8, and 9. These numbers are called composite numbers because they have more than two factors.
Which is the Smallest Odd Composite Number?
The smallest odd composite number is 9 because if we start checking the composite numbers from 4, 6 and 8, we see that these are even numbers. Hence the smallest odd composite number is 9.
Which is the First Composite Number?
The first composite number is 4 because if we start checking the natural numbers we get 1 which is neither prime nor composite, then 2 and 3 are prime numbers. Therefore, the first composite number is 4.
What are the First Five Composite Numbers?
The first five composite numbers can be listed as 4, 6, 8, 9, and 10. These numbers have more than 2 factors, therefore they are called composite numbers. The natural numbers before 4 are 1, 2 and 3 in which we know that 1 is neither prime nor composite, then 2 and 3 are prime numbers. Therefore, the first five composite numbers are 4, 6, 8, 9, and 10.
Download FREE Study Materials
Download FREE Concept E-book
Download FREE Tips and Tricks
Download FREE Worksheets
Math worksheets andvisual curriculum
FOLLOW CUEMATH
Facebook
Youtube
Instagram
Twitter
LinkedIn
Tiktok
MATH PROGRAM
Online math classes
Online Math Courses
online math tutoring
Online Math Program
After School Tutoring
Private math tutor
Summer Math Programs
Math Tutors Near Me
Math Tuition
Homeschool Math Online
Solve Math Online
Curriculum
NEW OFFERINGS
Coding
SAT
Science
English
MATH ONLINE CLASSES
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
ABOUT US
Our Mission
Our Journey
Our Team
QUICK LINKS
Maths Games
Maths Puzzles
Our Pricing
Math Questions
Events
MATH TOPICS
Algebra 1
Algebra 2
Geometry
Calculus math
Pre-calculus math
Math olympiad
MATH TEST
Math Kangaroo
AMC 8
MATH CURRICULUM
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
FOLLOW CUEMATH
Facebook
Youtube
Instagram
Twitter
LinkedIn
Tiktok
MATH PROGRAM
Online math classes
Online Math Courses
online math tutoring
Online Math Program
After School Tutoring
Private math tutor
Summer Math Programs
Math Tutors Near Me
Math Tuition
Homeschool Math Online
Solve Math Online
Curriculum
NEW OFFERINGS
Coding
SAT
Science
English
MATH CURRICULUM
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
MATH TEST
CAASPP
CogAT
STAAR
NJSLA
SBAC
Math Kangaroo
AMC 8
ABOUT US
Our Mission
Our Journey
Our Team
MATH TOPICS
Algebra 1
Algebra 2
Geometry
Calculus math
Pre-calculus math
Math olympiad
Numbers
Measurement
QUICK LINKS
Maths Games
Maths Puzzles
Our Pricing
Math Questions
Blogs
Events
FAQs
MATH ONLINE CLASSES
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
Terms and ConditionsPrivacy Policy |
7761 | https://pubchem.ncbi.nlm.nih.gov/compound/Hydrogen | Hydrogen | H2 | CID 783 - PubChem
An official website of the United States government
Here is how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
NIH National Library of Medicine NCBI
PubChem
About
Docs
Submit
Contact
Search PubChem
compound Summary
Hydrogen
PubChem CID
783
Structure
Primary Hazards
Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula
H 2
Synonyms
Hydrogen
Molecular hydrogen
Dihydrogen
1333-74-0
hydrogen-1
View More...
Molecular Weight
2.016 g/mol
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Element Name
Hydrogen
Dates
Create: 2004-09-16
Modify: 2025-09-13
Description
Hydrogen is a colorless, odorless gas. It is easily ignited. Once ignited it burns with a pale blue, almost invisible flame. The vapors are lighter than air. It is flammable over a wide range of vapor/air concentrations. Hydrogen is not toxic but is a simple asphyxiate by the displacement of oxygen in the air. Under prolonged exposure to fire or intense heat the containers may rupture violently and rocket. Hydrogen is used to make other chemicals and in oxyhydrogen welding and cutting.
CAMEO Chemicals
Hydrogen, refrigerated liquid (cryogenic liquid) appears as a colorless very cold liquid shipped in special heavily insulated containers. Boils at -442 °F. Contact causes severe frostbite. Hydrogen gas at normal temperatures is lighter than air but until vapors boiling off from the liquid warm up they are heavier than air. Very easily ignited. Burns with a pale blue, almost invisible flame. A leak could be either a vapor or a liquid leak. The container is designed to vent the vapor as it is boils off. Under prolonged exposure to intense heat the containers may rupture violently and rocket.
CAMEO Chemicals
Dihydrogen is an elemental molecule consisting of two hydrogens joined by a single bond. It has a role as an antioxidant, an electron donor, a fuel, a human metabolite and a member of food packaging gas. It is an elemental hydrogen, a gas molecular entity and an elemental molecule.
ChEBI
View More...
See also: Hydrogen, refrigerated liquid (cryogenic liquid) (narrower).
1 Structures
1.1 2D Structure
Structure Search
Get Image
Download Coordinates
Chemical Structure Depiction
Full screen Zoom in Zoom out
PubChem
2 Names and Identifiers
2.1 Computed Descriptors
2.1.1 IUPAC Name
molecular hydrogen
Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14)
PubChem
2.1.2 InChI
InChI=1S/H2/h1H
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.3 InChIKey
UFHFLCQGNIYNRP-UHFFFAOYSA-N
Computed by InChI 1.07.2 (PubChem release 2025.04.14)
PubChem
2.1.4 SMILES
[HH]
Computed by OEChem 2.3.0 (PubChem release 2025.04.14)
PubChem
2.2 Molecular Formula
H 2
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; EU Food Improvement Agents; PubChem
H 2
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3 Other Identifiers
2.3.1 CAS
1333-74-0
Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; DHS Chemical Facility Anti-Terrorism Standards (CFATS) Chemicals of Interest; DrugBank; EFSA OpenFoodTox; EPA Chemical Data Reporting (CDR); EPA Chemicals under the TSCA; EPA DSSTox; European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB); Human Metabolome Database (HMDB); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA); NJDOH RTK Hazardous Substance List
12184-96-2
CAS Common Chemistry
View More...
2.3.2 Deprecated CAS
725200-57-7
ChemIDplus; EPA Chemicals under the TSCA
2.3.3 European Community (EC) Number
215-605-7
EU Food Improvement Agents; European Chemicals Agency (ECHA)
2.3.4 UNII
7YNJ3PO35Z
FDA Global Substance Registration System (GSRS)
2.3.5 UN Number
1049 (HYDROGEN)
CAMEO Chemicals
1966 (HYDROGEN, REFRIGERATED LIQUID (CRYOGENIC LIQUID))
CAMEO Chemicals; Emergency Response Guidebook (ERG)
1049 (Hydrogen, compressed)
Emergency Response Guidebook (ERG)
1049
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3.6 ChEBI ID
CHEBI:18276
ChEBI
2.3.7 DrugBank ID
DB15127
DrugBank
2.3.8 DSSTox Substance ID
DTXSID9029643
EPA DSSTox
2.3.9 HMDB ID
HMDB0001362
Human Metabolome Database (HMDB)
2.3.10 ICSC Number
0001
ILO-WHO International Chemical Safety Cards (ICSCs)
2.3.11 KEGG ID
C00282
KEGG
2.3.12 NSC Number
356464
DTP/NCI
2.3.13 Wikidata
Q3027893
Wikidata
Q902508
Wikidata
2.3.14 Wikipedia
Hydrogen
Wikipedia
2.4 Synonyms
2.4.1 MeSH Entry Terms
Hydrogen
Hydrogen-1
Protium
Medical Subject Headings (MeSH)
2.4.2 Depositor-Supplied Synonyms
Hydrogen
Molecular hydrogen
Dihydrogen
1333-74-0
hydrogen-1
DTXSID9029643
7YNJ3PO35Z
CHEBI:18276
Hydrogen atoms
RefChem:389053
DTXCID00196437
215-605-7
H
Protium
Atomic hydrogen
Hydrogen gas
H2
1-Butanol, titanium(4+) salt, monohydrate, homopolymer (9CI)
Hydrogen, atomic
hidrogeno
hydrogene
Wasserstoff
Cellulose powder
o-Hydrogen
p-Hydrogen
hydrogen molecule
monoatomic hydrogen
3A molecular sieve
HYDROGEN [HSDB]
alpha-Cellulose, 90mum
HYDROGEN [MI]
T101 Monoclonal Antibody
HYDROGEN [WHO-DD]
COware 20ml with H-Caps
Hydrogen, >=99.99%
Hydrogen, >=99.999%
SCHEMBL29349524
SCHEMBL29357126
CHEBI:18140
CHEBI:33251
YZCKVEUIGOORGS-UHFFFAOYSA-N
BBL103822
NSC356464
STL557632
alpha-Cellulose, 25mum particle size
alpha-Cellulose, 65mum particle size
AKOS016038407
NSC-356464
UN 1049
UN 1966
E949
Hydrogen, Messer(R) CANgas, 99.999%
E 949
E-949
NS00080750
C00282
alpha-Cellulose, Partical size: d50,180-280mum
alpha-Cellulose, Partical size: d50,90-150mum
Q3027893
1-Butanol,titanium(4+)salt,monohydrate,homopolymer(9ci)
1H
H-H
PubChem
3 Chemical and Physical Properties
3.1 Computed Properties
Property Name
Property Value
Reference
Property Name
Molecular Weight
Property Value
2.016 g/mol
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
XLogP3-AA
Property Value
0
Reference
Computed by XLogP3 3.0 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Donor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Hydrogen Bond Acceptor Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Rotatable Bond Count
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Exact Mass
Property Value
2.0156500638 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Monoisotopic Mass
Property Value
2.0156500638 Da
Reference
Computed by PubChem 2.2 (PubChem release 2025.04.14)
Property Name
Topological Polar Surface Area
Property Value
0 Ų
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Heavy Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Formal Charge
Property Value
0
Reference
Computed by PubChem
Property Name
Complexity
Property Value
0
Reference
Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14)
Property Name
Isotope Atom Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Atom Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Defined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Undefined Bond Stereocenter Count
Property Value
0
Reference
Computed by PubChem
Property Name
Covalently-Bonded Unit Count
Property Value
1
Reference
Computed by PubChem
Property Name
Compound Is Canonicalized
Property Value
Yes
Reference
Computed by PubChem (release 2025.04.14)
PubChem
3.2 Experimental Properties
3.2.1 Physical Description
Hydrogen is a colorless, odorless gas. It is easily ignited. Once ignited it burns with a pale blue, almost invisible flame. The vapors are lighter than air. It is flammable over a wide range of vapor/air concentrations. Hydrogen is not toxic but is a simple asphyxiate by the displacement of oxygen in the air. Under prolonged exposure to fire or intense heat the containers may rupture violently and rocket. Hydrogen is used to make other chemicals and in oxyhydrogen welding and cutting.
CAMEO Chemicals
Hydrogen, refrigerated liquid (cryogenic liquid) appears as a colorless very cold liquid shipped in special heavily insulated containers. Boils at -442 °F. Contact causes severe frostbite. Hydrogen gas at normal temperatures is lighter than air but until vapors boiling off from the liquid warm up they are heavier than air. Very easily ignited. Burns with a pale blue, almost invisible flame. A leak could be either a vapor or a liquid leak. The container is designed to vent the vapor as it is boils off. Under prolonged exposure to intense heat the containers may rupture violently and rocket.
CAMEO Chemicals
Gas or Vapor; Dry Powder, Water or Solvent Wet Solid; Liquid; Gas or Vapor, Liquid
EPA Chemical Data Reporting (CDR)
Colourless, odourless, highly flammable gas
EU Food Improvement Agents
Colorless gas; [HSDB] Vapor density = 0.069 (lighter than air); [HSDB]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Liquid
Human Metabolome Database (HMDB)
ODOURLESS COLOURLESS COMPRESSED GAS.
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.2 Color / Form
Colorless gas
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-66
Hazardous Substances Data Bank (HSDB)
... hydrogen under ordinary conditions consists of two kinds of molecules, known as ortho- and para-hydrogen, which differ from one another by the spins of their electrons and nuclei. Normal hydrogen at room temperature contains 25% of the para form and 75% of the ortho form. The ortho form cannot be prepared in the pure state
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-17
Hazardous Substances Data Bank (HSDB)
3.2.3 Odor
Odorless
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 12th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2012., p. V4: 22431
Hazardous Substances Data Bank (HSDB)
3.2.4 Taste
Tasteless
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 12th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2012., p. V4: 2431
Hazardous Substances Data Bank (HSDB)
3.2.5 Boiling Point
-423 °F at 760 mmHg (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
-259.16 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-66
Hazardous Substances Data Bank (HSDB)
-253Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.6 Melting Point
-434 °F (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
-252.762 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-66
Hazardous Substances Data Bank (HSDB)
-259.2 °C
Human Metabolome Database (HMDB)
-259Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.7 Flash Point
Flammable gas
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.8 Solubility
In water, 1.62 mg/L at 21 °C
Venable CS, Fuwa T; Ind Eng Chem 14: 139-42 (1922)
Hazardous Substances Data Bank (HSDB)
Soluble in about 50 volumes of water at 0 degC
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
Very low solubility in most liquids
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 12th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2012., p. V4: 2431
Hazardous Substances Data Bank (HSDB)
0.00162 mg/mL at 21 °C
Human Metabolome Database (HMDB)
Solubility in water, mg/l at 21Â °C: 1.62 (very poor)
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.9 Density
0.071 at -423.4 °F (USCG, 1999) - Less dense than water; will float
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
0.082 g/L
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-66
Hazardous Substances Data Bank (HSDB)
Density: 0.069 (gas) (Air= 1); 0.0700 at BP (liq); 0.0763 at 13 K (solid). One liter of gas at 0 °C weighs 0.08987 g
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
Physical and Thermodynmic Properties Gaseous Hydrogen
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology. (1999-2017). New York, NY: John Wiley & Sons. Online Posting Date: 20 Dec 2001
Hazardous Substances Data Bank (HSDB)
Physical and Thermodynamic Properties of Solid Hydrogen
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology. (1999-2017). New York, NY: John Wiley & Sons. Online Posting Date: 20 Dec 2001
Hazardous Substances Data Bank (HSDB)
3.2.10 Vapor Density
Relative vapor density (air = 1): 0.07
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.11 Vapor Pressure
1.24X10+6 mm Hg at 25 °C
Ohe S; Computer Aided Data Book of Vapor Pressure. Data Book Publ. Co, Tokyo, Japan (1976)
Hazardous Substances Data Bank (HSDB)
Vapor pressure, kPa at 25Â °C: 165320
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.12 Stability / Shelf Life
Stable under recommended storage conditions.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
3.2.13 Autoignition Temperature
1065 °F (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
932 °F (500 °C)
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 325-72
Hazardous Substances Data Bank (HSDB)
560Â °C
ILO-WHO International Chemical Safety Cards (ICSCs)
3.2.14 Corrosivity
Noncorrosive
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
3.2.15 Heat of Combustion
-285.8 kJ/mol
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 5-68
Hazardous Substances Data Bank (HSDB)
3.2.16 Heat of Vaporization
0.90 kJ/mol at -252.87 °C
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 6-129
Hazardous Substances Data Bank (HSDB)
3.2.17 Other Experimental Properties
Atomic number 1; valence 1; elemental state: H2; isotopes: (1)H (protium 99.9844%), (2)H (deuterium 0.0156%), (3)H (tritium, traces only)
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
MP: -259.2 deg (13.96 K) at 54 mm (triple point); BP: -252.77 (20.39 K)
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
Ionization potential of H atom is 13.59 electron volts; 1 L of gas at 0 °C weighs 0.08987 g
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
Lightest of all gases; lifting power of 1 cu ft hydrogen gas is about 0.076 lb at 0 °C, 760 mm Hg
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-17
Hazardous Substances Data Bank (HSDB)
... Hydrogen gas under ordinary conditions is a mix of 2 kinds of molecules, known as ortho- & para-hydrogen, which differ from one another by the spins of their electrons & nuclei. Normal hydrogen at room temp contains 25% of para form & 75% ortho form. The ortho form cannot be prepd in the pure state.
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 94th Edition. CRC Press LLC, Boca Raton: FL 2013-2014, p. 4-18
Hazardous Substances Data Bank (HSDB)
In 1973, it was reported that a group of Russian experimenters may have produced metallic hydrogen at a pressure of 2.8 Mbar... earlier, in 1972, a Livermore (California) group also reported on a similar experiment in which they observed a pressure-volume point centered at 2 Mbar. It has been predicted that metallic hydrogen may be metastable; others have predicted it would be a superconductor at room temperature.
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 94th Edition. CRC Press LLC, Boca Raton: FL 2013-2014, p. 4-17
Hazardous Substances Data Bank (HSDB)
For more Other Experimental Properties (Complete) data for Hydrogen (15 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
3.3 SpringerMaterials Properties
Auger electron spectroscopy
Fourier transform infrared spectrum
Gibbs energy
Absorbance
Activation energy
Adsorbate coverage
Atomic radius
Avoided-level-crossing muon spin resonance
Band structure
Binding energy
Boiling point
Cell voltage
Chemical diffusion
Composition
Compressibility
Corrosion
Critical point
Crystal structure
Crystal structure type
Crystallization temperature
Density
Dielectric constant
Dielectricity
Diffusion
Diffusive flux
Dispersion
Elasticity
Electrode potential
Electromagnetic induction
Electron conductivity
Electron spin resonance
Enthalpy
Enthalpy change
Entropy
Excess enthalpy
Exchange current density
Fluorescence
Formation energy
Formation enthalpy
Formation entropy
Formula unit
Geothermal energy content
Heat capacity
Heat of combustion
Heat of solution
High frequency properties
Kinetic properties
Mass transfer coefficient
Melting curve
Migration energy
Miscibility gap
Mixing enthalpy
Nuclear magnetic resonance
Oxygen content
Phase diagram
Phase equilibrium
Phase transition
Phonon dispersion
Preexponential factor
Quasielastic neutron scattering
Reflection high-energy electron diffraction
Scanning electron microscopy
Self-diffusion
Sintering
Sound absorption
Sound propagation
Sound velocity
Space group
Spheroidization
Surface acoustic wave
Surface diffusion
Surface structure
Thermal grooving
Transition enthalpy
Transition entropy
Transmission electron microscopy
Unit cell
Unit cell parameter
Virial coefficient
Viscosity
Volumetric capacity
SpringerMaterials
3.4 Chemical Classes
Toxic Gases & Vapors -> Simple Asphyxiants
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Flammable agents - 4th degree
NJDOH RTK Hazardous Substance List
3.4.1 Food Contact Substances
FCS -> FDA Inventory of Food Contact Substances Listed in 21 CFR
FDA Packaging & Food Contact Substances (FCS)
4 Spectral Information
4.1 Mass Spectrometry
4.1.1 GC-MS
1 of 2 items
NIST Number
245692
Library
Main library
Total Peaks
2
m/z Top Peak
2
m/z 2nd Highest
1
m/z 3rd Highest
0
Thumbnail
NIST Mass Spectrometry Data Center
2 of 2 items
Source of Spectrum
Chemical Concepts, A Wiley Division, Weinheim, Germany
Copyright
Copyright © 2002-2025 Wiley-VCH GmbH. All Rights Reserved.
Thumbnail
SpectraBase
5 Related Records
5.1 Related Compounds with Annotation
Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here.
PubChem
5.2 Related Compounds
Same Connectivity Count
8
Mixtures, Components, and Neutralized Forms Count
319386
Similar Compounds (2D)
View in PubChem Search
Similar Conformers (3D)
View in PubChem Search
PubChem
5.3 Related Element
Element Name
Hydrogen
Element Symbol
H
Atomic Number
1
PubChem Elements
5.4 Substances
5.4.1 PubChem Reference Collection SID
504687494
PubChem
5.4.2 Related Substances
All Count
342660
Same Count
207
Mixture Count
342453
PubChem
5.4.3 Substances by Category
PubChem
5.5 Other Relationships
Hydrogen, refrigerated liquid (cryogenic liquid) (narrower)
PubChem
5.6 Entrez Crosslinks
PubMed Count
6
Protein Structures Count
1
Taxonomy Count
4
Gene Count
25
PubChem
6 Chemical Vendors
PubChem
7 Drug and Medication Information
7.1 Clinical Trials
7.1.1 ClinicalTrials.gov
ClinicalTrials.gov
7.1.2 NIPH Clinical Trials Search of Japan
NIPH Clinical Trials Search of Japan
7.2 Therapeutic Uses
/EXPL THER/ Hydrogen-rich saline (HRS) is a novel protection against various oxidative disorders and almost all types of inflammation. Moreover, its toxicity and side effects are rarely reported. We sought to clarify the protective effect of HRS against the oxygen-induced retinopathy (OIR) in C57BL/6 J model. The OIR in the HRS treated mice and the untreated controls were systematically compared. The retinas of both groups were analyzed using high-molecular-weight FITC-dextran staining of flat-mount preparations, hematoxylin and eosin (H&E) staining of cross-sections. The distribution and expression of the vascular endothelial growth factor (VEGF) were also evaluated by the immunohistochemical measurements between postnatal days 17 (P17) and P21. The leakage and non-perfusion areas of retinal blood vessels were not alleviated in the HRS treatment group. Moreover, the number of preretinal vascular endothelial cell in the HRS treatment group was similar to that in the untreated group after exposure to hyperoxia (P>0.05). The degree of OIR was positively correlated with the expression level of VEGF. Intriguingly, the preretinal vascular endothelial cell count in the retinas of pups reared in room air with HRS treatment was 15.21+/-2.98. The preretinal vascular endothelial cell count of the HRS treated mice was significantly higher than that of the untreated group reared in room air. In summary, HRS therapy (at the dose of 10mL/day, applied between P12 and P17) did not inhibit retinal neovascularization in OIR; On the contrary, it would induce the retinal neovascularization during the development of normal retinas. /Hydrogen-rich saline/
PMID:27091652
Zhang Q et al; Life Sci 153: 17-22 (2016)
Hazardous Substances Data Bank (HSDB)
/EXPL THER/ Acute pancreatitis (AP) is an inflammatory disease mediated by damage to acinar cells and pancreatic inflammation. In patients with AP, subsequent systemic inflammatory responses and multiple organs dysfunction commonly occur. Interactions between cytokines and oxidative stress greatly contribute to the amplification of uncontrolled inflammatory responses. Molecular hydrogen (H2) is a potent free radical scavenger that not only ameliorates oxidative stress but also lowers cytokine levels. The aim of the present study was to investigate the protective effects of H2 gas on AP both in vitro and in vivo. For the in vitro assessment, AR42J cells were treated with cerulein and then incubated in H2-rich or normal medium for 24 hr, and for the in vivo experiment, AP was induced through a retrograde infusion of 5% sodium taurocholate into the pancreatobiliary duct (0.1 mL/100 g body weight). Wistar rats were treated with inhaled air or 2% H2 gas and sacrificed 12 hr following the induction of pancreatitis. Specimens were collected and processed to measure the amylase and lipase activity levels; the myeloperoxidase activity and production levels; the cytokine mRNA expression levels; the 8-hydroxydeoxyguanosine, malondialdehyde, and glutathione levels; and the cell survival rate. Histological examinations and immunohistochemical analyses were then conducted. The results revealed significant reductions in inflammation and oxidative stress both in vitro and in vivo. Furthermore, the beneficial effects of H2 gas were associated with reductions in AR42J cell and pancreatic tissue damage. In conclusion, our results suggest that H2 gas is capable of ameliorating damage to the pancreas and AR42J cells and that H2 exerts protective effects both in vitro and in vivo on subjects with AP. Thus, the results obtained indicate that this gas may represent a novel therapy agent in the management of AP.
PMID:27115738
Full text:
Zhou HX et al; PLoS One 11 (4): e0154483 (2016)
Hazardous Substances Data Bank (HSDB)
/EXPL THER/ Hydrogen gas inhalation (HI) ameliorates cerebral and cardiac dysfunction in animal models of post-cardiac arrest syndrome (PCAS). HI for human patients with PCAS has never been studied. Between January 2014 and January 2015, 21 of 107 patients with out-of-hospital cardiac arrest achieved spontaneous return of circulation. After excluding 16 patients with specific criteria, 5 patients underwent HI together with target temperature management (TTM). No undesirable effects attributable to HI were observed and 4 patients survived 90 days with a favorable neurological outcome. HI in combination with TTM is a feasible therapy for patients with PCAS.
PMID:27334126
Tamura T et al; Circ J. 80 (8): 1870-3 (2016)
Hazardous Substances Data Bank (HSDB)
/EXPL THER/ Premature ovarian failure (POF) is a disease that affects female fertility but has few effective treatments. Ovarian reserve function plays an important role in female fertility. Recent studies have reported that hydrogen can protect male fertility. Therefore, we explored the potential protective effect of hydrogen-rich water on ovarian reserve function through a mouse immune POF model. To set up immune POF model, fifty female BALB/c mice were randomly divided into four groups: Control (mice consumed normal water, n = 10), hydrogen (mice consumed hydrogen-rich water, n = 10), model (mice were immunized with zona pellucida glycoprotein 3 [ZP3] and consumed normal water, n = 15), and model-hydrogen (mice were immunized with ZP3 and consumed hydrogen-rich water, n = 15) groups. After 5 weeks, mice were sacrificed. Serum anti-Mullerian hormone (AMH) levels, granulosa cell (GC) apoptotic index (AI), B-cell leukemia/lymphoma 2 (Bcl-2), and BCL2-associated X protein (Bax) expression were examined. Analyses were performed using SPSS 17.0 (SPSS Inc., Chicago, IL, USA) software. Immune POF model, model group exhibited markedly reduced serum AMH levels compared with those of the control group (5.41 +/- 0.91 ng/ml vs. 16.23 +/- 1.97 ng/mL, P = 0.033) and the hydrogen group (19.65 +/- 7.82 ng/mL, P = 0.006). The model-hydrogen group displayed significantly higher AMH concentrations compared with that of the model group (15.03 +/- 2.75 ng/mL vs. 5.41 +/- 0.91 ng/mL, P = 0.021). The GC AI was significantly higher in the model group (21.30 +/- 1.74%) than those in the control (7.06 +/- 0.27%), hydrogen (5.17 +/- 0.41%), and model-hydrogen groups (11.24 +/- 0.58%) (all P < 0.001). The GC AI was significantly higher in the model-hydrogen group compared with that of the hydrogen group (11.24 +/- 0.58% vs. 5.17 +/- 0.41%, P = 0.021). Compared with those of the model group, ovarian tissue Bcl-2 levels increased (2.18 +/- 0.30 vs. 3.01 +/- 0.33, P = 0.045) and the Bax/Bcl-2 ratio decreased in the model-hydrogen group. Hydrogen-rich water may improve serum AMH levels and reduce ovarian GC apoptosis in a mouse immune POF model induced by ZP3. /Hydrogen-rich water/
PMID:27647193
Full text:
He X et al; Chin Med J (Engl) 129 (19): 2331-7 (2016)
Hazardous Substances Data Bank (HSDB)
For more Therapeutic Uses (Complete) data for Hydrogen (24 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
8 Food Additives and Ingredients
8.1 FDA Generally Recognized as Safe - GRAS Notices
GRAS Notice Number(GRN)
520
Substance
Hydrogen gas
Intended Use
For use as an ingredient in drinking water, flavored beverages, and soda drinks at a level up to 2.14% by volume.
Basis
Scientific Procedures
FDA Letter
FDA has no questions
Notifier
MitoGene Beverage Company, LLC
Date of Filing
06/02/2014
FDA Generally Recognized as Safe (GRAS)
8.2 FDA Food Contact Substances (FCS)
Substance
HYDROGEN
Document Number (21 eCFR)
176.180
176.210
177.2800
FDA Packaging & Food Contact Substances (FCS)
8.3 Associated Foods
FooDB
9 Minerals
Name
hydrogen
Crystal Structure Data Link
American Mineralogist Crystal Structure Database
RRUFF Project
10 Pharmacology and Biochemistry
10.1 Absorption, Distribution and Excretion
Molecular hydrogen (H2) is an agent with potential applications in oxidative stress-related and/or inflammatory disorders. H2 is usually administered by inhaling H2-containing air (HCA) or by oral intake of H2-rich water (HRW). Despite mounting evidence, the molecular mechanism underlying the therapeutic effects and the optimal method of H2 administration remain unclear. Here, we investigated whether H2 affects signaling pathways and gene expression in a dosage- or dose regimen-dependent manner. We first examined the H2 concentrations in blood and organs after its administration and found that oral intake of HRW rapidly but transiently increased H2 concentrations in the liver and atrial blood, while H2 concentrations in arterial blood and the kidney were one-tenth of those in the liver and atrial blood. In contrast, inhalation of HCA increased H2 equally in both atrial and arterial blood ...
PMID:25707580
Sobue S et al; Mol Cell Biochem 403 (1-2): 231-41 (2015)
Hazardous Substances Data Bank (HSDB)
Hydrogen exerts beneficial effects in disease animal models of ischemia-reperfusion injury as well as inflammatory and neurological disease. Additionally, molecular hydrogen is useful for various novel medical and therapeutic applications in the clinical setting. In the present study, the hydrogen concentration in rat blood and tissue was estimated. Wistar rats were orally administered hydrogen super-rich water (HSRW), intraperitoneal and intravenous administration of hydrogen super-rich saline (HSRS), and inhalation of hydrogen gas. A new method for determining the hydrogen concentration was then applied using ... sensor gas chromatography, after which the specimen was prepared via tissue homogenization in airtight tubes. This method allowed for the sensitive and stable determination of the hydrogen concentration. The hydrogen concentration reached a peak at 5 minutes after oral and intraperitoneal administration, compared to 1 minute after intravenous administration. Following inhalation of hydrogen gas, the hydrogen concentration was found to be significantly increased at 30 minutes and maintained the same level thereafter. These results demonstrate that accurately determining the hydrogen concentration in rat blood and organ tissue is very useful and important for the application of various novel medical and therapeutic therapies using molecular hydrogen. /Hydrogen super-rich water or saline/
PMID:24975958
Full text:
Liu C et al; Sci Rep 4: 5485 (2014)
Hazardous Substances Data Bank (HSDB)
10.2 Metabolism / Metabolites
The ability of mammalian tissues to oxidize hydrogen under conditions similar to those encountered by deep divers breathing mixtures containing hydrogen was investigated. The kidneys, livers, spleen, heart, lungs, and quadriceps muscle were removed from guinea-pigs and rats. After mincing or homogenization, the tissues, along with myocytes prepared from rat hearts and porcine cerebral cortex capillary endothelial cells were placed in petri dishes and exposed to tritium tagged hydrogen at a pressure of 1 or 5 megapascals (MPa) for 1 hour in a specially designed exposure system. Helium at a pressure of 1 MPa was used as a carrier. Petri dishes filled with distilled water or saline served as negative controls. After decompression, the extent of hydrogen oxidized by the mammalian tissues and cells was determined by measuring the amounts of incorporated tritium by liquid scintillation counting. The tissues and cells incorporated tritium only at the rate of 10 to 50 nanomoles per gram per minute (nmol/g/min), rates that were similar to those of the negative controls. The authors conclude that mammalian tissues do not oxidize hydrogen under hyperbaric conditions. The small amounts of tritium label incorporation observed in the tissues is probably due to radioisotope phenomena, which sets the detection limit for determining hydrogen oxidation at 100 nmol/g/min.
Kayar SR et al; Undersea and Hyperbaric Medicine 21 (3): 265-275 (1994)
Hazardous Substances Data Bank (HSDB)
10.3 Mechanism of Action
Substantial evidence indicates that molecular hydrogen (H2) has beneficial vascular effects because of its antioxidant and/or anti-inflammatory effects. Thus, hydrogen-rich water may prove to be an effective anti-aging drink. This study examined the effects of H2 on endothelial senescence and clarified the mechanisms involved. Hydrogen-rich medium was produced by a high-purity hydrogen gas generator. Human umbilical vein endothelial cells (HUVECs) were incubated with 2,3,7,8-tetrachlorodibenzo-p-dioxin (TCDD) for various time periods in normal or hydrogen-rich medium. The baseline H2concentration in hydrogen-rich medium was 0.55 +/- 0.07 mmol/L. This concentration gradually decreased, and H2 was almost undetectable in medium after 12 hr. At 24 hr after TCDD exposure, HUVECs treated with TCDD exhibited increased 8OHdG and acetyl-p53 expression, decreased nicotinamide adenine dinucleotide (NAD(+))/NADH ratio, impaired Sirt1 activity, and enhanced senescence-associated beta-galactosidase. However, HUVECs incubated in hydrogen-rich medium did not exhibit these TCDD-induced changes accompanying Nrf2 activation, which was observed even after H2 was undetectable in the medium. Chrysin, an inhibitor of Nrf2, abolished the protective effects of H2 on HUVECs. H2 has long-lasting antioxidant and anti-aging effects on vascular endothelial cells through the Nrf2 pathway, even after transient exposure to H2. Hydrogen-rich water may thus be a functional drink that increases longevity. /Hydrogen-rich water/
PMID:27477846
Hara F et al; Circ J 80 (9): 2037-46 (2016)
Hazardous Substances Data Bank (HSDB)
Amyloid beta (Abeta) peptides are identified /as a/ cause of neurodegenerative diseases such as Alzheimer's disease (AD). Previous evidence suggests Abeta-induced neurotoxicity is linked to the stimulation of reactive oxygen species (ROS) production. The accumulation of Abeta-induced ROS leads to increased mitochondrial dysfunction and triggers apoptotic cell death. This suggests antioxidant therapies may be beneficial for preventing ROS-related diseases such as AD. Recently, hydrogen-rich water (HRW) has been proven effective in treating oxidative stress-induced disorders because of its ROS-scavenging abilities. However, the precise molecular mechanisms whereby HRW prevents neuronal death are still unclear. In the present study, we evaluated the putative pathways by which HRW protects against Abeta-induced cytotoxicity /in SK-N-MC cells/. Our results indicated that HRW directly counteracts oxidative damage by neutralizing excessive ROS, leading to the alleviation of Abeta-induced cell death. In addition, HRW also stimulated AMP-activated protein kinase (AMPK) in a sirtuin 1 (Sirt1)-dependent pathway, which upregulates forkhead box protein O3a (FoxO3a) downstream antioxidant response and diminishes Abeta-induced mitochondrial potential loss and oxidative stress. Taken together, our findings suggest that HRW may have potential therapeutic value to inhibit Abeta-induced neurotoxicity. /Hydrogen-rich water/
PMID:26271894
Lin CL et al; Chem Biol Interact 240: 12-21 (2015)
Hazardous Substances Data Bank (HSDB)
The NLRP3 inflammasome, an intracellular multi-protein complex controlling the maturation of cytokine interleukin-1beta, plays an important role in lipopolysaccharide (LPS)-induced inflammatory cascades. Recently, the production of mitochondrial reactive oxygen species (mtROS) in macrophages stimulated with LPS has been suggested to act as a trigger during the process of NLRP3 inflammasome activation that can be blocked by some mitochondria-targeted antioxidants. Known as a ROS scavenger, molecular hydrogen (H2) has been shown to possess therapeutic benefit on LPS-induced inflammatory damage in many animal experiments. Due to the unique molecular structure, H2 can easily target the mitochondria, suggesting that H2 is a potential antagonist of mtROS-dependent NLRP3 inflammasome activation. Here we have showed that, in mouse macrophages, H2 exhibited substantial inhibitory activity against LPS-initiated NLRP3 inflammasome activation by scavenging mtROS. Moreover, the elimination of mtROS by H2 resultantly inhibited mtROS-mediated NLRP3 deubiquitination, a non-transcriptional priming signal of NLRP3 in response to the stimulation of LPS. Additionally, the removal of mtROS by H2 reduced the generation of oxidized mitochondrial DNA and consequently decreased its binding to NLRP3, thereby inhibiting the NLRP3 inflammasome activation. Our findings have, for the first time, revealed the novel mechanism underlying the inhibitory effect of molecular hydrogen on LPS-caused NLRP3 inflammasome activation, highlighting the promising application of this new antioxidant in the treatment of LPS-associated inflammatory pathological damage.
PMID:26488087
Ren JD et al; Biochim Biophys Acta 1863 (1): 50-5 (2016)
Hazardous Substances Data Bank (HSDB)
... H2 decreased the tyrosine nitration level and suppressed oxidative stress damage in retinal cells. S-nitroso-N-acetylpenicillamine treatment decreased the cell numbers in the ganglion cell layer and inner nuclear layer, but the presence of H2 inhibited this reduction. These findings suggest that H2 has a neuroprotective effect against retinal cell oxidative damage, presumably by scavenging peroxynitrite. H2 reduces cellular peroxynitrite, a highly toxic reactive nitrogen species. Thus, H2 may be an effective and novel clinical tool for treating glaucoma and other oxidative stress-related diseases.
PMID:25801048
Yokota T et al; Clin Exp Ophthalmol 43 (6): 568-77 (2015)
Hazardous Substances Data Bank (HSDB)
Endothelial injury is a primary cause of sepsis and sepsis-induced organ damage. Heme oxygenase-1 (HO-1) plays an essential role in endothelial cellular defenses against inflammation by activating nuclear factor E2-related factor-2 (Nrf2). We found that molecular hydrogen (H2) exerts an anti-inflammatory effect. Here, we hypothesized that H2 attenuates endothelial injury and inflammation via an Nrf2-mediated HO-1 pathway during sepsis. First, we detected the effects of H2 on cell viability and cell apoptosis in human umbilical vein endothelial cells (HUVECs) stimulated by LPS. Then, we measured cell adhesion molecules and inflammatory factors in HUVECs stimulated by LPS and in a cecal ligation and puncture (CLP)-induced sepsis mouse model. Next, the role of Nrf2/HO-1 was investigated in activated HUVECs, as well as in wild-type and Nrf(-/-) mice with sepsis. We found that both 0.3 mmol/L and 0.6 mmol/L (i.e., saturated) H2-rich media improved cell viability and cell apoptosis in LPS-activated HUVECs and that 0.6 mmol/L (i.e., saturated) H2-rich medium exerted an optimal effect. H2 could suppress the release of cell adhesion molecules, such as vascular cell adhesion molecule-1 (VCAM-1) and intercellular cell adhesion molecule-1 (ICAM-1), and pro-inflammatory cytokines, such as tumor necrosis factor (TNF)-a, interleukin (IL)-1beta and high-mobility group box 1 protein (HMGB1). Furthermore, H2 could elevate anti-inflammatory cytokine IL-10 levels in LPS-stimulated HUVECs and in lung tissue from CLP mice. H2 enhanced HO-1 expression and activity in vitro and in vivo. HO-1 inhibition reversed the regulatory effects of H2 on cell adhesion molecules and inflammatory factors. H2 regulated endothelial injury and the inflammatory response via Nrf2-mediated HO-1 levels. These results suggest that H2 could suppress excessive inflammatory responses and endothelial injury via an Nrf2/HO-1 pathway. /Hydrogen-rich media/
PMID:26253656
Chen H et al; Int Immunopharmacol 28 (1): 643-54 (2015)
Hazardous Substances Data Bank (HSDB)
10.4 Human Metabolite Information
10.4.1 Cellular Locations
Cytoplasm
Endoplasmic reticulum
Extracellular
Golgi apparatus
Lysosome
Mitochondria
Nucleus
Peroxisome
Human Metabolome Database (HMDB)
10.4.2 Metabolite Pathways
Acute Intermittent Porphyria
Carnitine Synthesis
Clomipramine Metabolism Pathway
Congenital Erythropoietic Porphyria (CEP) or Gunther Disease
Degradation of Superoxides
Ethanol Degradation
Hereditary Coproporphyria (HCP)
Homocysteine Degradation
Leigh Syndrome
Malonic Aciduria
Total 37 pathways, visit the HMDB page for details
Human Metabolome Database (HMDB)
10.5 Biochemical Reactions
Rhea - Annotated Reactions Database
PubChem
11 Use and Manufacturing
11.1 Uses
EPA CPDat Chemical and Product Categories
The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125
EPA Chemical and Products Database (CPDat)
Sources/Uses
Used in welding (oxy-hydrogen and autogenous), petroleum refining, and as a coolant; Used to hydrogenate oils and other organic compounds and to reduce metal oxides; [Merck Index]
Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013.
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
Industrial Processes with risk of exposure
Petroleum Production and Refining [Category: Industry]
Gas Welding and Cutting [Category: Weld]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
In oxy-hydrogen blowpipe (welding) and limelight; autogenous welding of steel and other metals; manufacturing ammonia, synthetic methanol, HCl, NH3; hydrogenation of oils, fats, naphthalene, phenol; in balloons and airships; in metallurgy to reduce oxides to metals; in petroleum refining; in thermonuclear reactions (ionizes to form protons, deuterons (D) or tritons (T).
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
In bubble chambers to study subatomic particles; as a coolant. /Liquid hydrogen/
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
In cryogenics; in study of superconductivity /Liquid hydrogen/
Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 79th ed. Boca Raton, FL: CRC Press Inc., 1998-1999., p. 4-15
Hazardous Substances Data Bank (HSDB)
Production of ammonia, ethanol, and aniline; hydrocracking, hydroforming, and hydrofining of petroleum; hydrogenation of vegetable oils; hydrogenolysis of coal; reducing agent for organic synthesis and metallic ores; reducing atmosphere to prevent oxidation; as oxyhdrogen flame for high temperatures; atomic-hydrogen welding; instrument-carrying balloons; making hydrogen chloride and hydrogen bromide; production of high-purity metals; fuel for nuclear rocket engines for hypersonic transport; missile fuel; cryogenic research.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
For more Uses (Complete) data for Hydrogen (7 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
11.1.1 Use Classification
Food additives
EU Food Improvement Agents
11.1.2 Industry Uses
Processing aids, not otherwise listed
Odor agents
Fuels and fuel additives
Intermediates
Fuel agents
Fuel
Not Known or Reasonably Ascertainable
Functional fluids (closed systems)
Intermediate
Reducing agent
Other (specify)
EPA Chemical Data Reporting (CDR)
11.1.3 Consumer Uses
Other (specify)
Fuels and fuel additives
Not Known or Reasonably Ascertainable
Fuel agents
Processing aids, not otherwise listed
Flame retardants
Odor agents
EPA Chemical Data Reporting (CDR)
11.2 Methods of Manufacturing
The main commercial processes for the on-purpose production of hydrogen are steam reforming, partial oxidation (coal, coke, resid), or electrolysis of water. Hydrogen is also commercially produced as a by-product of chemical processes (ethylene crackers, styrene, MTBE etc) or gasoline manufacturing (catalytic reforming).
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2017). John Wiley & Sons, Inc. Online Posting Date: December 20, 2001
Hazardous Substances Data Bank (HSDB)
Obtained by passing water vapors over heated iron; by electrolysis of water or by action of hydrochloric acid or sulfuric acid on iron or zinc; by hydrolysis of metal hydrides.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
(1) Reaction of steam with natural gas (steam reforming) and susequent purification; (2) partial oxidation of hydrocarbons to carbon monoxide and interaction of carbon monoxide and steam; (3) gasification of coal; (4) dissociation of ammonia; (5) thermal or catalytic decomposition of hydrocarbons gases; (6) catalytic reforming of naphtha; (7) reaction of iron and steam; (8) catalytic reaction of methanol and steam; (9) electrolysis of water.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
Hydrogen can be produced using diverse, domestic resources including fossil fuels, such as natural gas and coal (preferentially with carbon capture, utilization, and storage); biomass grown from renewable, non-food crops; or using nuclear energy and renewable energy sources, such as wind, solar, geothermal, and hydroelectric power to split water. This diversity of potential supply sources is an important reason why hydrogen is such a promising energy carrier.
DOE; Fuel Cell technologies Office: Hydrogen Production (September, 2014). Available from, as of June 21, 2017:
Hazardous Substances Data Bank (HSDB)
Hydrogen is produced on a laboratory scale from the action of an aqueous acid on a metal or from the reaction of an alkali metal in water ... These reactions can be carried out at room temperature. Hydrogen gas can also be produced on a laboratory scale by the electrolysis of an aqueous solution. Production of hydrogen through electrolysis is also used industrially. ... Hydrogen atoms can be produced in significant quantities in the gas phase by the action of radiation on or by extreme heating of H2 (3000 K).
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2017). John Wiley & Sons, Inc. Online Posting Date: December 20, 2001
Hazardous Substances Data Bank (HSDB)
11.3 Formulations / Preparations
Grades: Technical, pure, from an electrolytic grade of 99.8% to ultra-pure, with less than 10 ppm impurities.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
11.4 Consumption Patterns
CHEMICAL INTERMEDIATE FOR AMMONIA, 56%; PETROLEUM REFINING, 25%; CHEMICAL INTERMEDIATE FOR METHANOL, 7%; OTHER, 12% (1980)
SRI
Hazardous Substances Data Bank (HSDB)
U.S. consumption of hydrogen in 1988: 49% as chemical intermediate for ammonia; 37% in petroleum refining; 8% as chemical intermediate for methanol; 6% as other uses
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V13 (95) 884
Hazardous Substances Data Bank (HSDB)
11.5 U.S. Production
Aggregated Product Volume
2019: 5,000,000,000 - <10,000,000,000 lb
2018: 10,000,000,000 - <20,000,000,000 lb
2017: 5,000,000,000 - <10,000,000,000 lb
2016: 10,000,000,000 - <20,000,000,000 lb
EPA Chemical Data Reporting (CDR)
(1977) 8.48X10+10 CU FT
SRI
Hazardous Substances Data Bank (HSDB)
(1979) 9.93X10+10 CU FT
SRI
Hazardous Substances Data Bank (HSDB)
Daily Capacity: 4,197,260 thousands of cu ft
SRI. 1999 Directory of Chemical Producers -United States. Menlo Park, CA. SRI Consulting 1999., p. 473
Hazardous Substances Data Bank (HSDB)
U.S. ... about 3 billion cu ft/yr
Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 79th ed. Boca Raton, FL: CRC Press Inc., 1998-1999., p. 4-14
Hazardous Substances Data Bank (HSDB)
For more U.S. Production (Complete) data for Hydrogen (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
11.6 U.S. Exports
(1977) 5.91X10+8 CU FT
SRI
Hazardous Substances Data Bank (HSDB)
(1979) 1.27X10+8 CU FT
SRI
Hazardous Substances Data Bank (HSDB)
11.7 General Manufacturing Information
Industry Processing Sectors
All Other Chemical Product and Preparation Manufacturing
Primary Metal Manufacturing
Petrochemical Manufacturing
Plastics Material and Resin Manufacturing
Utilities
All other Petroleum and Coal Products Manufacturing
All Other Basic Inorganic Chemical Manufacturing
All Other Basic Organic Chemical Manufacturing
Petroleum Refineries
Industrial Gas Manufacturing
Wholesale and Retail Trade
EPA Chemical Data Reporting (CDR)
EPA TSCA Commercial Activity Status
Hydrogen: ACTIVE
EPA Chemicals under the TSCA
Production of hydrogen in usa alone now amounts to about 3 billion cu ft/yr.
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 94th Edition. CRC Press LLC, Boca Raton: FL 2013-2014, p. 4-17
Hazardous Substances Data Bank (HSDB)
Method of purification: by scrubbing with various solutions (especially the Girbitol absorption process). For very pure hydrogen, by diffusion through palladium.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
More efficient methods than electrolysis for obtaining hydrogen from water are under investigation. One of these is thermochemical decomposition. Another is photochemical decomposition by solar radiation, either directly or via a solar power generator. Photolytic decomposition of water with platinum catalyst has been achieved. Hydrogen can also be obtained by photolytic decomposition of hydrogen sulfide with cadmium sulfide catalyst.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
Most U.S. hydrogen production ... is used captively.
Kirk-Othmer Encyclopedia of Chemical Technology. 4th ed. Volumes 1: New York, NY. John Wiley and Sons, 1991-Present., p. V13 (95) 878
Hazardous Substances Data Bank (HSDB)
For more General Manufacturing Information (Complete) data for Hydrogen (7 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
12 Identification
12.1 Analytic Laboratory Methods
For hydrogen sensors built with pure Pd nanowires, the instabilities causing baseline drifting and temperature-driven sensing behavior are limiting factors when working within a wide temperature range. To enhance the material stability, we have developed superlattice-structured palladium and copper nanowires (PdCu NWs) with random-gapped, screw-threaded, and spiral shapes achieved by wet-chemical approaches. The microstructure of the PdCu NWs reveals novel superlattices composed of lattice groups structured by four-atomic layers of alternating Pd and Cu. Sensors built with these modified NWs show significantly reduced baseline drifting and lower critical temperature (259.4 K and 261 K depending on the PdCu structure) for the reverse sensing behavior than those with pure Pd NWs (287 K). Moreover, the response and recovery times of the PdCu NWs sensor were of ~9 and ~7 times faster than for Pd NWs sensors, respectively.
PMID:24440892
Full text:
Yang D et al; Sci Rep 4: 3773 (2014)
Hazardous Substances Data Bank (HSDB)
Electroless templating on DNA is established as a means to prepare high aspect ratio nanowires via aqueous reactions at room temperature. In this report we show how Pd nanowires with extremely small grain sizes (< 2 nm) can be prepared by reduction of PdCl4(2-) in the presence of lambda-DNA. In AFM images the wires are smooth and uniform in appearance, but the grain size estimated by the Scherrer treatment of line broadening in X-ray diffraction is less than the diameter of the wires from AFM (of order 10 nm). Electrical characterization of single nanowires by conductive AFM shows ohmic behavior, but with high contact resistances and a resistivity (-10(-2) omega cm) much higher than the bulk value for Pd metal (-10(-5) cm at 20 °C). These observations can be accounted for by a model of the nanowire growth mechanism which naturally leads to the formation of a granular metal. Using a simple combing technique with control of the surface hydrophilicity, DNA-templated Pd nanowires have also been prepared as networks on an Si/SiO2 substrate. These networks are highly convenient for the preparation of two-terminal electronic sensors for the detection of hydrogen gas. The response of these hydrogen sensors is presented and a model of the sensor response in terms of the diffusion of hydrogen into the nanowires is described. The granular structure of the nanowires makes them relatively poor conductors, but they retain a useful sensitivity to hydrogen gas.
PMID:24466659
Al-Hinai MN et al; Faraday Discuss 164: 71-91 (2013)
Hazardous Substances Data Bank (HSDB)
The determination of hydrogen content of an organic compound consists of complete combustion of a known quantity of the material to produce water and carbon dioxide, and determination of the amount of water. The amount of hydrogen present in the initial material is calculated from the amount of water produced. This technique can be performed on macro (0.1-0.2 g), micro (2-10 mg), or submicro (0.02-0.2 mg) scale. ... The oldest and probably most reliable technique for water determination is a gravimetric one where the water is absorbed onto a desiccant, such as magnesium perchlorate. In the macro technique, which is the most accurate, hydrogen content of a compound can be routinely determined to within +/- 0.02%. Instrumental methods, such as gas chromatography and mass spectrometry, can also be used to determine water of combustion.
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2017). John Wiley & Sons, Inc. Online Posting Date: December 20, 2001
Hazardous Substances Data Bank (HSDB)
To determine quantitatively the amount of hydrogen present in a gas mixture, gas chromatography using a thermal conductivity detector is often employed.
Baade WF et al; Hydrogen. Kirk-Othmer Encyclopedia of Chemical Technology (1999-2017). John Wiley & Sons, Inc. Online Posting Date: December 20, 2001
Hazardous Substances Data Bank (HSDB)
12.2 Clinical Laboratory Methods
A standardized simple, indirect method for assessing the relative energy of dietary fiber carbohydrates is not yet established. There is a need for a standardized in vivo assay. The objective of the present study was to evaluate the relative available energy (RAE) for 9 major dietary fiber materials (DFMs) based on fermentability from breath hydrogen excretion (BHE) in subjects. Fructooligosaccharide (FOS) was used as a reference. The study was conducted using a within-subject, repeated measures design and approved by the Ethical Committee of University of Nagasaki. After DFM ingestion, end-expiratory gas (750-mL) was collected at 1-hr intervals for 8 hr, as well as at 2-hr intervals between 8 hr and 14 hr, and 30 min after waking up and 24 hr after DFM ingestion. Breath hydrogen concentration was assessed with a gas chromatograph. The RAE of DFMs tested was evaluated based on the area under the curve (AUC) of BHE of FOS. Based on the ratio of AUC for 8 hr, the RAE of polydextrose, partially hydrolysed guar gum, resistant maltodextrin and partially hydrolysed alginate was 1 kcal/g, and that of glucomannan, heat-moisture treatment and high-amylose cornstarch and cellulose was 0 kcal/g, while the RAE of all tested DEMs including cellulose and glucomannan was 1 kcal/g in the calculation based on AUCs for 14 hr and 24 hr in subjects. We suggest that a breath hydrogen collection period of 14 hr or more could be used to measure RAE for a range of fiber preparations in vivo.
PMID:25297613
Oku T, Nakamura S; J Nutr Sci Vitaminol (Tokyo) 60 (4): 246-54 (2014)
Hazardous Substances Data Bank (HSDB)
13 Safety and Hazards
13.1 Hazards Identification
13.1.1 GHS Classification
1 of 6 items
View All
Pictogram(s)
Signal
Danger
GHS Hazard Statements
H220: Extremely flammable gas [Danger Flammable gases]
Precautionary Statement Codes
P203, P210, P222, P280, P377, P381, and P403
(The corresponding statement to each P-code can be found at the GHS Classification page.)
Regulation (EC) No 1272/2008 of the European Parliament and of the Council
13.1.2 Hazard Classes and Categories
Flam. Gas 1 (99.8%)
Press. Gas (Comp.) (75.8%)
Press. Gas (Ref. Liq.) (13.5%)
European Chemicals Agency (ECHA)
View More...
13.1.3 NFPA Hazard Classification
NFPA 704 Diamond
NFPA Health Rating
0 - Materials that, under emergency conditions, would offer no hazard beyond that of ordinary combustible materials.
NFPA Fire Rating
4 - Materials that rapidly or completely vaporize at atmospheric pressure and normal ambient temperature or that are readily dispersed in air and burn readily.
NFPA Instability Rating
0 - Materials that in themselves are normally stable, even under fire conditions.
Hazardous Substances Data Bank (HSDB)
13.1.4 DOT Hazard Classification
1 of 3 items
View All
Substance (Descriptions/Shipping Name)
Hydrogen, compressed
DOT ID (UN/NA Number)
UN1049
Hazard Class/Label Code(s)
Div 2.1 Flammable gas (49 eCFR § 173.115)
Placard/Label(s)
US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172
13.1.5 Health Hazards
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
Vapors may cause dizziness or asphyxiation without warning, especially when in closed or confined areas. Some may be irritating if inhaled at high concentrations. Contact with gas, liquefied gas or cryogenic liquids may cause burns, severe injury and/or frostbite. Fire may produce irritating and/or toxic gases. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
If atmosphere does not contain enough oxygen, inhalation can cause dizziness, unconsciousness, or even death. Contact of liquid with eyes or skin causes freezing similar to burn. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· Vapors may cause dizziness or asphyxiation without warning, especially when in closed or confined areas.
· Some may be irritating if inhaled at high concentrations.
· Contact with gas, liquefied gas or cryogenic liquids may cause burns, severe injury and/or frostbite.
· Fire may produce irritating and/or toxic gases.
Emergency Response Guidebook (ERG)
13.1.6 Fire Hazards
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
EXTREMELY FLAMMABLE. Will be easily ignited by heat, sparks or flames. Will form explosive mixtures with air. Vapors from liquefied gas are initially heavier than air and spread along ground. CAUTION: Hydrogen (UN1049), Deuterium (UN1957), Hydrogen, refrigerated liquid (UN1966), Methane (UN1971) and Hydrogen and Methane mixture, compressed (UN2034) are lighter than air and will rise. Hydrogen and Deuterium fires are difficult to detect since they burn with an invisible flame. Use an alternate method of detection (thermal camera, broom handle, etc.) Vapors may travel to source of ignition and flash back. Cylinders exposed to fire may vent and release flammable gas through pressure relief devices. Containers may explode when heated. Ruptured cylinders may rocket. CAUTION: When LNG - Liquefied natural gas (UN1972) is released on or near water, product may vaporize explosively. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Behavior in Fire: Burns with an almost invisible flame. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· EXTREMELY FLAMMABLE.
· Will be easily ignited by heat, sparks or flames.
· Will form explosive mixtures with air.
· Vapors from liquefied gas are initially heavier than air and spread along ground.
CAUTION: Hydrogen (UN1049), Deuterium (UN1957), Hydrogen, refrigerated liquid (UN1966), Methane (UN1971) and Hydrogen and Methane mixture, compressed (UN2034) are lighter than air and will rise. Hydrogen and Deuterium fires are difficult to detect since they burn with an invisible flame. Use an alternate method of detection (thermal camera, broom handle, etc.)
· Vapors may travel to source of ignition and flash back.
· Cylinders exposed to fire may vent and release flammable gas through pressure relief devices.
· Containers may explode when heated.
· Ruptured cylinders may rocket.
CAUTION: When LNG - Liquefied natural gas (UN1972) is released on or near water, product may vaporize explosively.
Emergency Response Guidebook (ERG)
Extremely flammable. Many reactions may cause fire or explosion. Gas/air mixtures are explosive.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.1.7 Hazards Summary
Simple asphyxiant; Contact with evaporating liquid can cause frostbite; [HSDB]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
13.1.8 Fire Potential
Highly dangerous fire and severe explosion hazard when exposed to heat, flame, or oxidizers.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 1985
Hazardous Substances Data Bank (HSDB)
... Sudden release of hydrogen into the atmosphere from storage above 79 bars may cause spontaneous ignition. ...
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1186
Hazardous Substances Data Bank (HSDB)
13.2 Safety and Hazard Properties
13.2.1 Flammable Limits
Lower: 4.0%; upper: 75% (% by vol)
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 325-72
Hazardous Substances Data Bank (HSDB)
13.2.2 Lower Explosive Limit (LEL)
4 % (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
13.2.3 Upper Explosive Limit (UEL)
75 % (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
13.2.4 Critical Temperature & Pressure
Critical temperature: -239.9 °C; critical pressure: 12.8 atm
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
13.2.5 Physical Dangers
The gas mixes well with air, explosive mixtures are easily formed. The gas is lighter than air.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.2.6 Explosive Limits and Potential
Flammable or explosive when mixed with air, O2, chlorine.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 1985
Hazardous Substances Data Bank (HSDB)
Lower: 4.1%; upper: 74.2%
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 1985
Hazardous Substances Data Bank (HSDB)
Explosive limits , vol% in air: 4-75
ILO-WHO International Chemical Safety Cards (ICSCs)
13.3 First Aid Measures
Inhalation First Aid
Fresh air, rest.
ILO-WHO International Chemical Safety Cards (ICSCs)
Skin First Aid
ON FROSTBITE: rinse with plenty of water, do NOT remove clothes. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
Eye First Aid
ON FROSTBITE: rinse with plenty of water. Refer immediately for medical attention.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.3.1 First Aid
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
Refer to the "General First Aid" section. Specific First Aid: Clothing frozen to the skin should be thawed before being removed. In case of contact with liquefied gas, only medical personnel should attempt thawing frosted parts. In case of burns, immediately cool affected skin for as long as possible with cold water. Do not remove clothing if adhering to skin. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
The only effect of exposure to liquid hydrogen is that caused by its unusually low temperature and its action as a simple asphyxiant.
INHALATION: if victim is unconscious (due to oxygen deficiency), move him to fresh air and apply resuscitation methods; call physician.
EYES: treat for frostbite.
SKIN: treat for frostbite; soak in lukewarm water; get medical attention if burn is severe. (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
General First Aid:
· Call 911 or emergency medical service.
· Ensure that medical personnel are aware of the material(s) involved, take precautions to protect themselves and avoid contamination.
· Move victim to fresh air if it can be done safely.
· Administer oxygen if breathing is difficult.
· If victim is not breathing:
-- DO NOT perform mouth-to-mouth resuscitation; the victim may have ingestedor inhaled the substance.
-- If equipped and pulse detected, wash face and mouth, then give artificial respiration using a proper respiratory medical device (bag-valve mask, pocket mask equipped with a one-way valve or other device).
-- If no pulse detected or no respiratory medical device available, provide continuouscompressions. Conduct a pulse check every two minutes or monitor for any signs of spontaneous respirations.
· Remove and isolate contaminated clothing and shoes.
· For minor skin contact, avoid spreading material on unaffected skin.
· In case of contact with substance, remove immediately by flushing skin or eyes with running water for at least 20 minutes.
· For severe burns, immediate medical attention is required.
· Effects of exposure (inhalation, ingestion, or skin contact) to substance may be delayed.
· Keep victim calm and warm.
· Keep victim under observation.
· For further assistance, contact your local Poison Control Center.
· Note: Basic Life Support (BLS) and Advanced Life Support (ALS) should be done by trained professionals.
Specific First Aid:
· Clothing frozen to the skin should be thawed before being removed.
· In case of contact with liquefied gas, only medical personnel should attempt thawing frosted parts.
· In case of burns, immediately cool affected skin for as long as possible with cold water. Do not remove clothing if adhering to skin.
In Canada, an Emergency Response Assistance Plan (ERAP) may be required for this product. Please consult the shipping paper and/or the "ERAP" section.
Emergency Response Guidebook (ERG)
13.4 Fire Fighting
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED. CAUTION: Hydrogen (UN1049), Deuterium (UN1957), Hydrogen, refrigerated liquid (UN1966) and Hydrogen and Methane mixture, compressed (UN2034) will burn with an invisible flame. Use an alternate method of detection (thermal camera, broom handle, etc.).
SMALL FIRE: Dry chemical or CO2.
LARGE FIRE: Water spray or fog. If it can be done safely, move undamaged containers away from the area around the fire. CAUTION: For LNG - Liquefied natural gas (UN1972) pool fires, DO NOT USE water. Use dry chemical or high-expansion foam.
FIRE INVOLVING TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Cool containers with flooding quantities of water until well after fire is out. Do not direct water at source of leak or safety devices; icing may occur. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. For massive fire, use unmanned master stream devices or monitor nozzles; if this is impossible, withdraw from area and let fire burn. (ERG, 2024)
2024 Emergency Response Guidebook,
2024 Emergency Response Guidebook,
CAMEO Chemicals
Shut off supply; if not possible and no risk to surroundings, let the fire burn itself out. In other cases extinguish with water spray, powder, carbon dioxide. In case of fire: keep cylinder cool by spraying with water. Combat fire from a sheltered position.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.4.1 Fire Fighting Procedures
To fight fire, stop flow of gas.
Lewis, R.J. Sr. (ed) Sax's Dangerous Properties of Industrial Materials. 11th Edition. Wiley-Interscience, Wiley & Sons, Inc. Hoboken, NJ. 2004., p. 1985
Hazardous Substances Data Bank (HSDB)
Approach fire with caution as high-temperature flame is practically invisible. Stop flow of gas before extinguishing fire. Use water spray to keep fire-exposed containers cool. Use flooding quantities of water as fog or spray. /Hydrogen, refrigerated liquid/
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-86
Hazardous Substances Data Bank (HSDB)
Use water spray, alcohol-resistant foam, dry chemical or carbon dioxide. Wear self-contained breathing apparatus for firefighting if necessary. Use water spray to cool unopened containers.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
13.5 Accidental Release Measures
Public Safety: ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· CALL 911. Then call emergency response telephone number on shipping paper. If shipping paper not available or no answer, refer to appropriate telephone number listed on the inside back cover.
· Keep unauthorized personnel away.
· Stay upwind, uphill and/or upstream.
· Many gases are heavier than air and will spread along the ground and collect in low or confined areas (sewers, basements, tanks, etc.).
Emergency Response Guidebook (ERG)
Spill or Leak: ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area.
· All equipment used when handling the product must be grounded.
· Do not touch or walk through spilled material.
· Stop leak if you can do it without risk.
· If possible, turn leaking containers so that gas escapes rather than liquid.
· Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material.
· Do not direct water at spill or source of leak.
CAUTION: For LNG - Liquefied natural gas (UN1972), DO NOT apply water, regular or alcohol-resistant foam directly on spill. Use a high-expansion foam if available to reduce vapors.
· Prevent spreading of vapors through sewers, ventilation systems and confined areas.
· Isolate area until gas has dispersed.
CAUTION: When in contact with refrigerated/cryogenic liquids, many materials become brittle and are likely to break without warning.
Emergency Response Guidebook (ERG)
13.5.1 Isolation and Evacuation
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
LARGE SPILL: Consider initial downwind evacuation for at least 800 meters (1/2 mile).
FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 1600 meters (1 mile) in all directions; also, consider initial evacuation for 1600 meters (1 mile) in all directions. In fires involving Liquefied Petroleum Gases (LPG) (UN1075), Butane (UN1011), Butylene (UN1012), Isobutylene (UN1055), Propylene (UN1077), Isobutane (UN1969), and Propane (UN1978), also refer to the "BLEVE - Safety Precautions" section. (ERG, 2024)
2024 Emergency Response Guidebook,
2024 Emergency Response Guidebook,
CAMEO Chemicals
Evacuation: ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
Immediate precautionary measure
· Isolate spill or leak area for at least 100 meters (330 feet) in all directions.
Large Spill
· Consider initial downwind evacuation for at least 800 meters (1/2 mile).
Fire
· If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 1600 meters (1 mile) in all directions; also, consider initial evacuation for 1600 meters (1 mile) in all directions.
· In fires involving Liquefied Petroleum Gases (LPG) (UN1075), Butane (UN1011), Butylene (UN1012), Isobutylene (UN1055), Propylene (UN1077), Isobutane (UN1969), and Propane (UN1978), also refer to the "BLEVE - Safety Precautions" section.
Emergency Response Guidebook (ERG)
13.5.2 Spillage Disposal
Evacuate danger area! Consult an expert! Ventilation. Remove all ignition sources. Remove vapour with fine water spray.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.5.3 Cleanup Methods
ACCIDENTAL RELEASE MEASURES: Personal precautions, protective equipment and emergency procedures: Avoid breathing vapors, mist or gas. Ensure adequate ventilation. Remove all sources of ignition. Evacuate personnel to safe areas. Beware of vapors accumulating to form explosive concentrations. Vapors can accumulate in low areas. Environmental precautions: Prevent further leakage or spillage if safe to do so. Do not let product enter drains. Methods and materials for containment and cleaning up: Clean up promptly by sweeping or vacuum.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Eliminate all ignition sources. Approach release from upwind. Stop or control the leak, if this can be done without undue risk. Use water spray to disperse vapors and protect personnel. /Hydrogen, refrigerated liquid/
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-86
Hazardous Substances Data Bank (HSDB)
13.5.4 Disposal Methods
SRP: Recycle any unused portion of the material for its approved use or return it to the manufacturer or supplier. Ultimate disposal of the chemical must consider: the material's impact on air quality; potential migration in air, soil or water; effects on animal, aquatic and plant life; and conformance with environmental and public health regulations. If it is possible or reasonable use an alternative chemical product with less inherent propensity for occupational harm/injury/toxicity or environmental contamination.
Hazardous Substances Data Bank (HSDB)
Product: Contact a licensed professional waste disposal service to dispose of this material. Burn in a chemical incinerator equipped with an afterburner and scrubber but exert extra care in igniting as this material is highly flammable. Offer surplus and non-recyclable solutions to a licensed disposal company. Contaminated packaging: Dispose of as unused product
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
13.5.5 Preventive Measures
Handle in accordance with good industrial hygiene and safety practice. Wash hands before breaks and at the end of workday.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Gloves must be inspected prior to use. Use proper glove removal technique (without touching glove's outer surface) to avoid skin contact with this product. Dispose of contaminated gloves after use in accordance with applicable laws and good laboratory practices. Wash and dry hands.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Avoid inhalation of vapor or mist. Use explosion-proof equipment. Keep away from sources of ignition - No smoking. Take measures to prevent the build up of electrostatic charge.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
13.6 Handling and Storage
13.6.1 Nonfire Spill Response
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. All equipment used when handling the product must be grounded. Do not touch or walk through spilled material. Stop leak if you can do it without risk. If possible, turn leaking containers so that gas escapes rather than liquid. Use water spray to reduce vapors or divert vapor cloud drift. Avoid allowing water runoff to contact spilled material. Do not direct water at spill or source of leak. CAUTION: For LNG - Liquefied natural gas (UN1972), DO NOT apply water, regular or alcohol-resistant foam directly on spill. Use a high-expansion foam if available to reduce vapors. Prevent spreading of vapors through sewers, ventilation systems and confined areas. Isolate area until gas has dispersed. CAUTION: When in contact with refrigerated/cryogenic liquids, many materials become brittle and are likely to break without warning. (ERG, 2024)
2024 Emergency Response Guidebook,
2024 Emergency Response Guidebook,
CAMEO Chemicals
13.6.2 Safe Storage
Fireproof. Cool. Ventilation along the floor and ceiling. Separated from oxidizing materials.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.6.3 Storage Conditions
Store in a cool, dry, well-ventilated location. Outside or detached storage is preferred. Isolate from oxygen, halogens, other oxidizing materials. /Hydrogen, refrigerated liquid/
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-86
Hazardous Substances Data Bank (HSDB)
A safe storage method for hydrogen for possible use as automotive fuel involves the use of metal hydrides from which the hydrogen is released at specified temperatures. Iron titanium hydride has been found the most satisfactory.
Larranaga, M.D., Lewis, R.J. Sr., Lewis, R.A.; Hawley's Condensed Chemical Dictionary 16th Edition. John Wiley & Sons, Inc. Hoboken, NJ 2016., p. 733
Hazardous Substances Data Bank (HSDB)
Keep container tightly closed in a dry and well-ventilated place. Contents under pressure. Storage class (TRGS 510): Gases
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
13.7 Exposure Control and Personal Protection
Protective Clothing: ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· Wear positive pressure self-contained breathing apparatus (SCBA).
· Structural firefighters' protective clothing provides thermal protection but only limited chemical protection.
· Always wear thermal protective clothing when handling refrigerated/cryogenic liquids.
Emergency Response Guidebook (ERG)
13.7.1 Threshold Limit Values (TLV)
Simple asphyxiant; Explosion hazard: the substance is a flammable asphyxiant or excursions above the TLV could approach 10% of the lower explosive limit. /A simple asphyxiant may not be assigned a TLV because the limiting factor is the available oxygen./
American Conference of Governmental Industrial Hygienists TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. Cincinnati, OH 2017, p. 35
Hazardous Substances Data Bank (HSDB)
13.7.2 Emergency Response Planning Guidelines
Emergency Response: ERG 2024, Guide 115 (Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid))
· DO NOT EXTINGUISH A LEAKING GAS FIRE UNLESS LEAK CAN BE STOPPED.
CAUTION: Hydrogen (UN1049), Deuterium (UN1957), Hydrogen, refrigerated liquid (UN1966) and Hydrogen and Methane mixture, compressed (UN2034) will burn with an invisible flame. Use an alternate method of detection (thermal camera, broom handle, etc.)
Small Fire
· Dry chemical or CO2.
Large Fire
· Water spray or fog.
· If it can be done safely, move undamaged containers away from the area around the fire.
CAUTION: For LNG - Liquefied natural gas (UN1972) pool fires, DO NOT USE water. Use dry chemical or high-expansion foam.
Fire Involving Tanks
· Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles.
· Cool containers with flooding quantities of water until well after fire is out.
· Do not direct water at source of leak or safety devices; icing may occur.
· Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank.
· ALWAYS stay away from tanks in direct contact with flames.
· For massive fire, use unmanned master stream devices or monitor nozzles; if this is impossible, withdraw from area and let fire burn.
Emergency Response Guidebook (ERG)
13.7.3 Inhalation Risk
On loss of containment this substance can cause suffocation by lowering the oxygen content of the air in confined areas.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.4 Effects of Short Term Exposure
Asphyxiation. Exposure to cold gas could cause frostbite.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.5 Personal Protective Equipment (PPE)
Excerpt from ERG Guide 115 [Gases - Flammable (Including Refrigerated Liquids)]:
Wear positive pressure self-contained breathing apparatus (SCBA). Structural firefighters' protective clothing provides thermal protection but only limited chemical protection. Always wear thermal protective clothing when handling refrigerated/cryogenic liquids. (ERG, 2024)
2024 Emergency Response Guidebook,
CAMEO Chemicals
Safety goggles or face shield; insulated gloves and long sleeves; cuffless trousers worn outside boots or over high-top shoes to shed spilled liquid; self-contained breathing apparatus containing air (never use oxygen). (USCG, 1999)
U.S. Coast Guard. 1999. Chemical Hazard Response Information System (CHRIS) - Hazardous Chemical Data. Commandant Instruction 16465.12C. Washington, D.C.: U.S. Government Printing Office.
CAMEO Chemicals
Eye/face protection: Face shield and safety glasses. Use equipment for eye protection tested and approved under appropriate government standards such as NIOSH (US) or EN 166(EU).
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Wear special protective clothing and positive pressure self-contained breathing apparatus. /Hydrogen, refrigerated liquid/
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-86
Hazardous Substances Data Bank (HSDB)
Skin protection: Handle with gloves.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Body Protection: Impervious clothing. Flame retardant antistatic protective clothing. The type of protective equipment must be selected according to the concentration and amount of the dangerous substance at the specific workplace.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Respiratory protection: Where risk assessment shows air-purifying respirators are appropriate use a full-face respirator with multipurpose combination (US) or type AXBEK (EN 14387) respirator cartridges as a backup to engineering controls. If the respirator is the sole means of protection, use a full-face supplied air respirator. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU).
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
13.7.6 Fire Prevention
NO open flames, NO sparks and NO smoking. Closed system, ventilation, explosion-proof electrical equipment and lighting. Use non-sparking handtools. Do not handle cylinders with oily hands.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.7 Exposure Prevention
Use appropriate engineering controls.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.8 Inhalation Prevention
Use ventilation.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.9 Skin Prevention
Cold-insulating gloves.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.7.10 Eye Prevention
Wear face shield.
ILO-WHO International Chemical Safety Cards (ICSCs)
13.8 Stability and Reactivity
13.8.1 Air and Water Reactions
Highly flammable.
CAMEO Chemicals
13.8.2 Reactive Group
Reducing Agents, Weak
CAMEO Chemicals
13.8.3 Reactivity Alerts
Highly Flammable
CAMEO Chemicals
13.8.3.1 CSL Reaction Information
CSL No
CSL00194
Reactants/Reagents
Nitrous oxide + Nitric oxide + Sodium + Hydrogen + Sodium formate + Toluene
Warning Message
"A safety letter from Merck & Co. chemists titled “Nitric Oxide at High Pressure” (C&EN, Jan. 30, page 6) described two explosions during depressurization of a reaction between NO and methanol under basic conditions. The products in a model system with sodium methoxide were described as nitrous oxide and formic acid, presumably as sodium formate. A potential danger in this system should be pointed out: Sodium formate undergoes thermal decomposition to give hydrogen gas (J. Am. Chem. Soc.,DOI: 10.1021/ja02245a004), which explodes spontaneously in the presence of nitrous oxide above critical limits (J. Am. Chem. Soc., DOI: 10.1021/ja01179a036), even in the absence of a catalyst or source of ignition. The presence of hydrogen and nitrous oxide above a reaction mixture was undoubtedly the cause of an explosion and fire in my laboratory in 1981 during workup of a reaction between sodium and nitric oxide. The major product of the reaction is cis-sodium hyponitrite, which decomposes immediately in water to form sodium hydroxide and nitrous oxide. The employee, a biology major who was badly burned, had carried out the reaction a number of times without incident. This time he tried twice and failed to disperse about 30 g of sodium in toluene and, without consulting me, decided to continue the reaction. The explosion occurred as he was attempting to destroy the unreacted sodium, a lump too large to remove from the flask, by dropwise addition of water. Most of the sodium had reacted at the time of the explosion, and there was no indication of mechanical failure. At the time, I was unaware of the extreme incompatibility of the two gases, and the accident was extremely puzzling. The reaction mixture was close to room temperature and was stirred rapidly while the headspace was flushed with a stream of nitrogen. When I arrived at the laboratory a few minutes after the accident, nitrogen was still flowing from the burned-off end of the plastic tubing. Since that time, I noticed a reference to the “hydrogen explosion” in the ancient chemical literature as a way to identify nitrous oxide." (reprint of the full-text)
GHS Category
Explosive
Reaction Scale
Medium (up to 100g)
DOI Link
10.1021/cen-09013-letters
Reference Source
Literature Reference
Modified Date
10/15/2022
Create Date
10/14/2022
Pistoia Alliance Chemical Safety Library
13.8.4 Reactivity Profile
Finely divided platinum and some other metals will cause a mixture of hydrogen and oxygen to explode at ordinary temperatures. If a jet of hydrogen in air impinges on platinum black the metal surface gets hot enough to ignite the gases, [Mellor 1:325(1946-1947)]. Explosive reactions occur upon ignition of mixtures of nitrogen trifluoride with good reducing agents such as ammonia, hydrogen, hydrogen sulfide or methane. Mixtures of hydrogen, carbon monoxide, or methane and oxygen difluoride are exploded when a spark is discharged, [Mellor 2, Supp. 1:192(1956)]. An explosion occurred upon heating 1'-pentol and 1''-pentol under hydrogen pressure. It appears that this acetylenic compound under certain conditions suddenly breaks down to form elemental carbon, hydrogen, and carbon monoxide with the release of sufficient energy to develop pressures in excess of 1000 atmospheres, [AIChE Loss Prevention, p1, (1967)].
CAMEO Chemicals
Finely divided platinum and some other metals will cause a mixture of hydrogen and oxygen to explode at ordinary temperatures. If a jet of hydrogen in air impinges on platinum black the metal surface gets hot enough to ignite the gases, [Mellor 1:325(1946-1947)]. Explosive reactions occur upon ignition of mixtures of nitrogen trifluoride with good reducing agents such as ammonia, hydrogen, hydrogen sulfide or methane. Mixtures of hydrogen, carbon monoxide, or methane and oxygen difluoride are exploded when a spark is discharged, [Mellor 2, Supp. 1:192(1956)]. An explosion occurred upon heating 1'-pentol and 1''-pentol under hydrogen pressure. It appears that this acetylenic compound under certain conditions suddenly breaks down to form elemental carbon, hydrogen, and carbon monoxide with the release of sufficient energy to develop pressures in excess of 1000 atmospheres, [AIChE Loss Prevention, p1, (1967)]. Contact of very cold liquefied gas with water may result in vigorous or violent boiling of the product and extremely rapid vaporization due to the large temperature differences involved. If the water is hot, there is the possibility that a liquid "superheat" explosion may occur. Pressures may build to dangerous levels if liquid gas contacts water in a closed container, [Handling Chemicals Safely 1980].
CAMEO Chemicals
13.8.5 Hazardous Reactivities and Incompatibilities
Release of hydrogen @ 47.5 bar into a vented 17.5-l chromium-plated sphere caused explosive ignition. /Hydrogen/
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1186-7
Hazardous Substances Data Bank (HSDB)
Hydrogen ignites in bromine fluoride at ambient temp.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 90
Hazardous Substances Data Bank (HSDB)
Mixtures of iodine heptafluoride with hydrogen explode on heating or sparking.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1127
Hazardous Substances Data Bank (HSDB)
... Calcium, barium, and strontium react readily, sometimes igniting, in hydrogen above 300 °C. ...
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1189
Hazardous Substances Data Bank (HSDB)
For more Hazardous Reactivities and Incompatibilities (Complete) data for Hydrogen (19 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.9 Transport Information
13.9.1 DOT Emergency Guidelines
/GUIDE 115 GASES - FLAMMABLE (Including Refrigerated Liquids)/ Fire or Explosion: EXTREMELY FLAMMABLE. Will be easily ignited by heat, sparks or flames. Will form explosive mixtures with air. Vapors from liquefied gas are initially heavier than air and spread along ground. CAUTION: Hydrogen (UN1049), Deuterium (UN1957), Hydrogen, refrigerated liquid (UN1966) and Methane (UN1971) are lighter than air and will rise. Hydrogen and Deuterium fires are difficult to detect since they burn with an invisible flame. Use an alternate method of detection (thermal camera, broom handle, etc.) Vapors may travel to source of ignition and flash back. Cylinders exposed to fire may vent and release flammable gas through pressure relief devices. Containers may explode when heated. Ruptured cylinders may rocket. /Hydrogen; Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid)/
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
/GUIDE 115 GASES - FLAMMABLE (Including Refrigerated Liquids)/ Health: Vapors may cause dizziness or asphyxiation without warning. Some may be irritating if inhaled at high concentrations. Contact with gas or liquefied gas may cause burns, severe injury and/or frostbite. Fire may produce irritating and/or toxic gases. /Hydrogen; Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid)/
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
/GUIDE 115 GASES - FLAMMABLE (Including Refrigerated Liquids)/ Public Safety: CALL Emergency Response Telephone Number on Shipping Paper first. If Shipping Paper not available or no answer, refer to appropriate telephone number listed on the inside back cover. As an immediate precautionary measure, isolate spill or leak area for at least 100 meters (330 feet) in all directions. Keep unauthorized personnel away. Stay upwind, uphill and/or upstream. Many gases are heavier than air and will spread along ground and collect in low or confined areas (sewers, basements, tanks). /Hydrogen; Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid)/
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
/GUIDE 115 GASES - FLAMMABLE (Including Refrigerated Liquids)/ Protective Clothing: Wear positive pressure self-contained breathing apparatus (SCBA). Structural firefighters' protective clothing will only provide limited protection. Always wear thermal protective clothing when handling refrigerated/cryogenic liquids. /Hydrogen; Hydrogen, compressed; Hydrogen, refrigerated liquid (cryogenic liquid)/
U.S. Department of Transportation. 2016 Emergency Response Guidebook. Washington, D.C. 2016
Hazardous Substances Data Bank (HSDB)
For more DOT Emergency Guidelines (Complete) data for Hydrogen (8 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
13.9.2 Shipping Name / Number DOT/UN/NA/IMO
UN 1966; Hydrogen, refrigerated liquid (cryogenic liquid)
Hazardous Substances Data Bank (HSDB)
UN 1049; Hydrogen, compressed
Hazardous Substances Data Bank (HSDB)
IMO 2.1; Hydrogen, compressed; Hydrogen, refrigerated liquid
Hazardous Substances Data Bank (HSDB)
13.9.3 Shipment Methods and Regulations
No person may /transport,/ offer or accept a hazardous material for transportation in commerce unless that person is registered in conformance ... and the hazardous material is properly classed, described, packaged, marked, labeled, and in condition for shipment as required or authorized by ... /the hazardous materials regulations (49 CFR 171-177)./
49 CFR 171.2 (USDOT); U.S. National Archives and Records Administration's Electronic Code of Federal Regulations. Available from, as of June 27, 2017:
Hazardous Substances Data Bank (HSDB)
The International Air Transport Association (IATA) Dangerous Goods Regulations are published by the IATA Dangerous Goods Board pursuant to IATA Resolutions 618 and 619 and constitute a manual of industry carrier regulations to be followed by all IATA Member airlines when transporting hazardous materials. Hydrogen, compressed; and hydrogen, refrigerated liquid are included on the dangerous goods list. /Hydrogen, compressed; hydrogen, refrigerated liquid/
International Air Transport Association. Dangerous Goods Regulations. 57th Edition. Montreal, Quebec Canada. 2016., p. 263, 264
Hazardous Substances Data Bank (HSDB)
The International Maritime Dangerous Goods Code lays down basic principles for transporting hazardous chemicals. Detailed recommendations for individual substances and a number of recommendations for good practice are included in the classes dealing with such substances. A general index of technical names has also been compiled. This index should always be consulted when attempting to locate the appropriate procedures to be used when shipping any substance or article. Hydrogen, compressed; and hydrogen, refrigerated liquid are included on the dangerous goods list. /Hydrogen, compressed; hydrogen, refrigerated liquid/
International Maritime Organization. IMDG Code. International Maritime Dangerous Goods Code Volume 2 2014, p. 44, 95
Hazardous Substances Data Bank (HSDB)
13.9.4 DOT Label
Flammable Gas
CAMEO Chemicals
13.9.5 EC Classification
Symbol: F+; R: 12; S: (2)-9-16-33
ILO-WHO International Chemical Safety Cards (ICSCs)
13.9.6 UN Classification
UN Hazard Class: 2.1
ILO-WHO International Chemical Safety Cards (ICSCs)
13.10 Regulatory Information
The Australian Inventory of Industrial Chemicals
Chemical: Hydrogen
Australian Industrial Chemicals Introduction Scheme (AICIS)
EFSA Legal Basis
Regulation (EC) No 1935/2004 (amended)
EFSA OpenFoodTox
New Zealand EPA Inventory of Chemical Status
Hydrogen: HSNO Approval: HSR001002 Approved with controls
New Zealand Environmental Protection Authority (EPA)
New Jersey Worker and Community Right to Know Act
The New Jersey Worker and Community Right to Know Act requires public and private employers to provide information about hazardous substances at their workplaces. (N.J.S.A. 34:5A-1 et. seq.)
NJDOH RTK Hazardous Substance List
13.10.1 DHS Chemicals of Interest (COI)
Chemicals of Interest(COI)
Hydrogen
Release: Minimum Concentration (%)
1
Release: Screening Threshold Quantities (in pounds)
10000
Security Issue: Release - Flammables
Flammable chemical that can be released at a facility.
DHS Chemical Facility Anti-Terrorism Standards (CFATS) Chemicals of Interest
13.11 Other Safety Information
Chemical Assessment
IMAP assessments - Hydrogen: Human health tier I assessment
Australian Industrial Chemicals Introduction Scheme (AICIS)
13.11.1 Other Hazardous Reactions
Conditions to avoid: Heat, flames and sparks.
Sigma-Aldrich; Safety Data Sheet for Hydrogen. Product Number: , Version 3.10 (Revision Date 05/27/2016). Available from, as of June 16, 2017:
Hazardous Substances Data Bank (HSDB)
Palladium oxide glows in contact with hydrogen at ambient temperature.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1393
Hazardous Substances Data Bank (HSDB)
Contact with hydrogen causes unheated palladium trifluoride to be reduced incandescently.
Bretherick, L. Handbook of Reactive Chemical Hazards. 4th ed. Boston, MA: Butterworth-Heinemann Ltd., 1990, p. 1116
Hazardous Substances Data Bank (HSDB)
13.11.2 Special Reports
Harper SR, Pohland FG; Recent developments in hydrogen management during anaerobic biological wastewater treatment; Biotechnol Bioeng 28 (4): 585-602 (1986).
Hazardous Substances Data Bank (HSDB)
14 Toxicity
14.1 Toxicological Information
14.1.1 Toxicity Summary
IDENTIFICATION AND USE: Hydrogen is a colorless gas. Its many uses include the following: production of ammonia, ethanol, and aniline; hydrocracking, hydroforming, and hydrofining of petroleum; hydrogenation of vegetable oils; hydrogenolysis of coal; reducing agent for organic synthesis and metallic ores; reducing atmosphere to prevent oxidation; as oxyhydrogen flame for high temperatures; atomic-hydrogen welding; instrument-carrying balloons; making hydrogen chloride and hydrogen bromide; production of high-purity metals; fuel for nuclear rocket engines for hypersonic transport; missile fuel; cryogenic research. In addition, hydrogen is a versatile energy carrier that can be used to power nearly every end-use energy need (Fuel cells). Molecular hydrogen (H2) emerged as a novel therapeutic agent, with antioxidant, anti-inflammatory and anti-apoptotic effects demonstrated in plethora of animal disease models and human studies. HUMAN STUDIES: Hydrogen is a simple asphyxiant. Contact with liquid hydrogen will cause frostbite or severe burns of the skin. Hydrogen-rich water has been tested for treating oxidative stress-induced disorders because of its reactive oxygen species scavenging abilities. Hydrogen therapy may be an effective and specific innovative treatment for exercise-induced oxidative stress and sports injury, with potential for the improvement of exercise performance. ANIMAL STUDIES: A large bubble of the gas injected into anterior chamber of rabbit eyes was absorbed within three days and caused no injury. H2 was believed to be inert and nonfunctional in mammalian cells. More recently it was demonstrated that H2 reacts with highly reactive oxidants such as hydroxyl radical and peroxynitrite inside cells. Beneficial effects of molecular hydrogen in animal models were observed especially in oxidative stress-mediated diseases, such as diabetes mellitus, brain stem infarction, rheumatoid arthritis, or neurodegenerative diseases. H2 affects cell signal transduction.
Hazardous Substances Data Bank (HSDB)
14.1.2 Exposure Routes
Exposure mainly occurs via inhalation.
ILO-WHO International Chemical Safety Cards (ICSCs)
14.1.3 Signs and Symptoms
Inhalation Exposure
Dizziness. Headache. Lethargy. Suffocation.
ILO-WHO International Chemical Safety Cards (ICSCs)
Skin Exposure
ON CONTACT WITH GAS: FROSTBITE.
ILO-WHO International Chemical Safety Cards (ICSCs)
14.1.4 Adverse Effects
Other Poison - Simple Asphyxiant
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
14.1.5 Interactions
Substantial evidence indicates that molecular hydrogen (H2) has beneficial vascular effects because of its antioxidant and/or anti-inflammatory effects. Thus, hydrogen-rich water may prove to be an effective anti-aging drink. This study examined the effects of H2 on endothelial senescence and clarified the mechanisms involved. Hydrogen-rich medium was produced by a high-purity hydrogen gas generator. Human umbilical vein endothelial cells (HUVECs) were incubated with 2,3,7,8-tetrachlorodibenzo-p-dioxin (TCDD) for various time periods in normal or hydrogen-rich medium. The baseline H2concentration in hydrogen-rich medium was 0.55 +/- 0.07 mmol/L. This concentration gradually decreased, and H2 was almost undetectable in medium after 12 hr. At 24 hr after TCDD exposure, HUVECs treated with TCDD exhibited increased 8OHdG and acetyl-p53 expression, decreased nicotinamide adenine dinucleotide (NAD(+))/NADH ratio, impaired Sirt1 activity, and enhanced senescence-associated beta-galactosidase. However, HUVECs incubated in hydrogen-rich medium did not exhibit these TCDD-induced changes accompanying Nrf2 activation, which was observed even after H2 was undetectable in the medium. Chrysin, an inhibitor of Nrf2, abolished the protective effects of H2 on HUVECs. H2 has long-lasting antioxidant and anti-aging effects on vascular endothelial cells through the Nrf2 pathway, even after transient exposure to H2. Hydrogen-rich water may thus be a functional drink that increases longevity. /Hydrogen-rich water/
PMID:27477846
Hara F et al; Circ J 80 (9): 2037-46 (2016)
Hazardous Substances Data Bank (HSDB)
We observed the effect of hydrogen-rich medium on lipopolysaccharide (LPS)-induced human umbilical vein endothelial cells (HUVECs), hyaline leukocyte conglutination, and permeability of the endothelium. Endotheliocytes were inoculated on 6-well plates and randomly divided into 4 groups: control, H2, LPS, LPS+H2, H2, and LPS+H2 in saturated hydrogen-rich medium. We applied Wright's staining to observe conglutination of hyaline leukocytes and HUVECs, flow cytometry to determine the content of vascular cell adhesion protein 1 (VCAM-1) and intercellular adhesion molecule 1 (ICAM-1), enzyme-linked immunosorbent assay to measure the E-selectin concentration in the cell liquor, the transendothelial electrical resistance (TEER) to test the permeability of endothelial cells, and Western blot and immunofluorescence to test the expression and distribution of vascular endothelial (VE)-cadherin. Compared with control cells, there was an increase in endothelium-hyaline leukocyte conglutination, a reduction in VCAM-1, ICAM-1, and E-selectin, and the TEER value increased obviously. Compared with LPS, there was an obvious reduction in the conglutination of LPS+H2 cells, a reduction in VCAM-1, ICAM-1, and E-selectin levels, and a reduction in the TEER-resistance value, while the expression of VE-cadherin increased. Fluorescence results showed that, compared with control cells, the VE-cadherin in LPS cells was in-complete at the cell joints. Compared with LPS cells, the VE-cadherin in LPS+H2 cells was even and complete at the cell joints. Liquid rich in hydrogen could reduce LPS-induced production of adhesion molecules and endothelium-hyaline leukocyte conglutination, and influence the expression and distribution of VE-cadherin to regulate the permeability of the endothelium. /Hydrogen-rich medium/
PMID:26125821
Yu Y et al; Genet Mol Res 14 (2): 6202-12 (2015)
Hazardous Substances Data Bank (HSDB)
Maternal inflammation is associated with spontaneous preterm birth and respiratory impairment among premature infants. Recently, molecular hydrogen (H2) has been reported to have a suppressive effect on oxidative stress and inflammation. The aim of this study was to evaluate the effects of H2 on fetal lung injury caused by maternal inflammation. Cell viability and the production of interleukin-6 (IL-6) and reactive oxygen species (ROS) were examined by treatment with lipopolysaccharide (LPS) contained in ordinal or H2-rich medium (HM) using a human lung epithelial cell line, A549. Pregnant Sprague Dawley rats were divided into three groups: Control, LPS, and HW + LPS groups. Rats were injected with phosphate-buffered saline (Control) or LPS intraperitoneally (LPS) on gestational day 19 and provided H2 water (HW) ad libitum for 24 hr before LPS injection (HW + LPS). Fetal lung samples were collected on day 20, and the levels of apoptosis, oxidative damage, IL-6, and vascular endothelial growth factor (VEGF) were evaluated using immunohistochemistry. The number of apoptotic cells, and levels of ROS and IL-6 were significantly increased by LPS treatment, and repressed following cultured with HM in A549 cells. In the rat models, the population positive for cleaved caspase-3, 8-hydroxy-2'-deoxyguanosine, IL-6, and VEGF was significantly increased in the LPS group compared with that observed in the Control group and significantly decreased in the HW + LPS group. In this study, LPS administration induced apoptosis and oxidative damage in fetal lung cells that was ameliorated by maternal H2 intake. Antenatal H2 administration may decrease the pulmonary mobility associated with inflammation in premature infants. /Hydrogen-rich medium/
PMID:25947958
Hattori Y et al; Free Radic Res 49 (8): 1026-37 (2015)
Hazardous Substances Data Bank (HSDB)
An anesthetized rat preparation was used for breathing studies in hyperbaric H2/CO2 atmospheres in which total gaseous pressure was varied in the safe pressure range 8.3-19.1 atm. PCO2 (partial pressure of CO2) was studied at levels of 0.05 and 0.10 atm. The response variables measured were respiration rate, blood pH and an isotonic index of diaphragm twitch amplitude, in the presence of He/O2 gas mixtures used for compression and H2/O2/Co2 saturation gases. The rat preparation can be well maintained in H2/O2 environments at pressures in excess of 8 atm. Elevated PCO2 in these environments can evoke rapid increases in respiration rate and twitch amplitude, without any marked changes in blood pH. Graded increases in PH2 (partial pressure of H2) are able to lower and ultimately abolish the effects of either 0.05 or 0.10 atm levels of CO2 in breathing parameters. The intrinsic ability of hyperbaric H2 to effect a CO2-sparing action on this preparation is greater than that previously seen with hyperbaric Ar or He. The potency sequence for the sparing action by the inert component of the hyperbaric breathing gas is H2 > Ar > He.
Friess SL et al; Toxicol Appl Pharmacol; 46 (3): 717-726 (1978; RECD. 1979)
Hazardous Substances Data Bank (HSDB)
For more Interactions (Complete) data for Hydrogen (30 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.6 Antidote and Emergency Treatment
Immediate first aid: Ensure that adequate decontamination has been carried out. If patient is not breathing, start artificial respiration, preferably with a demand-valve resuscitator, bag-valve-mask device, or pocket mask, as trained. Perform CPR as necessary. Immediately flush contaminated eyes with gently flowing water. Do not induce vomiting. If vomiting occurs, lean patient forward or place on left side (head-down position, if possible) to maintain an open airway and prevent aspiration. Keep patient quiet and maintain normal body temperature. Obtain medical attention. /Flammable gases/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 140-1
Hazardous Substances Data Bank (HSDB)
Basic treatment: Establish a patent airway (oropharyngeal or nasopharyngeal airway, if needed). Suction if necessary. Watch for signs of respiratory insufficiency and assist ventilations as necessary. Administer oxygen by nonrebreather mask at 10 to 15 L/min. Monitor for pulmonary edema and treat if necessary ... . Monitor for shock and treat if necessary ... . Anticipate seizures and treat if necessary ... . For eye contamination, flush eyes immediately with water. Irrigate each eye continuously with 0.9% saline (NS) during transport ... . Treat frostbite with rapid rewarming ... . /Flammable gases/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 141
Hazardous Substances Data Bank (HSDB)
Advanced treatment: Consider orotracheal or nasotracheal intubation for airway control in the patient who is unconscious, has severe pulmonary edema, or is in severe respiratory distress. Positive-pressure ventilation techniques with a bag valve mask device may be beneficial. Consider drug therapy for pulmonary edema ... . Monitor cardiac rhythm and treat arrhythmias as necessary ... . Start IV administration of D5W /SRP: "To keep open", minimal flow rate/. Use 0.9% saline (NS) or lactated Ringer's (LR) if signs of hypovolemia are present. For hypotension with signs of hypovolemia, administer fluid cautiously. Watch for signs of fluid overload ... . Treat seizures with diazepam or lorazepam ... . Use proparacaine hydrochloride to assist eye irrigation ... . /Flammable gases/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 141
Hazardous Substances Data Bank (HSDB)
Immediate first aid: Ensure that adequate decontamination has been carried out. If patient is not breathing, start artificial respiration, preferably with a demand-valve resuscitator, bag-valve-mask device, or pocket mask, as trained. Perform CPR as necessary. Immediately flush contaminated eyes with gently flowing water. Do not induce vomiting. If vomiting occurs, lean patient forward or place on left side (head-down position, if possible) to maintain an open airway and prevent aspiration. Keep patient quiet and maintain normal body temperature. Obtain medical attention. /Simple asphyxiants and related compounds/
Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3rd revised edition, Elsevier Mosby, St. Louis, MO 2007, p. 449
Hazardous Substances Data Bank (HSDB)
For more Antidote and Emergency Treatment (Complete) data for Hydrogen (6 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.7 Human Toxicity Excerpts
/SIGNS AND SYMPTOMS/ No specific toxic action. In high concn can act as a simple asphyxiant.
The Merck Index. 10th ed. Rahway, New Jersey: Merck Co., Inc., 1983., p. 695
Hazardous Substances Data Bank (HSDB)
/SIGNS AND SYMPTOMS/ Contact with cryogenic liquid hydrogen will cause frostbite or severe burns of the skin. Simple asphyxiant.
National Fire Protection Association; Fire Protection Guide to Hazardous Materials. 14TH Edition, Quincy, MA 2010, p. 49-86
Hazardous Substances Data Bank (HSDB)
/SIGNS AND SYMPTOMS/ The relation between the /CNS depressant/ effect of nitrogen and that of hydrogen is 1:0.26.
International Labour Office. Encyclopedia of Occupational Health and Safety. Vols. I&II. Geneva, Switzerland: International Labour Office, 1983., p. 1087
Hazardous Substances Data Bank (HSDB)
/SIGNS AND SYMPTOMS/ Hydrogen has no known toxic effect on the eye.
Grant, W. M. Toxicology of the Eye. 2nd ed. Springfield, Illinois: Charles C. Thomas, 1974., p. 559
Hazardous Substances Data Bank (HSDB)
For more Human Toxicity Excerpts (Complete) data for Hydrogen (11 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.1.8 Non-Human Toxicity Excerpts
/LABORATORY ANIMALS: Acute Exposure/ The nuclear protein high-mobility group box 1 (HMGB1) is a key trigger for the inflammatory reaction during liver ischemia reperfusion injury (IRI). Hydrogen treatment was recently associated with down-regulation of the expression of HMGB1 and pro-inflammatory cytokines during sepsis and myocardial IRI, but it is not known whether hydrogen has an effect on HMGB1 in liver IRI. A rat model of 60 minutes 70% partial liver ischemia reperfusion injury was used. Hydrogen enriched saline (2.5, 5 or 10 mL/kg) was injected intraperitoneally 10 minutes before hepatic reperfusion. Liver injury was assessed by serum alanine aminotransferase (ALT) enzyme levels and histological changes. We also measured malondialdehyde (MDA), hydroxynonenal (HNE) and 8-hydroxy-guanosine (8-OH-G) levels as markers of the peroxidation injury induced by reactive oxygen species (ROS). In addition, pro-inflammatory cytokines including TNF-a and IL-6, and high mobility group box B1 protein (HMGB1) were measured as markers of post ischemia-reperfusion inflammation. Hydrogen enriched saline treatment significantly attenuated the severity of liver injury induced by ischemia-reperfusion. The treatment group showed reduced serum ALT activity and markers of lipid peroxidation and post ischemia reperfusion histological changes were reduced. Hydrogen enriched saline treatment inhibited HMGB1 expression and release, reflecting a reduced local and systemic inflammatory response to hepatic ischemia reperfusion. These results suggest that, in our model, hydrogen enriched saline treatment is protective against liver ischemia-reperfusion injury. This effect may be mediated by both the anti-oxidative and anti-inflammatory effects of the solution. /Hydrogen enriched saline/
PMID:24410860
Full text:
Liu Y et al; BMC Gastroenterol 14: 12 (2014)
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Subchronic or Prechronic Exposure/ Hydrogen has been reported to relieve damage in many disease models, and is a potential additive in drinking water to provide protective effects for patients as several clinical studies revealed. However, the absence of a dose-response relationship in the application of hydrogen is puzzling. We attempted to identify the dose-response relationship of hydrogen in alkaline electrolyzed drinking water through the aspirin induced gastric injury model. In this study, hydrogen-rich alkaline water was obtained by adding H2 to electrolyzed water at one atmosphere pressure. After 2 weeks of drinking, we detected the gastric mucosal damage together with MPO, MDA and 8-OHdG in rat aspirin induced gastric injury model. Hydrogen-dose dependent inhibition was observed in stomach mucosal. Under pH 8.5, 0.07, 0.22 and 0.84 ppm hydrogen exhibited a high correlation with inhibitory effects showed by erosion area, MPO activity and MDA content in the stomach. Gastric histology also demonstrated the inhibition of damage by hydrogen-rich alkaline water. However, 8-OHdG level in serum did not have significant hydrogen-dose dependent effect. pH 9.5 showed higher but not significant inhibitory response compared with pH 8.5. Hydrogen is effective in relieving the gastric injury induced by aspirin-HCl, and the inhibitory effect is dose-dependent. The reason behind this may be that hydrogen-rich water directly interacted with the target tissue, while the hydrogen concentration in blood was buffered by liver glycogen, evoking a suppressed dose-response effect. Drinking hydrogen-rich water may protect healthy individuals from gastric damage caused by oxidative stress. /Hydrogen water/
PMID:24589018
Full text:
Xue J et al; BMC Complement Altern Med 14: 81 (2014)
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Subchronic or Prechronic Exposure/ The prevalence of sleep apnea is very high in patients with heart failure (HF). The aims of this study were to investigate the influence of intermittent hypoxia (IH) on the failing heart and to evaluate the antioxidant effect of hydrogen gas. Normal male Syrian hamsters (n = 22) and cardiomyopathic (CM) hamsters (n = 33) were exposed to IH (repeated cycles of 1.5 min of 5% oxygen and 5 min of 21% oxygen for 8 hr during the daytime) or normoxia for 14 days. Hydrogen gas (3.05 vol/100 vol) was inhaled by some CM hamsters during hypoxia. IH increased the ratio of early diastolic mitral inflow velocity to mitral annulus velocity (E/e', 21.8 vs. 16.9) but did not affect the LV ejection fraction (EF) in normal Syrian hamsters. However, IH increased E/e' (29.4 vs. 21.5) and significantly decreased the EF (37.2 vs. 47.2%) in CM hamsters. IH also increased the cardiomyocyte cross-sectional area (672 vs. 443 sq um) and interstitial fibrosis (29.9 vs. 9.6%), along with elevation of oxidative stress and superoxide production in the left ventricular (LV) myocardium. Furthermore, IH significantly increased the expression of brain natriuretic peptide, beta-myosin heavy chain, c-fos, and c-jun mRNA in CM hamsters. Hydrogen gas inhalation significantly decreased both oxidative stress and embryonic gene expression, thus preserving cardiac function in CM hamsters. In conclusion, IH accelerated LV remodeling in CM hamsters, at least partly by increasing oxidative stress in the failing heart. These findings might explain the poor prognosis of patients with HF and sleep apnea.
PMID:25281567
Kato R et al; Am J Physiol Heart Circ Physiol 307 (11): H1626-33 (2014)
Hazardous Substances Data Bank (HSDB)
/LABORATORY ANIMALS: Developmental or Reproductive Toxicity/ Maternal inflammation is associated with spontaneous preterm birth and respiratory impairment among premature infants. Recently, molecular hydrogen (H2) has been reported to have a suppressive effect on oxidative stress and inflammation. The aim of this study was to evaluate the effects of H2 on fetal lung injury caused by maternal inflammation. Cell viability and the production of interleukin-6 (IL-6) and reactive oxygen species (ROS) were examined by treatment with lipopolysaccharide (LPS) contained in ordinal or H2-rich medium (HM) using a human lung epithelial cell line, A549. Pregnant Sprague Dawley rats were divided into three groups: Control, LPS, and HW + LPS groups. Rats were injected with phosphate-buffered saline (Control) or LPS intraperitoneally (LPS) on gestational day 19 and provided H2 water (HW) ad libitum for 24 hr before LPS injection (HW + LPS). Fetal lung samples were collected on day 20, and the levels of apoptosis, oxidative damage, IL-6, and vascular endothelial growth factor (VEGF) were evaluated using immunohistochemistry. The number of apoptotic cells, and levels of ROS and IL-6 were significantly increased by LPS treatment, and repressed following cultured with HM in A549 cells. In the rat models, the population positive for cleaved caspase-3, 8-hydroxy-2'-deoxyguanosine, IL-6, and VEGF was significantly increased in the LPS group compared with that observed in the Control group and significantly decreased in the HW + LPS group. In this study, LPS administration induced apoptosis and oxidative damage in fetal lung cells that was ameliorated by maternal H2 intake. Antenatal H2 administration may decrease the pulmonary mobility associated with inflammation in premature infants. /Hydrogen-rich medium/
PMID:25947958
Hattori Y et al; Free Radic Res 49 (8): 1026-37 (2015)
Hazardous Substances Data Bank (HSDB)
For more Non-Human Toxicity Excerpts (Complete) data for Hydrogen (16 total), please visit the HSDB record page.
Hazardous Substances Data Bank (HSDB)
14.2 Ecological Information
14.2.1 Natural Pollution Sources
... Hydrogen is the most abundant of all elements in the universe, and it is thought that the heavier elements were, and still are, being built from hydrogen and helium. It has been estimated that hydrogen makes up more than 90% of all the atoms or three quarters of the mass of the universe. It is found in the sun and most stars, and plays an important part in the proton-proton reaction and carbon-nitrogen cycle, which accounts for the energy of the sun and stars. It is thought that hydrogen is a major component of the planet Jupiter and that at some depth in the planet's interior the pressure is so great that solid hydrogen is converted into solid metallic hydrogen.
Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 95th Edition. CRC Press LLC, Boca Raton: FL 2014-2015, p. 4-17
Hazardous Substances Data Bank (HSDB)
Occurrence in earth's atmosphere 0.00005% H2.
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 887
Hazardous Substances Data Bank (HSDB)
Hydrogen from natural sources consists of more than 99.8% protium. ...
Osol, A. (ed.). Remington's Pharmaceutical Sciences. 16th ed. Easton, Pennsylvania: Mack Publishing Co., 1980., p. 346
Hazardous Substances Data Bank (HSDB)
14.2.2 Probable Routes of Human Exposure
According to the 2016 TSCA Inventory Update Reporting data, nn reporting facilities estimate the number of persons reasonably likely to be exposed during the manufacturing, processing, or use of hydrogen in the United States may be as low as <10 workers and as high as <10,000 workers per plant; the data may be greatly underestimated due to confidential business information (CBI) or unknown values(1).
(1) US EPA; Chemical Data Reporting (CDR). Non-confidential 2012 Chemical Data Reporting information on chemical production and use in the United States. Available from, as of Jun 22, 2017:
Hazardous Substances Data Bank (HSDB)
14.3 EFSA OpenFoodTox
14.3.1 EFSA Outputs
EFSA OpenFoodTox
14.3.2 EFSA Genotoxicity
EFSA OpenFoodTox
15 Associated Disorders and Diseases
Comparative Toxicogenomics Database (CTD)
Associated Occupational Diseases with Exposure to the Compound
Asphyxiation, simple [Category: Acute Poisoning]
Haz-Map, Information on Hazardous Chemicals and Occupational Diseases
16 Literature
16.1 Consolidated References
PubChem
16.2 NLM Curated PubMed Citations
Medical Subject Headings (MeSH)
16.3 Thieme References
Thieme Chemistry
16.4 Wiley References
Wiley
16.5 Chemical Co-Occurrences in Literature
PubChem
16.6 Chemical-Gene Co-Occurrences in Literature
PubChem
16.7 Chemical-Disease Co-Occurrences in Literature
PubChem
16.8 Chemical-Organism Co-Occurrences in Literature
PubChem
17 Patents
17.1 Depositor-Supplied Patent Identifiers
PubChem
Link to all deposited patent identifiers
PubChem
17.2 WIPO PATENTSCOPE
Patents are available for this chemical structure:
PATENTSCOPE (WIPO)
17.3 Chemical Co-Occurrences in Patents
PubChem
17.4 Chemical-Disease Co-Occurrences in Patents
PubChem
17.5 Chemical-Gene Co-Occurrences in Patents
PubChem
17.6 Chemical-Organism Co-Occurrences in Patents
PubChem
18 Interactions and Pathways
18.1 Protein Bound 3D Structures
View 1 protein in NCBI Structure
PubChem
18.2 Chemical-Target Interactions
Comparative Toxicogenomics Database (CTD)
18.3 Pathways
PubChem
19 Taxonomy
ECI Group, LCSB, University of Luxembourg
20 Classification
20.1 MeSH Tree
Medical Subject Headings (MeSH)
20.2 ChEBI Ontology
ChEBI
20.3 ChemIDplus
ChemIDplus
20.4 CAMEO Chemicals
CAMEO Chemicals
20.5 UN GHS Classification
GHS Classification (UNECE)
20.6 EPA CPDat Classification
EPA Chemical and Products Database (CPDat)
20.7 NORMAN Suspect List Exchange Classification
NORMAN Suspect List Exchange
20.8 EPA DSSTox Classification
EPA DSSTox
20.9 EPA TSCA and CDR Classification
EPA Chemicals under the TSCA
20.10 EPA Substance Registry Services Tree
EPA Substance Registry Services
20.11 MolGenie Organic Chemistry Ontology
MolGenie
20.12 Chemicals in PubChem from Regulatory Sources
PubChem
21 Information Sources
Filter by Source
Australian Industrial Chemicals Introduction Scheme (AICIS)LICENSE Hydrogen Hydrogen
CAMEO ChemicalsLICENSE CAMEO Chemicals and all other CAMEO products are available at no charge to those organizations and individuals (recipients) responsible for the safe handling of chemicals. However, some of the chemical data itself is subject to the copyright restrictions of the companies or organizations that provided the data. HYDROGEN HYDROGEN, REFRIGERATED LIQUID (CRYOGENIC LIQUID) CAMEO Chemical Reactivity Classification
CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. Hydrogen, mol. (H6) Hydrogen Hydrogen, ion (H21+) Atomic hydrogen Hydrogen, mol. (H4) Hydrogen, mol. (H12)
ChemIDplusLICENSE Hydrogen Hydrogen, atomic ChemIDplus Chemical Information Classification
DHS Chemical Facility Anti-Terrorism Standards (CFATS) Chemicals of InterestHydrogen
DrugBankLICENSE Creative Common's Attribution-NonCommercial 4.0 International License ( Hydrogen
EFSA OpenFoodToxLICENSE Hydrogen gas
EPA Chemical Data Reporting (CDR)LICENSE The U.S. Government retains a nonexclusive, royalty-free license to publish or reproduce these documents, or allow others to do so, for U.S. Government purposes. These documents may be freely distributed and used for non-commercial, scientific and educational purposes. Hydrogen
EPA Chemicals under the TSCALICENSE Hydrogen EPA TSCA Classification
EPA DSSToxLICENSE Hydrogen CompTox Chemicals Dashboard Chemical Lists
European Chemicals Agency (ECHA)LICENSE Use of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for non-commercial purposes provided that ECHA is acknowledged as the source: "Source: European Chemicals Agency, Such acknowledgement must be included in each copy of the material. ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. Hydrogen Hydrogen (EC: 215-605-7)
FDA Global Substance Registration System (GSRS)LICENSE Unless otherwise noted, the contents of the FDA website (www.fda.gov), both text and graphics, are not copyrighted. They are in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from FDA. Credit to the U.S. Food and Drug Administration as the source is appreciated but not required. HYDROGEN
Hazardous Substances Data Bank (HSDB)LICENSE Hydrogen
Human Metabolome Database (HMDB)LICENSE HMDB is offered to the public as a freely available resource. Use and re-distribution of the data, in whole or in part, for commercial purposes requires explicit permission of the authors and explicit acknowledgment of the source material (HMDB) and the original publication (see the HMDB citing page). We ask that users who download significant portions of the database cite the HMDB paper in any resulting publications. Hydrogen
ILO-WHO International Chemical Safety Cards (ICSCs)LICENSE Creative Commons CC BY 4.0 HYDROGEN
New Zealand Environmental Protection Authority (EPA)LICENSE This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International licence. Hydrogen
NJDOH RTK Hazardous Substance ListLICENSE hydrogen
EU Food Improvement AgentsHYDROGEN
Emergency Response Guidebook (ERG)LICENSE Hydrogen
Haz-Map, Information on Hazardous Chemicals and Occupational DiseasesLICENSE Copyright (c) 2022 Haz-Map(R). All rights reserved. Unless otherwise indicated, all materials from Haz-Map are copyrighted by Haz-Map(R). No part of these materials, either text or image may be used for any purpose other than for personal use. Therefore, reproduction, modification, storage in a retrieval system or retransmission, in any form or by any means, electronic, mechanical or otherwise, for reasons other than personal use, is strictly prohibited without prior written permission. Hydrogen
ChEBIDihydrogen ChEBI Ontology
RRUFF Projecthydrogen
ClinicalTrials.govLICENSE The ClinicalTrials.gov data carry an international copyright outside the United States and its Territories or Possessions. Some ClinicalTrials.gov data may be subject to the copyright of third parties; you should consult these entities for any additional terms of use.
Comparative Toxicogenomics Database (CTD)LICENSE It is to be used only for research and educational purposes. Any reproduction or use for commercial purpose is prohibited without the prior express written permission of NC State University. Hydrogen
DTP/NCILICENSE Unless otherwise indicated, all text within NCI products is free of copyright and may be reused without our permission. Credit the National Cancer Institute as the source. T101 Monoclonal Antibody
ECI Group, LCSB, University of LuxembourgLICENSE Data: CC-BY 4.0; Code: Artistic-2.0 Hydrogen
EPA Chemical and Products Database (CPDat)LICENSE Hydrogen EPA CPDat Classification
Regulation (EC) No 1272/2008 of the European Parliament and of the CouncilLICENSE The copyright for the editorial content of this source, the summaries of EU legislation and the consolidated texts, which is owned by the EU, is licensed under the Creative Commons Attribution 4.0 International licence. hydrogen
Hazardous Chemical Information System (HCIS), Safe Work Australiahydrogen
NITE-CMCHydrogen - FY2006 (New/original classication) Hydrogen - FY2018 (Revised classification)
FDA Generally Recognized as Safe (GRAS)LICENSE Hydrogen gas
FDA Packaging & Food Contact Substances (FCS)LICENSE HYDROGEN
FooDBLICENSE FooDB is offered to the public as a freely available resource. Use and re-distribution of the data, in whole or in part, for commercial purposes requires explicit permission of the authors and explicit acknowledgment of the source material (FooDB) and the original publication. Hydrogen
KEGGLICENSE Academic users may freely use the KEGG website. Non-academic use of KEGG generally requires a commercial license
NIPH Clinical Trials Search of Japan
NIST Mass Spectrometry Data CenterLICENSE Data covered by the Standard Reference Data Act of 1968 as amended. Hydrogen
SpectraBaseHYDROGEN
Pistoia Alliance Chemical Safety LibraryLICENSE Nitrous oxide + Nitric oxide + Sodium + Hydrogen + Sodium formate + Toluene
PubChem ElementsLICENSE Hydrogen
Rhea - Annotated Reactions DatabaseLICENSE Rhea has chosen to apply the Creative Commons Attribution License ( This means that you are free to copy, distribute, display and make commercial use of the database in all legislations, provided you credit (cite) Rhea.
SpringerMaterialsmolecular hydrogen
Thieme ChemistryLICENSE The Thieme Chemistry contribution within PubChem is provided under a CC-BY-NC-ND 4.0 license, unless otherwise stated.
US Code of Federal Regulations, Hazardous Materials, 49 CFR Part 172LICENSE Hydrogen, compressed
WikidataLICENSE CCZero dihydrogen cation
WikipediaHydrogen
Wiley
Medical Subject Headings (MeSH)LICENSE Works produced by the U.S. government are not subject to copyright protection in the United States. Any such works found on National Library of Medicine (NLM) Web sites may be freely used or reproduced without permission in the U.S. Hydrogen MeSH Tree
PubChem
GHS Classification (UNECE)GHS Classification
NORMAN Suspect List ExchangeLICENSE Data: CC-BY 4.0; Code (hosted by ECI, LCSB): Artistic-2.0 NORMAN Suspect List Exchange Classification
EPA Substance Registry ServicesLICENSE EPA SRS List Classification
MolGenieLICENSE CC-BY 4.0 MolGenie Organic Chemistry Ontology
PATENTSCOPE (WIPO)SID 403450257
Cite
Download
CONTENTS
Title and Summary
1 Structures Expand this menu
2 Names and Identifiers Expand this menu
3 Chemical and Physical Properties Expand this menu
4 Spectral Information Expand this menu
5 Related Records Expand this menu
6 Chemical Vendors
7 Drug and Medication Information Expand this menu
8 Food Additives and Ingredients Expand this menu
9 Minerals
10 Pharmacology and Biochemistry Expand this menu
11 Use and Manufacturing Expand this menu
12 Identification Expand this menu
13 Safety and Hazards Expand this menu
14 Toxicity Expand this menu
15 Associated Disorders and Diseases
16 Literature Expand this menu
17 Patents Expand this menu
18 Interactions and Pathways Expand this menu
19 Taxonomy
20 Classification Expand this menu
21 Information Sources
Connect with NLM
Twitter
Facebook
YouTube
National Library of Medicine 8600 Rockville Pike, Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
7762 | https://boredofstudies.org/threads/parabola-x-2-4ay.48933/ | Parabola? x^2=4ay | Bored Of Studies
Menu
Home
About UsMission StatementThe FoundersContact Us
ForumsNew posts
Notes & ResourcesAdd ResourceLatest reviews
What's newNew postsNew resourcesLatest activity
Advertise
Revision & ExamsStudy Tips
Tertiary Education GuidesUniversity GuideTAFE Guide
MembersCurrent visitors
Log inRegister
What's new
New posts
Menu
Log in
Register
Want to level up your HSC prep on those harder Maths questions?Register now for the BoS Trials (7th October)!
Make a Difference – Donate to Bored of Studies!
Students helping students, join us in improving Bored of Studies by donating and supporting future students!
Donate Now →
Home
Forums
Secondary Education
Maths
Mathematics Advanced
Parabola? x^2=4ay (1 Viewer)
Thread starterrsingh
Start dateNov 7, 2004
R
rsingh
Member
Joined Oct 2, 2004 Messages 186 Gender Male HSC 2005
Nov 7, 2004
#1
Hey guys,
You know the parabola x^2=4ay .. could someone explain why we put the parabola in this form, and all of its features, like focus, directrix, focal length, latus rectum?
I just don't get it. It's weird.
Thanks.
Xayma
Lacking creativity
Joined Sep 6, 2003 Messages 5,953 Gender Undisclosed HSC N/A
Nov 7, 2004
#2
The parabola x2=±4ay, or the general from (x-j)2=4a(y-k) is put in that form because it becomes easier to recognise the features.
For the parabola (x-j)2=4a(y-k), [x2=4ay being where j=k=0]
Its vertex lies at (j,k)
It has a focal length of a units and a focus at (j,k+a), the focus is the point at which any lines coming into the parabola parallel to the y-axis will be reflected into.
The directix (the directix and the focus being equal length away from any point on the parabola) is at y=k-a.
Latus Rectum is the focal chord (a chored passing through the focus) that is parallel to the directix.
R
rsingh
Member
Joined Oct 2, 2004 Messages 186 Gender Male HSC 2005
Nov 7, 2004
#3
Thanks a lot.
So we use this form because it's easier to recgonise the features than ax^2+bx+c=0?
But they're both the same?
Trev
stix
Joined Jun 28, 2004 Messages 2,035 LocationPine Palace, St. Lucia, Brisbane.Gender Male HSC 2005
Nov 7, 2004
#4
yea its both the same, but if u need to draw it or work out applications in relation to the parabola its easier ps. the x2=±4ay or (x-j)2=4a(y-k) form both have 'y' in it, the form u just mentioned doesnt - so u cant write parabolas in that form then (correct me if im wrong)
M
mojako
Active Member
Joined Mar 27, 2004 Messages 1,333 Gender Male HSC 2004
Nov 7, 2004
#5
Trev said:
form both have 'y' in it, the form u just mentioned doesnt
Click to expand...
it should have y.
it should be y=ax^2+bx+c,
NOT ax^2+bx+c=0
since ax^2+bx+c=0 is NOT a parabola AT ALL
(well... one can say that it's a kind of parabola.. it's a matter of interpretation)
EDIT:
whoops.. I didn't read this line:
Trev said:
so u cant write parabolas in that form then
Click to expand...
sorry
Last edited: Nov 7, 2004
withoutaface
Premium Member
Joined Jul 14, 2004 Messages 15,098 Gender Male HSC 2004
Nov 7, 2004
#6
I think you can write parabolas in that form, it just moves the parabola up the axes, or to the side. I think the definition of a parabola is the graph of a quadratic equation.
You must log in or register to reply here.
Share:
FacebookTwitterRedditPinterestTumblrWhatsAppEmailLink
Users Who Are Viewing This Thread (Users: 0, Guests: 1)
Home
Forums
Secondary Education
Maths
Mathematics Advanced
Default Style
Contact us
Terms and rules
Privacy policy
Help
Home
RSS
© Copyright 2002-2025 iStudy Australia Pty Ltd
Forum software by XenForo®© 2010-2019 XenForo Ltd.
Top |
7763 | https://math.stackexchange.com/questions/2123450/nonorthogonal-decomposition-of-a-vector-in-terms-of-orthogonal-projections | linear algebra - Nonorthogonal decomposition of a vector in terms of orthogonal projections - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Nonorthogonal decomposition of a vector in terms of orthogonal projections
Ask Question
Asked 8 years, 8 months ago
Modified8 years, 8 months ago
Viewed 287 times
This question shows research effort; it is useful and clear
0
Save this question.
Show activity on this post.
In R 2, given one-dimensional disjoint subspaces V 1 and V 2, and orthogonal projections P 1 and P 2, I can write the decomposition for v in the conic combination of V 1 and V 2 as
v=‖v−P 2 v‖2 c 1‖P 1 v‖2 P 1 v+‖v−P 1 v‖2 c 2‖P 2 v‖2 P 2 v
where c 1=c 2=1√1−(⟨P 1 v,P 2 v⟩)2‖P 1 v‖2 2‖P 2 v‖2 2
This was obtained by doing some basic high school geometry. I have a few questions about this.
When v is not in the conic combination of V 1 and V 2 it seems like c i is replaced by −c i in one of the two summands of the decomposition. Clearly this is from c i looking like the sin of the angle V 1 and V 2 make. I wanted an expression with P i and v only so I used the cosine expression in terms of the dot product. c i is replaced in the V i for which v is "farther" away from and gets projected to the negative part. Is there any way to quickly check that a vector v would be in the conic combination or outside of it?
I also am curious about ways to generalize this decomposition to higher dimensions.
Thank you, I have some quick and dirty matlab code if anyone wants to test this out.
b1 = [2;3]; %basis for V_1
b2 = [-1;-3]; %basis for V_2
v = [-1;0];
P1 = b1inv(b1'b1)b1';
P2 = b2inv(b2'b2)b2';
c = sqrt(1- (dot(P1v,P2v)/norm(P1v)/norm(P2v))^2);
v1 = norm(v-P2v)/norm(P1v)/cP1v;
v2 = norm(v-P1v)/norm(P2v)/cP2v;
linear-algebra
plane-geometry
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Feb 1, 2017 at 4:00
bringingdownthegaussbringingdownthegauss
asked Feb 1, 2017 at 3:48
bringingdownthegaussbringingdownthegauss
33 3 3 bronze badges
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
2
Save this answer.
Show activity on this post.
Compute the following angles:
θ 1=arccos(⟨v 1,v⟩‖v 1‖‖v‖)θ 2=arccos(⟨v 2,v⟩‖v 2‖‖v‖)θ=arccos(⟨v 1,v 2⟩‖v 1‖‖v 2‖)
If |θ 1|+|θ 2|=|θ|, then v∈Cone(v 1,v 2).
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Feb 1, 2017 at 4:50
bringingdownthegauss
33 3 3 bronze badges
answered Feb 1, 2017 at 4:29
DanDan
21 2 2 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
linear-algebra
plane-geometry
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
15Showing vectors span a vector space by definition
0how to show that the set of mutually perpendular unit vectors is basis of vector space
6Finding the Equations of Motion in a Classical Mechanics Problem
0General Basis Proof
0Linear combinations over the complex field, why complex coefficients?
0Inner product - orthogonal basis
1A is a linearly dependent set iff there are some vectors in A can be expressed as a linear combination of the others.
Hot Network Questions
What is this chess h4 sac known as?
Suspicious of theorem 36.2 in Munkres “Analysis on Manifolds”
Change default Firefox open file directory
Calculating the node voltage
Cannot build the font table of Miama via nfssfont.tex
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
Analog story - nuclear bombs used to neutralize global warming
What's the expectation around asking to be invited to invitation-only workshops?
Matthew 24:5 Many will come in my name!
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Direct train from Rotterdam to Lille Europe
alignment in a table with custom separator
What meal can come next?
How to rsync a large file by comparing earlier versions on the sending end?
How to home-make rubber feet stoppers for table legs?
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
How to use \zcref to get black text Equation?
Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road?
Identifying a movie where a man relives the same day
With with auto-generated local variables
What is a "non-reversible filter"?
For every second-order formula, is there a first-order formula equivalent to it by reification?
Bypassing C64's PETSCII to screen code mapping
Is direct sum of finite spectra cancellative?
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7764 | https://dspillustrations.com/pages/posts/misc/decibel-conversion-factor-10-or-factor-20.html | Decibel conversion: Factor 10 or Factor 20? - DSPIllustrations.com
Toggle navigationDSP Illustrations
Home
About
Tutorials
Hire me!
Login
Decibel Conversion: Factor 10 or 20?¶
The decibel is used in a wide range of applications. Decibels are especially used, when a referring to power or a derived measure, which values can vary in a wide range. The most prominent usage of decibels is in sound volume. So, for example a sound of 0dB is barely hearable, whereas a vaccum cleaner on average has 75dB and a rock concert reaches about 110dB.
In class, you often find the following definitions for something in decibels
X d B=10 log 10(X l i n X r e f).X d B=10 log 10(X l i n X r e f).
This equation transforms quantity X l i n X l i n from linear scale to a quantity in dB scale X d B X d B. In order to do that, first the linear quantity is related to a reference quantity X r e f X r e f and the ratio of both is transformed into the log-domain. Apparently, X d B X d B does actually have no unit and we artificially add dB to make clear we are in logarithmic scale. When X l i n X l i n equals the reference level, the dB-scale becomes zero:
X d B=10 log 10(X l i n X r e f)∣∣∣X l i n=X r e f=0 d B.X d B=10 log 10(X l i n X r e f)|X l i n=X r e f=0 d B.
Furthermore, when X l i n>X r e f X l i n>X r e f, X d B X d B is positive and if $X_{lin}
So far, so good. But, on the other hand, more often than not, you also see the following definition:
Y d B=20 log 10(Y Y r e f),Y d B=20 log 10(Y Y r e f),
i.e. the factor before the logarithm is 20 instead of 10. Which version is correct? Both formulas are correct, but you need to take care when you use which formula. So, here's the rule:
If you transform a quantity that relates to power or energy, the factor is 10.
If you transform a quantity that relates to amplitude, the factor is 20.
How do you know, if a quantity relates to power or amplitude? Check if squaring the quantity makes sense! If squaring makes sense, your quantity is likely to be an amplitude. If it does not make sense, you probably face a power-like quantity. For example, it does not make sense to square the power of a signal. But, it makes sense to square the amplitude of a signal, since this yields the power of the signal.
Another indicator for when to use 10 or 20 is the following:
If your quantity X l i n X l i n follows from some computation that involves squaring some other quantity, use factor 10.
Some examples¶
In the following we calculate some examples of linear scale to dB scale conversions:
Transform 100W into the dB scale. Reference X r e f=1 m W X r e f=1 m W.
print ("X_dB = %.1fdB" % (10np.log10(100/1e-3)))
X_dB = 50.0dB
We have to use factor 10, because we are clearly dealing with powers of a signal.
Transform 100V into the dB scale Reference X r e f=1 m V X r e f=1 m V.
print ("X_dB = %.1fdB" % (20np.log10(100/1e-3)))
X_dB = 100.0dB
Here, we use factor 20, because we deal with voltage amplitudes of a signal.
Transform an air pressure of 5 m P a 5 m P a into the dB scale. Reference X r e f=20 μ P a X r e f=20 μ P a.
print ("X_dB = %.1fdB" % (20np.log10(5e-3/20e-6)))
X_dB = 48.0dB
This one is a bit more tricky. Is air pressure an amplitude or power? If you consider that an acoustic wave propagates through the air by a periodic change of the air pressure, it becomes apparent that pressure is an amplitude-like value. So, we use factor 20.
Transform the quantity X l i n=E b N 0 X l i n=E b N 0 into the frequency domain, where E b E b is the average energy per bit and N 0 N 0 is the noise energy per bit.
Here, we have no values given, so we can only give an expression. First note, that this question is different from before, in the sense that here the reference level is explicitely given as the noise power. Second, we see we are working with powers here, so we have to use factor 10:
S N R d B=10 log 10(E b N 0).S N R d B=10 log 10(E b N 0).
This is clearly a sigal-to-noise ratio, since we relate signal power to noise power here. So, we can use the name SNR.
Why both 10 and 20?¶
Even though choosing different factors for amplitudes and powers seems a bit strange, it is indeed quite elegant.
Let us now think of unit-less signals and let our reference values be always 1. Given a signal x x that at time t t has amplitude x(t)x(t), hence its power at time t t is given by P(t)=x 2(t)P(t)=x 2(t). Now, let's convert both amplitude and power into the dB-scale:
x d B(t)P d B(t)=20 log 10(x(t))=10 log 10(P(t))=10 log 10(x 2(t))=20 log 10(x(t))=x d B(t)(1)(2)(3)(4)(1)x d B(t)=20 log 10(x(t))(2)P d B(t)=10 log 10(P(t))(3)=10 log 10(x 2(t))(4)=20 log 10(x(t))=x d B(t)
In dB-scale the power and the amplitude of a signal have the same value! Hence, if somebody says "This signal is 10dB above that signal", there is no ambiguity wether she meant 10dB in amplitude or power.
Summary¶
The dB scale is a logarithmic, unitless scale. It always requires a reference quantity to be related against. The dB is calculated via two different expressions
X d B=10 log 10(X l i n X r e f)or Y d B=20 log 10(Y l i n Y r e f).X d B=10 log 10(X l i n X r e f)or Y d B=20 log 10(Y l i n Y r e f).
If you convert a quantity X X that relates to power or energy, the factor is 10.
If you convert a quantity Y Y that relates to amplitude, the factor is 20.
If you convert an expression that contains squares, use factor 10.
If you convert an expression that does not contain squares, use factor 20.
Do you have questions or comments? Let's dicuss below!
Share this article
Related Affiliate Products
Related posts
Python OFDM Example
Quantization and Quantization noise
Next: Group delay and phase delay example
Previous: Python OFDM Example
Introduction
About Me
Imprint
Privacy
DSPIllustrations.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to amazon.com, amazon.de, amazon.co.uk, amazon.it. |
7765 | https://math.stackexchange.com/questions/922896/divisible-by-19-induction-proof | divisibility - Divisible by 19 Induction Proof - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Divisible by 19 Induction Proof
Ask Question
Asked 11 years ago
Modified11 years ago
Viewed 3k times
This question shows research effort; it is useful and clear
4
Save this question.
Show activity on this post.
Prove by induction that for all natural numbers n n, 5 4 8 n+3 3 n−1 5 4 8 n+3 3 n−1 is divisible by 19 19.
I'm running into trouble at the inductive part of the step, I am currently attempting to add/subtract the inductive hypothesis but I end up with two different coefficients that are seemingly unrelated to 19 19. I've been stuck on this for days, thanks for the help!
induction
divisibility
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Sep 7, 2014 at 20:31
Benjamin
1,044 1 1 gold badge 7 7 silver badges 13 13 bronze badges
asked Sep 7, 2014 at 20:24
JakeJake
41 1 1 silver badge 2 2 bronze badges
1
Hint:8 3≡−1 mod 19 8 3≡−1 mod 19 and 27 3≡−1 mod 19 27 3≡−1 mod 19.Lucian –Lucian 2014-09-07 20:39:02 +00:00 Commented Sep 7, 2014 at 20:39
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
7
Save this answer.
Show activity on this post.
Rewrite as f(n)=10⋅8 n−1+3 3 n−1 f(n)=10⋅8 n−1+3 3 n−1 to clear the fraction.
Then can you see that f(n+1)=10⋅8 n+3 3 n+2=8 f(n)+(27−8)3 3 n−1 f(n+1)=10⋅8 n+3 3 n+2=8 f(n)+(27−8)3 3 n−1
We choose the factor 8 8 (we could also have chosen 27 27) since we want to get rid of one of the exponentials altogether in the f(n)f(n) term, and to find that the other one ends up with a coefficient divisible by 19 19.
Note that if f(n)=A a n+B b n f(n)=A a n+B b n then f(n+1)=(a+b)f(n)−a b f(n−1)f(n+1)=(a+b)f(n)−a b f(n−1) so if a,b a,b are integers and some integer c c is a factor of both f(n−1)f(n−1) and f(n)f(n) it will also be a factor of f(n+1)f(n+1), which is another way of solving this kind of problem (requires two base cases rather than one, but then the induction goes through).
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Sep 7, 2014 at 20:43
answered Sep 7, 2014 at 20:37
Mark BennetMark Bennet
102k 14 14 gold badges 119 119 silver badges 232 232 bronze badges
0
Add a comment|
This answer is useful
5
Save this answer.
Show activity on this post.
For n+1 n+1 step, we have
5 4⋅8 n+1+3 3(n+1)−1=5 4⋅8 n+1+8⋅3 3 n−1−8⋅3 3 n−1+3 3 n+2=8(5 4⋅8 n+3 3 n−1)+(3 3−8)⋅3 3 n−1=8(5 4⋅8 n+3 3 n−1)+19⋅3 3 n−1.5 4⋅8 n+1+3 3(n+1)−1=5 4⋅8 n+1+8⋅3 3 n−1−8⋅3 3 n−1+3 3 n+2=8(5 4⋅8 n+3 3 n−1)+(3 3−8)⋅3 3 n−1=8(5 4⋅8 n+3 3 n−1)+19⋅3 3 n−1.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Sep 7, 2014 at 20:35
mathlovemathlove
154k 10 10 gold badges 126 126 silver badges 313 313 bronze badges
2
Ah I see! I was previously trying to add/subtract the entire hypothesis rather than just the second half. Could you maybe explain your thought process that led you to add/subtract 83^{3n-1} ? Once I've seen it then it makes sense of course, I'm just thinking about what first led you to that. Thank you again!Jake –Jake 2014-09-07 20:42:53 +00:00 Commented Sep 7, 2014 at 20:42
@Jake: I wanted the form of f(n+1)=8 f(n)+19⋅g(n)f(n+1)=8 f(n)+19⋅g(n) for some g(n)g(n). And 8 f(n)=5 4⋅8 n+1+8⋅3 3 n−1 8 f(n)=5 4⋅8 n+1+8⋅3 3 n−1. Here, 5 4⋅8 n+1 5 4⋅8 n+1 is needed for f(n+1)f(n+1), but 8⋅3 3 n−1 8⋅3 3 n−1 is an 'extra' term, so... make sense?mathlove –mathlove 2014-09-07 20:57:28 +00:00 Commented Sep 7, 2014 at 20:57
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
H i n t m o d 19:3 3≡8⇒3 3 n a n d 4≡−5⋅3⇒1 3 m u l t i p l y i n g y i e l d s 3 3 n−1≡≡≡8 n−5 4−5 4 8 n Q E D H i n t m o d 19:3 3≡8⇒3 3 n≡8 n a n d 4≡−5⋅3⇒1 3≡−5 4 m u l t i p l y i n g y i e l d s 3 3 n−1≡−5 4 8 n Q E D
Remark As explained here, the proofs in the other answers can be discovered from the above in the following mechanical manner: take the standard proof of the Congruence Product Rule, then substitute the specific numbers in this problem, i.e. simply repeat the proof for the special case at hand. Though many such inductive proofs appear at first glance to be pulled out of a hat like magic, in fact they are, at heart, instances of congruence arithmetic - exactly as above. After one has learned congruence arithmetic the problem can be easily transformed into the triviality (27/8)n≡1 n≡1(27/8)n≡1 n≡1 since 27≡8(mod 19).27≡8(mod 19).
See also this answer on fractions in modular (congruence) arithmetic.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Apr 13, 2017 at 12:21
CommunityBot
1
answered Sep 7, 2014 at 22:19
Bill DubuqueBill Dubuque
284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
induction
divisibility
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
107Congruence Arithmetic Laws, e.g. in divisibility by 7 7 test
9Modular Fraction Arithmetic
1Mathmatical Induction
Related
4Prove by induction that (x+1)n−n x−1(x+1)n−n x−1 is divisible by x 2 x 2
2Using mathematical induction prove that 11⋅3 n+3⋅7 n−6 11⋅3 n+3⋅7 n−6 is divisible by 8
0Proof by Induction involving divisibility
1proving the divisibility rule for 11 using induction
3Prove by mathematical induction that 2 3 n+1 2 3 n+1 is divisible by 3 n+1 3 n+1
0Finding that a statement is divisible by a greatest n n through induction
0Proof by induction n x n<1 n x n<1
5Demonstrating that an Induction Hypothesis is too weak.
Hot Network Questions
Why are LDS temple garments secret?
With with auto-generated local variables
ConTeXt: Unnecessary space in \setupheadertext
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
I have a lot of PTO to take, which will make the deadline impossible
Storing a session token in localstorage
Matthew 24:5 Many will come in my name!
What meal can come next?
Discussing strategy reduces winning chances of everyone!
Why include unadjusted estimates in a study when reporting adjusted estimates?
Passengers on a flight vote on the destination, "It's democracy!"
Calculating the node voltage
Riffle a list of binary functions into list of arguments to produce a result
Change default Firefox open file directory
Xubuntu 24.04 - Libreoffice
Why is the definite article used in “Mi deporte favorito es el fútbol”?
What happens if you miss cruise ship deadline at private island?
в ответе meaning in context
Are there any world leaders who are/were good at chess?
Analog story - nuclear bombs used to neutralize global warming
What is this chess h4 sac known as?
My dissertation is wrong, but I already defended. How to remedy?
Checking model assumptions at cluster level vs global level?
Lingering odor presumably from bad chicken
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7766 | https://www.yourdictionary.com/snapper | Snapper Definition & Meaning | YourDictionary
Dictionary
Thesaurus
Sentences
Grammar
Vocabulary
Usage
Reading & Writing
Articles
Vocabulary
Usage
Reading & Writing
Sign in
Menu
Word Finder
Words with Friends Cheat
Wordle Solver
Word Unscrambler
Scrabble Dictionary
Anagram Solver
Wordscapes Answers
Sign in with Google
Dictionary
Thesaurus
Sentences
Grammar
Vocabulary
Usage
Reading & Writing
Word Finder
Word Finder
Words with Friends Cheat
Wordle Solver
Word Unscrambler
Scrabble Dictionary
Anagram Solver
Wordscapes Answers
Dictionary
DictionaryThesaurusSentencesArticlesWord Finder
Make Our Dictionary Yours
Sign up for our weekly newsletters and get:
Grammar and writing tips
Fun language articles
WordOfTheDay and quizzes
Sign in with Google
By signing in, you agree to our
Terms and Conditions and Privacy Policy.
Success!
We'll see you in your inbox soon.
Thank you!
Undo
Home
Dictionary Meanings
Snapper Definition
Snapper Definition
snăpər
snappers
Meanings
Synonyms
Sentences
Definition Source
[x] All sources
[x] Webster's New World
[x] American Heritage
[x] Wiktionary
Word Forms Noun
Filter(0)
noun
snappers
A person or thing that snaps.
Webster's New World
Similar definitions
Snapping turtle.
Webster's New World
A snapping turtle.
American Heritage
Similar definitions
Any of a family (Lutjanidae) of percoid fishes inhabiting most warm seas; esp., the red snapper.
Webster's New World
Similar definitions
A porgy (Pagrus auratus) of the Indo-Pacific region, having a pinkish body with spiny fins and valued as a food and game fish.
American Heritage
More Noun Definitions (7)
Synonyms:
Synonyms:
center
cracker-bonbon
cracker
Chrysophrys auratus
Chelydra serpentina
common snapping turtle
Other Word Forms of Snapper
Noun
Singular:
snapper
Plural:
snappers
Snapper Sentence Examples
The snapper is at once the handsomest and most palatable of a good variety of sea fish.
The Snapper Rock set offers UV 50 sun protection, is lightweight and breathable and features a soft elastic waistband for increased comfort.
Fishy dishes include whole tail scampi, breaded cod, red snapper and.. .
There is also a wide range of fish available, including snapper, grouper and John Dory.
Try your hand at catching yellowtail snapper, grouper, porgy or any member of the fierce fighting jack family.
More Sentences
Related Articles
Examples of Overfishing
Football Positions Abbreviations and Meanings
Snapper Is Also Mentioned In
mutton snapper
muttonfish
blackfin
schoolmaster
lutjanid
yellowtail
chrysophrys
ciguatoxin
shotgun
blacksnake
red snapper
schnapper
Advertisement
Find Similar Words
Find similar words to snapper using the buttons below.
Words Starting With
SSNSNA
Words Ending With
RERPER
Unscrambles
snapper
Words Starting With S and Ending With R
Starts With S& Ends With RStarts With SN& Ends With RStarts With S& Ends With ER
Word Length
7 Letter Words7 Letter Words Starting With S7 Letter Words Ending With R
Words Near Snapper in the Dictionary
snap pea
snap-one-s-fingers
snap-out-of
snappable
snapped
snapped up
snapper
snapperhead
snappest
snappier
snappily
snappiness
Filter
Random Word
Learn a new word now!
Get a Random Word
Copyright © 2025 LoveToKnow Media. All Rights Reserved
Features
Dictionary
Thesaurus
Sentences
Grammar
Vocabulary
Usage
Reading & Writing
Company
About Us
Contact Us
Privacy Policy
Editorial Policy
Cookie Settings
Terms of Use
Suggestion Box
Do Not Sell My Personal Information
Random Word
Learn a new word now!
Get a Random Word
Follow Us
Connect
Contact Us
Suggestion Box
Follow Us
LinkedIn
Facebook
Instagram
TikTok
Do Not Sell My Personal Information
Copyright © 2025 LoveToKnow Media. All Rights Reserved |
7767 | https://www.youtube.com/watch?v=IfP2K9Gdakk | Calculus 1 - Intro to Derivatives - Tangent Lines
Matt Charnley's Math Videos
1310 subscribers
Description
16 views
Posted: 26 Sep 2021
Video introducing how the derivative at a point can be used to find the tangent line to the graph of a function.
Transcript:
another follow-on question that can be asked in involving that efficient derivative is that you find the equation of a tangent line to a graph at a given point the thing to keep in mind here is what do we need to determine the equation of a line well there are a few ways to do this you could use two points you could use the slope and the intercept and the one that's of most interest to us right now you can use the slope and any point on the line why is this the most interesting well because we know how to get these two things already if we want to look at the tangent line to the graph of f at a point x equals a well what do we know well we already know a point that is on this line we know we have the point that is that a f of a is on this line because if the line is going to be tangent to the graph it's going to have to touch it at that point but now we have a way to also find the slope the slope is given by the derivative of f at a which we can compute by our limit formula once we have a point and a slope we can then write out the equation of the line that is tangent to the graph at this point and the easiest way to do that is in point slope form let's see this in the concept of an example i want to find the equation of the tangent line to the graph of y equals x squared plus 3x minus 2 at x equals 1. let's first look for a point that is on this line we know a point on this line is going to be 1 and f of 1. graph of one is if i plug 1 into here and f of 1 is going to be 1 squared is 1 plus 3 minus 2 is 2. another point 1 2 is on this line what about the slope slope is found to be the derivative which we compute as limit h goes to 0 f of one plus h minus f of one all over h i can now plug in what f is limit h goes to zero i have a one plus h squared plus three times a one plus h minus two and i know f of one is two i just did that step up here so this then becomes minus two all over h i'm gonna expand out all the terms on the top the quadratic becomes one plus two h plus h squared plus three plus three h minus four over h we'll see as expected the one and three combine to cancel this 4 meaning i am left with limit h goes to 0 5 h plus h squared over h able to cancel an h from every term to get limit h goes to zero five plus h which equals five therefore my slope is five and the point is one comma two i can write the equation of the tangent line from that y minus two gets the y coordinate of the point i know is on the line equals slope times x minus 1 the x points on the line and that's how we can get the equation of the tangent line to a graph using this limit definition of the derivative |
7768 | https://math.stackexchange.com/questions/3770155/how-to-seat-n-people-in-n-seats-if-each-person-must-take-a-seat-next-to-an-a | combinatorics - How to seat $n$ people in $n$ seats if each person must take a seat next to an already sitting person? - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How to seat n n people in n n seats if each person must take a seat next to an already sitting person?
Ask Question
Asked 5 years, 2 months ago
Modified5 years, 2 months ago
Viewed 414 times
This question shows research effort; it is useful and clear
3
Save this question.
Show activity on this post.
The question is exactly the same as in Permutations - n people and n seats
Copied here for your reference:
Imagine we have a room containing n n seats in a row and n n people waiting in front of the room. The first person that enters the room can decide where he wants to sit. The remaining (n−1)(n−1) people must take a seat next to an already sitting person. What is the number of ways to sit all the people in the room?
The solution that I was provided is also identical to the accepted answer in the above link, but it's still not intuitive to me and I'm not sure if I'm thinking about it the right way.
Essentially, if, say, the first person takes the k k th, why does (n−1 k−1)(n−1 k−1) guarantee the constraint (that each person must sit next to another already seated person) is satisfied?
I understand that there are (n−1 k−1)(n−1 k−1) to choose a distinct set of k−1 k−1 people to be seated on the left, but it's not clear to me why this guarantees the adjacency.
One way that I've thought about this that seems to make sense is consider person #1 sitting at the k k th seat. For each unique set of k−1 k−1 people, there’s a corresponding set of n−k n−k people. There is exactly 1 1 way to arrange them on the left and right, respectively, to guarantee the adjacency. Basically if we label the n n people as 1,2,3,4,…,n 1,2,3,4,…,n. The k−1 k−1 people on the left would be sorted in decreasing order, and the people on the right would be sorted in increasing order. Is this the right idea, or is there a much simpler way to think about this? I feel like I'm making this more difficult.
combinatorics
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
edited Jul 26, 2020 at 19:23
N. F. Taussig
79.3k 14 14 gold badges 62 62 silver badges 77 77 bronze badges
asked Jul 26, 2020 at 18:50
DavidDavid
736 7 7 silver badges 19 19 bronze badges
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
2
Save this answer.
Show activity on this post.
You have the right idea. Imagine that the n n people are queued up, and we number them by their positions in the queue. Any k−1 k−1 of them could be the k−1 k−1 that end up sitting to the left of the first person, so there are (n−1 k−1)(n−1 k−1) possible sets of people on that side, but as you say, there is only one possible order in which they can be seated: the one with the smallest number must have sat immediately to the left of the first person, the one with the next smallest number immediately to that person’s left, and so on, so that their numbers are decreasing from left to right. The other n−k n−k people can also be seated in only one order, since the first one of them to take a seat — the one with the lowest number — must be immediately to the right of the first person, the one with the next lowest number must be immediately to that person’s right, and so on. Thus, the entire arrangement is completely determined by which k−1 k−1 people are on the left of the first person, and there are therefore (n−1 k−1)(n−1 k−1) arrangements with the first person in seat k k.
It’s actually the adjacency requirement that guarantees that we know the arrangement once we know which k−1 k−1 people are to the left of the first person: without it, there would be (k−1)!(k−1)! possible different arrangements of those k−1 k−1 people.
By the way, if you look at it in these terms, what we’re counting here are permutations ⟨a 1,a 2,…,a n⟩⟨a 1,a 2,…,a n⟩ of {1,…,n}{1,…,n} such that a k=1 a k=1, a i>a i+1 a i>a i+1 for 1≤i<k 1≤i<k, and a i<a i+1 a i<a i+1 for k≤i<n k≤i<n.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Jul 26, 2020 at 19:23
Brian M. ScottBrian M. Scott
633k 57 57 gold badges 824 824 silver badges 1.4k 1.4k bronze badges
4
I have a question regarding your last paragraph. Could you explain why the problem is equivalent to the permutations idea? I'm having trouble seeing the connection roulette01 –roulette01 2020-09-09 16:07:11 +00:00 Commented Sep 9, 2020 at 16:07
1 @roulette01: Think of a i a i as the position in the queue of the person who ends up sitting in seat i i.Brian M. Scott –Brian M. Scott 2020-09-09 16:28:11 +00:00 Commented Sep 9, 2020 at 16:28
Oh right. That's the "sorted order" that OP alluded to right?roulette01 –roulette01 2020-09-09 18:25:35 +00:00 Commented Sep 9, 2020 at 18:25
@roulette01: Yes.Brian M. Scott –Brian M. Scott 2020-09-09 18:26:42 +00:00 Commented Sep 9, 2020 at 18:26
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
Each person that enters the room has exactly two possible sits (at the end of the row from both sides). Once you chose which people will choose to sit on the smaller than k k sits, it is clear what is the sitting arrangement.
Suppose n=10 n=10 and #1 sits on sit number 5. You need to choose 4 more to sit on sits 1-4, say you chose 2,3,6,8. Then: When 2 enters the room, he is chosen to sit on sits 1-4 so he must sit on 4. 3 enters, he is chosen so he must sit on 3. 4 enters. He is not chosen so he must sit on 6. 5 enters. He is not chosen, so 7 for him. 6 enters. He is chosen so he'll sit at 2. 7 will occupy 8, 8 will sit in the place labeled 1 and 9,10 in 9,10, respectively. Once the set 2,3,6,8 2,3,6,8 was chosen, the arrangement is fixed.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Jul 26, 2020 at 19:19
YJTYJT
4,674 1 1 gold badge 8 8 silver badges 26 26 bronze badges
2
There is not always 2 2 possible seats, one side might already be saturated (no available seat left), for instance if #1 sits on seat 9 9, then right side will be saturated quickly.zwim –zwim 2020-07-26 19:30:55 +00:00 Commented Jul 26, 2020 at 19:30
1 Sure, but that just means that all the people that are assigned to one of the sides are seated, and the remaining should anyhow sit on the other side. Like #9 and #10 in the above example.YJT –YJT 2020-07-26 19:43:49 +00:00 Commented Jul 26, 2020 at 19:43
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
So, lets make a string of left or right for the people to follow in that way the choice of the i−i−th person is marked. For example L R R L L L R R L L and so you will sequentially doing 1→L 21→R 213→R 2134→L 52134→L 652134.1→L 21→R 213→R 2134→L 52134→L 652134. Now because exactly k−1 k−1 people has to be in the left, then you choose the k−1 k−1 positions in which you will place the L.L.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Jul 26, 2020 at 19:21
PhicarPhicar
14.8k 2 2 gold badges 22 22 silver badges 28 28 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
combinatorics
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
4Permutations - n people and n seats
0COMBINATORICS: Number of parking possibilities.
Related
14Expected number of people sitting in the right seats.
10There are 100 people in a queue waiting to enter a hall. The hall has exactly 100 seats numbered from 1 to 100. The first person in the queue...
7Number of circular seating arrangements if each person can't sit next to two other people
1Number of ways of seating people on chairs such that one seat among each pair is empty
0How many ways can six people sit around a table, if person 1 isn't next to person 4 and person 5 isn't next to person 6?
2How many seating arrangements of m m people in a row of n n seats are there if k k of the people must sit together?
5A Combinatorics Question: Why Is My Solution Wrong, And The Given Solution Correct?
52 n 2 n people around the table, ways to place exactly r r out of n n men next to their wives given that n n women are already sitting at the table
0In how many different ways can 7 friends be seated since Jundullah will always sit on a 3-person seat?
0Find the total number of ways in which 100 100 seats can be filled up, provided the 100 100 th person occupies seat number 100 100
Hot Network Questions
Lingering odor presumably from bad chicken
On being a Maître de conférence (France): Importance of Postdoc
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Why do universities push for high impact journal publications?
Are there any world leaders who are/were good at chess?
The geologic realities of a massive well out at Sea
How long would it take for me to get all the items in Bongo Cat?
Matthew 24:5 Many will come in my name!
In Dwarf Fortress, why can't I farm any crops?
Does a Linux console change color when it crashes?
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
For every second-order formula, is there a first-order formula equivalent to it by reification?
Origin of Australian slang exclamation "struth" meaning greatly surprised
Cannot build the font table of Miama via nfssfont.tex
Clinical-tone story about Earth making people violent
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Storing a session token in localstorage
Alternatives to Test-Driven Grading in an LLM world
Calculating the node voltage
Is direct sum of finite spectra cancellative?
Triangle with Interlacing Rows Inequality [Programming]
My dissertation is wrong, but I already defended. How to remedy?
Checking model assumptions at cluster level vs global level?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7769 | https://www.cuemath.com/geometry/properties-of-rectangle/ | LearnPracticeDownload
Properties of Rectangle
The properties of a rectangle distinguish it from the other quadrilaterals. A rectangle is an equiangular quadrilateral in which the opposite sides are parallel and equal to each other and all four angles are right angles. The longer side of a rectangle is called its length and the shorter side is the width. Let us learn more about the characteristics of a rectangle in this article.
| | |
--- |
| 1. | What are the Properties of a Rectangle? |
| 2. | Formulas of a Rectangle |
| 3. | Types of Rectangles |
| 4. | FAQs on Properties of Rectangle |
What are the Properties of a Rectangle?
All properties of rectangle help us identify the figure at one glance. A rectangle is a two-dimensional figure with four sides, four vertices, and four angles. The opposite sides of a rectangle are equal in length and are parallel to each other. Since a rectangle is a quadrilateral in which all four angles are equal to each other, the angle formed by its adjacent sides is 90°. Observe the rectangle given below to see that the four sides of a rectangle are not equal, only the opposite sides are equal. Some of the real-life examples of a rectangle that we see in our daily life are kites, paintings, slabs, storage boxes, and so on.
In order to understand the rectangle better, observe the rectangle given above and relate to the following properties of a rectangle.
A rectangle is a quadrilateral with four equal interior angles.
The opposite sides of a rectangle are equal and parallel to each other.
The interior angle of a rectangle at each vertex measures 90°.
The sum of all the interior angles of a rectangle is 360°.
The diagonals bisect each other.
The length of the diagonals is equal.
The length of the diagonals can be obtained using the Pythagoras theorem. The length of the diagonal with sides a and b is √( a² + b²).
Since the sides of a rectangle are parallel, it is also called a parallelogram.
All rectangles are parallelograms but all parallelograms are not rectangles.
Formulas of a Rectangle
There are three main formulas of a rectangle that need to be remembered. They are related to the area of a rectangle, the perimeter of a rectangle, and the length of the diagonal.
Area of a Rectangle: A = l × w, where 'l' and 'w' are the length and width of the rectangle, respectively.
Perimeter of a Rectangle: P = 2(l + w), where 'l' is the length and 'w' is the width of the rectangle.
Diagonal of Rectangle (d) = √(l² + w²), where 'l' is the length and 'w' is the width of the rectangle. The formula for the diagonal of a rectangle is derived from the Pythagoras theorem.
Types of Rectangles
A rectangle has four sides with the opposite sides equal to each other and with the adjacent sides meeting at 90°. These properties are seen in the two types of rectangles - the Square and the Golden Rectangle.
Square
A square is a type of rectangle with four equal sides and four equal angles. It is a two-dimensional shape where the interior angles at each vertex are 90°. Along with these properties, the opposite sides of a square are equal and parallel and the diagonals bisect each other at 90°. It can be said that all squares are rectangles but all rectangles cannot be squares.
Golden Rectangle
The golden rectangle is a rectangle whose sides are in the golden ratio, that is, (a + b)/a = a/b, where 'a' is the width and (a + b) is the length of the rectangle. In other words, a golden rectangle is a rectangle whose 'length to width ratio' is similar to the golden ratio, 1: (1+⎷ 5)/2. For example, if the length is around 1 foot long then the width will be 1.168 feet long or vice-versa where the Golden Ratio = 1: 1.618. Observe the following figure which shows the golden rectangle and its length and width.
☛ Related Links
Properties of Parallelograms
Polygon Shape
Difference Between Square and Rectangle
Examples on the Properties of a Rectangle
Example 1: State true or false using all properties of rectangle.
a.) A rectangle is a quadrilateral with four equal interior angles.
b.) The interior angle of a rectangle at each vertex measures 60°.
Solution:
Using the features of a rectangle, we can state true or false as follows:
a.) True, a rectangle is a quadrilateral with four equal interior angles.
b.) False, the interior angle of a rectangle at each vertex measures 90°.
2. Example 2: Use the features of a rectangle to fill in the blanks:
a.) Since the sides of a rectangle are parallel, it is also called a __________.
b.) The sum of all the interior angles of a rectangle is ___°.
Solution:
Using the characteristics of a rectangle, we can fill in the blanks as follows:
a.) Since the sides of a rectangle are parallel, it is also called a parallelogram.
b.) The sum of all the interior angles of a rectangle is 360°.
3. Example 3: If the length of a rectangle is 40 meters and its width is 3 meters, then find its perimeter using the properties of a rectangle.
Solution:
The formula to calculate the perimeter is P = 2 (length + width);
Given that, the length is 40 meters and the width is 3 meters. Substituting these values into the formula.
Perimeter of rectangle = 2(40 + 3)
Perimeter of rectangle = 2 × (43)
Perimeter of rectangle = 86 meters
Therefore, the Perimeter of rectangle = 86 meters
View Answer >
go to slidego to slidego to slide
Great learning in high school using simple cues
Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes.
Book a Free Trial Class
Practice Questions on Properties of Rectangle
Check Answer >
go to slidego to slidego to slide
FAQs on Properties of Rectangle
What are the Properties of a Rectangle?
The basic properties of a rectangle are that its opposite sides are parallel and equal and its interior angles are equal to 90°. Its diagonals are also equal and they bisect each other.
What are the Properties of the Diagonals of a Rectangle?
The properties of the diagonal of a rectangle are as follows:
The two diagonals of a rectangle are equal.
The diagonals bisect each other, but not at right angles.
The length of the diagonals can be obtained using the Pythagoras theorem.
Since the diagonals divide the rectangle into two right-angled triangles, they are considered to be the hypotenuse of these triangles.
Is a Square a Rectangle?
Yes, a square is considered as a rectangle because it contains the properties of a rectangle, like, all the four interior angles are 90°, the opposites sides of a square are parallel and equal to each other, and two diagonals of the square are equal and bisect each other.
What is the Difference Between a Square and a Rectangle?
Squares have some additional properties which do not apply to rectangles. A square has four equal sides, whereas, in a rectangle, only the opposite sides are equal. The diagonals of a square bisect at 90°, but the diagonals of a rectangle do not bisect at 90°.
What are the Various Types of Quadrilaterals other than Rectangles?
The various types of quadrilaterals other than rectangles are squares, rhombus, kite, parallelogram, and a trapezoid.
Why is a Rectangle not a Square?
Although many properties of a rectangle are similar to a square but a rectangle is not a square because it does not have all four sides of equal measure. This is the reason that a rectangle is not a square.
What is a Rectangle?
A rectangle is a two-dimensional shape (2D Shape) with four sides, four angles, and four vertices. The opposite sides of a rectangle are equal and parallel to each other. The interior angles of a rectangle are equal and measure 90°.
Write any Two Properties of Rectangle.
Although there are many properties of a rectangle, two properties of rectangle that distinguish it from the rest are as follows:
The opposite sides of a rectangle are parallel and equal.
All the interior angles of a rectangle are equal to 90°.
Download FREE Study Materials
Geometry Worksheets
Math worksheets andvisual curriculum
FOLLOW CUEMATH
Facebook
Youtube
Instagram
Twitter
LinkedIn
Tiktok
MATH PROGRAM
Online math classes
Online Math Courses
online math tutoring
Online Math Program
After School Tutoring
Private math tutor
Summer Math Programs
Math Tutors Near Me
Math Tuition
Homeschool Math Online
Solve Math Online
Curriculum
NEW OFFERINGS
Coding
SAT
Science
English
MATH ONLINE CLASSES
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
ABOUT US
Our Mission
Our Journey
Our Team
MATH TOPICS
Algebra 1
Algebra 2
Geometry
Calculus math
Pre-calculus math
Math olympiad
Numbers
Measurement
QUICK LINKS
Maths Games
Maths Puzzles
Our Pricing
Math Questions
Events
MATH WORKSHEETS
Kindergarten Worksheets
1st Grade Worksheets
2nd Grade Worksheets
3rd Grade Worksheets
4th Grade Worksheets
5th Grade Worksheets
6th Grade Worksheets
7th Grade Worksheets
8th Grade Worksheets
9th Grade Worksheets
10th Grade Worksheets
FOLLOW CUEMATH
Facebook
Youtube
Instagram
Twitter
LinkedIn
Tiktok
MATH PROGRAM
Online math classes
Online Math Courses
online math tutoring
Online Math Program
After School Tutoring
Private math tutor
Summer Math Programs
Math Tutors Near Me
Math Tuition
Homeschool Math Online
Solve Math Online
Curriculum
NEW OFFERINGS
Coding
SAT
Science
English
MATH WORKSHEETS
Kindergarten Worksheets
1st Grade Worksheets
2nd Grade Worksheets
3rd Grade Worksheets
4th Grade Worksheets
5th Grade Worksheets
6th Grade Worksheets
7th Grade Worksheets
8th Grade Worksheets
9th Grade Worksheets
10th Grade Worksheets
ABOUT US
Our Mission
Our Journey
Our Team
MATH TOPICS
Algebra 1
Algebra 2
Geometry
Calculus math
Pre-calculus math
Math olympiad
Numbers
Measurement
QUICK LINKS
Maths Games
Maths Puzzles
Our Pricing
Math Questions
Blogs
Events
MATH ONLINE CLASSES
1st Grade Math
2nd Grade Math
3rd Grade Math
4th Grade Math
5th Grade Math
6th Grade Math
7th Grade Math
8th Grade Math
Terms and ConditionsPrivacy Policy |
7770 | https://www.aofoundation.org/trauma/courses/basic-principles-fracture-management-course | AO Trauma Basic Principles of Fracture Management Course
The fundamental course for doctors in surgical training, and doctors interested in furthering their knowledge and skills in operative fracture management.
The AO Trauma course Basic Principles of Fracture Management teaches fundamental principles and current concepts in the treatment of injuries, incorporating the latest techniques in operative fracture management. The AO Trauma Basic Principles course is the initial step along the path of lifelong learning in the area of operative fracture management.
The course is available in a standard format (all face-to-face) and a blended version (with a 3-week online preparation phase).
Find this course near you
Why you should choose this course
Top national and regional faculty
2-4 days in duration
For surgeons in training and residents
Network with colleagues from all over the world
World-class curriculum
Find this course near you
Course content
Course modules
Module 1: Injury pattern (soft-tissue and bone) and biology of bone healing
Module 2: Stability and biomechanics of bone healing
Module 3: Principles and management of diaphyseal fractures
Module 4: Principles and management of articular fractures
Module 5: Emergency management, minimally invasive surgery, and special fractures
AO Skills Lab – Simulations
Torque measurement of bone screws
Soft-tissue penetration during drilling
Heat generation during drilling
Mechanics of bone fractures
Techniques of reduction
Mechanics of intramedullary fixation
Mechanics of plate fixation
Fracture healing and plate fixation
Damaged implant removal
Practical exercises – Dry lab
Internal fixation with screws and plates—absolute stability
Principles of the internal fixator using the locking compression plate (LCP)
Application of a modular large external fixator
Intramedullary (IM) nailing (tibial or femoral shaft)
Cerclage compression wiring of the olecranon
Management of a type 44C malleolar fracture
Management of a trochanteric fracture
Preoperative planning—plan your operation
Operate your plan
Management of a femoral neck fracture using 7.3 mm cannulated screws
Application of the large distractor to comminuted femoral fracture
Stabilization of the pelvic ring using a large external fixator
Patient positioning workshop (in lecture room)
Optional content: Only in selected courses. Check your chosen date and location for the full program.
Small group discussions
General principles, classification, concepts of stability, their influence on bone healing, and how to apply implants to achieve appropriate stability
Management principles for the treatment of diaphyseal fractures
Management principles for the treatment of articular fractures
Polytrauma, complications, and fragility fractures
Optional content: Only in selected courses. Check your chosen date and location for the full program.
Course details may be subject to change. Please check your chosen date and location for the detailed program.
Download brochure (Standard course)
Download brochure (Blended course)
Target audience
Doctors in surgical training
Doctors interested in furthering their knowledge and skills in operative fracture management
Course impressions
Load more
Competencies
Understand the concepts of stability, their influence on bone healing, and how to apply implants to achieve the appropriate stability
Plan a treatment based on assessment, imaging, classification, and decision making
Apply reduction techniques in fracture management with attention to the soft tissue
Apply implants according to their properties utilizing different application techniques
Assess and treat diaphyseal fractures
Assess and treat articular and periarticular fractures
Demonstrate strategies for assessing and treating open fractures and soft-tissue injuries
Manage the polytrauma patient
Evaluate, classify, and formulate a treatment plan for pelvic injuries and acetabular fractures
Recognize risk factors and complications and manage accordingly
Recognize and treat bone union disorders
Recognize and manage special fracture circumstances
What does competency-based curriculum development mean?
Learn more
Resources
AO Surgery Reference
Explore the award-winning online repository for surgical knowledge, describing the complete management process from diagnosis to aftercare for fractures in a given anatomical region.
Access now
Residents education taskforce
The Residents education taskforce is a group of experts who build and continuously improve our educational program. It consists of three international program editors (IPEs).
Jason A Lowe (USA)
Term: 2025–2027
Sue Deakin (UK)
Term: 2023–2026
Khairul Faizi Mohammad (MY)
Term: 2023–2025
See a full list of all regional program contributors and past international program editors.
Not the right course?
Find our list of all curricula courses here.
Browse the course finder
By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Social Media Cookies
These cookies are set by a range of social media services that we have added to the site to enable you to share our content with your friends and networks. They are capable of tracking your browser across other sites and building up a profile of your interests. This may impact the content and messages you see on other websites you visit. If you do not allow these cookies you may not be able to use or see these sharing tools. |
7771 | https://www.ieice.org/~isap/ISAP_Archives/2005/pdf/1B1-4.pdf | PARTIALLY–DIELECTRIC-FILLED OVERSIZED RECTANGULAR WAVEGUIDE FOR SUPRESSION OF SIDE LOBES IN SLOT ARRAYS Miroslav Samardzija, Takafumi Kai, Jiro Hirokawa, Makoto Ando Department of Electrical and Electronic Engineering, Tokyo Institute of Technology S3-19, 2-21-1, O-okayama, Meguro-ku, Tokyo, 152-8552 Japan E-mail: miro@antenna.ee.titech.ac.jp 1. Introduction We propose a hollow-waveguide-fed parallel plate slot array antenna in which two dielectric rods are installed at the both side walls of an oversized rectangular waveguide as shown in Fig.1(a). The size of the proposed antenna is assumed to be small, around 2×2 – 3×3 λ. The specific purpose is to achieve a uniform field distribution and a low side lobe level of radiation pattern. The slot pairs in the antenna are arrayed in spacing of one guided wavelength to be excited in phase. Conventionally, the guided wavelength in an oversized rectangular waveguide is reduced by filling the whole region with dielectric material to suppress the grating lobes in longitudinal direction , . In the proposed structure, the guided wavelength is controlled and reduced by changing the width of the two dielectric rods installed at the side walls of the oversized rectangular waveguide. Since the oversized rectangular waveguide is only partially filled with dielectric material, the power loss due to the dielectric will be noticeably less than that of the overall-filled waveguide . In this paper, we estimate the reduction of the guided wavelength of the lowest mode and the field distribution in the oversized rectangular waveguide due to the installation of the dielectric rods. Based upon these phenomena, we calculate and investigate array factors of 4×4 array for various widths of the dielectric rods to suppress the side lobes in both E and H plane. 2. Configuration and Analysis Fig.1(a) shows the configuration of the proposed waveguide-fed oversized rectangular slot array antenna. The radiating units are slot pairs etched on the top plate. In the radiating unit, two slots are spaced approximately by a quarter of a guide wavelength to cancel the reflections from the two slots. Two dielectric rods are installed at the both narrow walls of the radiating waveguide. A feed waveguide is placed on the same layer as the radiating waveguide and arranged as a cascade of several coupling windows. An analysis model for a partially–dielectric-filled oversized rectangular waveguide is shown in Fig.1(b). The air region is in the middle of the rectangular waveguide and is the region of interest. In partially filled waveguide, the hybrid modes that are the solution and satisfy the boundary conditions are transverse electric TEx and transverse magnetic TMx, for which Ex=0 and Hx=0, respectively . In this paper, we deal with only TEx modes uniform along y-direction in a thin oversized rectangular waveguide. For the symmetry in the x=0 plane, a perfect magnetic conducting (PMC) wall in the x=0 plane assumes only propagating modes that are odd and symmetric. A transcendental equation is derived by enforcing the condition of continuous admittance along the interface surface, i.e. along x=d plane. Solving the transcendental equation numerically the propagation constant (2π/λg) is obtained as function of the width of the air region and the width of dielectric rods. 3. Design for suppression of grating lobes The spacing of the radiating units should be set to less than the free space wavelength (λo) to suppress the side lobes in the longitudinal direction. The width of a partially–dielectric-filled waveguide is set to 2.8λo, which provides space for four radiating units in x-direction, spaced at the distance of 0.7λo. In z-direction, the guided wavelength is originally reduced due to the coupling of non-resonant slots with an incident field . The radiating units are arranged with 0.84λo spacing by PROCEEDINGS OF ISAP2005, SEOUL, KOREA - 89 -WB1-4 ISBN: 89-86522-77-2 94460 KEES taking in to account the average phase delay of 60 degrees due to the slot coupling. Considering this slow wave effect, the reduction of the guided wavelength by the dielectric rods is utilized actively to get the desired grating lobe suppression. On the other hand, the increase of the width of dielectric rods causes the inner field degradation in transverse direction and thereby increases the level of side lobes in x-direction. As a preliminary step, we calculate an array factor of the radiating units both in transverse and longitudinal direction for various widths of the dielectric rod. We evaluate the grating lobe variation caused by both guided wavelength reduction effect and the degradation of the field distribution in x-direction. The optimal width of the dielectric rods is determined to minimize the grating lobe. 4. Results In partially–dielectric-filled oversized rectangular waveguide, the number of propagating modes increases with the total width of the waveguide as shown in Fig.2. In the waveguide of 2.8λo air-region width and 0.23λo dielectric width there are three propagating modes. Only the lowest mode is discussed throughout this paper. The guided wavelength as a function of the width of dielectric rods for various widths of the air region is presented in Fig.3. The dielectric material is PTFE (Teflon) and the dielectric constant is 2.17. The operating frequency is 12GHz. The width of dielectric rods is independent of the air region width for guided wavelengths shorter than free space wavelength. It is of interest to note that for certain width of dielectric rods (0.23λo) the guided wavelength becomes equal to the free space wavelength (λo). For the width of dielectric rods above 0.23λo the guided wavelength (λg) becomes shorter than the free space wavelength (λo). In Figure 4, the width of the dielectric rods is given as a function of the air-region width for a desired guided wavelength. It converges to a certain value for increase of the air-region width. The magnitude of the electric field component in the y-direction (Ey) throughout the cross section of the air region is shown in Fig.5. When the guided wavelength is equal to the free space wavelength the magnitude of Ey becomes uniform across the air region independent of the air-region width. In case of the guided wavelength shorter than that of free space, the Ey in the waveguide becomes a cosine-hyperbolic function depending on the propagating constant. The difference in the magnitude of the Ey in the centre of the waveguide and at the air-dielectric interface increases exponentially as the guided wavelength decreases. This is taken in account when considering the radiation pattern in x-direction. An array factor of a 4 by 4 array is calculated in z- and x- directions and shown in Fig.6. The array factor in the z-direction shows the grating lobe suppression as the guided wavelength decreases, i.e. the width of dielectric rods increases above 0.23λo, as is expected. The radiation pattern in Fig.6(b) is calculated by allocating the field distribution in x-direction to each radiating unit. In addition, the side lobe level in the x-direction increases from -25dB to -7dB at ±45deg. as the guided wavelength decreases from 1.0λo to 0.97λo. A proper guided wavelength in a 2.8λo air-region wide partially–dielectric-filled waveguide for which the highest grating lobe suppression of -9.0dB is 0.978λo with respect to lowest side lobe level in x-direction as shown in Fig.7. 5. Discussion In the present analysis model of the partially–dielectric-filled oversized waveguide, only the lowest mode with cosine-hyperbolic field distribution is considered. As shown in Fig.2, number of propagating modes increases with the increase of the width of partially-filled waveguide. Propagating modes of higher order have cosinusoidal distributions while the lowest mode cosine-hyperbolic distribution, in x-direction. The uniformity of the inner field in x-direction would be improved by summing up the lowest mode and a few higher modes with proper ratio by the feed waveguide. The degradation of the inner field of the lowest mode in the waveguide generates higher side lobes in the x-direction due to the non-uniform excitation of the radiation units in transverse direction. The feed waveguide structure with coupling windows is to be included and designed to excite the desired excitation into the oversized radiating waveguide region in the analysis model. - 90 -6. Conclusion A partially–dielectric-filled rectangular waveguide has been proposed to suppress the side lobes in the small-size waveguide fed parallel plate slot array. The transcendental equation has been derived and thereby the guided wavelength as a function of the geometry of the partially-filled rectangular waveguide. For a partially–dielectric-filled waveguide of a size 2.8×2.8λo with slot array of 4 by 4 radiating units, a proper guided wavelength, i.e. width of the dielectric rods is given with respect to maximal grating lobe suppression of -9.0dB in longitudinal direction and minimal side lobe level of the worst case, in transverse direction in Fig.7. References J.Hirokawa, M.Ando and N.Goto, “Waveguide-Fed Parallel Plate Slot Array Antenna,” IEEE Trans. Antennas & Propagat., vol.40, no.2, pp.218-223, Feb. 1992. J.Hirokawa and M.Ando, “Single-Layer Feed Waveguide Consisting of Posts for Plane TEM Wave Excitation in Parallel Plates,” IEEE Trans. Antennas & Propagat., vol.46, no.5, pp.625-630, May 1998. C.A. Balanis, Advanced Engineering Electromagnetics, New York: Wiley, 1989. ( (a) Figure 1: (a) Antenna outline, (b) Analysis model of a partially dielectric filled rectangular waveguide. Figure 3: Guided wavelength as a function of dielectric width for various air region widths. ( (b) Figure 2: Number of propagating modes as a function of the air region width for various widths of dielectric rods. 1 2 3 4 5 0.5 1.5 2.5 3.5 Normalized width of air region [2d/λo] Number of propagating modes a-d=0.35λo a-d=0.23λo a-d=0 mm 0.8 0.85 0.9 0.95 1 1.05 1.1 0 0.1 0.2 0.3 0.4 0.5 Normalized dielectric width [(a-d)/λo] Normalized guided wavelength [λg/λo] 2d=1.4λo 2d=2.8λo 2d=5.6λo - 91 - Figure 5: Field distribution throughout the cross section of the air region in the waveguide. Figure 6: Radiation pattern of the array factor for 4 radiating units at various desired guided wavelengths a) in the z-direction b) in the x-direction. Figure 7: Optimal guided wavelength of 0.978λo (dielectric width of 0.26λo) for the waveguide of 2.8λo air-region width and the 4x4 radiating-unit array. Side lobe level is the worst case scenario due to the analysis of the lowest mode only. -14 -12 -10 -8 -6 0.95 0.96 0.97 0.98 0.99 1 Normalized guided wavelength [λg/λo] Side lobe level in x-direction [dB] Grating lobe level in z-direction [dB] -25 -20 -15 -10 -5 0 -90 -60 -30 0 30 60 90 Angle in H-plane [deg] Amplitude [dB] λg/λo decreases ( (a) ( (b) -25 -20 -15 -10 -5 0 -90 -60 -30 0 30 60 90 Angle in E-plane [deg] Amplitude [dB] λg/λo=1.0 λg/λo=0.99 λg/λo=0.98 λg/λo=0.97 λg/λo decreases 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Position along x-axis [x/λo] Normalized Ey-magnitude [V/m] λg/λo decreases λg/λo=1.1 λg/λo=1.0 λg/λo=0.99 λg/λo=0.98 λg/λo=0.97 0.2 0.3 0 1 2 Normalized width of air region [2d/λo] Normalized dielectric width [(a-d)/λo] λg/λo decreases λg/λo=1.0 λg/λo=0.95 λg/λo=0.90 Figure 4: Design geometry for the partially filled rectangular waveguide at the desired guided wavelength. - 92 - |
7772 | https://www.ibisworld.com/united-states/bed/total-vehicle-miles/4149/ | Total vehicle miles - Business Environment Profile Report | IBISWorld
Solutions
Solutions
For Your Role Learn how peers in your industry are using us to get results.
Academics
Accounting
Business Valuations
Commercial Banking
Consulting
Government Agencies
Investment Banks
Law Firms
Marketing
Private Equity
Procurement
Sales
Platforms & Integrations Discover multiple ways to access and integrate key industry data and insights.
Platform & Integrations Overview
API Solutions
Salesforce App
Snowflake
Phil, Your AI Economist
Products
Products
Industry & Market Research Analyze industry trends, market size, and key drivers to make informed decisions.
Industry Research Reports
iExpert Summary Reports
Business Environment Profiles
State & Provincial Reports
Australia SME Industry Reports
Competitive Benchmarking Compare competitors' performance, strengths, and market positioning.
Industry Financial Ratios
US Segment Benchmarking
AU & NZ Enterprise Profiles
Risk Assessments Assess the source and severity of risks, and stay ahead with early warnings when risk is on the rise.
ESG Risk Assessments
Early Warning System
Industry Risk Rating Reports
Data Analysis Tools Compare and analyze data across industries for a holistic view of your portfolio.
Industry Data Wizard
Tariff Exposure Tool
View all products
Industries
Industries
By Country
United States
Canada
Mexico
Australia
New Zealand
China
Global
Europe
France
Germany
Italy
Ireland
Spain
United Kingdom
View all (37 countries)
By Sector
Administration & Business Support Services
Arts, Entertainment & Recreation
Construction
Educational Services
Finance & Insurance
Healthcare and social Assistance
Information Sector
Manufacturing
Accommodation & Food Services
Mining
Personal Services
Professional, Scientific and Technical Services
Public Administration & Safety
Real Estate, Rental & Leasing
Retail Trade
Thematic Reports
Transportation and Warehousing
Utilities
Wholesale Trade
By Company
AU & NZ Enterprise Profiles
US Company Benchmarking
By State & Province
US States
Canadian Provinces
View All Industry Reports
Resources
Resources
About the Company
Our Team
Our Story
Our Founder
Careers
Contact Us
Additional Insights
Industry Insider Blog
Product Hub
Industry Classifications
Help Center
Trends & Statistics
Industry Statistics
Industry Trends
US State Economic Profiles
Canada Province Economic Profiles
Business Environment Profiles
Membership Benefits
Client Success Stories
Partnerships Overview
Newsletter
Membership Inquiry
Book a free demo
Contact Us
Log in
Contact Us
Log in
Mobile Menu
Home
United States
Other Indicators
Total vehicle miles
Business Environment Profiles - United States
Total vehicle miles
Published: 21 July 2025
Key Metrics
Total vehicle miles
Total (2025)
3 Trillion
Annualized Growth 2020-25
2.6 %
Definition of Total vehicle miles
This driver represents the total miles driven by all motor vehicles over the calendar year. This includes cars, trucks, motorcycles and buses, but not trains or planes. Data is sourced from the Bureau of Transportation Statistics (BTS).
Analyze the wider world in which businesses operate
We measure the upstream and downstream ramifications on thousands of industries so businesses can monitor their external operating environment. Explore membership options today.
Purchase options
Included in an IBISWorld Membership
Our industry reports include 35+ pages of data, analysis and charts, including:
Industry Financial Ratios
Historical and Forecast Growth
Industry Market Size
Industry Major Players
Profitability Analysis
SWOT Analysis
Industry Trends
Industry Operating Conditions
Total Total vehicle miles (Trillion)
Line chart with 35 data points.
View as data table, Total Total vehicle miles (Trillion)
The chart has 1 X axis displaying Year. Data ranges from 1990-01-12 00:00:00 to 2024-01-04 00:00:00.
The chart has 1 Y axis displaying Values (Trillion). Data ranges from 2.1 to 3.3.
End of interactive chart.
Recent Trends – Total vehicle miles
The total vehicle miles traveled in the United States is expected to reach 3.3 trillion in 2025, indicating modest expansion as pandemic-related volatility subsides. Year-on-year growth is projected at 0.4% in 2025, following trends established in the previous two years. The gradual recovery in vehicle miles in recent years has been supported by a return to workplace commuting as COVID-19 restrictions have been lifted. However, the continuation of hybrid work models and remote arrangements by some companies has moderated potential gains. In addition, a shift in travel preferences, with more consumers choosing international flights, is expected to temper further increases in domestic vehicle miles in the current year.
Total vehicle miles have demonstrated a pattern of recovery and stabilization during the past five years in the wake of considerable disruption in 2020. The onset of the COVID-19 pandemic in 2020 led to a historic drop of 11.0% in total vehicle miles due to widespread lockdowns, remote work mandates, and business closures. The distribution of vaccines and easing pandemic-related restrictions in 2021 enabled a sharp rebound, with total vehicle miles rising by 7.9% as more individuals returned to office settings and resumed travel. Total miles rose 2.0% in 2022 and 2.1% in 2023, as pandemic effects lessened; some remote work habits persisted. In-person requirements for essential activities, such as medical appointments, have also continued to generate demand for vehicle travel. Additionally, travel peaks during popular months in 2024 and 2025 have contributed to growth, although this has been offset by the resurging popularity of international air travel.
Broader macroeconomic factors have influenced total vehicle miles over this period. Labor market recovery and consumer spending growth have raised mobility needs, while demographic factors, such as population increases and urbanization patterns, also play a role. The continued popularity of ride-sharing services has increased the number of cars on the road and total miles traveled. However, the expansion of remote work and telecommuting acts as a structural restraint on total growth. Gasoline and energy prices, while not a major factor in this specific period, historically influence driving behavior but have remained relatively stable since 2021.
Between 2020 and 2025, total vehicle miles are forecast to grow at a compound annual rate of 2.6%. The transportation sector remains sensitive to changes in public health guidance, commuting patterns, business operations, and macroeconomic events. While the rebound from pandemic lows was rapid, ongoing structural shifts, such as hybrid work models and evolving travel preferences, are expected to keep the growth of total vehicle miles moderate relative to pre-pandemic norms.
Show more
Total vehicle miles (Annual Change)
Bar chart with 34 bars.
View as data table, Total vehicle miles (Annual Change)
The chart has 1 X axis displaying Year. Data ranges from 1991-01-12 00:00:00 to 2024-01-04 00:00:00.
The chart has 1 Y axis displaying Annual Change (%). Data ranges from -10.99 to 7.87.
End of interactive chart.
5-Year Outlook – Total vehicle miles
Total vehicle miles are anticipated to continue their gradual upward trajectory, projected to inc...
Looking for IBISWorld Industry Reports?
Gain strategic insight and analysis on thousands of industries.
Browse industries
Trusted by More Than 10,000 Clients Around the World
Global
Australia
Canada
China
Europe
Germany
Mexico
New Zealand
United Kingdom
United States
Deutsch
Products
API Solutions
AU & NZ Enterprise Profiles
Business Environment Profiles
Integrations Overview
Phil, Your AI Economist
Call Prep Insights for Salesforce
Early Warning System
ESG Risk Assessments
iExpert Summary Reports
Industry Research Reports
Industry Risk Rating Reports
Industry Financial Ratios
Industry Data Wizard
State & Provincial Reports
US Segment Benchmarking
US Company Benchmarking
US Benchmarking Pro
Industry Insider
SME Industry Reports
Company
Our Story
Our Team
Our Partners
Client Success Stories
Careers
Contact Us
FAQ - Help Center
Industry Classifications
Modern Slavery Statement
Membership
Membership Inquiry
Request a Demo
Free Sample Report
Newsletter
Log in
Follow us
© 2025 IBISWorld. All rights reserved.
Terms of Use
Cookie Policy
Cookie Preferences
Privacy Policy
Sitemap |
7773 | https://www.aafp.org/pubs/afp/issues/2019/0301/p301.html | Gas, Bloating, and Belching: Approach to Evaluation and Management
picture_as_pdfPDFcommentComments
JOHN M. WILKINSON, MD, ELIZABETH W. COZINE, MD, AND CONOR G. LOFTUS, MD
info
Am Fam Physician. 2019;99(5):301-309
share
Patient information: See related handout on gas, bloating, and belching, written by the authors of this article.
assignment
Author disclosure: No relevant financial affiliations.
Gas, bloating, and belching are associated with a variety of conditions but are most commonly caused by functional gastrointestinal disorders. These disorders are characterized by disordered motility and visceral hypersensitivity that are often worsened by psychological distress. An organized approach to the evaluation of symptoms fosters trusting therapeutic relationships. Patients can be reliably diagnosed without exhaustive testing and can be classified as having gastric bloating, small bowel bloating, bloating with constipation, or belching disorders. Functional dyspepsia, irritable bowel syndrome, and chronic idiopathic constipation are the most common causes of these disorders. For presumed functional dyspepsia, noninvasive testing for Helicobacter pylori and eradication of confirmed infection (i.e., test and treat) are more cost-effective than endoscopy. Patients with symptoms of irritable bowel syndrome should be tested for celiac disease. Patients with chronic constipation should have a rectal examination to evaluate for dyssynergic defecation. Empiric therapy is a reasonable initial approach to functional gastrointestinal disorders, including acid suppression with proton pump inhibitors for functional dyspepsia, antispasmodics for irritable bowel syndrome, and osmotic laxatives and increased fiber for chronic idiopathic constipation. Nonceliac sensitivities to gluten and other food components are increasingly recognized, but highly restrictive exclusion diets have insufficient evidence to support their routine use except in confirmed celiac disease.
Patients with symptoms of gas, bloating, and belching often consult family physicians, particularly when milder chronic symptoms of abdominal pain or altered bowel habits acutely flare up and become less tolerable. Most often, these symptoms are attributable to one or more of the functional gastrointestinal disorders (FGIDs), including functional dyspepsia, irritable bowel syndrome (IBS), and chronic idiopathic constipation.
| Clinical recommendation | Evidence rating | References | Comments |
--- --- |
| Functional dyspepsia, IBS, and chronic idiopathic constipation can be diagnosed using symptom-based clinical criteria. | C | 2, 3, 7–9, 16 | Excluding organic disease through exhaustive investigation is not necessary; usually only limited testing is needed. |
| Noninvasive testing for Helicobacter pylori, and eradication therapy if positive (test-and-treat strategy), should be used for the initial evaluation of dyspepsia without alarm symptoms in younger patients. | C | 7, 8, 19 | See Table 1 for a list of alarm symptoms; urea breath testing is preferred (Table 4); endoscopy is recommended as the initial test in patients older than 55 years (Table 4). |
| Part of the initial evaluation of patients with diarrhea-predominant or mixed-presentation IBS symptoms should include testing for celiac disease. | C | 9, 20, 21 | If the incidence of celiac disease is known to be less than 1%, testing can be deferred. |
| Empiric proton pump inhibitor therapy is moderately effective for treating functional dyspepsia in patients who are H. pylori negative or who remain symptomatic after H. pylori eradication. | C | 8 | Patients with functional dyspepsia may have increased acid sensitivity. |
| Highly restrictive gluten-free diets and diets restricted in fermentable oligosaccharides, disaccharides, monosaccharides, and polyols have insufficient evidence to be routinely recommended for IBS management. | C | 16, 32, 35 | Questions remain about safety, effectiveness, cost, and practicality of long-term implementation. |
IBS = irritable bowel syndrome.
A = consistent, good-quality patient-oriented evidence; B = inconsistent or limited-quality patient-oriented evidence; C = consensus, disease-oriented evidence, usual practice, expert opinion, or case series. For information about the SORT evidence rating system, go to
Classified primarily in terms of symptoms, FGIDs are separated into discrete syndromes and are diagnosed by specific criteria. In clinical practice, many patients may not meet all criteria or may have symptoms with significant overlap among syndromes.1 The FGIDs are not diagnoses of exclusion; certain clinical features may require limited testing to exclude other conditions, but exhaustive testing is not necessary before making a diagnosis.2,3
Definitions and Pathophysiology
The FGIDs are characterized as disorders of gut-brain interaction.2 Symptoms, including bloating and abdominal distention, are thought to result from disturbances in intestinal transit and motility, gut microflora, immune function, gas production, visceral hypersensitivity, and central nervous system processing.
Bloating is a sense of gassiness or of being distended, with or without a visible increase in abdominal girth. Bloating is primarily a sensory phenomenon in the small intestine; patients experiencing bloating usually do not produce excess gas but may have lower pain thresholds and increased sensitivity.4
Belching is the expulsion of excess gas from the stomach; it may or may not coexist with bloating and distention. Belching occurs because of an excess of swallowed air and is caused by processes often unrelated to those causing bloating.4
Flatulence is the expulsion of excess colonic gas and is usually related to diet.4 The colon is relatively insensitive to increased gas and distention; excess flatulence does not usually cause symptoms of bloating.
Coexisting anxiety and depression may increase the severity of symptoms of FGIDs, and stressful life events or illness may be related to acute flare-ups; however, these conditions do not specifically cause FGIDs.2
Evaluation
Acute flare-ups or psychological distress related to uncontrolled symptoms may lead patients to seek urgent medical evaluation. A framework for categorizing symptoms and a structured approach to evaluation, particularly when the patient and physician may be new to one another, help to establish an effective relationship.
ESTABLISHING A TRUSTING RELATIONSHIP
A trusting therapeutic relationship is essential for patients to understand and accept the biopsychosocial model of FGIDs, to be confident that the evaluation for other conditions has been adequate, to accept the limitations of therapy and incremental improvements of symptoms, and to engage in effective self-management.2,5
Patients often believe that their symptoms are not appreciated2 because FGIDs are often perceived as less legitimate than structural conditions such as inflammatory bowel disease or infections. Listening carefully and acknowledging patients' symptoms are essential. Patients should be asked about chronicity, waxing and waning of symptoms, temporal relationships, precipitating or aggravating events, recent illness, and psychosocial stressors.
ALARM SYMPTOMS
Alarm symptoms such as weight loss, fever, gastrointestinal bleeding, unusually severe symptoms (Table 14,6–12), and new-onset symptoms in older adults or in patients with previous cancers or abdominal surgery often require additional testing.
TABLE 1
Alarm Symptoms Suggesting Potentially Serious Causes of Gas, Bloating, and Belching
zoom_out_mapEnlargeprintPrint
| |
| Abdominal mass |
| Dysphagia (difficulty swallowing) |
| Extreme diarrhea symptoms (large volume, bloody, nocturnal, progressive pain, does not improve with fasting) |
| Fever |
| Gastrointestinal bleeding (melena or hematochezia) |
| Jaundice |
| Lymphadenopathy |
| New-onset symptoms in patients 55 years and older |
| Odynophagia (painful swallowing) |
| Symptoms of chronic pancreatitis† |
| Symptoms of gastrointestinal cancer, including family history‡ |
| Symptoms of ovarian cancer, including family history§ |
| Tenesmus (rectal pain or feeling of incomplete evacuation) |
| Unintentional weight loss |
| Vomiting |
—Patients do not typically develop functional gastrointestinal disorders later in life. Patients 55 years and older who report new-onset symptoms should be considered for more in-depth evaluation to exclude alternative diagnoses.
†—Patients with recurrent episodes of acute or chronic disabling pain, especially in the setting of many years of alcohol abuse, should be considered for evaluation for chronic pancreatitis (may rarely be confused with irritable bowel syndrome, gastroparesis, or small bowel obstruction).11
‡—Patients with a family history of gastrointestinal malignancy, particularly pancreatic or colorectal cancer, should be considered for evaluation for gastrointestinal cancers.
§—Women 55 years and older who report new-onset bloating, increased abdominal size, difficulty eating, or early satiety with abdominal, pelvic, or back pain should be considered for evaluation for ovarian cancer.12
Information from references 4, and 6 through 12.
DIETARY HISTORY
Physicians should ask about the timing and patterns of meals (e.g., larger or less frequent meals, gulping of food, poorly chewed food) and their content (e.g., gas-producing foods, artificial sweeteners, caffeinated or carbonated beverages; Table 24).
TABLE 2
Effect of Dietary Choices When Determining Possible Causes of Functional Gastrointestinal Disorders
zoom_out_mapEnlargeprintPrint
| Possible causes of functional gastrointestinal disorders | Dietary choices | Results |
---
| Artificial sweeteners | Sugar-free gum (especially gum containing sorbitol or mannitol) | Bloating, often with diarrhea |
| Caffeinated beverages | Coffee, soda | Can cause belching and diarrhea but less likely to cause bloating |
| Can decrease lower esophageal sphincter pressure, which causes belching |
| Carbonated drinks | Soda, other carbonated beverages | Excess gas and bloating |
| Eating habits | Bolting or gulping food, eating quickly, not thoroughly chewing food, routinely chewing gum | Belching, bloating, gas |
| Over-the-counter medications | Antacids containing magnesium | Diarrhea (bloating less likely) |
| Size and timing of meals | Large or frequent meals, meals eaten late in the day | Bloating with distention, dyspepsia |
| Specific foods | Beans, fiber, fructans, fructose, lactose, legumes | Gas |
Information from reference 4.
CATEGORIZE SYMPTOMS
Symptoms can be categorized as representing gastric bloating, small bowel bloating, bloating with constipation, or belching (Figure 113). Two particularly useful questions help localize which distinct level of the gastrointestinal tract is involved: Can you eat a full plate of food? Do you regularly have a good bowel movement?
FIGURE 1
zoom_out_mapEnlargeprintPrint
EXAMINATION
Although abnormal findings are uncommon, a careful abdominal examination is a valuable ritual and helps convey empathy.2,14 A rectal examination may provide valuable clues to pelvic floor dysfunction and can guide further testing6,15 (Table 36,15–17).
TABLE 3
Rectal Examination to Evaluate for Dyssynergic Defecation
zoom_out_mapEnlargeprintPrint
| |
| Patient lying on left side Inspect for anal fissure, external hemorrhoids, or other perineal abnormalities. Check perineal sensation and anocutaneous reflex (“anal wink”) using a cotton swab. |
| Patient bearing down, simulating defecation The perineum should relax and descend 1 to 3.5 cm; a minimal descent or paradoxical perineal rise suggests an inability to relax the pelvic floor muscles during defecation; an exaggerated descent suggests perineal laxity (e.g., childbirth, excessive straining attributable to chronic constipation). The anal sphincter should relax; paradoxical anal sphincter contraction suggests elevated sphincter pressure and anal stricture. Perineal “ballooning,” rectal prolapse, and prolapse of internal hemorrhoids are abnormal findings. Palpate the abdominal wall; excessive contraction suggests the Valsalva maneuver during defecation and ineffective effort. |
| Digital rectal examination while patient is relaxed Assess for increased anal sphincter tone, which may contribute to difficulty with evacuation. Palpate for anal fissure, tenderness, mass, stricture, rectocele, and hard stool. |
| Digital rectal examination while patient is instructed to try to expel finger Internal sphincter and puborectalis muscle should be felt to relax; tightening or lack of perineal descent suggests pelvic floor dyssynergia. Rectal propulsive force should be sufficient to expel finger. |
| Patient squatting, simulating defecation Rectal prolapse may not always be evident with the patient lying on his or her side, even when bearing down. |
Information from references 6, and 15 through 17.
TESTING
There are no definitive tests for FGIDs. Patients' fears about specific diagnoses, particularly cancer or infection, should be elicited and addressed2; testing should be done to exclude specific diagnoses, not simply to reassure18 (Table 44,6–9,15,16,19–24).
TABLE 4
Initial Testing for Patients with Gas, Bloating, and Belching
zoom_out_mapEnlargeprintPrint
| Suspected diagnosis | Testing to consider | Clinical features of patient |
---
| Gastric bloating | | |
| Functional dyspepsia | Helicobacter pylori testing (urea breath testing most cost-effective relative to endoscopy; stool antigen testing is less expensive and also a reasonable option; serologic tests are least accurate)7,8,19 | If younger than 55 years with no alarm symptoms, test-and-treat strategy for H. pylori detection and eradication is safe and cost-effective8 |
| EGD | If 55 years or older or with alarm symptoms (Table 1), EGD is indicated |
| Disorders of accommodation | Gastric accommodation study (only available at specialized centers) | Diagnosis is difficult, and treatments are only minimally effective |
| Testing is of limited utility |
| Gastroparesis | Gastric emptying study8 (recommended only if gastroparesis is strongly suspected; should not be routinely obtained in functional dyspepsia) | If diabetes mellitus or recent viral illness and negative EGD, consider gastroparesis |
| Gastric outlet obstruction | EGD | If early satiety or vomiting, or if suspected or known history of peptic ulcer disease, rule out gastric outlet obstruction with EGD |
| Small bowel bloating | | |
| IBS | Celiac serology | Consider testing for celiac disease in patients with diarrhea-predominant or mixed-presentation IBS, or if local prevalence of celiac disease > 10%20,21 |
| Colonoscopy | If 55 years or older, alarm symptoms (Table 1), or routine screening is due, colonoscopy indicated16 |
| Celiac disease | Celiac serology | Intestinal symptoms of celiac disease (diarrhea, weight loss, abdominal bloating and distention, gas) are less common than extraintestinal symptoms (anemia, dermatitis herpetiformis, oral lesions, osteoporosis/osteopenia)22 |
| Must include tissue transglutaminase and total IgA (to rule out IgA deficiency) |
| If IgA deficiency is present, test with deamidated gliadin |
| Serologic testing should ideally be confirmed by biopsy |
| SIBO | Lactulose hydrogen breath testing (appropriate only for patients with risk factors for SIBO) | Risk factors for SIBO: structural abnormalities (small bowel diverticula, strictures, surgical blind loops, ileocecal valve resection); disordered motility (scleroderma, type 1 diabetes, use of opioids); acid suppression (chronic proton pump inhibitor use, achlorhydria, gastric resection)4 |
| Functional abdominal distention | Testing usually not necessary | Subjective symptoms of recurrent abdominal pressure with objective increases in abdominal girth |
| More likely related to abdominal wall muscle relaxation than to retained gas9 |
| Functional abdominal bloating | Testing usually not necessary | Subjective symptoms of recurrent abdominal pressure, sensation of trapped gas |
| Typically worsens throughout day and after meals, improves overnight9 |
| Bloating with constipation | | |
| Chronic idiopathic constipation (functional constipation, constipation-predominant IBS) | Rule out dyssynergic defecation (see example in this table) and secondary constipation | Incomplete evacuation |
| Straining with defecation |
| Manual removal of stool |
| History of sexual or physical abuse6,15,16 |
| Secondary constipation | Hypothyroidism, diabetic neuropathy, hypomagnesemia, hypokalemia, and hypercalcemia | Inquire about medications (calcium-containing antacids, iron supplements, anticholinergics, opioids) |
| Dyssynergic defecation (pelvic floor dysfunction) | Careful perineal and rectal examination (often sufficient to guide further testing) | Incomplete evacuation |
| Anorectal manometry (may be required to justify insurance coverage of treatment) | Straining with defecation |
| Manual removal of stool |
| History of sexual or physical abuse6,15,16 |
| Slow transit constipation | Colonic transit study (consider only after ruling out dyssynergic defecation) | Rare condition |
| False-positive results not uncommon |
| Belching | | |
| Gastric belching | Testing usually not necessary but can be reliably diagnosed by manometry and impedance testing7 | Rapid eating and gum chewing, which may cause excessive air swallowing; lower esophageal sphincter relaxes with belching7 |
| Supragastric belching | Testing usually not necessary but can be reliably diagnosed by manometry and impedance testing7 | Patients who are belching during conversation; worse when discussing symptoms and better when distracted; lower esophageal sphincter does not relax with belching7 |
| Aerophagia (“air swallowing”—now considered a historical term)† | Testing not necessary | Anxiety, rapid eating, and chewing gum may cause excessive air swallowing, both supragastric and gastric (see above)23 |
EGD = esophagogastroduodenoscopy; IBS = irritable bowel syndrome; IgA = immunoglobulin A; SIBO = small intestinal bacterial overgrowth.
—Patients cannot be on a gluten-free diet at the time of testing; serologic tests and biopsies may be falsely negative.20,21
†—Continuous positive airway pressure (CPAP) may result in gas and belching, although this is more likely to increase nocturnal symptoms of gastroesophageal reflux disease than to cause gastric distention.24
Information from references 4, 6 through 9, 15, 16, and 19 through 24
Gastric Bloating
Symptoms occurring within 30 minutes after eating or the inability to finish a meal are attributable to upper gastrointestinal disorders, usually functional dyspepsia. Other conditions, including gastroesophageal reflux disease (GERD), Helicobacter pylori infection, gastroparesis, impaired gastric accommodation, and gastric outlet obstruction, must also be considered, although definitive testing may be deferred in favor of empiric treatment.
FUNCTIONAL DYSPEPSIA
Patients with functional (nonulcer) dyspepsia typically report postprandial fullness, bloating, or early satiation; however, some patients with functional dyspepsia may instead report epigastric pain or burning unrelated to meals.7,8 In addition, some patients with GERD may report symptoms that also occur in patients with functional dyspepsia, including nausea, vomiting, early satiety, bloating, and belching. This suggests that functional dyspepsia and GERD may coexist in some patients or that others thought to have GERD may instead have functional dyspepsia.25
The relationship between functional dyspepsia and H. pylori infection is unclear. H. pylori eradication results in functional dyspepsia symptom resolution in some patients. Consequently, a test-and-treat strategy (noninvasive testing for H. pylori [e.g., urea breath testing] and treatment of confirmed infection) is recommended rather than expensive and invasive tests, such as endoscopy8,19 (Table 44,6–9,15,16,19–24).
The relationship between functional dyspepsia and acid secretion is also unclear; empiric proton pump inhibitor therapy reduces functional dyspepsia symptoms in some patients, even when acid reflux cannot be demonstrated. Therefore, a trial of antisecretory therapy is recommended for patients who are H. pylori negative or for those who remain symptomatic after H. pylori eradication8,25 (eTable A).
eTABLE A
Initial Therapy of Functional Gastrointestinal Disorders
zoom_out_mapEnlargeprintPrint
| |
| Intestinal lumen |
| Diet General dietary guidance: Eat in moderation; get adequate but not excessive fiber; decrease consumption of fatty and spicy foods; avoid caffeine, soft drinks, carbonation, and artificial sweeteners.A1,A2 Restricted diet: Diets restricted in fermentable oligosaccharides, disaccharides, monosaccharides, and polyols may improve IBS symptoms in some patients, but questions remain regarding their long-term safety, effectiveness, and practicality.A3,A4 Gluten-free diet: The long-term effects of gluten-free diets on the microbiome and general nutrition are uncertain; they may lead to increased fat and sugar consumption. Strict gluten-free diets are expensive and difficult to follow and should be advised only for patients with proven celiac disease.A1,A5–A7 |
| Probiotics Probiotics, especially Bifidobacterium infantis, improve bloating, gas, and pain in IBS and may also help chronic idiopathic constipation.A1,A8 |
| Exercise Regular daily exercise decreases symptoms of IBS and chronic idiopathic constipation.A1 |
| Fiber Soluble fiber (psyllium; 25 to 30 g daily) provides a small benefit for some patients with IBS,A1,A8,A9 but it can also worsen bloating. Insoluble fiber (bran, methylcellulose, calcium polycarbophil; two heaped teaspoons daily) is somewhat more effective for chronic idiopathic constipation than soluble fiber, but it can worsen bloating and flatulence in many patients.A8 Increase fiber gradually to minimize bloating, distention, flatulence, and cramping.A1,A8,A9 Worsening of gas and bloating suggests underlying dyssynergic defecation (pelvic floor dysfunction).A9 |
| Osmotic laxatives Polyethylene glycol (Miralax; 17 g once daily) is the osmotic laxative best studied and most effective for chronic idiopathic constipation.A8,A10 Other osmotic laxatives include lactulose, sorbitol, and mannitol. |
| Proton pump inhibitors Proton pump inhibitors are recommended as empiric therapy in patients with functional dyspepsia who are Helicobacter pylori negative or who remain symptomatic after H. pylori eradication.A11,A12 |
| Bowel wall |
| Opioid agonists Loperamide (2 to 4 mg up to four times daily) decreases colonic transit and increases water absorption, improving many symptoms of diarrhea-predominant IBS.A8 Diphenoxylate/atropine (Lomotil) has not been studied in IBS.A8 |
| Antispasmodics Hyoscyamine (Levsin; 0.125 to 0.25 mg three times daily, 30 minutes before meals) is moderately effective for IBS.A9 Hyoscyamine, extended release (Levbid; 0.375 to 0.75 mg two times daily; maximum of 1.5 mg daily) is moderately effective for IBS.A9 Dicyclomine (20 mg four times daily, may increase to a maximum of 40 mg four times daily, 30 minutes before meals) is moderately effective for IBS.A9 Peppermint oil (200 to 750 mg two to three times dailyA8) is moderately effective for IBS.A9 |
| Stimulant laxative Stimulant laxatives (bisacodyl, cascara, senna) increase motility in chronic idiopathic constipation but may cause pain and diarrheaA8,A13,A14; they are usually reserved for patients with dysmotility disorders or opioid-induced constipation. |
| Prokinetic agents Buspirone (Buspar; 5 to 10 mg three times daily, 30 minutes before meals), in addition to its antidepressant benefit, relaxes the gastric fundus and is effective for symptoms of functional dyspepsia.A15 Metoclopramide (Reglan), although approved for gastroparesis, has not been well studied in functional dyspepsia. Irreversible tardive dyskinesia can occur even at low dosages. It should not be used routinely for functional dyspepsia symptoms but may be considered after test-and-treat therapy, proton pump inhibitor therapy, and tricyclic antidepressant therapy.A11,A16 |
| Neuromodulatory gut-brain axis |
| Tricyclic antidepressants Amitriptyline (10 to 50 mg at bedtime; increase by 10 mg every one to two weeks) and other tricyclic antidepressants are moderately effective for symptom relief in diarrhea-predominant IBS and functional dyspepsia (histaminergic properties are moderately sedating; anticholinergic properties are moderately constipating).A10,A11,A17,A18 |
| Central nervous system |
| Psychotropic drugs (e.g., selective serotonin reuptake inhibitors, serotonin-norepinephrine reuptake inhibitors) treat symptoms of underlying anxiety or depression but are less effective for associated pain and bloating.A17,A18 Psychological therapies (e.g., cognitive behavior therapy, hypnotherapy) improve quality of life and overall function in IBS and functional dyspepsia but are less effective for associated pain and bloating.A17,A18 |
IBS = irritable bowel syndrome.
Information from:
A1. Ford AC, Moayyedi P, Lacy BE, et al.; Task Force on the Management of Functional Bowel Disorders. American College of Gastroenterology monograph on the management of irritable bowel syndrome and chronic idiopathic constipation. Am J Gastroenterol. 2014;109(suppl 1):S2–S26, quiz S27.
A2. Wilder-Smith CH, Materna A, Wermelinger C, Schuler J. Fructose and lactose intolerance and malabsorption testing: the relationship with symptoms in functional gastrointestinal disorders. Aliment Pharmacol Ther. 2013;37(11):1074–1083.
A3. Dionne J, Ford AC, Yuan Y, et al. A systematic review and meta-analysis evaluating the efficacy of a gluten-free diet and a low FODMAPs diet in treating symptoms of irritable bowel syndrome. Am J Gastroenterol. 2018;113(9):1290–1300.
A4. Böhn L, Störsrud S, Liljebo T, et al. Diet low in FODMAPs reduces symptoms of irritable bowel syndrome as well as traditional dietary advice: a randomized controlled trial. Gastroenterology. 2015;149(6):1399–1407.e2.
A5. van der Windt DA, Jellema P, Mulder CJ, Kneepkens CM, van der Horst HE. Diagnostic testing for celiac disease among patients with abdominal symptoms: a systematic review. JAMA. 2010;303(17):1738–1746.
A6. Rubio-Tapia A, Hill ID, Kelly CP, Calderwood AH, Murray JA; American College of Gastroenterology. ACG clinical guidelines: diagnosis and management of celiac disease. Am J Gastroenterol. 2013;108(5):656–676.
A7. Leonard MM, Sapone A, Catassi C, Fasano A. Celiac disease and nonceliac gluten sensitivity: a review. JAMA. 2017;318(7):647–656.
A8. Lacy BE, Mearin F, Chang L, et al. Bowel disorders. Gastroenterology. 2016;150(6):1393–1407.
A9. Ford AC, Talley NJ, Spiegel BM, et al. Effect of fibre, antispasmodics, and peppermint oil in the treatment of irritable bowel syndrome: systematic review and meta-analysis [published correction appears in BMJ. 2009;338:b1881]. BMJ. 2008;337:a2313.
A10. Lacy BE, Gabbard SL, Crowell MD. Pathophysiology, evaluation, and treatment of bloating: hope, hype, or hot air? Gastroenterol Hepatol (N Y). 2011;7(11):729–739.
A11. Moayyedi PM, Lacy BE, Andrews CN, Enns RA, Howden CW, Vakil N. ACG and CAG clinical guideline: management of dyspepsia. Am J Gastroenterol. 2017;112(7):988–1013.
A12. Malfertheiner P, Megraud F, O'Morain CA, et al.; European Helicobacter and Microbiota Study Group and Consensus panel. Management of Helicobacter pylori infection—the Maastricht V/Florence Consensus Report. Gut. 2017;66(1):6–30.
A13. Camilleri M, Ford AC, Mawe GM, et al. Chronic constipation. Nature Reviews Disease Primers. 2017;3:17095. Accessed June 15, 2018.
A14. Bharucha AE, Pemberton JH, Locke GR III. American Gastroenterological Association technical review on constipation. Gastroenterology. 2013;144(1):218–238.
A15. Tack J, Janssen P, Masaoka T, Farré R, Van Oudenhove L. Efficacy of buspirone, a fundus-relaxing drug, in patients with functional dyspepsia. Clin Gastroenterol Hepatol. 2012;10(11):1239–1245.
A16. Camilleri M, Parkman HP, Shafi MA, Abell TL, Gerson L; American College of Gastroenterology. Clinical guideline: management of gastroparesis. Am J Gastroenterol. 2013;108(1):18–37, quiz 38.
A17. Ford AC, Quigley EM, Lacy BE, et al. Effect of antidepressants and psychological therapies, including hypnotherapy, in irritable bowel syndrome: systematic review and meta-analysis. Am J Gastroenterol. 2014;109(9):1350–1365, quiz 1366.
A18. Ford AC, Luthra P, Tack J, Boeckxstaens GE, Moayyedi P, Talley NJ. Efficacy of psychotropic drugs in functional dyspepsia: systematic review and meta-analysis. Gut. 2017;66(3):411–420.
GASTROPARESIS
Gastroparesis is a chronic disorder of delayed gastric emptying unrelated to mechanical obstruction. Most patients with gastroparesis experience nausea and vomiting in addition to dyspeptic symptoms. Idiopathic gastroparesis is most common in young or middle-aged women and sometimes develops after viral gastroenteritis; resolution of the disorder may take a year or more.26 Diabetic gastroparesis is a relatively rare complication (occurring in only 1% of patients with type 2 diabetes mellitus)27 closely related to the degree of hyperglycemia; improved glycemic control often results in improved symptoms. 28 Bariatric surgery or fundoplication may occasionally cause postsurgical gastroparesis.26
IMPAIRED GASTRIC ACCOMMODATION
Reduced gastric accommodation, a disorder of the vagally mediated reflex that permits the stomach to adapt to food as it enters, has been recognized as distinct from delayed gastric emptying, but its exact role in dyspeptic symptoms is unclear. Testing is done only in specialized centers, and no specific treatments exist.29
GASTRIC OUTLET OBSTRUCTION
Gastric outlet obstruction may also cause bloating. Most gastric outlet obstruction is attributable to chronic peptic ulcer disease and scarring; in patients without alarm symptoms (Table 14,6–12), the risk of malignancy is low. Eradication of H. pylori infection often results in long-term improvement of gastric outlet obstruction.19,30
Small Bowel Bloating
Symptoms, including abdominal distention, occurring more than 30 minutes after eating originate in the small bowel and proximal colon. IBS is the most common cause of small bowel bloating, but celiac disease should be excluded. Nonceliac food sensitivities and other conditions must also be considered. Functional abdominal bloating and functional abdominal distention are characterized by subjective symptoms of abdominal pressure, with or without objective increases in abdominal girth; they are more likely related to abdominal wall muscle relaxation than to retained gas9 (Table 44,6–9,15,16,19–24).
IRRITABLE BOWEL SYNDROME
IBS is characterized by pain, usually with bloating or abdominal distention, and associated with disordered bowel habits. Subtypes of IBS depend on the predominant bowel habit: diarrhea (IBS-D), constipation (IBS-C), or mixed (IBS-M).9,10,16 Fiber, antispasmodics, and peppermint oil are moderately effective for IBS symptoms in some patients16,31 (eTable A).
CELIAC DISEASE
Celiac disease may present with bloating, flatulence, diarrhea or constipation, weight loss, anemia attributable to malabsorption of iron or folic acid, or osteoporosis attributable to calcium malabsorption. If serologic testing is positive, duodenal biopsy should be performed. False-negative results may occur in serologic testing if the patient has already been on a gluten-free diet (Table 44,6–9,15,16,19–24). Strict gluten-free diets are expensive and difficult to follow and should be advised only for patients with proven celiac disease.20,21
FOOD SENSITIVITIES
In addition to malabsorption syndromes such as celiac disease or lactase deficiency, various foods can also induce or exacerbate symptoms in patients with various FGIDs, particularly IBS. Strict elimination diets are usually not necessary, but restriction of identified foods, particularly at times of symptom flare-ups, is often helpful.
Gluten. The clinical entity currently referred to as nonceliac gluten sensitivity is incompletely understood but seems to closely overlap with other FGIDs.32
Lactose. IBS may be exacerbated by milk or other dairy products, and lactose intolerance attributable to lactase deficiency may mimic the symptoms of IBS. In addition, at least 25% of patients with FGIDs also have lactase deficiency. Lactose intolerance may be diagnosed using hydrogen breath testing, but this is not always accurate in patients with FGIDs; careful monitoring of symptoms in relation to ingestion of dairy products may be equally helpful33 (Table 44,6–9,15,16,19–24).
FODMAPs. Foods containing a variety of fermentable oligosaccharides, disaccharides, monosaccharides, and polyols (FODMAPs) may precipitate symptoms in certain individuals (Table 54,34). Some patients find that avoiding certain FODMAP-containing foods may reduce IBS symptoms, but the routine use of highly restrictive exclusion diets has not been well studied and is not recommended.35
TABLE 5
Foods Containing Fermentable Oligosaccharides, Disaccharides, Monosaccharides, and Polyols
zoom_out_mapEnlargeprintPrint
| | |
--- |
| Oligosaccharides (fructans) | Wheat (large amounts), rye (large amounts), onions, leeks, zucchini |
| Disaccharides (lactose) | Dairy products, cheese, milk, yogurt |
| Monosaccharides (excess fructose) | Honey, apples, pears, peaches, mangoes, fruit juice, dried fruit |
| Polyols (sorbitol) | Apricots, peaches, artificial sweeteners, sugar-free gums |
| Galactose (raffinose) | Lentils, cabbage, brussels sprouts, asparagus, green beans, legumes |
Information from references 4 and 34.
SMALL INTESTINAL BACTERIAL OVERGROWTH
A variety of conditions (Table 44,6–9,15,16,19–24) are thought to predispose to overgrowth of colonic bacteria in the distal small intestine, resulting in gas production, malabsorption, and inflammation.4
Constipation with Bloating
Patients with difficult, infrequent, or incomplete bowel movements, usually with lower but sometimes with upper abdominal pain, typically have either IBS-C or functional constipation; dyssynergic defecation related to pelvic floor dysfunction, as well as secondary causes, must also be considered.9,16
CHRONIC IDIOPATHIC CONSTIPATION
Chronic idiopathic constipation includes IBS-C (predominantly pain) and functional constipation (predominantly constipation), both of which are probably part of the same condition. The term “normal transit constipation” also refers to functional constipation. Stool transit is normal, but bowel movements are considered unsatisfactory. Symptoms often worsen with psychosocial stress and usually respond to fiber supplementation or osmotic laxatives.6,15,16
DYSSYNERGIC DEFECATION
Dyssynergic defecation, caused by poor coordination of the pelvic floor, anal sphincter, and abdominal wall muscles during attempted defecation, results in prolonged or excessive straining even with relatively soft stools.9 Rectal examination is important17 (Table 36,15–17) to guide further testing (Table 44,6–9,15,16,19–24). Structured biofeedback-aided pelvic floor retraining is often successful.6,15
SLOW TRANSIT CONSTIPATION
Truly prolonged colonic transit times are relatively rare. Dyssynergic defecation can appear to affect colonic transit, so this condition must be excluded before considering a diagnosis of slow transit constipation. Symptoms tend not to respond to fiber or laxatives, although biofeedback has been reported to be effective in one trial.6,15
SECONDARY CONSTIPATION
Various medications and metabolic abnormalities may also cause constipation (Table 44,6–9,15,16,19–24 ).
Belching
Belching prevents gas accumulation and distention. Typically occurring 25 to 30 times daily, it is usually not perceived and is rarely excessive or troublesome.
SUPRAGASTRIC BELCHING
Troublesome, repetitive belching, sometimes occurring up to 20 times per minute, is an involuntary but learned behavior, often in response to stress, anxiety, or unpleasant gastrointestinal symptoms. Air is sucked into the esophagus and immediately expelled without ever reaching the stomach.36 Symptoms worsen while they are being discussed and abate with distraction and sleep.37 Treatment of the underlying anxiety, as well as biofeedback, may be helpful.38
GASTRIC BELCHING
Increased gas and belching may be caused by excessive gum chewing, drinking carbonated beverages, or eating too quickly. These may also occur with GERD and functional dyspepsia, but other symptoms usually predominate.7,39
Initial Treatment
Treatment of FGIDs begins with reassurance about the generally benign course of these conditions. The concept of the biopsychosocial model should be introduced as appropriate; it is important to stress that although anxiety, depression, and psychosocial stressors do not cause FGIDs, they can worsen symptoms and should be addressed if present.
Self-managed combinations of medications and dietary interventions are most effective. Several safe and inexpensive drugs are available, often over the counter (eTable A); newer agents are generally more appropriate for patients with complicated or intractable symptoms.2,31 Most patients will have incremental improvement over time with occasional flare-ups; approximately 50% of patients will have resolution of symptoms, 30% will have fluctuating symptoms, and 20% will develop new symptoms.40
Data Sources: We searched PubMed and Google Scholar using the search terms gas, bloating, belching, functional gastrointestinal disorders, FGID, IBS, functional dyspepsia, constipation, celiac disease, FODMAP, and gluten-free, alone and in combination with one another. We examined clinical trials, meta-analyses, review articles, and clinical guidelines, as well as the bibliographies of selected articles. Cochrane and Essential Evidence Plus were also searched. Search dates: June through October 2018.
JOHN M. WILKINSON, MD, is a consultant in the Department of Family Medicine and an associate professor in the Mayo Clinic Alix School of Medicine, Mayo Clinic College of Medicine and Science, Rochester, Minn.
ELIZABETH W. COZINE, MD, is a consultant in the Department of Family Medicine and an assistant professor in the Mayo Clinic Alix School of Medicine, Mayo Clinic College of Medicine and Science, Rochester.
CONOR G. LOFTUS, MD, is a consultant in the Division of Gastroenterology and Hepatology and an associate professor in the Mayo Clinic Alix School of Medicine, Mayo Clinic College of Medicine and Science, Rochester.
Address correspondence to John M. Wilkinson, MD, Mayo Clinic Alix School of Medicine, Mayo Clinic College of Medicine and Science, 200 1st St. SW, Rochester, MN 55905 (e-mail: wilkinson.john@mayo.edu). Reprints are not available from the authors.
Author disclosure: No relevant financial affiliations.
Vakil N, Halling K, Ohlsson L, Wernersson B. Symptom overlap between postprandial distress and epigastric pain syndromes of the Rome III dyspepsia classification. Am J Gastroenterol. 2013;108(5):767-774.
Drossman DA. Functional gastrointestinal disorders: history, pathophysiology, clinical features and Rome IV. Gastroenterology. 2016;150(6):1262-1279.
Ford AC, Talley NJ, Veldhuyzen van Zanten SJ, et al. Will the history and physical examination help establish that irritable bowel syndrome is causing this patient’s lower gastrointestinal tract symptoms? [published correction appears in JAMA. 2009;301(15):1544]. JAMA. 2008;300(15):1793-1805.
Lacy BE, Gabbard SL, Crowell MD. Pathophysiology, evaluation, and treatment of bloating: hope, hype, or hot air?. Gastroenterol Hepatol (N Y). 2011;7(11):729-739.
Owens DM, Nelson DK, Talley NJ. The irritable bowel syndrome: long-term prognosis and the physician-patient interaction. Ann Intern Med. 1995;122(2):107-112.
Camilleri M, Ford AC, Mawe GM, et al. Chronic constipation. Nature Reviews Disease Primers. 2017;3:17095. Accessed June 15, 2018.
Stanghellini V, Chan FK, Hasler WL, et al. Gastroduodenal disorders. Gastroenterology. 2016;150(6):1380-1392.
Moayyedi PM, Lacy BE, Andrews CN, et al. ACG and CAG clinical guideline: management of dyspepsia. Am J Gastroenterol. 2017;112(7):988-1013.
Lacy BE, Mearin F, Chang L, et al. Bowel disorders. Gastroenterology. 2016;150(6):1393-1407.
Chey WD, Kurlander J, Eswaran S. Irritable bowel syndrome: a clinical review. JAMA. 2015;313(9):949-958.
Barry K. Chronic pancreatitis: diagnosis and treatment. Am Fam Physician. 2018;97(6):385-393.
Yawn BP, Barrette BA, Wollan PC. Ovarian cancer: the neglected diagnosis. Mayo Clin Proc. 2004;79(10):1277-1282.
Cotter TG, Gurney M, Loftus CG. Gas and bloating-controlling emissions: a case-based review for the primary care provider. Mayo Clin Proc. 2016;91(8):1105-1113.
Verghese A, Brady E, Kapur CC, Horwitz RI. The bedside evaluation: ritual and reason. Ann Intern Med. 2011;155(8):550-553.
Bharucha AE, Pemberton JH, Locke GR III. American Gastroenterological Association technical review on constipation. Gastroenterology. 2013;144(1):218-238.
Ford AC, Moayyedi P, Lacy BE, et al.; Task Force on the Management of Functional Bowel Disorders. American College of Gastroenterology monograph on the management of irritable bowel syndrome and chronic idiopathic constipation. Am J Gastroenterol. 2014;109(suppl 1):S2-S26 , quiz S27.
Talley NJ. How to do and interpret a rectal examination in gastroenterology. Am J Gastroenterol. 2008;103(4):820-822.
Rolfe A, Burton C. Reassurance after diagnostic testing with a low pre-test probability of serious disease: systematic review and meta-analysis. JAMA Intern Med. 2013;173(6):407-416.
Malfertheiner P, Megraud F, O’Morain CA, et al.; European Helicobacter and Microbiota Study Group and Consensus panel. Management of Helicobacter pylori infection—the Maastricht V/Florence Consensus Report. Gut. 2017;66(1):6-30.
van der Windt DA, Jellema P, Mulder CJ, Kneepkens CM, van der Horst HE. Diagnostic testing for celiac disease among patients with abdominal symptoms: a systematic review. JAMA. 2010;303(17):1738-1746.
Rubio-Tapia A, Hill ID, Kelly CP, Calderwood AH, Murray JA American College of Gastroenterology. ACG clinical guidelines: diagnosis and management of celiac disease. Am J Gastroenterol. 2013;108(5):656-676.
Rampertab SD, Pooran N, Brar P, Singh P, Green PH. Trends in the presentation of celiac disease. Am J Med. 2006;119(4):355.e9-355.e14.
Hyams JS, Di Lorenzo C, Saps M, et al. Childhood functional gastrointestinal disorders: child/adolescent. Gastroenterology. 2016;150(6):1456-1468.
Shepherd K, Hillman D, Eastwood P. Symptoms of aerophagia are common in patients on continuous positive airway pressure therapy and are related to the presence of nighttime gastroesophageal reflux. J Clin Sleep Med. 2013;9(1):13-17.
Gerson LB, Kahrilas PJ, Fass R. Insights into gastroesophageal reflux disease-associated dyspeptic symptoms. Clin Gastroenterol Hepatol. 2011;9(10):824-833.
Camilleri M, Parkman HP, Shafi MA, Abell TL, Gerson L American College of Gastroenterology. Clinical guideline: management of gastroparesis. Am J Gastroenterol. 2013;108(1):18-37 , quiz 38.
Choung RS, Locke GR, Schleck CD, et al. Risk of gastroparesis in subjects with type 1 and 2 diabetes in the general population. Am J Gastroenterol. 2012;107(1):82-88.
Bytzer P, Talley NJ, Hammer J, et al. GI symptoms in diabetes mellitus are associated with both poor glycemic control and diabetic complications. Am J Gastroenterol. 2002;97(3):604-611.
Bredenoord AJ, Chial HJ, Camilleri M, Mullan BP, Murray JA. Gastric accommodation and emptying in evaluation of patients with upper gastrointestinal symptoms. Clin Gastroenterol Hepatol. 2003;1(4):264-272.
Gisbert JP, Pajares JM. Review article: Helicobacter pylori infection and gastric outlet obstruction—prevalence of the infection and role of antimicrobial treatment. Aliment Pharmacol Ther. 2002;16(7):1203-1208.
Ford AC, Talley NJ, Spiegel BM, et al. Effect of fibre, antispasmodics, and peppermint oil in the treatment of irritable bowel syndrome: systematic review and meta-analysis [published correction appears in BMJ. 2009;338:b1881]. BMJ. 2008;337:a2313.
Leonard MM, Sapone A, Catassi C, Fasano A. Celiac disease and nonceliac gluten sensitivity: a review. JAMA. 2017;318(7):647-656.
Wilder-Smith CH, Materna A, Wermelinger C, Schuler J. Fructose and lactose intolerance and malabsorption testing: the relationship with symptoms in functional gastrointestinal disorders. Aliment Pharmacol Ther. 2013;37(11):1074-1083.
Böhn L, Störsrud S, Liljebo T, et al. Diet low in FODMAPs reduces symptoms of irritable bowel syndrome as well as traditional dietary advice: a randomized controlled trial. Gastroenterology. 2015;149(6):1399-1407.e2.
Dionne J, Ford AC, Yuan Y, et al. A systematic review and meta-analysis evaluating the efficacy of a gluten-free diet and a low FODMAPs diet in treating symptoms of irritable bowel syndrome. Am J Gastroenterol. 2018;113(9):1290-1300.
Kessing BF, Bredenoord AJ, Smout AJ. Mechanisms of gastric and supragastric belching: a study using concurrent high-resolution manometry and impedance monitoring. Neurogastroenterol Motil. 2012;24(12):e573-e579.
Bredenoord AJ, Weusten BL, Timmer R, Smout AJ. Psychological factors affect the frequency of belching in patients with aerophagia. Am J Gastroenterol. 2006;101(12):2777-2781.
Hemmink GJ, Ten Cate L, Bredenoord AJ, et al. Speech therapy in patients with excessive supragastric belching—a pilot study. Neurogastroenterol Motil. 2010;22(1):24-28.
Bredenoord AJ. Excessive belching and aerophagia: two different disorders. Dis Esophagus. 2010;23(4):347-352.
Ford AC, Forman D, Bailey AG, Axon AT, Moayyedi P. Fluctuation of gastrointestinal symptoms in the community: a 10-year longitudinal follow-up study. Aliment Pharmacol Ther. 2008;28(8):1013-1020.
Continue Reading
Mar 1, 2019
Mar 1, 2019
Previous: Adhesive Capsulitis: Diagnosis and Management
Next: Hepatitis B: Screening, Prevention, Diagnosis, and Treatment
View the full table of contentschevron_right
Advertisement
More in AFP
Editor's Collections
Dyspepsia
Related Content
Celiac Disease
Constipation
Dyspepsia
More in PubMed
Citation
Related Articles
Most Recent IssueJul 2025
AFP Email Alerts
Free e-newsletter and email table of contents.
SIGN UP NOW
Copyright © 2019 by the American Academy of Family Physicians.
This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. |
7774 | https://www.wyzant.com/resources/answers/688740/minimum-slope-of-a-line-tangent-to-the-curve | Minimum slope of a line tangent to the curve HELP PLEASE | Wyzant Ask An Expert
Log inSign up
Find A Tutor
Search For Tutors
Request A Tutor
Online Tutoring
How It Works
For Students
FAQ
What Customers Say
Resources
Ask An Expert
Search Questions
Ask a Question
Wyzant Blog
Start Tutoring
Apply Now
About Tutors Jobs
Find Tutoring Jobs
How It Works For Tutors
FAQ
About Us
About Us
Careers
Contact Us
All Questions
Search for a Question
Find an Online Tutor Now
Ask a Question for Free
Login
WYZANT TUTORING
Log in
Sign up
Find A Tutor
Search For Tutors
Request A Tutor
Online Tutoring
How It Works
For Students
FAQ
What Customers Say
Resources
Ask An Expert
Search Questions
Ask a Question
Wyzant Blog
Start Tutoring
Apply Now
About Tutors Jobs
Find Tutoring Jobs
How It Works For Tutors
FAQ
About Us
About Us
Careers
Contact Us
Subject
ZIP
Search
SearchFind an Online Tutor NowAsk
Ask a Question For Free
Login
Calculus
Hiuofhgkjem F.
asked • 06/13/19
Minimum slope of a line tangent to the curve HELP PLEASE
Find the minimum slope of the line tangent to the curve y = x 3-3x 2+2x+9
Follow •2
Add comment
More
Report
3 Answers By Expert Tutors
Best Newest Oldest
By:
Jim S.answered • 06/13/19
Tutor
4.8(516)
Physics (and math) are fun, really
About this tutor›
About this tutor›
Hi,
If you plot this function you will see that it has two places where the slope is zero.
We can find those values of x where this is true by setting the derivative (the slope) equal to 0.
y(x)=x 3-3x 2+2x+9
dy/dx=3x 2-6x+2
Now if we set dy/dx=0 we have a quadratic equation to solve
3x 2-6x+2=0 find the roots and you will have found which values of x will locate the minimum slope of y(x)
Let me know if you have ??
Jim
Upvote • 1Downvote
Add comment
More
Report
Mark H.answered • 06/13/19
Tutor
4.9(72)
Tutoring in Math and Science at all levels
See tutors like this
See tutors like this
By definition. the slope is the 1st derivative of a function.
SO:
y = x 3 - 3x 2 + 2x + 9
y' = 3x 2 -6x + 2
Then, the minimum slope is when the 2nd derivative is zero:
y" = 6x - 6 = 0
x = 1
Thus, the minimum slope is 31 2 - 61 + 2 = 3 - 6 + 2 = -1
I am assuming that "minimum" means most negative. To find the minimum absolute value, set the 1st derivative = 0. This gives us x = 1 ± (√3 / 3)
Plot the curves to see how this works---use an online plotting tool such as this:
Upvote • 0Downvote
Add comment
More
Report
Kevin B.answered • 06/13/19
Tutor
4.9(281)
Former Teacher and Math Expert
See tutors like this
See tutors like this
Whenever you hear "slope of the tangent line", think derivative. The derivative of y = x 3-3x 2+2x+9 is
y' = 3x 2 - 6x + 2
The problem asks to find the minimum value of y'.There are several ways to do this. The quickest way is to recognize that f(x) = 3x 2 - 6x + 2 is the equation of a parabola that opens upward. Its minimum is at the vertex (-b/2a, f(-b/2a)) where a = 3 and b = -6:
-b/2a = - (-6)/2(3) = 1
f(-b/2a) = 3(1)2 - 6(1) + 2 = -1
Therefore, the minimum slope of y = x 3-3x 2+2x+9 is y' = -1, which occurs at x = 1.
Upvote • 0Downvote
Comment •1
More
Report
Jim S.
tutor
Kevin, My bad, you solved the correct problem. Nice! Jim
Report
06/13/19
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
¢€£¥‰µ·•§¶ß‹›«»<>≤≥–—¯‾¤¦¨¡¿ˆ˜°−±÷⁄׃∫∑∞√∼≅≈≠≡∈∉∋∏∧∨¬∩∪∂∀∃∅∇∗∝∠´¸ª º†‡À Á Â Ã Ä Å Æ Ç È É Ê Ë Ì Í Î Ï Ð Ñ Ò Ó Ô Õ Ö Ø Œ Š Ù Ú Û Ü Ý Ÿ Þ à á â ã ä å æ ç è é ê ë ì í î ï ð ñ ò ó ô õ ö ø œ š ù ú û ü ý þ ÿ Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ Σ Τ Υ Φ Χ Ψ Ω α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω ℵ ϖ ℜ ϒ℘ℑ←↑→↓↔↵⇐⇑⇒⇓⇔∴⊂⊃⊄⊆⊇⊕⊗⊥⋅⌈⌉⌊⌋〈〉◊
RELATED TOPICS
MathAlgebra 1PhysicsPrecalculusTrigonometryAlgebraPre CalculusLimitsFunctionsMath Help...DerivativeAp CalcAp CalculusIntegral CalculusCalcIntegrationDerivativesCalculus 3Calculus 2Calculus 1
RELATED QUESTIONS
##### CAN I SUBMIT A MATH EQUATION I'M HAVING PROBLEMS WITH?
Answers · 3
##### If i have rational function and it has a numerator that can be factored and the denominator is already factored out would I simplify by factoring the numerator?
Answers · 7
##### how do i find where a function is discontinuous if the bottom part of the function has been factored out?
Answers · 3
##### find the limit as it approaches -3 in the equation (6x+9)/x^4+6x^3+9x^2
Answers · 8
##### prove addition form for coshx
Answers · 4
RECOMMENDED TUTORS
Kubrat D. 5.0(1,104)
Harry O. 4.9(833)
Zachary C. 5.0(362)
See more tutors
find an online tutor
Calculus tutors
Multivariable Calculus tutors
Business Calculus tutors
AP Calculus tutors
Differential Equations tutors
Precalculus tutors
AP Calculus BC tutors
College Algebra tutors
Download our free app
A link to the app was sent to your phone.
Please provide a valid phone number.
App StoreGoogle Play
##### Get to know us
About Us
Contact Us
FAQ
Reviews
Safety
Security
In the News
##### Learn with us
Find a Tutor
Request a Tutor
Online Tutoring
Learning Resources
Blog
Tell Us What You Think
##### Work with us
Careers at Wyzant
Apply to Tutor
Tutor Job Board
Affiliates
Download our free app
App StoreGoogle Play
Let’s keep in touch
Need more help?
Learn more about how it works
##### Tutors by Subject
Algebra Tutors
Calculus Tutors
Chemistry Tutors
Computer Tutors
Elementary Tutors
English Tutors
Geometry Tutors
Language Tutors
Math Tutors
Music Lessons
Physics Tutors
Reading Tutors
SAT Tutors
Science Tutors
Spanish Tutors
Statistics Tutors
Test Prep Tutors
Writing Tutors
##### Tutors by Location
Atlanta Tutors
Boston Tutors
Brooklyn Tutors
Chicago Tutors
Dallas Tutors
Denver Tutors
Detroit Tutors
Houston Tutors
Los Angeles Tutors
Miami Tutors
New York City Tutors
Orange County Tutors
Philadelphia Tutors
Phoenix Tutors
San Francisco Tutors
Seattle Tutors
San Diego Tutors
Washington, DC Tutors
Making educational experiences better for everyone.
##### IXL Comprehensive K-12 personalized learning
##### Rosetta Stone Immersive learning for 25 languages
##### Education.com 35,000 worksheets, games, and lesson plans
##### TPT Marketplace for millions of educator-created resources
##### Vocabulary.com Adaptive learning for English vocabulary
##### ABCya Fun educational games for kids
##### SpanishDictionary.com Spanish-English dictionary, translator, and learning
##### Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje
##### Emmersion Fast and accurate language certification
SitemapTerms of UsePrivacy Policy
© 2005 - 2025 Wyzant, Inc, a division of IXL Learning - All Rights Reserved
Privacy Preference Center
Your Privacy
Strictly Necessary Cookies
Performance Cookies
Functional Cookies
Targeting Cookies
Your Privacy
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
[x] Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Clear
[x] checkbox label label
Apply Cancel
Confirm My Choices
Allow All |
7775 | https://math.stackexchange.com/questions/1001064/avoiding-the-integration-constant | calculus - Avoiding the Integration Constant - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Avoiding the Integration Constant
Ask Question
Asked 10 years, 11 months ago
Modified10 years, 10 months ago
Viewed 997 times
This question shows research effort; it is useful and clear
12
Save this question.
Show activity on this post.
I sometimes find writing and keeping track of the constants of integration a somewhat messy job. Yes, sometimes it's necessary but in many situations that I come across in my level of mathematics, it is a waste of time and space.
An exagerated example:
f′′(x)=g′′(x)⟹∫ing f′(x)+C 1=g′(x)+C 2⟹∫ing f(x)+C 3+C 1 x+C 4=g(x)+C 5+C 2 x+C 6⟹f(x)=g(x)+C a x+C b f″(x)=g″(x)⟹∫ing f′(x)+C 1=g′(x)+C 2⟹∫ing f(x)+C 3+C 1 x+C 4=g(x)+C 5+C 2 x+C 6⟹f(x)=g(x)+C a x+C b
where C a=C 2−C 1 C a=C 2−C 1 and C b=C 6+C 5−C 4−C 3 C b=C 6+C 5−C 4−C 3
This just seems like a ridiculous amount of tracking in certain cases and I often just combine as many constants in one shot without mercy in most problems I face and never explain the sources of the constants (because it doesn't really matter)
×× Some people I know avoid distinguishing the constants by just marking them all as C C and only distinguish between coefficients and constants as I've illustrated in the last step of my example.
×× Some other students leave out the constants all together and write a " + C " only in last step but they don't realize that they are often neglecting terms where the constants turn into coefficients.
⋆⋆ It isn't a mystery as to why the constant tickles people's lazy bones. The lethargic attitude almost everyone shows toward it is because that " + C " is just annoying. Sure, it may seem necessary but what about in simplifications?
∫f′(x)d x=f(x)+C=f 1(x)+C=f 2(x)+C=f 3(x)+C=…=f n(x)+C∫f′(x)d x=f(x)+C=f 1(x)+C=f 2(x)+C=f 3(x)+C=…=f n(x)+C
where f k∈N(x)f k∈N(x) is a simplified form of f(x)f(x)
It's usage is monotonous and seems absolutely unnecessary during simplification.
My question is hence this:
How can one safely hide the constant of integration during simplification of equations and what are the restrictions in such hiding?
Thank you in advance.
Edit: I am seeking a notation which would validly allow for a neglection of the constant during simplification of differential equations. Imagine 100 steps of simplification. The +C would be annoying. Apologies if this point was not clear through my rant.
calculus
notation
indefinite-integrals
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Nov 10, 2014 at 14:02
NickNick
asked Nov 1, 2014 at 10:53
NickNick
7,054 10 10 gold badges 46 46 silver badges 70 70 bronze badges
6
2 Can't you use one constant?user171358 –user171358 2014-11-01 11:01:13 +00:00 Commented Nov 1, 2014 at 11:01
@DigitalBrain: ...I often just combine as many constants in one shot without mercy... Yes, that's what I do. But that one constant bothers me during simplification.Nick –Nick 2014-11-01 11:02:38 +00:00 Commented Nov 1, 2014 at 11:02
When you integrate both sides of an equation, you need one new constant of integration. Period. You have written 4 in a single step.Mike –Mike 2014-11-01 14:11:46 +00:00 Commented Nov 1, 2014 at 14:11
@Mike: My peers keep bugging me for a 'real way' to exclude writing the annoying " +C+C " during simplification of differential equations. Is there no way to write ∫f′(x)d x≡f(x)∫f′(x)d x≡f(x) ?Nick –Nick 2014-11-01 14:16:42 +00:00 Commented Nov 1, 2014 at 14:16
1 To be clear: When you're talking about application to diff-eqs, do you mean in the context of finding the general solution, or to solving initial value problems specifically?Semiclassical –Semiclassical 2014-11-06 02:14:05 +00:00 Commented Nov 6, 2014 at 2:14
|Show 1 more comment
8 Answers 8
Sorted by: Reset to default
This answer is useful
8
Save this answer.
+150
This answer has been awarded bounties worth 150 reputation by Nick
Show activity on this post.
I don't think that there is a standard way of doing it but of course, you can invent your on. In this case, an equivalence relation or rather a sequence of equivalence relations:
For functions f,g f,g, let f≡d g f≡d g if there exists a polynomial p p of degree at most d d, such that f=g+p f=g+p. In this case, you want deg 0=−1 deg0=−1.
We have f≡−1 g f≡−1 g iff f=g f=g and integrating raises the degree d d by one.
Your example becomes
∫f′(x)d x≡0 f(x)=f 1(x)=⋯=f n(x)∫f′(x)dx≡0 f(x)=f 1(x)=⋯=f n(x)
and in a second step
∫f′(x)d x=f n(x)+C.∫f′(x)dx=f n(x)+C.
Edit: In this answer, I defined the symbol ≡≡ (if you don't like it, you can replace it by something else, say ∼∼). Another way to phrase the definition is the following: For functions f,g f,g and d∈{−1,0,1,2,…}d∈{−1,0,1,2,…}, let f≡d g f≡d g if the difference f−g f−g is a polynomial of degree at most d d. For every d d, ≡d≡d is an equivalence relation – the proof is easy.
Slightly abusing notation (I will identify a function f f with its term f(x)f(x) for brevity), we have for example
f(x)≡−1 g(x)f(x)≡−1 g(x) if and only if f=g f=g, because the only degree-(−1)(−1) polynomial is the zero polynomial, thus f−g=0 f−g=0, i.e. f=g f=g.
sin x≡0 sin x+15 sinx≡0 sinx+15, because 15 15 is a degree-0 0 polynomial
f(x)≡d f(x)+c d x d+c d−1 x d−1+⋯+c 0 f(x)≡d f(x)+c d x d+c d−1 x d−1+⋯+c 0, because c d x d+c d−1 x d−1+⋯+c 0 c d x d+c d−1 x d−1+⋯+c 0 is a degree-d d polynomial
Now, whenever f′′(x)=g′′(x)f″(x)=g″(x), we know that f′(x)+C=g′(x)+D f′(x)+C=g′(x)+D with constants C,D∈R C,D∈R. Thus, f′(x)−g′(x)=D−C f′(x)−g′(x)=D−C which is constant, i.e. a degree-0 0 polynomial. Using the notation defined above, we can write f≡0 g f≡0 g, eliminating the explicit constant.
This also works for higher degrees. Say we know f′(x)≡d g′(x)f′(x)≡d g′(x). Then there exists a degree-d d polynomial p p or more explicitly real numbers c 0,…,c d c 0,…,c d, such that f′(x)−g′(x)=p(x)=c d x d+⋯+c 0 f′(x)−g′(x)=p(x)=c d x d+⋯+c 0. Integration:
f(x)−g(x)=c d d+1 x d+1+⋯+c 0 x+C f(x)−g(x)=c d d+1 x d+1+⋯+c 0 x+C
which is a degree-(d+1)(d+1) polynomial. Thus f≡d+1 g f≡d+1 g.
The nice thing about an equivalence relation is that it behaves in many ways like equality. For example, you can use it in an equality chain like
f 0(x)=f 1(x)≡d f 2(x)=f 3(x)f 0(x)=f 1(x)≡d f 2(x)=f 3(x)
if you keep in mind that you can only deduce f 0(x)≡d f 3(x)f 0(x)≡d f 3(x) from that and not f 0(x)=f 3(0)f 0(x)=f 3(0).
Also note that this notation is not standard. If you want to use it and others to understand it, you will have to give the definition.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Nov 3, 2014 at 15:28
answered Nov 3, 2014 at 6:13
Eike SchulteEike Schulte
3,497 1 1 gold badge 17 17 silver badges 21 21 bronze badges
3
I love this answer but I am not able to understand what f≡d g f≡d g means exactly and I would also like to know where you have borrowed this notation from, if you have?Nick –Nick 2014-11-03 11:44:37 +00:00 Commented Nov 3, 2014 at 11:44
2 Defining an equivalence relation is the natural thing to do in situations like this. Probably somebody did this before but I don't have a source for this. I will also try do add some detail to the answer.Eike Schulte –Eike Schulte 2014-11-03 14:56:46 +00:00 Commented Nov 3, 2014 at 14:56
Congrats, you've won the bounty for your splendid style but I've found something more standard that serves my needs better (see my answer) :D Nick –Nick 2014-11-10 13:00:41 +00:00 Commented Nov 10, 2014 at 13:00
Add a comment|
This answer is useful
4
Save this answer.
Show activity on this post.
For the easy situation of one integration, you can write
∫f(x)d x∋F(x)=F 1(x)=…=F n(x)hence∫f(x)d x=F n(x)+C∫f(x)d x∋F(x)=F 1(x)=…=F n(x)hence∫f(x)d x=F n(x)+C
It's because you can say that primitive operator function really gives you the set of all primitive functions, all of which differs by a constant. Then, to be consistent, you can say that the second equality is a set equality and C C is a set of all constants.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 4, 2014 at 10:58
Adam BartošAdam Bartoš
9,423 1 1 gold badge 20 20 silver badges 35 35 bronze badges
1
exactly the type of answer i'm looking for :D Wonderful.Nick –Nick 2014-11-04 14:45:27 +00:00 Commented Nov 4, 2014 at 14:45
Add a comment|
This answer is useful
3
Save this answer.
Show activity on this post.
You could reduce the number of your constants by half by using only one constant each time you integrate. For example, when solving
r′(x)=s′(x)r′(x)=s′(x)
write the answer as
r(x)=s(x)+C 1 r(x)=s(x)+C 1
rather than (as you were doing)
r(x)+C 1=s(x)+C 2 r(x)+C 1=s(x)+C 2
Using only one constant when integrating both sides of an equation is perfectly rigorous and the proof is obvious.
ADDED:
1) Another possibility is to use constants like c,d,e,…c,d,e,… (or k,l,m,…k,l,m,… or p,q,r,…p,q,r,… or α,β,γ,…α,β,γ,…) rather than C 1,C 2,C 3,…C 1,C 2,C 3,…. This does not reduce the number of constants but it does reduce the tiresome subscripts and make it look like there are fewer constants.
2) When you wrote "I often just combine as many constants in one shot without mercy..." I thought you meant things like changing ±C 1 e x+C 2±C 1 e x+C 2 to C 1 e x C 1 e x, which is less rigorous. Using only one constant per integration is a no-brainer.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Nov 1, 2014 at 11:16
answered Nov 1, 2014 at 11:01
Rory DaultonRory Daulton
32.9k 6 6 gold badges 48 48 silver badges 65 65 bronze badges
3
Yes, that was exactly what I was doing: ...I often just combine as many constants in one shot without mercy.... My true problem is the boring repetition of the constant during simplification of equations Nick –Nick 2014-11-01 11:05:12 +00:00 Commented Nov 1, 2014 at 11:05
2 @Nick, not everything is fun in life. If you want to be correct, then just show the minimum number of constants that are necessary.k170 –k170 2014-11-01 11:11:39 +00:00 Commented Nov 1, 2014 at 11:11
@k170: Imagine 100 steps of simplification, the +C will drive you mad. That's my real problem.Nick –Nick 2014-11-01 12:37:07 +00:00 Commented Nov 1, 2014 at 12:37
Add a comment|
This answer is useful
3
Save this answer.
Show activity on this post.
Here is one way around your problem:
∫f′(x)d x=f(x)+C∫f′(x)d x=f(x)+C for some C C. So
∫f′(x)d x−C=f(x)=f 1(x)=f 2(x)=f 3(x)=…=f n(x)∫f′(x)d x−C=f(x)=f 1(x)=f 2(x)=f 3(x)=…=f n(x)
Therefore ∫f′(x)d x=f n(x)+C∫f′(x)d x=f n(x)+C for some C C.
I don't think this is exactly what you were looking for, but then I suspect that what you are looking for doesn't exist.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 3, 2014 at 6:23
TonyKTonyK
68.3k 5 5 gold badges 95 95 silver badges 189 189 bronze badges
1
lol, good solution. But a notation can be 'invented' as @Eike suggests. Don't you think? Also, imagine simplifying both sides of a differential equation simultaneously, there isn't an escape from the +Cs. (PS: Should I tag this as a soft question?)Nick –Nick 2014-11-03 06:30:21 +00:00 Commented Nov 3, 2014 at 6:30
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
In certain circumstances, Landau notation (aka Big-O notation) can be used for this. So, for example, your first example would look something like:
f′′(x)=g′′(x)⟹∫ing f′(x)+O(1)=g′(x)+O(1)⟹∫ing f(x)+O(x)=g(x)+O(x)⟹f(x)=g(x)+O(x)f″(x)=g″(x)⟹∫ing f′(x)+O(1)=g′(x)+O(1)⟹∫ing f(x)+O(x)=g(x)+O(x)⟹f(x)=g(x)+O(x)
and your second example would look like (maybe not such a great improvement):
∫f′(x)d x=f(x)+O(1)=f 1(x)+O(1)=f 2(x)+O(1)=f 3(x)+O(1)=…=f n(x)+O(1)∫f′(x)d x=f(x)+O(1)=f 1(x)+O(1)=f 2(x)+O(1)=f 3(x)+O(1)=…=f n(x)+O(1)
The caveat would be that this should probably only be used when you don't care too much about what is elided by the Big O (e.g.: x+1 l o g(x)=O(x)x+1 l o g(x)=O(x)). And, yes, there can be some slipperiness with the use of the equals sign with Big O notation in these contexts (see the Wikipedia section on this) and if you're not careful you might end up tripping yourself up. But, I think if your uses are mainly like your first example, I think some sparing and/or careful use of Big O's might help.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 6, 2014 at 2:58
mhummhum
2,577 16 16 silver badges 13 13 bronze badges
1
Although this may not have exactly been what I was looking for, it is impressively relevant and useful. Thank you for this answer :D Nick –Nick 2014-11-06 11:08:41 +00:00 Commented Nov 6, 2014 at 11:08
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
My solution to this problem (nuisance rather) that I actually make extensive use of is very similar to @EikeShulte's answer.
For an ordinary differential equation, suppose you have the following:
∫f′(x)d x=f(x)+C∫f′(x)d x=f(x)+C
This can be written in 'Don Juan' notation as
∫f′(x)d x=0 f(x)∫f′(x)d x=0 f(x)
Which translates as "f(x) has a polynomial of degree 0 after it with unknown coefficients". If you were to integrate that 30 more times and obtain a function g(x)g(x) with, as you may have guessed, a polynomial of degree 30 30 after with unknown coefficients, you could write it very simply with 'Don Juan' notation as
30 g(x)30 g(x)
And for what other purposes does anybody really use that pre-superscript? I don't know of any, so I think this notation solves the problem pretty nicely.
But what if you know what f(x)f(x) or g(x)g(x) actually are with the exception of the polynomial of unknown coefficients? No problem! Suppose g(x)g(x), the same one mentioned above, has sin(3 x)+3 x 2−e x sin(3 x)+3 x 2−e x before the polynomial. Then it can be written in 'Don Juan' notation as
30(sin(3 x)+3 x 2−e x)30(sin(3 x)+3 x 2−e x)
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 7, 2014 at 0:55
Arturo don JuanArturo don Juan
4,118 2 2 gold badges 34 34 silver badges 52 52 bronze badges
4
Dear Don Giovanni, your notation is as profound as your history.Nick –Nick 2014-11-07 10:19:16 +00:00 Commented Nov 7, 2014 at 10:19
I really can't tell whether you're being sarcastic or not.Arturo don Juan –Arturo don Juan 2014-11-07 14:14:26 +00:00 Commented Nov 7, 2014 at 14:14
T'was not intended to be sarcasm whence it was written. Btw, my remark was on the basis of the interpretation of Don Juan as a story of salvation. Yes, it was profound and creative.Nick –Nick 2014-11-07 14:28:05 +00:00 Commented Nov 7, 2014 at 14:28
Ah, I see. Don Giovanni has a great history. However, I actually got 'Don Juan' from a different set literature - The Teachings of Don Juan. en.wikipedia.org/wiki/The_Teachings_of_Don_JuanArturo don Juan –Arturo don Juan 2014-11-07 14:41:05 +00:00 Commented Nov 7, 2014 at 14:41
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
Because a sum of constants can be any combination of constants, then we can substitute ∑i=0∞C i=C 1∑i=0∞C i=C 1
Similarily, x(∑i=0∞C i)=C 2 x(∑i=0∞C i)=C 2.
We have then, ∫∫∫∫∫∫∫d x∫∫∫∫∫∫∫d x having 6 constants, but that is unavoidable. We can forcefully combine constants with the same factor, because a constant+constant2 is just a constant.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 6, 2014 at 1:41
TeocTeoc
8,913 4 4 gold badges 26 26 silver badges 67 67 bronze badges
1
1 You and @Rory seem to be under the impression that I am not able to digest the fact that the sum of constants is constant. this is not true. I do not blame you for this misinterpretation because a large section of my post is just a rant irellevant to the point which all other answers have addressed. I added those extra lines written in small font for the purpose of dramatically projecting my irritation with the constant. My sincerest Apologies.Nick –Nick 2014-11-06 11:14:44 +00:00 Commented Nov 6, 2014 at 11:14
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
Use modular arithmetic:
∫f′(x)d x≡f(x)mod R∫f′(x)d x≡f(x)mod R
But be warned, it has limitations during manipulation of the equation.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered Nov 10, 2014 at 12:59
NickNick
7,054 10 10 gold badges 46 46 silver badges 70 70 bronze badges
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
calculus
notation
indefinite-integrals
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
2Indefinite Integrals - Constant of Integration Question
2What does it mean mathematically to set some of the integration constants in the general solution to a linear differential equation, equal to zero?
0Partial integration (not integration by parts), how is the following result obtained?
0How to find velocity v→(t)v→(t) and position r→(t)r→(t) from acceleration a→(t)a→(t)
0Optimizing a function of two variables under constraint
5Freaky Polynomial: P n(x)=(x d d x)n f(x)P n(x)=(x d d x)n f(x)
0How do I solve this kind of 3rd order differential equation?
2Order of Differential Equation using number of arbitrary constants
Hot Network Questions
Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice?
Direct train from Rotterdam to Lille Europe
How to home-make rubber feet stoppers for table legs?
How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender?
Storing a session token in localstorage
Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-?
Lingering odor presumably from bad chicken
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
в ответе meaning in context
What were "milk bars" in 1920s Japan?
Discussing strategy reduces winning chances of everyone!
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
In the U.S., can patients receive treatment at a hospital without being logged?
how do I remove a item from the applications menu
How can the problem of a warlock with two spell slots be solved?
Alternatives to Test-Driven Grading in an LLM world
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
Cannot build the font table of Miama via nfssfont.tex
Why include unadjusted estimates in a study when reporting adjusted estimates?
Can a cleric gain the intended benefit from the Extra Spell feat?
Program that allocates time to tasks based on priority
Repetition is the mother of learning
Proof of every Highly Abundant Number greater than 3 is Even
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7776 | https://pubmed.ncbi.nlm.nih.gov/25905367/ | Evaluation of Amenorrhea, Anovulation, and Abnormal Bleeding - PubMed
Clipboard, Search History, and several other advanced features are temporarily unavailable.
Skip to main page content
An official website of the United States government
Here's how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Search: Search
AdvancedClipboard
User Guide
Save Email
Send to
Clipboard
My Bibliography
Collections
Citation manager
Display options
Display options
Format
Save citation to file
Format:
Create file Cancel
Email citation
Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page.
To:
Subject:
Body:
Format:
[x] MeSH and other data
Send email Cancel
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Add to My Bibliography
My Bibliography
Unable to load your delegates due to an error
Please try again
Add Cancel
Your saved search
Name of saved search:
Search terms:
Test search terms
Would you like email updates of new search results? Saved Search Alert Radio Buttons
Yes
No
Email: (change)
Frequency:
Which day?
Which day?
Report format:
Send at most:
[x] Send even when there aren't any new results
Optional text in email:
Save Cancel
Create a file for external citation management software
Create file Cancel
Your RSS Feed
Name of RSS Feed:
Number of items displayed:
Create RSS Cancel
RSS Link Copy
Actions
Cite
Collections
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Permalink
Permalink
Copy
Display options
Display options
Format
Page navigation
Title & authors
Excerpt
Sections
Similar articles
Publication types
Related information
Review
Evaluation of Amenorrhea, Anovulation, and Abnormal Bleeding
Robert Rebar1
Kenneth R Feingold1,S. Faisal Ahmed2,Bradley Anawalt3,Marc R Blackman4,Alison Boyce5,George Chrousos6,Emiliano Corpas7,Wouter W de Herder8,Ketan Dhatariya9,Kathleen Dungan10,Johannes Hofland11,Sanjay Kalra12,Gregory Kaltsas13,Nitin Kapoor14,Christian Koch15,Peter Kopp16,Márta Korbonits17,Christopher S Kovacs18,Wendy Kuohung19,Blandine Laferrère20,Miles Levy21,Elizabeth A McGee22,Robert McLachlan23,Radhika Muzumdar24,Jonathan Purnell25,Rodolfo Rey26,Rakesh Sahay27,Amy S Shah28,Frederick Singer29,Mark A Sperling30,Constantine A Stratakis31,Dace L Trence32,Don P Wilson33
, editors.
In: Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000.
2018 Jan 15.
Affiliations Expand
Affiliation
1 Professor and Chair, Department of Obstetrics and Gynecology, Western Michigan University Homer Stryker M.D. School of Medicine, Kalamazoo, MI
Book Affiliations
1 Professor of Medicine Emeritus, University of California, San Francisco, CA
2 Developmental Endocrinology Research Group, University of Glasgow, Glasgow, UK
3 Chief of Medicine at the University of Washington Medical Center and Professor and Vice Chair of the Department of Medicine, University of Washington
4 Sr. Physician Scientist, Washington DC VA Medical Center; Professor of Medicine & Rehabilitation Medicine, Georgetown University; Clinical Professor of Medicine, Biochemistry and Molecular Medicine, George Washington University; and Professor of Medicine (Part-time), Johns Hopkins University
5 Pediatric Endocrinologist and Associate Research Physician in the Skeletal Diseases and Mineral Homeostasis Section, National Institute of Dental and Craniofacial Research, National Institutes of Health
6 Professor of Pediatrics and Endocrinology, Division of Endocrinology, Metabolism and Diabetes, First Department of Pediatrics, National and Kapodistrian University of Athens Medical School, "Aghia Sophia" Children's Hospital, Athens, Greece
7 M.D. Ph.D in Gerontology. Honorary Professor of Medicine, Universidad de Alcalá, Madrid. Consultant in Endocrinology, Hospital HLA Guadalajara (Spain).
8 Professor of Endocrine Oncology, Erasmus MC and Erasmus MC Cancer Center, Rotterdam, the Netherlands
9 Consultant in Diabetes, Endocrinology and General Medicine, Norfolk and Norwich University Hospitals NHS Foundation Trust and University of East Anglia, Norwich, UK.
10 Professor of Medicine, Division of Endocrinology, Diabetes, and Metabolism, Ohio State University
11 Consultant Endocrinologist, Erasmus MC and Erasmus MC Cancer Center, Rotterdam, the Netherlands
12 Consultant Endocrinologist, Department of Endocrinology, Bharti Hospital, Karnal, India
13 Professor of General Medicine-Endocrinology, 1st Department of Propaedeutic Medicine, National and Kapodistrian University of Athens, Athens, Greece
14 Professor of Endocrinology, Department of Endocrinology, Diabetes and Metabolism, Christian Medical College & Hospital, Vellore, Tamil Nadu, India, Melbourne School of Population and Global Health, Faculty of Medicine, Dentistry and Health Science, The University of Melbourne, Australia.
15 Professor, The University of Tennessee Health Science Center, Memphis, Tennessee
16 Professor of Medicine and Chief of the Division of Endocrinology, Diabetology and Metabolism, University of Lausanne, Switzerland
17 Professor of Endocrinology and Metabolism, Centre Lead for Endocrinology and Deputy Institute Director, William Harvey Research Institute, Barts and the London School of Medicine and Dentistry, Queen Mary University of London, London, England
18 University Research Professor and Professor of Medicine (Endocrinology and Metabolism), Obstetrics & Gynecology, and BioMedical Sciences, at Memorial University of Newfoundland in St. John’s, Newfoundland, Canada.
19 Director of the Division of Reproductive Endocrinology at Boston Medical Center and an Associate Professor of Obstetrics and Gynecology at the Boston University School of Medicine
20 Professor of Medicine, New York Nutrition Obesity Research Center, Division of Endocrinology, Department of Medicine, Columbia University Irving Medical Center, New York, NY, USA.
21 Consultant endocrinologist at University Hospitals of Leicester and Honorary Associate Professor at Leicester University
22 Professor of Obstetrics and Gynecology at the University of Vermont and Director of the Division of Reproductive Endocrinology and Infertility. Burlington, Vermont
23 Director of Clinical Research, Hudson Institute of Medical Research; Consultant Endocrinologist, Monash Medical Centre, Melbourne, Australia
24 Allan L. Drash Endowed Chair, Chief, Division of Endocrinology and Diabetes, UPMC Children’s Hospital of Pittsburgh, Professor of Pediatrics and Cell Biology, University of Pittsburgh School of Medicine, Pittsburgh, PA
25 Professor of Medicine, Knight Cardiovascular Institute and the Division of Endocrinology, and Associate Director, Bob and Charlee Moore Institute for Nutrition and Wellness, Oregon Health and Science University, Portland, OR
26 Centro de Investigaciones Endocrinológicas “Dr. César Bergadá” (CEDIE), CONICET-FEI-División de Endocrinología, Hospital de Niños Ricardo Gutiérrez, Gallo 1330, C1425EFD Buenos Aires; and Departamento de Histología, Biología Celular, Embriología y Genética, Facultad de Medicina, Universidad de Buenos Aires, C1121ABG Buenos Aires, Argentina
27 Professor and Head of Department of Endocrinology, Osmania Medical College and Osmania General Hospital, Hyderabad, India.
28 Professor of Pediatrics, The University of Cincinnati, Department of Pediatrics and Cincinnati Children’s Hospital Medical Center, Division of Endocrinology, Cincinnati, OH, USA
29 Director of the Endocrine/Bone Disease Program, Saint Johns Cancer Institute at Saint John’s Health Center, Santa Monica, CA; Clinical Professor of Medicine, UCLA School of Medicine, Los Angeles, CA
30 Professorial Lecturer, Division of Pediatric Endocrinology and Diabetes, Icahn School of Medicine at Mount Sinai, New York, NY. Emeritus Professor and Chair, Department of Pediatrics, University of Pittsburgh.
31 CSO, ELPEN, Inc. & Director, Research Institute, Athens, Greece & Senior Investigator, Human Genetics & Precision Medicine, FORTH (ITE), Heraklion, Greece. Emeritus Scientific Director & Senior Investigator, NICHD, NIH, Bethesda, MD, USA
32 Professor of Medicine, Emeritus, University of Washington, Seattle, WA
33 Endowed Chair, Cardiovascular Health and Risk Prevention, Pediatric Endocrinology and Diabetes, Cook Children's Medical Center, Fort Worth, TX
PMID: 25905367
Bookshelf ID: NBK279144
Item in Clipboard
Review
Evaluation of Amenorrhea, Anovulation, and Abnormal Bleeding
Robert Rebar.
Show details
Display options
Display options
Format
In: Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000.
2018 Jan 15.
Author
Robert Rebar1
Book Editors
Kenneth R Feingold1,S. Faisal Ahmed2,Bradley Anawalt3,Marc R Blackman4,Alison Boyce5,George Chrousos6,Emiliano Corpas7,Wouter W de Herder8,Ketan Dhatariya9,Kathleen Dungan10,Johannes Hofland11,Sanjay Kalra12,Gregory Kaltsas13,Nitin Kapoor14,Christian Koch15,Peter Kopp16,Márta Korbonits17,Christopher S Kovacs18,Wendy Kuohung19,Blandine Laferrère20,Miles Levy21,Elizabeth A McGee22,Robert McLachlan23,Radhika Muzumdar24,Jonathan Purnell25,Rodolfo Rey26,Rakesh Sahay27,Amy S Shah28,Frederick Singer29,Mark A Sperling30,Constantine A Stratakis31,Dace L Trence32,Don P Wilson33
Affiliation
1 Professor and Chair, Department of Obstetrics and Gynecology, Western Michigan University Homer Stryker M.D. School of Medicine, Kalamazoo, MI
Book Affiliations
1 Professor of Medicine Emeritus, University of California, San Francisco, CA
2 Developmental Endocrinology Research Group, University of Glasgow, Glasgow, UK
3 Chief of Medicine at the University of Washington Medical Center and Professor and Vice Chair of the Department of Medicine, University of Washington
4 Sr. Physician Scientist, Washington DC VA Medical Center; Professor of Medicine & Rehabilitation Medicine, Georgetown University; Clinical Professor of Medicine, Biochemistry and Molecular Medicine, George Washington University; and Professor of Medicine (Part-time), Johns Hopkins University
5 Pediatric Endocrinologist and Associate Research Physician in the Skeletal Diseases and Mineral Homeostasis Section, National Institute of Dental and Craniofacial Research, National Institutes of Health
6 Professor of Pediatrics and Endocrinology, Division of Endocrinology, Metabolism and Diabetes, First Department of Pediatrics, National and Kapodistrian University of Athens Medical School, "Aghia Sophia" Children's Hospital, Athens, Greece
7 M.D. Ph.D in Gerontology. Honorary Professor of Medicine, Universidad de Alcalá, Madrid. Consultant in Endocrinology, Hospital HLA Guadalajara (Spain).
8 Professor of Endocrine Oncology, Erasmus MC and Erasmus MC Cancer Center, Rotterdam, the Netherlands
9 Consultant in Diabetes, Endocrinology and General Medicine, Norfolk and Norwich University Hospitals NHS Foundation Trust and University of East Anglia, Norwich, UK.
10 Professor of Medicine, Division of Endocrinology, Diabetes, and Metabolism, Ohio State University
11 Consultant Endocrinologist, Erasmus MC and Erasmus MC Cancer Center, Rotterdam, the Netherlands
12 Consultant Endocrinologist, Department of Endocrinology, Bharti Hospital, Karnal, India
13 Professor of General Medicine-Endocrinology, 1st Department of Propaedeutic Medicine, National and Kapodistrian University of Athens, Athens, Greece
14 Professor of Endocrinology, Department of Endocrinology, Diabetes and Metabolism, Christian Medical College & Hospital, Vellore, Tamil Nadu, India, Melbourne School of Population and Global Health, Faculty of Medicine, Dentistry and Health Science, The University of Melbourne, Australia.
15 Professor, The University of Tennessee Health Science Center, Memphis, Tennessee
16 Professor of Medicine and Chief of the Division of Endocrinology, Diabetology and Metabolism, University of Lausanne, Switzerland
17 Professor of Endocrinology and Metabolism, Centre Lead for Endocrinology and Deputy Institute Director, William Harvey Research Institute, Barts and the London School of Medicine and Dentistry, Queen Mary University of London, London, England
18 University Research Professor and Professor of Medicine (Endocrinology and Metabolism), Obstetrics & Gynecology, and BioMedical Sciences, at Memorial University of Newfoundland in St. John’s, Newfoundland, Canada.
19 Director of the Division of Reproductive Endocrinology at Boston Medical Center and an Associate Professor of Obstetrics and Gynecology at the Boston University School of Medicine
20 Professor of Medicine, New York Nutrition Obesity Research Center, Division of Endocrinology, Department of Medicine, Columbia University Irving Medical Center, New York, NY, USA.
21 Consultant endocrinologist at University Hospitals of Leicester and Honorary Associate Professor at Leicester University
22 Professor of Obstetrics and Gynecology at the University of Vermont and Director of the Division of Reproductive Endocrinology and Infertility. Burlington, Vermont
23 Director of Clinical Research, Hudson Institute of Medical Research; Consultant Endocrinologist, Monash Medical Centre, Melbourne, Australia
24 Allan L. Drash Endowed Chair, Chief, Division of Endocrinology and Diabetes, UPMC Children’s Hospital of Pittsburgh, Professor of Pediatrics and Cell Biology, University of Pittsburgh School of Medicine, Pittsburgh, PA
25 Professor of Medicine, Knight Cardiovascular Institute and the Division of Endocrinology, and Associate Director, Bob and Charlee Moore Institute for Nutrition and Wellness, Oregon Health and Science University, Portland, OR
26 Centro de Investigaciones Endocrinológicas “Dr. César Bergadá” (CEDIE), CONICET-FEI-División de Endocrinología, Hospital de Niños Ricardo Gutiérrez, Gallo 1330, C1425EFD Buenos Aires; and Departamento de Histología, Biología Celular, Embriología y Genética, Facultad de Medicina, Universidad de Buenos Aires, C1121ABG Buenos Aires, Argentina
27 Professor and Head of Department of Endocrinology, Osmania Medical College and Osmania General Hospital, Hyderabad, India.
28 Professor of Pediatrics, The University of Cincinnati, Department of Pediatrics and Cincinnati Children’s Hospital Medical Center, Division of Endocrinology, Cincinnati, OH, USA
29 Director of the Endocrine/Bone Disease Program, Saint Johns Cancer Institute at Saint John’s Health Center, Santa Monica, CA; Clinical Professor of Medicine, UCLA School of Medicine, Los Angeles, CA
30 Professorial Lecturer, Division of Pediatric Endocrinology and Diabetes, Icahn School of Medicine at Mount Sinai, New York, NY. Emeritus Professor and Chair, Department of Pediatrics, University of Pittsburgh.
31 CSO, ELPEN, Inc. & Director, Research Institute, Athens, Greece & Senior Investigator, Human Genetics & Precision Medicine, FORTH (ITE), Heraklion, Greece. Emeritus Scientific Director & Senior Investigator, NICHD, NIH, Bethesda, MD, USA
32 Professor of Medicine, Emeritus, University of Washington, Seattle, WA
33 Endowed Chair, Cardiovascular Health and Risk Prevention, Pediatric Endocrinology and Diabetes, Cook Children's Medical Center, Fort Worth, TX
PMID: 25905367
Bookshelf ID: NBK279144
Item in Clipboard
Cite
Display options
Display options
Format
Excerpt
Amenorrhea not due to pregnancy, lactation, or menopause is a relatively common abnormality of the reproductive years and indicative of a defect somewhere in the hypothalamic-pituitary-ovarian-uterine axis. This chapter considers the various causes of amenorrhea and their treatment. It also considers the diagnosis and treatment of abnormal uterine bleeding at all stages of life.
Copyright © 2000-2025, MDText.com, Inc.
PubMed Disclaimer
Sections
ABSTRACT
AMENORRHEA
HYPERGONADOTROPIC AMENORRHEA (Primary Hypogonadism; Gonadal Failure; Primary Ovarian Insufficiency)
TYPES OF PREMATURE OVARIAN FAILURE
DIAGNOSIS AND THERAPY OF PREMATURE OVARIAN FAILURE
CHRONIC ANOVULATION
CHRONIC ANOVULATION OF CENTRAL ORIGIN
ANOREXIA NERVOSA, BULIMIA NERVOSA AND ATYPICAL EATING DISORDERS.
SIMPLE WEIGHT LOSS AND AMENORRHEA
EXERCISE-ASSOCIATED AMENORRHEA
PSYCHOGENIC HYPOTHALAMIC AMENORRHEA
DIMINISHED GONADOTROPIN-RELEASING HORMONE AND LUTEINIZING HORMONE SECRETION IN ALL FORMS
TREATMENT
CHRONIC ANOVULATION DUE TO INAPPROPRIATE FEEDBACK IN POLYCYSTIC OVARY SYNDROME (PCOS)
CHRONIC ANOVULATION DUE TO OTHER ENDOCRINE AND METABOLIC DISORDERS
ABNORMAL UTERINE BLEEDING IN WOMEN OF REPRODUCTIVE AGE
ABNORMAL UTERINE BLEEDING IN POSTMENOPAUSAL WOMEN
REFERENCES
Similar articles
Adolescent menstrual irregularity.Mansfield MJ, Emans SJ.Mansfield MJ, et al.J Reprod Med. 1984 Jun;29(6):399-410.J Reprod Med. 1984.PMID: 6379175 Review.
Women with functional hypothalamic amenorrhea but not other forms of anovulation display amplified cortisol concentrations.Berga SL, Daniels TL, Giles DE.Berga SL, et al.Fertil Steril. 1997 Jun;67(6):1024-30. doi: 10.1016/s0015-0282(97)81434-3.Fertil Steril. 1997.PMID: 9176439
Dysfunctional uterine bleeding and amenorrhea. Differential diagnosis and management.Clark-Coller T.Clark-Coller T.J Nurse Midwifery. 1991 Jan-Feb;36(1):49-62. doi: 10.1016/0091-2182(91)90020-p.J Nurse Midwifery. 1991.PMID: 2002395 Review.
Amenorrhea (Nursing).Nawaz G, Rogol AD, Jenkins SM, Kuehl D.Nawaz G, et al.2024 Feb 25. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.2024 Feb 25. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.PMID: 33760532 Free Books & Documents.
A practical guide to the diagnosis and management of amenorrhoea.Crosignani PG, Vegetti W.Crosignani PG, et al.Drugs. 1996 Nov;52(5):671-81. doi: 10.2165/00003495-199652050-00005.Drugs. 1996.PMID: 9118817 Review.
See all similar articles
Publication types
Review
Actions
Search in PubMed
Search in MeSH
Add to Search
Related information
MedGen
[x]
Cite
Copy Download .nbib.nbib
Format:
Send To
Clipboard
Email
Save
My Bibliography
Collections
Citation Manager
[x]
NCBI Literature Resources
MeSHPMCBookshelfDisclaimer
The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited.
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
7777 | https://www.quora.com/If-the-rain-is-falling-vertically-wrt-the-ground-and-I-am-running-in-the-rain-do-I-really-see-the-rain-falling-at-an-angle-with-the-vertical-relative-to-me-I-have-a-hard-time-visualising-this | If the rain is falling vertically wrt the ground and I am running in the rain, do I really see the rain falling at an angle with the vertical relative to me? I have a hard time visualising this. - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Physics
Running (activity)
Visual Processing
Weather Conditions
Human Activities
Relative Motion
Rain
Perception
Motion (physics)
5
If the rain is falling vertically wrt the ground and I am running in the rain, do I really see the rain falling at an angle with the vertical relative to me? I have a hard time visualising this.
Ad by Grammarly
Stuck on the blinking cursor?
Move your great ideas to polished drafts without the guesswork. Try Grammarly today!
Download
All related (44)
Sort
Recommended
Assistant
Bot
·
1y
Yes, when you run through the rain that is falling vertically with respect to the ground, you will perceive the rain as falling at an angle relative to your body. This happens due to your motion through the rain.
Here's a breakdown to help you visualize it:
Rain's Motion: The rain is falling straight down (vertically) with respect to the ground.
Your Motion: As you run forward, you are moving through the rain. Your speed creates a relative motion between you and the rain.
Resulting Perception:
Imagine you are standing still. You would see the rain falling straight down.
However, if you s
Continue Reading
Yes, when you run through the rain that is falling vertically with respect to the ground, you will perceive the rain as falling at an angle relative to your body. This happens due to your motion through the rain.
Here's a breakdown to help you visualize it:
Rain's Motion: The rain is falling straight down (vertically) with respect to the ground.
Your Motion: As you run forward, you are moving through the rain. Your speed creates a relative motion between you and the rain.
Resulting Perception:
Imagine you are standing still. You would see the rain falling straight down.
However, if you start running forward, the rain appears to come at you from an angle. This is because, while you are moving forward, the rain is still falling down. The combination of your forward motion and the downward motion of the rain creates the perception of an angled trajectory.
Visualizing the Effect:
Imagine a Coordinate System:
If you draw a vertical line for the rain falling down and a horizontal line for your running direction, the rain's path relative to you forms a diagonal line.
Angle of Perception: The faster you run, the steeper the angle at which you perceive the rain to be falling. If you were to run very fast, the rain would seem to be coming more horizontally toward you.
Conclusion:
So, while the rain itself is indeed falling vertically, your movement through it causes you to see it at an angle. This is a common experience and illustrates how relative motion affects perception.
Upvote ·
Related questions
More answers below
Why does rain sometimes fall at an angle instead of falling straight down?
Is there an exact rain falling angle?
Why does rain fall at an inclined angle?
How does rain come down diagonally with no wind?
What causes rain to fall diagonally?
Mahesh Prakash
Author has 3.1K answers and 331.5K answer views
·Jul 16
rain is falling vertically wrt the ground
the answer is contained in this statement itself
description of an event depends on the state of the observer himself
rain is falling vertically with velocity R. but, this is the observation of a bystander, a person standing stationary by the roadside
if the observer himself is moving with velocity V, this motion will necessarily influence the observation
you go over to a frame of reference where this observer would be stationary
add a velocity -V to ALL the velocities
V becomes zero - the observer has been equivalently brought to rest
however, R becomes ( R -
Continue Reading
rain is falling vertically wrt the ground
the answer is contained in this statement itself
description of an event depends on the state of the observer himself
rain is falling vertically with velocity R. but, this is the observation of a bystander, a person standing stationary by the roadside
if the observer himself is moving with velocity V, this motion will necessarily influence the observation
you go over to a frame of reference where this observer would be stationary
add a velocity -V to ALL the velocities
V becomes zero - the observer has been equivalently brought to rest
however, R becomes ( R - V )
the observer will now observe the vector ( R - V )
this is a vector wetting him front
the faster he walks, the more frontal assault of the rain he will feel
if he stops in the middle of the road, he will find the rain hammering vertically down
Upvote ·
9 1
Brian Kinsey
IT Professional (2000–present) · Author has 1.7K answers and 2.8M answer views
·5y
Whether you would actually “see” it depends on how fast you’re running, to an extent. And on how much it’s raining. But yes, if the rain is vertical from a stationary perspective, it will seem to be at an angle when you are moving. From that same stationary perspective, you are actually running into the falling raindrops, so the motion is yours, not the rains.
If you stood still, the rain would hit you on the head, shoulders, and any other parts that stick out. As you move forward, you begin to “intercept” drops that would have fallen in front of you, but now hit you in the chest, stomach, and
Continue Reading
Whether you would actually “see” it depends on how fast you’re running, to an extent. And on how much it’s raining. But yes, if the rain is vertical from a stationary perspective, it will seem to be at an angle when you are moving. From that same stationary perspective, you are actually running into the falling raindrops, so the motion is yours, not the rains.
If you stood still, the rain would hit you on the head, shoulders, and any other parts that stick out. As you move forward, you begin to “intercept” drops that would have fallen in front of you, but now hit you in the chest, stomach, and legs, because you ran into them.
The reason you have a hard time visualizing this is most likely because we’re talking about running. And in running or walking, we don’t tend to really change our perspective completely. Easier to visualize if you think of the same situation when you are driving. Because you are more isolated from the outside environment, your perspective shifts more easily and completely, and you will more likely actually see the rain as falling at an angle. Walking or running, you will experience it as though it is, but it’s harder to actually “see” it. (There’s a discussion to be had there about inertial and non-inertial frames of reference, but its’ really beyond the scope of this question, so we won’t go there.)
Upvote ·
9 2
Sponsored by All Out
Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out.
Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association.
Shop Now
999 623
Oleg Shigiltchoff
PhD from University of Pittsburgh · Author has 149 answers and 45.3K answer views
·5y
To visualize it you need to move with constant speed and do not see any other objects around, because those objects are telling that it’s you who is moving, not the rain. Then your brain will tell you that you don’t move but rain goes on you with an angle. Similar effect is in Orlando, a Harry Potter attraction. You enter an elevator and the whole side wall starts moving up so your brain tells you you are going down. In fact the whole side wall is a big tv screen and the image goes up, but the elevator doesn’t actually move but rumble just to increase your feeling of moving down.
Upvote ·
Related questions
Why does rain sometimes fall at an angle instead of falling straight down?
Is there an exact rain falling angle?
Why does rain fall at an inclined angle?
How does rain come down diagonally with no wind?
What causes rain to fall diagonally?
Which rain drop, if either fall faster; small ones or large ones?
Why do the rain drops falling from the sky neither injure us nor make holes on the ground?
Is rain falling at 0.7 mm very light?
Why do sheets of falling rain in the distance appear still?
How long does it usually take CPS to place a child with a relative?
Sometimes when rain falls, fish and worms sometimes fall with rain water. How is it possible?
Why does it sometimes rain heavily for an hour or so and then it keeps drizzling for hours? What makes it rain heavy or slow?
Why does rain always fall in a straight line when wind is absent, not including rain falling on a sunny day?
Can we calculate the approximate number of rain drops that have fallen since the beginning of time?
A man moving in rain holds his umbrella inclined to the vertical even though the rain drops are falling vertically downwards. why?
Related questions
Why does rain sometimes fall at an angle instead of falling straight down?
Is there an exact rain falling angle?
Why does rain fall at an inclined angle?
How does rain come down diagonally with no wind?
What causes rain to fall diagonally?
Which rain drop, if either fall faster; small ones or large ones?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
Always Active
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
Always Active
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
Always Active
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
7778 | https://www.indeed.com/career-advice/career-development/operating-net-income | What Is Net Operating Income? (And How To Calculate It) | Indeed.com
Skip to main content
Home
Company reviews
Find salaries
Sign in
Sign in
Employers / Post Job
1 new update
Home
Company reviews
Find salaries
Employers
Create your resume
Resume services
Change country 🇺🇸 United States
Help
Start of main content
Career Guide
Finding a job
Resumes & cover letters
Resumes & cover letters articles
Resume samples
Cover letter samples
Interviewing
Pay & salary
Career development
Career development articles
Starting a new job
Career paths
News
Career development
What Is Net Operating Income? (And How To Calculate It)
What Is Net Operating Income? (And How To Calculate It)
Written by
Andrew Juma
Updated July 26, 2025
Companies use different calculations to determine their success, but some standard metrics are net operating income, operating income and net income. These figures provide information about the company's earnings by including and excluding different figures to assess financial and operational health. Understanding how to calculate each and their differences may help you assist your employer, business owner or investor.In this article, we define net operating income, operating income and net income, the differences between them and the documents and information needed to calculate these metrics.Key takeaways:
Net operating income measures the profitability of an income-producing property.
Operating income measures a company's income after accounting for operating expenses only.
Net income measures a company's total income after accounting for all business expenses.
This article is for informational purposes only and does not constitute financial advice. Consult with a licensed financial professional for any issues you may be experiencing.
Related jobs on Indeed
Property leasing & management jobs
Part-time jobs
Full-time jobs
Remote jobs
View more jobs on Indeed
What is net operating income (NOI)?
Net operating income is the property's generated revenue minus the operating expenses, excluding taxes and debt service. Real estate investors and their industry use this term frequently, but it applies to any company or business that earns income from rental properties. To determine net operating income, calculate all the revenue from the property and subtract any related operational expenses. Property revenue includes items like:
Rent
Parking
Vending
Laundry
Operating expenses associated with properties often include items like:
Maintenance
Insurance
Utilities
Repairs
Payroll for any staff
The net operating income calculation doesn't include capital expenses, like purchasing new appliances or updating the heating system. You can calculate net operating income over any set time frame, but it's most frequently done annually.Read more:What Is Net Operating Income?
How to calculate NOI
Here are the steps you can use to calculate NOI:
Add all gross income within the period you wish to track.
Sum up all operating expenses.
Subtract the operating expenses from the gross income to get your net operating income.
Total gross income - Operating expenses = Net operating income Businesses outside the real estate industry often refer to net operating income as earnings before interest and taxes (EBIT). EBIT includes the same types of revenue and expenses in its calculation as net operating income without property specificity.
Why is this metric significant?
This metric is significant for several reasons, including:
Estimating the income potential of a property
Calculating the property's debt service coverage ratio
Helping lenders with deciding whether to accept or reject a mortgage application
Calculating the property's capitalization rate
What is operating income?
Operating income or operating profit is an expression of a company's income after deducting operating expenses only. Operating income helps individuals see how successful their business operations are without considering additional income from unrelated revenue sources or expenses like taxes or interest expenses on loans. Usually, operating expenses include business activities like:
Rent
Payroll
Utilities
Office supplies
Benefits like health insurance
Marketing and sales materials
Related:Operating Income vs. EBITDA: Differences and Examples
How to calculate operating income
You can calculate operating income in one of two ways:
Revenue minus operating expenses
Revenue is the total sales minus any returns from products sold. To calculate operating income using this formula, look at the total revenue on your income statement, usually the top line, and subtract all operating expenses from that number.Total revenue - Operating expenses = Operating income
Gross profit minus operating expenses
Gross profit is total revenue minus the cost of goods sold. Total revenue includes all income from the business and not just the income generated from sales. The cost of goods sold is any expenses directly tied to the production of the product. To calculate operating income using these formulas, take gross profit and subtract operating expenses from that figure.Operating income = Gross income − Operating expenses
Additional operating income formulas
While the other two formulas go into further detail about the operating income:Operating income = Revenue − Cost of goods sold − Cost of labor − Other daily expenses Operating income = Gross income – Operating expenses – Depreciation – Amortization Many businesses use a yearly income statement to compare their operating income annually. Others calculate operating income more frequently, from quarter to quarter or month to month. If you have the data, you can calculate operating income over any time frame.Read more:How To Calculate Operating Income (Formula and Examples)
Why is this metric significant?
The significance of the operating income metric includes the following:
Showing how a company generates income from its operational activities
Helping show investors and creditors how well the business performs
Identifying where a company can improve specific inefficiencies
What is net income?
Net income is the total income remaining after accounting for all business expenses, interests and taxes. Individuals also refer to this term as the bottom line since net income is usually at the bottom of the income statement. In addition to accounting for operational expenses, net earnings account for all other expenses to determine how much pure profit the company has earned.
How to calculate net income
To calculate net income, subtract all non-operating expenses from operating income. This means you first calculate operating income before determining net income. Here's the net income formula to help you calculate it:Operating income - Non-operating expenses = Net income Non-operating expenses might include items like:
Restructuring
Interest payments
Taxes
Inventory
Lawsuit settlements
Derivatives
The final net income figure represents the company's total earnings or profits. You can calculate net income over any period, such as a month, quarter or year, like operating income.Read more:How To Calculate Net Income: Formula, Definition and FAQs
Why is this metric significant?
You may find the net income significant for many reasons, such as:
Identifying the company's profit within a certain timeframe
Allowing you to understand how much of a company's earnings can get distributed among the owners
Helping accountants and financial analysts advise investors to invest in the company's stock
Net income vs. net operating income
These terms differ from each other. Net income is revenue minus all operating and non-operating expenses. Net operating income, in contrast, is a company's revenue minus all its operating expenses.Related:Your Guide To Understanding Net Earnings vs. Net Income
Operating income vs. net operating income
These terms also have significant differences in meaning. Operating income is how much a company earns minus its operating expenses and includes selling, general & administrative expenses (SG&A). Operating income differs from net operating income because NOI accounts for the operating income minus non-operating expenses.Related:How To Calculate Net Operating Income (With Example)
Operating income vs. net income
While operating income and net income provide earnings figures, the formulas evaluate unique aspects of the business. Operating income shows you how successful a business is at operating and producing. Some businesses might have a substantial number of loans with high-interest payments that negatively affect their bottom line. Despite this, the operations of the company might be highly successful.Operating income helps you and your stakeholders see how effective the company's core is without deciphering other income or expenses. Net income, in contrast, shows a company's total earnings after accounting for every business expense. This figure helps you and your stakeholders see if the business is profitable after administrative costs like paying bills, fees and taxes. Unlike operating income, it doesn't indicate the company's operational performance but offers an earnings evaluation.Related:Net Income vs. Net Profit: How Do They Differ?
Search jobs and companies hiring now
Job title, keywords or company
Location
Search
Resources for calculating the incomes
Calculating the types of business income can require various resources and documents, like financial statements and balance sheets. When you're ready to determine your operating income, net income or net operating income, gather this information to help you:
Income statement
A company'sincome statement, a document that records a business's revenues and expenses over a specific time, is a valuable tool for calculating all three incomes. The figures required to calculate operating income and net income appear on an income statement, including revenue, itemized operating expenses, non-operating expenses, interest payments and taxes.
Tax documents
Depending on the structure of the income statement, you may consult the tax returns or other tax documents to find the figures for some net income and net operating income calculations. For example, an investor might seek to know how much they paid in property taxes rather than all other taxes when considering their total operating expenses for their business.
Property management documents
The operating expenses category of an NOI calculation can vary depending on the type of property and the included amenities. Review contracts and management documents to ensure you report all expenses and revenue streams.Related: Guide to Types of Commercial Real Estate
The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers
Property leasing & management
Share:
Related Articles
How To Use the Net Income Formula (Steps and Examples)
Your Guide To Understanding Net Earnings vs. Net Income
Net Operating Profit After Tax (NOPAT): A Definitive Guide
Explore more articles
10 Types of Prototypes (With Explanations and Tips)
12 Approaches To Problem-Solving for Every Situation
Guide To Customer Service
Top 10 HR Assistant Interview Questions and Answers
10 of the Highest-Paying Jobs in Psychology (With Salaries)
Audit Report Types (With Example Report)
26 High-Paying Marketing Jobs in 2024 (With Salary Info)
9 Commission-Based Jobs to Consider
12 High-Paying Entry-Level Jobs Requiring No Experience
How to Tender Your Resignation (With Example Letters)
How To Become a Tutor
Websites for Resume Templates You Can Build Online for Free
Hiring LabCareer adviceBrowse jobsBrowse companiesSalariesIndeed EventsWork at IndeedCountriesAboutHelpESG at Indeed
© 2025 IndeedYour Privacy ChoicesAccessibility at IndeedPrivacy Center and Ad ChoicesTerms |
7779 | https://researchbasics.education.uconn.edu/calculatingmeanstandarddev/ | UConn
University of Connecticut
Search University of Connecticut
Site A-Z
UConn A-Z
Neag School of Education
Educational Research Basics by Del Siegle
Calculating the Mean and Standard Deviation with Excel
Finding the Mean
Enter the scores in one of the columns on the Excel spreadsheet (see the example below). After the data have been entered, place the cursor where you wish to have the mean (average) appear and click the mouse button. Select Insert Function (fx) from the FORMULAS tab. A dialog box will appear. Select AVERAGE from the Statistical category and click OK. (Note: If you want the Median, select MEDIAN. If you want the Mode, select MODE.SNGL. Excel only provides one mode. If a data set had more than one mode, Excel would only display one of them.)
Enter the cell range for your list of numbers in the Number 1 box. For example, if your data were in column A from row 1 to 13, you would enter A1:A13. Instead of typing the range, you can also move the cursor to the beginning of the set of scores you wish to use and click and drag the cursor across them. Once you have entered the range for your list, click on OK at the bottom of the dialog box. The mean (average) for the list will appear in the cell you selected.
Finding the Standard Deviation
Place the cursor where you wish to have the standard deviation appear and click the mouse button.Select Insert Function (fx)from the FORMULAS tab. A dialog box will appear. Select STDEV.S (for a sample) from the the Statistical category. (Note: If your data are from a population, click on STDEV.P). After you have made your selections, click on OK at the bottom of the dialog box.
Enter the cell range for your list of numbers in the Number 1 box. For example, if your data were in column A from row 1 to 13, you would enter A1:A13. Instead of typing the range, you can also move the cursor to the beginning of the set of scores you wish to use and click and drag the cursor across them. Once you have entered the range for your list, click on OK at the bottom of the dialog box. The standard deviation for the list will appear in the cell you selected.
Del Siegle, Ph.D.
Neag School of Education – University of Connecticut del.siegle@uconn.edu
www.delsiegle.com |
7780 | https://mathbitsnotebook.com/Algebra2/Quadratics/QDfundamentalThm.html | | | | | | | | | | | | | | |
--- --- --- --- --- ---
| | | | | | | | | | | | | | --- --- --- --- --- --- | | | | | | --- | | | Fundamental Theorem of Algebra MathBitsNotebook.com Topical Outline | Algebra 2 Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | | | | | | | | | --- --- --- | If you know that a quadratic equation (degree 2) has exactly 2 roots, then you already know (and have used) the Fundamental Theorem of Algebra. While there are various statements of this theorem, the interpretation that we are interested in for this course is the following: | | | | --- | | | | --- | | | The Fundamental Theorem of Algebra: If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots. | | In plain English, this theorem says that the degree of a polynomial equation tells you how many roots the equation will have. A linear equation (degree 1) will have one root. A quadratic equation (degree 2) will have two roots. A cubic equation (degree 3) will have three roots. An nth degree polynomial equation will have n roots. Read carefully to fully understand what this theorem is saying. • It mentions "multiple" roots . . . so, if a polynomial has a repeated root, each repetition of the root is counted. • It mentions "complex" roots . . . so, if a polynomial has a complex root, then the conjugate is also counted since complex roots come in conjugate pairs. This theorem deals with complex roots, both real and non-real. Remember that all Real Numbers are a subset of the Complex Numbers. Dealing with Quadratics: Let's verify that the Fundamental Theorem of Algebra holds for quadratic polynomials. A quadratic polynomial is a second degree polynomial. According to the Fundamental Theorem of Algebra, the quadratic set = 0 has exactly two roots. As we have seen, factoring a quadratic equation will result in one of three possible situations. | | | | --- | The quadratic may have 2 distinct real roots. This graph crosses the x-axis in two locations. These graphs may open upward or downward. | It may appear that the quadratic has only one real root. But, it actually has one repeated root. This graph is tangent to the x-axis in one location (touching once). | The quadratic may have two non-real complex roots called a conjugate pair. This graph will not cross or touch the x-axis, but it has two complex (conjugate) roots. | In all three situations, there are two solutions to the quadratic equations, as stipulated by the Fundamental Theorem of Algebra. | Historical note: Mathematicians have been unable to devise an exact general method for factoring a polynomial. Factoring techniques (formulas) are known for "special" polynomials such as linear (degree 1), quadratic (degree 2), cubic (degree 3), and quartic (degree 4). Mathematician Evariste Galois (1811-1832) proved that there will never be a general formula to factor polynomials of degree five or higher. See more about this theorem under Polynomials. | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | Algebra 2 Outline| MathBitsNotebook.com | MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts | | |
7781 | https://english.stackexchange.com/questions/8680/how-much-exactly-is-increased-when-it-is-increased-by-1-1 | meaning - How much exactly is increased when it is "increased by 1.1"? - English Language & Usage Stack Exchange
Join English Language & Usage
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
English Language & Usage helpchat
English Language & Usage Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How much exactly is increased when it is "increased by 1.1"?
Ask Question
Asked 14 years, 8 months ago
Modified14 years, 8 months ago
Viewed 13k times
This question shows research effort; it is useful and clear
8
Save this question.
Show activity on this post.
I saw it in a text book, and a similar problem that also appeared in the book is "3 times faster", which is already asked. Simply speaking, the book says:
Unfortunately, it increases the CPI by 1.1.
Where CPI is for clock cycles per instruction, if that matters. According to the choices the question provided, it actually means the increased value is 1.1 times of the original one (10% increase). It's the first time I have seen "by" meaning "to". Is it acceptable in daily English?
meaning
numbers
ambiguity
Share
Share a link to this question
Copy linkCC BY-SA 2.5
Improve this question
Follow
Follow this question to receive notifications
edited Jan 19, 2011 at 1:50
Jimi Oke
27.5k 3 3 gold badges 80 80 silver badges 106 106 bronze badges
asked Jan 13, 2011 at 14:56
LLSLLS
1,103 4 4 gold badges 10 10 silver badges 19 19 bronze badges
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
13
Save this answer.
Show activity on this post.
The sentence really should have read:
Unfortunately, it increases the CPI by a factor of 1.1.
One would probably not find this sort of ambiguity in a math text, but it is certain that a factor of was implied.
Note, however, that
Unfortunately, it increases the CPI to 1.1
does not mean the same thing! This implies, the final value of the CPI, after the increase, is 1.1.
by is used to indicate a product, but usually in an explicit manner, e.g.
Taxes increased by 10%.
Multiply your answer by 3.
The new energy rating has decreased, by a factor of 1.5.
Indeed, in the absence of any context, I could be tempted to take
Unfortunately, it increases the CPI by 1.1
to mean 1.1 is added to the CPI, but I would really hard-pressed to do that! In a mathematical context, though, when one sees by, one should begin to think multiplication or division.
Share
Share a link to this answer
Copy linkCC BY-SA 2.5
Improve this answer
Follow
Follow this answer to receive notifications
answered Jan 13, 2011 at 15:07
Jimi OkeJimi Oke
27.5k 3 3 gold badges 80 80 silver badges 106 106 bronze badges
5
2 +1 for the last sentence. In this particular context, I would understand "increased by" as a multiplication. In general, however, I would understand it as an addition.b.roth –b.roth 2011-01-13 15:52:37 +00:00 Commented Jan 13, 2011 at 15:52
1 “but I would really hard-pressed to do that” – why? Doesn’t “increase x by y” mean exactly that? As in, “Christmas increased my weight by fifty pounds.”Konrad Rudolph –Konrad Rudolph 2011-01-13 17:05:03 +00:00 Commented Jan 13, 2011 at 17:05
@Konrad Rudolph: I included the hard-pressed part a few minutes after posting, just to counter some peer pressure :) After some thought, I now realize that, while in your excellent example, by implies an addition, it only emphatically does so when a unit is specified, e.g. increased...by 50 pounds/meters/watts, etc. A number on its own would be most likely read as a factor of multiplication. From my science background, I would battle hard to come to terms with, for instance, increases the CPI by 1.1. Units are very important! Ultimately, this ambiguity should have been avoided.Jimi Oke –Jimi Oke 2011-01-13 23:50:10 +00:00 Commented Jan 13, 2011 at 23:50
It make more sense to interpret the increase as the absolute numerical value in this case. It should be caused by the carelessness of the authors. It's also the first time I consider "increase by a factor of" seriously. I looked up in dictionary; the word factor has a meaning of "a quantity by which a given quantity is multiplied or divided in order to indicate a difference in measurement".LLS –LLS 2011-01-14 03:06:33 +00:00 Commented Jan 14, 2011 at 3:06
@Jimi: not only units: “five minutes later, the number of people in the room had increased by 10” – though I would use a completely different formulation (“… there were 10 more people in the room” or something like that), it’s nevertheless clear what is meant and as far as I know it’s correct – if unusual – English.Konrad Rudolph –Konrad Rudolph 2011-01-14 10:29:12 +00:00 Commented Jan 14, 2011 at 10:29
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
X increased by 1.1 → X is now 1.1 times the previous value of X (10% increase)
X increased to 1.1 → X is now 1.1 (before it was less than 1.1)
Note that this is my interpretation of increased by in this particular context. In general, however, I would interpret increased by as an addition, i.e. 1.1 added to the original value.
Share
Share a link to this answer
Copy linkCC BY-SA 2.5
Improve this answer
Follow
Follow this answer to receive notifications
edited Jan 13, 2011 at 15:49
answered Jan 13, 2011 at 15:00
b.rothb.roth
22k 22 22 gold badges 87 87 silver badges 129 129 bronze badges
0
Add a comment|
This answer is useful
0
Save this answer.
Show activity on this post.
The use of "by" isn't indicating "to" in this case. It's indicating that the current CPI would be increased by another 10%, not to a maximum threshold of 1.1.
See Wikipedia's info on percentages. It may help explain further.
Share
Share a link to this answer
Copy linkCC BY-SA 2.5
Improve this answer
Follow
Follow this answer to receive notifications
answered Jan 13, 2011 at 15:01
Bob GBob G
130 5 5 bronze badges
1
Agreed. If I see "to" separately, I will probably think the same thing. But it's not my point in this question.LLS –LLS 2011-01-14 02:58:30 +00:00 Commented Jan 14, 2011 at 2:58
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
meaning
numbers
ambiguity
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
9"much too [something]" vs "too much [something]"
0How to refer to a previous number by "that much"?
1How much to wait or how long to wait?
0how much bleeding was there VS How much bleed you got?
1What's the difference between ‘A increased with B’ and ’A increased with increasing B?
Hot Network Questions
Numbers Interpreted in Smallest Valid Base
What's the expectation around asking to be invited to invitation-only workshops?
Exchange a file in a zip file quickly
Why include unadjusted estimates in a study when reporting adjusted estimates?
Matthew 24:5 Many will come in my name!
What were "milk bars" in 1920s Japan?
How to rsync a large file by comparing earlier versions on the sending end?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Do sum of natural numbers and sum of their squares represent uniquely the summands?
Cannot build the font table of Miama via nfssfont.tex
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
Gluteus medius inactivity while riding
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
Calculating the node voltage
Drawing the structure of a matrix
Can a cleric gain the intended benefit from the Extra Spell feat?
Stress in "agentic"
Should I let a player go because of their inability to handle setbacks?
Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-?
An odd question
Does the curvature engine's wake really last forever?
Childhood book with a girl obsessed with homonyms who adopts a stray dog but gives it back to its owners
How do you create a no-attack area?
Languages in the former Yugoslavia
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
English Language & Usage
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
7782 | https://www.chegg.com/homework-help/questions-and-answers/problem-calculation-mass-rod-variable-mass-density-rod-total-length-l-2m-position-dependen-q62906885 | Solved Problem: Calculation of mass of a Rod with variable | Chegg.com
Skip to main content
Books
Rent/Buy
Read
Return
Sell
Study
Tasks
Homework help
Understand a topic
Writing & citations
Tools
Expert Q&A
Math Solver
Citations
Plagiarism checker
Grammar checker
Expert proofreading
Career
For educators
Help
Sign in
Paste
Copy
Cut
Options
Upload Image
Math Mode
÷
≤
≥
o
π
∞
∩
∪
√
∫
Math
Math
Geometry
Physics
Greek Alphabet
Science
Physics
Physics questions and answers
Problem: Calculation of mass of a Rod with variable Mass Density A rod of total length L = 2m has position dependent linear mass density (mass per unit length) given by the formula p(x) = 2.c +1 where x is the distance in the rod from one of its ends designated as x =0. We want to calculate the mass of the rod. The only formula we know for calculating the
Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
See Answer See Answer See Answer done loading
Question: Problem: Calculation of mass of a Rod with variable Mass Density A rod of total length L = 2m has position dependent linear mass density (mass per unit length) given by the formula p(x) = 2.c +1 where x is the distance in the rod from one of its ends designated as x =0. We want to calculate the mass of the rod. The only formula we know for calculating the
Show transcribed image text
Here’s the best way to solve it.Solution Share Share Share done loading Copy link View the full answer Previous questionNext question
Transcribed image text:
Problem: Calculation of mass of a Rod with variable Mass Density A rod of total length L = 2m has position dependent linear mass density (mass per unit length) given by the formula p(x) = 2.c +1 where x is the distance in the rod from one of its ends designated as x =0. We want to calculate the mass of the rod. The only formula we know for calculating the mass is p= which applies only when the linear mass density is constant. We assume that we have forgotten our calculus for the time being and that we have never seen the formula: p = lim Sam dm ax-0 AX Critical Thinking and Solution Methodology: • We observe, that if the rod is sufficiently short we can assume that the mass density does not appreciably vary across its length and more or less is constant equal to some "average value the mass density in the middle of the bar. We can therefore assume that the rod has constant mass density and its mass is M Pay L. • Some may argue that this approximation is a very crude one and the error is very large. The length of the bar is not short enough to get an accurate result, so we can do better by dividing the bar into two equal parts and assume that each part has constant mass density (with value the mass density at the center of the piece) and therefore the mass of each piece is its length times the constant mass density). Then the mass of the rod is the sum of its two parts. • If we wish to get an even more accurate result we can divide the bar into three equal length pieces and assume that each piece is short enough to have more or less constant mass density across it (the mass density in the middle of each piece). Now the bar can be approximated as a collection of three pieces each having constant mass density and the mass of each piece is its length times its mass density. The mass of the bar is just the sum of the masses of its three parts. • A division of the bar into four pieces will give us an even more accurate result of its mass. • Perfectionists and mathematically inclined students might protest that the method is still inaccurate and the obtained result is away from the correct one. That is why we will try a little more sophisticated method. We will keep splitting the rod into a larger number of smaller and smaller pieces always assuming that the mass density of each piece is constant equal to an average value the value of the mass density computed at the center of the piece. • If we keep dividing more and more, at the end the bar will be approximated with an infinite number of infinitesimally small pieces next to each other (in tandem) where each piece will essentially be a point with O length and therefore 0 mass. However the mass density of such piece will be the mass density value calculated at that point using the formula: p(x) = 2x2 +1 [ ] . We will then calculate the mass of the bar as a Riemann Sum (a sum of an infinite number of infinitesimally small terms) which is an Integral. • In your calculations be as accurate as possible by keeping as many decimal points as possible, but when you enter your results keep 3 significant decimal digits in tour answer for best results. Assume that the rod is short enough so that its mass density per unit length is constant and equal to an average mass density. As average mass density we consider the value at the center of the rod. With these assumptions in mind answer the following questions. Remember that the length of the rod is L=2m. L M • What is the x coordinate of the rod center? m • What is the average value of the mass density kg/m • What is the mass of the rod? kg
Not the question you’re looking for?
Post any question and get expert help quickly.
Start learning
Chegg Products & Services
Chegg Study Help
Citation Generator
Grammar Checker
Math Solver
Mobile Apps
Plagiarism Checker
Chegg Perks
Company
Company
About Chegg
Chegg For Good
Advertise with us
Investor Relations
Jobs
Join Our Affiliate Program
Media Center
Chegg Network
Chegg Network
Busuu
Citation Machine
EasyBib
Mathway
Customer Service
Customer Service
Give Us Feedback
Customer Service
Manage Subscription
Educators
Educators
Academic Integrity
Honor Shield
Institute of Digital Learning
© 2003-2025 Chegg Inc. All rights reserved.
Cookie NoticeYour Privacy ChoicesDo Not Sell My Personal InformationGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
Do Not Sell My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Sale of Personal Data
[x] Sale of Personal Data
Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link.
If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences.
Targeting Cookies
[x] Switch Label label
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices
mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
7783 | https://fiveable.me/key-terms/physical-chemistry-ii/wavenumber | Wavenumber - (Physical Chemistry II) - Vocab, Definition, Explanations | Fiveable | Fiveable
new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom
ap study content toolsprintablespricing
my subjectsupgrade
All Key Terms
Physical Chemistry II
Wavenumber
🧂physical chemistry ii review
key term - Wavenumber
Citation:
MLA
Definition
Wavenumber is a measurement used to describe the spatial frequency of a wave, defined as the number of waves per unit distance, typically expressed in reciprocal centimeters (cm^-1). In the context of spectroscopy, wavenumbers provide a direct correlation to the energy of transitions between different rotational and vibrational states in molecules, making it an essential parameter for analyzing molecular spectra.
5 Must Know Facts For Your Next Test
Wavenumber is inversely proportional to wavelength; as wavelength increases, wavenumber decreases and vice versa.
In rotational and vibrational spectroscopy, wavenumber is often preferred over frequency because it provides a more direct relationship to energy levels.
The wavenumber can be calculated using the formula: $$ ilde{ u} = \frac{1}{\lambda}$$, where $$\tilde{ u}$$ is the wavenumber and $$\lambda$$ is the wavelength.
Different molecular vibrations correspond to specific wavenumbers, allowing for the identification of functional groups in infrared (IR) spectroscopy.
The standard unit for wavenumber, cm^-1, makes it convenient for comparison across various spectroscopic techniques.
Review Questions
How does the concept of wavenumber relate to molecular energy transitions observed in spectroscopy?
Wavenumber is crucial in spectroscopy as it provides a direct measure of energy transitions between molecular states. Each transition corresponds to a specific wavenumber value, which reflects the energy difference between two states according to the relationship $$E = h u$$, where $$E$$ is energy, $$h$$ is Planck's constant, and $$\nu$$ is frequency. By analyzing these wavenumbers in a spectrum, scientists can identify the types of molecular vibrations or rotations occurring within a sample.
Discuss how wavenumber can be used to distinguish between different functional groups in IR spectroscopy.
In IR spectroscopy, different functional groups exhibit characteristic absorption bands at specific wavenumbers. For instance, O-H bonds typically absorb around 3200-3600 cm^-1 while C=O bonds show strong absorption near 1700 cm^-1. By comparing observed wavenumbers in a spectrum to known values for various functional groups, chemists can identify the presence of those groups within a molecule. This method leverages the unique relationships between molecular structure and vibrational frequencies represented by wavenumbers.
Evaluate the significance of wavenumber in understanding rotational and vibrational spectra and how this knowledge can be applied in real-world scenarios.
Understanding wavenumber is fundamental for interpreting rotational and vibrational spectra because it connects molecular structure with energetic transitions. The ability to quantify these transitions via wavenumber allows chemists to analyze molecular interactions and bonding characteristics. This knowledge is applied in various fields such as material science for developing new materials based on their spectral properties or in pharmaceuticals for identifying active compounds within complex mixtures. By analyzing how molecules behave at specific wavenumbers, researchers can derive insights into molecular dynamics that are crucial for advancements in technology and medicine.
Related terms
Spectroscopy:A technique that involves the interaction of light with matter, used to identify and analyze the composition of substances based on their spectral properties.
Energy Levels:
Quantized states of energy that molecules can occupy, which dictate how they absorb or emit light during transitions.
Quantum Mechanics:
The branch of physics that deals with the behavior of particles at atomic and subatomic levels, providing the framework for understanding molecular interactions in spectroscopy.
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
every AP exam is fiveable
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
every AP exam is fiveable
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
every AP exam is fiveable
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
every AP exam is fiveable
Study Content & Tools
Study GuidesPractice QuestionsGlossaryScore Calculators
Company
Get $$ for referralsPricingTestimonialsFAQsEmail us
Resources
AP ClassesAP Classroom
history
🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history
social science
✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾⚖️ ap us government
english & capstone
✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar
arts
🎨 ap art & design🖼️ ap art history🎵 ap music theory
science
🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics
math & computer science
🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p
world languages
🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature
go beyond AP
high school exams
✏️ PSAT🎓 Digital SAT🎒 ACT
honors classes
🍬 honors algebra II🐇 honors biology👩🏽🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history
college classes
👩🏽🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽🔬 science💶 social science
RefundsTermsPrivacyCCPA
© 2025 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
0 |
7784 | https://www.quora.com/Derive-g-GM-r-2-from-Newtons-universal-law-of-gravitation-and-Newtons-second-law | Derive g=GM/r^2 from Newton's universal law of gravitation and Newton's second law? - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Physics
Derivations
Newton's Law of Universal...
Masses
Gravitational Force
Physics Formulas
Newton's Second Law of Mo...
Law of Gravitation
Elementary Derivation
5
Derive g=GM/r^2 from Newton's universal law of gravitation and Newton's second law?
All related (35)
Sort
Recommended
Kaiser Tarafdar
Physician at Covenant Medical Group · Author has 2.6K answers and 5.2M answer views
·Updated 4y
That is rather straightforward really. Let’s do it with your mass and the earth’s mass in mind, and in the final equation your mass will not matter at all ! I will use this line of thinking as a springboard to discuss other remarkable aspects of gravitational physics as related to black holes later in this article. As you are on the surface of the earth your weight which is gravity’s pull on you is given as : W = mg which is from Newton’s second law of motion : F = ma, where F is force in Newtons, m is the mass, and a is the acceleration, here ‘a’ being equal to g which is the acceleration due
Continue Reading
That is rather straightforward really. Let’s do it with your mass and the earth’s mass in mind, and in the final equation your mass will not matter at all ! I will use this line of thinking as a springboard to discuss other remarkable aspects of gravitational physics as related to black holes later in this article. As you are on the surface of the earth your weight which is gravity’s pull on you is given as : W = mg which is from Newton’s second law of motion : F = ma, where F is force in Newtons, m is the mass, and a is the acceleration, here ‘a’ being equal to g which is the acceleration due to gravity. Also from Newton’s law of universal gravitation the force of gravity on you is given as noted : F = GmM/r^2 where G is the universal gravitational constant, m is your mass(or any mass on earth), M is earth’s mass, and r is the distance between you and the earth’s center, which is effectively its radius where you are. Here W, your weight in Newtons is the same as F in the above equation. So let’s see what happens now when we equate : mg = GmM/r^2. Here m cancels out and you end up with the following : g = GM/r^2. We have just derived the equation for acceleration due to gravity on earth in this case which is independent of the smaller mass or the free-falling mass, and is dependent on earth’s mass in direct proportionality, and inversely to the square of the radius. This also tells you that the acceleration due to gravity will be somewhat more in the polar areas than at the equatorial areas as the earth is an oblate spheroid with shorter radius at the poles, and somewhat longer radius at the equator. Similarly this equation can be used for all celestial bodies no matter how massive. Now let’s look at something really enormously gravitationally powerful : A black hole ! The acceleration due to gravity of a black hole, let’s call it g’, as you get near a black hole will be in the hundreds of thousands going into millions… Now let’s look at something profound out of all this : We have to use a little bit of calculus here. If the black hole has a radius of R from its singularity to the event horizon, a mass of M, and let’s say “magically” you are at the event horizon (still alive and kicking) then the gravitational force you will experience will be F = GmM/R^2. What would be your potential energy at R theoretically speaking if you were magically to rise from the inside to the event horizon in an attempt to escape ? It would be given by the work you did to get up there from inside the black hole’s center. This is W = F • d or F • R (as here d is your Radius R). Now you have to integrate the above from 0 to R assuming an average F over that distance which will give you the total work done, or the gravitational potential energy (GPE) as : GPE = ✔️F• R = ✔️(GmM/R^2) • R, going from 0 to R as your range of integration. (I am using the check sign as the integration symbol here). This will give you GPE = -GmM/R, the absolute value of which is GmM/R. The earlier negative sign is derived from integration and is a convention denoting potential energy going away from the center of gravity. Now you would also have to match the speed of light as your maximum possible speed, or your escape velocity, though you will not be able to do that as photons or gravitons can. So let’s say you were able to do that hypothetically, then your kinetic energy would have to equal the GPE at the event horizon level to be able to escape. So let’s look at it : your kinetic energy would have to be K.E. = 1/2mv^2 where v is c, the speed of light theoretically. That makes your K.E. = 1/2mc^2. Let’s now equate GPE and KE as the following shows : GmM/R = 1/2mc^2. We want to solve for R as the event-horizon radius the black hole must have to be a black hole with the gravitational and escape-velocity kinetic energy mechanics in place right there. Solving for R now gives us this beauty : R = 2GM/c^2, as m cancels out from both sides. So we have now derived the famous Schawrzchild equation from 1914 that he worked out from Einstein’s general relativity concepts with Newtonian equations in place for essentially non-rotating black holes. This equation is for the Schwarschild radius which is the radius of the event horizon of a black hole beyond which not even light can escape the infinite gravity of a black hole and exit the event horizon from inside the black hole. The Schwarschild radius can be essentially calculated for any mass no matter how large or small as a theoretical radius for any mass that could turn into a black hole ! Very interestingly a black hole’s average density(excluding the singularity which approaches infinite density) approaches the order of 10^19 kg/m^3 and higher which is two orders(around 100 times more) above the density of the nucleus of an atom with higher atomic number(the density of the atomic nucleus being in the order of 10^17 kg/m^3). Stephen Hawking during his university days once entertained the concept of atoms being like very tiny black hole equivalents ! In reality though black holes are known to only result from the stellar collapse of stars. So we have taken somewhat of a roller-coaster of a ride here starting from more straightforward gravitational physics, to more profound implications, involving concepts of light, gravitation in the extreme environment of black holes involving both Newtonian concepts, and Einstein’s Special and general relativity, and we have landed back on earth safely during this conceptual journey though gravitational physics… Kaiser T, MD (Life long physics, math, cosmology, and science proponent).
Upvote ·
9 3
9 1
Sponsored by Avnet Silica
We're at the Pulse of the Market.
Explore the trends shaping real innovation in AI, automotive & ADAS, 5G, renewables, power, and more.
Learn More
9 5
Wardell Lindsay
Former Electrical Engineer · Author has 492 answers and 116.1K answer views
·6y
Newton’s 2nd Law
Energy E=[c,V][ ,P]=[-vp,cP]
Force F=XE=[d/dr,Del][-mGM/r,cP]
F=[-d/dr mGM/r -cDel.P, cdP/dr -Del mGMR/dr + cDelxP]
F=[mGM/r^2 — cp/r cosV, dPdt + mGMR/r^3 + cDelxPsinV]
These are the five forces:
Centrepetal mGM/r^2 = mg scalar
Centrifugal cp/r cosV scalar
Tangent dP/dt vector
Gravitational force mGMR/r^3 vector
Curl cDel.P sinV rotation vector.
Newton’s 3rd Law is the Stationary Condition!
At Stationary Condition, the Curl =0 and the other vectors are opposite!
Upvote ·
Niels Abildskov
Studied Physics&Mathematics at Gefion Gymnasium (Graduated 2021) · Author has 66 answers and 74.7K answer views
·6y
Related
How did Newton derive the universal law of gravitation?
He considered the question: “If objects on the Earth fall, then does the moon also fall around the Earth?”
He started by thinking about an object falling to Earth. By his second law, the force that the Earth exerts on the object is,
F g=m g F g=m g
Where m is the object’s mass and g is the acceleration towards the Earth. It had earlier been observed that the acceleration towards the Earth was independent of the mass of an object, so we can state that,
F g∝m(1)(1)F g∝m
He then said, by his third law, that the force of gravity is also dependent on the mass of the Earth. This is because, if the Earth ac
Continue Reading
He considered the question: “If objects on the Earth fall, then does the moon also fall around the Earth?”
He started by thinking about an object falling to Earth. By his second law, the force that the Earth exerts on the object is,
F g=m g F g=m g
Where m is the object’s mass and g is the acceleration towards the Earth. It had earlier been observed that the acceleration towards the Earth was independent of the mass of an object, so we can state that,
F g∝m(1)(1)F g∝m
He then said, by his third law, that the force of gravity is also dependent on the mass of the Earth. This is because, if the Earth acts on the object, then the object also acts on the Earth. So we see that,
F g∝M(2)(2)F g∝M
Then he considered that if gravity works on the surface of the Earth, it must also work on the moon. So he said that, for the Moon to stay in orbit, the force of gravity would have to be balanced by the centripetal force, so we see that,
F g=F c e n F g=F c e n
⇒F g=m r(v)2=m r(2 π r T)⇒F g=m r(v)2=m r(2 π r T)
⇒F g=4 π 2 m r T 2⇒F g=4 π 2 m r T 2
Both sides of the equation can now be multiplied by, T 2 r T 2 r, which yields,
T 2 r F g=4 π 2 m T 2 r F g=4 π 2 m
T 2 r T 2 r looks a lot like Kepler’s third law which is,
T 2 r 3=c o n s t a n t≈0.040 T 2 r 3=c o n s t a n t≈0.040
So in order to make that true, the equation would have to be,
T 2 r(k r 2)=4 π 2 m T 2 r(k r 2)=4 π 2 m
Which means that,
F g=k r 2⇒F g∝1 r 2(3)(3)F g=k r 2⇒F g∝1 r 2
Where k is a constant. We see that in (1) and (2), the two masses are constants, so we can collect (1), (2) and (3) to one big proportionality,
F g∝M m r 2 F g∝M m r 2
⇒F g=G M m r 2⇒F g=G M m r 2
Where G is the Universal Constant of Gravitation, and it has the value, G=6.67408⋅10−11 m 3 k g−1 s−2 G=6.67408⋅10−11 m 3 k g−1 s−2.
I hope this answer helped. Just comment if there is any questions.
Upvote ·
99 50
99 10
9 1
Related questions
More answers below
How did Newton derive the universal law of gravitation?
Is Newton’s 2nd law actually equivalent to his gravitational law?
How did Einstein disprove Newton's Universal Law of Gravitation?
Which of Newton's Laws was wrong?
Why newtons second law is fundamental law?
Terry Moore
Former Statistics lecturer at Massey University (1978–2000) · Author has 16.6K answers and 29.4M answer views
·6y
Derive g=GM/r^2 from Newton's universal law of gravitation and Newton's second law?
This looks like an order rather than a question. But anyway…
Let a point object with mass m m be at a distance r r from a point mass M M.
Then by Newton’s second law, m g=G m M/r 2 m g=G m M/r 2, so g=G M/r 2 g=G M/r 2.
Note that you can use the centre of mass for an extended symmetrical mass, but that requires proof. So this would roughly apply to small bodies on the surface of the Earth. A proof of that is not so trivial.
Upvote ·
9 3
9 1
Sponsored by YouTrack JetBrains
YouTrack. Flexible project management for each of your teams.
Cloud or server – YouTrack your way. Free for teams of up to 10.
Learn More
99 40
Assistant
Bot
·
1y
To derive the formula g=G M r 2 g=G M r 2 for gravitational acceleration, we start with two fundamental principles: Newton's universal law of gravitation and Newton's second law of motion.
Newton's Universal Law of Gravitation
Newton's law of gravitation states that the force F F between two masses m 1 m 1 and m 2 m 2 is given by:
F=G m 1 m 2 r 2 F=G m 1 m 2 r 2
where:
F F is the gravitational force between the two masses,
G G is the gravitational constant,
r r is the distance between the centers of the two masses.
Newton's Second Law of Motion
Newton's second law states that the force acting on an object is e
Continue Reading
To derive the formula g=G M r 2 g=G M r 2 for gravitational acceleration, we start with two fundamental principles: Newton's universal law of gravitation and Newton's second law of motion.
Newton's Universal Law of Gravitation
Newton's law of gravitation states that the force F F between two masses m 1 m 1 and m 2 m 2 is given by:
F=G m 1 m 2 r 2 F=G m 1 m 2 r 2
where:
F F is the gravitational force between the two masses,
G G is the gravitational constant,
r r is the distance between the centers of the two masses.
Newton's Second Law of Motion
Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration a a:
F=m a F=m a
Applying the Laws
In the context of gravitational acceleration near a massive body (like the Earth), we can consider one of the masses to be the mass of the Earth M M and the other mass to be an object of mass m m that is experiencing the gravitational force.
Using Newton's law of gravitation for the force acting on the mass m m:
F=G M m r 2 F=G M m r 2
Now, according to Newton's second law, the gravitational force acting on mass m m can also be expressed as:
F=m g F=m g
where g g is the acceleration due to gravity at distance r r from the center of the Earth.
Setting the Forces Equal
Since both expressions represent the same gravitational force acting on the mass m m, we can set them equal to each other:
m g=G M m r 2 m g=G M m r 2
Solving for g g
We can simplify this equation by dividing both sides by m m (assuming m≠0 m≠0):
g=G M r 2 g=G M r 2
Conclusion
Thus, we have derived that the gravitational acceleration g g at a distance r r from the center of a massive body (like the Earth) is given by:
g=G M r 2 g=G M r 2
This formula indicates that the gravitational acceleration is directly proportional to the mass of the body creating the gravitational field and inversely proportional to the square of the distance from its center.
Upvote ·
Alejandro Jenkins
PhD in Physics, California Institute of Technology (Caltech) (Graduated 2006) · Featured on Forbes · Upvoted by
JP Seet
, M.A. Physics & Engineering, University of Cambridge (1996) and
Antariksha Das
, BS-MS Physics, Indian Institute of Science Education and Research, Kolkata (2018) · Author has 498 answers and 14.6M answer views
·Updated 9y
Related
How did Newton derive the universal law of gravitation?
According to a famous anecdote, when Newton was old and famous and someone asked him that question, how he’d arrived at his law of universal gravitation, his response was: “By thinking on it continually.”
It was, in fact, the fruit of many years of remarkable intellectual efforts, born out of Newton’s desire to explain, in terms of simple dynamical principles, why planets going around the Sun are seen to obey Kepler's three laws of planetary motion:
The orbit of a planet is an ellipse, with the Sun at one of the two foci.
A line segment joining a planet and the Sun sweeps out equal areas during e
Continue Reading
According to a famous anecdote, when Newton was old and famous and someone asked him that question, how he’d arrived at his law of universal gravitation, his response was: “By thinking on it continually.”
It was, in fact, the fruit of many years of remarkable intellectual efforts, born out of Newton’s desire to explain, in terms of simple dynamical principles, why planets going around the Sun are seen to obey Kepler's three laws of planetary motion:
The orbit of a planet is an ellipse, with the Sun at one of the two foci.
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
Johannes Kepler had formulated those three laws based on the painstaking astronomical observations carried out by Tycho Brahe, whom Kepler had once worked for as an assistant. But first Kepler had to re-interpret Brahe’s observations in terms of Nicolaus Copernicus’s cosmology, in which the Earth orbits around the Sun in the same way as the other planets. Thus, Newton was working from the astronomical observations of Brahe, as re-interpreted in light of Copernicus’s heliocentric system and summarized in the form of Kepler’s three laws.
At the time, Newton wasn’t the only one eager to account for Kepler’s kinematical laws in terms of a dynamical theory of gravity. Indeed, several people before him had realized that if the planet’s orbit were a perfect circle (which is the limiting case in which the two foci of the ellipse coincide, with the semi-major axis being simply the circle’s radius and with the planet going around at a constant angular velocity), then Kepler’s third law could be explained by an attractive force between the Sun and the planet inversely proportional to the square of their separation. In his Philosophiæ Naturalis Principia Mathematica (“Mathematical Principles of Natural Philosophy”), first published in 1687, Newton mentions Christopher Wren, Robert Hooke, and Edmond Halley as having realized this independently.
But Newton had a couple of aces up his sleeve, relative to others interested in the same problem. For starters, he had a greater conceptual clarity about how to reduce mechanical questions in general to the application of a simple set of dynamical principles. Now we call these principles “Newton's laws of motion”, although Newton didn’t invent them from scratch (nor did he claim to have done so). For example, Hooke's law for the restoring force of an elastic spring was worked out by Hooke in 1660, well before Newton published his general laws of mechanics.
Even more importantly, Newton could use the infinitesimal calculus, having developed it (but not published it) in the 1660s. Moreover, Newton had framed the calculus in terms of orbital geometry and was therefore able to describe mathematically the motion of bodies acted upon by various force laws. This allowed him to demonstrate rigorously that all three of Kepler’s laws follow from an inverse-square law for gravity of the form:
F=G m 1 m 2/r 2,F=G m 1 m 2/r 2,
where m 1 m 1 and m 2 m 2 are the respective masses of any two bodies, r r is the separation between them, and G G is the universal gravitational constant.
Newton also had compared the rate at which objects near the surface of the Earth (e.g., an apple) accelerate downwards with the motion of the Moon around the Earth. He was able to show that both phenomena could be explained in terms of the law of universal gravitation. To make all of this work out, he had had to show (by a brilliant argument) that a uniform spherical body gravitates as if all of its mass were concentrated at its center (see shell theorem).
Halley travelled from London to Cambridge in August of 1684 because he’d heard that Newton, a reclusive Cambridge professor who’d published important work on optics twenty years before and was rumored to be a mathematical genius, might have something interesting to tell him about celestial mechanics. Halley asked Newton what kind of orbit a planet would follow if it were subject to a attractive force towards the Sun that were inversely proportional to the square of the distance between to the two bodies. Newton calmly replied that he’d shown that the orbit would be an ellipse. Astounded, Halley asked to see the proof. Newton looked for his notes on the subject but couldn’t find them, so he offered to forward them to Halley later. Newton ended up having to re-derive the result, and when Halley saw Newton’s proof he insisted that he communicate it to the Royal Society. Halley then also pushed Newton to write the Principia, the single most important work in the history of physics. Halley even ended up paying for the book’s printing with his own money, as the Royal Society was low on funds.
Illustration from Newton’s “The System of the World” (appended to the English translation of the Principia), corresponding to a thought experiment intended to explain how a projectile near the Earth’s surface obeys the same mathematical laws as a body (such as the Moon) orbiting around the Earth. This is now called “Newton's cannonball”.
In the Principia, Newton went on to show that the same law for gravity could explain the orbits of comets, the tides, and the Earth’s axial precession. He even made progress towards explaining the irregularities in the motion of the Moon in terms of the perturbations exerted upon it by the Sun’s gravitational pull. (That was the first instance of the famously difficult three-body problem, which occupies mathematicians and physicists even today.)
Newton also used the law of universal gravitation to predict how the shape of the Earth deviates from perfect sphericity (see figure of the Earth). In the 1710s and 1720s, the Cassinis, a family of distinguished astronomers originally from Italy (see Giovanni Domenico Cassini) who’d settled in France and who controlled the Paris Observatory, claimed to have determined by their measurements that the Earth was prolate (shaped like a rugby ball), rather than oblate (flattened at the poles) as predicted by Newton; see spheroid. French savants invoked the Cassinis’ measurements in defense of René Descartes’s physical system against Newtonianism.
Newton’s predictions were decisively vindicated around 1740 by two French-led geodesic expeditions. One went to Lapland, in northern Scandinavia, and was led by Pierre Louis Maupertuis. The other traveled to Quito, in what is now Ecuador and was then part of the Spanish Empire in South America, and was led by Pierre Bouguer and Charles Marie de La Condamine.
Geodetic measurements carried out in Meänmaa during the expedition led by Maupertuis. The caption translates as “Map of the meridian arc measured at the polar circle”.
The writer and philosopher Voltaire, an early advocate of Newtonianism in France, had been a friend of Maupertuis and when he returned from Lapland Voltaire announced: Il a aplati la terre et les Cassinis (“He has flattened the Earth and the Cassinis”). But Voltaire later fell out with Maupertuis and famously mocked him in verse:
Héros de la Physique, argonautes nouveaux, / Qui franchissez les monts, qui traversez les eaux, / Ramenez des climats soumis aux trois couronnes, / Vos perches, vos secteurs et surtout deux Lapones, / Vous avez confirmé dans ces lieux pleins d'ennui, / Ce que Newton connut sans sortir de chez lui / Vous avez arpenté quelque faible partie / Des flancs toujours glacés de la Terre aplatie.
(“Heros of physics, new Argonauts, who cross mountains, who traverse the seas, bring back from climates ruled by the three crowns your staffs, your [astronomical] sectors, and most of all two Lapp women. You have confirmed in those frozen wastes what Newton knew without leaving his home. You have surveyed some small part of the ever frozen flanks of the flattened Earth.”)
This is amusing but unfair. It was the experimental proof, procured by the great efforts of Maupertuis and others, that established Newton’s law of universal gravitation as true scientific knowledge. (Voltaire also ridicules Maupertuis for reportedly taking a Lapp mistress, whom he brought to Paris along with another local woman.)
Actually, Maupertuis’s Arctic mission was a cakewalk compared to the equatorial expedition, which ended up taking ten years and faced all kinds of financial, personal, political, and practical difficulties. One of the members of the expedition married a woman of the local Creole high society, Isabel Godin des Odonais, but was soon separated from her and stranded in French Guiana. Isabel’s incredible odyssey through the Amazon to rejoin her husband is told in Robert Whitaker’s The Mapmaker's Wife: A True Tale Of Love, Murder, And Survival In The Amazon (2004), a real-life adventure story with a fascinating historical and scientific background.
Another spectacular vindication of Newton’s theory was that in 1705 Halley used it to determine the orbit of a comet and to predict its return in 1758. The comet did in fact return nearly as predicted (with a minor delay produced by the interference of Jupiter’s and Saturn’s gravity), by which time Newton and Halley were long dead. It has been known since as Halley's Comet.
Memorial to Edmond Halley (1656–1742) in Westminster Abbey, London. Source: The Scientific Tourist in London: #8 Edmond Halley Memorial
Upvote ·
999 171
9 4
99 15
Related questions
More answers below
What is the law of universal gravitation by Newton? Describe it clearly. Use examples.
Does Newton's Universal Law of Gravitation ever fail?
Can you derive Newton's law of universal gravitation from Kepler's second law?
Why is Newton’s gravitational law wrong?
What is the mathematical derivation of Newton's universal law of gravitation?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Related
Can you derive this equation, Fg = Gm1m2/r^2, and how did Newton come up with this equation?
In my opinion, I would say that this equation is the result of experiments he did.
I expect he considered two masses hanging vertically and found that the attractive force between them was proportional to the product of their masses. This is expressed by this equation…
Then by varying the distance between the masses, he would have found that the force is also inversely proportional to the square of the distance, r, between them. This is expressed by this equation…
He would have expected this to apply not just to masses in his experiments but also to the planets! So this constant of proportion is
Continue Reading
In my opinion, I would say that this equation is the result of experiments he did.
I expect he considered two masses hanging vertically and found that the attractive force between them was proportional to the product of their masses. This is expressed by this equation…
Then by varying the distance between the masses, he would have found that the force is also inversely proportional to the square of the distance, r, between them. This is expressed by this equation…
He would have expected this to apply not just to masses in his experiments but also to the planets! So this constant of proportion is called the Universal Gravitational constant!
Upvote ·
99 12
9 2
9 3
Sponsored by All Out
Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out.
Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association.
Shop Now
999 621
Hannibal
40+years of study in Physics, Aviation, Meteorology, Chemistry · Author has 10.1K answers and 1.1M answer views
·2y
Related
According to Newton's universal gravitational law, F=Gm1m2/r2. What are the origins of these forces?
Using the earth, its mass, and its radius, and applying it to all other objects in the universe.
Double the mass and you double the gravitational force.
Gravitational force is inversely proportional to the square of the distance from the center of any 2 massive objects like earth/sun
earth/moon.
These calculations are then applied to all other objects in the
solar system, provided the radius is known. All mass warps the fabric of space-time. Thus producing gravity, thus producing an acceleration DUE to gravity, and on earth???, that acceleration is 9.80665m/s²
Moon?? 1.67m/s²
sun? 275m/s²
Black hole?
Continue Reading
Using the earth, its mass, and its radius, and applying it to all other objects in the universe.
Double the mass and you double the gravitational force.
Gravitational force is inversely proportional to the square of the distance from the center of any 2 massive objects like earth/sun
earth/moon.
These calculations are then applied to all other objects in the
solar system, provided the radius is known. All mass warps the fabric of space-time. Thus producing gravity, thus producing an acceleration DUE to gravity, and on earth???, that acceleration is 9.80665m/s²
Moon?? 1.67m/s²
sun? 275m/s²
Black hole? Infinity
FASCINATING!😎👍
Upvote ·
9 2
9 1
Ron Brown
Decades of teaching physics to undergrads · Author has 13.6K answers and 84.3M answer views
·2y
Related
How is Newton's second law of motion a simplification of Newton's law of universal gravitation. Universal gravitation: F=a m1m2/r^2. Second law of motion: F= m x a?
It isn’t. The two laws are independent of one another.
Newton’s second law simply says that objects accelerate in proportion to the net force that acts on them. So a=F n e t/m a=F n e t/m.
Newton’s law of universal gravitation says that the gravitational force between any two masses is proportional to the product of the masses divided by the square of their separation.
Now gravity, of course, can be one of the forces that acts on an object. And if the gravitational force is the only force that acts on some mass m 1 m 1 due to its interaction with mass m 2 m 2, then one could combine the two laws to determine the acc
Continue Reading
It isn’t. The two laws are independent of one another.
Newton’s second law simply says that objects accelerate in proportion to the net force that acts on them. So a=F n e t/m a=F n e t/m.
Newton’s law of universal gravitation says that the gravitational force between any two masses is proportional to the product of the masses divided by the square of their separation.
Now gravity, of course, can be one of the forces that acts on an object. And if the gravitational force is the only force that acts on some mass m 1 m 1 due to its interaction with mass m 2 m 2, then one could combine the two laws to determine the acceleration of m 1 m 1. That is, m 1 a=G m 1 m 2/r 2 m 1 a=G m 1 m 2/r 2 would show that the acceleration of m 1 m 1 would equal G m 2/r 2 G m 2/r 2.
Upvote ·
9 5
9 1
9 1
Sponsored by State Bank of India
This Festive Season, Make Your Dream Home a Reality!
Apply for an SBI Home Loan—easy, quick & hassle-free. Celebrate in a home you can call your own.
Learn More
99 66
Ron Brown
Decades of teaching physics to undergrads · Upvoted by
Florens de Wit
, M.S. Physics, Eindhoven University of Technology (1999) · Author has 13.6K answers and 84.3M answer views
·2y
Related
How does "a" in Newton's second law of motion relate to "g" (9.8m/s^2)?
It sounds as if the question is coming from beginning student. Here’s a suggestion - focus on the ideas, not the equations. The equations are just ways of writing the ideas in a form that can be used in problem solving.
Now to your question: Newton’s second law states that the net force on an object causes it to accelerate proportional to the net force and inversely proportional to its mass. That is, a=F n e t/m a=F n e t/m. Or written more commonly, F n e t=m a F n e t=m a. When the only force on an object is the gravitational force, the acceleration of the mass is usually expressed as g g (which equals 9.8 m/s 2 m/s 2, appro
Continue Reading
It sounds as if the question is coming from beginning student. Here’s a suggestion - focus on the ideas, not the equations. The equations are just ways of writing the ideas in a form that can be used in problem solving.
Now to your question: Newton’s second law states that the net force on an object causes it to accelerate proportional to the net force and inversely proportional to its mass. That is, a=F n e t/m a=F n e t/m. Or written more commonly, F n e t=m a F n e t=m a. When the only force on an object is the gravitational force, the acceleration of the mass is usually expressed as g g (which equals 9.8 m/s 2 m/s 2, approximately). So most textbooks will then just write the gravitational force as m g m g. What that really is saying is, that when the net force is only the gravitational force, F n e t=F g F n e t=F g then m a=m g m a=m g, so the acceleration is g g, the free fall acceleration in a gravitational field near the surface of the Earth.
And here is another suggestion: When solving problems, just leave that symbol g g rather than substituting in the numerical value - up until you do a final calculation to get a numerical answer. It will help you understand the problem and where everything came from.
Upvote ·
9 7
9 1
Smitaj
Bsc in Physics&Mathematics, Monash University (Graduated 1973) · Author has 25.6K answers and 18.6M answer views
·3y
Related
Can you derive Newton's law of universal gravitation from Kepler's second law?
If you read “Principia” Newton derived his law of Gravitation from Kepler’s THIRD law (T is proportional to R^(3/2) and from Newtons law of circular motion F is proportional to V^2/R And from the obvious that V is proportional to R/T
therefore F is proportional to (R / R^(3/2) )^2 / R -> R ^ ( 2 - 6/2 -1) -> R^-2 ie 1/ R^2
The other part was from the knowledge that F is proportional to the mass m of a dropped mass. Otherwise the acceleration in a gravitational field would differ for different masses.
And finally due to symmetry ( every action has an equal and opposite reaction ) that by simply ch
Continue Reading
If you read “Principia” Newton derived his law of Gravitation from Kepler’s THIRD law (T is proportional to R^(3/2) and from Newtons law of circular motion F is proportional to V^2/R And from the obvious that V is proportional to R/T
therefore F is proportional to (R / R^(3/2) )^2 / R -> R ^ ( 2 - 6/2 -1) -> R^-2 ie 1/ R^2
The other part was from the knowledge that F is proportional to the mass m of a dropped mass. Otherwise the acceleration in a gravitational field would differ for different masses.
And finally due to symmetry ( every action has an equal and opposite reaction ) that by simply changing which one is the dropped mass it must also be proportional to M
hence F is proportional to Mm/ r^2
there is the law of universal gravitation as explained by Newton.
Upvote ·
9 4
Satya Parkash Sud
M.Sc. in Physics&Nuclear Physics, University of Delhi (Graduated 1962) · Author has 8.1K answers and 27.4M answer views
·1y
Related
What is the formula of the law of Newton's universal gravitation?
The Newton's law of Gravitation states, “The force of attraction between masses m and M is directly proportional to the product of their masses and inversely proportional to the square of distance separating the two masses.
The magnitude of force of attraction between m and M, with a seperation of r between the centre of mass of m and centre of mass of M is according to the Newton’s law is
F = G [(m
Continue Reading
The Newton's law of Gravitation states, “The force of attraction between masses m and M is directly proportional to the product of their masses and inversely proportional to the square of distance separating the two masses.
The magnitude of force of attraction between m and M, with a seperation of r between the centre of mass of m and centre of mass of M is according to the Newton’s law is
F = G [(mM)/r²]
where G = constant of gravitation.
Units of G
G = (force × r²)/(mass²) = Newton - m²/kg²
Value of G in SI units = 6.67 × 10^(- 11) Newton - meter²/kilogram² = 6.67 × 10^(- 11) N - m²/kg².
The same force acts on m an...
Upvote ·
9 1
9 1
Ron Brown
Decades of teaching physics to undergrads · Author has 13.6K answers and 84.3M answer views
·3y
Related
How can we calculate the weight of a planet using Newton's second law and F=ma?
You can’t. For one thing F=m a F=m a is Newton’s second law - not a different idea. But if you had said Newtons gravitational force law and Newton’s second law, then one can determine the mass of a planet from the motion of a satellite’s orbit about the planet - that is, the radius of the orbit and the time it takes to complete an orbit.
That is, the gravitational force law says the force acting on an orbiting object to pull it into a circular orbit is just F g=G(m M/r 2)F g=G(m M/r 2),where M M is the mass of the planet, m m is the mass of the orbiting object (a moon or satellite), and r r is the radius of the (assumed)
Continue Reading
You can’t. For one thing F=m a F=m a is Newton’s second law - not a different idea. But if you had said Newtons gravitational force law and Newton’s second law, then one can determine the mass of a planet from the motion of a satellite’s orbit about the planet - that is, the radius of the orbit and the time it takes to complete an orbit.
That is, the gravitational force law says the force acting on an orbiting object to pull it into a circular orbit is just F g=G(m M/r 2)F g=G(m M/r 2),where M M is the mass of the planet, m m is the mass of the orbiting object (a moon or satellite), and r r is the radius of the (assumed) circular orbit. G G in that equation is the universal gravitational constant which has been experimentally determined. But if the satellite or moon is in a circular orbit, its acceleration is just given by the centripetal acceleration equation, a c=v 2/r a c=v 2/r, where v v is its orbital speed.
So now enters Newton’s second law: That is, the gravitational force results in that centripetal acceleration so F=m a F=m a becomes F g=G(m M/r 2)=m v 2/r F g=G(m M/r 2)=m v 2/r. But the orbital speed of the satellite is just 2 π r/T 2 π r/T, where T T is the orbital period. So the mass of the planet can be solved for by combining those equations. That is, M=(4 π 2/G)(r 3/T 2)M=(4 π 2/G)(r 3/T 2). So knowing the orbit radius and the period of the orbit allows one to determine the mass of the planet. Notice it doesn’t depend on the mass of the orbiting object.
Are there assumptions in that derivation? Of course. It assumes a circular orbit. It assumes the mass of the satellite or moon (m m) was small compared to the mass of the planet (M M), so that the center of the orbit is essentially the center of the planet. The equation obtained for M M is the equivalent of Kepler’s third law, showing the relationship between the radius and the period for an orbiting object - although Kepler deduced that by observing the planetary orbits about the Sun. So the result is equally valid for objects orbiting a planet as it is for objects orbiting a star or a star orbiting a galaxy, and that is one of the ways masses of stars can be determined. That is, we can know the mass of the Sun, the mass of the Earth, the masses of the other planets, even the masses of the moons of the other planets if we have put satellites in orbit about them from that relationship.
Upvote ·
9 2
Related questions
How did Newton derive the universal law of gravitation?
Is Newton’s 2nd law actually equivalent to his gravitational law?
How did Einstein disprove Newton's Universal Law of Gravitation?
Which of Newton's Laws was wrong?
Why newtons second law is fundamental law?
What is the law of universal gravitation by Newton? Describe it clearly. Use examples.
Does Newton's Universal Law of Gravitation ever fail?
Can you derive Newton's law of universal gravitation from Kepler's second law?
Why is Newton’s gravitational law wrong?
What is the mathematical derivation of Newton's universal law of gravitation?
What is an intuitive explanation of Newton's second law of motion?
Is Newton law of gravitation called universal law?
Can you explain Newton's universal law of gravitation and his second law? How do they differ?
What is Newton's Law of Gravitation?
What is force according to Newton's second law?
Related questions
How did Newton derive the universal law of gravitation?
Is Newton’s 2nd law actually equivalent to his gravitational law?
How did Einstein disprove Newton's Universal Law of Gravitation?
Which of Newton's Laws was wrong?
Why newtons second law is fundamental law?
What is the law of universal gravitation by Newton? Describe it clearly. Use examples.
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
Always Active
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
Always Active
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
Always Active
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
7785 | https://www.aopa.org.au/documents/item/694 | What is an acute spinal column injury?
• An acute spinal column injury occurs when there is damage to the structures of the spinal column, including bone (vertebrae), ligaments and soft tissue. Typically, a spinal column injury occurs due to trauma. • The most common type of spinal column injury is a compression fracture of the thoracic spine. • Spinal fractures can also occur due to distraction, translation or rotational forces, or any combination of these.
What is the role of orthoses in managing acute spinal column injuries?
The immediate goal of orthotic management of acute spinal column injury is to immobilise and stabilise the spine through an externally applied force, to prevent further injury, aid healing and minimise pain. Orthotists use biomechanical principles and knowledge of anatomy, physiology and materials science to adapt orthotic designs to individual patient requirements and ensure the correct positioning and application. Correct positioning and application is essential to minimise the risk of further collapse or mal-alignment of the spinal column. Spinal orthoses may be used as part of non-operative or post-operative management.
What are spinal orthoses?
A spinal orthosis is an externally applied device, designed and fitted to the body to protect and support the spine after injury or following surgery. Spinal orthoses can be applied to any section of the spine (cervical, thoracic, lumbar, sacral or all of the above)
Clinical Specialties in Orthotics and Prosthetics
Orthoses to immobilise and heal acute spinal column injuries
www.aopa.org.au 1 of 2
and are named by the anatomical body part over which they are acting. Spinal orthoses may be commercially produced or custom made. They are prescribed with consideration of the location and severity of the injury, how the orthosis is intended to act and individual patient characteristics. Spinal orthoses for management of acute spinal column injuries include: • Cervical orthoses (CO) • Cervico-thoracic orthoses (CTO) including halo thoracic orthoses (HTO) • Thoraco-lumbo-sacral orthoses (TLSO) • Lumbo-sacral orthoses (LSO).
Cervical orthosis (CO)
COs (or cervical collars) are often fitted pre-hospital admission by emergency staff, including ambulance officers and paramedics to immobilise the cervical spine in cases of suspected spinal column injury. After a patient is admitted to hospital, orthotists are involved in fitting custom-made or customised orthoses to provide immobilisation at any spinal level. COs are classified as either ‘soft’ – providing comfort and proprioception (awareness of body position) but offering minimal immobilisation; or rigid – providing moderate immobilisation of the mid-cervical spine. Most rigid COs are made of bi-valved plastic shells with removable pads. Examples of rigid COs include the ‘Philadelphia’ collar, the ‘Miami J’ collar, the ‘Aspen’ cervical orthosis and the ‘Malibu’ collar.
Cervico-thoracic orthosis (CTO)
CTOs generally consist of supports to the chin and base of the skull which are attached to a thoracic vest. Compared with rigid COs, these orthoses improve control of the spine, particularly the middle to lower cervical spine and upper thoracic spine. Examples include the sterno-occipito-mandibular immobiliser, or ‘SOMI’, the ‘Minerva’ CTO and the ‘Miami’ JTO.
The different regions of the spine or vertebral column. A ‘Philadelphia’ Collar, a type of Cervical Collar.
Orthoses for people with a history of polio –
providing stability and mobility
Clinical Specialties in Orthotics and Prosthetics
What is polio?
• Polio, the common name for poliomyelitis, is a neuromuscular
condition caused by infection with the polio virus.
• Poliomyelitis can damage the nerves that control muscles,
causing total or partial weakness. Polio most often affects leg
muscles, but it is not unusual to also experience weak muscles
in the arms or chest/back. The nerves controlling involuntary
muscles, such as the heart, are not affected by polio.
• Initial recovery from polio generally involves return of some
strength, followed by a period of stability. Later in life it is very
common for people with a history of polio to experience the
‘late effects of polio’.
• The polio vaccine was introduced to Australia in 1956 and since
the year 2000, Australia has been declared polio-free.
What are the late effects of polio?
The ‘late effects of polio’ is a term used to describe the physical
changes that occur in an ageing, polio-affected body. There is a
gradual return of symptoms such as weakness, fatigue and pain,
typically at around 50 to 60 years of age. These new symptoms
usually reflect ‘wear and tear’ activity. A contributing factor is often
the effect of lifelong ‘compensations’ for weakness, for example
by limping, which causes changes in body posture and results in
body structures wearing-out prematurely. Because of the changes
in their body, these compensations are no longer a suitable way
for a person affected by polio to manage their current physical
condition. It is at this time, and for this reason, that orthoses are
most useful to provide a new way to compensate for the new
challenges of late effects of polio. Post-polio syndrome is the
most well-known of the late effects of polio, characterised by the
sudden onset of new or increased weakness, pain and/or fatigue.
Post-polio syndrome can occur between 15 and 30 years after
recovering from acute polio.
Who is affected by the late effects of polio?
Most polio survivors with residual weakness will experience the
late effects of polio to varying degrees; however, it is more likely
to affect people who were severely paralysed by polio as a child.
Muscles weakened by polio are expected to gradually deteriorate
in strength by 1-2% each year from the age of 65.
What are orthoses?
An orthosis (often called a splint, brace or calliper) is a supportive
device fitted to the body to achieve one or more of the following:
protect and support a body part, compensate for changes in
muscle length or function, re-align skeletal joints, redistribute
pressure or optimise walking pattern. Orthoses are generically
named for the body part over which they act. Orthoses comprise
many types of materials and may be commercially produced and
customised, or custom-made for the individual.
How do orthoses help people
experiencing the late effects of polio?
The primary aim of an orthosis for someone who is experiencing
the late effects of polio is to compensate for weak muscles and
protect painful joints in order to improve mobility and function.
Other benefits include reducing the likelihood of a trip or fall,
and conserving energy by making activities such as walking,
more efficient.
People with a history of polio often experience weakness in the
muscles that control the knee. Both weak thigh and calf muscles
may result in knee instability which can result in a fall. A common
compensation that stops the knee collapsing is to thrust the knee
backwards; however, this movement can damage the knee joint
and cause further pain and joint problems.
People with polio related weakness also experience weakness
in the muscles controlling the foot and ankle. Because they are
unable to lift their toes they must compensate by lifting their leg
higher to make sure they don’t catch their toes on the ground.
A knee-ankle-foot orthosis (KAFO) can help compensate for these
weaknesses. Conventional KAFOs restrict knee movement to
improve stability for walking.
www.aopa.org.au 1 of 2 What is an acute spinal column injury?
• An acute spinal column injury occurs when there is damage to the structures of the spinal column, including bone (vertebrae), ligaments and soft tissue. Typically, a spinal column injury occurs due to trauma. • The most common type of spinal column injury is a compression fracture of the thoracic spine. • Spinal fractures can also occur due to distraction, translation or rotational forces, or any combination of these.
What is the role of orthoses in managing acute spinal column injuries?
The immediate goal of orthotic management of acute spinal column injury is to immobilise and stabilise the spine through an externally applied force, to prevent further injury, aid healing and minimise pain. Orthotists use biomechanical principles and knowledge of anatomy, physiology and materials science to adapt orthotic designs to individual patient requirements and ensure the correct positioning and application. Correct positioning and application is essential to minimise the risk of further collapse or mal-alignment of the spinal column. Spinal orthoses may be used as part of non-operative or post-operative management.
What are spinal orthoses?
A spinal orthosis is an externally applied device, designed and fitted to the body to protect and support the spine after injury or following surgery. Spinal orthoses can be applied to any section of the spine (cervical, thoracic, lumbar, sacral or all of the above)
Clinical Specialties in Orthotics and Prosthetics
Orthoses to immobilise and heal acute spinal column injuries
www.aopa.org.au 1 of 2
and are named by the anatomical body part over which they are acting. Spinal orthoses may be commercially produced or custom made. They are prescribed with consideration of the location and severity of the injury, how the orthosis is intended to act and individual patient characteristics. Spinal orthoses for management of acute spinal column injuries include: • Cervical orthoses (CO) • Cervico-thoracic orthoses (CTO) including halo thoracic orthoses (HTO) • Thoraco-lumbo-sacral orthoses (TLSO) • Lumbo-sacral orthoses (LSO).
Cervical orthosis (CO)
COs (or cervical collars) are often fitted pre-hospital admission by emergency staff, including ambulance officers and paramedics to immobilise the cervical spine in cases of suspected spinal column injury. After a patient is admitted to hospital, orthotists are involved in fitting custom-made or customised orthoses to provide immobilisation at any spinal level. COs are classified as either ‘soft’ – providing comfort and proprioception (awareness of body position) but offering minimal immobilisation; or rigid – providing moderate immobilisation of the mid-cervical spine. Most rigid COs are made of bi-valved plastic shells with removable pads. Examples of rigid COs include the ‘Philadelphia’ collar, the ‘Miami J’ collar, the ‘Aspen’ cervical orthosis and the ‘Malibu’ collar.
Cervico-thoracic orthosis (CTO)
CTOs generally consist of supports to the chin and base of the skull which are attached to a thoracic vest. Compared with rigid COs, these orthoses improve control of the spine, particularly the middle to lower cervical spine and upper thoracic spine. Examples include the sterno-occipito-mandibular immobiliser, or ‘SOMI’, the ‘Minerva’ CTO and the ‘Miami’ JTO.
The different regions of the spine or vertebral column. A ‘Philadelphia’ Collar, a type of Cervical Collar.
Orthotist/Prosthetists – Supporting the Australian community
Sometimes these KAFOs have a ‘stiff-knee’ and sometimes they
have a ‘free moving’ knee, depending on the requirements and
types of muscle weakness experienced by the individual. Other
KAFOs can include technology that ensures the knee is stiff when
weight is being put through the leg and moves freely when the
leg needs to bend. These KAFOs include an ankle section which
prevents the toes from dragging. KAFOs are usually made from
thermoplastics or carbon fibre material. These devices are always
custom made for the needs of the individual.
Who provides orthoses for post-polio
syndrome?
Orthoses are provided by orthotists. Orthotists are the only
specialist tertiary qualified allied health practitioners in Australia
who prescribe the full range of orthoses. Orthotists are trained at
either a Bachelor or Master’s level and may work autonomously or
within the multidisciplinary team.
If you need to use the services of an orthotist they will:
• Perform a thorough clinical assessment
• Discuss the most suitable orthotic options to meet your goals
and requirements and support you in decision making
• Complete the digitisation/measurement/casting process and
oversee the manufacturing or procurement of the orthosis
• Provide clinical services associated with fitting, education of
use, regular reviews for functional effectiveness and adjustment
for fit, as well as liaising with other relevant members of the
healthcare team
How do I access orthotic treatment?
The Australian Orthotic Prosthetic Association (AOPA) is the
peak body regulating orthotists/prosthetists in Australia.
Membership is conditional upon tertiary training at University
level and meeting minimum professional competencies.
Members are required to abide by a standards including a code
of ethics and continuing
professional development.
Certified Orthotist/
Prosthetists (c-OP AOPA)
can be located using
the ‘Find a practitioner’
search function on
the AOPA website
(www.aopa.org.au ).
Orthotists working in
both the public and
private settings are listed.
If you require the services
of an orthotist you may be
referred by your specialist
clinic, physiotherapist or
other health professional
to one of our members
or you may contact them
directly yourself.
Disclaimer – This fact sheet does not replace clinical advice. If you require orthotic services AOPA
reccomends speaking to your practitioner. This fact sheet was developed based on interpretation
of current evidence as of May 2018. References available on request.
www.aopa.org.au 2 of 2 |
7786 | https://www.cis.upenn.edu/~cis5150/cis515-11-sl4.pdf | Chapter 4 Vector Norms and Matrix Norms 4.1 Normed Vector Spaces In order to define how close two vectors or two matrices are, and in order to define the convergence of sequences of vectors or matrices, we can use the notion of a norm.
Recall that R+ = {x ∈R | x ≥0}.
Also recall that if z = a + ib ∈C is a complex number, with a, b ∈R, then z = a −ib and |z| = √ a2 + b2 (|z| is the modulus of z).
207 208 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Definition 4.1. Let E be a vector space over a field K, where K is either the field R of reals, or the field C of com-plex numbers. A norm on E is a function : E →R+, assigning a nonnegative real number uto any vector u ∈E, and satisfying the following conditions for all x, y, z ∈E: (N1) x≥0, and x= 0 iffx = 0.
(positivity) (N2) λx= |λ| x.
(scaling) (N3) x + y≤x+ y.
(triangle inequality) A vector space E together with a norm is called a normed vector space.
From (N3), we easily get |x−y| ≤x −y.
4.1. NORMED VECTOR SPACES 209 Example 4.1.
1. Let E = R, and x= |x|, the absolute value of x.
2. Let E = C, and z= |z|, the modulus of z.
3. Let E = Rn (or E = Cn). There are three standard norms.
For every (x1, . . . , xn) ∈E, we have the 1-norm x1, defined such that, x1 = |x1| + · · · + |xn|, we have the Euclidean norm x2, defined such that, x2 = |x1|2 + · · · + |xn|21 2 , and the sup-norm x∞, defined such that, x∞= max{|xi| | 1 ≤i ≤n}.
More generally, we define the p-norm (for p ≥1) by xp = (|x1|p + · · · + |xn|p)1/p.
There are other norms besides the p-norms; we urge the reader to find such norms.
210 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Some work is required to show the triangle inequality for the p-norm.
Proposition 4.1. If E is a finite-dimensional vector space over R or C, for every real number p ≥1, the p-norm is indeed a norm.
The proof uses the following facts: If q ≥1 is given by 1 p + 1 q = 1, then (1) For all α, β ∈R, if α, β ≥0, then αβ ≤αp p + βq q .
(∗) (2) For any two vectors u, v ∈E, we have n i=1 |uivi| ≤up vq .
(∗∗) 4.1. NORMED VECTOR SPACES 211 For p > 1 and 1/p + 1/q = 1, the inequality n i=1 |uivi| ≤ n i=1 |ui|p 1/p n i=1 |vi|q 1/q is known as H¨ older’s inequality.
For p = 2, it is the Cauchy–Schwarz inequality.
Actually, if we define the Hermitian inner product −, − on Cn by u, v= n i=1 uivi, where u = (u1, . . . , un) and v = (v1, . . . , vn), then |u, v| ≤ n i=1 |uivi| = n i=1 |uivi|, so H¨ older’s inequality implies the inequality |u, v| ≤up vq also called H¨ older’s inequality, which, for p = 2 is the standard Cauchy–Schwarz inequality.
212 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS The triangle inequality for the p-norm, n i=1 (|ui+vi|)p 1/p ≤ n i=1 |ui|p 1/p + n i=1 |vi|q 1/q , is known as Minkowski’s inequality.
When we restrict the Hermitian inner product to real vectors, u, v ∈Rn, we get the Euclidean inner product u, v= n i=1 uivi.
It is very useful to observe that if we represent (as usual) u = (u1, . . . , un) and v = (v1, . . . , vn) (in Rn) by column vectors, then their Euclidean inner product is given by u, v= uv = vu, and when u, v ∈Cn, their Hermitian inner product is given by u, v= v∗u = u∗v.
4.1. NORMED VECTOR SPACES 213 In particular, when u = v, in the complex case we get u2 2 = u∗u, and in the real case, this becomes u2 2 = uu.
As convenient as these notations are, we still recommend that you do not abuse them; the notation u, vis more intrinsic and still “works” when our vector space is infinite dimensional.
Proposition 4.2. The following inequalities hold for all x ∈Rn (or x ∈Cn): x∞≤x1 ≤nx∞, x∞≤x2 ≤√nx∞, x2 ≤x1 ≤√nx2.
214 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Proposition 4.2 is actually a special case of a very impor-tant result: in a finite-dimensional vector space, any two norms are equivalent.
Definition 4.2. Given any (real or complex) vector space E, two norms a and b are equivalent iffthere exists some positive reals C1, C2 > 0, such that ua ≤C1 ub and ub ≤C2 ua , for all u ∈E.
Given any norm on a vector space of dimension n, for any basis (e1, . . . , en) of E, observe that for any vector x = x1e1 + · · · + xnen, we have x= x1e1 + · · · + xnen≤C x1 , with C = max1≤i≤n eiand x1 = x1e1 + · · · + xnen= |x1| + · · · + |xn|.
The above implies that | u−v| ≤u −v≤C u −v1 , which means that the map u →uis continuous with respect to the norm 1.
4.1. NORMED VECTOR SPACES 215 Let Sn−1 1 be the unit ball with respect to the norm 1, namely Sn−1 1 = {x ∈E | x1 = 1}.
Now, Sn−1 1 is a closed and bounded subset of a finite-dimensional vector space, so by Bolzano–Weiertrass, Sn−1 1 is compact.
On the other hand, it is a well known result of analysis that any continuous real-valued function on a nonempty compact set has a minimum and a maximum, and that they are achieved.
Using these facts, we can prove the following important theorem: Theorem 4.3. If E is any real or complex vector space of finite dimension, then any two norms on E are equivalent.
Next, we will consider norms on matrices.
216 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS 4.2 Matrix Norms For simplicity of exposition, we will consider the vector spaces Mn(R) and Mn(C) of square n × n matrices.
Most results also hold for the spaces Mm,n(R) and Mm,n(C) of rectangular m × n matrices.
Since n × n matrices can be multiplied, the idea behind matrix norms is that they should behave “well” with re-spect to matrix multiplication.
Definition 4.3. A matrix norm on the space of square n×n matrices in Mn(K), with K = R or K = C, is a norm on the vector space Mn(K) with the additional property that AB≤AB, for all A, B ∈Mn(K).
Since I2 = I, from I= I2 ≤I2, we get I≥1, for every matrix norm.
4.2. MATRIX NORMS 217 Before giving examples of matrix norms, we need to re-view some basic definitions about matrices.
Given any matrix A = (aij) ∈Mm,n(C), the conjugate A of A is the matrix such that Aij = aij, 1 ≤i ≤m, 1 ≤j ≤n.
The transpose of A is the n × m matrix Asuch that A ij = aji, 1 ≤i ≤m, 1 ≤j ≤n.
The conjugate of A is the n × m matrix A∗such that A∗= (A) = (A).
When A is a real matrix, A∗= A.
A matrix A ∈Mn(C) is Hermitian if A∗= A.
If A is a real matrix (A ∈Mn(R)), we say that A is symmetric if A= A.
218 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS A matrix A ∈Mn(C) is normal if AA∗= A∗A, and if A is a real matrix, it is normal if AA= AA.
A matrix U ∈Mn(C) is unitary if UU ∗= U ∗U = I.
A real matrix Q ∈Mn(R) is orthogonal if QQ= QQ = I.
Given any matrix A = (aij) ∈Mn(C), the trace tr(A) of A is the sum of its diagonal elements tr(A) = a11 + · · · + ann.
It is easy to show that the trace is a linear map, so that tr(λA) = λtr(A) and tr(A + B) = tr(A) + tr(B).
4.2. MATRIX NORMS 219 Moreover, if A is an m × n matrix and B is an n × m matrix, it is not hard to show that tr(AB) = tr(BA).
We also review eigenvalues and eigenvectors. We con-tent ourselves with definition involving matrices. A more general treatment will be given later on (see Chapter 8).
Definition 4.4. Given any square matrix A ∈Mn(C), a complex number λ ∈C is an eigenvalue of A if there is some nonzero vector u ∈Cn, such that Au = λu.
If λ is an eigenvalue of A, then the nonzero vectors u ∈ Cn such that Au = λu are called eigenvectors of A associated with λ; together with the zero vector, these eigenvectors form a subspace of Cn denoted by Eλ(A), and called the eigenspace associated with λ.
Remark: Note that Definition 4.4 requires an eigen-vector to be nonzero.
220 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS A somewhat unfortunate consequence of this requirement is that the set of eigenvectors is not a subspace, since the zero vector is missing!
On the positive side, whenever eigenvectors are involved, there is no need to say that they are nonzero.
If A is a square real matrix A ∈Mn(R), then we re-strict Definition 4.4 to real eigenvalues λ ∈R and real eigenvectors.
However, it should be noted that although every complex matrix always has at least some complex eigenvalue, a real matrix may not have any real eigenvalues. For example, the matrix A = 0 −1 1 0 has the complex eigenvalues i and −i, but no real eigen-values.
Thus, typically, even for real matrices, we consider com-plex eigenvalues.
4.2. MATRIX NORMS 221 Observe that λ ∈C is an eigenvalue of A iffAu = λu for some nonzero vector u ∈Cn iff(λI −A)u = 0 iffthe matrix λI −A defines a linear map which has a nonzero kernel, that is, iffλI −A not invertible.
However, from Proposition 2.10, λI −A is not invertible iff det(λI −A) = 0.
Now, det(λI −A) is a polynomial of degree n in the indeterminate λ, in fact, of the form λn −tr(A)λn−1 + · · · + (−1)n det(A).
Thus, we see that the eigenvalues of A are the zeros (also called roots) of the above polynomial.
Since every complex polynomial of degree n has exactly n roots, counted with their multiplicity, we have the fol-lowing definition: 222 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Definition 4.5. Given any square n × n matrix A ∈Mn(C), the polynomial det(λI −A) = λn −tr(A)λn−1 + · · · + (−1)n det(A) is called the characteristic polynomial of A. The n (not necessarily distinct) roots λ1, . . . , λn of the characteristic polynomial are all the eigenvalues of A and constitute the spectrum of A.
We let ρ(A) = max 1≤i≤n |λi| be the largest modulus of the eigenvalues of A, called the spectral radius of A.
4.2. MATRIX NORMS 223 Proposition 4.4. For any matrix norm on Mn(C) and for any square n × n matrix A, we have ρ(A) ≤A.
Remark: Proposition 4.4 still holds for real matrices A ∈Mn(R), but a different proof is needed since in the above proof the eigenvector u may be complex. (Hint: use Theorem 4.3.) Now, it turns out that if A is a real n × n symmetric matrix, then the eigenvalues of A are all real and there is some orthogonal matrix Q such that A = Qdiag(λ1, . . . , λn)Q, where diag(λ1, . . . , λn) denotes the matrix whose only nonzero entries (if any) are its diagonal entries, which are the (real) eigenvalues of A.
224 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Similarly, if A is a complex n × n Hermitian matrix, then the eigenvalues of A are all real and there is some unitary matrix U such that A = U ∗diag(λ1, . . . , λn)U, where diag(λ1, . . . , λn) denotes the matrix whose only nonzero entries (if any) are its diagonal entries, which are the (real) eigenvalues of A.
We now return to matrix norms. We begin with the so-called Frobenius norm, which is just the norm 2 on Cn2, where the n × n matrix A is viewed as the vec-tor obtained by concatenating together the rows (or the columns) of A.
The reader should check that for any n × n complex ma-trix A = (aij), n i,j=1 |aij|2 1/2 = tr(A∗A) = tr(AA∗).
4.2. MATRIX NORMS 225 Definition 4.6. The Frobenius norm F is defined so that for every square n × n matrix A ∈Mn(C), AF = n i,j=1 |aij|2 1/2 = tr(AA∗) = tr(A∗A).
The following proposition show that the Frobenius norm is a matrix norm satisfying other nice properties.
Proposition 4.5. The Frobenius norm F on Mn(C) satisfies the following properties: (1) It is a matrix norm; that is, ABF ≤AF BF, for all A, B ∈Mn(C).
(2) It is unitarily invariant, which means that for all unitary matrices U, V , we have AF = UAF = AV F = UAV F .
(3) ρ(A∗A) ≤AF ≤√n ρ(A∗A), for all A ∈ Mn(C).
226 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Remark: The Frobenius norm is also known as the Hilbert-Schmidt norm or the Schur norm. So many famous names associated with such a simple thing!
We now give another method for obtaining matrix norms using subordinate norms.
First, we need a proposition that shows that in a finite-dimensional space, the linear map induced by a matrix is bounded, and thus continuous.
Proposition 4.6. For every norm on Cn (or Rn), for every matrix A ∈Mn(C) (or A ∈Mn(R)), there is a real constant CA > 0, such that Au≤CA u, for every vector u ∈Cn (or u ∈Rn if A is real).
Proposition 4.6 says that every linear map on a finite-dimensional space is bounded.
4.2. MATRIX NORMS 227 This implies that every linear map on a finite-dimensional space is continuous.
Actually, it is not hard to show that a linear map on a normed vector space E is bounded iffit is continuous, regardless of the dimension of E.
Proposition 4.6 implies that for every matrix A ∈Mn(C) (or A ∈Mn(R)), sup x∈Cn x=0 Ax x≤CA.
Now, since λu= |λ| u, it is easy to show that sup x∈Cn x=0 Ax x= sup x∈Cn x=1 Ax.
Similarly sup x∈Rn x=0 Ax x= sup x∈Rn x=1 Ax.
228 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Definition 4.7. If is any norm on Cn, we define the function on Mn(C) by A= sup x∈Cn x=0 Ax x= sup x∈Cn x=1 Ax.
The function A →Ais called the subordinate matrix norm or operator norm induced by the norm .
It is easy to check that the function A →Ais indeed a norm, and by definition, it satisfies the property Ax≤Ax, for all x ∈Cn.
This implies that AB≤AB for all A, B ∈Mn(C), showing that A →Ais a matrix norm.
4.2. MATRIX NORMS 229 Observe that the subordinate matrix norm is also defined by A= inf{λ ∈R | Ax≤λ x, for all x ∈Cn}.
The definition also implies that I= 1.
The above show that the Frobenius norm is not a subor-dinate matrix norm (why?).
230 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Remark: We can also use Definition 4.7 for any norm on Rn, and define the function R on Mn(R) by AR = sup x∈Rn x=0 Ax x= sup x∈Rn x=1 Ax.
The function A →AR is a matrix norm on Mn(R), and AR ≤A, for all real matrices A ∈Mn(R).
However, it is possible to construct vector norms on Cn and real matrices A such that AR < A.
In order to avoid this kind of difficulties, we define sub-ordinate matrix norms over Mn(C).
Luckily, it turns out that AR = Afor the vector norms, 1 , 2, and ∞.
4.2. MATRIX NORMS 231 Proposition 4.7. For every square matrix A = (aij) ∈Mn(C), we have A1 = sup x∈Cn x1=1 Ax1 = max j n i=1 |aij| A∞= sup x∈Cn x∞=1 Ax∞= max i n j=1 |aij| A2 = sup x∈Cn x2=1 Ax2 = ρ(A∗A) = ρ(AA∗).
Furthermore, A∗2 = A2, the norm 2 is unitar-ily invariant, which means that A2 = UAV 2 for all unitary matrices U, V , and if A is a normal matrix, then A2 = ρ(A).
The norm A2 is often called the spectral norm.
232 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Observe that property (3) of proposition 4.5 says that A2 ≤AF ≤√n A2 , which shows that the Frobenius norm is an upper bound on the spectral norm. The Frobenius norm is much easier to compute than the spectal norm.
The reader will check that the above proof still holds if the matrix A is real, confirming the fact that AR = A for the vector norms 1 , 2, and ∞.
It is also easy to verify that the proof goes through for rectangular matrices, with the same formulae.
Similarly, the Frobenius norm is also a norm on rectan-gular matrices. For these norms, whenever AB makes sense, we have AB≤AB.
4.2. MATRIX NORMS 233 The following proposition will be needed when we deal with the condition number of a matrix.
Proposition 4.8. Let be any matrix norm and let B be a matrix such that B< 1.
(1) If is a subordinate matrix norm, then the ma-trix I + B is invertible and (I + B)−1 ≤ 1 1 −B.
(2) If a matrix of the form I + B is singular, then B≥1 for every matrix norm (not necessarily subordinate).
234 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Remark: Another result that we will not prove here but that plays a role in the convergence of sequences of pow-ers of matrices is the following: For every matrix A ∈Matn(C) and for every > 0, there is some subor-dinate matrix norm such that A≤ρ(A) + .
Note that equality is generally not possible; consider the matrix A = 0 1 0 0 , for which ρ(A) = 0 < A, since A = 0.
4.3. CONDITION NUMBERS OF MATRICES 235 4.3 Condition Numbers of Matrices Unfortunately, there exist linear systems Ax = b whose solutions are not stable under small perturbations of either b or A.
For example, consider the system 10 7 8 7 7 5 6 5 8 6 10 9 7 5 9 10 x1 x2 x3 x4 = 32 23 33 31 .
The reader should check that it has the solution x = (1, 1, 1, 1). If we perturb slightly the right-hand side, obtaining the new system 10 7 8 7 7 5 6 5 8 6 10 9 7 5 9 10 x1 + ∆x1 x2 + ∆x2 x3 + ∆x3 x4 + ∆x4 = 32.1 22.9 33.1 30.9 , the new solutions turns out to be x = (9.2, −12.6, 4.5, −1.1).
236 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS In other words, a relative error of the order 1/200 in the data (here, b) produces a relative error of the order 10/1 in the solution, which represents an amplification of the relative error of the order 2000.
Now, let us perturb the matrix slightly, obtaining the new system 10 7 8.1 7.2 7.08 5.04 6 5 8 5.98 9.98 9 6.99 4.99 9 9.98 x1 + ∆x1 x2 + ∆x2 x3 + ∆x3 x4 + ∆x4 = 32 23 33 31 .
This time, the solution is x = (−81, 137, −34, 22).
Again, a small change in the data alters the result rather drastically.
Yet, the original system is symmetric, has determinant 1, and has integer entries.
4.3. CONDITION NUMBERS OF MATRICES 237 The problem is that the matrix of the system is badly conditioned, a concept that we will now explain.
Given an invertible matrix A, first, assume that we per-turb b to b + δb, and let us analyze the change between the two exact solutions x and x + δx of the two systems Ax = b A(x + δx) = b + δb.
We also assume that we have some norm and we use the subordinate matrix norm on matrices. From Ax = b Ax + Aδx = b + δb, we get δx = A−1δb, and we conclude that δx≤ A−1 δb b≤Ax.
238 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Consequently, the relative error in the result δx/ x is bounded in terms of the relative error δb/ bin the data as follows: δx x≤ A A−1 δb b.
Now let us assume that A is perturbed to A + δA, and let us analyze the change between the exact solutions of the two systems Ax = b (A + ∆A)(x + ∆x) = b.
After some calculations, we get ∆x x + ∆x≤ A A−1 ∆A A.
Observe that the above reasoning is valid even if the ma-trix A + ∆A is singular, as long as x + ∆x is a solution of the second system.
Furthermore, if ∆Ais small enough, it is not unreason-able to expect that the ratio ∆x/ x + ∆xis close to ∆x/ x.
4.3. CONDITION NUMBERS OF MATRICES 239 This will be made more precise later.
In summary, for each of the two perturbations, we see that the relative error in the result is bounded by the relative error in the data, multiplied the number A A−1 .
In fact, this factor turns out to be optimal and this sug-gests the following definition: Definition 4.8. For any subordinate matrix norm , for any invertible matrix A, the number cond(A) = A A−1 is called the condition number of A relative to .
The condition number cond(A) measure the sensitivity of the linear system Ax = b to variations in the data b and A; a feature referred to as the condition of the system.
Thus, when we says that a linear system is ill-conditioned, we mean that the condition number of its matrix is large.
240 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS We can sharpen the preceding analysis as follows: Proposition 4.9. Let A be an invertible matrix and let x and x + δx be the solutions of the linear systems Ax = b A(x + δx) = b + δb.
If b = 0, then the inequality δx x≤cond(A)δb b holds and is the best possible. This means that for a given matrix A, there exist some vectors b = 0 and δb = 0 for which equality holds.
4.3. CONDITION NUMBERS OF MATRICES 241 Proposition 4.10. Let A be an invertible matrix and let x and x + ∆x be the solutions of the two systems Ax = b (A + ∆A)(x + ∆x) = b.
If b = 0, then the inequality ∆x x + ∆x≤cond(A)∆A A holds and is the best possible. This means that given a matrix A, there exist a vector b = 0 and a matrix ∆A = 0 for which equality holds.
Furthermore, if ∆Ais small enough (for instance, if ∆A< 1/ A−1 ), we have ∆x x≤cond(A)∆A A(1 + O(∆A)); in fact, we have ∆x x≤cond(A)∆A A 1 1 −A−1∆A .
242 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Remark: If A and b are perturbed simultaneously, so that we get the “perturbed” system (A + ∆A)(x + δx) = b + δb, it can be shown that if ∆A< 1/ A−1 (and b = 0), then ∆x x≤ cond(A) 1 −A−1∆A ∆A A+ δb b .
We now list some properties of condition numbers and figure out what cond(A) is in the case of the spectral norm (the matrix norm induced by 2).
4.3. CONDITION NUMBERS OF MATRICES 243 First, we need to introduce a very important factorization of matrices, the singular value decomposition, for short, SVD.
It can be shown that given any n×n matrix A ∈Mn(C), there exist two unitary matrices U and V , and a real diagonal matrix Σ = diag(σ1, . . . , σn), with σ1 ≥σ2 ≥· · · ≥σn ≥0, such that A = V ΣU ∗.
The nonnegative numbers σ1, . . . , σn are called the singular values of A.
If A is a real matrix, the matrices U and V are orthogonal matrices.
244 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS The factorization A = V ΣU ∗implies that A∗A = UΣ2U ∗ and AA∗= V Σ2V ∗, which shows that σ2 1, . . . , σ2 n are the eigenvalues of both A∗A and AA∗, that the columns of U are correspond-ing eivenvectors for A∗A, and that the columns of V are corresponding eivenvectors for AA∗.
In the case of a normal matrix if λ1, . . . , λn are the (com-plex) eigenvalues of A, then σi = |λi|, 1 ≤i ≤n.
4.3. CONDITION NUMBERS OF MATRICES 245 Proposition 4.11. For every invertible matrix A ∈Mn(C), the following properties hold: (1) cond(A) ≥1, cond(A) = cond(A−1) cond(αA) = cond(A) for all α ∈C −{0}.
(2) If cond2(A) denotes the condition number of A with respect to the spectral norm, then cond2(A) = σ1 σn , where σ1 ≥· · · ≥σn are the singular values of A.
(3) If the matrix A is normal, then cond2(A) = |λ1| |λn|, where λ1, . . . , λn are the eigenvalues of A sorted so that |λ1| ≥· · · ≥|λn|.
(4) If A is a unitary or an orthogonal matrix, then cond2(A) = 1.
246 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS (5) The condition number cond2(A) is invariant under unitary transformations, which means that cond2(A) = cond2(UA) = cond2(AV ), for all unitary matrices U and V .
Proposition 4.11 (4) shows that unitary and orthogonal transformations are very well-conditioned, and part (5) shows that unitary transformations preserve the condition number.
In order to compute cond2(A), we need to compute the top and bottom singular values of A, which may be hard.
The inequality A2 ≤AF ≤√n A2 , may be useful in getting an approximation of cond2(A) = A2 A−1 2, if A−1 can be determined.
Remark: There is an interesting geometric characteri-zation of cond2(A).
4.3. CONDITION NUMBERS OF MATRICES 247 If θ(A) denotes the least angle between the vectors Au and Av as u and v range over all pairs of orthonormal vectors, then it can be shown that cond2(A) = cot(θ(A)/2)).
Thus, if A is nearly singular, then there will be some orthonormal pair u, v such that Au and Av are nearly parallel; the angle θ(A) will the be small and cot(θ(A)/2)) will be large.
It should also be noted that in general (if A is not a normal matrix) a matrix could have a very large condition number even if all its eigenvalues are identical!
For example, if we consider the n × n matrix A = 1 2 0 0 . . . 0 0 0 1 2 0 . . . 0 0 0 0 1 2 . . . 0 0 .
.
. .
.
. ... ... ... .
.
. .
.
.
0 0 . . . 0 1 2 0 0 0 . . . 0 0 1 2 0 0 . . . 0 0 0 1 , it turns out that cond2(A) ≥2n−1.
248 CHAPTER 4. VECTOR NORMS AND MATRIX NORMS Going back to our matrix A = 10 7 8 7 7 5 6 5 8 6 10 9 7 5 9 10 , which is a symmetric matrix, it can be shown that λ1 ≈30.2887 > λ2 ≈3.858 > λ3 ≈0.8431 > λ4 ≈0.01015, so that cond2(A) = λ1 λ4 ≈2984.
The reader should check that for the perturbation of the right-hand side b used earlier, the relative errors δx/x and δx/xsatisfy the inequality δx x≤cond(A)δb b and comes close to equality. |
7787 | https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series?srsltid=AfmBOooMsNSdHfnvtcj1XfY4KkE24E3rKjNjBRHacekz7jpdjXQC8yhr | Art of Problem Solving
Arithmetico-geometric series - AoPS Wiki
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical CampusVisit the Virtual Campus
Sign In
Register
online school
Class ScheduleRecommendationsOlympiad CoursesFree Sessions
books tore
AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates
community
ForumsContestsSearchHelp
resources
math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten
contests on aopsPractice Math ContestsUSABO
newsAoPS BlogWebinars
view all 0
Sign In
Register
AoPS Wiki
ResourcesAops Wiki Arithmetico-geometric series
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Arithmetico-geometric series
An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: , where and are the th terms of arithmetic and geometric sequences, respectively.
Contents
1 Finite Sum
2 Infinite Sum
3 Example Problems
4 See Also
Finite Sum
The sum of the first terms of an is , where is the common difference of and is the common ratio of . Or, , where is the sum of the first terms of .
Proof:
Let represent the sum of the first terms.
Infinite Sum
The sum of an infinite arithmetico-geometric sequence is , where is the common difference of and is the common ratio of (). Or, , where is the infinite sum of the .
Example Problems
Mock AIME 2 2006-2007 Problem 5
1994 AIME Problem 4
See Also
Sequence
Arithmetic sequence
Geometric sequence
Retrieved from "
Art of Problem Solving is an
ACS WASC Accredited School
aops programs
AoPS Online
Beast Academy
AoPS Academy
About
About AoPS
Our Team
Our History
Jobs
AoPS Blog
Site Info
Terms
Privacy
Contact Us
follow us
Subscribe for news and updates
© 2025 AoPS Incorporated
© 2025 Art of Problem Solving
About Us•Contact Us•Terms•Privacy
Copyright © 2025 Art of Problem Solving
Something appears to not have loaded correctly.
Click to refresh. |
7788 | https://www.mathe2.uni-bayreuth.de/stoll/teaching/pAdicAnalysis-WS2015/Skript-pAdicAnalysis-pub-print.pdf | p-adic Analysis in Arithmetic Geometry Winter Semester 2015/2016 University of Bayreuth Michael Stoll Contents 1.
Introduction 2 2.
p-adic numbers 3 3.
Newton Polygons 14 4.
Multiplicative seminorms and Berkovich spaces 19 5.
The Berkovich affine and projective line 24 6.
Analytic spaces and functions 34 7.
Berkovich spaces of curves 39 8.
Integration 45 References 52 Print Version of March 29, 2020, 11:29.
§ 1. Introduction 2 1. Introduction We all know the fields R and C of real and complex numbers as the completion of the field Q of rational numbers and its algebraic closure. In particular, we have a canonical embedding Q , →R ⊂C, which we can sometimes use to get number theoretic results by applying analysis over R or C.
Now R is not the only completion of Q. Besides the usual absolute value, there are more absolute values on Q; to be precise, up to a natural equivalence (and except for the trivial one), there is one absolute value |·|p for each prime number p. (We will explain what an absolute value is in due course.) Completing Q with respect to |·|p leads to the field Qp of p-adic numbers; we can then take its algebraic closure ¯ Qp and the completion Cp of that (¯ Qp is, in contrast to ¯ R = C, not complete), which has similar properties as C: it is the smallest extension field of Q that is algebraically closed and complete with respect to the p-adic absolute value.
There is a general philosophy in Number Theory that ‘all completions are created equal’ and should have the same rights. In many situations, one gets the best results by considering them all together. Since Qp and Cp are complete metric spaces (and the former is, like R or C, locally compact), we can try to do analysis over them. Many concepts and results of ‘classical’ analysis carry over without great problems. But we will see that there is one feature of the p-adic topology that is a stumbling block for an easy transfer of certain parts of analysis (like for example line integrals) to the p-adic setting: this topology is totally discon-nected. One goal of this lecture course is to explain a way to resolve this problem, which is to embed Cp (say) into a larger ‘analytic space’ Can p that is (in this case, even uniquely) path-connected. This approach is due to Vladimir Berkovich; the V.G. Berkovich c ⃝MFO 2012 analytic spaces constructed in this way are also known as Berkovich Spaces.
§ 2. p-adic numbers 3 2. p-adic numbers In this section we will recall (or introduce) the field Qp of p-adic numbers and its properties. We begin with a rather general definition.
2.1. Definition.
Let K be a field. An absolute value on K is a map DEF absolute value |·|: K − →R≥0, x 7− →|x| with the following properties.
(1) |x| = 0 ⇐ ⇒x = 0.
(2) (Multiplicativity) |xy| = |x| · |y|.
(3) (Triangle Inequality) |x + y| ≤|x| + |y|.
If |·| satisfies the stronger inequality (3′) (Ultrametric Triangle Inequality) |x + y| ≤max{|x|, |y|}, then the absolute value is said to be ultrametric or non-archimedean, otherwise it is archimedean.
Two absolute values |·|1 and |·|2 on K are said to be equivalent, if there is a constant c > 0 such that |x|2 = |x|c 1 for all x ∈K.
♦ Properties (1), (2) and (3) imply that d(x, y) = |x −y| defines a metric on K.
((2) is needed for the symmetry: it implies that | −1| = 1, so that |y −x| = | −1| · |x −y| = |x −y|.) The absolute value is then continuous as a real-valued function on the metric space (K, d).
One can show that two absolute values on K are equivalent if and only if they induce the same topology on K (Exercise).
2.2. Lemma.
Let (K, |·|) be a field with a non-archimedean absolute value. Then LEMMA valuation ring R = {x ∈K : |x| ≤1} is a subring of K. R is a local ring (i.e., it has exactly one maximal ideal) and has K as its field of fractions.
Proof. Exercise!
K This ring is the valuation ring of (K, | · |).
DEF valuation ring Recall the following definition.
2.3. Definition.
A metric space (X, d) is said to be complete, if every Cauchy DEF complete metric space sequence in X converges in X: if (xn) is a sequence in X such that lim n→∞sup m≥1 d(xn, xn+m) = 0 , then there exists x ∈X such that limn→∞d(xn, x) = 0.
If K is a field with absolute value |·|, then (K, |·|) is said to be complete, if the metric space (K, d) is, where d is the metric induced by the absolute value.
♦ § 2. p-adic numbers 4 2.4. Examples.
EXAMPLES absolute values Every field K has the trivial absolute value |·|0 with |x|0 = 1 for all x ̸= 0.
The usual absolute value is an (archimedean) absolute value on Q, R and C. The latter two are complete, Q is not.
|f(x)/g(x)| = edeg(f)−deg(g) defines a non-archimedean absolute value on K(x), the field of rational functions in one variable over K. (Exercise!) If |·| is an absolute value and 0 < c ≤1, then |·|c is also an absolute value, which is equivalent to |·|. If |·| is non-archimedean, then this remains true for c > 1.
(Exercise!) ♣ We can now introduce the p-adic absolute values on Q.
2.5. Definition.
Let p be a prime number. If a ̸= 0 is an integer, we define its DEF p-adic absolute value p-adic valuation to be vp(a) = max{n ∈Z≥0 : pn | a} .
For a = r/s ∈Q× (with r, s ∈Z, s ̸= 0), we set vp(a) = vp(r) −vp(s). The p-adic absolute value on Q is given by |x|p = ( 0 if x = 0, p−vp(x) otherwise.
♦ So the p-adic absolute value of x ∈Q is small when (the numerator of) x is divisible by a high power of p, and it is large when the denominator of x is divisible by a high power of p. If x ∈Z, then we clearly have |x|p ≤1: contrary to the familiar situation with the usual absolute value, the integers form a bounded subset of Q with respect to |·|p! This explains the word ‘non-archimedean’ — the Archimedean Axiom states that if x, y ∈R>0, then there is n ∈Z such that nx > y; this is equivalent to saying that Z is unbounded.
2.6. Lemma.
The p-adic absolute value is a non-archimedean absolute value LEMMA |·|p is abs. value on Q.
Proof. We check the properties in Definition 2.1. Property (1) is clear from the definition. Property (2) follows from vp(ab) = vp(a)+vp(b), which is a consequence of unique factorization. Property (3′) follows from vp(a + b) ≥min{vp(a), vp(b)}, which is a consequence of the elementary fact that pn | a and pn | b together imply that pn | a + b.
K Note that Q is not complete with respect to |·|p (Exercise!).
2.7. Example.
Let (K, |·|) be a field with absolute value. Then for every x ∈K EXAMPLE geometric series with |x| < 1, the series ∞ X n=0 xn converges to the limit 1/(1 −x): we have N−1 X n=0 xn − 1 1 −x = −xN 1 −x = |1 −x|−1 · |x|N , § 2. p-adic numbers 5 which tends to zero, since |x| < 1 (note that |1 −x| ≥1 −|x| > 0, so the fraction makes sense).
For example, 1 + p + p2 + p3 + p4 + . . . = 1 1 −p in (Q, |·|p).
♣ From the viewpoint of analysis, it is desirable to work with a complete field.
Indeed, one possible construction of the field of real numbers is as the smallest complete field containing (Q, |·|∞), where (from now on) |·|∞denotes the usual absolute value. In fact, this construction works quite generally.
2.8. Theorem.
Let (K, |·|) be a field with an absolute value. Then there is a THM completion field (K′, |·|′) extending K such that |·|′ restricts to |·| on K, (K′, |·|′) is complete and K is dense in K′.
Proof. Let C(K) be the ring of Cauchy sequences over K (with term-wise addition and multiplication). The set N(K) of null sequences (i.e., sequences converging to zero) forms an ideal in the ring C(K). We show that this ideal is actually maximal. It does not contain the unit (all-ones) sequence, so it is not all of C(K).
If (xn) ∈C(K)\N(K), then it follows from the definition of ‘Cauchy sequence’ that there are n0 ∈Z>0 and c > 0 such that |xn| ≥c for all n ≥n0. Then the sequence (yn) given by yn = 0 for n < n0 and yn = 1/xn for n ≥n0 is a Cauchy sequence and (1) = (xn) · (yn) + (zn) with (zn) ∈N(K), so N(K) + C(K) · (xn) = C(K).
Since N(K) is a maximal ideal in C(K), the quotient ring K′ := C(K)/N(K) is a field. There is a natural inclusion K , →C(K) by mapping a ∈K to the constant sequence (a), which by composition with the canonical epimorphism C(K) →K′ gives an embedding i: K , →K′. We define |·|′ by |[(xn)]|′ = lim n→∞|xn| (where [(xn)] denotes the residue class mod N(K) of the sequence (xn) ∈C(K)).
The properties of absolute values and of Cauchy sequences imply that this is well-defined (i.e., the limit exists and does not depend on the choice of the represen-tative sequence). That |·|′ is an absolute value follows easily from the assumption that |·| is, and it is clear that |·|′ restricts to |·| on K.
It is also easy to see that K is dense in K′. Let x = [(xn)] ∈K′, then |i(xn)−x|′ = limm→∞|xn −xn+m| tends to zero as n tends to infinity, so we can approximate x arbitrarily closely by elements of K.
It remains to show that K′ is complete. So let (x(ν))ν be a Cauchy sequence in K′ and represent each x(ν) by a Cauchy sequence (x(ν) n )n in K. Since K is dense in K′, we can do this in such a way that |i(x(ν) n ) −x(ν)|′ ≤2−n for all ν and n.
(The reason for this requirement is that we need some uniformity of convergence to make the proof work.) Let yn = x(n) n . We claim that (yn) is a Cauchy sequence (in K) and that y = [(yn)] = limν→∞x(ν) (in K′). To see the first, pick ε > 0.
There is ν0 such that |x(ν) −x(ν+µ)|′ < ε for all ν ≥ν0 and µ ≥0. Then |yn −yn+m| = |x(n) n −x(n+m) n+m | ≤|i(x(n) n ) −x(n)|′ + |x(n) −x(n+m)|′ + |x(n+m) −i(x(n+m) n+m )|′ ≤2−n + ε + 2−n−m < 2ε § 2. p-adic numbers 6 when 2−n < ε/2 and n ≥ν0. To see the second, note that |x(ν) −y|′ = lim n→∞|x(ν) n −x(n) n | ≤lim sup n→∞ |i(x(ν) n ) −x(ν)|′ + |x(ν) −x(n)|′ + |x(n) −i(x(n) n )|′ ≤lim sup n→∞(2−n + |x(ν) −x(n)|′ + 2−n) = lim sup n→∞|x(ν) −x(n)|′ , which tends to zero as ν →∞.
K One can show that (K′, |·|′) is determined up to unique isomorphism (of extensions of K with absolute value), but we will not need this in the following.
Since |·|′ extends |·|, we will usually use the same notation for both. We will also consider K as a subfield of K′.
2.9. Definition.
Let p be a prime number. The completion Qp of Q with respect DEF field of p-adic numbers to the p-adic absolute value is called the field of p-adic numbers. Its valuation ring Zp is the ring of p-adic integers.
♦ Since |x|p takes a discrete set of values for x ̸= 0, it follows that Zp = {x ∈Qp : |x|p ≤1} = {x ∈Qp : |x|p < p} is closed and open in Qp. This implies that for any two distinct elements x, y ∈Qp, there are disjoint open neighborhoods X of x and Y of y such that Qp = X ∪Y : Qp is totally disconnected. To see this, let δ = |x −y|p > 0. Then the open ball X around x of radius δ (which is x + pn+1Zp, if δ = p−n) is also closed, so its complement Y = Qp \ X is open as well, and y ∈Y . One consequence of this is that any continuous map γ : [0, 1] →Qp is constant (otherwise let x and y be two distinct elements in the image and let X and Y be as above; then [0, 1] is the disjoint union of the two open non-empty subsets γ−1(X) and γ−1(Y ), contradicting the fact that intervals are connected).
Next we want to show that Zp is compact. For this we need a fact about approx-imation of p-adic numbers by rationals.
2.10. Lemma.
Let x ∈Qp {0} with |x|p = p−n. There is a ∈{0, 1, 2, . . . , p−1} LEMMA approximation of p-adic numbers such that |x −apn|p < |x|p.
Proof. Replacing x by p−nx, we can assume that |x|p = 1, in particular, x ∈Zp.
Since Q is dense in Qp, there is r/s ∈Q such that |x −r/s|p < 1.
By the ultrametric triangle inequality, |r/s|p ≤1, so that we can assume that p ∤s. Let a ∈{0, 1, . . . , p −1} be such that as ≡r mod p. Then |x −a|p = (x −r/s) + (r/s −a) p ≤max{|x −r/s|p, |r/s −a|p} < 1 , since |r/s −a|p = |r −as|p/|s|p < 1.
K It follows that that Z is dense in Zp: iterating the statement of the lemma, we find that for x ∈Zp and n > 0 there is a ∈Z (with 0 ≤a < pn) such that |x −a|p < p−n.
§ 2. p-adic numbers 7 2.11. Theorem.
The ring Zp is compact with respect to the metric induced by THM Zp is compact the p-adic absolute value. In particular, Qp is locally compact.
Proof. We show that any sequence (xn) in Zp has a convergent subsequence. Since Zp is complete (Zp = {x ∈Qp : |x|p ≤1} is a closed subset of the complete space Qp), it is enough to show that there is a subsequence that is a Cauchy sequence. We do this iteratively. Let n1 be the smallest index n such that there are infinitely many m > n with |xn −xm|p < 1. Such an n must exist by Lemma 2.10, which implies that there is some a such that |xn −a|p < 1 for infinitely many n.
Now let n2 be the smallest index n > n1 such that |xn −xn1|p < 1 and such that there are infinitely many m > n with |xn −xm| < p−1. This again exists by Lemma 2.10, where we restrict to the infinitely many n such that |xn −xn1|p < 1.
We continue in this way: nk+1 is the smallest n > nk such that |xn−xnk|p < p−(k−1) and such that there are infinitely many m > n with |xn −xm|p < p−k. Then |xnk+1 −xnk|p < p−(k−1) for all k ≥1, which by the ultrametric triangle inequality implies that (xnk)k is a Cauchy sequence: xnk+l −xnk p ≤max n xnk+m −xnk+m−1 p : 1 ≤m ≤l o < p−(k−1) .
That Qp is locally compact follows, since any x ∈Qp has the compact neighbor-hood x + Zp.
K 2.12. Definition.
Let (K, |·|) be a complete field with a non-archimedean abso-DEF residue field lute value. By Lemma 2.2 the valuation ring R = {x ∈K : |x| ≤1} is a local ring with unique maximal ideal M = {x ∈K : |x| < 1}. The quotient ring k = R/M is therefore a field, the residue field of K. The canonical map R →k is called the reduction map and usually denoted x 7→¯ x.
♦ Before we continue we note an easy but important fact on non-archimedean abso-lute values.
2.13. Lemma.
Let (K, |·|) be a field with a non-archimedean absolute value and LEMMA all triangles are isosceles let x, y ∈K with |x| > |y|. Then |x + y| = |x|.
If we have x1, x2, . . . , xn ∈K such that x1 + x2 + . . . + xn = 0 and n ≥2, then there are at least two indices 1 ≤j < k ≤n such that |xj| = |xk| = max{|x1|, |x2|, . . . , |xn|} .
Proof. We have |x + y| ≤max{|x|, |y|} = |x|. Assume that |x + y| < |x|. Then |x| = |(x + y) + (−y)| ≤max{|x + y|, | −y|} < |x| , a contradiction.
For the second statement observe that if we had just one j such that |xj| is maximal, then by the second statement we would have 0 = |x1 + x2 + . . . + xn| = xj + X k̸=j xk = |xj| , which is a contradiction, since xj ̸= 0 in this situation.
K If |·| is a non-archimedean absolute value on a field K, then we can extend it to the polynomial ring K[x].
§ 2. p-adic numbers 8 2.14. Lemma.
Let (K, |·|) be a field with a non-archimedean absolute value. For LEMMA extension of abs. value to K[x] f = a0 + a1x + a2x2 + . . . + anxn ∈K[x] we set |f| := max{|a0|, |a1|, |a2|, . . . , |an|} .
Then properties (1), (2) and (3 ′) of Definition 2.1 are satisfied for elements of K[x].
If K is complete with respect to |·|, then for each n ∈Z≥0, the space K[x]<n of polynomials of degree < n is a complete metric space for the metric induced by |·| on K[x].
Proof. Exercise.
K The following is an important tool when working in complete non-archimedean fields.
2.15. Theorem.
Let (K, |·|) be a complete non-archimedean field with valuation THM Hensel’s Lemma ring R and residue field k.
Let F ∈R[x] be a polynomial such that |F| = 1 and suppose that we have a factorization ¯ F = f1f2 in k[x] such that f1 is monic and f1 and f2 are coprime. (The notation ¯ F means the polynomial in k[x] whose coefficients are obtained by reduction from those of F.) Then there are unique polynomials F1, F2 ∈R[x] such that F = F1F2, F1 is monic and ¯ F1 = f1, ¯ F2 = f2.
Proof. Let n = deg F, n1 = deg f1, n2 = n −n1 ≥deg f2. We write R[x]<n for the R-module of polynomials of degree < n.
We choose polynomials F (0) 1 , F (0) 2 ∈R[x] such that deg F (0) 1 = n1, deg F (0) 2 = n2, ¯ F (0) 1 = f1, ¯ F (0) 2 = f2, F (0) 1 is monic and the leading coefficient of F (0) 2 is the same as that of F; then F −F (0) 1 F (0) 2 ∈R[x]<n and δ := |F −F (0) 1 F (0) 2 | < 1 .
We claim that the k-linear map k[x]<n1 × k[x]<n2 − →k[x]<n , (h1, h2) 7− →h1f2 + h2f1 is an isomorphism: it is injective, since h1f2 + h2f1 = 0 implies that f1 divides h1f2, which in turn implies that f1 divides h1 (since f1 and f2 are coprime). But deg h1 < n1 = deg f1, so h1 = 0. Since f1 ̸= 0, it follows that h2 = 0, too. Finally we observe that the dimensions of source and target are the same.
If M is the matrix representing this linear map with respect to the k-bases ((1, 0), (x, 0), . . . (xn1−1, 0), (0, 1), (0, x), . . . , (0, xn2−1)) and (1, x, x2, . . . , xn1+n2−1), then det(M) ̸= 0. Let now Φ: R[x]<n1 × R[x]<n2 − →R[x]<n , (H1, H2) 7− →H1F (0) 2 + H2F (0) 1 .
Its matrix ˜ M with respect to the ‘power bases’ reduces to M, so | det( ˜ M)| = 1, which means that ˜ M is in GL(n, R), and Φ is invertible. It also follows that if (H1, H2) = Φ−1(H), then |H1|, |H2| ≤|H|.
We now want to find F1 and F2 by adjusting F (0) 1 and F (0) 2 . So we would like to determine H1 ∈R[x]<n1 and H2 ∈R[x]<n2 with |H1|, |H2| < 1 such that F = (F (0) 1 + H1)(F (0) 2 + H2) = F (0) 1 F (0) 2 + H1F (0) 2 + H2F (0) 1 + H1H2 .
§ 2. p-adic numbers 9 We ignore the nonlinear term H1H2 and choose (H1, H2) such that the linear terms correct the mistake, i.e., (H1, H2) = Φ−1(F −F (0) 1 F (0) 2 ). From the above we know that |H1|, |H2| ≤δ and therefore F −(F (0) 1 + H1)(F (0) 2 + H2) = |H1H2| ≤δ2 .
Repeating this with the new approximations F (1) j = F (0) j +Hj we obtain F (2) j such that |F (1) j −F (2) j | ≤δ2 and |F −F (2) 1 F (2) 2 | ≤δ4. Iterating this procedure, we construct sequences (F (m) j )m≥0 in R[x]<nj such that |F (m) 1 −F (m+1) 1 | ≤δ2m , |F (m) 2 −F (m+1) 2 | ≤δ2m and |F −F (m) 1 F (m) 2 | ≤δ2m for all m. Since R[x]<nj is complete by Lemma 2.14, the sequences converge to polynomials F1 and F2 with F = F1F2 and ¯ F1 = f1, ¯ F2 = f2. This shows existence.
To show uniqueness, assume that ˜ F1 and ˜ F2 are another solution. Then 0 = F −F = F1F2 −˜ F1 ˜ F2 = (F1 −˜ F1)F2 + (F2 −˜ F2) ˜ F1 .
The map Φ as above, but using ˜ F1 and F2, is still invertible (this only uses that the reductions are f1 and f2), which immediately gives F1−˜ F1 = 0 and F2−˜ F2 = 0.
K We draw some conclusions from this.
2.16. Corollary.
Let (K, |·|) be a field that is complete with respect to a non-COR Hensel’s Lemma for roots archimedean absolute value, let R be its valuation ring and k its residue field and let f ∈R[x] be monic. Assume that ¯ f ∈k[x] has a simple root a ∈k. Then f has a unique root α ∈R such that ¯ α = a.
Proof. This is the case f1 = x −a of Theorem 2.15. Note that the assumption that a is a simple root of ¯ f implies that the cofactor f2 = ¯ f/(x −a) is coprime to f1.
K 2.17. Corollary.
Let (K, |·|) be a field that is complete with respect to a non-COR reducibility of certain polynomials archimedean absolute value and let f = a0 + a1x + . . . + anxn ∈K[x] with an ̸= 0.
Assume that there is 0 < m < n with |am| = |f| and that |a0| < |f| or |an| < |f|.
Then f is reducible.
Proof. After scaling f we can assume that |f| = 1. If |a0| < 1, then let m be minimal with |am| = 1. Then ¯ f = xmf2 with f2(0) ̸= 0, so that f1 = xm and f2 are coprime. By Theorem 2.15 there is a factorization f = F1F2 with deg F1 = m.
Since 0 < m < n = deg f, this shows that f is reducible.
If |an| < 1, then let m be maximal with |am| = 1. Then ¯ f = f1 · ¯ am with f1 monic and (trivially) coprime to f2 = ¯ am. Again by Theorem 2.15 there is a factorization f = F1F2 with deg F1 = m. Since again 0 < m < n = deg f, this shows that f is reducible also in this case.
K The following example really belongs right after Definition 2.12.
2.18. Example.
The residue field of Qp is Fp.
This essentially follows from EXAMPLE Qp has Fp as residue field Lemma 2.10; the details are left as an exercise.
♣ We will now look at field extensions of complete fields with absolute value. First we introduce the norm and trace of an element in a finite field extension.
§ 2. p-adic numbers 10 2.19. Definition.
Let K ⊂L be a finite field extension and let α ∈L. Then DEF norm and trace multiplication by α induces a K-linear map mα : L →L. We define the norm N(α) and trace Tr(α) of α to be the determinant and trace of mα, respectively.
If we want to make clear which fields are involved, we write NL/K(α) and TrL/K(α).
♦ Norms and traces have the following properties.
• If α ∈K, then clearly NL/K(α) = α[L:K] (since mα can be taken to be αI[L:K]).
• If K ⊂L is separable, then we can also write N(α) = Y σ : L, →¯ K σ(α) and Tr(α) = X σ : L, →¯ K σ(α) , where σ runs through all embeddings of L into a fixed algebraic closure ¯ K of K.
(This is because the σ(α) are the eigenvalues of mα.) • Norms and traces are transitive: if K ⊂L ⊂L′, then for α ∈L′ we have NL/K(NL′/L(α)) = NL′/K(α) and TrL/K(TrL′/L(α)) = TrL′/K(α).
• The norm is multiplicative and the trace is additive (even K-linear); this follows from the corresponding properties of determinants and traces of linear endomor-phisms.
• If L = K(α), then the characteristic polynomial of mα agrees with the minimal polynomial of α, so its constant term is ± N(α).
2.20. Lemma.
Let (K, |·|) be a field that is complete with respect to a non-LEMMA integrality archimedean absolute value and let R be its valuation ring. Let K ⊂L be a finite field extension and let α ∈L have norm N(α) ∈R. Then α is integral over R, i.e., α is a root of a monic polynomial with coefficients in R.
Proof. Let L′ = K(α) ⊂L. Then N(α) = NL′/K(α)[L:L′]. Let f ∈K[x] be the minimal polynomial of α. Its constant term a0 is ± NL′/K(α) and so the assumption implies |a0| ≤1. If we had |f| > 1, then f would be reducible by Corollary 2.17, a contradiction. So |f| = 1, meaning that f ∈R[x], and α is a root of the monic polynomial f.
K 2.21. Corollary.
In the situation of Lemma 2.20 we have | N(1 + α)| ≤1.
COR bound for N(1 + α) Proof. We know that α is a root of a monic polynomial f ∈R[x]. Then 1 + α is a root of the monic polynomial f(x −1) ∈R[x], and since a power of its constant term, which is in R, is ± N(1 + α), it follows that | N(1 + α)| ≤1.
K 2.22. Theorem.
Let (K, |·|) be a complete field with a nontrivial non-archimedean THM extensions of complete fields absolute value and let L be an algebraic extension of K. Then there is a unique absolute value |·|′ on L that extends |·|. If the extension K ⊂L is finite, then (L, |·|′) is complete.
§ 2. p-adic numbers 11 Proof. Since any algebraic extension of K can be obtained as an increasing union of finite extensions, it suffices to consider the finite case. So let [L : K] = n. We first show existence. To this end we define |α|′ = | N(α)|1/n. Then it is clear that |·|′ satisfies properties (1) and (2) of Definition 2.1. To show property (3′), let α, β ∈L and assume that |α|′ ≥|β|′. Then | N(β/α)| ≤1, hence by Corollary 2.21 |α + β|′n = | N(α + β)| = | N(α)| · | N(1 + β/α)| ≤| N(α)| = (max |α|′, |β|′)n .
It is also clear that |·|′ agrees with |·| on K.
To show uniqueness, let |·|′′ be another absolute value on L extending |·|. Let α ∈L with |α|′ ≤1, so that | N(α)| ≤1. Then the minimal polynomial f of α is in R[x]. If we had |α|′′ > 1, then by the ultrametric triangle inequality applied to f(α) = 0 we would obtain a contradiction, since then αn would be the unique term of maximal absolute value, compare Lemma 2.13. So we must have |α|′′ ≤1.
If |α|′ ≥1, then |α−1|′ ≤1, so |α−1|′′ ≤1 and |α|′′ ≥1.
This implies that |α|′ < 1 ⇐ ⇒|α|′′ < 1, from which it follows that the two absolute values on L are equivalent (compare Problem (1) on Exercise sheet 1). Since they have the same restriction to K, they must then be equal.
It remains to show that (L, |·|′) is complete.
Let (b1, b2, . . . , bm) be K-linearly independent elements of L. We claim that there are constants c, C > 0 (depending on b1, b2, . . . , bm) such that (2.1) c max{|a1|, . . . , |am|} ≤|a1b1 + . . . + ambm|′ ≤C max{|a1|, . . . , |am|} for all a1, a2, . . . , am ∈K. We prove this by induction on m. The cases m ≤1 are trivial. So let m ≥2. The upper bound is easy: |a1b1 + . . . + ambm|′ ≤max{|a1b1|′, . . . , |ambm|′} ≤max{|b1|′, . . . , |bm|′} · max{|a1|, . . . , |am|} .
To prove the lower bound, we argue by contradiction. If there is no lower bound, then there are a1, . . . , am with max{|a1|, . . . , |am|} = 1 and |a1b1 + . . . + ambm|′ arbitrarily small. Pick a sequence (a(k) 1 , . . . , a(k) m ) of such tuples so that (2.2) |a(k) 1 b1 + . . . + a(k) m bm|′ < 2−k .
By passing to a sub-sequence and scaling we can assume that a(k) j = 1 for all k and for some j, say j = 1. Taking differences, we see that (a(k+1) 2 −a(k) 2 )b2 + . . . + (a(k+1) m −a(k) m )bm ′ < 2−k for all k.
By induction, there is c > 0 such that |a2b2 +. . .+ambm|′ ≥c max{|a2|, . . . , |am|}.
This implies that |a(k+1) j −a(k) j | ≤c−12−k for all 2 ≤j ≤m, so (a(k) j )k is a Cauchy sequence and converges to a limit aj. Taking the limit in (2.2) we find that b1 + a2b2 + . . . + ambm = 0 , which contradicts the linear independence of b1, b2, . . . , bm.
Now let (b1, b2, . . . , bn) be a K-basis of L, and let c > 0 be the associated constant.
If (αm) is a Cauchy sequence in L, write αm = a1mb1 + . . . + anmbn. Then |ajm′ −ajm| ≤c−1|αm′ −αm|′ , so each sequence (ajm)m is a Cauchy sequence in K and so converges to some aj ∈K. But then αm →a1b1 + · · · + anbn converges as well.
K § 2. p-adic numbers 12 We remark that (2.1) implies that the topology on L induced by |·|′ is the same as the topology induced by any K-linear isomorphism L →Kn, where Kn has the product topology.
Since |·|′ is uniquely determined and extends |·|, one simply writes |·| for the absolute value on L.
We note that it can be shown that a field that is complete with respect to an archimedean absolute value must be isomorphic to R or C (with the usual absolute value) and that a locally compact (and then necessarily complete) non-archimedean field of characteristic 0 must be isomorphic to a finite extension of Qp for some prime number p. (In characteristic p, it will be isomorphic to a finite extension of the field Fp( (t) ) of formal Laurent series over Fp, which is the field of fractions of the ring of formal power series Fp[ [t] ].) 2.23. Corollary.
Let ¯ Qp be the algebraic closure of Qp. Then ¯ Qp has a unique COR ¯ Qp has unique abs. value absolute value extending |·|p on Qp. If σ ∈Aut(¯ Qp/Qp) is any automorphism, then |σ(α)|p = |α|p for all α ∈¯ Qp.
Proof. ¯ Qp is an algebraic extension of Qp, so Theorem 2.22 applies. The second statement follows from the uniqueness of the absolute value, since α 7→|σ(α)|p is another absolute value on ¯ Qp extending the p-adic absolute value on Qp.
K Next we want to show that ¯ Qp is not complete (contrary to the algebraic closure C of the completion R of Q with respect to the usual absolute value).
2.24. Lemma.
¯ Qp is not complete with respect to |·|p.
LEMMA ¯ Qp not complete Proof. For n ≥1, let ζn ∈¯ Qp be a primitive (p2n −1)-th root of unity. It follows from properties of finite fields that ζn is a root of a monic polynomial fn ∈Zp[x] of degree 2n that reduces to an irreducible polynomial in Fp[x]; in particular, fn is irreducible itself. Since ζn is a power of ζn+1, we obtain a tower of fields Qp = Qp(ζ1) ⊂Qp(ζ2) ⊂Qp(ζ3) ⊂. . . ⊂¯ Qp such that [Qp(ζn+1) : Qp(ζn)] = 2 for all n ≥1.
We now consider the series P∞ n=1 ζnpn. If ¯ Qp were complete, then series would converge (since |ζn|p = 1).
So we will show that the series does not converge in ¯ Qp. The proof will be by contradiction. So we assume that the series has a limit α ∈¯ Qp. Then m := [Qp(α) : Qp] is finite. Let σ ∈Aut(¯ Qp/Qp). By Corollary 2.23, σ is continuous with respect to the p-adic topology. This implies that σ(α) = ∞ X n=2 σ(ζn)pn .
Since [Qp(α) : Qp] = m, there are exactly m values that σ(α) can take (namely, the roots of the minimal polynomial of α over Qp). Now pick some n such that 2n > m. Note that sn = Pn k=1 ζkpk generates the field Qp(ζn) (this is easily seen by induction). So, as σ runs through the automorphisms of ¯ Qp over Qp, σ(sn) takes exactly 2n distinct values.
Since the various possible values of σ(ζn) all differ mod p (this is because fn has only simple roots in ¯ Fp), we conclude that the various values of σ(sn) all differ mod pn+1. On the other hand, σ(α) ≡σ(sn) mod pn+1 , § 2. p-adic numbers 13 and there are only m < 2n possibilities for the left hand side. This gives the desired contradiction.
K 2.25. Definition.
We define Cp to be the completion of ¯ Qp.
♦ DEF Cp 2.26. Lemma.
Let (K, |·|) be an algebraically closed field with a non-archimedean LEMMA completion stays alg. closed absolute value, and let K′ be its completion. Then K′ is also algebraically closed.
Proof. Let K′ ⊂L be a finite field extension and let α ∈L; after scaling α by an element of K, we can assume that |α| ≤1 (recall that L has a unique absolute value extending that of K′). We must show that α ∈K′. Let f ∈K′[x] be the minimal polynomial of α. Since K is dense in K′, there is a sequence (fn) of monic polynomials in K[x] such that |fn −f| < 2−n. Since K is algebraically closed, each fn splits into linear factors. By the ultrametric triangle inequality (and using |α| ≤1), we have Y α′ : fn(α′)=0 |α −α′| = |fn(α)| = |fn(α) −f(α)| ≤|fn −f| < 2−n .
There must then be a root αn of fn such that |α −αn| < 2−n/ deg(f). This implies that (αn) converges in L to α. But K′ is the closure of K, therefore the limit α must already be in K′.
K So Cp is the unique (up to isomorphism) minimal extension of Qp that is complete and algebraically closed. So in this sense, Cp is the p-adic analogue of the field of complex numbers.
The residue field of ¯ Qp and of Cp is ¯ Fp. Since this is infinite, it follows that nei-ther of these two fields is locally compact: the closed ball around zero of radius 1 contains infinitely many elements whose pairwise distance is 1 (a system of rep-resentatives of the residue classes), and so no sub-sequence of a sequence of such elements can converge.
§ 3. Newton Polygons 14 3. Newton Polygons Let (K, |·|) be a complete field with a nontrivial non-archimedean absolute value.
Consider a polynomial 0 ̸= f ∈K[x]. Let α ∈¯ K be a root of f. We have seen that there is a unique extension of |·| to ¯ K, so it makes sense to consider |α|. We will now describe a method that determines the absolute values of the roots of f and how often they occur.
In this context it is advantageous to switch to a ‘logarithmic’ version of the absolute value. We fix a positive real number c and set v(x) = −c log |x| for x ∈K× and v(0) = +∞.
3.1. Definition.
The map v: K →R ∪{+∞} is the (additive) valuation asso-DEF valuation ciated to |·|.
♦ Of course, changing c will scale v by a positive factor, so v is not uniquely deter-mined.
Corresponding to the properties of absolute values, the valuation satisfies (1) v(x) < ∞if x ∈K×; (2) v(xy) = v(x) + v(y) for all x, y ∈K; (3) v(x + y) ≥min{v(x), v(y)} for all x, y ∈K, with equality when v(x) ̸= v(y).
The adjective ‘additive’ refers to the second property.
When dealing with p-adic fields like Qp, ¯ Qp or Cp, we choose c = 1/ log p; then v = vp on Qp, so v(Q× p ) = Z and (as is easily seen) v(¯ Q× p ) = v(C× p ) = Q.
Now let α ∈¯ K be a root of f = a0 + a1x + . . . + anxn ∈K[x] with an ̸= 0. Then a0 + a1α + a2α2 + . . . + anαn = 0 , so by Lemma 2.13 there must be (at least) two terms in the sum whose absolute value is maximal, or equivalently, whose valuation is minimal. These valuations are v(a0), v(a1) + v(α), v(a2) + 2v(α), . . . , v(an) + nv(α) .
So we need to have 0 ≤k < m ≤n such that v(ak) + kv(α) = v(am) + mv(α) ≤v(aj) + jv(α) for all 0 ≤j ≤n.
This is equivalent to saying that the line of slope −v(α) through the points (k, v(ak)) and (m, v(am)) in the plane has no point (j, v(aj)) below it.
This prompts the following definition.
3.2. Definition.
Let 0 ̸= f = a0 + a1x + . . . + anxn ∈K[x] with K as above.
DEF Newton polygon The Newton polygon of f is the lower convex hull of the set of the points (j, v(aj)) for 0 ≤j ≤n such that aj ̸= 0, i.e., the union of all line segments joining two of these points and such that the line through these points does not run strictly above any of the other points. A maximal such line segment is a segment of the Newton polygon; it has a slope (which is just its usual slope) and a length, which is the length of its projection to the x-axis.
♦ § 3. Newton Polygons 15 3.3. Example.
Consider the polynomial EXAMPLE Newton polygon f = x5 + 3x4 + 4x3 + 6x2 + 8 ∈Q2[x] .
The points (j, v(aj)) with aj ̸= 0 are (0, 3), (2, 1), (3, 2), (4, 0), (5, 0) .
We find three segments, forming a broken line with vertices (0, 3), (2, 1), (4, 0), (5, 0). The slopes are −1, −1/2 and 0, and the lengths are 2, 2 and 1.
0 1 2 3 4 5 1 2 3 ♣ What we did above amounts to the statement that the valuations of the roots of f are among the slopes taken negatively of the Newton polygon. We now want to prove a converse and give a more precise statement. We first introduce a variation of the absolute value on the polynomial ring.
3.4. Definition.
Let (K, |·|) be a field with a nontrivial non-archimedean abso-DEF |·|r on K[x] lute value and let r > 0. For f = a0 + a1x + . . . + anxn ∈K[x] we define |f|r := max{|aj|rj : 0 ≤j ≤n} .
When f ̸= 0, then we set ℓr(f) = max{j : |aj|rj = |f|r} −min{j : |aj|rj = |f|r} ∈Z≥0 .
♦ We observe that ℓr(f) is strictly positive if and only if c log r is the slope of a segment of the Newton polygon of f, and that in this case ℓr(f) is the length of the corresponding segment.
One can easily adapt the proof for the case r = 1 to show that |·|r is an abso-lute value on K[x] (in the sense that it satisfies properties (1), (2) and (3′) of Definition 2.1). It also restricts to |·| on K.
3.5. Lemma.
Let f, g ∈K[x] be nonzero polynomials and let r > 0. Then we LEMMA additivity of ℓr have ℓr(fg) = ℓr(f) + ℓr(g) .
Proof. If f = a0 + a1x + . . . + anxn, write n−(f) = min{j : |aj|rj = |f|r} and n+(f) = max{j : |aj|rj = |f|r} and similarly for g and fg. Precisely as in the proof of property (2) for |·|r (i.e., basically Gauss’ Lemma) one sees that n−(fg) = n−(f) + n−(g) and n+(fg) = n+(f) + n+(g) , § 3. Newton Polygons 16 which implies the claim, since ℓr(f) = n+(f) −n−(f) and similarly for g and fg.
K 3.6. Theorem.
Let 0 ̸= f ∈K[x] and let r > 0. Then the number of roots α ∈¯ K THM roots and Newton polygon of f (counted with multiplicity) such that |α| = r is exactly ℓr(f).
In terms of the Newton polygon, this says that f has roots of valuation s if and only if its Newton polygon has a segment of slope −s, and the number of such roots (counted with multiplicity) is exactly the length of the segment.
Proof. We can assume that K = ¯ K. The proof is by induction on the degree of f.
If f is constant, there is nothing to show. If f = x −α, then ℓr(f) = 0 if |α| ̸= r and ℓr(f) = 1 if |α| = r.
Now assume f is not constant and let β ∈¯ K be a root of f. Then f = (x −β)f1 for some 0 ̸= f1 ∈K[x]. By the inductive hypothesis, the number of roots α of f1 such that |α| = r is ℓr(f1). If |β| = r, then the number of such roots of f is ℓr(f1) + 1 = ℓr(f1) + ℓr(x −β) = ℓr(f) (using Lemma 3.5). If |β| ̸= r, then the number of such roots of f is ℓr(f1) = ℓr(f1) + ℓr(x −β) = ℓr(f) again.
K 3.7. Lemma.
Let f ∈K[x] be irreducible. Then the Newton polygon of f consists LEMMA Newton polygon of irreducible polynomial of a single segment.
Proof. We assume for simplicity that K has characteristic zero. Then ¯ K is sepa-rable over K and the roots of f form one orbit under the Galois group Aut( ¯ K/K).
Since the absolute value is invariant under the action of this group, we see that all roots of f have the same absolute value. The claim then follows from Theorem 3.6.
(In characteristic p, let α be a root of f. Then for some n ≥0, αpn is separable over K and f = g(xpn) for an irreducible polynomial g ∈K[x]. The previous argument applies to g, but this then implies the claim also for f.) K This leads to the following consequence.
3.8. Lemma.
Let 0 ̸= f ∈K[x] with f(0) ̸= 0 and let σ1, . . . , σm be the segments LEMMA slope factorization of the Newton polygon of f. Then there is a factorization f = f1f2 · · · fm such that the Newton polygon of fj is a single segment with the same slope and length as σj.
Proof. We can assume that f is monic. Let f = h1 · · · hn be the factorization of f into monic irreducible polynomials over K. By Lemma 3.7, the Newton polygon of each hj consists of a single segment. Let s1, s2, . . . , sm be the distinct slopes that occur for these segments; we can number them so that sj is the slope of σj.
We then define fj to be the product of the hi whose slope is sj. The claim follows from the additivity of the lengths of segments of the same slope, Lemma 3.5.
K We extend |·|r to an absolute value on the field K(x) of rational functions in one variable over K in the usual way. For a given rational function f, we can then study how |f|r varies with r.
§ 3. Newton Polygons 17 3.9. Theorem.
Let f ∈K(x)×. Then the function THM variation of |f|r with r ϕf : R − →R , s 7− →−c log |f|e−s/c is piecewise affine with integral slopes. If ∂−ϕf(s) −∂+ϕf(s) = ν, then ν is the number of zeros α of f with v(α) = s minus the number of poles α of f with v(α) = s (each counted with multiplicity).
Here ∂+ϕf(s) = lim ε↘0 ϕf(s + ε) −ϕf(s) ε and ∂−ϕf(s) = lim ε↘0 ϕf(s) −ϕf(s −ε) ε are the right and left derivatives of the piecewise affine function ϕf.
Proof. We can write f = γ Qn j=1(x −αj)ej for some γ ∈K, αj ∈¯ K and ej ∈Z, so ϕf(s) = −c log |γ| −c n X j=1 ej log |x −αj|e−s/c = v(γ) + n X j=1 ejϕx−αj(s) , and it suffices to prove the claim for f = x −α. In this case, we have ϕx−α(s) = −c log max{e−s/c, |α|} = min{s, v(α)} (recall that v(0) = +∞). This is a piecewise affine function with slopes 1 and 0 (unless α = 0). If α = 0, the slope is constant, so ν = 0 for all s, and there are no zeros or poles α with v(α) = s. If α ̸= 0, then the slope changes from 1 to 0 at s = v(α), so ν = 1 there and ν = 0 for all other s.
K 3.10. Example.
For the polynomial EXAMPLE ϕf for f from Example 3.3 f = x5 + 3x4 + 4x3 + 6x2 + 8 ∈Q2[x] from Example 3.3, the graph of ϕf looks as follows.
5 4 2 0 0 1 2 1/2 3/2 1 2 3 -2 -1 § 3. Newton Polygons 18 The green lines are the graphs of s 7→v(aj) + js (for j = 0, 2, 3, 4, 5); ϕf is the minimum of these functions. The segments of the graph of ϕf correspond to the vertices of the Newton polygon of f and the vertices of the graph of ϕf correspond to the segments of the Newton polygon of f.
♣ We give another characterization of |f|r.
3.11. Lemma.
Let (K, |·|) be a complete non-archimedean field, let r > 0 and LEMMA |·|r as supremum norm 0 ̸= f ∈K[x]. Then |f|r = sup{|f(α)| : α ∈¯ K, |α| ≤r} .
Proof. Let f = a0 + a1x + . . . + anxn. If |α| ≤r, then |f(α)| = |a0 + a1α + . . . + anαn| ≤max{|a0|, |a1|r, . . . , |an|rn} = |f|r .
Conversely, for any ε > 0 there is r −ε < ρ ≤r such that ρ ∈| ¯ K×| and c log ρ is not a slope of the Newton polygon of f (recall that | ¯ K×| is dense in R>0; there are only finitely many slopes). Then for any α ∈¯ K such that |α| = ρ, there is a unique term ajαj of maximal absolute value and therefore we have |f(α)| = |f|ρ.
Letting ρ tend to r, we find |f|r ≤sup{|f(α)| : α ∈¯ K, |α| ≤r}.
K § 4. Multiplicative seminorms and Berkovich spaces 19 4. Multiplicative seminorms and Berkovich spaces The absolute values |·|r on the polynomial ring K[x] that we have studied in the last section are examples of ‘multiplicative seminorms’.
4.1. Definition.
Let K be a field with absolute value |·| and let A be a K-algebra DEF multiplicative seminorm Banach algebra (i.e., A is a ring with a ring homomorphism K →A that gives A a compatible structure as a K-vector space). A multiplicative seminorm on A is a map A − →R≥0 , a 7− →∥a∥ such that (1) ∥·∥restricts to |·| on K; (2) ∥a + b∥≤∥a∥+ ∥b∥for all a, b ∈A; (3) ∥ab∥= ∥a∥· ∥b∥for all a, b ∈A.
If in addition a = 0 is the only element with ∥a∥= 0, then ∥·∥is a multiplicative norm (which is the same as an absolute value on A extending |·|). In general, we call ker ∥·∥= {a ∈A : ∥a∥= 0} the kernel of ∥·∥; it is a prime ideal in A.
A K-algebra A with a fixed multiplicative norm such that A is complete with respect to this norm is a Banach algebra over K.
♦ A seminorm on A only needs to satisfy ∥ab∥≤∥a∥· ∥b∥with equality for a ∈K; similarly for a norm.
If |·| is non-archimedean, then ∥·∥also satisfies the ultrametric triangle inequality, since then ∥a + b∥n = ∥(a + b)n∥= n X j=0 n j ajbn−j ≤ n X j=0 n j ∥a∥j∥b∥n−j ≤ n X j=0 ∥a∥j∥b∥n−j ≤(n + 1)(max{∥a∥, ∥b∥})n and n √n + 1 →1 as n →∞.
We introduce the following notation.
4.2. Definition.
Let (K, |·|) be a field with absolute value, let a ∈K and r ≥0.
DEF D(a, r) Then D(a, r) := DK(a, r) := {ξ ∈K : |ξ −a| ≤r} is the closed disk of radius r around a.
♦ 4.3. Examples.
Let (K, |·|) be a complete non-archimedean field.
EXAMPLES multiplicative seminorms on K[x] (1) For any a ∈K, the map f 7− →∥f∥a,0 := |f(a)| is a multiplicative seminorm on K[x].
(2) For any a ∈K and any r > 0, the map f 7− →∥f∥a,r := |f(x + a)|r is a multiplicative (semi)norm on K[x].
§ 4. Multiplicative seminorms and Berkovich spaces 20 (3) Let (an) be a sequence in K and (rn) a strictly decreasing sequence in R>0 such that D(an+1, rn+1) ⊂D(an, rn) for all n. Then f 7− →∥f∥= lim n→∞∥f∥an,rn is a multiplicative seminorm on K[x].
(1) is clear and (2) follows from the properties of |·|r.
For (3) note that by Lemma 3.11, ∥f∥an+1,rn+1 = sup{|f(α)| : α ∈¯ K, |α −an+1| ≤rn+1} ≤sup{|f(α)| : α ∈¯ K, |α −an| ≤rn} = ∥f∥an,rn , so that the sequence (∥f∥an,rn) decreases, hence must have a limit. Properties (2) and (3) from Definition 4.1 then follow by taking the limit in the corresponding relations for the ∥·∥an,rn.
♣ In fact, this is essentially the full story, at least when K is algebraically closed and complete.
4.4. Theorem.
Let (K, |·|) be a complete and algebraically closed non-archimedean THM classification of mult.
seminorms on K[x] field and let ∥·∥be a multiplicative seminorm on K[x]. Then there is a decreasing nested sequence of disks D(an, rn) such that ∥f∥= lim n→∞∥f∥an,rn for all f ∈K[x].
Proof. Let D = {D(a, r) : a ∈K, r > 0, ∥·∥≤∥·∥a,r} be the set of all closed disks such that ∥·∥is bounded above by the corresponding seminorm. Then for all a ∈K we have D(a, ∥x −a∥) ∈D: if f(x + a) = a0 + a1x + . . . + anxn, then ∥f∥= ∥a0 + a1(x −a) + . . . + an(x −a)n∥ ≤max{|a0|, |a1|∥x −a∥, . . . , |an|∥x −a∥n} = |f(x + a)|∥x−a∥= ∥f∥a,∥x−a∥.
In particular, D is non-empty. Conversely, we have |x −a|a,r = r, which implies that if D(a, r) ∈D, then r ≥∥x −a∥. So D = {D(a, r) : a ∈K, r ≥∥x −a∥} .
We claim that D does not contain two disjoint disks: assume that D1 = D(a1, r1) and D2 = D(a2, r2) are in D. Then we have ∥x −a1∥≤r1 and ∥x −a2∥≤r2, so that |a1 −a2| = ∥(x −a2) −(x −a1)∥≤max{r1, r2} .
If r1 ≤r2, then this implies that a1 ∈D2 and then that D1 ⊂D2; if r2 ≤r1, then we see in the same way that D2 ⊂D1.
Now let ρ = inf{r > 0 : ∃a ∈K : D(a, r) ∈D} and choose sequences (an) in K and (rn) in R>0 such that (rn) is strictly decreasing with rn →ρ and D(an, rn) ∈D for all n. Then we have D(an+1, rn+1) ⊂D(an, rn). This implies ∥f∥≤lim n→∞∥f∥an,rn .
§ 4. Multiplicative seminorms and Berkovich spaces 21 We still have to show the reverse inequality. Since every nonzero f is a product of a constant and polynomials of the form x −a, it suffices to prove this for the latter. If |a −an0| > rn0 for some n0, then ∥x −a∥= ∥(x −an0) −(a −an0)∥= |a −an0| = lim n→∞max{|a −an|, rn} = lim n→∞∥x −a∥an,rn , since ∥x−an0∥≤rn0 < |a−an0|. Otherwise, we have |a−an| ≤rn for all n, which says that a ∈T n D(an, rn), so that D(an, rn) = D(a, rn) and we have lim n→∞∥x −a∥an,rn = lim n→∞∥x −a∥a,rn = ∥x −a∥a,ρ = ρ .
Since D(a, ∥x −a∥) ∈D and ρ is the smallest possible radius of a disk in D, we get ∥x −a∥≥ρ as required.
K Note that if a ∈T n D(an, rn) in the proof above, then D(a, ρ) ∈D, since ∥f∥≤lim n→∞∥f∥an,rn = lim n→∞∥f∥a,rn = ∥f∥a,ρ for all f ∈K[x]. Since ρ ≤r for any radius r of a disk in D and no two disks in D are disjoint, it follows that D(a, ρ) = T D. We can therefore distinguish the following four types of multiplicative seminorms on K[x].
4.5. Definition.
Let ∥·∥be a multiplicative seminorm on K[x], where K is a DEF types of mult.
seminorms complete and algebraically closed non-archimedean field. Let D be as in the proof of Theorem 4.4.
(1) ∥·∥is of type 1 if T D = {a} for some a ∈K. Then ∥f∥= |f(a)| and ker ∥·∥ is the kernel of the evaluation map f 7→f(a).
(2) ∥·∥is of type 2 if T D = D(a, r) for some a ∈K and r > 0 such that r ∈|K×|.
Then ∥f∥= |f|a,r.
(3) ∥·∥is of type 3 if T D = D(a, r) for some a ∈K and r > 0 such that r / ∈|K×|.
Then ∥f∥= |f|a,r.
(4) Finally, ∥·∥is of type 4 if T D = ∅.
In the last three cases, ∥·∥is actually a norm.
♦ We note that if ρ = 0 in the proof of Theorem 4.4, then the completeness of K implies that D = {a} for some a ∈K. If T D ̸= ∅for every decreasing nested sequence of disks in K, then K is said to be spherically complete.
(Then no DEF spherically complete multiplicative seminorms of type 4 exist.) For example, Qp is spherically complete, but Cp is not (Exercise).
For types 2, 3 and 4, ∥·∥defines an absolute value on K[x]. We extend it to an absolute value on the field of fractions K(x), which we can then complete to obtain H∥·∥, and we obtain a K-algebra homomorphism K[x] →H∥·∥. For example, if ∥·∥= |·|r is of type 2 or 3 with a = 0, then the corresponding completion of K[x] is the ring K⟨r−1x⟩:= n ∞ X j=0 ajxj ∈K[ [X] ] : lim j→∞|aj|rj = 0 o of power series converging on D(0, r) (Exercise), so H|·|r is the field of fractions of K⟨r−1x⟩.
If ∥·∥is of type 1, then the evaluation map at a gives us a K-algebra homo-morphism K[x] →K =: H∥·∥. In each case, we obtain ∥·∥as the pull-back of § 4. Multiplicative seminorms and Berkovich spaces 22 the absolute value of a complete field H that is a K-Banach algebra via a K-algebra homomorphism K[x] →H. Conversely, if we have such a homomorphism K[x] →H, then pulling back the absolute value of H to K[x] will give us a multiplicative seminorm on K[x].
4.6. Definition.
Let K be a complete and algebraically closed non-archimedean DEF Berkovich space field and let A be a finitely generated K-algebra, so that A is the coordinate ring of the affine K-variety X = Spec A. Then the Berkovich space associated to A or X is Berk A := Xan := {∥·∥: A →R multiplicative seminorm on A} , the set of multiplicative seminorms on A. The topology on Xan is the weakest topology that makes the maps Xan →R, ∥·∥7→∥f∥, continuous for all f ∈A.
(Concretely, this means that any open set is a union of finite intersections of sets of the form Uf,a,b = {∥·∥∈Xan : a < ∥f∥< b}.) ♦ We will usually call elements of Xan points and denote them by ξ or similar. In this context, the corresponding multiplicative seminorm will be written ∥·∥ξ and the K-Banach algebra obtained by completion, Hξ.
4.7. Lemma.
The topological space Xan as defined in Definition 4.6 is Hausdorff.
LEMMA Xan is Hausdorff Proof. Let ξ, ξ′ ∈Xan be distinct. We must show that there are disjoint open sets U, U ′ ⊂Xan with ξ ∈U and ξ′ ∈U ′. Since ξ ̸= ξ′, there must be f ∈A such that ∥f∥ξ ̸= ∥f∥ξ′. Assume without loss of generality that ∥f∥ξ < ∥f∥ξ′ and let ∥f∥ξ < a < ∥f∥ξ′. Then we can take U = Uf,−∞,a and U ′ = Uf,a,∞.
K Note that we can alternatively define Berk A as the set of all K-algebra homomor-phisms A →H into complete K-Banach algebras that are fields, up to an obvious equivalence. The topology is then that of pointwise convergence. This can be seen as similar to the definition of Spec A as the set of all K-algebra homomor-phisms into fields, up to an obvious equivalence. Here the topology is again that of pointwise convergence, but with the cofinal topology on the target fields (so that basic open sets are defined by relations f ̸= 0). Since a Banach algebra has more structure than just a K-algebra, there are fewer equivalences and therefore more points in Berk A than in Spec A. This is made precise by the following statements.
4.8. Lemma.
Let K, A and X be as in Definition 4.6. Then there is a canonical LEMMA X(K) , →Xan inclusion of X(K) into Xan. The inclusion is continuous when X(K) is given the topology induced by the absolute value on K.
Proof. Let ξ ∈X(K). Then ξ gives rise to a K-algebra homomorphism A →K, f 7→f(ξ).
We define ˜ ξ ∈Xan to correspond to the multiplicative seminorm f 7→|f(ξ)|. Its kernel is the ideal of A consisting of functions vanishing at ξ and so determines ξ. This gives us the desired inclusion.
To show that ξ 7→˜ ξ is continuous, consider a basic open set Uf,a,b in Xan. Its preimage is {ξ ∈X(K) : a < |f(ξ)| < b}.
Since every f ∈A is continuous as a map X(K) →K (with respect to the K-topology) and |·|: K →R is also continuous, this set is open.
K § 4. Multiplicative seminorms and Berkovich spaces 23 It is in fact the case that the image of X(K) in Xan is dense. We will see this later when X is the affine line.
There is also a natural map in the other direction.
4.9. Lemma.
Let K, A and X be as in Definition 4.6. Then there is a canonical LEMMA Berk A →Spec A continuous map Berk A →Spec A (where Spec A has the Zariski topology).
Proof. Let ξ ∈Berk A. Then ∥·∥ξ is the pull-back of the absolute value from some K-Banach algebra H that is a field under a homomorphism φ: A →H. Since H is a field, the kernel of φ must be a prime ideal of A and so defines an element of Spec A (or we just take H as a field, forgetting the absolute value).
It remains to show that the map Berk A →Spec A obtained in this way is continu-ous. Let Uf = {ξ ∈Spec A : f(ξ) ̸= 0} be a basic open set in Spec A. Its pull-back to Berk A is {ξ ∈Berk A : ∥f∥ξ ̸= 0} = Uf,0,∞and therefore open, too.
K The composition of the two maps X(K) →Xan →Spec A is the inclusion of X(K) (the set of maximal ideals of A) into Spec A (the set of prime ideals of A).
The map is actually surjective: let ξ ∈Spec A; this gives us a K-algebra homo-morphism A →R into an integral domain finitely generated over K. One can show that one can always define an absolute value on such an R that extends the absolute value on K. Pulling back to A, we obtain a multiplicative seminorm; this is a preimage of ξ.
It is also true that Xan is (even path-)connected when X is connected as an algebraic variety and that Xan is locally compact. We will see this concretely for X = A1 in the next section.
§ 5. The Berkovich affine and projective line 24 5. The Berkovich affine and projective line In the following, we will take a closer look at the Berkovich affine line over Cp, A1,an Cp = Berk Cp[x]. According to the classification of Definition 4.5, we have four types of points in A1,an Cp . The type 1 points recover the points in A1(Cp) = Cp as in Lemma 4.8. The type 2 and 3 points correspond to closed disks D(a, r) with r > 0 in, respectively, not in, the value group pQ of Cp. We will reserve the notation ζa,r or ζD where D = D(a, r) for these points. We will frequently identify a with ζa,0, however. Finally, the type 4 points correspond to nested sequences of disks with empty intersection; these points are somewhat annoying, but don’t usually give us problems. They are necessary to make the space locally compact. Two nested sequences of disks (D(an, rn))r and (D(a′ n, r′ n))n with empty intersection define the same type 4 point if and only if D(an, rn) ∩D(a′ n, r′ n) ̸= ∅for all n.
Now fix ξ ∈Cp and consider real numbers 0 ≤r0 < r1. Then there is a map [r0, r1] − →A1,an Cp , r 7− →∥·∥ξ,r = ζξ,r .
This map is continuous, since for any f ∈Cp[x], the map r 7− →∥f∥ξ,r = max{|aj|prj : 0 ≤j ≤n} is continuous, where f(x + ξ) = a0 + a1x + . . . + anxn. The map is also clearly injective. Since [r0, r1] is compact, the map is actually a homeomorphism onto its image. We write [ζξ,r0, ζξ,r1] for this image.
Now consider two points ξ, η ∈Cp and let δ = |ξ −η|p their distance. Then D(ξ, δ) = D(η, δ) (“every point in a disk is a center”). Define ξ ∨η := ζξ,δ = ζη,δ .
Then γξ,η : [0, 2δ] − →A1,an Cp , r 7− → ( ζξ,r if 0 ≤r ≤δ, ζη,2δ−r if δ ≤r ≤2δ is a continuous path in A1,an Cp joining ξ and η, whose image is [ξ, ξ ∨η] ∪[η, ξ ∨η].
We can extend the definition of ξ ∨η to arbitrary points in A1,an Cp . We first observe that ∥x −η∥ξ,r = ( ∥x −η∥η,r = r if η ∈D(ξ, r), |ξ −η| > r if η / ∈D(ξ, r).
This shows that D(ξ, r) is uniquely determined by ∥·∥ξ,r and also shows that D1 ⊂D2 holds for two closed disks if and only if ∥·∥D1 ≤∥·∥D2 (the ‘only if’ part follows also from the characterization of ∥·∥ξ,r as the sup norm on D(ξ, r)). This prompts us to define ξ ≤ξ′ ⇐ ⇒∀f ∈Cp[x]: ∥f∥ξ ≤∥f∥ξ′ for arbitrary points ξ, ξ′ ∈A1,an Cp . Furthermore, if ξ ≤ξ′, then we write [ξ, ξ′] for the set of points ξ′′ such that ξ ≤ξ′′ ≤ξ′. If ξ and ξ′ are points of type 1, 2 or 3, then it is clear by the above that [ξ, ξ′] is homeomorphic to a closed interval in R.
Regarding type 4 points, we have the following.
§ 5. The Berkovich affine and projective line 25 5.1. Lemma.
Let ξ ∈A1,an Cp be a type 4 point, represented by the nested sequence LEMMA comparison with type 4 points of disks D(an, rn) with empty intersection. If ξ′ ∈A1,an Cp with ξ ≤ξ′, then either ξ′ = ξ, or else ξ′ is a point of type 2 or 3, corresponding to a disk D(a, r) such that D(an, rn) ⊂D(a, r) for all sufficiently large n.
Let r∞= limn→∞rn. If ξ′ corresponds to D(a, r), then γξ,ξ′ : [r∞, r] − →A1,an Cp , ρ 7− → ( ζan,ρ if ρ ≥rn, ξ if ρ = r∞ is a continuous map with image [ξ, ξ′].
Proof. Let ξ′ ≥ξ and assume first that ξ′ is of type 2 or 3, so ξ′ = ζa,r for some a ∈Cp and r > 0. We claim that D(a, r) ∩D(an, rn) ̸= ∅for all n. Otherwise, we would have empty intersection for all sufficiently large n, and then ∥x −a∥ξ′ = r < lim n→∞|a −an| = lim n→∞∥x −a∥an,rn = ∥x −a∥ξ , which contradicts ξ ≤ξ′. If n is large enough so that rn ≤r, then we must therefore have D(an, rn) ⊂D(a, r).
Now assume that ξ′ is of type 1 or 4. If ξ′ is of type 1, then ∥x −η∥ξ′ is zero when η = ξ′, whereas ∥x−η∥ξ ≥r∞> 0 for all η, so this is not possible. If ξ′ is of type 4, say represented by the nested sequence of disks D(a′ n, r′ n), then ξ ≤ξ′ ≤ζa′ n,r′ n for all n. By the argument above, it follows that D(an, rn) ∩D(a′ m, r′ m) ̸= ∅for all m, n. But this exactly means that ξ = ξ′.
It remains to show that γξ,ξ′ is well-defined and continuous. For the first, note that ζan,ρ = ζan+1,ρ when ρ ≥rn, since D(an, ρ) = D(an+1, ρ). For the second, consider f ∈Cp[x]. We have to show that [r∞, r] ∋ρ 7− →∥f∥γξ,ξ′(ρ) is continuous. This is clear on (r∞, r], and it follows for the left endpoint by the definition of ∥·∥ξ, which implies that ∥f∥ξ = lim ρ↘r∞∥f∥an(ρ),ρ (with n(ρ) such that ρ ≥rn).
K For ξ′ ≤ξ, we define the path γξ,ξ′ as the reversal of the path γξ′,ξ.
5.2. Theorem.
Any two points ξ, ξ′ ∈A1,an Cp have a least upper bound ξ ∨ξ′ (with THM A1,an Cp is path-connected respect to the ordering introduced above). The path γξ,ξ∨ξ′ + γξ∨ξ′,ξ′ connects the two points.
Proof. This is clear if neither ξ nor ξ′ are of type 4: if ξ = ζa,r and ξ′ = ζa′,r′, then ξ ∨ξ′ = ζa,ρ = ζa′,ρ, where ρ = max{r, r′, |a −a′|}. If ξ ≤ξ′ or ξ′ ≤ξ, then the statement is also clear (with ξ ∨ξ′ = ξ or ξ′). Otherwise, we can represent ξ and ξ′ by nested sequences of disks D(an, rn) and D(a′ n, r′ n), respectively, such that D(an, rn) ∩D(a′ n, r′ n) = ∅for n large enough. For all such n, ζan,rn ∨ζa′ n,r′ n is the same and therefore agrees with ξ ∨ξ′.
The statement on path-connectedness is then clear.
K We define analogues of closed disks in A1,an Cp .
§ 5. The Berkovich affine and projective line 26 5.3. Definition.
Let a ∈Cp and r ≥0. We set DEF closed disk in A1,an Cp D(a, r) = {ξ ∈A1,an Cp : ξ ≤ζa,r} = {ξ ∈A1,an Cp : ∥x −a∥ξ ≤r} .
and call this a closed disk in A1,an Cp . We also define the corresponding open disk in A1,an Cp to be D(a, r)−= {ξ ∈A1,an Cp : ∥x −a∥ξ < r} = Ux−a,−∞,r .
♦ The equality in the definition of D(a, r) can be seen as follows. ‘⊂’ is clear. To show ‘⊃’, assume that ∥x −a∥ξ ≤r. Then for any b ∈Cp, we have ∥x−b∥ξ = ∥(x−a)+(a−b)∥ξ ≤max{∥x−a∥ξ, |a−b|} ≤max{r, |a−b|} = ∥x−b∥a,r .
Since every f ∈Cp[x] is a constant times a product of such terms, it follows that ξ ≤ζa,r.
5.4. Lemma.
LEMMA sub-basis of topology (1) Let f = (x −α1) · · · (x −αn) ∈Cp[x] be non-constant and let a ∈R. Then there are r1, . . . , rn ≥0 such that {ξ ∈A1,an Cp : ∥f∥ξ ≤a} = D(α1, r1) ∪. . . ∪D(αn, rn) and {ξ ∈A1,an Cp : ∥f∥ξ < a} = D(α1, r1)−∪. . . ∪D(αn, rn)− (2) The open disks D(a, r)−and the complements of closed disks D(a, r) generate the topology of A1,an Cp (i.e., every open set is a union of finite intersections of such sets).
(3) Two (open or closed) disks in A1,an Cp are either disjoint or one is contained in the other.
Proof.
(1) Exercise.
(2) We have Uf,a,b = Uf,−∞,b ∩Uf,a,∞. By part (1), we can write Uf,−∞,b as a union of open disks, and we can write Uf,a,∞, which is the complement of the set of ξ such that ∥f∥ξ ≤a, as the complement of a finite union of closed disks, which is the same as a finite intersection of complements of closed disks. So each basic open set Uf,a,b is a finite intersection of open disks and complements of closed disks; this implies the claim.
(3) Assume that D(a, r) and D(a′, r′) are not disjoint, so that there is ξ with ∥x −a∥ξ ≤r and ∥x −a′∥ξ ≤r′, w.l.o.g. such that r ≥r′. Then |a −a′| ≤max{∥x −a∥ξ, ∥x −a′∥ξ} ≤r .
Now let ξ′ ∈D(a′, r′) be arbitrary. Then ∥x −a∥ξ′ ≤max{∥x −a′∥ξ′, |a −a′|} ≤max{r′, r} = r , so ξ′ ∈D(a, r). Hence D(a′, r′) ⊂D(a, r). The case when one or both disks are open is similar.
K Together with (2) (and the fact that A1,an Cp is a union of open disks), statement (3) says that the open disks, from which a finite number of (pairwise disjoint) closed disks is removed, form a basis of the topology: every open set is a union of such sets.
§ 5. The Berkovich affine and projective line 27 5.5. Lemma.
Let a ∈Cp and r ≥0.
LEMMA D(a, r) is compact (1) The set of type 1 points in D(a, r) is D(a, r), and D(a, r) is dense in D(a, r).
(2) D(a, r) is compact.
Proof. The first claim in (1) is clear (and holds in a similar way for open disks). For the second claim, consider the intersection of a basic open set as above with D(a, r).
The set of type 1 points contained in this intersection is the intersection of D(a, r) with the open disk in Cp corresponding to the open disk, minus the union of the closed disks in Cp corresponding to the closed disks that were removed. Any such set is non-empty (unless the whole disk D(a, r) is removed, but then the basic open set has empty intersection with D(a, r)).
For (2), we can assume without loss of generality that a = 0. We know that ξ ∈D(0, r) ⇐ ⇒∥·∥ξ ≤|·|r, so the image of D(0, r) in RCp[x] under the map Φ: A1,an Cp − →RCp[x] , ξ 7− →(∥f∥ξ)f∈Cp[x] is the intersection of im(Φ) with the product C = Q f[0, |f|r]. The definition of the topology on A1,an Cp is equivalent to saying that it is the subspace topology induced by the map Φ above (with the product topology on RCp[x]). The product C of compact intervals is itself compact by Tychonoff’s Theorem. On the other hand, the conditions defining a multiplicative seminorm are closed conditions, so the image of Φ is closed. Now Φ(D(0, r)) is the intersection of a closed set and a compact set, so it is itself compact. Since Φ is a homeomorphism onto its image, it follows that D(0, r) is compact as well.
K 5.6. Corollary.
The space A1,an Cp is locally compact.
COR A1,an Cp is locally compact Proof. We must show that for every point ξ ∈A1,an Cp and every open subset U of A1,an Cp containing ξ, there is a compact neighborhood V of ξ contained in U.
Since U is open, U contains an open set of the form D(a, r)−\ S j D(aj, rj) that in turn contains ξ.
This means that ∥x −a∥ξ < r and ∥x −aj∥ξ > rj.
For r′ < r sufficiently close to r and for r′ j > rj sufficiently close to rj, we still have ∥x −a∥ξ < r′ and ∥x −aj∥ξ > r′ j. Then ξ ∈V := D(a, r′) \ [ j D(aj, r′ j)−⊂U .
Now V is compact (we intersect the compact set D(a, r′) with a closed set) and contains the open neighborhood D(a, r′)−\ S j D(aj, r′ j) of ξ.
K We now look a bit closer at the tree structure of A1,an Cp .
5.7. Definition.
The diameter of a point ξ ∈A1,an Cp is DEF diameter diam(ξ) := inf{∥x −a∥ξ : a ∈Cp} .
♦ Then diam(ζa,r) = r, and for a type 4 point ξ represented by a nested sequence of disks D(an, rn), we have diam(ξ) = limn→∞rn > 0.
We use this notion to define two metrics.
§ 5. The Berkovich affine and projective line 28 5.8. Definition.
For ξ, ξ′ ∈A1,an Cp , we define the small metric by DEF small and big metric d(ξ, ξ′) = 2 diam(ξ ∨ξ′) −diam(ξ) −diam(ξ′) = (diam(ξ ∨ξ′) −diam(ξ)) + (diam(ξ ∨ξ′) −diam(ξ′)) and in the case that ξ, ξ′ are both not of type 1, the big metric by ρ(ξ, ξ′) = c(2 log diam(ξ ∨ξ′) −log diam(ξ) −log diam(ξ′)) .
♦ So the small metric gives the total change of diameter as we move along the path joining ξ to ξ′, whereas the big metric does the same for the change in (additive) valuation (= −c log diam). Both are indeed metrics, d on all of A1,an Cp and ρ on the ‘hyperbolic part’ of A1,an Cp , which consists of the points of types 2, 3 and 4. The former has the advantage that it is also defined on the points of type 1, but the latter is in some sense more natural. It can be seen as analogous to the hyperbolic metric on the upper half plane in C (which is given by ds2 = (dx2 + dy2)/y2 when x and y are the real and imaginary parts), which is invariant under the action of PSL(2, R) by M¨ obius transformations. This analogous property will turn out to be satisfied (with respect to PGL(2, Cp)) by the big metric.
Now we introduce the kind of tree structure that shows up here.
(For more information, consult [BR, Appendix B].) 5.9. Definition.
An R-tree is a metric space (T, d) such that for any two points DEF R-tree x, y ∈T, there is a unique path [x, y] in T joining x to y, which is a geodesic segment (this means that the map γ : [a, b] →T giving the path can be chosen so that d(γ(u), γ(v)) = |u −v| for all u, v ∈[a, b]).
A point x ∈T is a branch point if T {x} has at least three connected components (in the metric topology). x is an endpoint if T \ {x} is connected. If T \ {x} has exactly two connected components, then x is said to be ordinary. The R-tree T is said to be finite, if it is compact and has only finitely many branch points and endpoints.
The strong topology on the R-tree is the metric topology on T. To define the weak topology, we define a tangent direction at x ∈T to be an equivalence class of paths [x, y] with y ̸= x, where two paths are equivalent when they share an initial segment.
The tangent directions at x are in one-to-one correspondence with the connected components of T \ {x}. For a tangent direction v at x, we set Bx,v = {y ∈T : y ̸= x, [x, y] ∈v} (which is the connected component of T \ {x} corresponding to v). Then the weak topology on T is the topology generated by the sets Bx,v.
♦ The set Bx,v can be interpreted as the set of points of T that can be ‘seen’ from x when looking into direction v. This is why the weak topology is also called the ‘observer’s topology’.
Now the following is a fact.
5.10. Theorem.
THM A1,an Cp as R-tree (1) The small metric d turns A1,an Cp into an R-tree. The Berkovich topology on A1,an Cp is the weak topology on this R-tree.
(2) The big metric ρ turns A1,an Cp \ Cp into an R-tree.
The subspace topology on A1,an Cp \ Cp is again the weak topology on this R-tree.
§ 5. The Berkovich affine and projective line 29 Proof. See [BR, Section 1.4], where the analogous statements are shown for D(0, 1).
For A1,an Cp the argument is the same (except that there is no natural way to pick a root).
K Compactifying the R-tree by adding a point at infinity (see below) we obtain the universal dendrite that was first constructed by Wa˙ zewski in his thesis in 1923. Recent work by Hrushovski, Loeser and Poonen1 shows that V an can be embedded in R2d+1 for any quasi-projective d-dimensional Cp-variety V (and P1,an Cp can be embedded in R2).
We can classify the points in the R-tree structure.
(1) Points of types 1 and 4 are endpoints.
(2) Points of type 2 are branch points, and the set of tangent directions corre-sponds (canonically up to an automorphism of P1 ¯ Fp) to P1(¯ Fp).
(3) Points of type 3 are ordinary.
Let ξ ∈A1,an Cp be any point. If ξ′ is another point, then there are three possibilities: ξ > ξ′, ξ < ξ′, or neither. In the last two cases, the path from ξ to ξ′ starts ‘going up’ to ξ ∨ξ′ (= ξ′ in the second case). So the points ξ′ ̸= ξ such that ξ ̸> ξ′ constitute the set Bξ,up. If ξ is of type 1 or 4, then there are no ξ′ with ξ > ξ′, so ‘up’ is the only direction.
Now consider ξ = ζa,r of type 3. If ξ′ < ξ, then there is some ξ′ ≤ζa′,r′ < ξ, and we have D(a′, r′) ⊂D(a, r), so |a′ −a|p ≤r. Since r / ∈|C× p |p, we can choose r′ such that |a′ −a|p < r′ < r, but then ζa′,r′ = ζa,r′, so the path from ξ to ξ′ shares an initial segment with [ζa,r, ζa,0]. This shows that other than ‘up’, there is exactly one tangent direction ‘down’.
Finally, when ξ = ζa,r is of type 2, then by a similar argument, for any ξ′ < ξ there is a′ ∈D(a, r) such that ξ′ ∈D(a′, r)−. If two points ξ′, ξ′′ are in the same open disk, then the paths from ξ to these two points share an initial segment; otherwise ξ′ ∨ξ′′ = ξ and they do not. So the directions other than ‘up’ at ξ are in one-to-one correspondence with open disks of radius r contained in D(a, r).
To show that the set of such disks corresponds to ¯ Fp, we assume that a = 0 and r = 1 (we can shift and scale to reduce to this case). Two disks D(a′, 1)−and D(a′′, 1)−contained in D(0, 1) are equal if and only if |a′ −a′′|p < 1, which means that a′, a′′ ∈R (the valuation ring of Cp) have the same image in the residue class field ¯ Fp. So we see that the ‘downward’ directions at a type 2 point correspond to the elements of ¯ Fp; together with the ‘up’ direction, which we can let correspond to ∞, we get P1(¯ Fp). (For ζ0,1 the correspondence is canonical. For other points it is so only up to an automorphism of P1 ¯ Fp, since there is a choice involved in the shifting and scaling.) If ξ = ζa,r is a point of type 2 or 3, then the basic open set Bξ,v of the weak topology is the complement of D(a, r) when v = ‘up’ and is D(a′, r)−with a′ as above when v is a ‘downward’ direction. For ξ of type 1 or 4, Bξ,v is just A1,an Cp \ {ξ}. This shows that the weak topology of the R-tree agrees with the Berkovich topology.
Here is a rough sketch of D(0, 1): 1E. Hrushovski, F. Loeser and B. Poonen, Berkovich spaces embed in Euclidean spaces, L’Enseignement Math. 60 (2014), no. 3–4, 273–292.
§ 5. The Berkovich affine and projective line 30 ζ0,1 ζ0,r ζ0,|p|1/2 ζ0,|p| ζ1,|p| ζ−1,|p| ζ0,|p|2 ζ1,0 ζ1+p,0 ζ−1,0 ζ0,0 ζp2,0 ζ−p,0 ζp,0 ξ Points of type 1, 2, 3 and 4 are green, red, purple and blue, respectively.
Let T be an R-tree and let S ⊂T be nonempty and finite. Then one can consider the convex hull Γ of S; this is the union of the paths joining the points in S.
Then Γ is a finite R-tree and there is a natural deformation retraction τΓ : T →Γ that sends any point x ∈T to the point of Γ that is hit first by the unique path from x to any fixed point in Γ. It can be shown that the weak topology on T is the weakest topology that makes all the maps τΓ continuous (where we take, say, the metric topology on Γ). (This is an exercise.) Now we want to introduce the Berkovich projective line, P1,an Cp . There are several ways of constructing it.
(1) We set P1,an Cp = A1,an Cp ∪{∞}, where ∞is a point of type 1, whose open neigh-borhoods are the complements of compact subsets of A1,an Cp together with the point ∞. (This is the one-point compactification from general topology.) (2) We observe that the map Cp →Cp, z 7→z−1 induces a continuous involution φ on A1,an Cp \ {0} (that is given on points not of type 4 by ζa,r 7→ζa−1,r/|a|2 when |a| > r and ζ0,r 7→ζ0,1/r for r > 0). One can then ‘glue’ two copies of A1,an Cp along A1,an Cp \ {0} via this map. This identifies a neighborhood of ∞with a neighborhood of 0 (and so shows in particular that ∞is in no way special).
One can also show that the big metric is invariant under φ and also under affine maps z 7→az + b (Exercise).
(3) One obtains the same result by gluing two copies of D(0, 1) along the ‘spheres’ D(0, 1) \ D(0, 1)−via φ. Since D(0, 1) is compact, this shows that P1,an Cp is compact.
(4) Similar to the Proj construction in algebraic geometry, there is a ‘Berkovich Proj’ taking a graded finitely generated algebra A over a complete non-archimedean field as input and having a Berkovich type space as output, whose points corre-spond to equivalence classes of multiplicative seminorms on A whose kernel doe § 5. The Berkovich affine and projective line 31 not contain the irrelevant ideal. Applying this to the polynomial ring Cp[x, y] with the standard grading, this results in P1,an Cp . We will discuss this in more detail below. An advantage of this approach is that it makes it easy to see that a morphism P1 Cp →P1 Cp induces a continuous map P1,an Cp →P1,an Cp .
Recall that a graded ring is a ring R coming with a direct sum decomposition R = L n≥0 Rn as an additive group such that for all f ∈Rm and g ∈Rn we have fg ∈Rm+n. The elements of Rn are said to be homogeneous of degree n. An ideal I of R is said to be homogeneous if it is generated by homogenous elements.
If K is a field and the decomposition is a direct sum of K-vector spaces, then R is a graded K-algebra. In algebraic geometry, Proj R is the set of all homogeneous prime ideals of R that do not contain the irrelevant ideal L n≥1 Rn, with the Zariski topology. For example, Pn K = Proj K[X0, X1, . . . , Xn] with the usual grading of the polynomial ring. The ‘Berkovich Proj’ construction does something similar for Berkovich spaces.
5.11. Definition.
Let A be a finitely generated graded K-algebra with A0 = DEF Berkovich Proj K, where K is a complete and algebraically closed non-archimedean field. We consider the set of multiplicative seminorms on A whose kernel does not contain the irrelevant ideal. We declare that two seminorms ∥·∥and ∥·∥′ are equivalent if there is C > 0 such that ∥F∥′ = Cd∥F∥for all F ∈Ad (i.e., F ∈A homogeneous of degree d). Then the projective Berkovich space associated to A, PBerk A, is the set of equivalence classes of these seminorms.
Let a1, . . . , am be homogeneous generators of the irrelevant ideal of A. Then in each equivalence class, there are normalized seminorms ∥·∥with the property that maxj ∥aj∥= 1, and all equivalent normalized seminorms agree on all homoge-neous elements. We define the topology on PBerk A to be the weakest one such that ∥·∥ 7→∥F∥is continuous for every homogeneous F ∈A, where ∥·∥is a normalized representative of the class ∥·∥ .
If V = Proj A is the projective K-variety defined by A, then we also write V an for PBerk A.
♦ The notation PBerk A is not standard.
In analogy with the usual construction in algebraic geometry, we can then define the Berkovich projective line to be DEF P1,an Cp P1,an Cp = PBerk Cp[X, Y ] , where the polynomial ring Cp[X, Y ] has its usual grading.
We find the usual two embeddings of A1,an Cp into P1,an Cp by pulling back seminorms under the two maps Cp[X, Y ] →Cp[x], F(X, Y ) 7→F(x, 1) and F(X, Y ) 7→ F(1, x).
The equivalence class we do not obtain under the first map satisfies ∥Y ∥= 0; it corresponds to the type 1 point 0 under the second map and represents the point at infinity.
It is then natural to consider sets of the form P1,an Cp \ D(a, r)−as closed and sets of the form P1,an Cp \ D(a, r) as open Berkovich disks (‘around infinity’) in P1,an Cp . Then the open Berkovich disks generate the topology of P1,an Cp , and a basis of the topology is given by the open disks (including the whole space) minus finitely many closed disks.
§ 5. The Berkovich affine and projective line 32 Since P1,an Cp without the points of type 1 is the same as A1,an Cp without the points of type 1, which is an R-tree with respect to the big metric, it follows by the results mentioned in (2) above that Aut(P1 Cp) = PGL(2, Cp) acts on this R-tree by isometries. This action is transitive on the set of type 2 points, and the stabilizer of the point ζ0,1 is PGL(2, R) (where R is the valuation ring of Cp), with the tangent directions at ζ0,1 being permuted transitively (by PGL(2, ¯ Fp) = Aut(P1 ¯ Fp) via the canonical map PGL(2, R) →PGL(2, ¯ Fp)).
A type 2 point ζ of P1,an Cp can then be interpreted as corresponding to a reduction map P1(Cp) →P1(¯ Fp), up to an automorphism of the target. The map is given by associating to z ∈P1(Cp) the tangent direction at ζ in which the type 1 point corresponding to z can be seen. For ζ = ζ0,1 we get the usual reduction map. Such a reduction map comes from a ‘model’ of P1 over R of the form P1 R; the reduction map depends on how we identify the generic fiber of P1 R with P1 Cp. More generally, we call any finite R-subtree of P1,an Cp \P1(Cp) with the big metric that is the convex hull of finitely many points of type 2 a skeleton of P1,an Cp . Any skeleton corresponds DEF Skeleton to a model of P1 Cp over R that is usually more complicated than P1 R in that its special fiber (i.e., the base change to ¯ Fp) is a configuration of several P1’s arranged in tree form. (Possibly we return to that later.) If we modify the definition of the diameter by defining it on P1,an Cp as diam′(ξ) = ( diam(ξ) if ξ ∈D(0, 1) diam(φ(ξ)) if ξ ∈P1,an Cp \ D(0, 1), where φ is the involution induced by z 7→1/z, and set ξ ∨′ η to be the point where [ξ, ζ0,1] and [η, ζ0,1] first meet, then we can define a small metric d′ on P1,an Cp in the same way as before. Then P1,an Cp can be identified with the R-tree given by the small metric. However, the small metric is not invariant under automorphisms in general (the point ζ0,1 plays a special role in its definition; any automorphism that moves ζ0,1 will change the metric).
To conclude this section, we give another interpretation of the seminorm associated to a point in P1,an Cp that (contrary to the interpretation as the supremum norm on the associated disk when applied to polynomials) also works for rational functions.
Consider first a point ξ = ζa,r of type 2. If f is a nonzero polynomial with roots α1, . . . , αm in D(a, r), then for all α in the closed set D(a, r) \ Sm j=1 D(αj, r)−we have |f(α)|p = ∥f∥ξ. To see this, we can assume that a = 0 and f is monic. Write f = (x −α1) · · · (x −αm)(x −αm+1) · · · (x −αn) , where αm+1, . . . , αn are the roots of f outside of D(0, r). Then ∥f∥ξ = n Y j=1 ∥x −αj∥ξ = n Y j=1 |α −αj|p = |f(α)|p , since for j ≤m, we have ∥x−αj∥ξ = r = |α−αj|p (here we use that α / ∈D(αj, r)−), and for j > m, we have ∥x −αj∥ξ = |αj|p = |α −αj|p. So |f|p is constant and equal to ∥f∥ξ on D(a, r) outside a finite union of open disks contained in D(a, r).
We can apply this separately to the numerator and the denominator of a rational function f = g/h ∈Cp(x), which gives the same statement for such an f. So for a type 2 point ξ = ζa,r, the induced seminorm ∥·∥ξ on Cp(x) is the absolute value taken by any given function f ∈Cp(x) on D(a, r) outside finitely many smaller open disks (the set of disks to be removed depends on f); this is also the same as § 5. The Berkovich affine and projective line 33 the absolute value taken on P1,an Cp \ D(a, r)−outside a finite union of smaller open disks (one of which is the ‘exterior’ P1,an Cp \ D(a, r)). Note that when considering rational functions f as quotients of two homogeneous polynomials in Cp[X, Y ] of the same degree, then ∥f∥ξ is independent of the choice of representative of the equivalence class of seminorms corresponding to ξ (the scaling factor Cd cancels).
For a point ξ = a ∈Cp of type 1, we have of course ∥f∥ξ = |f(a)|p unless a is a pole of f (this also makes sense when a = ∞, with f(∞) interpreted as the value at zero of f(1/x)).
If ξ = ζa,r is of type 3, then we have to use a ‘limit version’ of the above. Since r / ∈|C× p |, no two nonzero terms of the form rj|aj|p with aj ∈Cp can be equal and so we have ∥f∥ξ = maxj rj|aj|p when f(x + a) = a0 + a1x + a2x2 + . . . + anxn is a nonzero polynomial. If s ∈|C× p | is sufficiently close to r, then all terms sj|aj|p will still be pairwise distinct, hence for α ∈Cp such that |α −a|p = s we will have |f(α)|p = maxj sj|aj|p, which is close to ∥f∥ξ. So for any given ε > 0 there is a closed annulus A = D(a, r + δ) \ D(a, r −δ)−such that |f(α)|p −∥f∥ξ ≤ε for all α ∈A.
The general statement is as follows (see Section 2.4 in [BR]).
5.12. Lemma.
Let ξ ∈P1,an Cp and denote the corresponding multiplicative semi-LEMMA characteri-zation of ∥·∥ξ norm on Cp(x) as usual by ∥·∥ξ. Then for every f ∈Cp(x) (that does not have a pole at ξ when ξ is of type 1) we have that ∥f∥ξ is the unique ν ∈R≥0 such that for every ε > 0 there is a closed neighborhood U of ξ in P1,an Cp such that |f(α)|p−ν ≤ε for all α ∈P1(Cp)∩U. The set U can be taken to be a closed Berkovich disk minus a finite union of open Berkovich disks.
Proof. Fix f. Since η 7→∥f∥η is continuous by definition of the Berkovich topology, the set U = {η : |∥f∥η −∥f∥ξ| ≤ε} is a closed neighborhood of ξ. It contains an open neighborhood of ξ (for example, the set obtained by replacing ‘≤ε’ by ‘< ε’), which contains an open neighborhood V that is an open Berkovich disk minus finitely many closed Berkovich disks. Then U contains the closure of V , which is the corresponding closed Berkovich disk minus the corresponding open Berkovich disks. We can always make U smaller, so we can assume that U has this form. Then |f(α)|p−∥f∥ξ ≤ε for α ∈P1(Cp)∩U follows from the definition of U. Since P1(Cp) ∩U is dense in U, this property determines ∥f∥ξ to within ε.
Since ε > 0 was arbitrary, ∥f∥ξ is characterized by it.
K In a similar way as for Berkovich spaces associated to affine Cp-varieties, mor-phisms between projective Cp-varieties induce continuous maps between the as-sociated Berkovich spaces. For example, any rational function ϕ ∈Cp(x) can be considered as a morphism P1 Cp →P1 Cp and therefore induces a map P1,an Cp →P1,an Cp .
If ϕ is not constant (otherwise it is quite clear what happens), then it follows from the discussion above that ϕ(ξ) = η for type 2 points ξ = ζa,r and η = ζa′,r′ if and only if ϕ maps a suitable set D(a, r) \ S j D(aj, r)−to a set of the form D(a′, r′) \ S j D(a′ j, r′). This can be helpful when one is trying to figure out what ϕ is doing on P1,an Cp .
§ 6. Analytic spaces and functions 34 6. Analytic spaces and functions In this section, we follow roughly part of Brian Conrad’s text [AWS, Chapter 1].
Recall that a K-Banach algebra (for a complete field K with absolute value) is a K-algebra A with a (submultiplicative) norm |·|A that restricts to the absolute value on K and such that A is complete with respect to this norm. So we have |a|A ̸= 0 for a ̸= 0, |a + b|A ≤|a|A + |b|A and |ab|A ≤|a|A · |b|A .
If K is non-archimedean, then |·|A also satisfies the ultrametric triangle inequality.
6.1. Definition.
Let A be a K-Banach algebra; we write the norm on A as |·|A.
DEF bounded multiplicative seminorm A multiplicative seminorm ∥·∥on A is said to be bounded, if ∥f∥≤|f|A for all f ∈A.
♦ We note that it is sufficient to require ∥f∥≤C|f|A for some constant C > 0, since this implies ∥f∥= (∥f n∥)1/n ≤(C|f n|A)1/n ≤C1/n|f|A ; we obtain ∥f∥≤|f|A by letting n tend to infinity. So the notion of ‘bounded multiplicative seminorm’ does not change when we replace the norm on A by an equivalent one (i.e., one that is bounded below and above by some positive multiple of |·|A).
6.2. Definition.
Let A be a K-Banach algebra. The Berkovich space Berk A DEF Berkovich space associated to A associated to A is the set of all bounded multiplicative seminorms on A, with the weakest topology that makes the maps ∥·∥7→∥f∥continuous for all f ∈A.
♦ In a similar way as we did earlier, one shows the following.
6.3. Theorem.
Let A be a K-Banach algebra. Then Berk A is a compact Haus-THM Berk A is compact and Hausdorff dorffspace, which is empty if and only if A = {0}.
We note that one can define Berk A for any (commutative) ring A that is complete with respect to some absolute value (a Banach ring); it does not have to be an algebra over some complete field. One could take A = Z with the usual absolute value, for example (completeness follows trivially, since every Cauchy sequence must be eventually constant), or any ring with the trivial absolute value. It is an instructive exercise to work out the structure of Berk Z!
From now on, K will be, as usual, a complete and algebraically closed non-archimedean field, for example, K = Cp. There is an important class of K-Banach algebras.
6.4. Definition.
Let n ∈Z>0 and r1, . . . , rn > 0. We define DEF Tate algebra K⟨r−1 1 x1, . . . , r−1 n xn⟩:= n X i∈Zn ≥0 aixi : ri|ai| →0 as |i| →∞ o ⊂K[ [x1, . . . , xn] ] .
Here we use multi-index notation: for i = (i1, . . . , in) we set xi = xi1 1 · · · xin n and ri = ri1 1 · · · rin n ; also |i| = i1 + . . . + in. Then X i aixi := max i ri|ai| § 6. Analytic spaces and functions 35 defines a multiplicative norm on A = K⟨r−1 1 x1, . . . , r−1 n xn⟩turning A into a K-Banach algebra. We call A a Tate algebra over K when all rj ∈|K×|, otherwise A is a generalized Tate algebra over K.
More generally, if A is any K-Banach algebra, then we can define (generalized) relative Tate algebras A⟨r−1 1 x1, . . . , r−1 n xn⟩in the same way.
♦ Note that by scaling the variables suitably, any Tate algebra in n variables is isomorphic to the standard Tate algebra K⟨x1, . . . , xn⟩:= K⟨1−1x1, . . . , 1−1xn⟩.
This is not true for generalized Tate algebras.
6.5. Example.
We have Berk Cp⟨x⟩∼ = D(0, 1).
The map is induced by the EXAMPLE D(0, 1) inclusion Cp[x] , →Cp⟨x⟩; the image consists of all seminorms on Cp[x] that are bounded by the restriction of the norm on the Tate algebra to the polynomial ring, which is exactly |·|1 = ∥·∥0,1. By definition, this set is D(0, 1). It also follows easily from the definitions that the map induces a homeomorphism.
More generally, the same argument shows that Berk Cp⟨r−1x⟩∼ = D(0, r), for every r > 0. We see that there is some kind of correspondence between type 2 points and Tate algebras, and type 3 points and generalized Tate algebras. Quite explicitly, we obtain Cp⟨r−1x⟩as the completion of Cp[x] with respect to the norm corresponding to ζ0,r.
♣ We state the following facts without proof.
6.6. Lemma.
The Tate algebras over K are noetherian (every ideal is finitely LEMMA Properties of Tate algebras generated) unique factorization domains.
They are also Jacobson rings (every prime ideal is the intersection of the maximal ideals it is contained in). Every ideal is closed.
The norm is intrinsically defined as |f| = maxφ |φ(f)|, where φ runs over all continuous K-algebra homomorphisms to K.
The Tate algebra T = K⟨r−1 1 x1, . . . , r−1 n xn⟩has the following universal property.
Given any K-Banach algebra A and elements a1, . . . , an ∈A with |aj|A ≤rj, there is a unique continuous K-algebra homomorphism φ: T →A with φ(xj) = aj.
Let A be a Tate algebra and I ⊂A an ideal. Then by the above, I is closed.
Define a real-valued function on A/I by |a + I|A/I = inf{|a + b|A : b ∈I} .
6.7. Lemma.
|·|A/I is a norm on A/I turning A/I into a Banach algebra such LEMMA induced norm that the canonical homomorphism A →A/I is continuous.
Proof. We have to show a number of things.
(1) |a + I|A/I = 0 only for a ∈I.
If |a + I|A/I = 0, then there is a sequence (bn) in I such that |a + bn|A →0, hence −bn →a. Since I is closed, this implies a ∈I.
(2) |(a + a′) + I|A/I ≤max{|a + I|A/I, |a′ + I|A/I}.
This follows easily from the definition.
§ 6. Analytic spaces and functions 36 (3) |aa′ + I|A/I ≤|a + I|A/I · |a′ + I|A/I.
For b, b′ ∈I we have |aa′ + I|A/I ≤|(a + b)(a′ + b′)|A ≤|a + b|A · |a′ + b′|A; taking the infimum over all b and b′ yields the claim.
(4) A/I is complete.
Consider a Cauchy sequence (an + I) in A/I, so that |an+1 −an + I|A/I →0.
Use the definition of |·|A/I to pick recursively a sequence (bn) in I with b0 = 0 such that |(an+1 + bn+1) −(an −bn)|A →0.
Then (an + bn) is a Cauchy sequence in A; since A is complete, this sequence has a limit a. But then |an −a + I|A/I ≤|(an + bn) −a|A →0, so (an + I) converges to a + I in A/I.
(5) A →A/I is continuous.
This follows from |(a + I) −(b + I)|A/I ≤|a −b|A.
K 6.8. Definition.
A K-Banach algebra A is a K-affinoid algebra if there is a DEF affinoid algebra and domain generalized Tate algebra T and a surjective K-algebra homomorphism φ: T →A such that the norm of A is (equivalent to) the induced norm on T/ ker(φ) ∼ = A. If T can be taken to be a Tate algebra, then A is a strict K-affinoid algebra.
The space Berk A associated to a (strict) affinoid algebra is a (strict) affinoid domain.
♦ Simple examples of affinoid domains are the closed Berkovich disks D(0, r), where A is itself a (generalized) Tate algebra.
6.9. Example.
Let 0 < r ≤s. Then we can consider the algebra EXAMPLE annuli A = K⟨s−1X, rX−1⟩:= K⟨s−1X, rY ⟩ ⟨XY −1⟩ with its induced norm. Using the relation XY = 1, its elements can be written uniquely as the image of a series of the form ∞ X m=1 a−mY m + a0 + ∞ X n=1 anXn with sn|an| →0 as n →∞and rm|am| →0 as m →−∞. We can interpret this (via Y = X−1 in A) as the Laurent series ∞ X n=−∞ anXn ; the conditions on the coefficients then are equivalent to saying that this series converges whenever we plug in some element α such that r ≤|α| ≤s. For K = Cp, we see that via the map Cp[x] →A sending x to X, Berk A gets identified with A(r, s) := D(0, s) \ D(0, r)−, which is a closed Berkovich annulus.
♣ Most of the structure of affinoid algebras is intrinsic.
We state the following without proof.
§ 6. Analytic spaces and functions 37 6.10. Theorem.
Let A be a K-affinoid algebra.
THM Properties of affinoid algebras (1) Any two K-Banach algebra norms on A are equivalent.
In particular, the topology on A induced by the norm is intrinsic, and the same is true for notions of boundedness.
(2) Any K-algebra morphism A′ →A from another K-affinoid algebra is contin-uous (and hence a morphism of K-affinoid algebras).
There are more properties similar to those of algebras of finite type in algebraic geometry. For example, if A is K-affinoid and A′ is a K-algebra that is a finitely generated A-module, then A′ is also K-affinoid. Any K-affinoid algebra is a finitely generated module over some Tate algebra (this corresponds to the Noether nor-malization theorem).
One can then construct more general Berkovich K-analytic spaces by gluing affi-noid domains. There are some technical issues here, because the gluing is along compact ‘sub-affinoid domains’ and not along open sets as one does in algebraic or differential geometry.
For example, for fixed s > 0 one can glue the closed disks of radius r < s via the natural inclusions D(0, r) ⊂D(0, r′) whenever r < r′ < s to obtain the open disk D(0, s)−. In a similar way, one gets open annuli A(r, s)−(for r < s).
Considering the affine line as ‘the open disk of radius ∞’, we obtain A1,an Cp . We get the projective line P1,an Cp by gluing two closed disks D(0, 1) along the annulus A(1, 1) as discussed in the last section.
We also have the following (see the exercises). As usual, R denotes the valuation ring of K and k the residue field.
6.11. Theorem.
Let A be a K-Banach algebra. Then THM Reduction map A◦= a ∈A : {|an|A : n ≥0} is bounded is a closed subalgebra of A that contains (the image of) R and depends on the norm of A only up to equivalence (and so is completely intrinsic when A is affinoid).
The set A◦◦= {a ∈A : an →0} is an ideal of A◦, and the quotient ˜ A = A◦/A◦◦ is a k-algebra functorially associated to A. If A is strictly K-affinoid, then ˜ A is a finitely generated k-algebra. If ∥·∥∈Berk A, then the set {a ∈A◦: ∥a∥< 1} is a prime ideal of A◦containing A◦◦; in this way we obtain a canonical reduction map ρA : Berk A →Spec ˜ A. If A is strictly K-affinoid, then ρA is anti-continuous: preimages of closed sets are open and vice versa.
For example, taking A = Cp⟨x⟩, we find ˜ A = ¯ Fp[x], and the reduction map sends ζ0,1 to the generic point of A1 ¯ Fp = Spec ˜ A and the points in the open Berkovich disk D(α, 1)−(for α ∈D(0, 1)) are mapped to the closed point ¯ α ∈A1(¯ Fp). In particular, we see that the preimage of a closed point is an open Berkovich disk, and the preimage of the open set A1 ¯ Fp \ S, where S is a finite set of closed points, is the closed set D(0, 1) \ S s∈S D(αs, 1)−, where αs ∈D(0, 1) has image s.
6.12. Example.
Let 0 < r < s be in |K×| and consider as before the algebra EXAMPLE annuli A = K⟨s−1X, rX−1⟩. The induced norm on A is given by X n∈Z anXn A = max {rn|an| : n ≤0} ∪{sn|an| : n ≥0} § 6. Analytic spaces and functions 38 (this is because |XY | = s/r > 1, so the ‘smallest’ representative of 1 ∈A = K⟨X, Y ⟩/⟨XY −1⟩is 1). By choosing c, C ∈K such that |c| = r and |C| = s and scaling X and Y , we can write A as the quotient A ∼ = K⟨X, Y ⟩/⟨XY −cC−1⟩ of the standard two-variable Tate algebra T. From this it is fairly easy to see that ˜ A = ˜ T/⟨XY ⟩= k[X, Y ]/⟨XY ⟩, where k is the residue field. So Spec ˜ A is a union of two affine lines (corresponding to X = 0 and Y = 0) meeting transversally in one point (X = Y = 0). Let Γ = [ζ0,r, ζ0,s] ⊂A(r, s). One can check that ρA maps ζ0,r ∈A(r, s) to the generic point of the line X = 0, ζ0,s to the generic point of the line Y = 0, points retracting under τΓ to ζ0,r to the point on X = 0 corresponding to their tangent direction and similarly for points retracting to ζ0,s, and all points retracting to any other point of Γ (in the ‘open interval’ between the endpoints) to the point of intersection of the two lines.
♣ The anti-continuity of the reduction map ρA implies that the gluing together of strict affinoids along closed sub-affinoids induces a corresponding gluing of the Spec ˜ A’s along open sets. For example, gluing two copies of D(0, 1) along A(1, 1) to obtain P1,an Cp corresponds to gluing to copies of A1 ¯ Fp along A1{0} (which is Spec ˜ A for the annulus A(1, 1)), resulting in P1 ¯ Fp in the usual way. The reduction maps also glue, which in the example gives the map identifying the tangent directions at ζ0,1 ∈P1,an Cp with P1(¯ Fp) and sending ζ0,1 to the generic point of P1 ¯ Fp.
A given space can usually be obtained in many different ways by gluing. This can result in different ‘special fibers’ (the glued spectra of the algebras ˜ A) and therefore different reduction maps. For example, we can obtain P1,an Cp also by gluing D(0, 1) (via the inversion map) and D(0, 1/p) to A(1/p, 1) along the sub-annuli A(1, 1) and A(1/p, 1/p). The special fiber of this object consists of two copies of P1 ¯ Fp meeting transversally in one point. It is obtained from the special fiber of the annulus — two A1’s meeting in one point — by gluing an A1 to each of the lines (along the complement of the intersection point). In a similar way, we can glue together also more general affinoids (the most general of which is a closed disk minus a finite union of open ones) to obtain a ‘model’ of P1,an Cp .
The skeleton associated to such a model is the convex hull of the points mapping to generic points of the special fiber; in the preceding example, this would be the interval [ζ0,1/p, ζ0,1]. In general, it will be a finite sub-tree of P1,an Cp with endpoints of type 2.
There is a converse to this, valid for general smooth projective irreducible curves C over Cp: There is a bijective correspondence between finite subgraphs of Berk C with vertices of type 2 and semistable models of C over the valuation ring R of Cp (up to isomorphism), such that the vertices correspond to the generic points of the special fiber of the semistable model under the reduction map.2 2M. Baker, S. Payne, J. Rabinoff, On the structure of non-Archimedean analytic curves.
Tropical and non-Archimedean geometry, 93–121, Contemp. Math., 605, Amer. Math. Soc., Providence, RI, 2013; arXiv: 1404.0279.
§ 7. Berkovich spaces of curves 39 7. Berkovich spaces of curves In this section we will take a closer look at Berkovich spaces of (smooth projective irreducible) curves over Cp. We begin with some lemmas. Let Tr = Cp⟨r−1x⟩be the univariate (generalized) Tate algebra with parameter r > 0.
7.1. Lemma.
Let LEMMA roots of polynomials over Tate algebras F(x, y) = Fn(x)yn + Fn−1(x)yn−1 + . . . + F0(x) ∈Tr[y] be a polynomial such that there is δ < 1 with |F0|r ≤δ, |1 + F1|r ≤δ and |Fj|r ≤δ2−j for all j ≥2. Then there is h(x) ∈Tr such that |h(0)|p < 1 and F(x, h(x)) = 0.
Proof. The equation F(x, y) = 0 is equivalent to the fixed point equation y = Ψ(y) := F0(x) + (1 + F1(x))y + F2(x)y2 + . . . + Fn(x)yn .
For |y1|, |y2| ≤δ we have |Ψ(y2) −Ψ(y1)|r = |y2 −y1|r · |(1 + F1(x)) + F2(x)(y2 + y1) + . . . + Fn(x)(yn−1 2 + yn−2 2 y1 + . . . + yn−1 1 )|r ≤max{|1 + F1|r, |F2|rδ, . . . , |Fn|rδn−1} · |y2 −y1|r ≤δ · |y2 −y1|r .
Also, |Ψ(y)|r ≤max{|F0|r, |1 + F1|rδ, |F2|rδ2, . . .} ≤δ when |y|r ≤δ. So Ψ defines a contracting map on {y ∈Tr : |y|r ≤δ}. The Banach Fixed Point Theorem then gives us a unique solution.
K 7.2. Lemma.
Let F(x, y) ∈Tr[y] be monic of degree n such that F(0, y) ∈Cp[y] LEMMA ´ etale covering of disk has no multiple roots. Then there is 0 < r′ ≤r and there are h1, . . . , hn ∈Tr′ such that F(x, y) = (y −h1(x)) · · · (y −hn(x)).
Proof. Let η1, . . . , ηn ∈Cp be the n distinct roots of F(0, y).
By assumption, βj := F ′(0, ηj) ̸= 0 (where F ′ denotes the derivative of F as a polynomial in y).
Let ˜ Fj(x, y) := −β−1 j F(x, y + ηj) Note that ˜ Fj has no constant term and that the coefficient of y is −1. So, writing ˜ Fj(x, y) = f0(x) + f1(x)y + . . . and setting γ = max{|f0|r, |1 + f1|r, |f2|r, . . .}, we have for every 0 < r′ ≤r that |f0|r′ ≤γr′/r , |1 + f1(x)|r′ ≤γr′/r and |fm|r′ ≤γ for m ≥2.
Now pick λ ∈C× p with |λ|p ≤γ−1 and set ˜ ˜ Fj(x, y) = λ−1 ˜ Fj(x, λy) = λ−1f0(x) + f1(x)y + λf2(x)y2 + . . . ; we have |λ−1f0|r′ ≤|λ|−1 p γr′/r, |1+f1|r′ ≤γr′/r and |λm−1fm|r′ ≤|λ|m−1 p γ ≤γ2−m for m ≥2. Fix 0 < δ < 1 such that δ ≤γ−1. With r′ = r min{1, |λ|pγ−1δ, γ−1δ} the assumptions of Lemma 7.1 are then satisfied for ˜ ˜ Fj, so (translating back to the original coordinates) there is hj ∈Tr′ with hj(0) = ηj and F(x, hj(x)) = 0.
We can take r′ so that it works for all j; we then obtain n distinct roots of F in Tr′.
K § 7. Berkovich spaces of curves 40 I think that one can take any r′ ≤r here such that F(ξ, y) has no multiple roots for all ξ ∈Cp with |ξ|p ≤r′, at least when p > n, but so far I have found no proof for this more precise statement. We will see below that it is true when n = 2 and p > 2.
7.3. Example.
The example F(x, y) = y2 −y + x over C2 shows that the EXAMPLE necessity of ‘p > n’ condition p > n is necessary in general. The two roots of F are h1 = 1 −√1 −4x 2 = x + x2 + 2x3 + 5x4 + 14x5 + . . . = ∞ X n=1 1 n 2n −2 n −1 xn and h2 = 1 −h1. The coefficients are integers, but the coefficient of x2n is odd for all n, so the coefficients of the series are bounded but do not tend to zero 2-adically. This means that h1, h2 ∈Tr for all r < 1, but not for r = 1. On the other hand, the discriminant of F is 1 −4x, which does not vanish even for |x|2 < 4. This seems to be some issue related to ‘wild ramification’.
♣ 7.4. Theorem.
Let C be a smooth projective irreducible curve over Cp and let THM coverings of P1 φ: C →P1 be a morphism of degree n. Then the induced map φ∗: Can →P1,an has fibers of size at most n.
Proof. Let ζ ∈P1,an be some point; we can assume without loss of generality that ζ ̸= ∞. The statement is clear when ζ is a type 1 point, so we consider ζ not of type 1; in particular, the seminorm on Cp[x] associated to ζ is a norm. The point ζ also corresponds to a homomorphism Cp[x] →H into a complete field H with absolute value such that the absolute value pulls back to the seminorm given by ζ and the image of Cp[x] in H generates a dense subfield.
Let K be the function field of C; we have an inclusion Cp(x) ⊂K, which is a field extension of degree n. Then H ⊗Cp(x) K is an algebra over H of degree n, which splits as a direct product of finite field extensions of H (whose degrees sum to n). There is a unique extension of the absolute value of H to each of these fields, which by pulling back to K (and then restricting to the affine coordinate ring A of φ−1(A1)) give rise to the various multiplicative (semi)norms on A extending ∥·∥ζ. So the fiber φ−1 ∗(ζ) has as many points as there are fields in the splitting of H ⊗Cp(x) K; this number is at most n.
K This basically shows that we can glue together Can from n copies of P1,an. If the statement above on the choice of r′ is true and p > n, then it follows that φ∗can be branched (i.e., have fibers of size < n) only above points in the convex hull of the type 1 points corresponding to branch points of φ. This might be true even when this statement is wrong; in any case it would mean that Can deformation retracts to a finite graph that is glued from n copies of a tree (= the convex hull of the branch points).
We now consider hyperelliptic curves in more detail. We assume that p > 2. First we need two elementary lemmas.
7.5. Lemma.
Let f(x) ∈Tr with |f|r < 1. Then there is h(x) ∈Tr with |h|r < 1 LEMMA square roots such that 1 + f(x) = (1 + h(x))2.
More generally, the same result applies to any complete ring R in place of Tr as long as |2|R = 1 and 2 ∈R×.
§ 7. Berkovich spaces of curves 41 Proof. We want a root h of F(x, y) = −1 2y2 −y + 1 2f(x) with |h −1|r < 1. Since |2|p = 1 and |f|r < 1, we can directly apply Lemma 7.1.
The more general statement follows in the same way.
K 7.6. Lemma.
Let H be the completion of the field of fractions of Tr. Then x is LEMMA x not a square not a square in H.
Proof. First consider the case that r ∈|C× p |. Then we can assume r = 1, since Tr is isomorphic to T1 and the isomorphism only scales x. Assume that x = z2 is a square in H. Since the field of fractions of T1 is dense in H, there must be a, b ∈T1 such that |z −a/b| is small, which implies that |a(x)2 −xb(x)2| ≪|b(x)|2.
We can scale a and b so that |b| = 1; then also |a| ≤1. Looking at the reductions, we must have that a(x)2 −xb(x)2 reduces to zero in ¯ Fp[x], but this is impossible, since the first term has even degree and the second term is nonzero and has odd degree.
To deal with arbitrary r, we can enlarge Cp to obtain a field K (that is again complete and algebraically closed) such that r ∈|K×|. Then the argument above shows that x is not even a square in a larger field.
K The preceding lemma can be generalized: if f ∈T ◦ 1 has reduction in ¯ Fp[x] that is not a square, then f is not a square in H. Conversely (assuming p is odd as we do here), one can show using Lemma 7.5 that when the reduction of a polynomial f ∈Cp[x] ∩T ◦ 1 is a nonzero square, then f is a square in H. See the exercises.
A hyperelliptic curve C of genus g over Cp can be glued together from two affine pieces, the plane curves y2 = f2g+2x2g+2 + f2g+1x2g+1 + . . . + f1x + f0 and Y 2 = f0X2g+2 + f1X2g+1 + . . . + f2g+1X + f2g+2 with the identifications xX = 1 and Y = Xg+1y; we assume that the polynomial f(x) = P j fjxj has no multiple roots and degree at least 2g + 1. If we assume that f2g+2 ̸= 0, then C →P1 x is unbranched above infinity, so we can restrict our attention to the first affine patch. We denote the double cover C →P1 given by the x-coordinate by π, and we denote by Θ the set of branch points in P1(Cp) (i.e., the roots of f), which we also consider as type 1 points of P1,an Cp .
7.7. Lemma.
Let ζ = ζ0,r ∈A1,an be of type 2 or 3. We have a natural partition LEMMA branching of π of Θ into finitely many nonempty subsets Θv (indexed by the tangent direction v in which the points in Θv are visible). Then π−1 ∗(ζ0,r) consists of two points if all subsets Θv have even cardinality, and of one point otherwise.
Proof. If r / ∈|C× p | (then ζ is of type 3), then there are just two tangent directions, which we denote 0 and ∞. Otherwise we select one point a ∈P1(Cp) representing each occurring tangent direction different from ‘up’ and write Θa for the corre-sponding subset (and Θ∞for the subset of roots ξ such that |ξ|p > r). Then we can write f(x) = c Y ξ∈Θ∞ 1 −x ξ · Y a (x −a)#Θa Y ξ∈Θa 1 −ξ −a x −a § 7. Berkovich spaces of curves 42 with some c ∈C× p . By Lemma 7.5 (note that |x/ξ|r < 1 when |ξ|p > r) each factor 1 −x/ξ in the first product is a square in Tr. Now let H be the completion of the field of fractions of Tr. Then for ξ ∈Θa, we have |ξ −a|p < r and |x −a|r = r, so (ξ −a)/(x −a) ∈H has absolute value < 1, so 1 −(ξ −a)/(x −a) is also a square in H using Lemma 7.5 again. Since Cp is algebraically closed, c is of course also a square. So f(x) is a square in H times Q a(x −a)#Θa. By Lemma 7.6 and the remark following it, it follows that the latter and hence f(x) is a square in H if and only if all exponents #Θa are even (if ζ is of type 3, then we only need that x is not a square; if ζ is of type 2, we can assume that r = 1). On the other hand, the degree 2 algebra over H induced by π is H[y]/⟨y2 −f(x)⟩. This is a field if and only if f(x) is not a square in H; otherwise it splits as a product of two copies of H (by the Chinese Remainder Theorem). Since the number of fields in the product decomposition of this algebra is the number of points in the fiber of π∗above ζ, this completes the proof.
K 7.8. Corollary.
Let p > 2 and let C : y2 = f(x) be a hyperelliptic curve COR hyperelliptic Berkovich curve over Cp with hyperelliptic double cover π: C →P1.
Then the induced double cover π∗: Can →P1,an is branched exactly above points ζ such that there is at least one tangent direction at ζ that points to an odd number of roots of f. In particular, branching only occurs along the convex hull T of the branch points of π (i.e., the roots of f) considered as points of type 1, and Can can be obtained by gluing two copies of P1,an along parts of T .
Proof. Let ζ ∈P1,an. If ζ / ∈T , then there is some Berkovich disk D(a, r) con-taining ζ such that D(a, r) does not contain branch points of π. The proof of Lemma 7.7 in the case that Θ = Θ∞(or also Lemma 7.5 directly) then shows that f(x+a) is a square in Cp⟨r−1x⟩, which implies that π∗is unramified above D(a, r) and in particular above ζ. So branching can only occur along T ; in particular, no branching occurs at points of type 4 (which is consistent with the statement, since there is only one tangent directions in which all 2g +2 branch points can be seen).
For points ζa,r ∈T not of type 1, the statement on branching is Lemma 7.7 (after shifting x by a). For the points ζ of type 1, the statement is clear — note that when ζ is a branch point of π, then one can see the other 2g + 1 branch points in the unique tangent direction; otherwise one sees all 2g + 2 branch points in this direction.
K Since P1,an deformation retracts to T , we see that Can will deformation retract to what is obtained by gluing two copies of the tree T along the set of points that partition the set of branch points (which are the leaves of T ) into subsets at least one of which has an odd number of elements. For simplicity, let us say that a vertex or edge of T is even, if the partition of the leaves it induces results in sets with even cardinality, and odd otherwise. We will consider leaves as odd vertices.
Then the edge connecting a leaf to its adjacent vertex is always odd, so (as long as there is at least one non-leaf vertex of T ) at the cost of another deformation retraction, we can remove these edges from the graph obtained by gluing the trees.
We can continue this process as long as there are terminal odd edges (and at least one vertex remains). In particular, we see that Can can be contracted to a point (and so is simply connected) if and only if all edges are odd (equivalently, the branch locus is all of T ). Otherwise we can find two odd vertices connected by a chain of even edges and vertices, and the gluing produces a circle, so the resulting space is not simply connected.
§ 7. Berkovich spaces of curves 43 7.9. Examples.
The simplest case is when g = 0 (strictly speaking, a double EXAMPLES double covers of P1,an cover of P1 is a hyperelliptic curve only when g ≥2, but we can consider them also for g = 0 or g = 1). Up to a change of coordinates, we can take f(x) = x; then the branch locus in P1(Cp) consists of the two points 0 and ∞. Their convex hull T is simply the arc connecting the two (it consists of all points ζ0,r for r ≥0 together with ∞). So as a tree, T has two leaves connected by an odd edge. By the remarks above, Can is simply connected, which is no surprise, since C ∼ = P1.
In the sketches below, P1(Cp) is symbolized by the circle and P1,an by the disk it encloses; odd vertices and edges of T are red and even ones (which do not exist here) are green.
0 ∞ The next case is when g = 1; it is a bit more interesting. We have four branch points; by a change of coordinates, we can move three of them to 0, 1, ∞. Then f(x) = x(x −1)(x −λ) for some λ ∈Cp \ {0, 1}. There are four automorphisms of P1 that fix a set of four points, but there are 24 ways of selecting the three points that are mapped to 0, 1, ∞, so each curve C arises from (in general) six different choices of λ. If λ is one of them, then the others are 1 −λ , 1 λ , 1 −1 λ , 1 1 −λ and λ λ −1 .
In particular, we can assume that |λ|p ≤1. There are now two cases.
If |λ|p = |1 −λ|p = 1, then the four branch points lie in four different tangent directions at ζ0,1, since the reduction ¯ λ ∈P1(¯ Fp) is distinct from (the reductions of) 0, 1, ∞. So the tree T has one inner vertex of degree 4 and four edges connecting it to the four leaves. In particular, all vertices and edges are odd, and Can is simply connected. In this case C is an elliptic curve with good reduction.
In the second case, we can assume (perhaps after replacing λ by 1 −λ) that |λ|p < 1. Then at ζ0,1, we see one branch point each when looking to ∞or 1, but we see two (namely 0 and λ) when looking to 0. We get a similar partition at ζ0,|λ|p, which is now {0}, {λ}, {1, ∞}. So T has the two (odd) inner vertices ζ0,1 and ζ0,|λ|p, each of degree 3, there is one (even) edge connecting them, and each of the two vertices has two more (odd) edges connecting them to two of the branch points (1 and ∞, resp., 0 and λ). Gluing two copies of T along the odd edges and vertices then gives a graph that looks like T with the central edge doubled. This double edge forms a circle to which everything else can be retracted. Its length in the big metric is twice the length of the edge, so it is 2vp(λ), which is the same as § 7. Berkovich spaces of curves 44 −vp(j(C)), where j(C) is the j-invariant of the elliptic curve C, given by j(C) = 28 (λ2 −λ + 1)3 λ2(1 −λ)2 .
In any case, Can is not simply connected; its fundamental group is that of the circle, so ∼ = Z. Taking the universal covering space ‘unwraps’ the circle and replaces it by a line; the universal covering ˜ Can of Can is Gan m = P1,an \ {0, ∞}.
(Gm denotes the multiplicative group; its affine coordinate ring is Cp[x, x−1] and its Cp-points are Gm(Cp) = C× p .) Tate has shown (using different language) that the covering map Gan m →Can is actually a group homomorphism and that the group of deck transformations is generated by ‘multiplication by q’, where q ∈C× p satisfies |q|p < 1. Pulling back the coordinate functions to Gm gives an explicit analytic uniformization C(Cp) ∼ = C× p /qZ that is similar in spirit to the uniformization over the complex numbers: C(C) ∼ = C/(Z + Zτ) ∼ = C×/qZ, where the second isomorphism is induced by z 7→e2πiz and q = e2πiτ, with τ ∈C in the upper half-plane (so |q| < 1).
ζ0,1 0 ∞ 1 λ ζ0,1 ζ0,|λ| 0 ∞ 1 λ Left: |λ|p = |1 −λ|p = 1, right: |λ|p < 1.
♣ § 8. Integration 45 8. Integration In this final section we want to consider integration on Berkovich spaces. The goal is to have a ‘reasonable’ way of associating to a closed 1-form ω (meaning that dω = 0) on an analytic space X and a path γ : [0, 1] →X whose endpoints are ‘type 1 points’ (i.e., they correspond to seminorms whose kernel is a maximal ideal, so functions can be evaluated at such points to give a value in Cp) an integral Z γ ω ∈Cp .
By ‘reasonable’, we mean that this integral has the properties we expect from such a notion, for example: (1) Linearity in the 1-form.
Z γ (α1ω1 + α2ω2) = α1 Z γ ω1 + α2 Z γ ω2 for α1, α2 ∈Cp and ω1, ω2 closed 1-forms on X.
(2) Additivity on paths.
Z γ1+γ2 ω = Z γ1 ω + Z γ2 ω when γ1(1) = γ2(0) and γ1 + γ2 is the path obtained by traversing first γ1 and then γ2.
(3) Fundamental theorem of calculus.
Z γ d f = f(γ(1)) −f(γ(0)) when f is a function on X.
(4) Change of variables.
Z γ φ∗ω = Z φ◦γ ω when φ: Y →X is a morphism, γ is a path in Y and ω is a closed 1-form on X.
(5) Homotopy invariance.
Z γ1 ω = Z γ2 ω when γ1 and γ2 are homotopic with fixed endpoints. (This includes invariance under orientation-preserving re-parameterization of the path.) We note that Property (3) uniquely determines the integral when X = D(0, 1)− is an open disk, since then any 1-form on X has the form ω = f(t) dt where f(t) = P∞ n=0 antn converges on D(0, 1)−, and f(t) dt = d ∞ X n=1 an−1 n tn , where the series on the right converges, too (since for 0 ≤r < 1 we have that rn/|n|p ≤nrn →0). So every 1-form is exact and Property (3) applies. The same is true for open Berkovich ‘poly-disks’ (which are the analytic spaces associated to a product of open disks in Cp), when we deal with higher-dimensional spaces.
As a more interesting example, let us consider the logarithm, which should be defined on X = Gan m = P1,an \ {0, ∞} by the integral log x = x Z 1 dt t .
§ 8. Integration 46 Recall that Gan m is simply connected, so by Property (5) the integral depends only on the endpoints and not on the path (which is basically unique here anyway). If |x −1|p < 1, then we can apply Property (3) using Property (4) for the inclusion D(1, 1)−⊂Gan m ; this leads to the usual expression log(1 + z) = z −z2 2 + z3 3 ∓. . . = ∞ X n=1 (−1)n−1zn n for |z|p < 1.
We can do the same within any open disk not containing zero; using Property (2) this gives us log a(1 + z) = log a + log(1 + z) for a ∈C× p , |z|p < 1.
The question then is, how to fix the values of log a when |a −1|p ≥1?
If we use another important property of the logarithm, namely its functional equa-tion log(xy) = log x + log y , then this determines log on the subgroup R× = {a ∈C× p : |a|p = 1}: If a ∈R×, then there is a root of unity ζ ∈R× such that |a−ζ|p < 1. The functional equation forces us to set log ζ = 0, so log a = log ζ + log(ζ−1a) = log(1 + z) with |z|p = |ζ−1a −1|p < 1. Note that the functional equation follows from the properties an integral should have: log(xy) = xy Z 1 dt t (2) = x Z 1 dt t + xy Z x dt t (4) = x Z 1 dt t + y Z 1 dt t = log x + log y , where we use that dt/t is invariant under the scaling map t 7→xt.
This does not tell us, however, how to define log p, say. In fact, it turns out that we can give log p an arbitrary value λ, but then log is indeed uniquely determined as log(upv) = log u + vλ for |u|p = 1, v ∈Q.
We can take λ = 0, or (at the other extreme) treat λ as an indeterminate and consider our integrals to have values in Cp[λ]. We write logλ for the ‘branch of the logarithm’ with logλ p = λ.
As it turns out, fixing λ already fixes the integral. The following was proved by Berkovich [Ber, Thm. 9.1.1].
8.1. Theorem.
If we fix λ, then there is a unique integral satisfying properties THM Existence and uniqueness of integral (1) to (5) and (6) p Z 1 dt t = λ.
(In fact, Berkovich does more: he defines sheaves of functions on X that allow for iterated integration.) § 8. Integration 47 8.2. Example.
Consider the open Berkovich annulus A = A(r, R)−. In a similar EXAMPLE Integral on an annulus way as for an open disk, the analytic functions on A can be written as Laurent series f(t) = . . . + a−3t−3 + a−2t−2 + a−1t−1 + a0 + a1t + a2t2 + . . . = ∞ X n=−∞ antn that converge for all τ ∈Cp such that r < |τ|p < R. Then f(t) dt = dg(t)+a−1dt/t where g(t) = X n̸=0 an−1 n tn is again an analytic function on A. Recall that A is simply connected, so integrals depend only on the endpoints. If τ1, τ2 ∈Cp satisfy r < |τj|p < R, then we obtain τ2 Z τ1 f(t) dt = τ2 Z τ1 dg(t) + a−1 dt t (1) = τ2 Z τ1 dg(t) + a−1 τ2 Z τ1 dt t (3,4,6) = g(τ2) −g(τ1) + a−1 logλ τ2 τ1 .
Here we use the change of variables formula for the map A →Gan m that multiplies by τ −1 1 and (6). So if we want the integral to be well-defined (i.e., independent of the choice of λ), we need a−1 = 0, so that f(t) dt = dg(t) is exact.
♣ 8.3. Example.
Consider an elliptic curve E over Cp (say, p > 2, but this is EXAMPLE Integral on Tate ell. curve not really necessary) such that vp(j(E)) < 0. Then (as Tate has shown) there is q ∈C× p such that |q|p < 1 and E(Cp) ≃C× p /qZ; the latter extends to a covering map π: Gan m →Gan m /qZ ≃Ean that exhibits Gan m as the universal covering of Ean (recall that Gan m is simply connected).
Let 0 ̸= ω ∈Ω1(E) be a regular (and therefore invariant) differential on E and let γ : [0, 1] →Ean be the closed path that is the image under π of the (unique) path from 1 to q in Gan m . Then π∗ω is an invariant differential on Gm, so π∗ω = α dt/t for some α ∈C× p . Then Z γ ω (4) = q Z 1 αdt t (1) = α q Z 1 dt t (6) = α logλ q .
There is a unique λ ∈Cp such that logλ q = 0; if we choose this value, then integrals over ω on E will not depend on the path. But this will not work for two Tate curves simultaneously when their q parameters are not multiplicatively dependent. So we cannot avoid the path-dependence of the integral. In particular, there will not be a function f (in the sense of Berkovich’s integration theory) on all of Ean such that d f = ω. This is somewhat analogous with the situation over C, where there is no holomorphic function f on C× such that d f = dz/z.
♣ There is a different kind of integral that one can define on elliptic curves, and more generally, abelian varieties over Cp. We will do this here for regular (= invariant) 1-forms, but the theory can be extended to arbitrary algebraic 1-forms on analytic spaces associated to algebraic varieties, see [Col].
This is similar to the standard logarithm, which is related to the group structure on the multiplicative group Gm.
Let ω be an invariant 1-form on an abelian variety A over Cp (for example, an elliptic curve). There is an open (and closed) neighborhood U of the origin 0 ∈A(Cp) that is a subgroup and analytically isomorphic to an open polydisk D.
Every closed 1-form is exact on an open § 8. Integration 48 polydisk, so we obtain a function logω : U →Cp that satisfies logω x = R x 0 ω for x ∈U.
By the same argument as before, logω is a homomorphism on U.
Contrary to Gm, A is projective, which implies that A(Cp)/U is a torsion group (every element has finite order), so in this case there is a unique extension of logω to all of A(Cp) as a group homomorphism logω : A(Cp) →Cp: if x ∈A(Cp) is arbitrary, then there is some n ∈Z≥1 such that nx ∈U; we must then have logω x = (logω(nx))/n (and this does not depend on the choice of n with nx ∈U).
We define, for ω as above and x, y ∈A(Cp), Ab y Z x ω = logω y −logω x .
Then this integral satisfies Properties (1), (2) and (4) for abelian varieties and in-variant differentials, where in (4) we only consider morphisms as varieties (which are morphisms as abelian varieties composed with a translation). This integral depends only on the endpoints and does not involve a choice of path, so Prop-erty (5) is trivially satisfied. We can extend it to (the Berkovich spaces associated to) arbitrary smooth projective varieties and closed regular 1-forms as follows. Let V be such a variety and let A be its Albanese variety; write [y −x] for the image of (y, x) under the Albanese map V × V →A. Fixing x0 ∈V , set α: V →A, x 7→[x −x0]. Then if ω is a closed regular 1-form on V , there is an invariant 1-form ωA on A such that α∗ωA = ω (this does not depend on the choice of x0); in fact, α∗gives a canonical isomorphism of the space of invariant 1-forms on A with the space of closed regular 1-forms on V . We then define the ‘abelian integral’ as Ab y Z x ω = Ab [y−x] Z 0 ωA = logωA[y −x] .
It satisfies (1), (2), and (4) for morphisms of varieties. If one extends it to arbitrary algebraic closed 1-forms as in [Col], then it also satisfies (3) (for a suitable class of functions) and (6). Also this more general integral depends only on the endpoints, not on a path between them. This makes it clear that it differs from Berkovich’s integral, which in general does depend on the chosen path.
(This does not contradict the uniqueness statement in Theorem 8.1, because the abelian integral is restricted to algebraic objects and morphisms, whereas Berkovich’s theorem talks about integration for analytic 1-forms, spaces and maps.) The abelian integral has some useful applications, which derive from the following easy result. The rank of an abelian group G is defined as rk G = dimQ G ⊗Z Q, or DEF rank equivalently, the maximal number of elements of G that are linearly independent over Z (rk G := ∞if this number is unbounded).
8.4. Theorem.
Let A be an abelian variety over Cp and let Γ ⊂A(Cp) be a THM 1-forms killing Γ subgroup of rank rk Γ = r < g = dim A. Then there are at least g −r linearly independent (over Cp) invariant 1-forms ω such that logω γ = 0 for all γ ∈Γ.
Proof. Write Ω1(A) for the Cp-vector space of invariant 1-forms on A. This space has dimension g = dim A. The map Ω1(A) × A(Cp) − →Cp, (ω, x) 7− →logω x § 8. Integration 49 is Cp-linear in ω and Z-linear in x. Let γ1, . . . , γr ∈Γ be Z-linearly independent; then for each γ ∈Γ there are integers m > 0 and m1, . . . , mr such that (8.1) mγ = m1γ1 + . . . + mrγr .
Now let V ⊂Ω1(A) be the subspace given by V = {ω ∈Ω1(A) : logω γj = 0 for j = 1, . . . , r} .
Since we are imposing r linear conditions, we have dimCp V ≥g −r. The rela-tion (8.1) then implies that logω γ = 0 for all γ ∈Γ. So any Cp-basis of V does what is required.
K 8.5. Corollary.
Let C be a curve of genus g > 0 defined over a number field K, COR killing rational points on a curve with Jacobian variety J. Let P0 ∈C(K); denote by i the associated embedding C →J sending P0 to the origin.
Let Kv be the completion of K at some p-adic place v. If rk J(K) = r < g, then there are at least g −r > 0 Kv-linearly independent 1-forms ω on C over Kv such that Ab P Z P0 ω = 0 for all P ∈C(K).
Proof. We apply Theorem 8.4 and its proof to Γ = J(K) ⊂J(Kv) ⊂J(Cv). If we restrict to 1-forms defined over Kv, the proof goes through as before, working with vector spaces over Kv instead of over Cp. This gives us ω1, . . . , ωg−r ∈Ω1(J/Kv) such that logωj x = 0 for all j and all x ∈J(K). Pulling back 1-forms induces an isomorphism between Ω1(J/Kv) and Ω1(C/Kv), and we have for any j and any P ∈C(K) Ab P Z P0 i∗ωj = Ab i(P) Z 0 ωj = logωj i(P) = 0 , so i∗ω1, . . . , i∗ωg−r do what we want.
K We would like to obtain a uniform bound for the number of points in C(K). The following result allows us to reduce to disks and annuli.
8.6. Lemma.
Let C and Kv be as above. Then there is a finite collection of LEMMA partition of C(/Kv) analytic maps φm : D(0, 1)−→Can and ψn : A(ρn, 1)−→Can with ρn ∈|Kv|× (where Can is defined with respect to Kv ⊂Cp) defined over Kv such that C(Kv) = [ m φm D(0, 1)−∩Kv ∪ [ n ψn A(ρn, 1)−∩Kv , and the number of the φm and the ψn each are bounded in terms of g and Kv only.
Proof. See [Sto, Prop. 5.3]. The numer of annuli is ≤3g −3, and the number of disks is in O(qg), where q is the size of the residue field of Kv.
K § 8. Integration 50 Write Dm = φm(D(0, 1)−∩Kv) and An = ψn(A(ρn, 1)−∩Kv). It then suffices to bound # C(K) ∩Dm and C(K) ∩An for each m and n. Let V ⊂Ω1(C/Kv) be the space of regular 1-forms that kill C(K) as in Corollary 8.5. Then C(K) ∩Dm ⊂ n P ∈Dm : Ab P Z P0 ω = 0 o for every ω ∈V , and similarly for An. For a disk Dm we have for τ ∈Kv with |τ|p < 1 Ab φm(τ) Z P0 ω = Ab φm(0) Z P0 ω + Ab φm(τ) Z φm(0) ω = Ab φm(0) Z P0 ω + τ Z 0 φ∗ mω =: h(τ) (on a disk there is no difference between the two integrals; this comes from the definition of logω on U by formal integration of power series on the polydisk D).
Let π be a uniformizer of Kv (i.e., |π|p < 1 and generates the value group of Kv).
Then the number of zeros of h in Kv ∩D(0, 1)−⊂D(0, |π|p) is bounded in terms of the number of zeros of φ∗ mω in D(0, 1)−and |π|p (see Exercises). The number of zeros of φ∗ mω in D(0, 1)−is the same as the number of zeros of ω in φm(D(0, 1)−), which is at most the total number of zeros of ω, which in turn is 2g −2. This gives us a bound for #(C(K) ∩Dm) that only depends on g and Kv (via |π|p). Since the total number of disks Dm is bounded in terms of g and Kv, we get a bound on the total number of K-points of C contained in some Dm that only depends on g and Kv.
For annuli, the situation is more complicated. There is the following theorem that compares the two integrals on an annulus.
8.7. Theorem.
Fix a Kv-defined analytic map ψ: A(ρ, 1)−→Jan. There is a THM comparison of integrals on annuli linear form α: Ω1(J/Kv) →Kv such that for any x, y ∈A(ρ, 1)−we have y Z x ψ∗ω − Ab ψ(y) Z ψ(x) ω = α(ω) vp(y) −vp(x) for all ω ∈Ω1(J/Kv).
Proof. See [Sto, Prop. 7.3] or [KRZB, Prop. 3.29].
K If r ≤g −3, then for each annulus An there will be at least one 0 ̸= ω ∈V such that ψ∗ nω is exact and α(ω) = 0 (note that α depends on the annulus), since this imposes two linear conditions. If ψ∗ nω = dh is exact, then we can again bound the number of zeros of h in Kv ∩A(ρn, 1)−⊂A(ρn/|π|p, |π|p) in terms of g and Kv, under some additional assumption (the ‘relevant’ range of exponents in the Laurent series representation of h includes 0), which can be shown to always hold (this is done in [Sto, Cor. 6.7] for hyperelliptic curves and in [KRZB, Lemma 4.15] in general). This finally leads to the following result.
§ 8. Integration 51 8.8. Theorem.
Fix d ≥1 and g ≥3. There is some B = B(d, g) depending only THM uniform bound on rational points on d and g such that for any number field K with [K : Q] ≤d and any curve C of genus g over K, with Jacobian J such that rk J(K) ≤g−3, we have #C(K) ≤B.
Proof. We can assume that C(K) ̸= ∅; let P0 ∈C(K) and use it as the base-point for an embedding i: C →J as in Corollary 8.5. Fix some prime p; there are then only finitely many possibilities (up to isomorphism) for the completion Kv at a p-adic place v. Let B(d, g) be the maximum of the bounds for C(K) obtained by the reasoning above for g and each Kv. The claim follows.
K For K = Q, one can write down explicit bounds. For hyperelliptic curves (which allow for quite explicit computations) a possible bound is #C(Q) ≤33(g −1) + ( 1 if r = 0 (and g ≥3), 8rg −1 if 1 ≤r ≤g −3, where r = rk J(Q), see [Sto, Thm. 9.1] (this works with Q3). In the general case, one obtains a bound of the shape O(g2), see [KRZB, Thm. 5.1]. For arbitrary d, one still has a bound of the shape Od(g2), but the implied constant depends exponentially on d.
Literatur 52 References [BR] Matthew Baker, Robert Rumely: Potential theory and dynamics on the Berkovich projective line. Mathematical Surveys and Monographs 159. American Mathematical Society, Providence, RI, 2010.
[AWS] Matthew Baker, Brian Conrad, Samit Dasgupta, Kiran S. Kedlaya, Jeremy Teitelbaum: p-adic geometry. Lectures from the 10th Arizona Winter School held at the University of Arizona, Tucson, AZ, March 10–14, 2007. Edited by David Savitt and Dinesh S. Thakur. University Lecture Series, 45. American Mathematical Society, Providence, RI, 2008. xii+203 pages.
[Ber] Vladimir G. Berkovich: Integration of one-forms on p-adic analytic spaces. Annals of Mathematics Studies 162. Princeton University Press, Princeton, NJ, 2007. vi+156 pages.
[Col] Pierre Colmez: Int´ egration sur les vari´ et´ es p-adiques. Ast´ erisque 248, Soci´ et´ e Math´ ematique de France, Paris, 1998. viii+155 pages.
[KRZB] Eric Katz, Joseph Rabinoff and David Zureick-Brown: Uniform bounds for the number of rational points on curves of small Mordell-Weil rank, Preprint (2015), arXiv:1504.00694v2 [math.NT].
[Sto] Michael Stoll: Uniform bounds for the number of rational points on hyperelliptic curves of small Mordell-Weil rank, Preprint (2015), arXiv:1307.1773v6 [math.NT]. |
7789 | https://www.quora.com/How-many-multiples-of-5-are-there-between-1-to-100 | How many multiples of 5 are there between 1 to 100? - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Mathematics
Common Multiples
Count
Mathematics Homework Ques...
Arithmetic Number Theory
Counting in General
Integers
Arithmetic
Counting Rules
5
How many multiples of 5 are there between 1 to 100?
All related (71)
Sort
Recommended
Assistant
Bot
·
1y
There are 20 multiples of 5 between 1 and 100. The multiples of 5 in that range are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
Upvote ·
Bhaskar Reddy
M.Sc.Ed(maths) from Regional Institute of Education, Mysore (RIE Mysore)
·4y
Related
How many multiples of 3 are there between 1 to 100?
Hi, Here is a simple way to know it,
We can use the following method to know the number of multiplies of any number between any two numbers.
Here I go
No. of multiples of a number (x) = (last multiple of x - first multiple of x)/x
See the picture for more clarity
Here in place of x we have 3 so, 99–3=96
96/3=32
Now just add one to it
so, 32+1=33
It holds for any number between any numbers.
Thank you
Continue Reading
Hi, Here is a simple way to know it,
We can use the following method to know the number of multiplies of any number between any two numbers.
Here I go
No. of multiples of a number (x) = (last multiple of x - first multiple of x)/x
See the picture for more clarity
Here in place of x we have 3 so, 99–3=96
96/3=32
Now just add one to it
so, 32+1=33
It holds for any number between any numbers.
Thank you
Upvote ·
99 18
9 4
Sponsored by Grammarly
Stuck on the blinking cursor?
Move your great ideas to polished drafts without the guesswork. Try Grammarly today!
Download
99 34
Related questions
More answers below
How many multiples of 5 lie between 100 to 500?
How many multiples of 3 are there among from 100 to 200?
What are multiples of 100?
How many multiples of 4 between 1 and 100 have 4 as one of the digits?
How many multiples of 5 are there from 1 to 200 which are not multiples of 4?
Akshay Khandelwal
Lives in Mhow, India
·7y
This question can be solved by using the concept of Arithmetic Progressions (A.P.). The first multiple is 5. (a = first term, d = common difference, l = last term) and n = number of terms)
So, a = 5.
Now,d = 5.
The last multiple of 5 between 1 and 100 is 95.
So, l = 95.
A.P. = 5, 10, 15,……., 95
Using formula(l = a + d(n - 1)),we get
95 = 5 + 5(n-1)
n-1 = 90/5
n = 18 + 1
n = 19;
Hence,there are 19 multiples of 5 between 1 and 100.
Upvote ·
9 7
9 3
Laura Kay Posey
Education, K-6th from SWT (Graduated 1966) · Author has 6.5K answers and 2M answer views
·1y
MULTIPLES from 1 to 100. Don’t include 100: “ to 100, not through 100.”
ANSWER: See Below= A,B = 19
20—1=19 Multiples from 1 -100 =19
A.
100 ÷ 5 = 20 —1 = 19 Do not include 100
100 — 1 = 95 to 100, not through 100
95 ÷ 5 = 19
B.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 95, X
Upvote ·
Utkarsh Singh
Bachelor of arts from Indira Gandhi National Open University (IGNOU) (Graduated 2025)
·1y
Originally Answered: How many times is 5 in (1 to 100)? ·
You can do it by simple counting
Lets count from 1 to 10 = 5 will appear 1 time
Same from 10 to 100 = 5 will appear 19 times ( 3 from 11 to 40 and 2 in 41 to 50 and 10 in 51 to 60 and also 4 from 61 to 100 )
Now just add = 19+1 = 20
From 1 to 100 5 will appear 20 times
Just let me know if there is any other method to solve this question .☺️☺️
Upvote ·
Promoted by Betterbuck
Anthony Madden
Writer for Betterbuck
·Updated Aug 15
What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
Continue Reading
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
Upvote ·
999 557
99 60
9 4
Related questions
More answers below
How many 1 comes between 100 to 1000?
How many times is 5 in (1 to 100)?
What is the sum of the numbers multiple of 5 from 1 to 100?
How many 9 are there in 1 to 100?
How many ones are between 1 to 100?
Abhi Kumbhani
7y
Related
How many integers from 1 to 100 are either multiples of 3 or multiples of 5?
So first divide 100/3 = 33 1/3. That means there are 33 multiples of 3 from 1–100.
Do the same 100/5 = 20. So 20 multiples of 5 from 1–100.
Now don’t forget there are some numbers that are repeated by the 3s and the 5s, so multiply 3 5 = 15. Now do 100/15 = 6 2/3. There are 6 multiples of 15 under 100.
So now you add 33+20=53 and subtract the redundant numbers: 53 - 6 = 47.
Answer: 47
Upvote ·
99 18
Alvaro Arce
Studied Mathematics&Linguistics at UNIVERSIDAD MAYOR DE SAN ANDRES (Graduated 2002) · Author has 676 answers and 120.8K answer views
·Sep 8
There are
[100 5]=20[100 5]=20
multiples of 5, including 100.
Upvote ·
9 1
Sponsored by All Out
Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out.
Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association.
Shop Now
999 619
John Smith
Masters Degree in Mechanical and Aerospace Engineering&Mathematics, University of Missouri-Columbia (Graduated 2003) · Author has 649 answers and 567.6K answer views
·Updated 6y
Originally Answered: How many numbers between 1 and 100 have 5 ones? ·
The study of numbers is very interesting. Take a couple of days out of your life and study rational, irrational and imaginary numbers and it will teach you much about life and God.
That being said, in order to answer your question, one must define which category of numbers are being considered. Considering all types of numbers, the answer is infinity. For instance, let's just take several real numbers:
11.111 or 1.23451111 or 91.11119 or 3.56757575651474638477645741111
Clearly, there are infinite combinations of numbers with 5 ones between 1 and 100.
If you specify the number to be an integer (1,2
Continue Reading
The study of numbers is very interesting. Take a couple of days out of your life and study rational, irrational and imaginary numbers and it will teach you much about life and God.
That being said, in order to answer your question, one must define which category of numbers are being considered. Considering all types of numbers, the answer is infinity. For instance, let's just take several real numbers:
11.111 or 1.23451111 or 91.11119 or 3.56757575651474638477645741111
Clearly, there are infinite combinations of numbers with 5 ones between 1 and 100.
If you specify the number to be an integer (1,2,3,4, etc.) the answer is zero. If you don’t limit the integer between 1 and 100, there are also an infinite number of integers containing 5 ones.
In the number’s world, integers are extremely rare even though they are an infinite number of them.
Upvote ·
9 3
9 2
Calvin L.
18, Mathematics & Statistics major · Author has 10K answers and 2.2M answer views
·3y
There are 20 multiples of 5 between 1 and 100.
They are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. (If it is inclusive)
Upvote ·
9 4
Sponsored by LPU Online
Career Ka Turning Point with LPU Online.
100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support.
Apply Now
999 263
Related questions
How many multiples of 5 lie between 100 to 500?
How many multiples of 3 are there among from 100 to 200?
What are multiples of 100?
How many multiples of 4 between 1 and 100 have 4 as one of the digits?
How many multiples of 5 are there from 1 to 200 which are not multiples of 4?
How many 1 comes between 100 to 1000?
How many times is 5 in (1 to 100)?
What is the sum of the numbers multiple of 5 from 1 to 100?
How many 9 are there in 1 to 100?
How many ones are between 1 to 100?
How many integers from 1 to 100 are either multiples of 3 or multiples of 5?
Related questions
How many multiples of 5 lie between 100 to 500?
How many multiples of 3 are there among from 100 to 200?
What are multiples of 100?
How many multiples of 4 between 1 and 100 have 4 as one of the digits?
How many multiples of 5 are there from 1 to 200 which are not multiples of 4?
How many 1 comes between 100 to 1000?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
Always Active
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
Always Active
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
Always Active
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
7790 | https://prime-numbers.fandom.com/wiki/3-Digit_Primes | Skip to content
Prime Numbers Wiki
3,035
pages
Contents
1 100s
2 200s
3 300s
4 400s
5 500s
6 600s
7 700s
8 800s
9 900s
in:
3-Digit Prime Numbers, Prime Numbers, Prime numbers from 101-200,
and
8 more
Prime Numbers from 201-300
Prime Numbers from 301-400
Prime Numbers from 401-500
Prime Numbers from 501-600
Prime Numbers from 601-700
Prime numbers from 701-800
Prime Numbers from 801-900
Prime Numbers from 901-1000
3-Digit Primes
2
Sign in to edit
History
Purge
Talk (0)
Here are all the 3 digit prime numbers, i.e. all prime numbers between 101-1,000.
Contents
1 100s
2 200s
3 300s
4 400s
5 500s
6 600s
7 700s
8 800s
9 900s
100s[]
One Hundred One
One Hundred Three
One Hundred Seven
One Hundred Nine
One Hundred Thirteen
One Hundred Twenty-Seven
One Hundred Thirty-One
One Hundred Thirty-Seven
One Hundred Thirty-Nine
One Hundred Forty-Nine
One Hundred Fifty-One
One Hundred Fifty-Seven
One Hundred Sixty-Three
One Hundred Sixty-Seven
One Hundred Seventy-Three
One Hundred Seventy-Nine
One Hundred Eighty-One
One Hundred Ninety-One
One Hundred Ninety-Three
One Hundred Ninety-Seven
One Hundred Ninety-Nine
| | | | | |
--- ---
| 101 | 103 | 107 | 109 | 113 |
| 127 | 131 | 137 | 139 | 149 |
| 151 | 157 | 163 | 167 | 173 |
| 179 | 181 | 191 | 193 | 197 |
| 199 | | | | |
200s[]
Two Hundred Eleven
Two Hundred Twenty-Three
Two Hundred Twenty-Seven
Two Hundred Twenty-Nine
Two Hundred Thirty-Three
Two Hundred Thirty-Nine
Two Hundred Forty-One
Two Hundred Fifty-One
Two Hundred Fifty-Seven
Two Hundred Sixty-Three
Two Hundred Sixty-Nine
Two Hundred Seventy-One
Two Hundred Seventy-Seven
Two Hundred Eighty-One
Two Hundred Eighty-Three
Two Hundred Ninety-Three
| | | | | |
--- ---
| 211 | 223 | 227 | 229 | 233 |
| 239 | 241 | 251 | 257 | 263 |
| 269 | 271 | 277 | 281 | 283 |
| 293 | | | | |
300s[]
Three Hundred Seven
Three Hundred Eleven
Three Hundred Thirteen
Three Hundred Seventeen
Three Hundred Thirty-One
Three Hundred Thirty-Seven
Three Hundred Forty-Seven
Three Hundred Forty-Nine
Three Hundred Fifty-Three
Three Hundred Fifty-Nine
Three Hundred Sixty-Seven
Three Hundred Seventy-Three
Three Hundred Seventy-Nine
Three Hundred Eighty-Three
Three Hundred Eighty-Nine
Three Hundred Ninety-Seven
| | | | | |
--- ---
| 307 | 311 | 313 | 317 | 331 |
| 337 | 347 | 349 | 353 | 359 |
| 367 | 373 | 379 | 383 | 389 |
| 397 | | | | |
400s[]
Four Hundred One
Four Hundred Nine
Four Hundred Nineteen
Four Hundred Twenty-One
Four Hundred Thirty-One
Four Hundred Thirty-Three
Four Hundred Thirty-Nine
Four Hundred Forty-Three
Four Hundred Forty-Nine
Four Hundred Fifty-Seven
Four Hundred Sixty-One
Four Hundred Sixty-Three
Four Hundred Sixty-Seven
Four Hundred Seventy-Nine
Four Hundred Eighty-Seven
Four Hundred Ninety-One
Four Hundred Ninety-Nine
| | | | | |
--- ---
| 401 | 409 | 419 | 421 | 431 |
| 433 | 439 | 443 | 449 | 457 |
| 461 | 463 | 467 | 479 | 487 |
| 491 | 499 | | | |
500s[]
Five Hundred Three
Five Hundred Nine
Five Hundred Twenty-One
Five Hundred Twenty-Three
Five Hundred Forty-One
Five Hundred Forty-Seven
Five Hundred Fifty-Seven
Five Hundred Sixty-Three
Five Hundred Sixty-Nine
Five Hundred Seventy-One
Five Hundred Seventy-Seven
Five Hundred Eighty-Seven
Five Hundred Ninety-Three
Five Hundred Ninety-Nine
| | | | | |
--- ---
| 503 | 509 | 521 | 523 | 541 |
| 547 | 557 | 563 | 569 | 571 |
| 577 | 587 | 593 | 599 | |
600s[]
Six Hundred One
Six Hundred Seven
Six Hundred Thirteen
Six Hundred Seventeen
Six Hundred Nineteen
Six Hundred Thirty-One
Six Hundred Forty-One
Six Hundred Forty-Three
Six Hundred Forty-Seven
Six Hundred Fifty-Three
Six Hundred Fifty-Nine
Six Hundred Sixty-One
Six Hundred Seventy-Three
Six Hundred Seventy-Seven
Six Hundred Eighty-Three
Six Hundred Ninety-One
| | | | | |
--- ---
| 601 | 607 | 613 | 617 | 619 |
| 631 | 641 | 643 | 647 | 653 |
| 659 | 661 | 673 | 677 | 683 |
| 691 | | | | |
700s[]
Seven Hundred One
Seven Hundred Nine
Seven Hundred Nineteen
Seven Hundred Twenty-Seven
Seven Hundred Thirty-Three
Seven Hundred Thirty-Nine
Seven Hundred Forty-Three
Seven Hundred Fifty-One
Seven Hundred Fifty-Seven
Seven Hundred Sixty-One
Seven Hundred Sixty-Nine
Seven Hundred Seventy-Three
Seven Hundred Eighty-Seven
Seven Hundred Ninety-Seven
| | | | | |
--- ---
| 701 | 709 | 719 | 727 | 733 |
| 739 | 743 | 751 | 757 | 761 |
| 769 | 773 | 787 | 797 | |
800s[]
Eight Hundred Nine
Eight Hundred Eleven
Eight Hundred Twenty-One
Eight Hundred Twenty-Three
Eight Hundred Twenty-Seven
Eight Hundred Twenty-Nine
Eight Hundred Thirty-Nine
Eight Hundred Fifty-Three
Eight Hundred Fifty-Seven
Eight Hundred Fifty-Nine
Eight Hundred Sixty-Three
Eight Hundred Seventy-Seven
Eight Hundred Eighty-One
Eight Hundred Eighty-Three
Eight Hundred Eighty-Seven
| | | | | |
--- ---
| 809 | 811 | 821 | 823 | 827 |
| 829 | 839 | 853 | 857 | 859 |
| 863 | 877 | 881 | 883 | 887 |
900s[]
Nine Hundred Seven
Nine Hundred Eleven
Nine Hundred Nineteen
Nine Hundred Twenty-Nine
Nine Hundred Thirty-Seven
Nine Hundred Forty-One
Nine Hundred Forty-Seven
Nine Hundred Fifty-Three
Nine Hundred Sixty-Seven
Nine Hundred Seventy-One
Nine Hundred Seventy-Seven
Nine Hundred Eighty-Three
Nine Hundred Ninety-One
Nine Hundred Ninety-Seven
| | | | | |
--- ---
| 907 | 911 | 919 | 929 | 937 |
| 941 | 947 | 953 | 967 | 971 |
| 977 | 983 | 991 | 997 | |
All in all, there are 143 prime numbers from 101-1,000. This means that 143/900 or around 1 in 6 numbers from 101-1,000 are prime. 757 numbers are composite.
| | | |
---
| 1 or 2 Digits (1-100) | List of Prime Numbers (3 digits) | Go on to 4 Digits! |
Lorem ipsum dolor sit amet
Categories
Community content is available under CC-BY-SA unless otherwise noted.
2 Comments
Join the conversation
Sign in to read what others are saying and share your thoughts.
SIGN IN
Don't have account? Register now
Do Not Sell My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Targeting Cookies
These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising.
Social Media Cookies
These cookies are set by a range of social media services that we have added to the site to enable you to share our content with your friends and networks. They are capable of tracking your browser across other sites and building up a profile of your interests. This may impact the content and messages you see on other websites you visit. If you do not allow these cookies you may not be able to use or see these sharing tools.
Search this wiki
Search all wikis |
7791 | https://www.sciencedirect.com/science/article/abs/pii/0306454986900976 | Eigenvalues basic to diffusion in a hollow cylinder without azimuthal symmetry - ScienceDirect
Skip to main contentSkip to article
Journals & Books
Access throughyour organization
Purchase PDF
Search ScienceDirect
Article preview
Abstract
References (5)
Cited by (2)
Annals of Nuclear Energy
Volume 13, Issue 2, 1986, Pages 93-99
Eigenvalues basic to diffusion in a hollow cylinder without azimuthal symmetry
Author links open overlay panel R.M.Cotta, M.N.Özişik
Show more
Add to Mendeley
Share
Cite
rights and content
Abstract
The eigenvalues needed for the computation of transient heat or mass diffusion in a hollow cylinder without azimuthal symmetry are calculated. The first 10 eigenvalues are tabulated for the convection boundary condition at the outer and inner surfaces for several different values of the radius ratio, the Biot number and the separation constant associated with the azimuthal variation.
Access through your organization
Check access to the full text by signing in through your organization.
Access through your organization
Recommended articles
References (5)
R.M. Cotta et al.### Ann. Nucl. Energy (1984)
M.D. Mikhailov et al.### Int. J. Heat Mass Transfer (1983)
There are more references available in the full text version of this article.
Cited by (2)
Heat transfer-a review of 1986 literature
1987, International Journal of Heat and Mass Transfer
### On the eigenvalues basic to the analytical solution of convective heat transfer with axial diffusion effects
1995, Communications in Numerical Methods in Engineering
View full text
Copyright © 1986 Published by Elsevier Ltd.
Recommended articles
The “Vicious Circle of addictive Social Media Use and Mental Health” Model
Acta Psychologica, Volume 247, 2024, Article 104306 Julia Brailovskaia
### Enhancing bird conservation in tropical urban parks through land sparing and sharing strategies: Evidence from occupancy data
Urban Forestry & Urban Greening, Volume 98, 2024, Article 128415 Muhamad Amir Hadi, …, Badrul Azhar
### The casein in sheep milk processed by cold plasma technology: Phosphorylation degree, functional properties, oxidation characteristics, and structure
Food Chemistry, Volume 457, 2024, Article 140140 Jiaying Zhang, …, Wupeng Ge
### Identification, expression, and characterization of a marine-derived chitinase Ce 0303 from Chitiniphilus eburneus YS-30 with exo- and endo- hydrolytic properties
International Journal of Biological Macromolecules, Volume 276, Part 2, 2024, Article 133980 Yuan Yu, …, Kailiang Leng
### Preparation and performance of bicomponent polyacrylonitrile/polymethyl methacrylate lithium-ion battery separator by centrifugal spinning
Polymer, Volume 307, 2024, Article 127226 Quan Zheng, …, Baiwei Zhang
### Anthocyanin/halloysite hybrid pigments with enhanced thermal and acid stability
Dyes and Pigments, Volume 231, 2024, Article 112356 Amirreza Ghavami, …, Alireza Mohebali
Show 3 more articles
About ScienceDirect
Remote access
Contact and support
Terms and conditions
Privacy policy
Cookies are used by this site.Cookie settings
All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply.
We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy
Cookie Settings Accept all cookies
Cookie Preference Center
We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors.
You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings.
You may also be able to exercise your privacy choices as described in our Privacy Policy
Allow all
Manage Consent Preferences
Strictly Necessary Cookies
Always active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work.
Cookie Details List
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site.
Cookie Details List
Contextual Advertising Cookies
[x] Contextual Advertising Cookies
These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user.
Cookie Details List
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Confirm my choices |
7792 | https://www.connectionsacademy.com/support/resources/article/how-25-riddles-can-help-your-elementary-student/ | Enrollment for this school year is open! Learn how to apply
Skip Navigation
How 25 Riddles Can Help Your Elementary Student
Curriculum & Skills
byChristopher E. Nelson
6 min to read
Fun riddles for kids can serve as teaching tools for your student, whether they’re enrolled in an online school like Connections Academy, a traditional brick-and-mortar school, or they’re homeschooled.
Download Full List of 25 Kids' Riddles
Whether you’re arming your elementary school student with a list of riddles, challenging them with brainteasers, or writing short riddles together—riddles for kids are fun learning activities that you can do not just from home, but anywhere!
Riddles can potentially make kids smarter. They introduce children to functional thinking, which emphasizes seeing relationships between new ideas and previous knowledge to help them learn and remember faster and more easily. Functional thinking skills developed in the elementary grades are considered a gateway to algebra and other higher mathematics.
So What is a Riddle?
A riddle consists of a question and a surprise answer that relies on an unexpected interpretation of the question or a play on words (such as a pun):
RIDDLE: The more of them you take, the more you leave behind. What are they? ANSWER:Footsteps.
RIDDLE: What do you get when you cross an automobile with a domesticated animal? ANSWER:A carpet.
What is the Difference Between a Joke and Riddle?
A riddle is considered a joke when the person asked isn’t expected to know the answer, but instead the question is simply a set-up for the punch line. This is sometimes called a “conundrum” style of riddle. Elephant jokes are an example of conundrum riddles:
RIDDLE: What was the elephant doing on the freeway? ANSWER:About 5 mph.
RIDDLE: What gets wet when drying? ANSWER: A towel.
RIDDLE: Where do fish keep their money? ANSWER:In a riverbank, of course.
Why Are Riddles Great For Kids?
When age-appropriate, a riddle can help elementary students:
Sharpen their vocabulary/wordplay skills
Exercise critical thinking
Problem-solve
Improve their reading comprehension
Enhance their creativity
All children have a sense of humor and riddles are among the first forms of written humor that young children can truly understand, play along with, and initiate. Children who have a well-developed sense of humor are happier and more optimistic, have better self-esteem, and handle differences better (their own and others’).
Riddles also create a bond. Your child will always remember when they first heard that silly riddle from you. And because riddles are meant to be retold, they give your elementary schooler something to share with others.
Many children become friends through shared humor, and for some, being able to make other kids laugh is a talent they hone and find rewarding all their lives.
Easy Riddles For Kids
RIDDLE: What is orange and sounds like a parrot? ANSWER:A carrot.
RIDDLE: 100 feet in the air, but its back is on the ground. What is it? ANSWER:A centipede on its back.
RIDDLE: I will bring you down, but I will never lift you up. What am I? ANSWER:Gravity.
RIDDLE: What is something you always have with you, but you always leave behind? ANSWER: Fingerprints.
Rhyming Kids’ Riddles
RIDDLE: I can be cracked, I can be made. I can be told, I can be played. What am I? ANSWER:A joke.
RIDDLE: First, I was yellow, now I am white. Salty or sweet, I crunch with each bite. What am I? ANSWER:Popcorn.
RIDDLE: Glittering points that downward thrust, sparking spears that never rust. What are they? ANSWER:Icicles.
RIDDLE: I have two arms, but fingers I have none. I’ve got two feet, but I cannot run. I carry well, but I carry best with my feet off the ground. What am I? ANSWER:A wheelbarrow.
RIDDLE: If I have it, I shouldn’t share it, because if I share it, I won’t have it. What is it? ANSWER:A secret.
RIDDLE: I live off of a busy street, if you want you can stay for an hour or two, but if you don’t pay rent, I’ll tell on you. What am I? ANSWER: A parking meter.
Problem-Solving Riddles
Some riddles pose problems that teach deductive reasoning. They ask a reachable “How?” or “Why?” They are more about thinking than laughing.
RIDDLE: Two mothers and two daughters go to a pet store and buy three cats. Each gets her own cat. How is this possible? ANSWER:They are a grandmother, a mother, and a daughter. The grandmother is also the mother’s mother, so there are two daughters and two mothers, but only three people.
RIDDLE: While walking across a bridge I saw a boat full of people. Yet, there wasn't a single person on the boat. Why? ANSWER:Everyone on the boat was married.
Tricky Riddles For Kids
RIDDLE: I might be far from the point, but I’m not a mistake. In fact, I fix yours. What am I? ANSWER:An eraser.
RIDDLE: I can be long, I can be short. I can be grown, I can be bought. I can be painted or left bare. I can be round or a little square. What am I? ANSWER:A fingernail.
RIDDLE: I’m at the beginning of time and part of the past, present, and future. I’m part of history, but not of here and now. In a moment you’ll find me, if you know what I am. What am I? ANSWER:The letter T.
RIDDLE: You can swallow me, but I can consume you too. What am I? ANSWER:Pride.
RIDDLE: What is it that you can keep after giving it to someone else? ANSWER:Your word.
RIDDLE: It is the beginning of eternity, the end of time and space, the beginning of the end and the end of every space. What is it? ANSWER:The letter E.
RIDDLE: What goes around the house and in the house, but never touches the house? ANSWER:The sun.
RIDDLE: What comes once in a minute, twice in a moment, but never in a thousand years? ANSWER:The letter M.
How to Write a Riddle
Many children will try their own riddles unprompted and are often surprisingly clever. Even at its simplest, creating a brainteaser requires strong writing skills—expressing a thought that only makes sense if you make the wordplay or skewed connection between question and answer understandable. Here’s how to write a riddle:
Come up with a punchline—the answer to a question.
Brainstorm words and phrases that go with your punchline. List things it (the answer/punchline) does or that you can do with it.
Choose a few words or phrases from your list and search for synonyms online (or use a thesaurus). Note some surprising, interesting, or new words and phrases you find and look for synonyms to them. Make lists.
Make another list of synonyms for your punchline word or phrase.
Be your punchline. Imagine yourself as the punchline. Think about its place in the world, how it acts, or how it is acted upon (used).
Make connections between your lists. Think figuratively:
Use simile, comparisons that use “like” or “as.”
Use metaphors, phrases that describe the object symbolically, not literally.
Use onomatopoeia, words that sound like their meaning.
Before you know it, you’ll see some concepts that go together in a funny, odd, or unexpected way, and be crafting a fun brainteaser for kids. Now you get the last laugh!
Not all children excel at language arts. If the idea of writing riddles causes boredom, try whipping out some kitchen ingredients and build a fruit volcano instead to keep learning from home fun!
f
p
t
Link
Ins
Learning Activities
Learning Skills
Related Posts
How Clubs and Extracurricular Activities Help Virtual School Students Socialize
by Connections Academy
read more
Learning Environment & Community
Why Building Strong Teacher-Student Relationships Matters
by Valerie Kirk
read more
Learning Environment & Community
Home Classroom Setup: 8 Online School Essentials
by Kristina Cappetta
read more
Learning Environment & Community |
7793 | https://www.stat.berkeley.edu/~vadicgor/Random_tilings.pdf | Lectures on random lozenge tilings Vadim Gorin University of Wisconsin–Madison, Massachusetts Institute of Technology, and Institute for Information Transmission Problems of Russian Academy of Sciences.
vadicgor@gmail.com This material was published by Cambridge University Press as “Lectures on Random Lozenge Tilings” by Vadim Gorin. This pre-publication version is free to view and download for personal use only. Not for re-distribution, re-sale or use in derivative works.
This online version has several minor errors corrected, as compared to the printed edition.
Contents About this book page 6 1 Lecture 1: Introduction and tileability 7 1.1 Preface 7 1.2 Motivation 9 1.3 Mathematical questions 10 1.4 Thurston’s theorem on tileability 15 1.5 Other classes of tilings and reviews 20 2 Lecture 2: Counting tilings through determinants 22 2.1 Approach 1: Kasteleyn formula 22 2.2 Approach 2: Lindstr¨ om-Gessel-Viennot lemma 26 2.3 Other exact enumeration results 31 3 Lecture 3: Extensions of the Kasteleyn theorem 33 3.1 Weighted counting 33 3.2 Tileable holes and correlation functions 35 3.3 Tilings on a torus 36 4 Lecture 4: Counting tilings on large torus 41 4.1 Free energy 41 4.2 Densities of three types of lozenges 43 4.3 Asymptotics of correlation functions 46 5 Lecture 5: Monotonicity and concentration for tilings 48 5.1 Monotonicity 48 5.2 Concentration 50 5.3 Limit shape 52 6 Lecture 6: Slope and free energy 54 6.1 Slope in a random weighted tiling 54 6.2 Number of tilings of a fixed slope 56 6.3 Concentration of the slope 58 6.4 Limit shape of a torus 59 3 4 Contents 7 Lecture 7: Maximizers in the variational principle 60 7.1 Review 60 7.2 The definition of surface tension and class of functions 61 7.3 Upper semicontinuity 64 7.4 Existence of the maximizer 66 7.5 Uniqueness of the maximizer 67 8 Lecture 8: Proof of the variational principle 69 9 Lecture 9: Euler-Lagrange and Burgers equations 76 9.1 Euler-Lagrange equations 76 9.2 Complex Burgers equation via a change of coordinates 77 9.3 Generalization to qVolume–weighted tilings 80 9.4 Complex characteristics method 81 10 Lecture 10: Explicit formulas for limit shapes 83 10.1 Analytic solutions to the Burgers equation 83 10.2 Algebraic solutions 87 10.3 Limit shapes via quantized Free Probability 88 11 Lecture 11: Global Gaussian fluctuations for the heights 92 11.1 Kenyon-Okounkov conjecture 92 11.2 Gaussian Free Field 94 11.3 Gaussian Free Field in Complex Structures 98 12 Lecture 12: Heuristics for the Kenyon-Okounkov conjecture 100 13 Lecture 13: Ergodic Gibbs translation-invariant measures 108 13.1 Tilings of the plane 108 13.2 Properties of the local limits 110 13.3 Slope of EGTI measure 112 13.4 Correlation functions of EGTI measures 114 13.5 Frozen, liquid, and gas phases 115 14 Lecture 14: Inverse Kasteleyn matrix for trapezoids 119 15 Lecture 15: Steepest descent method for asymptotic analysis 126 15.1 Setting for steepest descent 126 15.2 Warm up example: real integral 126 15.3 One-dimensional contour integrals 127 15.4 Steepest descent for a double contour integral 129 16 Lecture 16: Bulk local limits for tilings of hexagons 132 17 Lecture 17: Bulk local limits near straight boundaries 141 5 18 Lecture 18: Edge limits of tilings of hexagons 148 18.1 Heuristic derivation of two scaling exponents 148 18.2 Edge limit of random tilings of hexagons 150 18.3 The Airy line ensemble in tilings and beyond 155 19 Lecture 19: The Airy line ensemble and other edge limits 157 19.1 Invariant description of the Airy line ensemble 157 19.2 Local limits at special points of the frozen boundary 159 19.3 From tilings to random matrices 160 20 Lecture 20: GUE–corners process and its discrete analogues 167 20.1 Density of GUE–corners process 167 20.2 GUE–corners process as a universal limit 171 20.3 A link to asymptotic representation theory and analysis 174 21 Lecture 21: Discrete log-gases 179 21.1 Log-gases and loop equations 179 21.2 Law of Large Numbers through loop equations 182 21.3 Gaussian fluctuations through loop equations 185 21.4 Orthogonal polynomial ensembles 188 22 Lecture 22: Plane partitions and Schur functions 191 22.1 Plane partitions 191 22.2 Schur polynomials 193 22.3 Expectations of observables 195 23 Lecture 23: Limit shape and fluctuations for plane partitions 202 23.1 Law of Large Numbers 202 23.2 Central Limit Theorem 208 24 Lecture 24: Discrete Gaussian component in fluctuations 212 24.1 Random heights of holes 212 24.2 Discrete fluctuations of heights through GFF heuristics.
213 24.3 Approach through log-gases 217 24.4 2d Dirichlet energy and 1d logarithmic energy 220 24.5 Discrete component in tilings on Riemann surfaces 227 25 Lecture 25: Sampling random tilings 229 25.1 Markov Chain Monte-Carlo 229 25.2 Coupling from the past [ProppWilson96] 233 25.3 Sampling through counting 237 25.4 Sampling through bijections 237 25.5 Sampling through transformations of domains 238 References 241 Index 254 About this book These are lecture notes for a one semester class devoted to the study of random tilings. It was 18.177 taught at Massachusetts Institute of Technology during Spring of 2019. The brilliant students who participated in the class1: Andrew Ahn, Ganesh Ajjanagadde, Livingston Albritten, Morris (Jie Jun) Ang, Aaron Berger, Evan Chen, Cesar Cuenca, Yuzhou Gu, Kaarel Haenni, Sergei Ko-rotkikh, Roger Van Peski, Mehtaab Sawhney, Mihir Singhal provided tremen-dous help in typing the notes.
Additional material was added to most of the lectures after the class was over. Hence, when using this review as a textbook for a class, one should not expect to cover all the material in one semester, something should be left out.
I would like to thank my wife Anna Bykhovskaya for her advice, love, and support. I also thank Amol Aggarwal, Alexei Borodin, Richard Kenyon, Chris-tian Krattenthaler, Arno Kuijlaars, Igor Pak, Jiaming Xu, Marianna Russkikh, and Semen Shlosman for their useful comments and suggestions. I am grateful to Christophe Charlier, Maurice Duits, Sevak Mkrtchyan, and Leonid Petrov for the help with simulations of random tilings.
Funding acknowledgements. The work of V.G. was partially supported by NSF Grants DMS-1664619, DMS-1949820, by NEC Corporation Fund for Research in Computers and Communications, and by the Office of the Vice Chancellor for Research and Graduate Education at the University of Wisconsin–Madison with funding from the Wisconsin Alumni Research Foun-dation. Chapters 6 and 13 of this work were supported by Russian Science Foundation (project 20-41-09009).
1 Last name alphabetic order.
6 Lecture 1: Introduction and tileability 1.1 Preface The goal of the lectures is to understand the mathematics of tilings. The general setup is to take a lattice domain and tile it with elementary blocks. For the most part, we study the special case of tiling a polygonal domain on the triangular grid (of mesh size 1) by three kinds of rhombi that we call “lozenges”.
The left panel of Figure 1.1 shows an example of a polygonal domain on the triangular grid. The right panel of Figure 1.1 shows the lozenges: each of them is obtained by gluing two adjacent lattice triangles. A triangle of the grid is surrounded by three other triangles, attaching one of them we get one of the three types of lozenges. The lozenges can be also viewed as orthogonal projections onto the x+y+z = 0 plane of three sides of a unit cube. Figure 1.2 provides an example of a lozenge tiling of the domain of Figure 1.1.
Figure 1.3 shows a lozenge tiling of a large domain, with the three types of lozenges shown in three different colors. The tiling here is generated uniformly at random over the space of all possible tilings of this domain. More precisely, it is generated by a computer that is assumed to have access to perfectly ran-dom bits. It is certainly not clear at this stage how such “perfect sampling” may be done computationally, in fact we address this issue in the very last lecture. Figure 1.3 is meant to capture a “typical tiling”, making sense of what this means is another topic that will be covered in this book. The simulation re-veals interesting features: one clearly sees next to the boundaries of the domain formation of the regions, where only one type of lozenges is observed. These regions are typically referred to as “frozen regions” and their boundaries are “artic curves”; their discovery and study has been one of the important driving forces for investigations of the properties of random tilings.
We often identify a tiling with a so-called “height function”. The idea is to think of a 2-dimensional stepped surface living in 3-dimensional space and treat tiling as a projection of such surface onto x + y + z = 0 plane along the 7 8 Lecture 1: Introduction and tileability A = 5 B = 5 C = 5 Figure 1.1 Left panel: A 5×5×5 hexagon with 2×2 rhombic hole. Right panel: three types of lozenges obtained by gluing two adjacent triangles of the grid.
A = 5 B = 5 C = 5 Figure 1.2 A lozenge tiling of a 5×5×5 hexagon with a hole.
(1,1,1) direction. In this way three lozenges become projections of three el-ementary squares in three-dimensional space parallel to each of the three co-ordinate planes. We formally define the height function later in this lecture.
We refer to [BorodinBorodinA] for a gallery of height functions in a 3d virtual reality setting.
9 Figure 1.3 A perfect sample of a uniformly random tiling of a large hexagon with a hole. (I thank Leonid Petrov for this simulation.) 1.2 Motivation Figure 1.3 is beautiful, and as mathematicians, this suffices for probing more deeply into it and trying to explain various features that we observe in the simulation.
There are also some motivations from theoretical physics and statistical me-chanics. Lozenge tilings serve as a “toy model” that helps in understanding the 3-dimensional Ising model (a standard model for magnetism). Configurations of the 3-dimensional Ising model are assignments of + ⃝and -⃝spins to lattice points of a domain in Z3. A parameter (usually called temperature) controls how much the adjacent spins are inclined to be oriented in the same direc-tion. The zero temperature limit leads to spins piling into as large as possible groups of the same orientation; these groups are separated with stepped sur-faces whose projections in (1,1,1) direction are lozenge tilings. For instance, if we start from the Ising model in a cube and fix boundary conditions to be + ⃝ along three faces (sharing a single vertex) of this cube and -⃝along other three 10 Lecture 1: Introduction and tileability faces, then we end up with lozenge tilings of a hexagon in zero temperature limit.1 Another deformation of lozenge tilings is the 6-vertex or square ice model, whose state space consists of configurations of the molecules H2O on the grid.
There are six weights in this model (corresponding to six ways to match an oxygen with two out of the neighboring four hydrogens), and for particular choices of the weights one discovers weighted bijections with tilings.
We refer to [Baxter82] for some information about the Ising and the six-vertex model, further motivations to study them and approaches to the analysis.
In general, both Ising and six-vertex model are more complicated objects than lozenge tilings, and they are much less understood. From this point of view, theory of random tilings that we develop in these lectures can be treated as the first step towards the understanding of more complicated models of statistical mechanics.
For yet another motivation we notice that the two dimensional stepped surfaces of our study have flat faces (these are frozen regions consisting of lozenges of one type, cf. Figures 1.3, 1.5), and, thus, are relevant for modelling facets of crystals. One example from the everyday life here is a corner of a large box of salt. For a particular (non-uniform) random tiling model, leading to the shapes reminiscent of such a corner we refer to Figure 10.1 in Lecture 10.
1.3 Mathematical questions We now turn to describing the basic questions that drive the mathematical study of tilings.
1 Existence of tilings: Given a domain R drawn on the triangular grid (and thus consisting of a finite family of triangles), does there exist a tiling of it?
For example, a unit sided hexagon is trivially tileable in 2 different ways and bottom-right panel in Figure 1.1 shows one of these tilings. On the other hand, if we take the equilateral triangle of sidelength 3 as our domain R, then it is not tileable. This can be seen directly, as the corner lozenges are fixed and immediately cause obstruction. Another way to prove non-tileability is by coloring unit triangles inside R in white and black colors in an alternating fashion. Each lozenge covers one black and one white tri-angle, but there are an unequal number of black and white triangles in the 1 See [Shlosman00, CerfKenyon01, BodineauSchonmannShlosman04] for the discussion of the common features between low-temperature and zero-temperature 3d Ising model, as well as the interplay between the topics of this book and more classical statistical mechanics.
11 No tilings Figure 1.4 A tileable and a non-tileable domain.
R: equilateral triangle of sidelength 3 has 6 triangles of one color and 3 triangles of another color.
Another example is shown in Figure 1.4: in the left panel we see two domains. The top one is tileable, while the bottom one is not.
More generally, is there a simple (and efficient from the computational point of view) criterion for tileability of R? The answer is yes by a theorem of [Thurston90]. We discuss this theorem in more detail in Section 1.4 later in this lecture.
2 How many tilings does a given domain R have? The quality of the an-swer depends on one’s taste and perhaps what one means by a “closed form”/“explicit answer”. Here is one case where a “good” answer is known by the following theorem due to MacMahon (who studied this problem in the context of plane partitions, conjectured a formula in [MacMahon1896], and proved it in [MacMahon1915, Art. 495]). Let R be a hexagon with side lengths A,B,C,A,B,C in cyclic order. We denote this henceforth by the A × B ×C hexagon (in particular, the left panel of Figure 1.1 was showing 5×5×5 hexagon with a rhombic hole).
Theorem 1.1 ([MacMahon1896]). The number of lozenge tilings of A×B× C hexagon equals A ∏ a=1 B ∏ b=1 C ∏ c=1 a+b+c−1 a+b+c−2.
(1.1) As a sanity check, one can take A = B = C = 1, yielding the answer 2 — and indeed one readily checks that there are precisely two tilings of 1×1×1 12 Lecture 1: Introduction and tileability hexagon. A proof of this theorem is given in Section 2.2. Another situation where the number of tilings is somewhat explicit is for the torus and we discuss this in Lectures 3 and 4. For general R, one can not hope for such nice answers, yet certain determinantal formulas (involving large matrices whose size is proportional to the area of the domain) exist, as we discuss in Lecture 2.
3 Law of large numbers. Each lozenge tilings is a projection of a 2d–surface, and therefore can be represented as a graph of a function of two variables which we call the height function (its construction is discussed in more detail in Section 1.4). If we take a uniformly random tiling of a given domain, then we obtain a random height function h(x,y) encoding a random stepped surface. What is happening with the random height function of a domain of linear size L as L →∞? As we will see in Lectures 5-10 and in Lecture 23, the rescaled height function has a deterministic limit: lim L→∞ 1 Lh(Lx,Ly) = ˆ h(x,y), An important question is how to compute and describe the limit shape ˆ h(x,y). One feature of the limit shapes of tilings is the presence of regions where the limiting height function is linear. In terms of random tilings, these are “frozen” regions, which contain only one type of lozenges. In particular, in Figure 1.3 there is a clear outer frozen region near each of the six vertices of the hexagon; another four frozen regions surround the hole in the middle.
Which regions are “liquid”, i.e. contain all three types of lozenges? What is the shape of the “Arctic curve”, i.e. the boundary between frozen and liquid regions? For example, with the aL×bL×cL hexagon setup, one can visually see from Figure 1.5 that the boundary appears to be an inscribed ellipse: Theorem 1.2 ([CohnLarsenPropp98, BaikKriecherbauerMcLaughlinMiller03, Gorin08, Petrov12a]). For aL × bL × cL hexagon, a uniformly random tiling is with high probability asymptotically frozen outside the inscribed ellipse as L →∞. In more detail, for each (x,y) outside the ellipse, with probability tending to 1 as L →∞, all the lozenges that we observe in a finite neighborhood of (xL,yL) are of the same type.
The inscribed ellipse of Theorem 1.2 is the unique degree 2 curve tangent to the hexagon’s sides. This characterization in terms of algebraic curves extends to other polygonal domains, where one picks the degree such that there is a unique algebraic curve tangent (in the interior) of R. Various ap-13 Figure 1.5 Arctic circle of a lozenge tiling proaches to Theorem 1.2, its relatives and generalizations are discussed in Lectures 7, 10, 16, 21, 23.
4 Analogs of the Central Limit Theorem: The next goal is to understand the random field of fluctuations of the height function around the asymptotic limit shape, i.e. to identify the limit lim L→∞(h(Lx,Ly)−E[h(Lx,Ly)]) = ξ(x,y).
(1.2) Note the unusual scaling; one may naively expect a need for dividing by √ L to account for fluctuations as in the classical central limit theorem for sums of independent random variables and many similar statements. But there turns out be some “rigidity” in tilings and the fluctuations are much smaller.
ξ(x,y) denotes the limiting random field, in this case it can be identified with the so-called “Gaussian free field”. The Gaussian free field is related to conformal geometry as it turns out to be invariant under conformal transfor-14 Lecture 1: Introduction and tileability Figure 1.6 Fluctuations of the centered height function for lozenge tilings of a hexagon with a hole. Another drawing of the same system is in Figure 24.2 later in the text.
mations. This topic will be explored in Lectures 11, 12, 21, and 23. For now we confine ourselves to yet another picture given in Figure 1.6.
5 Height of a plateau/hole: Consider Figures 1.2, 1.3. The central hole has an integer valued random height. What is the limiting distribution of this height? Note, that comparing with (1.2) we expect that no normalization is necessary as L →∞, and, therefore, the distribution remains discrete as L →∞. Hence, the limit cannot be Gaussian. You can make a guess now or proceed to Lecture 24 for the detailed discussion.
6 Local limit: Suppose we “zoom in” at a particular location inside a random tiling of a huge domain. What are the characteristics of the tiling there?
For example, consider a certain finite pattern of lozenges, call it P and see Figure 1.7 for an example. Asymptotically, what is the probability that P appears in the vicinity of (Lx,Ly)? Note that if P consists of a single lozenge, then we are just counting the local proportions for the lozenges of three types, hence one can expect that they are reconstructed from gradients of the limit shape ˜ h. However, for more general P it is not clear what to 15 Figure 1.7 An example of a local pattern P of lozenges. The bulk limit question asks about the probability of observing such (or any other) pattern in a vicinity of a given point (Lx,Ly) in a random tiling of a domain of linear scale L →∞.
expect. This is called a “bulk limit” problem and we return to it in Lectures 16 and 17.
7 Edge limit: How does the Arctic curve (border of the frozen region) fluctu-ate? What is the correct scaling? It turns out to be L 1 3 here, something that is certainly not obvious at all right now. The asymptotic law of rescaled fluc-tuations turns out to be given by the celebrated Tracy-Widom distribution from random matrix theory as we discuss in Lectures 18 and 19.
8 Sampling: How does one sample from the uniform distribution over tilings?
The number of tilings grows extremely fast (see e.g. the MacMahon for-mula (1.1)), so one can not simply exhaustively enumerate the tilings on a computer, and a smarter procedure is needed. We discuss several approaches to sampling in Lecture 25.
9 Open problem: Can we extend the theory to 3 dimensional tiles?
1.4 Thurston’s theorem on tileability We begin our study from the first question: given a domain R, is there at least one tiling? The material here is essentially based on [Thurston90].
Without loss of generality we may assume R is a connected domain; the question of tileability of a domain is equivalent to that of its connected compo-nents. We start by assuming that R is simply connected, and then remove this restriction.
We first discuss the notion of a height function in more detail, and how it relates to the question of tileability of a domain. There are 6 directions on the triangular grid, the unit vectors in those directions are: a = (0,1), b = − √ 3 2 ,−1 2 !
, c = √ 3 2 ,−1 2 !
, −a, −b, −c.
16 Lecture 1: Introduction and tileability a = (0, 1) b = − √ 3 2 , −1 2 c = √ 3 2 , −1 2 Figure 1.8 Out of the six lattice directions three are chosen to be positive (in bold).
We call a,b,c positive directions, and their negations are negative directions, as in Figure 1.8.
We now define an asymmetric nonnegative distance function d(u,v) for any two vertices u,v on the triangular grid (which is the lattice spanned by a and b) and a domain R. d(u,v) is the minimal number of edges in a positively oriented path from u to v staying within (or on the boundary) of R. This is well defined as we assumed R to be connected. The asymmetry is clear: consider R consisting of a single triangle, and let u,v be two vertices of it. Then d(u,v) = 1 and d(v,u) = 2 (or other way round).
We now formally define a height function h(v) for each vertex v ∈R, given a tiling of R. This is given by a local rule: if u →v is a positive direction, then h(v)−h(u) = ( 1, if we follow an edge of a lozenge −2, if we cross a lozenge diagonally.
(1.3) It may be easily checked that this height function is defined consistently. This is because the rules are consistent for a single lozenge (take e.g. the lozenge {0,a,b,a + b}), and the definition extends consistently across unions. Note that h is determined up to a constant shift. We may assume without loss of generality that our favorite vertex v0 has h(v0) = 0.
Let us check that our definition matches the intuitive notion of the height.
For that we treat positive directions a, b, c in Figure 1.8 as projections of coordinate axes Ox, Oy, Oz, respectively. Take one of the lozenges, say {0,a,b,a + b}. Up to rescaling by the factor q 2 3, it can be treated as a projection of the square {(0,0,0),(1,0,0),(0,1,0),(1,1,0)} onto the plane x + y + z = 0. Hence, our locally defined height function becomes the value of x + y + z. Similar observation is valid for two other types of lozenges. The conclusion is that if we identify lozenge tiling with a stepped surface in three-17 0 −1 −2 −3 −4 −5 −6 −7 −8 0 1 2 3 2 3 2 1 0 0 −1 −1 0 1 2 3 2 3 2 1 0 0 1 1 Figure 1.9 Three domains: the triangle on the left has no tiles, because there is no way to consistently define the heights along its boundary; the domain in the middle has no tilings because the inequality (1.4) fails for the encircled pair of points; the right domain is tileable.
dimensional space, then our height is (signed and rescaled) distance from the surface point to its projection onto x+y+z = const plane in (1,1,1) direction.
Note also that even if R is not tileable, but is simply connected, then our local rules uniquely (up to global shift) define h on the boundary ∂R: no lozenges are allowed to cross it and therefore the increments in positive di-rection all equal 1.
Exercise 1.3. Find the values of the height function along the two connected components of the boundary of the holey hexagon in the left panel of Figure 1.1. Note that there are two arbitrary constants involved: one per connected component.
With these notions in hand, we state the theorem of [Thurston90] on tileabil-ity.
Theorem 1.4. Let R be a simply connected domain, with boundary ∂R on the triangular lattice. Then R is tileable if and only if both conditions hold: 1 One can define h on ∂R, so that h(v)−h(u) = 1 whenever u →v is an edge of ∂R, such that v−u is a unit vector in one of the positive directions a,b,c.
2 The above h satisfies ∀u,v ∈∂R : h(v)−h(u) ≤d(u,v).
(1.4) Before presenting the proof of Theorem 1.4, let us illustrate its conditions.
For that we consider three domains drawn in Figure 1.9, set h = 0 in the left(-bottom) corner of each domain and further define h on ∂R by local rules fol-lowing the boundary ∂R in the clockwise direction. For the triangle domain there is no way to consistently define the height function along its boundary: when we circle around, the value does not match what we started from. And indeed, this domain is not tileable. For the middle domain of Figure 1.9, we can 18 Lecture 1: Introduction and tileability define the heights along the boundary, but (1.4) fails for the encircled points with height function values −1 and 3. On the other hand, this domain clearly has no tilings. Finally, for the right domain of Figure 1.9, both conditions of Theorem 1.4 are satisfied; a (unique in this case) possible lozenge tiling is shown in the picture.
Proof of Theorem 1.4 Suppose R is tileable and fix any tiling. Then it defines h on all of R including ∂R by local rules (1.3). Take any positively oriented path from u to v. Then the increments of d along this path are always 1, while the increments of h are either 1 or −2. Thus (1.4) holds.
Now suppose that h satisfies (1.4). Define for v ∈R: h(v) = min u∈∂R [d(u,v)+h(u)].
(1.5) We call this the maximal height function extending h ∂R and corresponding to the maximal tiling. As d is nonnegative, the definition (1.5) of h matches on ∂R with the given in the theorem h ∂R. On the other hand, since the inequality h(v)−h(u) ≤d(u,v) necessarily holds for any height function h and any u,v ∈ R (by the same argument as for the case u,v ∈∂R above), no height function extending h ∂R can have larger value at v than (1.5).
Claim. h defined by (1.5) changes by 1 or −2 along each positive edge.
The claim would allow us to reconstruct the tiling uniquely: each −2 edge gives a diagonal of a lozenge and hence a unique lozenge; each triangle has exactly one −2 edge, as the only way to position increments +1, −2 along three positive edges is +1 + 1 −2 = 0; hence, for each triangle we uniquely reconstruct the lozenge to which it belongs. As such, we have reduced our task to establishing the claim.
Let v →w be a positively oriented edge in R \∂R. We begin by establishing some estimates. For any u ∈∂R, d(u,w) ≤d(u,v)+1 by augmenting the u,v path by a single edge. Thus we have: h(w) ≤h(v)+1.
Similarly, we may augment the u,w path by two positively oriented edges (at least one of the left/right pair across v,w is available) to establish d(u,v) ≤d(u,w)+2, and so: h(w) ≥h(v)−2.
It remains to rule out the values 0 and −1 for h(v) −h(w). We achieve this through two observations: 19 • For any two positively oriented paths linking the same two vertices, their number of edges differs by a multiple of three.
• The function u 7→d(u,v) + h(u) changes by multiples of three as u varies over ∂R.
In order to prove the first observation, let n1 a, n1 b, and n1 c denote the numbers of edges of three types (according to Figure 1.8) for the first path and let n2 a, n2 b, and n2 c be the numbers for the second path. We have two identities, obtained by projecting the paths on a and b directions: 2n1 a −n1 b −n1 c = 2n2 a −n2 b −n2 c and −n1 a +2n1 b −n1 c = −n2 a +n2 b −n2 c.
Adding the identities together, we conclude that (n1 a +n1 b +n1 c)−(n2 a +n2 b +n2 c) equals 3(n1 c −n2 c), i.e. it is a multiple of three.
The second observation is proven by tracing d(u,v)+h(u) as u moves by one edge along ∂R; note that here we use that R is simply-connected, implying connectivity of ∂R.
Combining these two observations together with the small augmentations of the paths above, we conclude that necessarily h(w) ≡h(v)+1 (mod 3). This rules out h(v)−h(w) ∈{0,−1}.
How can we generalize this when R is not simply connected? We note two key issues above: 1 It is not clear how to define h on ∂R if one is not already given a tiling. On each closed loop piece of ∂R we see that h is determined up to a constant shift, but these constants may differ across loops.
2 The step of the above proof that shifts the argument of the function u 7→ d(u,v)+h(u), u ∈∂R, used the fact that we could simply move along ∂R edge by edge. This is not in general true for a multiply connected domain.
These issues can be addressed by ensuring that d(u,v)−h(v)+h(u) = 3k(u,v) for k(u,v) ∈N, ∀u,v, and simply leaving the ambiguity in h as it is: Corollary 1.5. Let R be a non-simply connected domain, with boundary ∂R on the triangular lattice. Define h along ∂R; this is uniquely defined up to constants c1,c2,...,cl corresponding to constant shifts along the l pieces of ∂R. Then R is tileable if and only if there exist c1,...,cl such that for every u and v in ∂R: d(u,v)−h(v)+h(u) ≥0, (1.6a) d(u,v)−h(v)+h(u) ≡0 (mod 3).
(1.6b) 20 Lecture 1: Introduction and tileability Figure 1.10 Left panel: Aztec diamond domain on the square grid. Middle panel: two types of dominos. Right panel: one possible domino tiling.
Proof of Corollary 1.5 ⇒: (1.6a) ensures that this part of the proof of The-orem 1.4 remains valid; (1.6b) is also true when we have a tiling, again by the above proof. ⇐: The shift from u to u′ is now valid, so the proof carries through.
We remark that Thurston’s theorem 1.4 and the associated height func-tion method provide a O(|R|ln(|R|)) time algorithm for tileability. There are recent improvements, see e.g. [PakShefferTassy16, Thm. 1.2] for an O(|∂R|ln(|∂R|)) algorithm in the simply-connected case. [Thiant03] pro-vides an O(|R|l + a(R)) algorithm, where l denotes the number of holes in R, and a(R) denotes the area of all the holes of R. We also refer to [PakShefferTassy16, Section 5] for the further discussion of the optimal al-gorithms for tileability from the computer science literature.
1.5 Other classes of tilings and reviews Throughout these lectures we are going to concentrate on lozenge tilings. How-ever, the theory of random tilings is not restricted to only them.
The simplest possible alternative version of the theory deals with domino tilings on square grid. In this setting we consider a domain drawn on the square grid Z2 and its tilings with horizontal and vertical 2 rectangles called dominos. Figure 1.10 shows a domain known as the Aztec diamond and one of its domino tilings. Most of the results that we discuss in the book have their counterparts for domino tilings. For instance, their height functions and the question of tileability are discussed in the same article [Thurston90] as for lozenges. The definitions become slightly more complicated and we refer to [BorodinBorodinB] for an appealing 3d visualization of the height func-tions of domino tilings. Some aspects of random tilings of the Aztec diamond are reviewed in [BaikDeiftSuidan16, Chapter 12] and in [Johansson17]. Two other lecture notes on tilings that we are happy to recommend, [Kenyon09] 21 and [Toninelli19], deal with dimers (or perfect matchings) which combine both lozenge and domino tilings.
Some additional classes of tilings and techniques for their enumeration can be found in [Propp14]. However, as we drift away from lozenges and dominos, the amount of available information about random tilings starts to decrease.
For instance, square triangle tilings have central importance in representation theory and algebraic combinatorics, since they appear in the enumeration of Littlewood-Richardson coefficients, see [Purbhoo07], [Zinn-Justin08]. Yet, at the time of writing this book, our understanding of the asymptotic behavior of random square triangle tilings is very limited.
Moving farther away from the topics of this book, an introductory ar-ticle [ArdilaStanley05] as well as two detailed textbooks [Golomb95] and [GrunbaumShephard16] show what else can be hiding under the word “tilings”.
Lecture 2: Counting tilings through de-terminants The goal of this lecture is to present two distinct approaches to the counting of lozenge tilings. Both yield different determinantal formulae for the number of tilings.
A statistical mechanics tradition defines a partition function as a total num-ber of configurations in some model or, more generally, sum of the weights of all configurations in some model. The usual notation is to use capital letter Z in various fonts for the partition functions. In this language, we are interested in tools for the evaluation of the partition functions Z for tilings. In Lectures 2 and 3 we present several approaches to computing the partition functions, whereas Lecture 4 contains the first asymptotic results for the partition functions for tilings of large domains – we deal with large torus there. In addition to being interesting on their own, these computations will form a base for establishing the Law of Large Numbers and the Variational Principle for random tilings in subsequent Lectures 6, 7, and 8.
2.1 Approach 1: Kasteleyn formula The first approach relies on what may be called “Kasteleyn theory” (or more properly “Kasteleyn-Temperley-Fisher (KTF) theory”), originally developed in [Kasteleyn61, TemperleyFisher61]. We illustrate the basic approach in Fig-ure 2.1 for lozenge tilings on the triangular grid. One may color the triangles in two colors in an alternating fashion similar to a checkerboard. Connecting the centers of adjacent triangles, one obtains a hexagonal dual lattice to the original triangular lattice, with black and white alternating vertices. The graph of this hexagonal dual is clearly bipartite by the above coloring. Consider a simply connected domain R that consists of an equal number of triangles of 22 23 Figure 2.1 The hexagonal dual and connection of tilings to perfect matchings two types (otherwise it can not be tiled), here we simply use a unit hexagon.
Then lozenge tilings of it are in bijection with perfect matchings of the associ-ated bipartite graph (G ) formed by restriction of the dual lattice to the region corresponding to R, here a hexagon. As such, our question can be rephrased as that of counting the number of perfect matchings of a (special) bipartite graph.
The solid lines in Figure 2.1 illustrate one of the two perfect matchings (and hence tilings) here.
The basic approach of KTF is to relate the number of perfect matchings of the bipartite G to the determinant of a certain matrix K (Kastelyn matrix) associated to G . Let there be N black vertices, and N white vertices. Then K is an N ×N matrix with entries: Ki j = ( 1, if white i is connected to black j, 0, otherwise.
Then we have the following theorem: Theorem 2.1 ([Kasteleyn61, TemperleyFisher61]). Let R be a simply con-nected domain on the triangular grid, and let G be its associated bipartite graph on the dual hexagonal grid. Then, the number of lozenge tilings of R is |det(K)|.
Before getting into the proof of this theorem, it is instructive to work out a simple example. Let R be a unit hexagon, label its vertices in the dual graph 24 Lecture 2: Counting tilings through determinants by 1,1′,2,2′,3,3′ in clockwise order. Then, K = 1 0 1 1 1 0 0 1 1 Then det(K) = 2, which agrees with the number of tilings. However, also note that one can swap 1′,2′ and label the graph by 1,2′,2,1′,3,3′ in clockwise order, resulting in: K′ = 0 1 1 1 1 0 1 0 1 and det(K′) = −2. Thus, the absolute value is needed. This boils down to the fact that there is no canonical order for the black vertices relative to the white ones.
Proof of Theorem 2.1 First, note that although there are N! terms in the de-terminant, most of them vanish. More precisely, a term in the determinant is nonzero if and only if it corresponds to a perfect matching: using two edges adjacent to a single vertex corresponds to using two matrix elements from a single row/column, and is not a part of the determinant’s expansion.
The essence of the argument is thus understanding the signs of the terms.
We claim that for simply connected R (the hypothesis right now, subsequent lectures will relax it), all the signs are the same. We remark that this is a crucial step of the theory; dealing with permanents (which are determinants without the signs) is far trickier if not infeasible.
Given its importance, we present two proofs of this claim.
1 The first one relies on the height function theory developed in Lecture 1.
Define elementary rotation E that takes the matching of the unit hexagon (1,1′),(2,2′),(3,3′) to (1,3′),(2,1′),(3,2′), i.e. swaps the solid and dash-dotted matchings in Figure 2.1. We may also define E−1. Geometrically, E and E−1 correspond to removing and addding a single cube on the stepped surface (cf. Lecture 1). We claim that any two tilings of R are connected by a sequence of elementary moves of form E,E−1. It suffices to show that we can move from any tiling to the maximal tiling, i.e. the tiling corresponding to the point-wise maximal height function1. For that, geometrically one can 1 The point-wise maximum of two height functions (that coincide at some point) is again a height function, since the local rules (1.3) are preserved — for instance, one can show this by induction in the size of the domain. Hence, the set of height functions with fixed boundary conditions has a unique maximal element. In fact, we explicitly constructed this maximal element in our proof of Theorem 1.4.
25 simply add one cube at a time until no more additions are possible. This process terminates on any simply connected domain R, and it can only end at the the maximal tiling.
It now remains to check that E does not alter the sign of a perfect match-ing. It is clear that the sign of a perfect matching is just the number of inversions in the black to white permutation obtained from the matching (by definition of K and det). In the examples above, these permutations are π(1,2,3) = (1,2,3) and π′(1,2,3) = (3,1,2) respectively. (3,1,2) is an even permutation, so composing with it does not alter the parity of the num-ber of inversions (i.e. the parity of the permutation itself). Thus, all perfect matchings have the same sign.
2 The second approach relies upon comparing two perfect matchings M1 and M2 of the same simply connected domain R directly. We work on the dual hexagonal graph. Consider the union of these two matchings. Each vertex has degree 2 now (by perfect matching hypothesis). Thus, the union con-sists of a bunch of doubled edges as well as loops. The doubled edges may be ignored (they correspond to common lozenges). M1 and M2 are obtained from each other by rotation along the loops, in a similar fashion to the opera-tion E above, except possibly across a larger number of edges. For a loop of length 2p (p black, p white), the sign of this operation is (−1)p−1 as it cor-responds to a cycle of length p which has p−1 inversions. Thus it suffices to prove that p is always odd.
Here we use the specific nature of the hexagonal dual graph. First, note that each vertex has degree 3. Thus, any loop that does not repeat an edge can not self intersect at a vertex, and is thus simple. Such a loop encloses some number of hexagons. We claim that p has opposite parity to the total (both black and white) number of vertices strictly inside the loop. We prove this claim by induction on the number of enclosed hexagons. With a single hexagon, p = 3, the number of vertices inside is 0 and the claim is trivial.
Consider a contiguous domain made of hexagons P with a boundary of length 2p. One can always remove one boundary hexagon such that it does not disconnect the domain P. Doing a case analysis on the position of the surrounding hexagons, we see that the parity of the boundary loop length (measured in terms of say black vertices) remains opposite to that of the total number of interior vertices when we remove this hexagon. Hence, the claim follows by induction.
It remains to note that the number of the vertices inside each loop is even.
26 Lecture 2: Counting tilings through determinants Indeed, otherwise, there would have been no perfect matchings of the inte-rior vertices2. Thus, p is always odd.
Remark 2.2. In general the permanental formula for counting perfect match-ings of a graph is always valid. However, to get a determinantal formula one needs to introduce signs/factors into K. It turns out that one can always find a consistent set of signs for counting matchings of any planar bipar-tite graph. This is quite nontrivial, and involves the Kasteleyn orientation of edges [Kasteleyn63, Kasteleyn67]. The hexagonal case is simple since one can use the constant signs by the above proof. For non-planar graphs, good choices of signs are not known. However, for special cases, such as the torus that will be covered in Lecture 3, small modifications of the determinantal formula still work. More generaly, on genus g surface, the number of perfect matchings is given by a sum of 22g signed determinants, cf. [CimasoniReshetikhin06] and reference therein. For non-bipartite graphs, one needs to replace determinants with Pfaffians. Further information is available in, for instance, the work of Kasteleyn [Kasteleyn67], as well as the lecture notes on dimers [Kenyon09].
Exercise 2.3. Consider the domino tilings of the Aztec diamond, as in Figure 1.10. Find out what matrix elements we should take for the Kasteleyn matrix K, so that its determinant gives the total number of tilings. (Hint: try to use the 4th roots of unity: 1, −1, i, −i).
Remark 2.4. The construction that we used in the second proof of Theo-rem 2.1 can be turned into an interesting stochastic system. Take two inde-pendent uniformly random lozenge tilings (equivalently, perfect matchings) of the same domain and superimpose them. This results in a collection of random loops which is known as the double-dimer model. What is happen-ing with this collection as the domain becomes large and the mesh size goes to 0? It is expected that one observes the Conformal Loop Ensemble CLEκ with κ = 4 in the limit. For tilings of general domains this was not proven at the time when this book was written. However, partial results exist in the literature and there is little doubt in the validity of this conjecture, see [Kenyon11, Dubedat14, BasokChelkak18].
2.2 Approach 2: Lindstr¨ om-Gessel-Viennot lemma Suppose we want to apply the above machinery to derive MacMahon’s formula for the number of tilings of a R = A × B ×C hexagon. In principle, we have 2 The assumption of the domain being simply-connected is used at this point.
27 A B C B+C A B Figure 2.2 Bijection with non-intersecting lattice paths reduced the computation to that of a rather sparse determinant. However, it is not clear how we can proceed further. The goal of this section is to describe an alternative approach.
We use a bijection of tilings with another combinatorial object, namely non-intersecting lattice paths as illustrated in Figure 2.2.
We describe the bijection as follows. Enumerate (without loss, consistent with above orientation of hexagon) the three fundamental lozenges la,lb,lc with their longer diagonals inclined at angles 0,−π 3 , π 3 , respectively. Leave la as is, and draw stripes at angles −π 6 , π 6 connecting the midpoints of op-posite sides of lozenges lb,lc respectively. There is thus a bijection between lozenge tilings of the hexagon and non-intersecting lattice paths connecting the two vertical sides of length A. We may apply an affine transformation to obtain non-intersecting lattice paths on Z2 connecting (0,0),...,(0,A−1) with (B +C,B),...,(B+C,B+A−1) by elementary steps ⃗ vb ≜(1,0),⃗ vc ≜(1,1), corresponding to the stripes across lb,lc respectively. For general domains one can still set up this bijection, the problem is that the starting and ending lo-cations may no longer be contiguous, something that plays a key role in the derivation of MacMahon’s formula by this approach.
Various forms of the following statement were used by many authors, see [KarlinMcGregor59b, KarlinMcGregor59a, GesselViennot85], with the most general result appearing in [Lindstrom73].
Theorem 2.5. The number of non-intersecting paths which start at x1 < x2 < ··· < xN and end at y1 < y2 < ··· < yN after T steps is: det T yi −xj N i, j=1 .
(2.1) 28 Lecture 2: Counting tilings through determinants Proof of Theorem 2.5 First, notice that for any permutation σ of [N] = {1,2,...,N}, we see that: N ∏ j=1 T yσ( j) −x( j) counts the total number of collections of paths linking xj to yσ(j) ∀j ∈[N].
The terms come with various signs, depending on the parity of σ. The goal is to show that the sum over the entangled paths is 0. By entangled, we mean any set of paths that are not non-intersecting. Let the collection of entan-gled paths be denoted E . We construct an involution f : E →E such that sign(σ( f(P))) = −sign(σ(P)) for all P ∈E , where σ(P) denotes the per-mutation corresponding to which xi gets connected to which yj, and sign(σ) denotes the parity of σ. This would complete the task, by the elementary: 2 ∑ P∈E sign(σ(P)) = ∑ P∈E sign(σ(P))+ ∑ P∈E sign(σ( f(P))) = ∑ P∈E sign(σ(P))+ ∑ P∈E −sign(σ(P)) = 0.
The involution is achieved by “tail-swapping”. Care needs to be taken to en-sure that it is well defined because there can be many intersections. We take the right-most intersection points of two paths (if there is more than one intersec-tion point with the same abscissa, then we take one with the largest ordinate) and swap their tails to the right from the intersection point. Thus, if before the involution we had these two paths linking xi to yj and xi′ to yi′, then after the involution xi is linked to yj′ and xi′ is linked to yj. If paths had no intersec-tions, then we do nothing. This is an involution because the chosen intersection point remains the same when we iterate f, and swapping twice gets us back to where we started. Furthermore, the parity of σ changes when we apply f. We note that other choices of the involution are possible as long as they are well defined.
After cancellation, what we have left are non-intersecting paths. In the spe-cific setting here, they can only arise from σ being the identity, since any other σ will result in intersections. The identity is an even permutation, so we do not need to take absolute values here unlike the Kasteleyn formula.
Exercise 2.6. For a collection of paths E , let w(E ) denote the sum of vertical coordinates of all vertices of all paths. Fix a parameter q, and using the same method find a q–version of (2.1). You should get a N × N determinantal for-mula for the sum of qw(E ) over all collections of non-intersecting paths starting 29 at x1 < x2 < ··· < xN and ending at y1 < y2 < ··· < yN after T steps. At q = 1 the formula should match (2.1).
We remark that for a general domain, we still do not know how to compute the determinant (2.1). However, if either xi or y j consist of consecutive integers, we can evaluate this determinant in “closed form”. This is true in the case of the A×B×C hexagon, and so we now prove Theorem 1.1.
By Theorem 2.5 and the bijection with tilings we described, we have reduced our task to the computation of: det 1≤i, j≤A B+C B−i+ j (2.2) Our proof relies upon the following lemma, which can be found in a very helpful reference for the evaluation of determinants [Krattenthaler99].
Lemma 2.7. Let X1,...,Xn,A2,...,An,B2,...,Bn be indeterminates. Then, det 1≤i, j≤n((Xi +An)(Xi +An−1)...(Xi +Aj+1)(Xi +Bj)(Xi +Bj−1)...(Xi +B2)) = ∏ 1≤i<j≤n (Xi −Xj) ∏ 2≤i≤j≤n (Bi −Aj).
Proof The proof is based on reduction to the standard Vandermonde determi-nant by column operations. First, subtract the n −1-th column from the n-th, the n−2-th from the n−1-th, ..., the first column from the second, to reduce the LHS to: " n ∏ i=2 (Bi −Ai) # det 1≤i, j≤n((Xi+An)(Xi+An−1)...(Xi+Aj+1)(Xi+Bj−1)...(Xi+B2)).
(2.3) Next, repeat the same process to the determinant of (2.3), factoring out n−1 ∏ i=2 (Bi −Ai+1).
We can clearly keep repeating the process, until we have reached the simplified form: " ∏ 2≤i≤j≤n (Bi −A j) # det 1≤i, j≤n((Xi +An)(Xi +An−1)...(Xi +Aj+1)).
At this stage we have a slightly generalized Vandermonde determinant which 30 Lecture 2: Counting tilings through determinants evaluates to the desired3: ∏ 1≤i<j≤n (Xi −Xj).
Proof of Theorem 1.1 Observe that: det 1≤i, j≤A B+C B−i+ j = " A ∏ i=1 (B+C)!
(B−i+A)!(C +i−1)!
× det 1≤i,j≤A (B−i+A)!
(B−i+ j)!
(C +i−1)!
(C +i−j)!
= (−1)(A 2) " A ∏ i=1 (B+C)!
(B−i+A)!(C +i−1)!
× det 1≤i, j≤A (i−B−A)(i−B−A+1)···(i−B−j −1) ×(i+C −j +1)(i+C −j +2)···(i+C −1) .
Now take Xi = i,A j = −B−j,B j =C−j+1 in Lemma 2.7 to simplify further.
We get: (−1)(A 2) " A ∏ i=1 (B+C)!
(B−i+A)!(C +i−1)!
" ∏ 1≤i< j≤A (i−j) #" ∏ 2≤i≤j≤A (C +B+1−i+ j) # = " A ∏ i=1 (B+C)!
(B−i+A)!(C +i−1)!
" ∏ 1≤i< j≤A ( j −i) #" ∏ 2≤i≤j≤A (C +B+1−i+ j) # = " A ∏ i=1 (B+C)!
(B−i+A)!(C +i−1)!
" ∏ 1≤j<A j!
" ∏ 2≤j≤A (C +B+ j −1)!
(B+C)!
= " A ∏ i=2 1 (B−i+A)!(C +i−1)!
" ∏ 1≤j<A j!
" ∏ 2≤j≤A (C +B+ j −1)!
(B+C)!
(B+A−1)!C!.
(2.4) Perhaps the easiest way to get MacMahon’s formula out of this is to induct on A. This is somewhat unsatisfactory as it requires knowledge of MacMahon’s formula a priori, though we remark that this approach is common.
First, consider the base case A = 1. Then MacMahon’s formula is B ∏ b=1 C ∏ c=1 b+c b+c−1 = B ∏ b=1 b+C b = B+C B .
3 Here is a simple way to prove the last determinant evaluation. The determinant is a polynomial in Xi of degree n(n−1)/2. It vanishes whenever Xi = Xj and hence it is divisible by each factor (Xi −Xj). We conclude that the determinant is C ·∏i 0 assigning to each edge ij ∈E of G with white i and black j some positive weight. Then we let the Kasteleyn matrix KR be an n×n matrix with entries KR(i, j) = ( w(ij) ij is an edge of G 0 otherwise.
The difference with the setting of the previous lecture is that w(·) was identical 1 there. When there is no risk of confusion we abbreviate KR to K. This matrix depends on the labeling of the vertices, but only up to permutation of the rows and columns, and hence we will not be concerned with this distinction.
By abuse of notation we will then denote to the weight of a tiling T by w(T) def = ∏ ℓ∈T w(ℓ).
An extension of Theorem 2.1 states that we can compute the weighted sum of tilings as the determinant of the Kasteleyn matrix.
Theorem 3.1. The weighted number of tilings of a simply-connected domain R is equal to ∑T w(T) = |detKR|.
33 34 Lecture 3: Extensions of the Kasteleyn theorem 1′ 2′ 3′ 2 1 3 w(2, 1) w(1, 2) w(3, 2) w(3, 1) w(2, 3) w(1, 3) Figure 3.1 A unit hexagon with weights of edges for the corresponding graph.
The proof of Theorem 3.1 is the same as the unweighted case of counting the number of tilings (obtained by setting w(ij) = 1 for each edge ij ∈E) in Theorem 2.1. Let us remind the reader that simply-connectedness of R is used in the proof in order to show that the signs of all the terms in the determinant expansion are the same.
Example 3.2 (Kasteleyn matrix of a hexagon). The unit hexagon with six tri-angles can be modelled by a cycle graph G on six vertices as shown in Figure 3.1 Then detKR = det 0 w(1,2) w(1,3) w(2,1) 0 w(2,3) w(3,1) w(3,2) 0 = w(1,2)w(2,3)w(3,1)+w(2,1)w(3,2)w(1,3) and as advertised the nonzero terms have the same sign and correspond to weights of the two obvious tilings. Note that detKR is only defined up to sign, since if we relabel the vertices in a different way (equivalently, change some rows and columns) then the determinant’s sign changes.
35 3.2 Tileable holes and correlation functions There are situations in which the result of Theorem 3.1 is still true, even if R is not simply connected.
Proposition 3.3. Suppose a region R is a difference of a simply connected do-main and a union of disjoint lozenges inside this domain. Let K be the Kaste-leyn matrix for R. Then the weighted sum of perfect matchings of R equals |detK|.
Proof Let e R denote the simply connected region obtained by adding in the removed lozenges. Then every nonzero term in detKR can be mapped into a corresponding nonzero term in detK e R by adding in the deleted lozenges.
We then quote the result that all terms in detK e R have the same sign, so the corresponding terms in detKR should all have the same sign, too.
There is an important probabilistic corollary of Proposition 3.3. Let R be as in the proposition, and fix a weight function w(•) > 0. We can then speak about random tilings by setting the probability of a tiling T to be P(tiling T) = 1 Z w(T) = 1 Z ∏ lozenge ℓ∈T w(ℓ).
The normalizing constant Z = ∑T w(T) is called the partition function.
Definition 3.4. Define the nth correlation function ρn as follows: given lozenges ℓ1, ..., ℓn we set ρn(ℓ1,...,ℓn) = P(ℓ1 ∈T, ℓ2 ∈T, ...,ℓn ∈T).
The following proposition gives a formula for ρn.
Theorem 3.5. Write each lozenge as ℓi = (wi,bi) for i = 1,...,n. Then ρn(ℓ1,...,ℓn) = n ∏ i=1 w(wi,bi) det i, j=1,...,n K−1(bi,wj) (3.1) Remark 3.6. The proposition in this form was stated in [Kenyon97]. How-ever, the importance of the inverse Kasteleyn matrix was known since 60s, cf.
[MontrollPottsWard63], [McCoyWu73].
36 Lecture 3: Extensions of the Kasteleyn theorem Proof of Theorem 3.5 Using Proposition 3.3, we have ρn (ℓ1,...,ℓn) = 1 Z ∑ tiling T T∋ℓ1,...,ℓn ∏ ℓ∈T w(ℓ) = 1 Z n ∏ i=1 w(wi,bi) det i′, j′∈R{ℓ1,...,ℓn} K(wi′,bj′) = n ∏ i=1 w(wi,bi) det i′, j′∈R{ℓ1,...,ℓn} K(wi′,bj′) det i′, j′∈R K(wi′,bj′) = n ∏ i=1 w(wi,bi) det i, j=1,...,n K−1(bi,w j) , where the last equality uses the generalizes Cramer’s rule (see e.g. [Prasolov94, Section 2.5.2]), which claims that a minor of a matrix is equal (up to sign) to the product of the complimentary–transpose minor of the inverse matrix and the determinant of the original matrix. In particular, the n = 1 case of this statement is the computation of the inverse matrix as the transpose cofactor matrix divided by the determinant: K−1(bi,wj) = (−1)i+j det i′, j′∈R{bi,w j}[K(wi′,bj′)] det i′, j′∈R[K(wi′,bj′)] .
Note that Proposition 3.3 involved the absolute value of the determinant. We leave it to the reader to check that the signs in the above computation match and (3.1) has no absolute value.
3.3 Tilings on a torus Our next stop is to count tilings on the torus. The main motivation comes from the fact that translation-invariance of the torus allows as to use the Fourier analysis to compute the determinants, which Kasteleyn theory outputs — this will be important for the subsequent asymptotic analysis. Yet, for non-planar domains (as torus), the Kasteleyn theorem needs a modification.
3.3.1 Setup We consider a hexagonal grid on a torus T = S1 × S1, again with the corre-sponding bipartite graph G = (W ⊔B,E). The torus has a fundamental domain 37 n1 = 3 n2 = 3 (0, 0) (0, 1) (0, 2) (1, 0) (2, 0) (2, 1) (2, 2) (1, 2) (1, 1) eβ eα 1 eβ(−1)a eβ(−1)a eβ(−1)a eα(−1)b eα(−1)b eα(−1)b Figure 3.2 An n1 × n2 = 3 × 3 torus and its coordinate system. Three types of edges have weights eα, eβ, and 1. When we loop around the torus, the weights get additional factors (−1)a or (−1)b.
F drawn as a rhombus with n1 and n2 side lengths, as shown in Figure 3.2.
We impose coordinates so that the black points are of the form (x,y) where 0 ≤x < n1 and 0 ≤y < n2; see Figure 3.2. The white points have similar co-ordinates, so that black and white points with the same coordinate are linked by a diagonal edge with white vertex being below. Our goal is to compute the weighted number of tilings, with a general weight function w(•).
As with the original Kasteleyn theorem, permK gives the number of perfect 38 Lecture 3: Extensions of the Kasteleyn theorem matchings, and our aim is to compute the signs of the terms when permanent is replaced by determinant.
To this end, we fix M0 the matching using only diagonal edges between vertices of the same coordinates — its edges are adjacent to coordinate labels ((0,0), (0,1), etc) in Figure 3.2, and choose the numbering of the white and black vertices so that the ith black vertex is directly above the ith white vertex.
We let M be a second arbitrary matching of the graph. We would like to under-stand the difference between signs of M and M0 in detK. As before, overlaying M and M0 gives a 2-regular graph, i.e. a collection of cycles.
3.3.2 Winding numbers The topological properties of the cycles (or loops) in M ∪M0 play a role in finding the signs of M and M0 in detM.
We recall that the fundamental group of the torus is Z × Z, which can be seen through the concept of the winding number of an oriented loop, which is a pair (u,v) ∈Z×Z associated to a loop and counting the number of times the loop traverses the torus in two coordinate directions. In order to speak about the winding numbers of the loops in M ∪M0, we need to orient them in some way. This orientation is not of particular importance, as eventually only the oddity of u and v will matter for our sign computations.
For the nontrivial loops, we need the following two topological facts: Proposition 3.7. [Facts from topology] On the torus, • a loop which is not self-intersecting and not nullhomotopic has winding number (u,v) ∈Z×Z with greatest common divisor gcd(u,v) = 1; • any two such loops which do not intersect have the same winding number.
Both facts can be proven by lifting the loops to R2 — the universal cover of the torus — and analyzing the resulting curves there, and we will not provide more details here.
Proposition 3.8. Take all loops from M ∪M0, which are not double edges.
Then each such loop intersects the vertical border of the fundamental domain (of Figure 3.2) u times and horizontal (diagonal) border of the fundamental domain v times for some (u,v) independent of the choice of the loop and with gcd(u,v) = 1. Further, the length of the loop, i.e. the number of black vertices on it, is n1u+n2v.
Proof Let γ denote one of the loops in question, and let us lift it to a path in 39 R2 via the universal cover R2 ↠S1 ×S1.
We thus get a path γ′ in R2 linking a point (x,y) to (x + n1u′,y + n2v′) for some integers u′ and v′. By Proposition 3.7, gcd(u′,v′) = 1 (unless u′ = v′ = 0) and u′, v′ are the same for all loops. Let us orient γ′ by requiring that all the edges coming from M0 are oriented from black to white vertex. Then, since every second edge of γ′ comes from M0, we conclude that the x–coordinate is increasing and the y–coordinate is decreasing along the path γ′, i.e. the steps of this path of one oddity are (x′,y′) →(x′,y′) and the steps of this path of another oddity are (x′,y′) →(x+1,y) or (x′,y′) →(x,y−1), (3.2) in the coordinate system of Figure 3.2. The monotonicity of path γ′ implies that u′ = v′ = 0 is impossible, i.e. non-trivial loop obtained in this way can not be null-homotopic. The same monotonicity implies u = |u′|, v = |v′| and that the length of the path is |n1u′|+|n2v′|. The claim follows.
3.3.3 Kasteleyn theorem on the torus In order to handle the wrap-around caused by the winding numbers u and v, we have to modify our Kasteleyn matrix slightly.
Fix the fundamental domain F as in Figure 3.2. We make the following definition.
Definition 3.9. Let (a,b) ∈{0,1}2. The Kasteleyn matrix Ka,b is the same as the original K, except that • For any edge crossing the vertical side of F with n1 vertices, we multiply its entry by (−1)a.
• For any edge crossing the horizontal/diagonal side of F with n2 vertices, we multiply its entry by (−1)b.
Theorem 3.10. For perfect matchings on the n1 ×n2 torus we have ∑ T w(T) = ε00 detK00 +ε01 detK01 +ε10 detK10 +ε11 detK11 2 (3.3) for some εab ∈{−1,1} which depend only on the parity of n1 and n2.
40 Lecture 3: Extensions of the Kasteleyn theorem Proof The correct choice of εab is given by the following table.
(n1 mod 2,n2 mod 2) (0,0) (0,1) (1,0) (1,1) ε00 −1 +1 +1 +1 ε01 +1 −1 +1 +1 ε10 +1 +1 −1 +1 ε11 +1 +1 +1 −1 (3.4) For the proof we note that each detKab is expanded as a signed sum of weights of tilings and we would like to control the signs in this sum. For the matching M0, the sign in all four Kab is +1. Since the sum over each column in (3.4) is 2, this implies that the weight of M0 enters into the right-hand side of (3.3) with the desired coefficient 1.
Any other matching M differs from M0 by rotations along k loops of class (u,v), as described in Proposition 3.8. Note that at least one of the numbers u, v, should be odd, since gcd(u,v) = 1. The loop has the length n1u + n2v and rotation along this loop contributes in the expansion of detKab the sign (−1)n1u+n2v+1+au+bv. Thus we want to show that the choice of εab obeys 1 ∑ a=0 1 ∑ b=0 εab ·(−1)k(n1u+n2v+au+bv+1) = 2.
If k is even, then all the signs are +1 and the check is the same as for M0. For the odd k case, we can assume without loss of generality that k = 1 and we need to show 1 ∑ a=0 1 ∑ b=0 εab ·(−1)n1u+n2v+au+bv = −2.
We use the fact that u,v are not both even to verify that the choice (3.4) works.
For example, if n1 and n2 are both even, then we get the sum −1+(−1)u +(−1)v +(−1)u+v.
Checking the choices (u,v) = (0,1),(1,0),(1,1) we see that the sum is −2 in all the cases. For other oddities of n1 and n2 the proof is the same.
Remark 3.11. For the perfect matchings on a genus g surface, one would need to consider a signed sum of 22g determinants for counting, as first noticed in [Kasteleyn67]. For the detailed information on the correct choice of signs, see [CimasoniReshetikhin06] and references therein.
Exercise 3.12. Find an analogue of Theorem 3.10 for perfect matchings on n1 ×n2 cylinder.
Lecture 4: Counting tilings on large torus This lecture is devoted to the asymptotic analysis of the result of Theorem 3.10.
Throughout this section we use the eα, eβ weight for tilings as in Figure 3.2 with a = b = 0.
4.1 Free energy From the previous class we know that on the torus with side lengths n1 and n2, the partition function is Z(n1,n2) = ∑ Tilings∏w(Lozenges) = ±detK0,0 ±detK1,0 ±detK0,1 ±detK1,1 2 where Ka,b are the appropriate Kasteleyn matrices. Note that we can view Ka,b as linear maps from CB to CW where B and W are the set of black and white vertices on the torus.
Our first task is to compute the determinants detKab. The answer is explicit in the case of a translation-invariant weight function w. Choose any real num-bers α and β, and write w(ij) ∈ n 1,eα,eβo according to the orientation of edge ij as in Figure 3.2. We evaluate detKab = ∏eigenvalues Kab.
We start with K00 and note that it commutes with shifts in both directions.
Because the eigenvectors of shifts are exponents, we conclude that so should 41 42 Lecture 4: Counting tilings on large torus be the eigenvectors of K00 (commuting family of operators can be diagonalized simultaneously). The eigenvalues are then computed explicitly, as summarized below.
Claim 4.1. The eigenvectors are given as follows: given input (x,y) ∈ {0,...,n1 −1}×{0,...,n2 −1} the eigenvector is (x,y) 7→exp i· 2πk n1 x+i· 2πℓ n2 y for each 0 ≤k < n1, 0 ≤ℓ< n2. The corresponding eigenvalue is 1+eβ exp −i2πk n1 +eα exp i2πℓ n2 .
Therefore the determinant of K0,0 is n1 ∏ k=1 n2 ∏ ℓ=1 1+eβ exp −i2πk n1 +eα exp i2πℓ n2 .
For Ka,b we can prove the analogous facts essentially by perturbing the eigenvalues we obtained in the K0,0 case. The proof is straightforward and we omit it.
Claim 4.2. Eigenvectors of Ka,b are (x,y) 7→exp i·π 2k +a n1 x+i·π 2ℓ+b n2 y and the determinant of Ka,b is n1 ∏ k=1 n2 ∏ ℓ=1 1+eβ exp −iπ 2k +a n1 +eα exp iπ 2ℓ+b n2 .
(4.1) Using this claim we find the asymptotic behavior of the (weighted) number of possible matchings in the large scale limit.
Theorem 4.3. For the weights of edges 1,eα,eβ , as in Figure 3.2, we have lim n1,n2→∞ ln(Z(n1,n2)) n1n2 = " |z|=|w|=1 ln(|1+eαz+eβw|) dw 2πiw dz 2πiz.
(4.2) Remark 4.4. The quantity ln(Z(n1,n2)) n1n2 is known as the free energy per site.
Exercise 4.5. Show that the value of the double integral remains unchanged if we remove the |·| in (4.2).
43 Proof of Theorem 4.3 We first consider the case when eα + eβ < 1. In this case the integral has no singularities and we conclude that lim n1,n2→∞ ln(|detKa,b|) n1n2 = " |z|=|w|=1 ln(|1+eαz+eβw|) dw 2πiw dz 2πiz, since the left-hand side is essentially a Riemann sum for the quantity on the right. When using this limit for the asymptotics of Z(n1,n2), the only possible issue is that upon taking the ± for each of a,b there may be a magic cancella-tion in Z(n1,n2) resulting in it being lower order. To see no such cancelation occurs note that max a,b |detKa,b| ≤Z(n1,n2) ≤2max a,b |detKa,b|.
The second inequality is immediate and follows from the formula for Z(n1,n2).
The first one follows from noting that |detKa,b| counts the sets of weighted matchings with ± values while Z(n1,n2) counts the weighted matchings un-signed. The result then follows in this case.
In the case when eα +eβ ≥1 we may have singularities; we argue that these singularities do not contribute enough to the integral to matter. The easiest way to see this is to consider a triangle with side lengths 1,eα,eβ. This gives two angles θ,φ which are the critical arguments (angles) of z and w for singularity to occurs. However note that πi( 2k+a n1 ) cannot simultaneously be close for some k to the critical angle φ for both a = 0 and a = 1. Doing similarly for θ it follows that there exists a choice of a,b for which one can bound the impact of the singularity on the convergence of the Riemann sum to the corresponding integral and the result follows. (In particular one notices that roughly a constant number of partitions in the Riemann sum have |1 + eαz + eβw| close to zero and the above guarantees that you are at least ≥ 1 max(n1,n2) away from zero so the convergence still holds.) 4.2 Densities of three types of lozenges We can use the free energy computation of the previous section to derive prob-abilistic information on the properties of the directions of the tiles in a random tiling (or perfect matching). Recall that the probability distribution in question assigns to each tiling (or perfect matching) T the probability P(T) = 1 Z(n1,n2) ∏ Lozenges in T w(lozenge).
44 Lecture 4: Counting tilings on large torus Theorem 4.6. For lozenges of weight eα, we have lim n1,n2→∞P(given lozenge is in tiling) = " |z|=|w|=1 eαz 1+eαz+eβw dw 2πiw dz 2πiz.
(4.3) Proof Using the shorthand # to denote the number of objects of a particular type in a tiling, we have Z(n1,n2) = ∑ Tilings exp(α ·# +β ·# ) and therefore ∂ ∂α ln(Z(n1,n2)) = ∑Tilings # ·exp(α ·# +β ·# ) ∑Tilings exp(α ·# +β ·# ) = E[# ] = n1n2 P(given lozenge is in tiling).
Therefore, using Theorem 3.10, we have P(given lozenge is in tiling) = lim n1,n2→∞ 1 n1n2 ∂ ∂α ln(Z(n1,n2)) = lim n1,n2→∞ 1 n1n2 ∂ ∂α ∑a,b=0,1 ±detKa,b ∑a,b=0,1 ±detKa,b = lim n1,n2→∞∑ a,b=0,1 ∂ ∂α detKa,b n1n2 detKa,b · ±detKa,b ∑a,b=0,1 ±detKa,b (4.4) Ignoring the possible issues arising from the singularities of the integral, we can repeat the argument of Theorem 4.3, differentiating with respect to α at each step, to conclude that lim n1,n2→∞ ∂ ∂α detKa,b n1n2 detKa,b = " |z|=|w|=1 eαz 1+eαz+eβw dw 2πiw dz 2πiz, (4.5) where the last expression can be obtained by differentiating the right-hand side of (4.2). On the other hand, the four numbers ±detKa,b ∑a,b=0,1 ±detKa,b have absolute values bounded by 1 and these numbers sum up to 1. Hence, (4.4) implies the validity of (4.3).
45 As in previous theorem, we have to address cases of singularities in the in-tegral in order to complete the proof of the statement. To simplify our analysis we only deal with a sub-sequence of n1,n2 tending to infinity, along which (4.5) is easy to prove. The ultimate statement certainly holds for the full limit as well.
We use the angles θ and φ from the proof of Theorem 4.3. Define Wθ ⊆N as Wθ = n : min 1≤j≤n θ −π j n > 1 n 3 2 .
Define Wφ similiarly Note here that θ and φ are the critical angles coming from triangle formed by 1,eα,eβ.
Lemma 4.7. For large enough n, either n or n+2 is in Wθ.
Proof Suppose that |θ −π j n | ≤ 1 n 3 2 and |θ −π j′ n+2| ≤ 1 (n+2) 3 2 . Note that for n sufficiently large this implies that j′ ∈{j, j +1, j +2}. However note that for such j and j′ it follows that π j n −π j′ n+2 = O 1 n and thus for n sufficiently large we have a contradiction.
Now take n1 ∈Wφ and n2 ∈Wθ. Then note that the denominator in (4.5) (which we get when differentiating in α the logarithm of (4.1)) satisfies 1+eβ exp −i·π 2k +a n1 +eα exp i·π 2ℓ+b n2 ≥C · 1 min(n1,n2) for all but a set of at most 4 critical (k,ℓ) pairs. For these four critical pairs we have that 1+eβ exp −i·π 2k +a n1 +eα exp i·π 2ℓ+b n2 ≥C · 1 min(n1,n2) 3 2 !
as we have taken n1 ∈Wφ and n2 ∈Wθ. This bound guarantees that the appro-priate Riemann sum converges to the integral expression shown in (4.5).
Remark 4.8. Using the same proof as for formula (4.3), we can show that lim n1,n2→∞P(given lozenge is in tiling) = " |z|=|w|=1 1 1+eαz+eβw dw 2πiw dz 2πiz and lim n1,n2→∞P(given lozenge is in tiling) = " |z|=|w|=1 eβw 1+eαz+eβw dw 2πiw dz 2πiz.
46 Lecture 4: Counting tilings on large torus We introduce the notation for these probabilities as p , p , and the one in the theorem statement as p .
Definition 4.9. The (asymptotic) slope of tilings on the torus is (p , p , p ).
4.3 Asymptotics of correlation functions The computation of Theorem 4.6 can be extended to arbitrary correlation func-tions. The extension relies on the following version of Theorem 3.5.
Theorem 4.10. Take a random lozenge tiling on the torus with arbitrary weights. Write n lozenges as ℓi = (wi,bi) for i = 1,...,n. Then P(ℓ1,...,ℓN ∈tiling) = ∑a,b=0,1 ±detKab ∏n i=1 Kab(wi,bi)det1≤i,j≤n K−1 ab (bi,w j) ∑a,b=0,1 ±detKab .
The proof here is nearly identical to the simply connected case, and we omit it. Then, using Theorem 4.10 as an input we can calculate the asymptotic nth correlation function.
Theorem 4.11. The nth correlation function in the n1,n2 →∞limit is lim n1,n2→∞P((x1,y1, ˜ x1, ˜ y1),...,(xn,yn, ˜ xn, ˜ yn) ∈tiling) = n ∏ i=1 K00(xi,yi, ˜ xi, ˜ yi) det 1≤i, j,≤n( ˜ Kα,β[ ˜ xi −xj, ˜ yi −y j]) (4.6) where ˜ Kα,β(x,y) = " |z|=|w|=1 wxz−y 1+eαz+eβw dw 2πiw dz 2πiz.
(4.7) Note that in the above theorem (xi,yi) correspond to the coordinates of the white vertices while ( ˜ xi, ˜ yi) correspond to the coordinates of the black vertices.
The coordinate system here is as in Figure 3.2. Therefore (xi,yi, ˜ xi, ˜ yi) is simply referring to a particular lozenge.
Sketch of the proof of Theorem 4.11 Here is the plan of the proof: • We know both the eigenvalues and eigenvectors for the matrix K and this immediately gives the same for the inverse matrix K−1.
• The key step in analyzing the asymptotic of the expression in Theorem 4.10 is to write down the elements of matrix inverse as a sum of the coordinates of eigenvectors multiplied by the inverses of the eigenvalues of K. As before this gives a Riemann sum approximation to the asymptotic formula shown 47 in (4.7). Note that one can handle singularities in a manner similar to that of Theorem 4.6 in order to get convergence along a sub-sequence.
• In more details, the normalized eigenvectors of Kab are (x,y) 7→ 1 √n1n2 exp i·π 2k+a n1 x+i·π 2ℓ+b n2 y and the corresponding eigenvalues are 1+eβ exp −i 2πk+a n1 +eα exp i 2πℓ+b n2 Now note that K−1 ab can be written as Q−1Λ−1Q where the columns of Q are simply the eigenvectors of Kab and Λ the corresponding diagonal matrix of eigenvalues. Note here Q−1 = Q∗as Q corresponds to a discrete 2-D Fourier transform. Substituting, it follows that K−1 ab (x,y;0,0) = 1 n1n2 n1 ∑ k=1 n2 ∑ ℓ=1 exp i·π 2k+a n1 x+i·π 2ℓ+b n2 y 1+eβ exp −i 2πk+a n1 +eα exp i 2πℓ+b n2 (4.8) and the expressions in (4.7) appears as a limit of the Riemann sum of the appropriate integral. (Note K−1 ab (x,y;0,0) = K−1 ab ((x + ˆ x,y + ˆ y; ˆ x, ˆ y)) as the torus is shift invariant.) Remark 4.12. The asymptotic of the free energy per site on n1 × n2 torus was investigated for numerous models of statistical mechanics going well be-yond weighted tilings (which were investigated in the largest generality in [KenyonOkounkovSheffield03]). A widely used approach is to treat the torus as a sequence of n2 strips of size n1 ×1. One then introduces the transfer ma-trix T encoding the transition from a strip to the adjacent one and computes the free energy as Trace(T n2). Hence, the computation reduces to the study of eigenvalues of T and systematic ways for evaluations of these eigenvalues are discussed in [Baxter82]. The answer is typically presented in terms of so-lutions to a system of algebraic equations known as Bethe equations. Finding analytic expressions for the solutions of these equations (in the limit of large torus) is a hard task, which is still mathematically unresolved in many situa-tions. From this perspective, tilings represent a specific “nice” case, in which relatively simple formulas of Theorems 4.3 and 4.11 are available.
Lecture 5: Monotonicity and concentra-tion for tilings In the last lecture we studied tilings on the torus, and now we go back to pla-nar domains. The setup for this lecture is slightly different from before. We fix a boundary of our domain and values of a height function on the boundary, and consider all height functions extending the given one. This is equivalent to studying random tilings because we have a correspondence between height functions (up to shifting by a constant) and tilings. The advantage of this setup is that we can compare height functions much easier than comparing tilings directly. The material of this lecture is based on [CohnElkiesPropp96] and [CohnKenyonPropp00]. Our aim here is to establish monotonicity and con-centration properties for random tilings, which will eventually lead to the Law of Large Numbers for tilings of large domains.
5.1 Monotonicity We start with a domain R with boundary ∂R on the triangular grid. Recall the three positive directions we defined in Lecture 1: (0,1), (− √ 3 2 ,−1 2), and ( √ 3 2 ,−1 2). Also recall that along an edge in the positive direction, the height function changes by +1 if the edge is an edge of a lozenge; and the height function changes by −2 if the edge crosses a lozenge. Fix a height function h on ∂R. Note that if h changes by −2 on the boundary, then a lozenge is half outside R. Let us emphasize that this is a slightly different situation com-pared to the previous lectures, as the lozenges are allowed to stick out of R; in other words, we are now considering tilings not of R, but of another domain obtained by adding to R some triangles along the boundary ∂R.
Let us consider a height function H on R chosen uniformly randomly from 48 49 all height functions extending h. In the following, ∂R ⊆R can be any set of vertices such that every vertex in R\∂R is adjacent only to vertices of R.
Proposition 5.1. Let R and ∂R be as above. Let h and g be two height func-tions on ∂R. Assume that h ≡g (mod 3), h ≥g and that h and g can be extended to height functions on R. Let H (resp. G) be a height function on R chosen uniformly randomly from all height functions extending h (resp. g).
Then we can couple (that is, define on the same probability space) H and G in such a way that H ≥G almost surely.
Remark 5.2. We require that h ≡g (mod 3) because all height functions are the same modulo 3 up to shifting by a global constant.
Proof of Proposition 5.1 We fix R and perform induction on |R\∂R|. Base case is R = ∂R, where there is nothing to prove.
Suppose ∂R ⊊R. Choose a vertex v ∈∂R adjacent to w ∈R\∂R, such that h(v) > g(v). (If no such v can be chosen, then we can couple H and G in such a way that H|R\∂R = G|R\∂R, and there is nothing to prove.) Let us sample H and G on R\∂R in two steps: First we sample H(w) and G(w), and then we sample the rest.
Because h(v) > g(v) and h(v) ≡g(v) (mod 3), we have h(v) ≥g(v)+3. Set x = h(v)−2 and y = g(v)−2. The definition of the height function then implies H(w) ∈{x,x+3} and G(w) ∈{y,y+3}. Since x ≥y+3, we can couple H(w) and G(w) so that H(w) ≥G(w).
It remains to couple the values of H and G on R(∂R ∪{w}) by induction hypothesis.
Corollary 5.3. In the setting of Proposition 5.1, for any w ∈R, we have EH(w) ≥EG(w).
Remark 5.4. The proof of Proposition 5.1 extends from the uniform distribu-tion on height functions to more general ones. The only feature of the distri-bution that we used is the possibility of sequential (Markovian) sampling of the values of the height function, and the fact that the set of admissible values for the height function at a point can be reconstructed by the value at a single neighbor. Here is one possible generalization1.
Exercise 5.5. Prove an analogue of Proposition 5.1 for the measure on lozenge 1 Another generalization of tilings is the six-vertex model. The monotonicity property extends only to some particular cases of the latter, but not to the model with arbitrary parameters. On the level of the proofs, the difficulty arises because the weights of the six-vertex model are determined by the local configurations in squares (x,y), (x,y+1), (x+1,y), (x+1,y+1) and hence it is not enough to know h(x,y+1) and h(x+1,y) in order to sample h(x+1,y+1) — we also need h(x,y).
50 Lecture 5: Monotonicity and concentration for tilings tilings with weights 1 Z ∏w(lozenges), where w(·) is a positive function of type of lozenge and its position; the product goes over all lozenges in a tilings and Z is a normalizing constant making the total mass of measure equal to 1.
Similar generalizations to non-uniform distributions are possible for the fur-ther results in this lecture.
Proposition 5.6. Let R1 and R2 be two domains with boundaries ∂R1 and ∂R2 resp. Let h be a height function on ∂R1 and let g be a height function on ∂R2 such that h|∂R1∩∂R2 ≡g|∂R1∩∂R2 (mod 3). Suppose that each vertex v of ∂R1 is within distance ∆1 from a vertex w in ∂R2, and vice versa, and for each such pair of vertices, we have |h(v) −g(w)| ≤∆2. Then there exists an absolute constant C > 0 such that |EH(x)−EG(x)| ≤C(∆1 +∆2) for all x ∈R1 ∩R2, where H (resp. G) is chosen uniformly randomly from height functions on R1 (resp. R2) extending h (resp. g).
Proof Let ˜ h be a minimal extension of h to R1, and ˜ g be a maximal ex-tension of g to R2. (They exist and are unique because pointwise maxi-mum or minimum of two height functions is again a height function.) For any vertex v ∈∂(R1 ∩R2) ⊆∂R1 ∪∂R2, the conditions imply that ˜ h(v) ≥ ˜ g(v)−C(∆1 +∆2) for some absolute constant C > 0. Hence the same inequal-ity is true for any extension of h to R1 and g to R2. By Corollary 5.3, we get EH(x) ≥EG(x)−C(∆1 +∆2) for all x ∈R1 ∩R2. By swapping R1 and R2, we get EG(x) ≥EH(x)−C(∆1 +∆2).
Remark 5.7. In the proofs of this section we silently assumed that the exten-sions of the height functions from the boundaries exist. As long as this is true, we do not need to assume that the domains are simply connected.
5.2 Concentration We proceed to the next question: how close is a random height function to its expectation?
Theorem 5.8. Let R be a connected domain with boundary ∂R. Let h be a height function on ∂R. Let H be a uniformly random extension of h to R.
Suppose that w,v ∈R are linked by a path of length m. Then E(H(w)−E[H(w)|H(v)])2 ≤9m 51 and for each c > 0, P |H(w)−E[H(w)|H(v)]| > c√m ≤2exp −c2 18 .
(5.1) Remark 5.9. If we choose as v a point for which the value of H(v) is determin-istically known (e.g., it can be a point on the boundary ∂R, then expectations become unconditional and we get: E(H(w)−E[H(w)])2 ≤9m and for each c > 0, P |H(w)−E[H(w)]| > c√m ≤2exp −c2 18 .
(5.2) Remark 5.10. If the linear size of the domain is proportional to L, then so is the maximum (over w) possible value of m. Hence, (5.2) implies that as L →∞ the normalized height function H(Lx,Ly) L concentrates around its expectation.
We will soon prove that it converges to a limit, thus showing the existence of a limit shape for tilings.
Remark 5.11. It is conjectured that the variance should be of order O(lnm), rather than O(m) which we prove. As of 2021, the tight bound was proved only for some specific classes of domains.
Remark 5.12. Similarly to Exercise 5.5, the result extends to the measures on tilings with weight 1 Z ∏w(lozenges). The proof remains the same.
Proof of Theorem 5.8 Let x0 = v,x1,...,xm = w be a path connecting v and w. Let Mk be the conditional expectation of H(w) given the values of H(x0), ..., H(xk): Mk = E H(w) | H(x0),H(x1),...,H(xk) In particular, M0 = E[H(w)|H(v)] and Mm = H(w). The tower property of con-ditional expectations implies that M0,...,Mm is a martingale.
For fixed values of H(x0), ..., H(xk), the random variable H(xk+1) takes at most two distinct values. Also, if H(xk+1) takes two values a < b, then we must have b−a = 3. By Proposition 5.1, we have E H(w) | H(x0),H(x1),...,H(xk); H(xk+1) = a ≤E H(w) | H(x0),H(x1),...,H(xk); H(xk+1) = b ≤E H(w)+3 | H(x0),H(x1),...,H(xk); H(xk+1) = a .
(We get the second inequality by looking at the height function shifted 52 Lecture 5: Monotonicity and concentration for tilings up by 3 everywhere.) Therefore E[H(w)|H(x0),H(x1),...,H(xk)] and E[H(w)|H(x0),H(x1),...,H(xk+1)] differ by at most 3. Hence E[(Mm −M0)2] = E ∑ 1≤k≤m (Mk −Mk−1) !2 = ∑ 1≤k,l≤m E[(Mk −Mk−1)(Ml −Ml−1)] = ∑ 1≤k≤m E[(Mk −Mk−1)2] ≤9m.
The third step is because E[(Mk −Mk−1)(Ml −Ml−1)] = 0 for k ̸= l, by martingale property (taking first the conditional expectation with respect to H(x0),H(x1),...,H(xmin(k,l))).
The concentration inequality (5.1) follows from Azuma’s inequality applied to the martingale Mk, see Lemma 5.13 below.
Lemma 5.13 (Azuma’s inequality [Azuma67]). Let X0,...,XN be a martin-gale with |Xk −Xk−1| ≤bk for all 1 ≤k ≤N. Then for each t > 0, P(|XN −X0| ≥t) ≤2exp − t2 2∑N k=1 b2 k !
.
5.3 Limit shape The results of the previous two sections say that random height functions con-centrates around their expectations, and the expectations continuously depends on the boundary conditions. The limiting profile, which we thus observe as the size of the domain L tends to infinity, is called the limit shape. So far, we do not have any description for it, and our next aim is to develop such a description.
Let us give a preview of the statement that we will be proving. The details are presented in the several following lectures.
We can encode the gradient to (asymptotic) height function via proportions of three types of lozenges , , . We then define (minus) surface tension to be S(∇h) = S(p , p , p ) = 1 π (L(π p )+L(π p )+L(π p )) where L(θ) = − ˆ θ 0 ln(2sint)dt is the Lobachevsky function.
53 Exercise 5.14. Assume that p , p , p are non-negative and satisfy p + p + p = 1. Show that S(p , p , p ) vanishes whenever one of the arguments vanishes and S(p , p , p ) > 0 otherwise.
Theorem 5.15 (Variational principle, [CohnKenyonPropp00]). Consider a do-main R∗with piecewise-smooth boundary ∂R∗. Let hb be a real function on ∂R∗. Take a sequence of tileable, lattice domains RL with boundaries ∂RL with scale depending linearly on L →∞. Fix a height function hL on ∂RL.
Suppose that ∂RL L →∂R∗and hL L converges to hb in sup norm.2 Then the height function HL of a uniformly random tiling of RL converges in probabil-ity, in sup norm. That is, H(Lx,Ly) L →hmax(x,y). Furthermore, hmax is the unique maximizer of the integral functional ¨ R∗S(∇h)dxdy, h|∂R∗= hb.
Simultaneously, we have 1 L2 ln(number of tilings of RL) → ¨ R∗S(∇hmax)dxdy.
(5.3) The proof of Theorem 5.15 is given at the end of Lecture 8 after we make some preparatory work in the next two lectures.
Historically, Theorem 5.15 first appeared with a full proof in mathemat-ical literature in [CohnKenyonPropp00]. Yet, some of its ingredients were implicitly known in theoretical physics earlier, see [NienhuisHilhorstBlote84, DestainvilleMosseruBailly97, Hoffe97].
2 Since the domain of definition of the function ∂hL L depends on L, one should be careful in formalizing the sup-norm convergence. One way is to demand that for each ε > 0 there exists δ and L0, such that for all L > L0 and all pairs (x,y) ∈∂RL L , (x∗,y∗) ∈∂R∗satisfying |x−x∗|+|y−y∗| < δ, we have 1 L hL(Lx,Ly)−hb(x∗,y∗) < ε.
Lecture 6: Slope and free energy In order to prove Theorem 5.15, we would like to asymptotically count the number of tilings for generic domains. The answer depends on the slope of the limit shape for random tilings, so we start with the fixed slope situation, where our computations for torus are helpful.
6.1 Slope in a random weighted tiling We work in the (α,β)-weighted setting, as in Lecture 4. Lozenges of type has weight eβ, type has weight eα, and type has weight 1, where α,β ∈R.
In Theorem 4.6 and Remark 4.8 we computed the asymptotic slope of random tilings on the torus: p = ‹ |z|=|w|=1 eβw 1+eαz+eβw dz 2πiz dw 2πiw, p = ‹ |z|=|w|=1 eαz 1+eαz+eβw dz 2πiz dw 2πiw, p = ‹ |z|=|w|=1 1 1+eαz+eβw dz 2πiz dw 2πiw.
Let us evaluate one of these integrals, say, the second one, p . Fix z and com-pute the w-integral ˛ |w|=1 eαz 1+eαz+eβw dw 2πiw.
The integrand has two poles, one at w = 0 and one at w = −1+eαz eβ . If both poles are in the contour, then the w-integral has value 0 because we can let the contour go to ∞, and the residue at ∞is 0 because of O( 1 w2 ) decay. If w = −1+eαz eβ is outside the contour, then the w-integral evaluates to eαz 1+eαz by 54 55 1 eα eβ θα θβ θ1 Figure 6.1 Triangle with sides eα, eβ, 1.
Cauchy’s integral formula. So we get p = ˛ |1+eαz|>eβ ,|z|=1 eαz 1+eαz dz 2πiz.
(6.1) Suppose there exists a triangle with side lengths eα, eβ, 1, as in Figure 6.1. Let θα and θβ be the angles opposite to the eα and eβ edges, respectively. Take θ = π −θβ, so that |1+eα exp(iθ)| = eβ. The integral (6.1) evaluates to p = 1 2πi ln(1+eαz) z=exp(iθ) z=exp(−iθ) = θα π .
We can compute p and p similarly, and get the following theorem.
Theorem 6.1. Suppose there exists a triangle with side lengths eα, eβ, 1 and let its angles be θα, θβ, θ1, as in Figure 6.1. Then p = θβ π , p = θα π , p = θ1 π .
In words, the density of a lozenge is encoded by the angle opposite to the side of its weight.
The case when a triangle does not exist is left as exercise.
Exercise 6.2. 1 If eα ≥eβ +1, then p = 1.
2 If eβ ≥eα +1, then p = 1.
3 If 1 ≥eα +eβ, then p = 1.
Corollary 6.3. Any slope (p , p , p ) with p + p + p = 1, p , p , p > 0 can be achieved from (α,β)-weighted tilings by choosing α and β appropri-ately.
The observation of Corollary 6.3 is the key idea for the Cohn-Kenyon-Propp variational principle.
56 Lecture 6: Slope and free energy 6.2 Number of tilings of a fixed slope Now we compute the number of tilings of a torus of a fixed slope. Theorem 4.3 computes the asymptotic of the number of tilings as a function of α and β, while Theorem 6.1 relates α and β to the slope. So essentially all we need to do is to combine these two theorems together. We present a way to do it through the Legendre duality. We start from the setting of the finite torus in order to see how this duality arises. (In principle, this is not strictly necessary, as one can analyze the formulas of Theorems 4.3, 6.1 directly.) Say the torus has dimensions n1 × n2. Somewhat abusing the notations, we define the slope of a fixed tiling as slope = 1 n1n2 (#lozenges of types( , , )).
(Previously, we thought of the slope only as a limit of this quantity as n1 and n2 tend to infinity.) We have already computed the asymptotics of the partition function Z as a function of α and β in Theorem 4.3. Z is related to the slope by 1 n1n2 lnZ = 1 n1n2 ln ∑ slope exp(α# +β# )·#tilings of this slope.
(6.2) Claim 6.4. 1 Under the (α,β)-measure, the distribution of slope as a random variable is concentrated as n1,n2 →∞around (p , p , p ), which we have computed. This is proved later in this lecture.
2 Asymptotics of the number of tilings depends continuously on the slope. This is proved in Lecture 8.
Using the claim, we have 1 n1n2 lnZ ≈α p +β p + 1 n1n2 ln#tilings of slope(p , p , p ).
(6.3) We define the surface tension as σ(slope) = lim n1,n2→∞−1 n1n2 ln#tilings of such slope.
(This will soon be shown to be the same as (−1)·S, with S defined at the end of the last lecture.) Let R(α,β) be the limit of 1 n1n2 lnZ as n1,n2 →∞. Then (6.3) gives R(α,β) = αs+βt −σ(s,t), (6.4) where (s,t) are asymptotic values for the first two coordinates of the slope: 57 Figure 6.2 The concave function S(s,t) = −σ(s,t) appearing in Theorems 5.15 and 6.5.
s = p , t = p . Comparing Theorem 4.3 and 4.6, we also have s = p = ∂R ∂α , t = p = ∂R ∂β .
(6.5) Here is another point of view on (6.4), (6.5): R(α,β) = max s,t (αs+βt −σ(s,t)).
(6.6) The last formula can be also obtained directly: indeed (s,t) = (p , p ) is (ap-proximately) the slope with the largest probability under the (α,β)-measure and R(α,β) is the asymptotic of the logarithm of this probability.
The formulas (6.4), (6.5), (6.6) mean that R(α,β) and σ(s,t) are Legendre duals of each other. Legendre duality is best defined for the convex functions, and in our situation R(α,β) is indeed convex in α and β. (This can be checked by directly by computing the second derivatives in the formula for R(α,β) of Theorem 4.3 and using the Cauchy-Schwarz inequality.) Since Legendre duality of R and σ is a symmetric relation, we also have ∂σ ∂s = α, ∂σ ∂t = β.
This gives a formula for σ(s,t).
58 Lecture 6: Slope and free energy Theorem 6.5. Under the identification s = p and t = p , we have σ(s,t) = −1 π (L(πs)+L(πt)+L(π(1−s−t))) (6.7) where L(θ) = − ´ θ 0 ln(2sint)dt is the Lobachevsky function.
Proof When s = t = 0, both sides are equal to 0. σ(0,0) = 0 because there is exactly one tiling with slope (0,0,1). right-hand side is 0 is because L(0) = L(π) = 0.
Exercise 6.6. Prove that L(0) = L(π) = 0.
Let us compare the derivatives.
∂RHS ∂s = ln 2sin(πs) −ln 2sin(π(1−t −s)) = α = ∂σ ∂s .
The second step is by the law of sines applied to the triangle with side lengths eα, eβ, 1. Similarly, we can show that ∂RHS ∂t = β = ∂σ ∂t .
For the graph of the σ(s,t) function, see Figure 6.2.
Proposition 6.7. σ(s,t) is a convex function of (s,t) with s ≥0, t ≥0, s+t ≤1.
Proof We can check this directly from formula 6.7. This also follows from the fact that σ(s,t) is the Legendre dual of R(α,β).
6.3 Concentration of the slope Now let us prove that the slope concentrates as n1,n2 →∞.
Theorem 6.8. For the (α,β)–weighted random tilings of n1 × n2 torus, lim n1,n2→∞ |# −E# | n1n2 = 0 in probability.
Proof Let us compute the variance.
Var(# ) = Var ∑ (x,y)∈torus I( at (x,y)) !
= ∑ (x,y)∈torus ∑ (x′,y′)∈torus Cov(I( at (x,y)),I( at (x′,y′))).
(I(·) denotes the indicator function of an event.) If we show that Cov →0 as 59 |x −x′|,|y −y′| →∞, then Var = o(n2 1n2 2) and concentration holds by Cheby-shev’s inequality. We have Cov = P( at (x,y) and at (x′,y′))−P( at (x,y))P( at (x′,y′)).
By Theorem 4.11 of Lecture 4, as n1,n2 →∞, the covariance goes to e2β det ˜ Kα,β(1,0) ˜ Kα,β(x−x′ +1,y−y′) ˜ Kα,β(x′ −x+1,y′ −y) ˜ Kα,β(1,0) −e2β ˜ Kα,β(1,0)2 (6.8) where ˜ Kα,β(x,y) = ‹ |z|=|w|=1 wxz−y 1+eαz+eβw dz 2πiz dw 2πiw.
The double contour integral is a Fourier coefficient of a reasonably nice func-tion, so it goes to 0 as |x|,|y| →∞. Therefore, (6.8) also goes to zero, as desired.
(Note that we use here slightly more than what is proved in Lecture 4. In Lec-ture 4, x−y′ and y−y′ are kept constant as n1,n2 →∞. Here they should grow with n1,n2, so formally we have to deal directly with discrete Fourier coeffi-cients of (4.8) — but those still go to 0.) 6.4 Limit shape of a torus We end this lecture by discussing how a random tiling of the torus looks like on the macroscopic scale.
Take a fundamental domain of the torus. Let H = 0 at the bottom-left corner, and use paths only inside the fundamental domain to define height function H(x,y) for the torus. By translation invariance, we have EH(x,y) −EH(x − 1,y) = a and EH(x,y)−EH(x,y−1) = b, for all (x,y). Here (a,b) is a vector which can be treated as a version of the slope. We conclude that EH(x,y) = ax+by.
Hence, using concentration inequality (5.2) from Lecture 5 (see Remark 5.12), as L →∞we have sup x,y H(Lx,Ly) L −(ax+by) →0 in probability. In other words, the limit shape for tilings is a plane.
Lecture 7: Maximizers in the variational principle 7.1 Review Recall the definitions of the Lobachevsky function, given by: L(θ) = − ˆ θ 0 ln(2sint) dt.
and of the (minus) surface tension associated to a triple (p , p , p ) of densi-ties given by S(p , p , p ) = 1 π L(π p )+L(π p )+L(π p ) .
The setup for the identification of the limit shape in the variational principle of Theorem 5.15 is as follows: • For L →∞we consider a sequence of finite domains RL, whose scale de-pends linearly in L, and fix a height function hL on each boundary ∂RL. We silently assume that hL is such that it has extensions to height functions on the whole RL.
• As our target, we take a continuous region R∗⊆R2 with piecewise smooth boundary ∂R∗, and let hb be a real function on this boundary.
We require the boundaries of RL to converge to those of the limiting region R∗ 1 L∂RL →∂R∗.
We do not need to specify the precise notion of convergence here, since the limit shapes do not depend on the small perturbations of the boundaries (see 60 61 Lecture 5 and the following Lecture 8). Moreover, we require that hL(Lx,Ly) L →hb(x,y) on ∂R∗, which means that for each ε > 0 there exists δ > 0 and L0, such that for all L > L0 and all pairs (x,y) ∈∂RL L , (x∗,y∗) ∈∂R∗satisfying |x−x∗|+|y−y∗| < δ, we have 1 LhL(Lx,Ly)−hb(x∗,y∗) < ε.
The variational principle asserts (among other things) that a uniformly ran-dom height function HL of RL extending hL on ∂RL (=height function of a uniformly random tiling) converges in probability to a function h∗given by h∗= argmaxh∈H ¨ R∗S(∇h) dx dy (7.1) In other words, the limit shape should be one that maximizes a certain surface tension integral.
The goal of this lecture is merely to make (7.1) precise (and subsequent lectures will actually prove the theorem). This means we need to make the following steps: • We define and motivate the class of functions H mentioned in maximiza-tion problem (7.1).
• We define S(∇h) for h ∈H by associating a triple (p , p , p ) of local densities to it.
• We prove that there exists a maximizing function h∗across the choice of h ∈H .
• We prove that the maximizing function is unique.
Throughout this lecture we retain R∗as our target region with piecewise smooth boundary ∂R∗. We fix the boundary function hb on ∂R∗as well.
7.2 The definition of surface tension and class of functions 7.2.1 Identifying the gradient with the probability values Let’s consider once again an h on a finite region. We can imagine walking in the three directions indicated in Figure 7.1. Recall that the way our height function h is defined, as we move in the +y direction we either get an increment of +1 or −2 according to whether we cross a lozenge or not. Accordingly, in the standard coordinate system with x-axis pointing to the right and y–axis 62 Lecture 7: Maximizers in the variational principle ∂h ∂y √ 3 2 ∂h ∂x −1 2 ∂h ∂y − √ 3 2 ∂h ∂x −1 2 ∂h ∂y Figure 7.1 Three coordinate directions correspond to three linear combinations of partial derivatives.
pointing up, we might colloquially write ∂h ∂y = −2p + p + p = 1−3p .
Similarly, moving in the other directions gives: + √ 3 2 ∂h ∂x −1 2 ∂h ∂y = p + p −2p = 1−3p − √ 3 2 ∂h ∂x −1 2 ∂h ∂y = p −2p + p = 1−3p .
In this way, we can now give a definition which gives a local density triple (p , p , p ) at a point for a differentiable function h: R∗→R.
Definition 7.1. Let h: R∗→R be differentiable. For a point w in the interior of R∗, we identify the gradient (∇h)w with the three local densities p = 1−∂h ∂y 3 p = 1− − √ 3 2 ∂h ∂x −1 2 ∂h ∂y 3 p = 1− √ 3 2 ∂h ∂x −1 2 ∂h ∂y 3 .
63 Naturally, we have p + p + p = 1. Thus, moving forward, we will just write S(∇h) for height functions h, with the above convention.
7.2.2 The desired set of functions Next, we define the set of functions H we wish to consider: they are the func-tions h: R∗→R which satisfy a Lipschitz condition so that p , p , p are in the interval [0,1].
Definition 7.2. Let δ1,δ2,δ3 be unit vectors in the coordinate directions of Figure 7.1. We denote by H the set of all functions h: R∗→R such that: (i) h agrees with hb (on ∂R∗) (ii) h satisfies the Lipschitz condition: for each i = 1,2,3 and ℓ> 0 −2 ≤h((x,y)+ℓδi)−h(x,y) ℓ ≤1, (7.2) where (x,y) and ℓ> 0 can be taken arbitrary as long as both (x,y) and (x,y)+ℓδi belong to R∗.
Let us recall a general result about differentiability of Lipshitz functions known as the Rademacher theorem.
Theorem 7.3. A Lipschitz function is differentiable almost everywhere.
Whenever a function h ∈H is differentiable at (x,y), the inequality (7.2) implies that the gradient encoded via Definition 7.1 by local densities (p , p , p ), satisfies 0 ≤p , p , p ≤1.
We henceforth take as a standing assumption that H ̸= ∅; this essentially means that the boundary function hb should satisfy a similar Lipschitz condi-tion. (Which is automatically true if hb is obtained as a limit of height functions of tileable regions.) 7.2.3 Goal We have made all necessary definitions and the rest of this lecture is devoted to proving that there is a unique function h∗∈H which maximizes the value of ¨ R∗S(∇h) dx dy.
64 Lecture 7: Maximizers in the variational principle 7.3 Upper semicontinuity Definition 7.4. Let M be a metric space. A real-valued function f : M →R is upper semicontinuous if for every point p ∈M, limsupx→p f(x) ≤f(p).
As the first step in proving the existence/uniqueness result, we prove the following: Theorem 7.5. Equip H with the uniform topology. Then over functions h ∈ H , the function h 7→ ¨ S(∇h) dx dy is upper semicontinuous.
The proof relies on several lemmas.
Lemma 7.6. Let h ∈H . Choose a mesh of equilateral triangles of side length ℓ. Take a function e h agreeing with h at vertices of each triangle, and linear on each triangle (hence piecewise linear).
For each ε > 0, if ℓ> 0 is smaller than ℓ0 = ℓ0(ε), then for at least 1 −ε fraction of the triangles, the following holds: • |e h−h| ≤ℓε at each point inside the triangle, • for (1−ε) fraction of the points of triangles, ∇h exists and ∇h−∇e h < ε.
Outline of proof We mirror the proof of [CohnKenyonPropp00, Lemma 2.2].
We first show that there is a 1 −ε/3 fraction of triangles contained strictly within the region satisfying the first condition, and then proceed to the second condition. In both parts we implicitly use the Rademacher theorem implying that h is differentiable almost everywhere.
• First part: For every point of differentiability w, there exists a small real number r(w) > 0 such that |h(w+d)−h(w)−(∇h)w ·d| < 1 10ε ·|d| for small displacements |d| ≤r(p). If our mesh ℓis so small that the set {w | r(w) > ℓ} (7.3) has measure at least (1 −ε/3) · (AreaR∗), then (ignoring the triangles cut off by the boundary), we can find a point w from the above set in each of the (1−ε/3) fraction of triangles. Thus, the linear approximation property holds on these triangles.
65 • For the second part we need to use the Lebesgue density theorem: Let A be any Lebesgue measurable set, define the local density of A at x as lim ε→0 mes(Bε(x)∩A) mes(Bε(x)) , with Bε(x) being a ball of radius ε around x. Then the theorem (which we do not prove here) claims that for almost every point x ∈A the local density equals 1.
Next, we choose finitely many sets U1, U2, ..., Un covering all R∗and such that whenever p,q ∈Ui, we have (∇h)p −(∇h)q < ε/2 (this is pos-sible since the space of possible gradients ∇h is compact). By applying Lebesgue’s density theorem to each of Ui, if ℓis small enough, for point w in all but ε/3 fraction of R∗, the value of ∇h in all but ε/3 fraction of the ball of radius ℓcentered at w differs from (∇h)w at most by ε/2.
If w additionally belongs to the set (7.3) from the first part, then on the triangle to which w belongs, ∇˜ h differs from (∇h)w at most by ε/2. Hence, the second property holds.
Here is an immediate consequence of Lemma 7.6.
Corollary 7.7. Retaining the notation of the lemma, lim ε→0 ¨ R∗S(∇h) dx dy− ¨ R∗S(∇e h) dx dy = 0.
The next lemma deals with very specific choices of R∗.
Lemma 7.8. Take as R∗an equilateral triangle T of side length ℓ. Let h ∈H and choose a linear function e h satisfying h−e h < εℓon ∂T. Then ˜ T S(∇h) dxdy AreaT ≤ ˜ T S(∇e h) dxdy AreaT +o(1) as ε →0.
In other words we can replace h by its linear approximation if they agree along the border of the domain T. Note that e h is linear, rather than piecewise linear — hence, this lemma is being applied to just one triangle T.
Proof of Lemma 7.8 By concavity of S, we have ˜ T S(∇h)dxdy AreaT ≤S(Avg(∇h)), (7.4) where Avg(∇h) is the average value of ∇h on T. Also, we have Avg(∇h)−Avg(∇e h) = O(ε), 66 Lecture 7: Maximizers in the variational principle which can be proven by reducing the integral over T in the definition of the average value to the integral over ∂T using the fundamental theorem of the calculus and then using h−e h < εℓ. Therefore, by continuity of S, S(Avg(∇h)) = S(Avg(∇e h))+o(1).
Since e h is linear, its gradient is constant on T and we can remove the average value in the right-hand side of the last identity. Combining with (7.4) we get the result.
Proof of Theorem 7.5 Let h ∈H . Then for each γ > 0 we want to show that there exists δ > 0 such that whenever g and h differ by at most δ everywhere, we should have ¨ R∗S(∇g)dxdy ≤ ¨ R∗S(∇h)dxdy+γ Take a piecewise linear approximation e h for h from Lemma 7.6, and choose δ < εℓ. Let us call a triangle T of mesh “good”, if two approximation proper-ties of Lemma 7.6 hold on it. Then as ε →0 we have by Lemma 7.8 ¨ T S(∇g) ≤ ¨ T S(∇e h)+Area(T)·o(1) and thus by summing over all the good triangles T ¨ good triangles in R∗S(∇g) ≤ ¨ good triangles in R∗S(∇e h)+Area(R∗)·o(1).
Because S is bounded, the bad triangles only add another O(ε). Thus, using also Corollary 7.7, we get ¨ R∗S(∇g) ≤ ¨ R∗S(∇h)+Area(R∗)·o(1).
7.4 Existence of the maximizer The development of the previous section leads to the existence of maximizers.
Corollary 7.9. There exists a maximizer h∗∈H : h∗= argmaxh∈H ¨ R∗S(∇h)dxdy.
67 Proof We equip the set H with the uniform topology. It is then compact by the Arzela-Ascoli theorem.
Let hn, n = 1,2..., be a sequence of functions such that the integrals ˜ S(∇hn)dxdy approach as n →∞the supremum suph∈H ˜ S(∇h)dxdy.
Passing to a convergent subsequence by compactness, we can assume with-out loss of generality that hn →h∗, for some h∗∈H . Then, by Theorem 7.5 sup h∈H ¨ S(∇h)dxdy = limsup n→∞ ¨ S(∇hn)dxdy ≤ ¨ S(∇h∗)dxdy.
On the technical level, the difficulty in the proof of Corollary 7.9 is to pass from uniform convergence of the functions to convergence of their gradients, which is necessary, since S is a function of the gradient. This was the main reason for the introduction of the mesh of small equilateral triangles T in the previous section.
7.5 Uniqueness of the maximizer We end this lecture with a discussion of the uniqueness of maximizers.
Let h1 and h2 be two maximizers (with the same value of the integral of S), and let h = h1+h2 2 ∈H . Then using the concavity of S, we have ¨ S(∇h)dxdy ≥1 2 ¨ S(∇h1)dxdy+ ¨ S(∇h2)dxdy = ¨ S(∇h1)dxdy.
(7.5) If S(•) were strictly concave, the inequality in (7.5) would have been strict and this would have been a contradiction with h1 ̸= h2 being two maximizers. An-noyingly, this is not quite true — S is only strictly concave in the region where p , p , p > 0 (i.e. the interior of the triangle of possible slopes). Nonetheless we would be done if there are points with ∇h1 ̸= ∇h2 with at least one of them strictly inside the triangle of slopes.
Hence, the only remaining case is that all gradients ∇h1 and ∇h2 lie on the boundary of the triangle of the slopes. Let us distinguish two kinds of boundary slopes: extreme slopes, such that (p , p , p ) is either (1,0,0), or (0,1,0), or (0,0,1) and non-extreme boundary slopes where proportions of two types of lozenges are non-zero. Note that if h1 ̸= h2 are two maximizers with all the slopes on the boundary, then passing to h = h1+h2 2 and using (7.5), if necessary, we would find a maximizer with non-extreme boundary slopes.
However, we did not see any such slopes in the simulations shown in Figures 68 Lecture 7: Maximizers in the variational principle 1.3, 1.5 of Lecture 1. In fact, the following statement explains that non-extreme boundary slopes never appear in the limit shapes of tilings, as long as we tile non-degenerate domains.
Proposition 7.10. If a domain R∗with hb on ∂R∗has at least one height function h◦∈H extending hb, such that all the slopes ∇h◦inside R∗are not on the boundary of the triangle of possible slopes, then h with non-extreme boundary slopes (i.e. boundary slopes with non-zero proportions of precisely two types of lozenges) cannot be a maximizer of ˜ S.
Remark 7.11. For all the domains that we encounter in these lectures, the exis-tence of h◦is immediate. For instance, this is the case for tilings of polygonal domains with sides parallel to the six grid directions of the triangular lattice.
The only possible “bad” situation that we need to exclude is when the bound-ary conditions make some part of the tiling frozen (i.e. having only one type of lozenges) for every extension of hb — in this case we should simply remove this frozen part from the consideration and study the height functions in the complement.
Outline of the proof of Proposition 7.10 For ε > 0 consider ¨ R∗[S(∇(εh◦+(1−ε)h))−S(∇h)] dxdy.
(7.6) We would like to study the small ε expansion of this difference by expanding the integrand at each (x,y).
The explicit formula for S implies that the partial derivatives of S are bounded inside the triangle of slopes, but grow logarithmically at the boundary (except for the vertices (1,0,0), (0,1,0), (0,0,1)), cf. Figure 6.2.
Hence, the contribution to (7.6) of (x,y) with non-boundary (or extreme boundary) slopes of ∇h is O(ε). On the other hand, the contribution of (x,y) with non-extreme boundary slopes is of order ε ln 1 ε ≫ε, since the definition of h◦implies that ∇(εh◦+(1−ε)h) is strictly inside the triangle of slopes for any ε > 0. Simultaneously, this contribution is positive for all (x,y), since S has a minimum on the border of the triangle, cf. Figure 6.2. Hence, (7.6) is positive for small ε contradicting the maximality of the integral for h.
Another proof for the uniqueness of maximizers can be found in [DeSilvaSavin08, Proposition 4.5].
Lecture 8: Proof of the variational prin-ciple In this lecture, using the machinery we have set up, prove the variational prin-ciple of Theorem 5.15 following [CohnKenyonPropp00].
Proposition 8.1. Let R be a convex region in the plane, and fix positive reals k and ε. Suppose that R fits within some L × L square, where L > ε−1, and suppose it has area A ≥kL2.
Let hb be a height function on ∂R which is within εL of the height function of a fixed plane Π. Then, the quantity: 1 A ln(# of extensions of hb to R) (8.1) is independent of the choice of hb (but depends on Π) up to an additive error of O(ε1/2 ln 1 ε ), which might depend on k.
Proof The proof splits into two cases depending on the slope of Π.
Case 1: The slope of Π is at least ε1/2 away from any boundary slope; in other words, the proportion of each lozenge corresponding to the slope of Π is at least ε1/2 and at most 1−ε1/2.
Take a height function gb (on ∂R) also agreeing with the plane Π; we will show that the values of (8.1) for gb and hb are within O(ε1/2 ln 1 ε ) of each other.
Assume gb ≥hb; if not, then we can compare both gb and hb with min(gb,hb) instead. Let g∗and h∗denote the minimal extension of gb and the maximal extension of hb to R, respectively (since maximum/minimum of two height functions is again a height function, such extensions are uniquely defined). We define two maps that operate on height functions on R. If f is an extension of gb to R, then we define H( f) to be the extension of hb to R given by the pointwise minimum: H( f) = min(f,h∗) 69 70 Lecture 8: Proof of the variational principle Similarly, if f is an extension of hb to R, then G( f) is the extension of gb to R given by: G( f) = max( f,g∗) Note that H(G( f)) = f at all points in R at which the value of f is between g∗and h∗(in that order).
Recall the formula for h∗in terms of hb from Leacture 1 (for g∗the formula is similar): h∗(x) = min u∈∂R(d(x,u)+hb(u)), where d is the distance function within R. Note that since R is convex, d is dif-fers (up to a small error) by a constant factor from the Euclidean distance in the plane. Thus, g∗and h∗differ from the height of Π by at least Ω(ε1/2d(x,∂R)), since the slope of Π is at least ε1/2 away from any boundary slope.
On the other hand, we claim that a typical random extension of hb to R is close to Π. Indeed, to see that we embed R into a torus and consider a random (α,β)–weighted tiling of the torus, where α and β are chosen so that the limit shape of such a random tiling has slope Π, cf. Lecture 6. The concentration of the height function of Lecture 5 applied to the torus means that hb is close on ∂R to the height function of random tiling of the torus. We can then com-pare random extension of hb to a random tiling of the torus conditional on the values of its height function on ∂R. Note that the conditioning removes non-uniformity, i.e. the conditional law of (α,β)–weighted tilings on the torus is the same as for the uniformly random tilings1 Hence, we can use the arguments of Lecture 5. In particular, Proposition 5.1 (see also Proposition 5.6) implies that with probability tending to 1 (as the size of the domain grows), the height function of random extension of hb is close to that of a random tiling of the torus, which is close to Π.
We conclude that with high probability, a random extension of hb to R is a distance at most εL from Π. Thus, we have that for random extension f of hb, with probability approaching 1, H(G( f)) = f at all points at a distance at least ε1/2(ln 1 ε )L from the boundary ∂R. This gives a partial bijection between extensions of hb and extensions of gb.
It remains to deal with the points at a distance less than ε1/2(ln 1 ε )L from the boundary ∂R. The possible values of the height function at these points 1 In order to see the conditional uniformity, notice that as soon as we know the values of the height function on ∂R, we know the total number of lozenges of each of the three types inside R. Indeed, all the tilings with fixed boundary condition can be obtained from each other by adding/removing unit cubes, as in the first proof of Theorem 2.1, and each such operation keeps the number of lozenges of each type unchanged. Hence, the each lozenge tiling of R with fixed values of the heights on ∂R has the same (α,β)-weight.
71 can be constructed sequentially: when we move from a point to its neighbor there are two possible choices for the value of the height function. Thus, noting that there are O(ε1/2(ln 1 ε )L2) points close to the boundary ∂R, the number of extensions of g and h differ by a multiplicative factor of at most 2O(ε1/2(ln 1 ε )L2), as desired.
Case 2: The slope of Π is within ε1/2 of some boundary slope; in other words, the proportion of some type of lozenges, p , p , or p , is at most ε1/2.
Without loss of generality, assume that the probability of a horizontal lozenge is p < ε1/2. Let us show that in this case, (8.1) is close to 0.
Consider some extension of hb, and consider all the vertical sections of R by straight lines. Note that we know the number of horizontal lozenges on these lines exactly, by calculating the change in the value of hb on each of these lines. This number is O(ε1/2L) due to hb being close to Π on ∂R. Thus, the total number of lozenges inside R is at most O(ε1/2L2). Note that the posi-tions of lozenges uniquely determine the rest of the tiling, i.e. the lozenges of the other two types. Thus the number of tilings is at most the number of arrangements of lozenges, which is at most (where c is the constant in the big O): A cε1/2L2 ≤ Acε1/2L2 cε1/2L2 e cε1/2L2 ≤ L2 cε1/2L2 e !cε1/2L2 = e cε1/2 cε1/2L2 = exp L2O(ε2 ln 1 ε ) .
Corollary 8.2. In the setting of Proposition 8.1, if R is an L by L square, then: 1 A ln(# of extensions of h to R) = S(slope of Π)+O(ε2 ln 1 ε ) (8.2) Proof Let us embed L × L square into L × L torus. If we deal with random tilings of this torus, then the height function on the boundary of our square is now random. However, in Section 6.4 we have shown this height function is asymptotically planar. Hence, using Proposition 8.1 we conclude that the number of tilings of the square is approximately the same as the number of the fixed slope tilings of the torus2 The asymptotic of the latter was computed in Theorem 6.5.3 2 We also need to sum over all possible choices of the approximately planar heights on the boundary of the square. However, the number of such choices is exp(O(L)) and we can ignore this factor in computation, since we only care about the leading O(L2) asymptotics of the logarithm of number of extensions.
3 Theorem 6.5 relies on Theorem 4.3, which was proven only for tori of specific sizes. However, the sizes form a positive-density subset and thus it suffices to use only those sizes.
72 Lecture 8: Proof of the variational principle Proposition 8.3. The statement (8.2) still holds when R is an equilateral tri-angle.
Proof Tile R with small squares, so that every square is completely contained within R (a small portion of R will be uncovered, but this is negligible so we ignore it). Fix a (consistent) height function that is within O(1) of the plane on the boundaries of the squares. Note that, given this height function on the boundaries of the squares, a tiling of each of the squares yields a valid tiling of R. Therefore, 1 A ln(# of tilings of R) ≥1 A ln(# of tilings of all squares) = 1 A ∑ squares s ln(# of tilings of s) ≥1 A ∑ squares s Area(s)(S(slope of Π)+O(ε2 ln 1 ε )) = S(slope of Π)+O(ε2 ln 1 ε ) Now, for the other direction of the inequality, we will use an identical argu-ment; tile a large square with (approximately) equal numbers of copies of R and R′ (a vertically flipped copy of R). Then, each tiling of the square yields a tiling of each triangle, so we have: ∑ copies of R,R′ ln(# of tilings) ≤(total area)·S(slope of Π)+O(ε2 ln 1 ε ).
Noting that the lower bound applies for both R and R′ parts of the last sum, we obtain the desired bound.
The final ingredient of the proof of the variational principle, is contained in the following theorem: Theorem 8.4. Let R∗be a region with a piecewise smooth boundary, and let h∗ b be a continuous height function on ∂R∗. Further, let R = RL be a lattice region with a height function hb on ∂R, and for some fixed δ, suppose that 1 LR is within δ of R∗and 1 Lhb is within δ of h∗ b. Suppose also that there exist extensions of hb to R.
Then, given a function h∗∈H which agrees with h∗ b on ∂R∗, we have: 73 1 L2 ln(number of extensions of hb to R within Lδ of Lh∗) = ¨ S(∇h∗)dxdy+o(1) (8.3) in the limit regime when we first send L →∞and then δ →0. The remainder o(1) is uniform over the choice of h∗.
Proof Pick arbitrary (small) ε > 0. Pick ℓ= ℓ(ε) as in Lemma 7.6 in Lecture 7 and create an ℓ-mesh of equilateral triangles on the region 1 LR. We will take δ < ℓε. Let ˜ h be a height function which agrees with h∗on the vertices of the mesh and is piecewise linear within each triangle of the mesh. By Lemma 7.6 on all but an ε fraction of the triangles, the function ˜ h is within εℓof h∗on the boundaries of the triangle.
Note that we can ignore the ε fraction of “bad” triangles, since the contri-bution from these to (8.3) is negligible as ε →0. The negligibility is based on the observation that for each bad triangle there is at least one discrete height function extending to the entire triangle the values of Lh∗on the vertices of this triangle in such a way that it stays within Lδ of Lh∗. On the other hand, the total number of such extensions is at most exp(O(area of the triangle)).
Now, in order to lower bound the LHS of (8.3), note that we can obtain a valid extension of hb to R by first fixing the values at the vertices of the mesh to agree with h∗, then choosing the values of the height function on the boundaries of the triangles in an arbitrary way (but keeping them close to the piecewise-linear approximation ˜ h), and finally, counting all possible extensions from the boundary of the triangles of the mesh to the interior. Using Proposi-tion 8.3 we get an estimate for the total number of tilings constructed in this way, which matches the right-hand side of (8.3). The overwhelming majority of the height functions of these tilings is in Lδ–neighborhood of Lh∗, since we chose them in such a way of the boundaries of the triangles and then can use the comparison with torus, as in the proof of Proposition 8.1 to conclude that the height function is close to ˜ h inside the triangles. Since ˜ h is close to h∗by the construction, the desired lower bound is obtained.
For the upper bound of the right-hand side of (8.3), we take an arbitrary tiling of R with height function h and restrict it to each triangle in the mesh; note that this must yield a valid tiling of each good triangle. Thus, summing over all possible boundary conditions on sides of the triangles of the mesh (whose normalized heights must be within ℓε of h∗, since δ < ℓε) and using Proposition 8.3 for each triangle, we obtain the desired upper bound. It is cru-cial at this step that the number of choices of all possible boundary conditions 74 Lecture 8: Proof of the variational principle on the sides of the triangles grows as exp(o(L2))4 and is therefore negligible for (8.3).
Remark 8.5. One might want to be slightly more careful near the boundaries of the domain, since h∗might be not defined in the needed points. We do not present these details here and refer to [CohnKenyonPropp00]. We also do not detail uniformity of the remainder o(1), leaving this as an exercise to the readers.
Proof of Theorem 5.15 For a function h∗∈H and δ > 0, let Uδ(h∗) denote the set of all functions h from H , for which sup(x,y)∈R∗|h(x,y)−h∗(x,y)| < δ.
Let hmax be the maximizer of Corollary 7.9. Choose an arbitrary small ε > 0.
Using uniqueness of maximizer, compactness of H , and Theorem 7.5, we can find δ0 (depending on ε), such that ¨ R∗S(∇h)dxdy < ¨ R∗S(∇hmax)dxdy−ε, for each h ∈H outside Uδ0(hmax).
Note that the choice δ0 = δ0(ε) can be made in such a way that δ0 →0 as ε →0. Next, choose another δ > 0, so that o(1) in (8.3) is smaller in magnitude than ε/2. This δ can be arbitrary small and therefore we can require without loss of generality that δ < δ0.
Using compactness of the space H , we can now choose finitely many elements h∗ 1,...,h∗ k ∈Uδ0(hmax) with h∗ 1 = hmax and finitely many elements h∗ k+1,...,h∗ m ∈H \Uδ0(hmax), so that (for large L) each height function HL of a tiling of RL is in δL–neighborhood from one of the functions Lh∗ i , 1 ≤i ≤m.
The numbers k, m and functions h∗ 1,...,h∗ m depend on ε and δ, but not on L.
Let Ni, 1 ≤i ≤m, denote the number computed in (8.3) for h∗= h∗ i . Then we have a bound exp(L2N1) ≤number of extensions of hb to RL ≤ m ∑ i=1 exp(L2Ni) (8.4) Sending L →∞using (8.3) and the fact that h∗ 1 = hmax is the maximizer, we 4 There are O(ℓ−2) triangles in the mesh and each of them has sides of length O(ℓL). Hence, the total length of all sides of the triangles of the mesh is O(ℓ−1 ·L). If we define heights along the sides of triangles sequentially, starting from the boundary of the domain (where they are fixed), then when we move to an adjacent vertex of the grid, we choose from at most 2 possible values of the height function. Hence, the total number of possible values for the height function on all sides of all triangles can be upper-bounded by 2O(ℓ−1·L).
75 get ¨ R∗S(∇hmax)dxdy−ε/2 ≤lim L→∞ 1 L2 ln(number of extensions of hb to RL) ≤ ¨ R∗S(∇hmax)dxdy+ε/2 (8.5) Since ε > 0 is arbitrary, this proves (5.3).
Further, using δ < δ0, we can write 1−P(HL is within 2Lδ0 from hmax) ≤∑m i=k+1 exp(L2Ni) exp(L2N1) (8.6) Using (8.3) and definition of δ0 we conclude that the numerator in (8.6) is upper-bounded by (m−k)exp L2 ¨ R∗S(∇hmax)dxdy−ε/2 +o(L2) .
Simultaneously, the denominator is lower-bounded using (8.3) by exp L2 ¨ R∗S(∇hmax)dxdy+o(L2) .
We conclude that the ratio in (8.6) goes to 0 as L →∞. Since δ0 can be made arbitrary small as ε →0, this implies convergence in probability 1 LH →hmax.
Lecture 9: Euler-Lagrange and Burgers equations In the previous lectures we reduced the problem of finding a limit shape of random tilings to a variational problem. Now we start discussing ways to solve the latter.
9.1 Euler-Lagrange equations Recall that the limit shape of a tiling, under certain conditions, is given by: argmaxh ¨ S(∇h)dxdy.
We will consider the general problem of finding argmaxh ¨ F(hx,hy,h)dxdy (9.1) for a general function F of three real arguments (and we are going to plug in hx as the first argument, hy as the second argument, and h as the third argument).
Proposition 9.1 (Euler-Lagrange equations). For maximal h(x,y) solving (9.1), at all points (x,y) where h is smooth (as a function of (x,y)) and F(hx,hy,h) is smooth (as a function of its argument in a neighborhood of (hx(x,y),hy(x,y),h(x,y))), we have: ∂ ∂x F1(hx,hy,h) + ∂ ∂y F2(hx,hy,h) −F3(hx,hy,h) = 0 (9.2) where Fi denotes the derivative of F in its i-th argument.
In the non-coordinate form, the first two terms can be written as div(∇F ◦ ∇h), representing the divergence of the vector field ∇F evaluated at ∇h(x,y).
76 77 Proof Note that since h is maximal, we must have, for all smooth g, ∂ ∂ε ¨ F((h+εg)x,(h+εg)y,(h+εg))dxdy ε=0 = 0 Using the Taylor series expansions of F, this reduces to: ¨ (F1(hx,hy,h)gx +F2(hx,hy,h)gy +F3(hx,hy,h)g) dxdy = 0 Now pick g to be smooth with compact support strictly inside the domain and integrate by parts in x in the first term and in y in the second term to get: ¨ ∂ ∂x(F1(hx,hy,h))+ ∂ ∂y(F2(hx,hy,h))−(F3(hx,hy,h)) gdxdy = 0.
Since g is arbitrary, the first factor in the integrand has to vanish.
Specializing Euler-Lagrange equations to our functional S(∇h) we see two different situations: • If ∇h at a point (x,y) lies strictly inside of the triangle of possible slopes, then (9.2) is a meaningful equation at this point.
• If ∇h is a boundary slope at (x,y), then (9.2) does not hold at such point, since S(∇h) fails to be a smooth function of ∇h — in fact it is not even defined if we deform ∇h outside the triangle of possible slopes.
In the first situation we say that (x,y) belongs to the liquid region, while the second one corresponds to the frozen region.
9.2 Complex Burgers equation via a change of coordinates If we try to write down the Euler-Lagrange equations directly for the functional of Theorem 5.15 we get a heavy and complicated expression.
[KenyonOkounkov05] found a way to simplify it by encoding the slope of the limit shape by a complex number. We now present their approach.
First, we wish to fix a coordinate system so that the gradient of h repre-sents the probabilities of encountering each type of lozenge. Rather than using rectangular coordinate system, we follow the conventions of Figure 3.2: the y–coordinate axis points up, while the x–coordinate axis points down–right. In other words, we use up and right vectors from Figure 7.1 as two coordinate directions.
78 Lecture 9: Euler-Lagrange and Burgers equations ℜ(z) ℑ(z) 0 1 z eα = |1 −z| eβ = |z| Figure 9.1 Triangle defining the complex slope z.
After fixing the x and y directions, we further modify the definition of the height function to ˜ h = x+y−h 3 .
(9.3) Recall that by definition of a height function (see discussion in Section 7.2.1), we have hx = p + p −2p , and hy = p + p −2p . Thus, since the local densities of three types of lozenges sum to 1, we have for the modified height function ∇˜ h = (p , p ).
(9.4) Recall the results of Lecture 6, namely Theorems 6.1 and 6.5 and Legendre duality used to prove them. An important role is played by a triangle with side-lengths eα, eβ, 1. Its angles are π(p , p , p ) with correspondence shown in Figure 9.1. Let us position this triangle in the complex plane is such a way that two its vertices are 0 and 1, while the third one is a complex number z in the upper half-plane. Following Kenyon and Okounkov we call z the complex slope encoding the triplet (p , p , p ).
Using identification of (9.4) and recalling that σ = −S, the Legendre duality of Lecture 6 can be now restated as −∂S ∂˜ hx = α = ln|1−z|, −∂S ∂˜ hy = β = ln|z|.
(9.5) It turns out that z satisfies a first order differential equation.
Theorem 9.2 ([KenyonOkounkov05]). In the liquid region (i.e., where p , p , p > 0), there exists a function z(x,y) taking values in the complex 79 upper half plane such that (˜ hx, ˜ hy) = 1 π argz,−arg(1−z) (9.6) and −zx 1−z + zy z = 0.
(9.7) Proof It can be shown using general results on variational problems and PDEs that ˜ h is analytic in the liquid region — this is a particular case of the Hilbert’s nineteenth problem, and we refer to the introduction of [DeSilvaSavin08] for references and discussion in the context of random sur-faces.
Then ˜ h is differentiable and (9.6) is a restatement of (9.4) and geometric definition of the complex slope z above.
By the Euler-Lagrange equations for the variational problem of Theorem 5.15, we obtain using (9.5): 0 = ∂ ∂y ∂S ∂˜ hy + ∂ ∂x ∂S ∂˜ hx = −∂ ∂y ln|z|−∂ ∂x ln|1−z| Note that this is the real part of (9.7). For the imaginary part, we need to show ∂ ∂y argz+ ∂ ∂x arg(1−z) = 0, which is the identity ˜ hxy = ˜ hyx rewritten using (9.6).
We remark that the definition of the complex slope z is not canonical, since we need to somehow identify three angles of the (0,1,z) triangle with three proportions (p , p , p ) and there are six ways to do it. In particular, if we perform the substitution u = z 1−z (this corresponds to a rearrangement of the vertices of the triangle of Figure 9.1), then the chain rule gives: ux = zx (1−z)2 , uy = zy (1−z)2 .
In u–coordinate (9.7) becomes: ux ·u = uy.
(9.8) This equation is well-known, it is called the complex inviscid Burgers equation.
Exercise 9.3. There are 6 ways to identify three angles of the (0,1,z) triangle with three proportions (p , p , p ). Two of these ways lead to equations (9.7) and (9.8). Find the equations corresponding to four other ways.
80 Lecture 9: Euler-Lagrange and Burgers equations 9.3 Generalization to qVolume–weighted tilings We now consider a generalization of the uniformly random tilings by changing the weight of the measure to a qVolume, where the volume is the number of unit cubes that must be added to the minimial tiling to obtain a given tiling.1 Addi-tion of one cube increases the height function h(x,y) of Section 1.4 by 3 at a single point, hence, volume is given by the double integral 1 3 ˜ h(x,y)dxdy up to a constant shift. In terms of the modified height function ˜ h of (9.3) addition of one cube decreases the height by 1 and, therefore, volume is equal to minus the double integral: Volume = − ˜ ˜ h(x,y)dxdy+const.
Let us investigate the limit shape of the random height function under the measure qVolume with q changing together with the linear size L of the domain by q = ec/L.
Proposition 9.4. Consider random tilings under the qVolume measure with q = exp(c/L) as the linear size of the domain L grows. In the setting of Theorem 5.15, the limit shape for the height function still exists in this case, it is given by: h∗= argmaxh ¨ R∗ S(∇h)+ c 3h dxdy.
(9.9) Proof The proof is identical to the proof of the variational principle of The-orem 5.15 in the uniform case; we only need to multiply by qVolume when we count tilings.
Corollary 9.5. The limit shape for qVolume–random lozenge tilings with q = exp(c/L) satisfies: div(∇S◦∇h) = c 3 (9.10) where ◦is the evaluation operator; i.e., ∇S is being evaluated at ∇h and then div is applied to the resulting function of (x,y).
Proof This is the Euler-Lagrange equation for the variational problem (9.9).
Theorem 9.2 extends to the q–weighted case with a similar proof.
Corollary 9.6. For qVolume–weighted tilings with q = exp(c/L) the statement of Theorem 9.2 remains valid with (9.7) replaced by −zx 1−z + zy z = c.
(9.11) 1 The qVolume measure will appear again in Lectures 22-23 and then in Lecture 25. In particular, the effect of adding one cube on the height function is shown in Figure 25.2.
81 Proof The imaginary part of (9.11) is the same as that of (9.7) and its validity follows from ˜ hxy = ˜ hyx relation. For the real part, we need to add an additional term in the Euler–Lagrange equations, which produces the right-hand side of (9.11). Note that c is being multiplied by (−3) when we pass from h to ˜ h by (9.3), hence, the final form of (9.11).
9.4 Complex characteristics method Since (9.11) is a first order partial differential equation, it can be solved using the method of characteristics. The only tricky point is that the coordinate z is complex, while the textbooks usually present the method in the real case. Yet, as we will see in this and the next lecture, all the principles continue to work in the complex case.
For simplicity, imagine first that z were real. Consider any curve x(t),y(t),z(t) satisfying the following differential equations: dx dt (z−1) = dy dt z = dz dt · 1 c We call such a curve a characteristic curve.
Proposition 9.7. A surface z = z(x,y) solving (9.11) must be a union of char-acteristic curves.
Proof The normal vector to such a surface is (zx,zy,−1). The tangent to a characteristic curve is, up to scaling by a function of t, ( 1 z−1, 1 z ,c). Note that these two vectors have dot product 0 by (9.11), so the tangent vector to the characteristic curve is tangent to the surface. Thus, the characteristic curve remains on the surface.
Next, we solve the equation determining the characteristic curves. Choose z = t. Then: dy dz = 1 cz, dx dz = 1 c(z−1).
The solutions to these equations are: cy = lnz+k1, cx = ln(z−1)+k2.
Rearranging, we get the following form for a characteristic curve: ze−cy = ˜ k1, (1−z)e−cx = ˜ k2, 82 Lecture 9: Euler-Lagrange and Burgers equations where ˜ k1 and ˜ k2 are arbitrary constants. The choice of these constants is re-lated to the specification of the boundary conditions for the PDE (9.11), and we need to specify which curves to use to form the surface. In general a two-dimensional surface is a union of a one-parameter family of the curves, hence, we can parameterize the desired ˜ k1, ˜ k2 by an equation Q(˜ k1, ˜ k2) = 0. We con-clude that the solution to (9.11) must be given by Q(ze−cy,(1−z)e−cx) = 0 for some function Q.
We proceed to the complex case and seek for a solution in the same form.
The only difference is that Q is no longer arbitrary, but it is an analytic function.
We have arrived at the following statement from [KenyonOkounkov05]: Theorem 9.8. Let c ̸= 0 and take any analytic function of two variables Q(u,v). The surface given by: Q(ze−cy,(1−z)e−cx) = 0 (9.12) solves (9.11) Proof Taking the derivatives in x and y of both sides of (9.12), we get Q1zxe−cy +Q2(−zx)e−cx −cQ2(1−z)e−cx = 0, Q1zye−cy −cQ1ze−cy +Q2(−zy)e−cx = 0, where Qi denotes the derivative of Q in its i-th argument.
Note that we can isolate Q1e−cy Q2e−cx in both these expressions to obtain the equal-ity: c(1−z)+zx zx = zy zy −cz Clearing denominators and rearranging, we obtain precisely (9.11).
A remark is in order. The Burgers equation (9.11) is only valid in the liq-uid region. In the frozen part of the limit shape, only one type of lozenges is present and the definition of the complex slope z is meaningless. On the other hand, the limit shape is still analytic — it is in fact linear. However, non-analyticity appears on the Arctic boundary — separation curve between liquid and frozen phases; the behavior of the limit shape on two sides of this curve is very different.
We still have not discussed how to find Q based on the boundary conditions for lozenge tilings. We examine this question in the next lecture.
Lecture 10: Explicit formulas for limit shapes In the previous lectures we discussed two properties of the limit shapes of tilings: they solve a variational problem and in liquid regions they solve the complex Burgers equation. In this lecture we concentrate on algorithmic ways to explicitly identify the limit shapes.
10.1 Analytic solutions to the Burgers equation First, we continue our study of the solutions to the complex Burgers equation appearing as Euler-Lagrange equation for limit shapes of random tilings. The equation reads −zx 1−z + zy z = c.
(10.1) In the last lecture we have shown in Theorem 9.8 that an analytic function of two variables Q(u,v) gives rise to a solution to (10.1). However, we have not explained whether every solution are obtained in this way, i.e. whether we can use the formulation in terms of Q(u,v) as an ansatz for finding all possible solutions to the Burgers equation and, hence, for finding the limit shapes of tilings. We proceed by showing that this is indeed the case (still in the situation c ̸= 0).
Theorem 10.1. If c ̸= 0, given a solution to (10.1) there exists an analytic Q such that Q(ze−cy,(1−z)e−cx) = 0.
(10.2) Remark 10.2. In the present formulation Theorem 10.1 is a bit vague and we are not going to make it more precise here. Of course, one can always choose Q to be identical zero, but this is not what we want. The theorem rather says 83 84 Lecture 10: Explicit formulas for limit shapes that the first and second arguments of Q, ze−cy and (1−z)e−cx depend on each other (locally near each point (x0,y0) of the liquid region) in an analytic way.1 The last property is easy to make precise in the case when the dependencies (x,y) →ze−cy and (x,y) →(1−z)e−cx are non-degenerate; this is what we do in the proof.
For the applications, the statement of Theorem 10.1 should be treated as an ansatz: one should try to find a solution to (10.1) by searching for an analytic function Q. Often, Q will end up being very nice, like a polynomial in Theorem 10.7 below.
If we knew the exact shape of the liquid region, then once Q is found, it automatically provides the desired limit shape (maximizer) due to convexity of the functional in the variational problem of Theorem 5.15, which guar-antees that a solution to the Euler-Lagrange equations is the global maxi-mizer. However, the Euler–Lagrange equations or the complex Burgers equa-tion do not tell anything about the behavior (extremality) in the frozen regions, and therefore, additional checks are necessary there after Q is identified, see [AstalaDusePrauseZhong20] and, in particular, Theorem 8.3 there for a more detailed discussion.
Sketch of the proof of Theorem 10.1 Take z(x,y) solving (10.1) and define W = ze−cy and V = (1 −z)e−cx. What we need to prove is that W(V) is an analytic function, in other words, W(V) should satisfy the Cauchy-Riemann equations. This is equivalent to showing ∂W ∂¯ V = 0.
Here we use the following notation: for a function f(z) : C →C with z = x+iy we denote ∂f ∂z = 1 2 ∂f ∂x −i∂f ∂y and ∂f ∂¯ z = 1 2 ∂f ∂x +i∂f ∂y .
The following lemma is the first step in computation.
Lemma 10.3. The Jacobian of the map (x,y) →(V,W) vanishes, i.e.
WxVy = VxWy.
1 Formally, this is a weaker property since, in principle, local analytic dependence does not imply the existence of a global dependence as in (10.2), see the discussion in [AstalaDusePrauseZhong20, Remark 3.10].
85 Proof We have WxVy = (zxe−cy)(−zye−cx) and VxWy = (−zxe−cx −c(1−z)e−cx)(zye−cy −cze−cy).
Therefore, subtracting and using (10.1) we find VxWy −WxVy = e−cxe−cy(c2z(1−z)−czy(1−z)+czxz) = 0, as desired. The cancellation may seem (almost) magical; however the defini-tions of W and V were chosen by the method of characteristic essentially to have this property.
Now consider W(V) = W(V(x,y)) = W(x,y). Using the chain rule, we find ∂W ∂x = ∂W ∂V ∂V ∂x + ∂W ∂¯ V ∂¯ V ∂x and ∂W ∂y = ∂W ∂V ∂V ∂y + ∂W ∂¯ V ∂¯ V ∂y .
Multiplying the first equation by ∂V ∂y and subtracting the second equation mul-tiplied by ∂V ∂x , we get a restatement of Lemma 10.3: 0 = ∂W ∂x ∂V ∂y −∂W ∂y ∂V ∂x = ∂W ∂¯ V ∂¯ V ∂x ∂V ∂y −∂¯ V ∂y ∂V ∂x .
This proves the desired Cauchy–Riemann relation ∂W ∂¯ V = 0 unless we have that ∂¯ V ∂x ∂V ∂y −∂¯ V ∂y ∂V ∂x = 0.
(10.3) The expression (10.3) coincides with the Jacobian of the transformation (x,y) →V from (two-dimensional) liquid region to complex numbers (another two-dimensional real space) so as long as the mapping is not degenerate we are done. Note that in a similar manner we can demonstrate that ∂V ∂¯ W = 0 given a similar non-degeneracy for the map (x,y) →W.
In other words, we have found Q(·,·), which is analytic at any pairs of com-plex points (V(x,y),W(x,y)), where at least one of the maps (x,y) →V(x,y), (x,y) →W(x,y) is non-degenerate. In a similar way, if we assume that (x,y) → αV(x,y) + βW(x,y) is non-degenerate near (x0,y0) for some numbers α and β, then the possibility of choosing an analytic Q would also follow.
86 Lecture 10: Explicit formulas for limit shapes Theorems 9.8, 10.1 say that the map (x,y) 7→(ze−cy,(1−z)e−cx) sends the liquid region to a one-dimensional complex Riemann surface in C2. In fact, this map is a bijection with its image, and therefore, it can be used to equip the liquid region with the structure of a one-dimensional complex manifold.
Lemma 10.4. In the liquid region, given (W,V) = (ze−cy,(1−z)e−cx) one can recover (x,y) Proof Note that given W,V one can immediately recover arguments of the complex numbers z and (1 −z) as x,y are real. Given this, one can recover z and since z ̸= 0,1, as we are in the liquid region, we can therefore recover x,y.
Next, we examine the c = 0 case.
Theorem 10.5. For a solution z(x,y) of the complex Burgers equation with c = 0 as in (9.7) there exists an analytic Q0 such that Q0(z) = yz+x(1−z).
(10.4) Remark 10.6. The discussion of Remark 10.2 also applies to Theorem 10.5.
In addition, a solution z = const is possible for (9.7) with specific boundary conditions. This corresponds to sending Q0 to infinity in (10.4), so that the right-hand side becomes irrelevant.
The arguments leading to Theorem 10.5 are similar to the c ̸= 0 case and we omit all of this analysis except for the basic computation through the charac-teristics method.
The characterstic curves for −zx 1−z + zy z = 0 solve dx dt (z−1) = dy dt (z) and z is being kept constant. This gives dx dy = z z−1 leading to the parameteriza-tion of the curves by two arbitrary constant k1, k2 via x = y z z−1 +k1, z = k2.
This is equivalent to x(1 −z) + yz = k1(1 −k2) = Q0(z) of (10.4), where Q0 expresses the dependence between k1 and k2.
87 Figure 10.1 A qVolume–random tiling of 70×90×70 hexagon with q = 0.96.
10.2 Algebraic solutions Theorems 10.1 and 10.5 claim the existence of an abstract Q which describes the liquid region. But how do we determine such Q? The following theorem of Kenyon and Okounkov shows that in the case of polygonal domains that Q is a polynomial.
Theorem 10.7 ([KenyonOkounkov05]). Take a polygonal region R with 3d sides parallel to the coordinate direction repeated in cyclic order. (R is simply connected.) Then Q in Theorem 10.1 is a polynomial of degree (at most) d with real coefficients.
We sketch a plan of the proof and leave to the reader to figure out the neces-sary (highly non-trivial) details, following [KenyonOkounkov05].
• Formulate the necessary properties that Q should have. “Q is given by a cloud curve”. One of the properties is that the tangent vector should rotate d times as we move around the boundary of the frozen region.
• Study c = −∞case. In this case the limit shape is given by the tiling with the minimal height function.
• Obtain a solution (i.e. the function Q) for finite c through a continuous de-formation from the c = −∞case, preserving the properties from the first step.
• Prove that the resulting Q is not only a solution to the Euler–Lagrange equa-tions but a genuine maximizer. This check is tricky in the frozen regions, as the Burgers equation no longer holds there.
Let us explain how the boundary of the frozen region can be computed using Q. In order to do this, we note that the boundary corresponds to real z such 88 Lecture 10: Explicit formulas for limit shapes that Q(ze−cy,(1 −z)e−cx) = 0. Since Q has real coefficients, z is a solution, so is ¯ z. Hence, the frozen boundary corresponds to the double real roots of Q. We conclude that the frozen boundary is the algebraic curve in e−cx, e−cy describing the values when Q(ze−cy,(1−z)e−cx) has a double root. We expect this curve to be tangent to all sides of the polygon that we tile — this is often enough to uniquely fix Q.
Let us now specialize to the case of a hexagon. Then, by Theorem (10.7), Q must be quadratic; that is, it has the form Q(V,W) = a+bV +cW +dV 2 +eVW + fW 2, and we need to substituteW = ze−cy andV = (1−z)e−cx. This gives a quadratic equation in z. The boundary curve is then the curve where discriminant of Q vanishes, which is a fourth order polynomial equation in e−cx,e−cy. Since this curve needs to be tangent to the six sides of the hexagon2 this gives 6 equations in 6 unknowns a,b,c,d,e, f. Solving them, one finds Q.
Exercise 10.8. Find Q explicitly for given proportions of the hexagon.
The result of a computer simulation of the q–weighted random tilings of the hexagon and corresponding theoretical boundary of the frozen boundary are shown in Figure 10.1. Random tilings were sampled using the algorithm from [BorodinGorinRains09]; another algorithm will be discussed in Lecture 25.
10.3 Limit shapes via quantized Free Probability Let us now present another approach to computations of the limit shapes of uniformly random lozenge tilings. Take any large domain, which does not have to be polygonal, but we assume that its boundary has one vertical segment with two adjacent diagonal segments forming 2π/3 angle with the vertical one, as in Figure 10.2.
The combinatorics implies that in a vicinity of this vertical segment the hor-izontal lozenges interlace and at a vertical line at distance N from this seg-ment we observe N lozenges . Let their (random) positions be x1,...,xN in the coordinate system with zero at the lower boundary of the domain. We en-code these lozenges through their empirical measure µN and its Stieltjes 2 This tangency can be clearly seen in computer simulations; it is rigorously proven by a different method in [BorodinGorinRains09]. Also the tangency condition is an essential component of the construction of Q in [KenyonOkounkov05], see Theorem 3 there.
89 0 3 N Figure 10.2 A part of a domain consisting of a vertical segment and two adjacent straight diagonal segments. At distance N from the left boundary we observe N horizotnal lozenges.
transform GµN(z): µN = 1 N N ∑ i=1 δxi/N, GµN(z) = ˆ R 1 z−xµN(dx) = 1 N N ∑ i=1 1 z−xi/N .
Now assume that the domains start to grow and set N = ⌊αL⌋. Here L is the linear size of the domain and parameter α > 0 is chosen so that αL remains smaller than the length of the diagonal sides adjacent to the vertical segment — this choise guarantees that we still have N horizontal lozenges at distance N from the left boundary.
The limit shape theorem for uniformly random lozenge tilings (which we established through the variational principle in Lectures 5-8) implies the con-vergence lim L→∞µ⌊Lα⌋= µα, lim L→∞Gµ⌊Lα⌋(z) = Gµα(z) = ˆ R 1 z−xµα(dx), where µα is a deterministic measure, whose density encodes the asymptotic proportion of lozenges along the vertical line at distance α from the left vertical boundary of the rescaled domain. Note that µα can be directly related to the gradient of the limiting height function in vertical direction. Complex 90 Lecture 10: Explicit formulas for limit shapes variable z in Gµα(z) should be chosen outside the support of µα, so that the integral is non-singular.
Let us emphasize that xi and the total weight of the measure are rescaled by N, rather than by L in the definition of µN — in other words, this rescaling depends on the horizontal coordinate α, which is slightly unusual.
It turns out that the measures µα are related to each other in a simple way.
Theorem 10.9 ([BufetovGorin13]). Define the quantized R–transform through Rquant µα (z) = G(−1) µα (z)− 1 1−e−z .
Then for all α1,α2 > 0, for which the measures µα1,µα2 are defined, we have α1Rµα1(z) = α2Rµα2(z).
(10.5) Remark 10.10. By G(−1) µα (z) we mean the functional inverse: the function Gµα(z) behaves as 1 z as z →∞, hence, we can define an inverse function in a neighborhood of 0, which has a simple pole at 0. Subtraction of 1 1−e−z can-cels this pole and Rquant µα (z) is an analytic function in a complex neighborhood of 0.
A modification of Rquant µ (z) given by Rµ(z) := G(−1) µ (z) −1 z is known in the Free Probability theory as the Voiculescu R–transform, see [Voiculescu91, VoiculescuDykemaNica92, NicaSpeicher06]. One standard use of the R– transform is to describe how the spectrum of large random Hermitian matri-ces changes when we add independent matrices or multiply them by projec-tors. [BufetovGorin13] demonstrated that Rquant µ (z) replaces Rµ(z) when we deal with discrete analogues of these operations: compute tensor products and restrictions of the irreducible representations of unitary groups. The latter op-eration turns out to intimately related to combinatorics of lozenge tilings (cf.
Proposition 19.3 in Lecture 19 below), hence, the appearance of Rquant µ (z) in Theorem 10.9. We discuss the link between tilings and random matrices in more details in Lectures 19 and 20.
From the computational point of view, Theorem 10.9 leads to reconstruction of all µα (and hence of the limit shape) once such measure is known for some fixed choice of α. The following exercise provides an example where such approach is useful.
Exercise 10.11. Consider the domain of Figure 10.3, which is a trapezoid with N teeth sticking out of the right boundary at every second positions. By definition as N →∞the empirical measures on the right boundary converge to µ1 which is a uniform measure on [0,2]. Use this observation together with (10.5) to compute the limit shape as N →∞.
91 N −1 N N 2N −1 Figure 10.3 The half-hexagon domain: trapezoid with N teeth sticking to the right at every second positions.
In particular, Exercise 10.11 recovers the result of [NordenstamYoung11] which shows that the frozen boundary for this domain is a parabola.
The complex Burgers equation can be also extracted from the result of The-orem 10.9. The following exercise is close in spirit to the computations in [Gorin16, Section 4].
Exercise 10.12. Treating the formula (10.5) as a definition of the measures µα (assuming one of them to be given) show that the limit shape of tilings encoded by these measures satisfies the complex Burgers equation (9.7).
Hint: The density of the measure µα can be reconstructed as the imagi-nary part of the Stieltjes transform Gµα(z) as z approaches the real axis. On the other hand, in the definition of the Kenyon–Okounkov complex slope the density p becomes the argument of a certain complex number. This suggest reconstruction of complex slope by exponentiating the Stieltjes transform.
We end this lecture by a remark that in the random matrix theory to which we got linked through R–transforms, the complex Burgers equation has also shown up, see [Matytsin93, Guionnet02], which were even before the work of [KenyonOkounkov05].
Lecture 11: Global Gaussian fluctuations for the heights 11.1 Kenyon-Okounkov conjecture In previous lectures we found that the height function of random tilings for an arbitrary domain R converges in probability to a deterministic limit shape as the size of the domain grows to infinity 1 LH(Lx,Ly) L→∞ − − − →h(x,y).
(11.1) The limit shape h(x,y) is a maximizer of a functional of the form ¨ R S(∇h)dxdy, (11.2) and can be also identified with solutions to the complex Burgers equation on a complex function z(x,y), where z encodes the gradient ∇h through a certain geometric procedure.
In this section we start discussing the asymptotic fluctuations of the random height function around its expectation. The fluctuations are very different in the liquid and frozen regions. In the latter, the height function is flat with over-helming probability already for finite values of L and hence, there are essen-tially no fluctuations. For the former, the situation is much more interesting.
Conjecture 11.1 ([KenyonOkounkov05], [Kenyon04]). Consider qVolume– weighted random tilings in domain R of linear size L and with q = exp(c/L).
In the liquid region L , H(Lx,Ly)−E[H(Lx,Ly)] d − − − → L→∞GFF, (11.3) Where GFF is the Gaussian Free Field on L with respect to the complex struc-92 93 ture given by ze−cy and with Dirichlet boundary conditions. The GFF is nor-malized, so that Cov H(Lx,Ly),H(Lx′,Ly′) ≈−1 2π2 lnd (x,y),(x′,y′) (11.4) for (x,y) ≈(x′,y′), with d(·,·) denoting the distance in the local coordinates given by the complex structure, and with normalization for the definition of the height function as in the discrete version of (9.3), (9.4).
In this lecture we aim to define and understand the limiting object — the Gaussian Free Field. The next lecture contains a heuristic argument towards the validity of Conjecture 11.1. Later on, in Lectures 22 and 23 we present a proof of a particular case of this conjecture for random plane partitions.
For general domains R, Conjecture 11.1 is a major open problem in the area. Yet, there are several approaches to its proofs for particular classes of domains.
• One approach was suggested in [Kenyon99], [Kenyon00], [Kenyon04]. The correlation functions of tilings are minors of the inverse Kasteleyn ma-trix, which can be thought of as a discrete harmonic function. On the other hand, the covariance of GFF is a continuous harmonic function. Thus, one can be treated as a discretization of the other, which allows to use various convergence results from the analysis. Techically, two challenging points of such approach is to deal with non-trivial complex structure and with frozen boundaries. Despite some progress (cf. [Russkikh16], [Russkikh18], [BerestyckiLaslierRay16]), the proofs along these lines until very recently were restricted to a special class of domains, which, in particular, have no frozen regions. At the time of writing of this book new developments ap-peared in [ChelkakLaslierRusskikh20], [ChelkakRamassamy20]: the cen-tral idea is to construct a discrete approximation of the limiting complex structure using the origami maps. There is a hope to reach a resolution of Conjecture 11.1 through this approach in the future.
• The second approach (see [BorodinFerrari08], [Petrov12b]) is to use a dou-ble contour integral representation for the inverse Kasteleyn matrix, and then apply the steepest descent method for the asymptotic analysis. We will dis-cuss some components of this method in Lectures 14-16. This method is naturally restricted to the domains, where such contour integral representa-tions are available. Yet, such domains include a class of specific polygons with arbitrary many sides.
• The third approach combines a gadget called “Discrete Loop Equations” 94 Lecture 11: Global Gaussian fluctuations for the heights with the idea of applying differential/difference operators to inhomoge-neous partition functions to extract asymptotic information. We discuss dis-crete loop equations (cf. [BorodinGorinGuionnet15]) in Lecture 21 and the use of difference operators for GFF asymptotics (cf. [BorodinGorin13], [BufetovGorin16], [Ahn18]) in Lectures 22-23. Tilings of non-simply connected domains with holes were analyzed through this approach in [BufetovGorin17].
• In some case, the marginal distributions of random tilings can be identified with discrete log-gases, as we will compute in Lecture 19. Yet another ap-proach to global fluctuations exists in this case and links it to the asymptotic of recurrence coefficients of orthogonal polynomials, see [BreuerDuits13], [Duits15].
We remark that our toy example of the uniform measure on tilings of the hexagon is covered by each of the last three methods.
11.2 Gaussian Free Field We start our discussion of Conjecture 11.1 by defining the (two-dimensional) Gaussian Free Field in a domain D ⊂C in the standard complex structure of C.
Take the Laplace operator on D △= ∂2 ∂x2 + ∂2 ∂y2 (x,y)∈D .
We consider inverse operator −△−1 (note the minus sign) taken with Dirich-let (i.e., identical zero) boundary conditions on the boundary ∂D. △−1 is an integral operator (with respect to the Lebesgue measure on D), and its kernel is known as the Green’s function. Equivalently, we define: Definition 11.2. The Green’s function GD(z,z′) on D with Dirichlet boundary condition is a function G : D × D →R of two complex arguments z = x + iy, z′ = x′ +iy′, such that △GD(z,z′) = −δ(z = z′), where △can be applied either in z or in z′ variables, and GD(z,z′) = 0, if either z ∈∂D or z′ ∈∂D.
We remark without proof that for (non-pathological D) the Green’s func-tion exists, is unique and symmetric, i.e., GD(z,z′) = GD(z′,z). When D is 95 unbounded, one should add a (at most) logarithmic growth condition at ∞to the definition of the Green functions.
Exercise 11.3. Suppose that D is the upper halfplane H = {z ∈C | ℑz > 0}.
Show that in this case GH(z,z′) = −1 2π ln z−z′ z−¯ z′ .
(11.5) For general domains, there is often no explicit formula for the Green’s func-tion, but locally such function always grows like a logarithm of the distance: GD(z,z′) ∼−1 2π ln|z−z′| for z ≈z′.
(11.6) We can now present an informal definition of the Gaussian Free Field.
Definition 11.4. The Gaussian Free Field on D is a random Gaussian function GFFD : D →R such that E GFFD = 0 E GFFD(z)GFFD(z′) = GD(z,z′) However, there is a problem with this definition, if z = z′, then GD(z,z′) = +∞, hence, the values of GFFD have infinite variance. This means that GFFD is not a function, but a generalized function: we can not ask about the values of this random function at a point, but we can compute its integrals (or pairings) with reasonably smooth test functions.
Definition 11.5. For a test-function u : D →R, such that ¨ D u(z)G(z,z′)u(z′)dxdydx′ dy′ < ∞, z = x+iy, z′ = x′ +iy′, we define a pairing of GFFD with u, as a mean zero Gaussian random variable ⟨GFFD,u⟩with variance E ⟨GFFD,u⟩⟨GFFD,u⟩ = ¨ D×D u(z)G(z,z′)u(z′)dxdydx′ dy′.
If we take several such u’s, then the pairings are jointly Gaussian with covari-ance E ⟨GFFD,u⟩⟨GFFD,v⟩ = ¨ D×D u(z)G(z,z′)v(z′)dxdydx′ dy′.
(11.7) Informally, one should think about the pairings as integrals ⟨GFFD,u⟩= ¨ D GFFD(z,·)u(z)dxdy.
(11.8) 96 Lecture 11: Global Gaussian fluctuations for the heights From this point of view, one can take as u a (reasonably smooth) measure rather than a function.
By invoking the Kolmogorov consistency theorem, GFFD is well-defined as a stochastic process through Definition 11.5, if we show that the joint distri-butions appearing in this definition are consistent. Since they are all mean 0 Gaussians, it suffices to check that the covariance is positive-definite.
Lemma 11.6. The covariance of ⟨GFFD,u⟩given by (11.7) is positive-definite.
Proof We need to check that the operator with kernel GD is positive-definite, which is equivalent to −△being positive definite. The latter can be checked integrating by parts to reduce to the norm of the gradient: ⟨−△u,u⟩= ¨ D −△u(z) u(z)dxdy = ¨ D ||∇u(z)||2 L2 dxdy ≥0.
Remark 11.7. A better alternative to Definition 11.5 would be to define a func-tional space (like the space of continuous functions in the definition of the Brownian motion) to which the GFFD belongs. We refer to [Sheffield03b], [Dubedat07, Section 4], [HuMillerPeres09, Section 2], [WernerPowell20] for more details on the definition of the Gaussian Free Field.
An important property of the Gaussian Free Field is its conformal invari-ance.
Recall that a map φ : D →D′ is called conformal if it is holomorphic, bijec-tive and its inverse is also holomorphic. A direct computation with Laplacian △leads to the conclusion that the Green’s function is preserved under confor-mal bijections, thus implying the following result: Exercise 11.8. If φ is a conformal bijection of D with D′, then {GFFD′(φ(z))}z∈D′ d = {GFFD(z)}z∈D.
For instance, in the case when the domain is upper half-plane, D = D′ = H, the conformal bijections are Moebius transformations z 7→az+b cz+d , a,b,c,d ∈R, and one readily checks that the formula (11.5) is unchanged under such trans-formations.
Another remark is that if we follow the same recipe for defining the Gaussian Free Field in dimension 1, rather than 2, then we get the Brownian bridge.
Lemma 11.9. Consider the standard Brownian bridge, B : [0,1] →R , B(0) = B(1) = 0, (11.9) 97 which can be defined as the Brownian motion conditioned to be at 0 at time 1, or as a centered Gaussian process with covariance function EBsBt = C (s,t) = min(t,s)(1−max(t,s)), 0 ≤t,s ≤1.
(11.10) Then the function C (s,t) is the kernel of the (minus) inverse Laplace operator on the interval [0,1] with Dirichlet boundary conditions at 0 and 1.
Proof Since the Laplace operator in dimension 1 is simply the second deriva-tive, and the vanishing of C (s,t) at t = 0, s = 0, t = 1, and s = 1 is clear, it suffices to check that ∂2 ∂t2 C (s,t) = −δ(s = t).
(11.11) We compute ∂C (s,t) ∂t = ( 1−s , t < s, −s , t ≥s.
Differentiating, we get (11.11).
We now give another informal definition of the Gaussian Free Field. Accord-ing to it, GFFD is a probability measure on functions f in D with 0 boundary conditions with density ρ( f) ∼exp −1 2 ˆ D ||∇f||2 L2 dxdy .
(11.12) The formula (11.12) is based on the well-known computation for the finite-dimensional Gaussian vectors: Exercise 11.10. For a Gaussian vector v ∈RN with probability density ρ(v) = √ detB p (2π)N exp −1 2⟨v,Bv⟩ , the covariance matrix E[vvT] is equal to B−1.
Since the covariance for the GFFD is given by the (minus) inverse Laplace operator, the matrix B appearing in its density should be identified with −△.
Integrating by parts, we then arrive at the formula (11.12). We will repeatedly use (11.12) in the heuristic arguments of the following lectures. However, let us emphasize, that the precise mathematical meaning of (11.12) is very tricky: it is unclear with respect to which underlying measure the density is computed, and also the symbol ||∇f||2 L2 is not well-defined for non-smooth f (and we already know that GFF is not smooth, even its values at points are not properly defined).
98 Lecture 11: Global Gaussian fluctuations for the heights 11.3 Gaussian Free Field in Complex Structures The Gaussian Free Field appearing in Conjecture 11.1 is defined using the complex structure ze−cy rather than the standard complex structure of the plane.
In this section we explain the meaning of this statement in two ways.
For the first approach, recall the results of Theorems 9.8, 10.1 and Lemma 10.4. There exists an analytic function Q(v,w), such that the map (x,y) 7→(ze−cy,(1−z)e−cx) (11.13) is a bijection of the liquid region L with (a part) of the curve Q(v,w) = 0 in C2. This part, being embedded in C2 has a natural local coordinate system on it; in particular, we can define the gradient and the Laplace operator on the curve Q(v,w) = 0 and then use one of the definitions of GFF from the previous section directly on this curve. Then we pullback the resulting field onto the liquid region L using the map (11.13). This is the desired field on Conjecture 11.1.
For the second approach, given the liquid region L and complex slope z = z(x,y) on it we define a class of analytic functions. We give two closely related definitions: Definition 11.11. f : L →C is analytic with respect to ze−cy if for each (x0,y0) ∈L , there exists a holomorphic function g of complex variable, such that f(x,y) = g(ze−cy) (11.14) in a small neighborhood of (x0,y0).
Definition 11.12. f : L →C is analytic with respect to ze−cy if f satisfies the first order PDE: fy z − fx 1−z = 0 (11.15) The complex Burgers equation of Corollary 9.6 implies that ze−cy itself sat-isfies (11.15): ze−cy y z − ze−cy x 1−z = 0.
(11.16) This can be used to show that (11.14) implies (11.15). If ze−cy is locally in-jective, then we can also argue in the opposite direction: (11.15) implies the Cauchy–Riemann relations for the function g in (11.14) and two definitions become equivalent.
All the above approaches accomplish the same goal: they create a structure 99 of a one-dimensional complex manifold (Riemann surface) on the liquid region L . The general Riemann Uniformization Theorem says that all Riemann sur-faces of the same topology are conformally equivalent. In our case it specifies to: Theorem 11.13. There exists a conformal bijection Ω: L 7→D with some domain D ⊂C, where D is equipped with the standard complex structure of C, and the liquid region L is equipped with the complex structure of ze−cy, i.e., Ωis analytic by Definitions 11.11, 11.12.
In other words, Ωis an identification of (part of) the curve {Q(v,w) = 0} ⊂ C2 with a part of the plane C.
Definition 11.14. The Gaussian Free Field in the liquid region L is defined through GFFL (x,y) = GFFD(Ω(x,y)).
This means, that for test functions u on L , its pairings with the Gaussian Free Field ⟨GFFL ,u⟩are jointly Gaussian with covariance E ⟨GFFL ,u⟩⟨GFFL ,v⟩ = E ⟨GFFD,[u◦Ω−1]·JΩ−1⟩⟨GFFD,[u◦Ω−1]·JΩ−1⟩ = ˆ ˆ D×D u◦Ω−1GD(z,z′)v◦Ω−1JΩ−1(z)JΩ−1(z′)dxdydx′ dy′, (11.17) where JΩ−1 is the Jacobian of the map Ω−1 and z = x+iy, z′ = x′ +iy′.
Let us emphasize that under the map Ω, GFF is transformed as a function (rather than a distribution or measure); this again indicates that the test func-tions u should be treated as measures u(x+iy)dxdy.
One might be worried that the map Ωin Theorem 11.13 is not unique, as it can be composed with any conformal bijection φ : D 7→D′. However, the conformal invariance of the Gaussian Free Field of Exercise 11.8 guarantees that the distribution of GFFL does not depend on the choice of Ω.
Finally, note that (11.4) differs from (11.6) by a factor of π. This is not a typo, but rather a universal constant appearing in tilings. We will see this constant appearing in the computation in the next section — the conceptual reasons for the particular value of the constant for random lozenge tilings are not completely clear to us at this point. The same factor π appears in [KenyonOkounkovSheffield03, Theorem 4.5, see also Section 5.3.2] as a uni-versal normalization prefactor for random tilings with periodic weights.
Lecture 12: Heuristics for the Kenyon-Okounkov conjecture This lecture focuses on a heuristic argument for the Kenyon-Okounkov con-jecture on the convergence of the centered height function to the Gaussian Free Field. We concentrate only on the case c = 0, i.e., the uniform mea-sure, here. By the variational principle [CohnKenyonPropp00], in the limit most height functions will be close to the unique maximizer h∗, which we have pinned down (somewhat) explicitly in Lecture 9 by showing that the nor-malized shifted height function ˜ h∗satisfies ( ˜ h∗x, ˜ h∗y) = 1 π (arg(z),−arg(1−z)) where z = z(x,y) satisfies the transformed Burgers equation − zx 1−z + zy z = 0.
(12.1) The Kenyon-Okounkov conjecture focuses on the fluctuations about this limit shape and we repeat it now in the c = 0 case.
Conjecture 12.1 ([KenyonOkounkov05], [Kenyon04]). Take the same setup as in Theorem 5.15: let R∗be a domain in R2 with piecewise-smooth boundary ∂R∗and a specified boundary height function hb. Take a sequence of domains RL for a sequence L →∞, such that ∂RL L →∂R∗and the boundary height func-tions converge as well. Now for each L consider the uniform probability mea-sure on tilings of RL. Finally, let H(Lx,Ly) be the value of the height function of a random tiling of RL at (Lx,Ly). Then for the points (x,y) ∈R∗in the liquid region, as L →∞, the centered heights H(Lx,Ly) −E[H(Lx,Ly)] converge in distribution to the Gaussian free field in complex structure given by1 z.
Remark 12.2. Let us clarify the normalizations. If we choose the (modi-fied) definition of the height function as in (9.3), (9.4), then √π(H(Lx,Ly) − 1 See previous lecture for a precise description of what this means.
100 101 E[H(Lx,Ly)]) should asymptotically behave as GFF of Definition 11.14 with short-scale behavior of the covariance as in (11.6).
From here on, we will denote the liquid region by L . Recall from previous lectures (see Theorem 8.4) that for large L, for a height function H on RL, Pr(H) ≈exp L2 ¨ L S ∇H L dxdy (12.2) (note that Pr(H) is really describing not a probability but a probability density at H in the space of height functions, i.e. the probability of being close to H).
Since we would like to study limiting fluctuations about h∗, we can let H = L h∗+L−1g , where g is a random function supported in the liquid re-gion which represents the fluctuations about the limiting height function h∗.
Plugging this H into (12.2) yields Pr(h∗+L−1g) ≈exp L2 ¨ L S(∇h∗+L−1∇g)dxdy .
(12.3) We now Taylor expand the above integrand f(L−1) := S(∇h∗+ L−1∇g) with respect to the variable L−1, f(L−1) = f(0)+ f ′(0)L−1 + 1 2 f ′′(0)L−2 +(lower order terms).
Using the chain rule, we have f(0) = S(∇h∗), f ′(0) = S1(∇h∗)gx +S2(∇h∗)gy, where S1,S2 are the partial derivatives of S with respect to its first and second coordinate, viewing S as a function of two arguments h∗ x,h∗ y–the two coordi-nates of ∇h∗. We also have f ′′(0) = S11(∇h∗)(gx)2 +2S12(∇h∗)gxgy +S22(∇h∗)(gy)2.
The first term yields an L-dependent normalizing constant exp L2 ˜ L S(∇h∗) which does not affect the probabilities. Because H = Lh∗minimizes ¨ R∗S ∇H L dxdy, it follows that d dt t=0 ¨ S(∇(h∗+tg))dxdy = 0; (12.4) this is exactly what was used to derive the Euler-Lagrange equations. Hence the integral of the f ′(0)L−1 term vanishes. The L−2 in the quadratic term cancels 102 Lecture 12: Heuristics for the Kenyon-Okounkov conjecture with the L2 outside the integral in (12.3), yielding a nonzero probability density function at g.
Thus Pr(h∗+L−1g) ≈exp L2 ¨ L S(∇h∗+L−1∇g)dxdy = exp 1 2 ¨ L S11(∇h∗)(gx)2 +2S12(∇h∗)gxgy +S22(∇h∗)(gy)2 dxdy +(lower order) !
.
(12.5) Having completed this heuristic derivation, we can now (rigorously) show that the right-hand side of (12.5) is exactly what we would get if the height func-tion fluctuations converged to a Gaussian free field. This is the content of the following lemma.
Lemma 12.3. With the setup above, we have ¨ L S11(∇h∗)(gx)2 +2S12(∇h∗)gxgy +S22(∇h∗)(gy)2 dxdy = −2πi ¨ L dg dz dg d¯ z dz∧d¯ z.
(12.6) Before proving this lemma, we show why when combined with (12.5) it shows that the fluctuations converge to a Gaussian free field. Recall from (11.12) in Lecture 11 (with additional π–factor, as discussed at the end of that lecture) that the Gaussian free field on a domain D may be thought of as a probability measure on functions on D which vanish on ∂D, with density ρ(f) ∼exp −π 2 ¨ D ||∇f||2d ˜ xd ˜ y .
(12.7) Since dz = d ˜ x+id ˜ y and d¯ z = d ˜ x−id ˜ y, dz∧d¯ z = −2id ˜ x∧d ˜ y.
(12.8) Similarly, dg dz = 1 2 ∂g ∂˜ x −i∂g ∂˜ y , and dg d¯ z = 1 2 ∂g ∂˜ x +i∂g ∂˜ y , (12.9) so that dg dz dg d¯ z = 1 4||∇g||2.
(12.10) Making these substitutions in Lemma 12.3 and combining with (12.5), we have 103 ℜ(z) ℑ(z) 0 1 z πp2 |1 −z| |z| πp1 πp3 Figure 12.1 Correspondence between complex slope z and local proportions of lozenges identified with angles of a triangle.
that the probability density on height function fluctuations g converges to the one in (12.7). This completes the heuristic argument for Gaussian free field fluctuations; the remainder of this lecture will be spent proving Lemma 12.3.
Proof of Lemma 12.3 First, recall from Lecture 9 that S1(p1, p2, p3) = ln|1−z|, (12.11) S2(p1, p2, p3) = ln|z|, (12.12) where z is the complex number encoding the probabilities p1, p2, and p3 = 1 −p1 −p2, which we use as shorthand for the three lozenge probabilities p (·), p (·), and p (·).
From Figure 12.1 clearly z = |z|eiπ p1, and by the law of sines |z| sinπ p2 = 1 sinπ(1−p1 −p2) = 1 sinπ(p1 + p2).
Therefore, z = sinπ p2 sinπ(p1 + p2)eiπ p1.
It follows that S12 = ∂ln|z| ∂p1 = ∂ ∂p1 (ln(sinπ p2)−ln(sinπ(p1 + p2))) = −π cotπ(p1 + p2), S22 = ∂ln|z| ∂p2 = π cotπ p2 −π cotπ(p1 + p2), S11 = π cotπ p1 −π cotπ(p1 + p2), 104 Lecture 12: Heuristics for the Kenyon-Okounkov conjecture where the formula for S11 follows by symmetry of p1 and p2 from the formula for S22.2 Recall that g is a function of x,y and also of (z, ¯ z), hence, ∂g ∂x = dg dz zx + dg d¯ z ¯ zx, (12.13) ∂g ∂y = dg dz zy + dg d¯ z ¯ zy.
(12.14) Therefore, combining the equations above, we have S11(gx)2 +2S12gxgy +S22(gy)2 = π −cotπ(p1 + p2) dg dz (zx +zy)+ dg d¯ z (¯ zx + ¯ zy) 2 +cotπ p1 dg dz zx + dg d¯ z ¯ zx 2 +cotπ p2 dg dz zy + dg d¯ z ¯ zy 2 !
, (12.15) which is a quadratic form in dg dz and dg d¯ z . We split our further calculation into three claims.
Claim 12.4. The coefficient of dg dz 2 in (12.15) is 0.
Proof The coefficient is π(−(zx +zy)2 cotπ(p1 + p2)+z2 x cotπ p1 +z2 y cotπ p2).
(12.16) Since zy z −zx 1−z = 0 by (12.1) we may write zx = 1−z z zy in the above, yielding πz2 y −1 z2 cotπ(p1 + p2)+ 1−z z 2 cotπ p1 +cotπ p2 !
.
(12.17) Substituting z = sinπ p2 sinπ(p1 + p2)eiπ p1, 1−z = sinπ p1 sinπ(p1 + p2)e−iπ p2 (12.18) (the latter of which is easy to see by a similar law of sines argument on Figure 2 Explicit formulas make is straightforward to show that det S11 S12 S21 S22 = π2, which actually holds for the surface tensions in greater generality, see [KenyonOkounkov03, Section 2.2.3].
105 12.1) to (12.17) yields πz2 y z2 sin2 π(p1 + p2) −cosπ(p1+ p2)sinπ(p1+ p2)+e−2πip2 cosπ p1 sinπ p1 +e2πip1 cosπ p2 sinπ p2 = πz2 y 2z2 sin2 π(p1 + p2)(−sin2π(p1+ p2)+e−2πip2 sin2π p1+e2πip1 sin2π p2).
Splitting both complex exponentials into real and imaginary parts, the imagi-nary parts on the second and third of the above terms cancel, and the real part is −sin2π(p1 + p2)+sin2π p1 cos2π p2 +cos2π p1 sin2π p2 = 0 by the identity for sin(A+B). This proves the claim.
Claim 12.5. The coefficient of dg d¯ z 2 in (12.15) is 0.
Proof The proof is exactly the same as for Claim 12.4.
Claim 12.6. The coefficient of dg dz dg d¯ z in (12.15), multiplied by dxdy as in the integral in (12.6), is equal to −2i dg dz dg d¯ z dz∧d¯ z.
Proof First, since dz = zxdx+zydy and d¯ z = ¯ zxdx+ ¯ zydy, dz∧d¯ z = (zx¯ zy −¯ zxzy)dx∧dy = zy¯ zy z¯ z (¯ z−z)dx∧dy, (12.19) where in the last step we use (12.1) to get zx¯ zy −¯ zxzy = zy¯ zy 1−z z −1−¯ z ¯ z = zy¯ zy z¯ z (¯ z−z).
Thus it suffices to show the following: π(−cotπ(p1 + p2)·2(zx +zy)( ¯ zx + ¯ zy)+cotπ p1 ·2zx ¯ zx +cotπ p2 ·2zy ¯ zy) ?
= −2πizy¯ zy z¯ z (¯ z−z).
(12.20) Again using zx = 1−z z zy to get a common factor of zy ¯ zy, and multiplying (12.20) by z¯ z and dividing by 2πzy ¯ zy, we see (12.20) is equivalent to −i(¯ z−z) ?
= −cotπ(p1 + p2)+(1−z)(1−¯ z)cotπ p1 +z¯ zcotπ p2.
(12.21) Substituting (12.18), the LHS of (12.21) equals 2sinπ p1 sinπ p2 sinπ(p1 + p2) (12.22) 106 Lecture 12: Heuristics for the Kenyon-Okounkov conjecture and the right-hand side equals −cotπ(p1 + p2)+ sin2 π p1 sin2 π(p1 + p2) cotπ p1 + sin2 π p2 sin2 π(p1 + p2) cotπ p2.
(12.23) Multiplying through by sin2 π(p1 + p2), we have that (12.20) is equivalent to 2sinπ(p1 + p2)sinπ p1 sinπ p2 ?
= −cosπ(p1 + p2)sinπ(p1 + p2) +sinπ p1 cosπ p1 +sinπ p2 cosπ p2.
Moving all terms with sinπ(p1 + p2) to one side, this is the same as sinπ(p1 + p2)(cosπ(p1 + p2)+2sinπ p1 sinπ p2) ?
= sinπ p1 cosπ p1 +sinπ p2 cosπ p2.
Using the sine double angle identity and the formula cos(A + B), the desired identity becomes sinπ(p1 + p2)(cosπ p1 cosπ p2 −sinπ p1 sinπ p2 +2sinπ p1 sinπ p2) ?
= 1 2 sin2π p1 + 1 2 sin2π p2.
(12.24) Since cosπ p1 cosπ p2 −sinπ p1 sinπ p2 +2sinπ p1 sinπ p2 = cosπ p1 cosπ p2 +sinπ p1 sinπ p2 = cosπ(p1 −p2), (12.24) becomes 2sinπ(p1 + p2)cosπ(p1 −p2) ?
= sin2π p1 +sin2π p2.
(12.25) This is true by simple trigonometry. We proved Claim 12.6 and hence Lemma 12.3.
Exercise 12.7. Repeat the heuristic derivation of the GFF–fluctuations for general c ̸= 0, i.e. for the asymptotic of the qVolume–weighted lozenge tilings.
Remark 12.8.
One expects that a similar computation (leading to GFF heuristics) can be done for dimers with periodic weights of [KenyonOkounkovSheffield03], however, this is not explicitly present in the literature as of 2021. The arguments of [BorodinToninelli18, Section 3] might turn out to be helpful for simplifying computations for such generalizations.
107 Remark 12.9. Can we turn the heuristics of this lecture into a rigorous proof of convergence of fluctuations to the GFF? This is very difficult and has not been done. In particular, the very first step (12.2) had o(L2) error in the exponent, while we would need o(1) error for a rigorous proof.
Lecture 13: Ergodic Gibbs translation-invariant measures So far we were discussing global limits of random tilings. In other words, we were looking at a tiling of a large domain from a large distance, so that its rescaled size remained finite. We now start a new topic of local limits. The aim is to understand structure of a random tiling of a large domain near a given point and at the local scale so that we can observe individual lozenges.
We consider a uniformly random tiling of a domain of linear size L and would like to understand the limit as L →∞of the probability measure on tilings near (0,0), cf. Figure 13.1. There are two basic questions: 1 How do you think mathematically about the limiting object?
2 What are all possible objects (probability measures?) appearing in the limit.
13.1 Tilings of the plane As our domain becomes large, its boundary is no longer visible on the local scale and we deal with lozenge tilings of the whole plane.
Let us identify a tiling with a perfect matching of black and white vertices of the infinite regular hexagonal lattice (cf. Figure 13.2). Let us denote the set of all perfect matchings through Ω.
By its nature, Ωis a subset of the set of all configurations of edges on the lattice, Ω⊂2edges. If we equip 2edges with the product topology and corre-sponding Borel σ–algebra, then Ωis a measurable subset.
Hence, we can speak about probability measures on Ω, which are the same as probability measures P on the Borel σ-algebra of 2edges, such that P(Ω) = 1.
We describe measures on Ωthrough their correlation functions.
108 109 (0, 0) L →∞ Figure 13.1 Random tiling of a large domain seen locally near the point (0,0).
Definition 13.1. For n = 1,2,..., the n-th correlation function of a probability measure P on Ωis a function of n edges of the grid: ρn(e1,e2,...,en) = P(e1 ∈random matching, ..., en ∈random matching).
Lemma 13.2. The collection of all correlation functions ρn, n = 1,2,...,, uniquely determines the measure P.
Proof By the Caratheodory theorem, any probability measure on 2edges (and hence, on Ω) is uniquely fixed by probabilities of cylinders, which are proba-bilities of the kind P(a collection of edges ∈matching, another collection of edges / ∈matching).
These probabilities can be expressed as finite signed sums of the correlation functions by using the inclusion-exclusion principle.
Exercise 13.3. Consider the following probability measure P: we choose one of the three types of lozenges , , with probability 1/3 each and then take the (unique) tiling of the plane consisting of only the lozenges of this type.
Compute correlation functions of P.
110 Lecture 13: Ergodic Gibbs translation-invariant measures Figure 13.2 Infinite regular hexagonal graph.
13.2 Properties of the local limits When a measure P on Ωappears as a local limit of uniform measures on tilings of planar domains, it is not an arbitrary measure. Let us list the main properties that we expect such measure to have.
Definition 13.4. A measure P on Ωis called Gibbs, if for any finite subdomain R (=collection of vertices), the conditional measure on perfect matching of R given that all vertices of R are matched with vertices of R, but not outside (and in addition one can condition on any event outside R), is the uniform measure.
The Gibbs property is a way to say that a measure is uniform for the situation when the state space Ωis infinite. Clearly, uniform measure on tilings of any finite domain is Gibbs, as the conditionings do not change uniformity. Hence, we expect every local limit of tilings also to be Gibbs.
Definition 13.5. A measure P on Ωis called translation–invariant, if the cor-relation functions are unchanged upon translations of all arguments by a same vector.
Why do we expect local limits to be translation–invariant? This property should be treated as some kind of a continuity statement. In general, the local limit might depend on the macroscopic zoom-in position inside the domain.
However, if we believe that this dependence is continuous, then a shift by a finite vector should become negligible in the large domain limit, and, hence, should not change the local limit.
111 Let us take two Gibbs translation–invariant measures µ1 and µ2. For every 0 < α < 1 the (convex) linear combination αµ1 + (1 −α)µ2 is also a Gibbs translational–invariant measure. Hence, such measures form a convex set.
Definition 13.6. A Gibbs translation–invariant measure is called ergodic, if it is an extreme point of the set of all such measures, i.e. if it can not be de-composed into convex linear combination of two distinct measures from this class.
We abbreviate EGTI for ergodic Gibbs translation–invariant measures.
Lemma 13.7. If P is an EGTI probability measure, and A is a translation– invariant event, then either P(A) = 0 or P(A) = 1.
Remark 13.8. By definition, a translation–invariant event in Ωis a measurable subset of Ω, which is preserved under shifts of the entire grid. For example, if Rn is a growing sequence of convex domains (say, rectangles) exhausting the plane, then limsup n→∞ number of horizontal edges of matching inside Rn area of Rn ≤1 7 is a translationa–invariant event.
Proof of Lemma 13.7 We argue by contradiction. Assume that 0 < P(A) < 1 and let ¯ A be the complementary event. Define P1 to be the normalized restric-tion of P onto the set A, i.e., P1(B) = P(A∩B) P(A) .
Also define P2 to be the normalized restriction of P onto the set ¯ A, i.e., P2(B) = P( ¯ A∩B) P( ¯ A) .
Then both P1 and P2 are Gibbs translation–invariant measures and P(·) = P(A)P1(·)+P( ¯ A)P2(·).
Therefore, P is not ergodic.
Exercise 13.9. Show that the measure P of Exercise 13.3 is Gibbs and translation–invariant, but not ergodic. Represent it is as a convex linear com-bonation of EGTI measures.
112 Lecture 13: Ergodic Gibbs translation-invariant measures 13.3 Slope of EGTI measure Let us introduce an important numeric characteristic of an EGTI measure.
Definition 13.10. Slope of an EGTI probability measure on Ωis a triplet (p , p , p ) of non-negative numbers summing up to 1, defined in either of two equivalent ways: 1 For an arbitrary triangle of the grid (=vertex of the hexagonal lattice), p , p , p are probabilities that this trangle belongs to a lozenge of a corre-sponding type (is matched to one of the three corresponding adjacent ver-tices of another color).
2 If Rn is a growing sequence of convex domains (say, rectangles) exhausting the plane, then p = lim n→∞ number of triangles inside Rn covered by lozenges of type number of triangles inside Rn , (13.1) and similarly for p and p .
Remark 13.11. The probabilities in the first definition of the slope do not de-pend on the choice of the triangle due to the translation invariance of the mea-sure.
The limits in (13.1) exist almost surely, which is a property that needs to be proven. (If we know that the limits exist, they have to be deterministic due to ergodicity.) One way to prove that the limits in (13.1) exist in probability (rather than the stronger condition of being almost sure limits) is by bounding the variance of the right-hand side, using the explicit description of the EGTI measures given later in this lecture.
As soon as we know that a deterministic limit in (13.1) exists, it has to be equal to p from the first definition, as this is the expectation of the right-hand side in (13.1).
The possible slopes (p , p , p ) form a triangle: p > 0, p > 0, p > 0, p + p + p = 1. We say that the slope is extreme, if one of the proportions of lozenges, p , p , or p is zero, and non-extreme otherwise.
Theorem 13.12 ([Sheffield03a]). For each non-extreme slope there exists a unique EGTI probability measure of such slope.
Idea of the proof Suppose that there are two different EGTI measures of the same slope and let π1, π2 be corresponding random perfect matchings from Ωsampled independently. Superimpose π1 and π2, i.e. consider the union of all edges of the hexagonal lattice appearing in either of the matchings. Then 113 π1 ∪π2 is a collection of double edges (if the same edge belonged to both π1 and π2), finite cycles (of alternating edges from π1 and π2), and infinite curves extending to infinity (again of alternating edges from π1 and π2).
Claim. Almost surely, π1 ∪π2 has double edges and finite cycles, but no infinite curves.
The claim is a difficult statement and we refer to [Sheffield03a] for the proof.
Note that for measures of different slopes it no longer holds.
Given the claim, let us show that the distributions of π1 and π2 coincide.
For that we demonstrate a sampling procedure, which is the same for both measures.
Note that given a cycle of π1 ∪π2, half of its edges belong to π1 and another half belongs to π2. There are two options: either all even edges belong to π1 (and all odd edges belong to π2) or all even edges belong to π2. Applying Gibbs property to this cycle separately for π1 and for π2, we see that these two options arise with equal probability 1 2. Hence, we can sample π1 by the following procedure: first sample π1 ∪π2, and then flip an independent coin for each cycle to decide which of its edges belong to π1. Since the procedure for π2 is the same, we are done.
Theorem 13.12 has an interesting corollary. First, it explains why we ex-pect local limits of random tilings to be ergodic: indeed, we want them to have fixed slopes matching the slope of the global macroscopic limit shape. This ar-gument can be also reversed: if we knew a priori that the local limits of random tilings should be given by ergodic Gibbs translation invariant measures, then Theorem 13.12 would uniquely fix such measure. Unfortunately, in practice, checking the EGTI property is hard, with, perhaps, the most complicated part being translation invariance.
Nevertheless, it was conjectured in [CohnKenyonPropp00, Conjecture 13.5] that all the limits are EGTI, which was recently proven in [Aggarwal19].
Theorem 13.13 ([Aggarwal19]). Locally, near any point of a (proportionally growing) domain with non-extreme slope of the limit shape, the uniform mea-sure on tilings converges to EGTI measure of the same slope.
In Lecture 17 we present a partial result in the direction of Theorem 13.13 from [Gorin16] after some preparations in Lectures 14, 15, and 16. The proof of the general case in [Aggarwal19] starts with the same approach and then adds to it several additional ideas which we are not going to present in these lectures.
114 Lecture 13: Ergodic Gibbs translation-invariant measures 13.4 Correlation functions of EGTI measures Theorem 13.13 provides a simple way to identify the EGTI measures for each slope. We need to take any domain where such slope appears and compute the local limit. In fact, we already made such computation when analyzing tilings on the torus in Lecture 4 and the result is given by Theorem 4.11 there. Let us recast the result here.
Corollary 13.14. For an EGTI measure of slope (p , p , p ), the nth correlation function computing the probability of observing the edges (x1,y1, ˜ x1, ˜ y1),...,(xn,yn, ˜ xn, ˜ yn) between white vertices (xi,yi) and black ver-tices (˜ xi, ˜ yi) in a random perfect matching is ρn((x1,y1, ˜ x1, ˜ y1),...,(xn,yn, ˜ xn, ˜ yn)) = n ∏ i=1 K00(xi,yi, ˜ xi, ˜ yi) det 1≤i, j,≤n( ˜ Kα,β[ ˜ xi −xj, ˜ yi −yj]) (13.2) where ˜ Kα,β(x,y) = " |z|=|w|=1 wxz−y 1+eαz+eβw dw 2πiw dz 2πiz, we use the coordinate system of Figure 3.2, K00(xi,yi, ˜ xi, ˜ yi) are weights from that figure, and the correspondence between (p , p , p ) and (α,β) is as in Theorem 6.1.
Remark 13.15. At this point the Gibbs property of the measure might seem to be very far away from the determinantal formulas for the correlation func-tions. However, the Gibbsianity can actually be directly seen from (13.2), as explained in [BorodinShlosman08].
Let us get comfortable with formula (13.2) by computing the correlation functions of horizontal lozenges along a vertical line. This corresponds to taking ˜ xi = 1, xi = 0 and yi = ˜ yi. Hence, absorbing the K0,0 prefactor (which is eβ in this case) into the determinant, we conclude that the correlation functions of horizontal lozenges at positions yi are minors of the matrix (yi,yj) 7→ " |z|=|w|=1 eβwzyi−y j 1+eαz+eβw dw 2πiw dz 2πiz, (13.3) We further argue as in Section 6.1, and compute the w–integral as a residue at the unique pole of the integrand to get 1 2πi ˆ z0 ¯ z0 zyi−yj−1dz = z yi−y j 0 −¯ z yi−y j 0 2πi(yi −yj) = |z0|yi−y j π(yi −yj) sin Arg(z0)(yi −yj) .
115 Note that the |z0|yi−y j prefactor cancels out when we compute the minors. By looking at yi −yj = 0 case, we also match Arg(z0) with π p . Hence, we reach a conclusion.
Corollary 13.16. For the EGTI probability measure of slope (p , p , p ) P( at positions (0,y1),...,(0,yn)) = det[K(yi −y j)]n i, j=1, with K(y) = sin(π p y) πy , y ̸= 0, p , y = 0.
(13.4) Remark 13.17. Rotating by 120 degrees, we can also get similar statements for the and lozenges — their distributions along appropriate sections will be described by (13.4) with parameter p replaced by p and p , respectively.
Exercise 13.18. Perform the computation of Remark 13.17 in details.
The kernel of (13.4) has the name discrete sine kernel and the point process of Corollary 13.16 is called the discrete sine process.
The continuous counterpart of the sine process (when y’s become real num-bers rather than integers) is a universal object in the random matrix theory1, where it describes local limits for the eigenvalues in the bulk of the spectrum of large complex Hermitian matrices, see [Kuijlaars11, TaoVu12, Lubinsky16, ErdosYau17] for recent reviews. Conjecturally, sine process also appears in many other settings such as large eigenvalues of Laplace operator in various domains or even spacings between zeros of Riemann zeta-function.
13.5 Frozen, liquid, and gas phases An important property of the EGTI measure with non-degenerate slope (p , p , p ) satisfying p , p , p > 0 is the polynomial decay of the correla-tions between lozenges with distance. This can be directly seen from the result of Corollary 13.14, since ˜ Kα,β(x,y) decays polynomially in x and y. This type of EGTI measures is called the liquid phase in [KenyonOkounkovSheffield03] with an alternative name being the rough unfrozen phase.
Extreme slopes (when either p , or p , or p vanishes) lead to very different structure of the EGTI measure: the random tiling becomes degenerate and has non-decaying correlations. This is the frozen phase.
1 We return to the connection between tilings and random matrix theory in Lectures 19 and 20.
116 Lecture 13: Ergodic Gibbs translation-invariant measures 1 1 1 1 1 1 1 1 1 1 1 α α α α 1 1 1 Figure 13.3 3×2–periodic weights used in Figure 13.4 with α = 10.
For uniformly random lozenge tilings the liquid and frozen phases exhaust the list of possible types of EGTI measures appearing in local limits. How-ever, we can allow more general non-uniform weights of tilings. One natural choice of non-uniformity is by declaring that whenever a cube is being added to the stepped surface encoding tiling, i.e. whenever one tiling differs from another one by increase by 3 of the value of the height function (in terminol-ogy of Lecture 1) at a single point (x,y), the weight of the tiling is multiplied by w(x,y). If we require w(x,y) to be a periodic function of x and y, then we arrive at periodically weighted lozenge tilings. In particular, if the period is 1 and w(x,y) = q, then these are the qVolume–weighted tilings, which we have al-ready encountered in the previous lectures. A simpler, but slightly less general way to introduce periodic weighting is by putting the weights on edges of the hexagonal grid for perfect matchings, as we did in Lectures 3 and 4 (cf. Fig-ure 3.2). The second approach does not allow one to obtain qVolume–weighting with q ̸= 1; but modulo this fact the two approaches are equivalent.
[KenyonOkounkovSheffield03] produced a complete classification of possi-ble EGTI measures for periodically weighted tilings. In particular, they have shown that for certain choices of weights a third type of EGTI measures appears. They called these new measures the gas phase, with an alternative name being the smooth unfrozen phase. In this phase the correlations between lozenges decays exponentially and the height function is almost linear with small isolated defects. [KenyonOkounkovSheffield03] constructed EGTI mea-sures by a limit transition from the torus, similarly to what we did in Lecture 4.
Obtaining the gas phase as a local limit of tilings of planar polygonal domains is a more complicated task. It was first achieved for 2 × 2–periodic domino 117 Figure 13.4 A 3×2–periodic random lozenge tiling of 100×100×100 hexagon.
Two gas phases are visible adjacent to top and bottom green frozen regions. Let us note that asymptotically a gas phase cannot be adjacent to a frozen phase (this would contradict [DeSilvaSavin08, Theorem 4.1], which says that discontinuity of lozenge densities can happen only at boundary slopes), hence, there is, in fact, a very narrow liquid region separating the gas and frozen phases. (I thank Christophe Charlier and Maurice Duits for the simulation.) tilings of the Aztec diamond in [ChhitaYoung13, ChhitaJohansson14]. A sys-tematic approach to periodic situations was introduced in [DuitsKuijlaars17] and there is a hope that the gas phase in periodically weighted lozenge tilings of polygons can be rigorously studied using this approach, yet, it was not done at the time when this book was finished.
Figure 13.4 shows a simulation2 of 3 × 2–periodically weighted lozenge tilings of the hexagon with weights assigned to edges in a periodic way with fundamental domain shown in Figure 13.3, see also [BorodinBorodinA] for a 3d virtual reality version of the same simulation. As a remark, the results 2 It was produced using an adaptation of the algorithm from [Propp01].
118 Lecture 13: Ergodic Gibbs translation-invariant measures Figure 13.5 A staircase–like frozen phase with two types of lozenges.
of [KenyonOkounkovSheffield03] imply that 3 × 2 is the minimal size of the fundamental domain, which makes the appearance of a gas phase possible.
The phenomenology for periodically weighted tilings becomes richer than for the uniform case, which we mostly study in this book and we refer to [Mktchyan19, CharlierDuitsKuijlaarsLenells19, BerggrenDuits19, Charlier20, BeffaraChhitaJohansson20] for some results. Just as an example, while for the uniformly random lozenge tilings we could observe only three frozen phases corresponding to 3 types of lozenges, for 2 × 2 periodic case another frozen phase is possible, in which densities of two types of lozenges are equal to 1/2 and the corresponding stepped surface looks like a staircase, see Figure 13.5.
Lecture 14: Inverse Kasteleyn matrix for trapezoids Our goal over the next few lectures is to discuss a local limit theorem for the trapezoid (or sawtooth) domain of Figure 14.1. It is so named because of the “dents” along its right edge. We are going to prove that the local structure of uniformly random lozenge tilings for such domains is asymptotically governed by ergodic Gibbs translation–invariant (EGTI) measures introduced in the last lecture.
Why should we study this particular class of domains?
• The sawtooth domains are integrable in the sense that they have explicit inverse Kasteleyn matrix; • This type of domains has links to representation theory. In particular, con-sider a sawtooth domain with width N and dents at locations t1 > ··· > tN.
Writing λj = tj −(N −j), we see that λ1 ≥··· ≥λN. Each irreducible rep-resentation of the unitary group U(N) has a signature, and it turns out that the number of tilings of the sawtooth domain is given by the (vector space) dimension of the representation having signature (λ1,...,λN), and by the Weyl dimension formula this is given by ∏i<j (λi−i)−(λj−j) j−i . We come back to this computation in Lecture 19, see Proposition 19.3.
• The boundary of every domain locally looks like a trapezoid; eventually, the results for sawtooth domains and their extensions can be used as a building block for proving the local limit theorem for general domains, see [Aggarwal19] and [Gorin16].
In Lecture 3 (see Theorem 3.5) we expressed the correlation functions of ran-dom tilings through the inverse Kasteleyn matrix. The goal of this lecture is to compute this inverse matrix explicitly, which will be the main tool for our analysis in the subsequent lecture.
Following the notations of [Petrov12a] (which this lecture is based on), we 119 120 Lecture 14: Inverse Kasteleyn matrix for trapezoids Figure 14.1 Left: A trapezoid/sawtooth domain. Right: Observe that if the width of the domain is N, then the domain is tileable by lozenges if and only if its right border has N dents. Indeed, it is clear by inspection that the first layer has exactly one horizontal lozenge, the second has two, and so forth.
apply a rotation and skew transformation to the sawtooth domain, as in Fig-ure 14.2. This is helpful, as the triangles of the grid are now parameterized by pairs of integers, the correspondence is shown in the bottom part of Figure 14.2.
The Kasteleyn (adjacency) matrix of the dual graph of triangles, restricted to the sawtooth domain is as follows: if (x,n) and (y,m) are both inside the domain, then K( (x,n); (y,m)) = 1, if (y,m) = (x,n), 1, if (y,m) = (x,n−1), 1, if (y,m) = (x+1,n−1), 0, otherwise.
If one of the trigangles (x,n) or (y,m) is outside the domain, then K( (x,n); (y,m) vanishes.
The following theorem is the main result of this lecture. We use the Pochhammer symbol (u)a = u(u+1)···(u+a−1).
Theorem 14.1 ([Petrov12a, Theorem 1]). As in Figure 14.2, consider a saw-tooth domain with height N and dents at (t1,N),...,(tN,N). The inverse Kaste-121 n N 0 x tN · · · · · · t1 Figure 14.2 Top: After rotating and shearing the diagram, all lattice lines are either grid-aligned or at a 45◦angle. We set coordinates (x,n), where x is the horizontal coordinate and n the vertical. We only care about x modulo additive constant so we don’t specify it fully, and n starts at 0 on the bottom boundary and increases to N at the top boundary. We understand the domain to be the sawtooth region with its N dents removed (white portion of figure). Bottom left: There are now two types of lattice triangles. We place a root on the horizontal edge of each triangle, and define the coordinate of the triangle to be the coordinate of its root. Bottom right: There are three types of lozenges we can make out of triangles (x,n) and (y,m). From left to right: (y,m) = (x,n), (y,m) = (x,n−1), (y,m) = (x+1,n−1).
leyn matrix K−1( (y,m); (x,n)) is given by K−1( (y,m); (x,n)) = (−1)y−x+m−n+11m<n1y≤x (x−y+1)n−m−1 (n−m−1)!
(14.1) +(−1)y−x+m−n (2πi)2 (N −n)!
(N −m−1)!
˛ {y,...,t1} ˛ ∞ dwdz w−z (z−y+1)N−m−1 (w−x)N−n+1 N ∏ r=1 w−tr z−tr , (14.2) where the z–contour, Cz, is a loop around {y,y + 1,...,t1} (and enclosing no poles of the integrand outside this set) and the w–contour, Cw, is a large loop around Cz and all poles of the integrand.
Remark 14.2. We define our domain to be the trapezoid with dents re-122 Lecture 14: Inverse Kasteleyn matrix for trapezoids moved, and therefore, the above formula is applicable only for triangles (y,m), (x,n) which lie entirely in the white region of Figure 14.2 (Top).
In particular, we do not allow (x,n) ∈{ (t1,N),..., (tN,N)} as these cor-respond to the dents.
Remark 14.3. Our proof of Theorem 14.1 is through verification and before proceeding to it, let us discuss how the statement can be obtained.
Historically, determinantal formulas for the two-dimensional correlation functions were first deduced by Eynard and Mehta [EynardMehta97] in the study of coupled random matrices, with the particular case given by the Dyson Brownian Motion. The latter is the evolution of the eigenvalues of a Hermi-tian matrix, whose matrix elements evolve in time as independent Brownian motions. Another equivalent definition identifies the Dyson Brownian Mo-tion with N independent Brownian motions conditioned to have no collisions, which makes analogy with tilings in the form of Section 2.2 clear.
Simultaneously, Brezin and Hikami [BrezinHikami96a, BrezinHikami96b] introduced double contour integral formulas for the correlation kernel for the fixed time distribution of the Dyson Brownian Motion started from an arbitrary initial condition.
During the next 15 years, double contour integrals were found in numer-ous settings of random matrices and random tilings. The closest to the context of Theorem 14.1 instances are the formulas for the random plane partitions in [OkounkovReshetikhin01]1 and for continuous Gelfand–Tsetlin patterns in [Metcalfe11].
Thus, the appearance of the double contour integrals in the setting of The-orem 14.1 was not completely unexpected. The arguments in [Petrov12a] (see also [DuseMetcalfe14]) use the Eynard-Mehta theorem to produce determi-nantal formulas and then introduce a nice trick by inverting a matrix, which enters into the correlation kernel, by using contour integrals.
Proof of Theorem 14.1 Throughout the proof, by K−1( (y,m); (x,n)) we mean the expression of (14.1), (14.2). We would like to evaluate K−1K( (y,m); (y′,m′)) for each pair (y,m),(y′,m′), and show that it is the delta function δ(y = y′;m = m′).
Case 1: Triangle (y′,m′) has all its neighbors inside the domain.
We need to show that K−1( (y,m); (y′,m′))+K−1( (y,m); (y′,m′ +1)) +K−1( (y,m); (y′ −1,m′ +1)) ?
= δ(y = y′;m = m′).
(14.3) 1 We will return to this model in Lectures 22-23.
123 We first compute the double integral term (14.2) for the LHS of (14.3), and verify that it is zero. Indeed, the integrand is a multiple of (N −m′)!
(w−y′)N−m′+1 −(N −m′ −1)!
(w−y′)N−m′ + (N −m′ −1)!
(w−y′ +1)N−m′ = (N −m′ −1)!
(w−y′)N−m′+1 (N −m′)−(w−y′ +N −m′)+(w−y′) = 0.
Next, we check that the first term (14.1) after summation in the LHS of (14.3) gives δ(y = y′;m = m′). We have four subcases: • (y,m) = (y′,m′). Then a direct evaluation yields −0 + (1)0 0! −0 = 1, as needed.
• m′ < m. All the indicators are zero, so this evaluates to zero.
• m′ = m,y′ ̸= y. Then the sum reduces to 0+1y≤y′ −1y≤y′, which is zero.
• m′ > m. In this case, 1m y′, all terms are zero since the indicators are zero.
– When y = y′, we indeed get (1)m′−m−1 (m′ −m−1)! −(1)m′−m (m′ −m)! +0 = 0, since (1)a = a!.
– When y < y′ we have (y′ −y+1)m′−m−1 (m′ −m)!
(m′ −m)−(y′ −y+m′ −m)+(y′ −y) = 0, as needed.
This finishes the proof for the case when triangle (y′,m′) has all its neighbors inside the domain. Note that the situations when (y′,m′) is on a vertical boundary of the trapezoid, or on its top boundary, but not adjacent to dents, are included in this case.
Case 2: (y′,m′) lies on the bottom boundary; i.e. m′ = 0.
In this case, we claim that K−1((y,m);(y′,m′)) = 0, so the analysis of the previous case is sufficient. The first term (14.1) is zero since 1m 0 and ∆x, the desired statement can be written as A!
B!
1 2πi ˛ e Cz (z+∆x+1)B (z)A+1 dz = 1∆x≥0 (B−A+1)∆x (∆x)!
, (14.4) where the contour e Cz surrounds the integers −∆x,...,0. We prove (14.4) by induction. We need the following base cases where (14.4) is easy to check: • Case ∆x < 0. Then right-hand side is clearly zero, and the left-hand side is also zero since there are no singularities enclosed by the contour.
• Case ∆x ≥0 and A = B. Then the right-hand side is clearly one, and the integrand of the left-hand side has no poles outside of e Cz, so we may enlarge the contour to infinity, and thus see that the residue is 1 (there’s one more linear in z factor in the denominator than in the numerator).
We also need the following linear relations. Write L(A,B,∆x) and R(A,B,∆x) 125 for the left-hand side and right-hand side of (14.4), respectively. Then we have L(A,B+1,∆x)−L(A,B+1,∆x−1) = L(A,B,∆x), (14.5) R(A,B+1,∆x)−R(A,B+1,∆x−1) = R(A,B,∆x).
(14.6) Indeed, to prove the first of these relations, we observe that the integrand in the left-hand side is equal to that in the right-handside: z+∆x+B+1 B+1 −z+∆x B+1 A!
B!
1 2πi (z+∆x+1)B (z)A+1 = A!
B!
1 2πi (z+∆x+1)B (z)A+1 .
Likewise, we can check the second linear relation (14.6): B−A+1+∆x B−A+1 − ∆x B−A+1 1∆x≥0 (B−A+1)∆x (∆x)!
= 1∆x≥0 (B−A+1)∆x (∆x)!
.
Consequently, if we know that (14.4) holds for two of the three triples (A,B+ 1,∆x), (A,B+1,∆x−1), (A,B,∆x), then it also holds for the third one.
We now prove (14.4) by induction on ∆x. We know that (14.4) holds for all triples (A,B,∆x) with ∆x < 0. Now suppose that for some ∆x ≥0 we know for all A,B that (14.4) holds for (A,B,∆x−1). For any A, we also know that (14.4) holds for the triple (A,A,∆x). Repeatedly using (14.5), (14.6), we extend from B = A to arbitrary B case and conclude that (14.4) holds for all (A,B,∆x).
Exercise 14.5. Consider a macroscopic rhombus or, equivalently, A × B × 0 hexagon (which is a particular case of a trapezoid). It has a unique lozenge tiling and, therefore, the correlation functions for uniformly random tilings of this domain are straightforward to compute. Check that they match the expres-sion given by the combination of Theorems 3.5 and 14.1.
Lecture 15: Steepest descent method for asymptotic analysis 15.1 Setting for steepest descent In this lecture we discuss how to understand asymptotic behavior of the inte-grals similar to that of the correlation function for trapezoids in Theorem 14.1.
A general form of the integral is ˛ Cw ˛ Cz exp N(G(z)−G(w)) f(z,w) dz dw z−w .
(15.1) This sort of integral arises frequently in integrable probability, and as such there is a well developed machinery to deal with it. The key features of (15.1) are: • The integrand grows exponentially with N. In (14.2) this is due to ∏N r=1 product and growing Pochhammer symbols.
• The leading contributions in z and w are inverse to each other, hence G(z)− G(w) in the exponent of (15.1).
• The integrand has a simple pole at z = w.
• Additional factors in the integrand denoted f(z,w) grow subexponentially in N.
We start with simpler examples and work our way towards (15.1).
15.2 Warm up example: real integral Let us compute the large N asymptotics of the real integral ˆ 1 0 exp(N(x−x2)) dx.
126 127 We use the Laplace method. Write f(x) = x−x2; this is a concave function on [0,1] with maximum at 1 2. The integral is dominated by a neighborhood of 1 2 (indeed, exp(N f(x)) is exponentially smaller away from 1 2), so we do a Taylor expansion there: f(x) = 1 4 − x−1 2 2 .
Making the substitution y = √ N(x−1 2), we get ˆ 1 0 exp N 1 4 − x−1 2 2!!
dx = e N 4 1 √ N ˆ √ N/2 − √ N/2 e−y2 dy = e N 4 1 √ N ˆ ∞ −∞ e−y2 dy· 1+o(1) = e N 4 r π N · 1+o(1) .
The conclusion is that the integral is dominated by the neighborhood of the point where f(x) is maximized. This point can be found solving the critical point equation f ′(x) = 0.
Exercise 15.1. Prove the Stirling’s formula by finding the large N asymptotics of the integral N! = ˆ ∞ 0 xN exp(−x)dx.
(Hint: start by changing the variables x = Ny.) 15.3 One-dimensional contour integrals For the next step we take K = αN and study for large N the binomial coefficient N K = 1 2πi ˛ (1+z)N zK+1 dz, where the contour encloses the origin. (Since (1+z)N zK+1 has only a single pole at z = 0, any contour surrounding 0 gives the same result.) Inspired by the previous case, we might think that the integral is dominated by a neighborhood of the maximizer of |1+z|N/|z|K+1 on the integration con-tour. This turns out to be false; indeed, if this worked, one could vary the contour and get a different answer! The problem is that there could be huge oscillations in the complex argument (i.e., angle) of the integrand, causing can-cellations. We generalize the previous approach in a different way.
128 Lecture 15: Steepest descent method for asymptotic analysis We write f(z) = ln(1+z)−α lnz, so that the integral can be written as 1 2πi ˛ exp(N f(z))dz z .
Notice that f ′(z) = 1 1+z −α z Hence, f has a unique critical point zc = α 1−α and f ′′(zc) = (1−α)3 α > 0.
f admits the Taylor expansion in a neighborhood of zc f(z) = f(zc)+ 1 2 f ′′(zc)(z−zc)2 +....
Notice that along a vertical segment through zc, since (z −zc)2 < 0, the real part of f(z) is maximized at zc (it would have been minimized for the horizon-tal contour). Thus, we want to choose a contour through zc which is locally vertical at zc. But how do we specify the contour globally? The trick is to use the level lines of ℑf — imaginary part of the complex function f.
Lemma 15.2. Take a function f(z) with a critical point zc such that f ′′(zc) ̸= 0.
Suppose that f(z) is holomorphic in a neighborhood of zc. Then there are two curves γ passing transversally to each other through zc and such that ℑf(γ) is constant along each curve. Along one of these curves γ, the function ℜf decreases as you move away from zc, until you reach either another critical point of f or a point of nonanalyticity. The same is true for the other curve except that ℜf instead increases as you move away from zc.
Proof Using Taylor expansions in a neighborhood of zc, one sees four level lines ℑf(z) = const staring at z = zc in four different directions; they form two smooth curves passing through zc. Consider the curve for which ℜf decreases locally as we move away from zc, and suppose that when moving along the curve we reach a local minimum (along the curve) of ℜf at a point w. If f is not analytic at w we are done, so assume it is. Since w is a minimizer of ℜf along γ we have d dγ wℜf = 0, but since γ is a level line of ℑf we also have d dγ wℑf = 0. Combining these yields f ′(w) = 0, so w is a critical point of f.
The same argument holds for the other curve.
129 In our case f(z) = ln(1 + z) −α lnz, so the function is holomorphic in the slit domain C(−∞,0]. Let’s take the locally vertical contour ℑf(z) = const passing through zc, and try to understand its shape. By symmetry it suffices to understand the half-contour γ starting at zc and going upward.
• γ cannot escape to infinity. Indeed, ℜf is bounded from above on γ (since it is decreasing away from zc), but exp(Nℜf(z)) = |1 + z|N/|z|K grows as |z| →∞.
• Similarly γ cannot hit 0 since |1+z|N/|z|K explodes near 0.
• Finally, γ cannot hit a point in R+ = (0,+∞). If it did, γ and its conjugate would together form a closed contour in the slit domain C(−∞,0) along which the harmonic function ℑf is constant. By the maximum principle of harmonic functions we conclude that ℑf is constant on the whole domain, which is false.
We conclude that γ starts at zc and exits the upper half plane somewhere along R−= (−∞,0). We take for our contour γ ∪γ — we can deform the orig-inal contour of integration into γ ∪γ without changing the value of the integral.
At this point by the same analysis as for the warm up example above (the in-tegral is dominated by a neighborhood of zc since exp(N f(z)) is exponentially smaller away from zc), we see that N K = 1+o(1) 2πi (1+zc)N zK+1 c ˆ zc+i∞ zc−i∞ exp 1 2N (1−α)3 α (z−zc)2 dz = (1+o(1)) ( 1 α )N 1−α α K+1 1 √ 2πN r α (1−α)3 = 1+o(1) √ 2πN αK+1/2(1−α)N−K+1/2 .
Exercise 15.3. Prove the Stirling’s formula by finding the large N asymptotics of the contour integral around 0 1 N! = 1 2πi ˛ ez dz zN+1 .
15.4 Steepest descent for a double contour integral Now we turn to the double contour integral (15.1). A typical situation is that G(z) is real on the real line, which implies G(z) = G(z). Hence, the solutions to G′(z) = 0 come in complex-conjugate pairs. For now we do not discuss the question of choosing the correct critical point and assume that G(z) has a unique pair of critical points: zc, zc. From these critical points, we construct level lines of ℑG to obtain a pair of contours e Cw, e Cz (see Figure 15.1 for a 130 Lecture 15: Steepest descent method for asymptotic analysis schematic drawing), different from the original Cw, Cz. Note that near the crit-ical point zc, two level lines intersect orthogonally; ℜG has a maximum along one of them at zc and a minimum along another. This is good for us, since we would like to have a maximum both for ℜG(z) and ℜ(−G(w)).
f Cw f Cz zc zc Figure 15.1 A different pair of contours for (15.1). Here, zc and zc are critical points of G. The dashed curves are the level lines of ℜG, and the black and gray contours are the level lines of ℑG.
In two dimensions, it turns out that the singularity 1 z−w is integrable since e Cz, e Cw are transverse. Thus we can use the methods of the previous case to understand the integral ˛ e Cw ˛ e Cz f(z,w)exp(N(G(z)−G(w)))dz dw z−w , (15.2) by expanding around zc and zc for the level lines e Cw, e Cz. The exponentially large prefactor vanishes (since it’s reciprocal for the w and z integrals), leaving only the polynomially decaying term. (Indeed, we only care about (z,w) close to (zc,wc), more precisely, we should set z−zc = ˜ z √ N and w−wc = ˜ w √ N . After this change of variables the integrand behaves as 1 √ N 2 1 √ N →0.). We conclude that this integral goes to zero as N →∞.
Does this mean that asymptotically the integrals of the kind (15.1) always vanish? Note quite, since (15.1) differs from (15.2) by the choice of contours.
Hence, we should account for additional residues which might arise in the con-tour deformation. If we go back to the setting of Theorem 14.1, when we see 131 that the original contours are disjoint, while the new ones intersect. Hence, in the deformation we pick up the residue at z = w arising from the 1 z−w factor in the integrand. In principle, we might have needed to cross other poles, coming from the remaining factors of the integrand. Yet, let us assume that this is not happening. Then we can conclude ˛ Cw ˛ Cz f(z,w)exp(N(G(z)−G(w)))dz dw z−w = ˆ zc zc Resz=w dz+O 1 √ N , where the integral is over the part of e Cz between zc and zc. If the factor f(z,w) in (15.1) is non-singular, then the residue at z = w is equal to f(z,z), which needs to be integrated from zc to zc. This integral gives the leading asymptotic of (15.1).
Lecture 16: Bulk local limits for tilings of hexagons In this lecture, we use the material of the previous few lectures to find the asymptotic for the correlation functions of uniformly random tilings near a point of a large hexagonal domain. Specifically, we apply the steepest descent techniques of the previous lecture to find the asymptotic of the double contour integral formula for the entries of the inverse Kasteleyn matrix from Lecture 14.
Recall that we use shifted coordinates, so that the hexagonal domain has two horizontal and two vertical edges. For simplicity we consider a hexagon with all side lengths N/2. Let us relate the trapezoids (or sawtooth domains) studied in previous lectures to the hexagons. If we consider a trapezoid with all dents located in the extreme rightmost and left-most positions along the top side (shown as solid black lozenges in Figure 16.1), then the lozenges below the dents are forced to be vertical (shown as outlined lozenges), and once all of these forced lozenges are removed, the domain becomes a hexagon.
For the desired N/2×N/2×N/2 hexagon, we choose its location such that the coordinates of the dents are −N/2,−N/2+1,...,−1 for the dents on the right, and −3N/2,...,−N −1 for the dents on the left.
Theorem 14.1 from Lecture 14 specializes to the following statement for the hexagon.1 Theorem 16.1 ([Petrov12a]). For even N consider the N/2 × N/2 × N/2 1 Another approach to explicit inversion of the Kasteleyn matrix for the hexagon is in [Gilmore17].
132 133 n N 0 x N/2 N/2 N/2 Figure 16.1 A trapezoid/sawtooth domain with the dents chosen such that once the ‘forced’ vertical lozenges are removed, the remainder of the domain is an N/2 × N/2 × N/2 hexagon. Note that while the diagonal sides appear longer by a factor of √ 2, length should be measured in terms of the sides of our sheared lozenges.
hexagon. The inverse Kasteleyn matrix K−1( (y,m); (x,n)) is given by (−1)y−x+m−n+1 " 1m<n1y≤x (x−y+1)n−m−1 (n−m−1)!
(16.1) − (N −n)!
(N −m−1)!
˛ Cz ˛ Cw (z−y+1)N−m−1 (w−x)N−n+1 (w+1)N/2(w+N +1)N/2 (z+1)N/2(z+N +1)N/2 dzdw w−z # , (16.2) where the contour Cz is a loop around {y,y + 1,...,−1} (and enclosing no other integers) and the contour Cw is a large loop around Cz and all poles of the integrand.
Since we are sending N to ∞and studying the local behavior near a point (N ˜ x,N ˜ n) in the bulk, we should work in coordinates which reflect this. Namely, we consider the asymptotic of the inverse Kasteleyn matrix evaluated in two points (x,n) and (y,m) given by x := N ˜ x+∆x, (16.3) y := N ˜ x+∆y, (16.4) n := N ˜ n+∆n, (16.5) m := N ˜ n+∆m.
(16.6) 134 Lecture 16: Bulk local limits for tilings of hexagons The fact that x and y are both close to N ˜ x, and similarly m,n are close to N ˜ n, guarantees that the points (x,n) and (y,m) stay close to (N ˜ x,N ˜ n).
∆x,∆y,∆n,∆m control exactly where the two points sit relative to (N ˜ x,N ˜ n) on a local scale.
Here is our aim for this lecture: Theorem 16.2. For (˜ x, ˜ n) inside the liquid region (i.e. inside the ellipse in-scribed into the hexagon), the correlation functions of uniformly random tilings of the hexagon converge to those of the ergodic Gibbs translation-invariant measure (see Lecture 13) associated to the slope of the limit shape at (˜ x, ˜ n).
Remark 16.3. Since probabilities of the cylindrical events for random tilings can be expressed as finite linear combinations of the correlation functions, the convergence of the latter implies weak convergence of measures on tilings.
Proof of Theorem 16.2 Theorem 3.5 expresses the correlation functions of tilings through the inverse Kasteleyn matrix which is computed in Theorem 16.1. We would like to apply the steepest descent method to the double con-tour integral in (16.2). For that we want to write the integrand of (16.2) as exp(N(G(z) −G(w))) · O(1) for some function G, where (z,w) = (N˜ z,N ˜ w).
We set G(z) = 1 N N(1−˜ n) ∑ i=1 ln(z−N ˜ x+i)−1 N N/2 ∑ i=1 ln(z+i)−1 N N/2 ∑ i=1 ln(z+N +i) (16.7) and note that 1 N N(1−˜ n) ∑ i=1 ln(z−N ˜ x+i)−lnN = 1 N N(1−˜ n) ∑ i=1 ln(˜ z−˜ x+ i N ) (16.8) is a Riemann sum for ´ 1−˜ n 0 ln(˜ z−˜ x+u)du. Hence by subtracting one lnN term and adding two more to (16.7) to normalize the three sums as in (16.8), we obtain three Riemann sums corresponding to the three integral summands in ˜ G(˜ z) := ˆ 1−˜ n 0 ln(˜ z−˜ x+u)du− ˆ 1/2 0 ln(˜ z+u)du− ˆ 1/2 0 ln(˜ z+1+u)du.
(16.9) It follows that G(z) = ˜ G(˜ z)+o(1)−˜ nlnN, (16.10) and the lnN term will cancel in G(z)−G(w) so we may ignore it.
For steepest descent analysis, we need the critical points of ˜ G. In order to compute the derivative ∂˜ G ∂˜ z , we differentiate under the integral sign, then take 135 the definite integral with respect to u which gives us back the logarithm, yield-ing ∂˜ G ∂˜ z = ln(˜ z−˜ x+(1−˜ n))−ln(˜ z−˜ x)−ln(˜ z+1/2) +ln(˜ z)−ln(˜ z+3/2)+ln(˜ z+1).
(16.11) Exponentiating we get a quadratic equation in ˜ z for the critical point: ∂˜ G ∂˜ z = 0 ⇐ ⇒(˜ z−˜ x+(1−˜ n))(˜ z)(˜ z+1) = (˜ z−˜ x)(˜ z+1/2)(˜ z+3/2) (16.12) ⇐ ⇒˜ n˜ z2 −(˜ x+ 1 4 −˜ n)˜ z−3 4 ˜ x = 0.
(16.13) This equation has a pair of complex conjugate roots if and only if the discrim-inant is negative: D = (˜ x+1/4−˜ n)2 +3˜ n˜ x < 0.
Remark 16.4. The region on the ˜ x, ˜ n coordinate plane given by (˜ x+1/4−˜ n)2+ 3˜ n˜ x < 0 is exactly the interior of the inscribed ellipse of the hexagon of side lengths 1/2 which is the scaled version of Figure 16.1. Clearly the equation corresponds to a ellipse, and we can check that (˜ x+1/4−˜ n)2 +3˜ n˜ x = 0 has a single solution (−3/4,1) when ˜ n = 1 and a single solution (−1/4,0) when ˜ n = 0, corresponding to the tangency points at the top and bottom of the hexagon.
The tangency points to other sides of the hexagon can be found similarly. Since an ellipse is determined by five points, these tangency points fix the ellipse uniquely.
We assume that D < 0 and consider two conjugate critical points, zc and ¯ zc; this corresponds to (˜ x, ˜ n) lying in the liquid region.
Next, we wish to find the steepest-descent contours for z and w. For that we consider level lines of real and imaginary parts (ℜ˜ G(˜ z) = const and ℑ˜ G(˜ z) = const) passing through zc, zc. The latter will be the steepest descent contours, while the former will help us to understand their geometry. The schematic drawing of the configuration of the level lines is shown in Figure 16.2. The level lines of the real part, ℜ˜ G(˜ z) = const, partition the Riemann sphere into regions where ℜ˜ G(˜ z) −ℜ˜ G(zc) is < 0 and > 0, represented by + and −in the figure. Because ℜ˜ G(˜ z) goes to −∞as z →∞, the ‘outer’ region is a “−” region, and this determines all the others since crossing a contour changes the sign.
There are two contours passing through zc and ¯ zc along which ˜ G(˜ z) has constant imaginary part; one lies in the region of negative real part, the other in 136 Lecture 16: Bulk local limits for tilings of hexagons − − − − − − + + + + ℑ˜ G(z) = ℑ˜ G(zc) ℜ˜ G(z) = ℜ˜ G(zc) zc zc ℜ˜ G(z) = ℜ˜ G(zc) ℑ˜ G(z) = ℑ˜ G(zc) Figure 16.2 Level lines of ℜ˜ G(z) and ℑ˜ G(z) passing through zc. The w–contour in (16.2) is to be deformed to the closed loop of the constant imaginary part in “+” region and the z–controur is to be deformed to the curve of the constant imaginary part extending to infinity in “−” region.
the region of positive real part. However, the geometry of these contours is not obvious: the one that lies partially in the outer “−” region may either intersect with the positive real axis, or the negative real axis, or just go off to infinity. In order to distinguish between the three possible scenarios, we use the explicit formula for ˜ G(˜ z). First, note that for a complex, which very close to the real axis, the value of 1 π ℑln(z −a) is very close to 0 for z > a and close to 1 for z < a, with some smoothed-out step function-like behavior near a. Hence each of the three ´ ln terms in ˜ G(˜ z) cause the slope of the function 1 π ˜ G(x + i0) to change at various points: 1 π ´ 1−˜ n 0 ln(˜ z −˜ x + u)du gives slope −1 contribution on the segment [˜ x + ˜ n −1, ˜ x], −1 π ´ 1/2 0 ln(˜ z + u)du gives slope 1 contribution on the segment [−1 2,0], and −1 π ´ 1/2 0 ln(˜ z+1+u)du gives slope 1 contribution on the segment [−3 2,−1 2]. Adding the three terms, we arrive at the graph of 1 π ˜ G(x+i0) schematically shown in Figure 16.3.
We are seeking for the possible points of intersections of the curve ℑ˜ G(˜ z) = ℑ˜ G(zc) with the real axis. They correspond to the intersections of the graph in Figure 16.3 with a horizontal line. Note that the curve ℑ˜ G(˜ z) = ℑ˜ G(zc) can not intersect the real axis in a point inside a horizontal segment of the graph of Figure 16.3, since that would have to be another critical point other than zc, and we know there is no other critical points. Hence, there are either 1, 3, or 0 points where ℑ˜ G(˜ z) = ℑ˜ G(zc) can intersect the real axis.
137 1 πℑ˜ G(x + i0) x −3 2 −1 −1 2 ˜ x ˜ x + ˜ n −1 0 Figure 16.3 The graph of ℑ˜ G(x+i0).
Recall that there are four different curves ℑ˜ G(˜ z) = ℑ˜ G(zc) starting at zc.
If less than three of them reach the real axis, then at least two should escape to infinity. However, this would contradict the configuration of Figure 16.2, as the outer “−” region for the real part can have only one ℑ˜ G(˜ z) = ℑ˜ G(zc) curve inside (in the upper halfplane). Hence, there are precisely three points of intersection of ℑ˜ G(˜ z) = ℑ˜ G(zc) with the real axis (each one inside the cor-responding slope ±1) and the configuration should necessary resemble Figure 16.2.
Because we want the integral of exp(N(G(z) −G(w))) to go to 0, the con-stant imaginary part contour in the region of positive real part should be the w contour Cw, and the contour in the negative real part region should be the z contour Cz. Recall that in Theorem 16.1, the two contours look as in Figure 16.4, i.e. one is inside another.
Hence deforming the integration contour in (16.2) to be as in Figure 16.2 requires passing Cz through Cw, which picks up the residue at w = z given by (−1)y−x+m−n 2πi (N −n)!
(N −m−1)!
˛ Nzc N ¯ zc (z−y+1)N−m−1 (z−x)N−n+1 dz, (16.14) with integration contour going to the right from the singularities of the inte-grand. As we discussed in the previous lecture, the double contour integral on the deformed contours of Figure 16.2 vanishes asymptotically. Hence, the lead-ing contribution of (16.2) is given by (16.14). Note that we should also check that in the contour deformation we do not cross any z–poles arising from the denominator (z+1)N/2(z+N +1)N/2 or w–poles arising from the poles from the denominator (w −x)N−n+1; this check can be done by locating the inter-138 Lecture 16: Bulk local limits for tilings of hexagons Cz Cw Figure 16.4 In (16.2) the z–contour is inside the w–contour.
section points of the desired steepest descent contours with the real line using Figures 16.2 and 16.3.
Sending N →∞and substituting (16.3), (16.4), (16.5), (16.6), the integral (16.14) becomes (−1)∆x+∆n−∆y−∆m 2πi (1−˜ n)∆m−∆n+1 × ˛ zc ¯ zc (˜ z−˜ x)∆y−∆x−1(˜ z+ ˜ x+1−˜ n)∆x+∆n−∆y−∆m−1d˜ z, (16.15) and the contour again passes to the right from the singularities. We would like to change the variables to simplify (16.15). Set v = ˜ z−˜ x ˜ z−˜ x+1−˜ n, (16.16) so that 1−v = 1−˜ n ˜ z−˜ x+1−˜ n, (16.17) and let vc be the result when zc is plugged into the above. Then changing vari-ables to v in (16.15) yields (−1)∆x−∆y+∆n−∆m 2πi ˛ vc ¯ vc v∆y−∆x−1(1−v)∆m−∆ndv, (16.18) where the contour is one for which the intersection with the real axis lies in (0,1).
It remains to deal with the indicator term in (16.1). Lemma 14.4 says that 139 this term can be identified with the full residue of the double contour integral at z = w. One can check that this remains true as N →∞. Therefore, if we add (16.18) to the indicator function in (16.1), the result is the same integral as in (16.18), except that the integration contour intersects the real axis on (0,1) for ∆m ≥∆n, and intersects on (−∞,0) for ∆m < ∆n, which is when the indicator function is nonzero. This gives us the limiting value of K−1.
Remark 16.5. The limiting kernel which we just found, was first introduced by Okounkov and Reshetikhin in [OkounkovReshetikhin01]. It is called the incomplete beta kernel (since the integrand is the same as in the classical Beta integral) or extended sine kernel. Here is the reason for the latter terminology.
If we set ∆m = ∆n in (16.18), then it becomes (−1)∆x−∆y 2πi ˛ vc ¯ vc v∆y−∆x−1dv.
(16.19) Since the inverse Kasteleyn matrix is encoding probabilities of observing lozenges of different types, for any set of k points with the same n-coordinate and x-coordinates x1,...,xn, the probability of observing a vertical lozenge (i.e., of left-most type in bottom-right panel of Figure 14.2) at each of the k points is det(K(xi −xj))1≤i, j≤k where K(x) = (−1)x 2πi ˆ vc ¯ vc vx−1dv = (−1)x (vc)x −¯ vcx 2πix = (−|vc|)x πx sin(xArg(vc)).
(16.20) The (−|vc|)x factor cancels in det(K(xi−xj))1≤i, j≤k and therefore, we can omit it. The conclusion is that the correlation functions of vertical lozenges along a horizontal slice are given by the minors of the matrix K(x) = ( sin p1πx πx , x ̸= 0, p1, x = 0, where p1 is the asymptotic density of the vertical lozenges. We recognize the discrete sine-kernel, which already appeared before in Corollary 13.16. Hence the kernel of (16.18) yields the discrete sine kernel when restricted to a line, which is the origin of the name extended sine kernel.
Exercise 16.6. Show that the measure on lozenge tilings of the plane with cor-relation functions given by the minors of the extended sine kernel (16.18) (with the choice of integration contour described in the paragraph after the formula) is the same as one coming from the ergodic Gibbs translation-invariant mea-sure of Lecture 13 with slope corresponding to vc. Note that our ways of draw-ing lozenges differ between this lecture and Lecture 13, since we used affine-140 Lecture 16: Bulk local limits for tilings of hexagons transformed coordinate system starting from Figure 14.2 in order to match the notations of [Petrov12a].
This check is a more general version of the argument in Remark 16.5.
Coming back to the proof of Theorem 16.2, at this point we have established the local convergence of the correlation functions to those of an EGTI measure of Lecture 13. It remains to identify the slope of the local measure with the one of the limit shape for the height function (found through the variational prob-lem and Burgers equation) at (˜ x, ˜ y). Indeed, the local slope gives the average proportion of lozenges of each of the three types locally around (˜ x, ˜ y). Because the asymptotics are uniform over all points (˜ x, ˜ y), they yield asymptotics for the expected increments of the limiting height function or, equivalently, for the in-tegral of the slope along macroscopic domains. By concentration of the height function, expected increments are asymptotically the same as limit shape in-crements. Hence, integrals of the local slope and of the limit shape slope along macroscopic domains are the same. Therefore, these two kinds of slopes have to coincide.
Exercise 16.7. Show that outside the inscribe ellipse there are six regions (adjacent to six vertices of the hexagon) where one observes only one type of lozenges. Find out which type of lozenges appears in each region, thus explain-ing Figure 1.5.
Lecture 17: Bulk local limits near straight boundaries In the last lecture we studied the asymptotic behaviour of correlation functions for tilings of hexagons. It turned out that in the bulk the correlation functions converge to those of EGTI (ergodic Gibbs translation invariant) measure. The proof consisted of the steepest descent method applied to the inverse of the Kasteleyn matrix.
We would like to extend this approach to more general regions. To describe the situations where our approach still works, we need the following definition.
Recall that a trapezoid is a quadrilateral drawn on the triangular grid (with one pair of parallel sides and one pair of non-parallel sides), with the longer base being “dashed” as in Figure 17.1.
Definition 17.1. Let R be a tileable region on the triangular grid. A trapezoid T is covering a part of R if 1) R\T touches the boundary of T only along the dashed side.
2) T\R consists of big disjoint triangles along the dashed part of the bound-ary, that is, T\R can be obtained as the region of T frozen by sequences of consecutive ”teeth” along the dashed side.
The definition above ensures the following property: every tiling of a region R gives a tiling of a trapezoid T covering R. Note that here by a tiling of R we mean a strict tiling, forbidding lozenges to stick out of the region R, while a tiling of trapezoid T is allowed to have lozenges sticking out of T (called ”teeth”), but only along the dashed side (actually, any tiling of T will have exactly N teeth, where N is the height of T), see Figure 17.1.
Theorem 17.2. Take a sequence of tileable domains L · R with L →∞, and a point (x,y) ∈R such that 141 142 Lecture 17: Bulk local limits near straight boundaries N C Figure 17.1 A trapezoid with a tiling 1) proportions p , p , p encoding the gradient to limit shape of uniformly random tilings of L·R are continuous and ̸= 0 in a neighbourhood of (x,y).
2) (x,y) is covered by a trapezoid.
Then the correlation functions near (Lx,Ly) converge to those of EGTI mea-sure of slope (p , p , p ).
Remark 17.3. The continuity assumption for p , p , p ensures that their value at a point is well defined. Also, recall that for polygons limit shapes are algebraic, hence the proportions are continuous in the liquid region.
Remark 17.4. For some regions (including a hexagon with a hole) the second condition holds everywhere inside (that is, the region R can be completely covered by trapezoids). But we cannot completely cover a general region (even a polygon) by trapezoids, see Figure 17.2.
Remark 17.5. The theorem remains valid without the second condition, but the argument becomes much more elaborate (although, the steepest de-scent analysis of double contour integrals still remains a key ingredient), see [Aggarwal19].
We will derive Theorem 17.2 as a corollary of the following statement.
Theorem 17.6. Take a sequence of trapezoids TN of height N and with teeth t1 < ··· < tN. Let µN = 1 N N ∑ i=1 δti/N be a scaled empirical measure encoding {t1,...,tN}. Assume that 1) the support Supp[µn] ⊂[−C;C] for a constant C > 0 independent of N.
2) µN − − − → weak µ as N →∞for some probability measure µ.
143 Figure 17.2 Examples of various domains covered by trapezoids. The right-most region cannot be completely covered Then: 1) Uniformly random tilings of TN have limit shape as N →∞, de-pending only on µ.
2) For (x,y) s.t. p , p , p are continuous and nonzero in a neighbourhood of (x,y), correlation functions near (Nx,Ny) converge to those of EGTI mea-sure with slope p , p , p .
Remark 17.7. In Theorem 17.6 one can take random empirical measures µN, converging in probability to a deterministic µ. This is exactly the form of The-orem 17.6 used in the proof of Theorem 17.2.
Proof of Theorem 17.2 using Theorem 17.6 Let (x,y) be a point of R satis-fying conditions of the theorem, T be a trapezoid covering (x,y). As noted before, tilings of L · R give tilings of TN(L) (for some N(L) tending to infinity as L →∞) with random teeth t1 < ··· < tN(L). Moreover, by Gibbs property the restriction of a uniformly random tiling on L · R is a uniformly random tiling on TN(L).
Note that the locations of teeth t1 < ··· < tN(L) determines a height function 144 Lecture 17: Bulk local limits near straight boundaries hL along the dashed side of TN up to a constant, and µN(L) is the distribution of hL(Lx,Ly) L . Since uniformly random tilings of R converge to a deterministic limit shape, random functions hL(Lx,Ly) L converge to a deterministic function ˆ h and hence the corresponding measures µN(L) converge to a deterministic measure µ. Also, note that the length of the dashed side of TN(L) grow linearly as L →∞, hence the support of measures µN(L) is inside [−C;C] for some C > 0.
So, both assumptions of Theorem 17.6 hold, hence the correlation functions of uniformly random tilings near (Lx,Ly) converge to those of EGTI with the same slope.
Proof of Theorem 17.6 For the first part note that the convergence of mea-sures µn imply the convergence of height functions on the dashed side of the boundary, hence height functions on the boundaries of TN converge and we have a limit shape.
The proof of the second part is similar to the hexagon case (which is a partic-ular case where all teeth are concentrated near end-points of the dashed side): we study the asymptotic behaviour of the correlation kernel K−1((y,m),(x,n)) using the steepest descent method. Here we cover only the main differences from the hexagon case, for the full detailed exposition see [Gorin16].
Recall that the correlation kernel is given by the inverse Kasteleyn matrix (−1)y−x+m−nK−1((y,m),(x,n)) = −1m 0) in a neigh-borhood of the edge of the liquid region, where x is the distance from the frozen 148 149 L1/3 L2/3 Figure 18.1 A tiling of a hexagon, corresponding non-intersecting paths, and a L2/3 ×L1/3 window for observing the edge limit.
Figure 18.2 Left panel: unit hexagon and corresponding frozen boundary. Right panel: asymptotic density of horizontal lozenges along the vertical line x = 0.
We use formulas from [CohnLarsenPropp98, Theorem 1.1].
region re-scaled by the linear size L of the hexagon. One way to see this √x behavior is through the relation of densities to the complex slope. The latter for the hexagon is found as a solution to a quadratic equation, see Lectures 10 or 16. The square root in the formula for the solution of a quadratic equa-tion eventually gives rise to the desired C√x behavior of the densities. Figure 18.2 shows the plot of the complimentary density of the lozenges of the third type along one particular vertical section of the hexagon. We remark that the same square root behavior at the edges also appears in lozenge tilings of more complicated domains.
If we (non-rigorously) assume that the √x behavior of the density extends 150 Lecture 18: Edge limits of tilings of hexagons up to the local scales, then we can deduce one the transversal edge scaling as a corollary. Indeed, if the k-th path is at (vertical) distance xk from the frozen boundary, then we expect ˆ xk/L 0 C√xdx ≈k/L, implying xk =C′ ·L1/3. Thus, we expect the k–th path to be at distance of order L1/3 from the frozen boundary and the transversal rescaling should be L1/3.
For the second heuristic, note that, locally, paths look like random walks with up-right and down-right steps (these steps correspond to lozenges of types and , respectively). Moreover, there are rare interactions between paths, since the spacings are large (L1/3 due to the first step). Hence, on the local scale we expect to see some version of a random walk, and therefore, zooming further away, in a proper rescaling we should locally see some “Brownian-like behavior”. For the standard Brownian bridge B(t), 0 ≤t ≤T, the variance of B(αT) (for 0 < α < 1) is α(1−α)T, so for large T one needs a √ T transversal rescaling to see non-trivial fluctuations. Since in our case we already fixed the transversal rescaling to be of order L1/3, then the horizontal rescaling should be given by a square of that, i.e. L2/3, as in Figure 18.1.
18.2 Edge limit of random tilings of hexagons For the rigorous part of the lecture, we consider a regular hexagon of side-length N/2 (for N ranging over even positive integers) and recall the notation and the result of Theorem 16.1 (yet again we switch from lozenges of Figure 18.1 to those of Figures 14.2, 16.1). We know an explicit expression for the probabilities of the kind Prob(there are lozenges at (x1,n1),...,(xk,nk)) = det K−1((xi,ni);(xj,nj)) 1≤i, j≤k .
(18.1) Let us look at the paths formed by two other (not ) types of the lozenges.
The correlation functions for points of such paths can be found through the following complementation principle.
Lemma 18.1. If P is a determinantal point process on a countable set L (e.g., one can take L = Z) whose correlation functions are given by the minors of 151 the correlation kernel C(x,y), then the point process of complementary config-urations P = L \ P is a determinantal point process with correlation kernel C(x,y) = δx,y −C(x,y).
Proof Let k ∈N be any positive integer, then Prob x1,...,xk ∈P = Prob(x1,...,xk / ∈P) = 1−∑ 1≤i≤k Prob(xi ∈P)+ ∑ 1≤i, j≤k i̸=j Prob(xi,xj ∈P)−··· = 1−∑ 1≤i≤k C(xi,xi)+ ∑ 1≤i, j≤k i̸= j det C(xi,xi) C(xi,xj) C(xj,xi) C(xj,xj) −··· = det[δi,j −C(xi,x j)]1≤i, j≤k , where the first equality is obvious, the second is the principle of inclusion-exclusion, the third is because P is a determinantal point process, and the last equality follows from using the multilinearity (and definition) of the determi-nant.
From (18.1) and the previous lemma, we have Prob(non-intersecting paths go through (x1,n1),...,(xk,nk)) = det δi, j −K−1((xi,ni);(xj,nj)) 1≤i,j≤k .
(18.2) Thus we would like to compute the limit of δ(x1,n1),(x2,n2) − K−1((x1,n1);(x2,n2)) when (x1,n1) and (x2,n2) are close to the bound-ary of the frozen region.
Recall the result of Theorem 16.1: K−1 ((x1,n1);(x2,n2)) = (−1)x1−x2+n1−n2+11{n1<n2}1{x1≤x2} (x2 −x1 +1)n2−n1−1 (n2 −n1 −1)!
+(−1)x1−x2+n1−n2 (N −n2)!
(N −n1 +1)!
˛ C(x1,x1+1,...,t1−1) dz 2πi ˛ C(∞) dw 2πi × 1 w−z (z−x1 +1)N−n1−1 (w−x2)N−n2+1 (w+1)N/2(w+N +1)N/2 (z+1)N/2(z+N +1)N/2 .
(18.3) We will zoom in near the right part of the frozen boundary, circled in Figure 18.3. From our initial heuristics, we want the points (x1,n1) and (x2,n2) to be separated by the distance O(N2/3) in the tangential direction to the frozen boundary and by O(N1/3) in the normal direction.
152 Lecture 18: Edge limits of tilings of hexagons n N 0 x N/2 N/2 N/2 Figure 18.3 Hexagonal domain, boundary of the frozen region, and the part of the boundary where we zoom in to see the edge limit.
For that we specify the points via n1 = e nN +(∆n1)N2/3, n2 = e nN +(∆n2)N2/3, x1 = x(n1)+(∆x1)N1/3, x2 = x(n2)+(∆x2)N1/3.
(18.4) At this moment we need to be careful in specifying x(n1),x(n2) — the curva-ture of the frozen boundary is important (because of the different tangential and normal scalings) and we are not allowed to simply use linear approximation.
Taking logarithm of the z–part of the integrand in (18.3), we define Gn1 N (z) through Gn1 N (z) := N−n1 ∑ i=1 ln(z−x+i)− N/2 ∑ i=1 ln(z+i)− N/2 ∑ i=1 ln(z+N +i).
The equation ∂ ∂zGn1 N (z) = 0 is ∂ ∂z N−n1 ∑ i=1 ln(z−x+i)− N/2 ∑ i=1 ln(z+i)− N/2 ∑ i=1 ln(z+N +i) !
= 0.
Instead of this equation, it is easier to consider its N →∞approximation: ∂ ∂z ˆ 1−n1 N 0 ln z N −x N +u du − ˆ 1/2 0 ln z N +u du− ˆ 1/2 0 ln z N +1+u du !
= 0.
153 The above equation is quadratic in z and one needs to take x = x(n1) so that this equation has a double real root zc. The value of x(n2) is obtained through the same procedure.
The steepest descent analysis of the double contour integral is done similarly to the bulk case, except that now two complex conjugate critical points merge into a single double critical point on the real axis. In the neighborhood of the critical point we have Gn1 N (z) ≈Gn1 N (zc)+(z−zc)3α, (18.5) for some explicit α.
Previously, in the bulk analysis the steepest descent contours of integration passed through two critical points, but now there is only one critical point zc, so the situation is degenerate. One can obtain the configuration of the contours for the present case by continuously deforming Figure 16.2 into the situation of merging critical points. The z contour should pass through zc and w contour should pass through similarly defined wc (which is close to zc, since n2 is close to n2) under such angles as to guarantee the negative real part of Gn1 N (z) − Gn2 N (w) after using (18.5) in a neighborhood of the critical point.
For the limit, we zoom-in near the critical points, where the contours can be chosen to look like in Figure 18.4. As in the bulk case, one also needs to make sure that this local picture can be extended to the global steepest descent contours (such that the real part of the exponent is still maximized near the critical point) — we omit this part of the argument. We reach the following theorem and refer to [Petrov12a, Section 8] for more details.
Theorem 18.2. In the limit regime (18.4), which describes the random lozenge tiling near the boundary of the frozen region, we have lim N→∞N1/3 δ(x1,n1),(x2,n2) −K−1 ((x1,n1);(x2,n2)) = A ((∆x1,∆n1);(∆x2,∆n2)), where A ((∆x1,∆n1);(∆x2,∆n2)) := −1∆n1<∆n2(residue of the integral at its pole) + 1 (2πi)2 ˆ ˆ exp e w3 −e z3 3 exp(−e w∆x2β +e z∆x1β) β d e wde z e w−e z+γ(∆n1 −∆n2), (18.6) and where β,γ are explicit (by rescaling ∆xi,∆nj properly, these constants can actually be removed from the formula). The contours are as in Figure 18.4.
Remark 18.3. The z and w integration variables were shifted by different amounts zc and wc when getting (18.6). Hence 1 w−z turned into 1 e w−e z+γ(∆n1−∆n2).
154 Lecture 18: Edge limits of tilings of hexagons w-contour z–contour π 3 Figure 18.4 Local configuration of the steepest descent integration contours for the edge limit. The figure is centered at the critical point.
The tips of the contours (where they intersect the real axis) should have suf-ficient distance between themselves, so that the denominator does not vanish on the contours. The factor exp(−e w∆x2β +e z∆x1β) in (18.6) appears because Gn1 N (z) has x = x(n1) in its definition, but in the integrand of (18.3) we rather have x = x1 (and similarly for Gn2 N (w)).
Remark 18.4. The convergence of the correlation kernels implies conver-gence of the paths in the sense of finite-dimensional distributions. The lim-iting object is defined through its correlation function given by the minors of (18.6). The definition of the correlation functions is slightly more complicated for the continuous (rather than discrete) state space, see, e.g., [Johansson06], [Borodin11], or [BorodinGorin12, Section 3] for the details in determinantal contexts and [DaleyVere-Jones03] for the general theory.
Exercise 18.5. Finish the proof of Theorem 18.2 and compute the values of β and γ.
A rescaled version of the kernel A above is known as the extended Airy kernel in the literature. Here is its standard form: KAiry extended ((x,t);(y,s)) := (´ ∞ 0 e−λ(t−s)Ai(x+λ)Ai(y+λ)dλ, if t ≥s, − ´ 0 −∞e−λ(t−s)Ai(x+λ)Ai(y+λ)dλ, if t < s, (18.7) 155 where Ai(z) := 1 2πi ˆ exp v3 3 −zv dv, (18.8) is the Airy function1; the contour in the integral is the upwards-directed con-tour which is the union of the lines {e−iπ/3t : t ≥0} and {eiπ/3t : t ≥0} (it looks like the e w–contour in Figure 18.4).
Exercise 18.6. Match KAiry extended((x,t);(y,s)) with A ((x,t);(y,s)) in Theorem 18.2 by substituting (18.8) into (18.7), and explicitly evaluating the λ-integral.
The conclusion is that the ensembles of extreme paths (edge limit) in ran-dom tilings of large hexagons converge to the random path ensemble whose correlation functions are minors of the extended Airy kernel.
18.3 The Airy line ensemble in tilings and beyond It is believed that Theorem 18.2 is not specific to the hexagons.
Conjecture 18.7. The extended Airy kernel arises in the edge limit of the uni-form lozenge tilings of arbitrary polygonal domains, as the mesh size of the grid goes to zero.
The conjecture has been proved in certain cases (for trapezoids) in [Petrov12a], [DuseMetcalfe17]. The first appearance of the extended Airy ker-nel is in [PrahoferSpohn01] in the study of the polynuclear growth model (PNG). Closer to the tilings context, [FerrariSpohn02] were first to find the extended Airy kernel in the edge limit of the qVolume–weighted plane parti-tions (we will discuss this model of random plane partitions in Lecture 22) and [Johansson05] found it in the domino tilings of the Aztec diamons.
The random two-dimensional ensemble (a random ensemble of paths) aris-ing from the extended Airy kernel is called the Airy line ensemble. It is a translation-invariant ensemble (in the space direction). In the next lecture, we will learn more about it and tie it to other subjects in probability. As a warm-up, let us mention that the one-point distribution of the “top-most” path is known as the Tracy-Widom distribution2. This distribution also appears in var-ious other contexts: 1 Airy function solves the second order differential equation ∂2 ∂z2 Ai(z)−zAi(z) = 0.
2 The wikipedia page is a good initial source of the information 156 Lecture 18: Edge limits of tilings of hexagons 1 In random matrix theory: the TW distribution is (the limit of) the dis-tribution of the largest eigenvalue of a large Hermitian random matrices of growing sizes, see reviews [Forrester10, AndersonGuionnetZeitouni10, AkemannBaikFrancesco11]. We are going to explain in more details the link between tilings and random matrices in the next two lectures.
2 A closely related applied point of view says that TW distribution appears as an asymptotic law in statistical hypothesis testing procedures involving sample covariance matrices, sample canonical correlations between various types of data, multivariate analysis of variance (MANOVA). Note that in order to see our TW distribution in such a context one needs to deal with complex data, which might be less common. A more widespread setting of real data leads to the appearance of the closely related β = 1 Tracy-Widom distribution TW1, see, e.g., [Johnstone08], [BykhovskayaGorin20] for the examples of such results.
3 In interacting particle systems: the TW distribution arises as the limit-ing particle current for various systems, including the Totally Asymmetric Simple Exclusion Process (TASEP). We do not have time to discuss the direct links between tilings and interacting particle systems in these lec-tures, and we instead only refer to reviews [Johansson06, BorodinGorin12, BorodinPetrov13, Romik15].
4 In directed polymers: the TW distribution governs the asymptotic law of the partition function for directed polymers in random media and related solution of the Kardar–Parisi–Zhang stochastic partial differential equation, see reviews [Corwin11, QuastelSpohn15] 5 In asymptotic representation theory: the TW distribution is the asymp-totic law for the first row of random Young diagrams encoding decompo-sition of representations (typically of unitary or symmetric groups) into ir-reducible components. The most celebrated result of this kind deals with Plancherel measure for the symmetric groups, see reviews [Johansson06, BorodinGorin12, Romik15, BaikDeiftSuidan16].
Lecture 19: The Airy line ensemble and other edge limits 19.1 Invariant description of the Airy line ensemble In the previous lecture we found Airy line ensemble as the edge limit of the uniformly random tilings of hexagons. We described it in terms of its correla-tion functions, which are minors of the extended Airy kernel of (18.7). On the other hand, in the bulk, we could describe the limiting object without formulas: it was a unique Gibbs, translation–invariant, ergodic measure of a given slope.
Is there a similar description for the Airy line ensemble?
Conjecturally, the answer is yes, based on the following three facts: (1) The Airy line ensemble is translation–invariant in the t (i.e., horizontal) direction — this immediately follows from (18.7).
(2) The Airy line ensemble possesses an analogue of the Gibbs property known as the Brownian Gibbs property. Let {Li(t)}i≥1 be an ensemble of random paths such that L1(t) ≥L2(t) ≥..., for all t ∈R. It satisfies the Brownian Gibbs property if the following is true. Fix arbitrary k = 1,2,..., two reals a < b, k reals x1 > ··· > xk, k reals y1 > ··· > yk and a trajectory f(t), a ≤t ≤b, such that f(a) < xk and f(b) < yk. Then the conditional distribution of the first k paths L1,...,Lk conditioned on Li(a) = xi, Li(b) = yi, for 1 ≤i ≤k, and on Lk+1(t) = f(t), for a ≤t ≤b, is the same as that of k independent Brownian bridges conditioned to have no intersection and to stay above of f(t), a ≤t ≤b, see Figure 19.1.
Theorem 19.1 ([CorwinHammond11]). For the Airy line ensemble Ai(t), i = 1,2,..., with correlation functions given by the minors of (18.7), let Li(t) := 1 √ 2(Ai(t)−t2), for all i ≥1. Then the random path ensemble {Li}i≥1 satisfies the Brownian Gibbs property.
157 158 Lecture 19: The Airy line ensemble and other edge limits x1 x2 y2 y1 f(t) a b x t Figure 19.1 The Brownian Gibbs property: given x1, x2, y1, y2 and f(t), a ≤t ≤ b, the distribution of the dashed portion of the paths is that of non-intersecting Brownian bridges staying above f(t).
Roughly speaking, the proof is based on the Gibbs resampling property for the tilings: clearly, the non-intersecting paths coming from uniformly ran-dom tilings, as in Figure 18.3, satisfy a similar Gibbs property with Brownian bridges replaced by simple random walks. We know that these paths converge to the Airy line ensemble and simultaneously non-intersecting random walks converge to non-intersecting Brownian bridges in the same regime. What re-mains to show (and what is the essense of the [CorwinHammond11] argument) is that the topology of the convergence toward the Airy line ensemble can be made strong enough, so that to guarantee that the Gibbs property survives in the limit. The subtraction of t2 in the statement comes from the curvature of the frozen boundary near the edge, which was a part of x(n) in (18.4).
(3) [CorwinSun14] proved that the sequence {Ai(t) +C}i≥1 is a translation– invariant, ergodic line ensemble, for any constant C.
The additive constant C mentioned in the last point plays the role of the slope (s,t) for the corresponding result at the bulk. Taking it into account, the following conjecture has been formulated by Okounkov and Sheffield in 00’s: Conjecture 19.2. {Ai(t)}i and its deterministic shifts (in the vertical direc-159 tion) are the unique translation–invariant, ergodic line ensembles such that the sequence { 1 √ 2(Ai(t)−t2)}i≥1 has the Brownian Gibbs property.
Conceptual conclusion. The Airy line ensemble should appear whenever we deal with a scaling limit of random ensembles of non-intersecting paths at the points where their density is low.
19.2 Local limits at special points of the frozen boundary In the previous lecture we saw the appearance of the Airy line ensemble at a generic point where the frozen boundary is smooth. However, as we tile vari-ous complicated polygons, the frozen boundary develops various singularities, where the asymptotic behavior changes.
• One type of special points of the boundary arises in any polygons. These are the points, where the frozen boundary is tangent to a side of the polygon.
For instance, in the hexagon we have six such points. In this situation the correct scaling factor in the tangential direction to the frozen boundary is L1/2 (where L is the linear size of the domain) and we do not need any rescaling in the normal direction. The limiting object GUE–corners process connects us to the random matrix theory and we are going to discuss it in the next section and continue in the next lecture.
• The frozen boundary can develop cusp singularities. For instance, they are visible in the connected component of the boundary surrounding the hole in our running example of the holey hexagon, see Figure 24.2 in Lecture 1 and Figures 24.1, 24.5 in Lecture 24. There are two types of these cusps: one type is presented by two vertical cusps in the above figure, which separate two types of frozen regions; the limit shape develops a corner at such cusp.
In this situation the scaling factors are L1/3 in the direction of cusp and finite in the orthogonal direction. The scaling limit is known as Cusp–Airy pro-cess, see [OkounkovReshetikhin06], [DuseJohanssonMetcalfe15]. We also observe two cusps of another type in the picture: these cusps are adjacent to a single type of the frozen region each; the limit shape is C1–smooth near these cusps. The fluctuations of the discrete boundaries of such cusp are of order L1/4 and one needs to travel the distance of order L1/2 along the cusp to see a non-trivial 2d picture. The limit is known as the (extended) Pearcey process, see [OkounkovReshetikhin05, BorodinKuan08] for its appearances in tilings.
• Another possible situation is when the frozen boundary (locally near a point) has two connected components, which barely touch each other.
160 Lecture 19: The Airy line ensemble and other edge limits As in the cusp case, there are two subcases depending on how many types of frozen regions we see in a neighborhood and we refer to [AdlerJohanssonMoerbeke16, AdlerJohanssonMoerbeke17] and references therein for exact results. The limiting processes which appear in this situa-tion share the name Tacnode process.
Together with generic points the above list gives six different scaling limits, corresponding to six ways the frozen boundary can look like near a given point.
As [AstalaDusePrauseZhong20, Section 9] suggests, this list corresponds to six types of local behaviors of the frozen boundaries and, as long as we deal with uniformly random tilings of polygons with macroscopic side lengths, no other types of behavior for the frozen boundary should occur.
In all the above situations the processes which appear in the limit are be-lieved to be universal (and their universality extends beyond random lozenge tilings), but such asymptotic behavior was proven rigorously only for specific situations.
Let us also remark that if we allow non-uniform measures on tilings1, then the number of possibilities starts to grow and we can encounter more and more exotic examples. Just to give one example: [BorodinGorinRains09] stud-ied tilings with weight proportional to the product w( ) over all horizontal lozenges in tilings. The possible choices for w(·) included the linear function of the vertical coordinate of the lozenge . Tuning the parameters, one can make this weight to vanish near one corners of the hexagon, leading to the frozen boundary developing a node, see Figure 19.2. Again one expects to see a new scaling limit near this node.
19.3 From tilings to random matrices Our next topic is the link between random tilings and random matrix theory.
Indeed, historically the distribution A1(0) of the “top-most” path of the Airy line ensemble at time t = 0, known as the Tracy-Widom distribution, first ap-peared in the random matrix theory. At this point, one might be wondering, what do tilings have in common with matrices and why do the same limiting objects appear in both topics and we are going to explain that.
The general philosophy comes from the notion of semiclassical limit in rep-resentation theory: large representations of a Lie group G behave like orbits of the coadjoint action on the corresponding Lie algebra g.
1 Alternatively, one can tile more complicated domains than macroscopic polygons, e.g., by allowing various tiny defects to the domain boundaries.
161 Figure 19.2 Random lozenge tiling of a 70×90×70 hexagon and corresponding limit shape. Weight of tiling is proportional to the product of the vertical coordi-nates of all its horizontal lozenges (shown in gray).
When G = U(N), the Lie algebra is g = iHerm(N) ∼ = Herm(N) (the latter is the space of complex Hermitian matrices). We care about this case, because tilings in the hexagonal lattice are linked to the representation theory of U(N).
This is illustrated by the following proposition, which claims that the number of tilings of certain domain can be calculated via Weyl’s formula for dimen-sions of U(N) irreducible representations.
Proposition 19.3. The number of tilings of the trapezoid with teeth at positions t1 < t2 < ··· < tN, as in Figures 14.1, 14.2, is ∏ 1≤i< j≤N tj −ti j −i and coincides with the dimension of the irreducible representation of U(N) with highest weight λ1 ≥λ2 ≥··· ≥λN such that tN+1−i = λi + N −i, i = 1,...,N.
Proof This is left as an exercise, we only mention three possible approaches: • One can identify the terms in the combinatorial formula for the expansion of Schur polynomials sλ(x1,x2,...,xN) into monomials with tilings of the trapezoid. Then use known formulas for evaluations of Schur polynomials sλ(1,1,...,1). We refer to [Macdonald95, Chapter I] for the extensive infor-mation on the Schur polynomials and to [BorodinPetrov13, Section 2.2] for the review of the interplay between tilings and symmetric polynomials. This interplay appears again in Lecture 22 and is explained there in more details.
162 Lecture 19: The Airy line ensemble and other edge limits • Identify tilings of the trapezoid with basis elements of the irreducible rep-resentation Vλ of U(N) associated to λ. Then use Weyl’s formula to find the dimension of the irreducible representation of U(N) corresponding to λ.
See, e.g., [BorodinPetrov13, Section 2.2] for more details on this approach and further references.
• Identify tilings of the trapezoid with families of non-intersecting paths as in the previous lecture. Then use Theorem 2.5 from Lecture 2 to count the number of such non-intersecting paths via determinants, and use the evalu-ation formulas from [Krattenthaler99] to compute these determinants.
Exercise 19.4. Identifying hexagon with a trapezoid as in Figure 18.3, use Proposition 19.3 to give another proof of MacMahon formula (1.1) for the total number of tilings of A×B×C hexagon.
Remark 19.5. Enumeration of tilings of trapezoids possessing axial symme-try also admits a connection to representation theory, this time of orthogonal and symplectic groups. We refer to [BorodinKuan10] and [BufetovGorin13, Section 3.2] for some details.
We now explain an instance of the semiclassical limit by making a down-to-earth computation for random lozenge tilings of a hexagon. For that let us investigate the law of the N horizontal lozenges at distance N from the left edge in a uniformly random tiling of the A×B×C hexagon, as in Figure 19.3. The vertical line at distance N from the left intersects two sides of the hexagon and splits it into two parts. The left part has a geometry of a trapezoid. The right part can be also identified with trapezoid, if we extend it to the continuation of two sides of the trapezoid, as in the right panel of Figure 19.3.
Using Proposition 19.3 and denoting the positions of the desired N hori-zontal lozenges through t1,...,tN, we compute for the case N < A = B = C (general case can be obtained similarly) Prob(t1,...,tN) = #tilings of left trapezoid×#tilings of right trapezoid #tilings of the hexagon ∝ ∏ 1≤i<j≤N (tj −ti)· ∏ 1≤a<b≤2A−N (e tb −e ta) ∝ ∏ 1≤i<j≤N (tj −ti)2 · N ∏ i=1 (ti +1)A−N(A+N −ti)A−N.
(19.1) (Recall the Pochhammer symbol notation: (x)n := x(x + 1)···(x + n −1) = Γ(x+n)/Γ(x).) Above, 0 ≤t1 < ··· < tN ≤A + N −1 are the positions of the horizontal lozenges at level N. Equivalently, they are the teeth of the left trapezoid. On 163 N = 3 A = 3 B = 4 C = 5 Figure 19.3 Three horizontal lozenges at the vertical line at distance 3 of the left edge in 3×4×5 hexagon. In order to find the law of these lozenges, one needs to count tilings of two trapezoids: to the left and to the right from the cut.
the other hand, e t1 < e t2 < ... < e t2A−N are the positions of the teeth of the right trapezoid. That is, {e t1,...,e tA−N} = {−1,...,−(A −N)}, {e tA+1,...,e t2A−N} = {A+N,...,2A−1} and {e tA−N+1,...,e tA} = {t1,...,tN}.
Exercise 19.6. Generalize the computation (19.1) to arbitrary A × B × C hexagons and arbitrary values of 0 ≤N ≤B+C.
Exercise 19.7. Find an analogue of (19.1) for qVolume–weighted lozenge tilings of the hexagon.2 We would like to study the limit A →∞of the distribution (19.1), while keeping N fixed. It is clear that E[ti] ≈A/2 when A is large compared to N, because of the symmetry of the model. Hence, recentering by A 2 is necessary.
Let us guess the magnitude of the fluctuations by looking at the simplest N = 1 case. From (19.1), we have Prob(t) ∝(t + 1)A−1(A + 1 −t)A−1, which is pro-portional to the product of two binomial coefficients. Recall the De Moivre– Laplace Central Limit Theorem theorem, which says that if Prob(t) was pro-portional to a single binomial coefficient, then the order of the fluctuations, as A →∞, would be √ A. It turns out that the same is true in our case of the prod-uct of two binomial coefficients. Thus we expect to see a non-trivial scaling 2 The computation is possible for more complicated distributions on tilings as well. In particular, [BorodinGorinRains09, Section 10] studies the situation when weights of lozenges are given by elliptic functions in their coordinates.
164 Lecture 19: The Airy line ensemble and other edge limits limit in the following regime: ti = A 2 + √ Axi, i = 1,...,N.
(19.2) We plug (19.2) into (19.1) and send A →∞. With (19.2), we have ∏ 1≤i<j≤N (ti −tj)2 ∝ ∏ 1≤i<j≤N (xi −x j)2, (19.3) where the hidden constant is independent of the xi’s. Moreover, (ti +1)A−N(A+N −ti)A−N = (ti +A−N)!
ti!
(2A−ti −1)!
(A+N −ti −1)!
= 3A 2 −N + √ Axi !
A 2 + √ Axi !
3A 2 − √ Axi −1 !
A 2 +N − √ Axi −1 !
.
We can use Stirling’s formula M! = √ 2πM(M/e)M(1+o(1)), M →∞, in the formula above. It is safe to ignore N and other finite order constants for the computation, as only growing terms will give a non-trivial contribution. Using the asymptotic expansion ln(1+x) = x−x2/2+O(x3), x →0, we get (ti +1)A−N(A+N −ti)A−N ≈exp 3A 2 + √ Axi ln 3A 2 + √ Axi −1 + 3A 2 − √ Axi ln 3A 2 − √ Axi −1 exp − A 2 + √ Axi ln A 2 + √ Axi −1 − A 2 − √ Axi ln A 2 − √ Axi −1 ≈exp A 3 2 + xi √ A ln 1+ 2xi 3 √ A + 3 2 −xi √ A ln 1−2xi 3 √ A exp A − 1 2 + xi √ A ln 1+ 2xi √ A − 1 2 −xi √ A ln 1−2xi √ A ≈exp −4 3x2 i .
As a result, in the regime (19.2), we have N ∏ i=1 (ti +1)A−N(A+N −ti)A−N = N ∏ i=1 exp −4 3x2 i ·(1+o(1)).
(19.4) From (19.1), (19.3), and (19.4), we get the following proposition, whose proof first appeared in [Nordenstam09].
Proposition 19.8. Consider uniformly random tilings of A × A × A hexagon and let t1,...,tN denote the (random) coordinates of N lozenges on the vertical 165 line at distance N from the left edge. The N–dimensional vectors ti −A 2 q 3A 8 1≤i≤N converge in distribution to the absolutely continuous probability measure on {(x1,...,xN) ∈RN : x1 ≤··· ≤xN} with density proportional to ∏ 1≤i<j≤N (xi −xj)2 N ∏ i=1 exp −x2 i 2 .
(19.5) We can go further and look not only at level N (at distance N from the base), but at the first N levels simultaneously. In the tilings model, let t j = (t j 1 < ··· < t j j) be the positions of horizontal lozenges at the j-th level, counted so that we have 0 ≤t j 1 < ··· < t j j ≤A + j −1. The combinatorics of tilings implies that the interlacing3 inequalities t j+1 i+1 > t j i ≥t j+1 i , for all relevant i, j.
Proposition 19.9. The N(N +1)/2–dimensional vectors t j i −A 2 q 3A 8 1≤i≤j≤N (19.6) converge in distribution as A →∞to the absolutely continuous probability measure on {(x j i )1≤i≤j≤N ∈RN(N+1)/2 : x j+1 i+1 ≥x j i ≥x j+1 i for all relevant i, j} with density proportional to ∏ 1≤i<j≤N (xN j −xN i )· N ∏ i=1 exp −(xN i )2 2 .
(19.7) Proof We set t j i = A 2 + r 3A 8 x j i .
3 An interlacing triangular array of integers (or, sometimes, reals) satisfying such inequalities is often called a Gelfand–Tsetlin pattern. The name originates from the labeling with such patterns of the Gelfand–Tsetlin basis in irreducible representations of unitary groups U(N), first constructed in [GelfandTsetlin50].
166 Lecture 19: The Airy line ensemble and other edge limits From Proposition 19.8, the law of tN i −A 2 q 3A 8 1≤i≤N (19.8) as A →∞has density proportional to ∏ 1≤i<j≤N (xN j −xN i )2 · N ∏ i=1 exp −(xN i )2 2 ·(1+o(1)), A →∞.
By the Gibbs property (following from the uniformity of random tilings), the law of the N(N +1)/2–dimensional vectors (19.6) is obtained from the law of tN 1 < ··· < tN N dividing by the number of lozenge tilings of the trapezoid with teeth at the positions tN i . Therefore, by Proposition 19.3, we need to multiply the law of the rescaled vector (19.8) by a constant (independent of the tN i ) multiple of 1 ∏1≤i<j≤N (tN j −tN i ) ∝ 1 ∏1≤i<j≤N (xN j −xN i ).
The result follows.
So far, we have only discussed random tilings of hexagons and a scaling limit of their marginals. What is the relation with random matrices, that we promised? In the next lecture we are going to show that the distribution (19.7) has random matrix origin: its name is GUE–corners process and it appears in the celebrated Gaussian Unitary Ensemble.
Lecture 20: GUE–corners process and its discrete analogues At the very end of Lecture 19 we obtained an interesting distribution on inter-lacing arrays of reals {x j i }1≤i≤j≤N as a scaling limit for positions of lozenges in uniformly random tilings of the hexagon near one of the sides of the hexagon.
Our next task is to identify this distribution with the so–called GUE–corners process.1 20.1 Density of GUE–corners process For k dimensional real vector u = (u1,...,uk) and (k + 1)–dimensional real vector v = (v1,...,vk+1) we say that u interlaces with v and write u ≺v, if v1 ≤u1 ≤v2 ≤u2 ≤··· ≤uk ≤vk+1.
Theorem 20.1. Let X be an N ×N matrix of i.i.d. complex Gaussian random variables of the form N(0,1)+iN(0,1) and set M = X+X∗ 2 to be its Hermitian part. Define xm 1 ≤xm 2 ≤··· ≤xm m to be the eigenvalues of the principal top-left corner of M as in Figure 20.1. Then the array {xm i }1≤i≤m≤N has the density (with respect to the N(N +1)/2–dimensional Lebesgue measure) proportional to 1x1≺x2≺···≺xN ∏ 1≤i<j≤N (xN j −xN i ) N ∏ i=1 exp −(xN i )2 2 .
(20.1) Our proof closely follows the exposition of [Baryshnikov01], [Neretin03].
One earlier appearance of such arguments is [GelfandNaimark50, Section 9.3].
1 Other name used in the literature is the GUE–minors process. Since, by the definition a minor is the determinant of a submatrix (rather than the submatrix itself appearing in the definition of our process), we choose to use the name GUE–corners instead.
167 168 Lecture 20: GUE–corners process and its discrete analogues Figure 20.1 Principal submatrices of M.
Proof of Theorem 20.1 We proceed by induction in N with base case N = 1 being the computation of the probability density of the Gaussian law.
For the induction step, we need to verify the following conditional density computation Prob xm+1 i = yi +dyi, 1 ≤i ≤m+1 | x j i ,1 ≤i ≤j ≤m ?
∼1xm≺y ∏1≤i<j≤m+1(y j −yi) ∏1≤i 0. Since the characteristic polynomial of ˜ Y can be alternatively computed as ∏m+1 i=1 (y−yi), (20.3) implies m+1 ∏ i=1 (y−yi) m ∏ i=1 (y−xi) = y−v′ − m ∑ i=1 ξi y−xi .
(20.4) Here is a corollary of the last formula.
Lemma 20.2. We have ξa = − ∏m+1 b=1 (xa −yb) ∏1≤c≤m;c̸=a(xa −xc), 1 ≤a ≤m, and v′ = m+1 ∑ b=1 yb − m ∑ a=1 xa.
(20.5) Proof Multiplying (20.4) with y−xa and then setting y = xa, we get the for-mula for ξa is obtained. Looking at the coefficient of 1 y in the expansion of (20.4) into a series in 1 y for large y, we get the formula for v′.
In addition we have the following statement, whose proof we leave as an exercise.
170 Lecture 20: GUE–corners process and its discrete analogues Exercise 20.3. In formula (20.4) all the numbers ξa, 1 ≤a ≤m are non-negative if and only if the sequences {xi} and {yi} interlace, i.e.⃗ x ≺⃗ y.
We continue the proof of Theorem 20.1. Since we know the joint distribution of ξa and v′, and we know their link to {yi} through (20.4), it only remains to figure out the Jacobian of the change of variables {ξa, 1 ≤a ≤m; v′} 7→ {yi, 1 ≤i ≤m+1}. Straightforward computation based on Lemma 20.2 shows that ∂ξa ∂yb = ξa · 1 xa −yb , ∂v′ ∂yb = 1.
Hence, the desired Jacobian is ξ1 ···ξm ·det 1 1 xa−yb .
.
.
1 .
(20.6) We can simplify the last formula using the following determinant computation.
Lemma 20.4 (Cauchy determinant formula). For any u1,...,uN, v1,...,vN we have det 1 ua −vb N a,b=1 = ∏a<a′(ua −ua′)∏b<b′(vb′ −vb) ∏N a,b=1(ua −vb) (20.7) Proof Multiply (20.7) by ∏N a,b=1(ua −vb) and notice that the left-hand side becomes a polynomial in ui and vj of degree N(N −1). This polynomial is skew symmetric both in variables (u1,...,uN) and in variables (v1,...,vN).
Therefore, it has to be divisible by ∏a<a′(ua −ua′)∏b<b′(vb′ −vb). Compar-ing the degrees, we conclude that (20.7) has to hold up to multiplication by a constant. Comparing the coefficients of the leading monomials we see that this constant is 1.
Sending one of the variables in (20.7) to infinity, we compute the determi-nant in (20.6). Hence, the desired Jacobian is ξ1 ···ξm · ∏ 1≤a<a′≤m (xa −xa′) ∏ 1≤b<b′≤m+1 (yb −yb′) m ∏ a=1 m+1 ∏ b=1 (xa −yb) = ∏ 1≤b<b′≤m+1 (yb −yb′) ∏ 1≤a<a′≤m (xa −xa′) .
Using the explicit formula for the joint density of {ξa, 1 ≤a ≤m; v′}, we conclude that the conditional density of y1,...,ym+1 given x1,...,xm is propor-171 tional to ∏ 1≤b<b′≤m+1 (yb −yb′) ∏ 1≤a0 and an infinite matrix X of i.i.d. N(0,1)+iN(0,1) matrix elements. Consider a random matrix γ2 X+X∗ 2 .
3 Wishart (or Laguerre) ensemble. Take α ∈R and an infinite vector V of i.i.d. N(0,1) + iN(0,1) components. Consider a rank 1 random matrix αVV ∗.
A theorem of Pickrell and Olshanski-Vershik claims that sums of indepen-dent matrices from these three ensembles exhaust the list of ergodic U(∞)– invariant measures.
Theorem 20.7 ([Pickrell91], [OlshanskiVershik96]). Ergodic U(∞)–invariant measures on H are parameterized by γ1 ∈R, γ2 ∈R≥0, and a collection of real numbers {αi}, such that ∑i α2 i < ∞. The corresponding random matrix is given by γ1 ·Id+γ2 X +X∗ 2 +∑ i αi ViV ∗ i 2 −Id , (20.8) where all terms are taken to be independent.
Note that, in principle, the subtraction of ∑i αiId could have been absorbed into the γ1 term. However, in this case the sum over i would have failed to converge (on the diagonal) for sequences αi such that ∑i α2 i < ∞, but ∑i |αi| = ∞.
One sees that the GUE–corners process appears in a particular instance of Theorem 20.7 with γ1 = 0 and αi = 0. Let us explain how to link this appear-ance to the limits of random tilings.
173 Lemma 20.8. Let M ∈H be a random U(∞)–invariant infinite hermitian matrix and denote mN 1 ≤mN 2 ≤··· ≤mN N the eigenvalues of its principle N ×N top-left corner. Then the eigenvalues mj i interlace, which means mN+1 i ≤mN i ≤mN+1 i+1 , 1 ≤i ≤N.
Moreover, they satisfy the Gibbs property, which says that conditionally on the values of mN 1 ,mN 2 ,...,mN N, the N(N −1) eigenvalues mj i , 1 ≤i ≤j < N, have uniform distribution on the polytope determined by the interlacing conditions.
Sketch of the proof The interlacement is a general linear algebra fact about eigenvalues of corners of a hermitian matrix. The conditional uniformity can be proven by a version of the argument that we used in determining the density of GUE–corners process in Theorem 20.1.
In fact, one can show that classification of ergodic U(∞)–invariant random hermitian matrices of Theorem 20.7 is equivalent to the classification of all ergodic Gibbs measures on infinite arrays mi j of interlacing real numbers.
Lemma 20.8 explains why we expect the scaling limit of uniform measure on tilings near a straight boundary of the domain to be one of the measures in Theorem 20.1: horizontal lozenges in tiling near the boundary satisfy the discrete versions of the interlacement and Gibbs property. Hence, we expect continuous analogues to hold for the limiting object. But why do we observe only the GUE–corners process, but no deterministic or Wishart components?
This can be explained by very different Law of Large Numbers for these en-sembles. Take a random infinite hermitian matrix M and consider the empirical measure of the eigenvalues of its corner: δ N = 1 N N ∑ i=1 δmN i .
The N →∞behavior of the empirical measure is very different for the three basic examples: 1 For the diagonal matrix γ1 ·Id, δ N is a unit mass at γ1 for all N.
2 For the GUE ensemble with γ2 = 1, the eigenvalues as N →∞fill the seg-ment [−2 √ N,2 √ N], after normalization by √ N, the measure δ N converges to the Wigner semicircle law.
3 For the rank 1 Wishart matrix 1 2VV ∗, the empirical measures δ N has mass N−1 N at 0 with an outlier of mass 1 N at point ≈N as N →∞.
Exercise 20.9. Prove the last statement about rank 1 Wishart matrices.
174 Lecture 20: GUE–corners process and its discrete analogues On the other hand, in the tilings picture, we deal with a point where the frozen boundary is tangent to the straight boundary of the domain. Hence, the frozen boundary locally looks like a parabola, which is consistent only with [−2 √ N,2 √ N] segment filled with eigenvalues in the GUE case.
Conceptual conclusion. The GUE–corners process is expected to appear in the scaling limit near the frozen boundary tangency points whenever we deal with interlacing triangular arrays of particles satisfying (asymptotically) conditional uniformity.
This prediction matches rigorous results going beyond random tilings. For instance, [GorinPanova13, Gorin13, Dimitrov16] demonstrated the appearance of the GUE–corners process as the scaling limit of the six-vertex model.
20.3 A link to asymptotic representation theory and analysis In Theorem 20.7 and Lemma 20.8 we saw how GUE–corners process gets linked to the classification problem of ergodic theory or asymptotic represen-tation theory. But in fact, the same connection exists already on the discrete level of random lozenge tilings.
Let us present a discrete space analogue of Theorem 20.7. For that we let T denote the set of all lozenge tilings of the (right) half–plane, such that far up there are only lozenges of type and far down there are only lozenges of type , cf. Figure 20.2. This constraint implies that there is precisely one horizontal lozenge on the first vertical line, precisely two on the second one, precisely three on the third one, etc. Their coordinates interlace, as in Sections 10.3 or in Proposition 19.9. When one forgets about tilings considering only these coordinates, then the resulting object is usually called an (infinite) Gelfand– Tsetlin pattern or a path in the Gelfand–Tsetlin graph.
We aim to study the probability measures on T , which are Gibbs in the same sense as in Definition 13.4 of Lecture 13. Due to our choice of boundary conditions, the interlacing can be equivalently restated in terms of horizontal lozenges : given the positions of N horizontal lozenges on Nth vertical, the conditional distribution of N(N −1)/2 horizontal lozenges on the first N −1 verticals is uniform on the set defined by the interlacing conditions.
Theorem 20.10 ([AissenSchoenbergWhitney52], [Edrei53, Voiculescu76, VershikKerov82, BorodinOlshanski12, Petrov12c, GorinPanova13]). The er-godic Gibbs measures on T are parameterized by two reals γ+ ≥0, γ−≥0 175 0 N x Figure 20.2 The first four verticals of a tiling of the right half–plane, which has prescribed lozenges far up and far down. Horizontal lozenges interlace and are shown in dark gray. The coordinate system is shown by dashed lines.
and four sequences of non–negative reals α+ 1 ≥α+ 2 ≥··· ≥0, α− 1 ≥α− 2 ≥··· ≥0, β + 1 ≥β + 2 ≥··· ≥0, β − 1 ≥β − 2 ≥··· ≥0, such that ∑∞ i=1(α+ i +β + i +α− i +β − i ) < ∞and β + 1 +β − 1 ≤1.
One exciting feature of Theorem 20.10 (which is also a reason for so many authors producing different proofs of it) is that it is equivalent to two other important theorems, one from the representation theory and another one from the classical analysis. Let us present them.
Recall that infinite-dimensional unitary group U(∞) is defined as the in-ductive limit, or union of finite-dimensional unitary groups U(1) ⊂U(2) ⊂ 176 Lecture 20: GUE–corners process and its discrete analogues U(3) ⊂··· ⊂U(∞). A (normalized) character χ(u) of U(∞) is a continuous complex function on the group, which is: 1 central, i.e. χ(ab) = χ(ba) for all a,b ∈U(∞), 2 positive–definite, i.e. for each k = 1,2,..., complex numbers z1,...,zk, and elements of the group u1,...,uk we have k ∑ i,j=1 ziz jχ(uiu−1 j ) ≥0, 3 Normalized, i.e. χ(e) = 1, where e is the unit element (diagonal matrix of 1s) in U(∞).
The characters ofU(∞) form a convex set, i.e. the convex linear combination of characters is again a character, and we are interested in extreme ones, which can not be decomposed into linear combination of others. As it turns out, the complete list of extreme characters is given by Theorem 20.10. The value of an extreme character corresponding to the parameters of Theorem 20.10 on the matrix U ∈U(∞) is given by χ(U) = ∏ u∈Spectrum(U) eγ+(u−1)+γ−(u−1−1) ∞ ∏ i=1 1+β + i (u−1) 1−α+ i (u−1) · 1+β − i (u−1 −1) 1−α− i (u−1 −1).
(20.9) Note that all but finitely many eigenvalues of a matrix U ∈U(∞) are equal to 1, therefore, the product over u in (20.9) is actually finite. The double multi-plicativity (in u and in i) of the extreme characters is a remarkable feature of U(∞), which has analogues for other infinite–dimensional groups. However, nothing like this is true for the finite–dimensional unitary groups U(N), whose extreme characters are much more complicated (normalized) Schur polynomi-als sλ(u1,...,uN) (see Definition 22.3 in Lecture 22).
The correspondence between Theorem 20.10 and characters of U(∞) goes through the expansion of the restrictions of characters. Take any character χ of U(∞) and consider its restriction to U(N). This restriction is a symmetric func-tion in N eigenvalues u1,...,uN of N ×N unitary matrix and can be expanded in Schur polynomials basis as follows: χ U(N)(u1,...,uN) = ∑ λ=(λ1≥λ2≥···≥λN)∈ZN PN(λ)sλ(u1,...,uN) sλ(1,...,1) .
(20.10) One proves that the coefficients PN(λ) are non–negative and sum up to 1: the first condition follows from positive-definiteness and the second one is implied by the normalization χ(e) = 1. By setting µi = λi + N −i, these coefficients thus define a probability distribution on N tuples of horizontal lozenges as at 177 the Nth vertical line of the lozenge tiling of the right halfplane, cf. Figure 20.2.
With a bit more work one proves that the correspondence χ ↔{P1,P2,P3,...} of (20.10) is a bijection between characters of U(∞) and Gibbs measures on T .
For another equivalent form of Theorem 20.10 consider a doubly–infinite sequence of reals ...d−2,d−1,d0,d1,d2,... and introduce a doubly–infinite Toeplitz matrix D by D[i, j] = dj−i, i, j = ...,−1,0,1,.... The matrix D is called totally–positive, if all (finite) minors of D are non–negative. For a nor-malization, we further assume that the sum of elements in each row of D is 1, i.e. ∑∞ i=−∞di = 1 3. Then the classification of totally positive doubly–infinite Toeplitz matrices is equivalent to Theorem 20.10, with bijection given by the generating series (convergent for u in the unit circle on complex plane): ∞ ∑ i=−∞ uidi = eγ+(u−1)+γ−(u−1−1) ∞ ∏ i=1 1+β + i (u−1) 1−α+ i (u−1) · 1+β − i (u−1 −1) 1−α− i (u−1 −1).
The classification of totally–positive Toeplitz matrices was originally moti-vated by a data smoothing problem. Suppose that we have a real data sequence xi, i ∈Z and a pattern di, i ∈Z, then we can construct a smoothed data sequence yi through yi = ∑ j xi−jdj.
One shows that the amount of sign–changes (“oscillations”) in yi is always smaller than that in xi if and only if the corresponding Toeplitz matrix D[i, j] = dj−i is totally–positive.
The reason why Theorem 20.10 is linked to totally positive Toeplitz matrices lies in the formulas for the coefficients in the decomposition (20.10). From one side, they must be non–negative, as they define probability measures, on the other hand, they essentially are the minors of a certain Toeplitz matrix, as the following exercise shows: Exercise 20.11. Consider the following decomposition of symmetric two-sided power series in variables u1,...,uN: N ∏ i=1 ∑ k∈Z ck(ui)k !
= ∑ λ=(λ1≥λ2≥···≥λN)∈ZN cλ · det h u λj+N−j i iN i, j=1 ∏1≤i<j≤N(ui −u j).
(20.11) Multiplying both sides by the denominator show that the coefficients cλ are 3 A relatively simple argument shows that if D is a totally–positive Toeplitz matrix, then either di = CRi for all i, or replacing di by CRidi with appropriate C and R, we can guarantee the condition ∑∞ i=−∞di = 1. Therefore, the normalization condition does not reduce the generality.
178 Lecture 20: GUE–corners process and its discrete analogues computed as cλ = det cλi−i+j N i, j=1 .
Using the Weyl character formula (cf. [Weyl39]) one identifies the Schur functions sλ(u1,...,uN) with the ratios in the right-hand side of (20.11).
Hence, the coefficients cλ of Exercise 20.11 differ from PN(λ) in (20.10) by the factor sλ(1,...,1). This factor is a positive integer (it counts the dimension of a representation of U(N) as well as lozenge tilings of a trapezoid domain, see Proposition 19.3 and Section 22.2) — hence, non-negativity of cλ and PN(λ) is the same condition.
The character theory for U(∞) also admits a q–deformation, which turns to be closely related both to the study of qVolume–weighted random tilings and to the quantum groups, see [Gorin10, GorinOlshanski15, Sato19].
We end this lecture by a remark that both Theorems 20.7 and 20.10, as well as their q–deformations fit into a general framework for studying Gibbs measures on branching graphs. These graphs can be very different and we refer to [Kerov03], [BorodinOlshanski13], [BorodinOlshanski16, Section 7] for some details and examples.
Lecture 21: Discrete log-gases 21.1 Log-gases and loop equations Recall that in Lecture 19 we computed in (19.1) the law of horizontal lozenges on N-th vertical of randomly tiled equilateral hexagon with side length A: P(x1,...,xN) ∝ ∏ 1≤i<j≤N (xi −xj)2 N ∏ i=1 w(xi), (21.1) where w(x) = (x+1)A−N(A+N −x)A−N. The distribution of form (21.1) with arbitrary w is usually called “log–gas” in statistical mechanics. The name orig-inates from the traditional rewriting of (21.1) as exp β ∑ 1≤i<j≤N H(xi,xj)+ N ∑ i=1 lnw(xi) !
, H(x,y) = ln|x−y|, β = 2, (21.2) where H(x,y) represents the functional form of the interaction between parti-cles (which is logarithmic in our case) and β is the strength of interaction.
The distributions of the form (21.2) with xi ∈R or xi ∈C are widespread in the random-matrix theory, where they appear as distributions of eigenval-ues for various ensembles of matrices, see [Forrester10]. In this setting the parameter β is the dimension of the (skew)–field to which the matrix elements belong. Here β = 1,2,4 corresponds to reals, complex numbers, and quater-nions, respectively. General real values of parameter β also have relevance in theoretical physics and statistical mechanics, but we are not going to discuss this here.
One powerful tool for the study of log-gases as the number of particles N goes to infinity is based on certain exact relations for the expectation of observ-ables1 in the system. The usefulness of these relations was understood by the 1 By an observable we mean a function f(X) of the random system state X, such that the expectation E f(X) is accessible either through explicit formulas or equations fixing it.
179 180 Lecture 21: Discrete log-gases mathematical community since [Johansson98]; before that they were known to theoretical physicists under the name of loop or Dyson–Schwinger equations; we refer to [Guionnet19] for a recent review.
These equations can be treated as far–reaching generalizations of the follow-ing characterization of the Gaussian law. Let ξ be a standard Normal N(0,1) random variable and let f(x) be a polynomial. Then straightforward integration by parts shows that E ξ f(ξ) = E f ′(ξ) (21.3) and the relation (21.3) can be used to compute all the moments Eξ k and, hence, to reconstruct the distribution of ξ.
In a similar spirit the loop equations for (continuous space) log-gases can be obtained by a smart integration by parts and further used to extract probabilistic information.
In the context of this book we are interested in the discrete setting when the particles xi in (21.1) are constrained to the lattice Z. It took a while to generalize the loop equations to such setting, since the naive replacement of integration by parts with summation by parts did not seem to work. The correct form of equations was recently introduced in [BorodinGorinGuionnet15], it was guided by the notion of qq–characters of [Nekrasov15].
Theorem 21.1 (Nekrasov equation, discrete loop equation). Consider a point process (x1 < x2 < ··· < xN) ⊂{0,...,M} with law of the form (21.1). Suppose that there exist functions ϕ+(z),ϕ−(z) such that 1 Functions ϕ+(z) and ϕ−(z) are holomorphic in a complex neighbourhood of [0,M +1] 2 For any x ∈{1,...,M} we have w(x) w(x−1) = ϕ+(x) ϕ−(x).
3 ϕ+(M +1) = ϕ−(0) = 0.
Then the function RN(z) defined by RN(z) = ϕ−(z)E N ∏ i=1 1− 1 z−xi +ϕ+(z)E N ∏ i=1 1+ 1 z−xi −1 (21.4) is holomorphic in the neighbourhood of [0,M +1].
Remark 21.2. The theorem admits a generalization to distributions involving 181 an additional parameter β similarly to (21.2). However, rather than straight-forward ∏i<j |xi −xj|β one has to use a more delicate deformation involving Gamma-functions, see [BorodinGorinGuionnet15] for the details.
Proof of Theorem 21.1 Note that RN(z) is a meromorphic function in the neighbourhood of [0,M] with possible poles only at {0,1,...,M + 1}. More precisely, the first term of (21.4) might have poles only at z = xi ∈{0,...,M} and the second term only at z = xi +1 ∈{1,...,M +1}.
Thus, in order to show that RN(z) is holomorphic, it is enough to prove that RN(z) has no poles at {0,...,M+1}. Note that in the point process (x1,...,xN) all particles are almost surely at distinct positions, hence all poles of RN must be simple.
For the point z = 0, we have ϕ−(0) = 0, hence, in the first term of (21.4) a possible simple pole at z = 0 cancels out with the zero of ϕ−, while the second term has no pole at 0. Therefore, RN(z) has no pole at 0. Similarly for z = M+1 a possible pole in the second term cancels out, while the first term has no poles.
Thus, RN(z) has no pole at M +1.
Let m ∈{1,...,M}. Since all poles of RN(z) are simple, it is enough to prove that the residue at m vanishes. Residue at m results from configurations with xi = m for some i (first term) or configurations with xi = m + 1 for some i.
Expanding E in (21.4) we get resmRN = ∑ i res(i) m RN, where res(i) m RN = −∑ ⃗ x:xi=m ϕ−(m)P(⃗ x)∏ j̸=i 1− 1 m−xj + ∑ ⃗ x:xi=m−1 ϕ+(m)P(⃗ x)∏ j̸=i 1+ 1 m−xj −1 .
Note that we can pair configurations⃗ x in the first term and⃗ x−(i) = (x1,...,xi − 1,...,xN) in the second term. The poles cancel out leading to res(i) m RN = 0, once we prove ϕ−(m)P(x1,...,xi−1,m,xi+1,...,xN)∏ j̸=i 1− 1 m−xj = ϕ+(m)P(x1,...,xi−1,m−1,xi+1,...,xN)∏ j̸=i 1+ 1 m−x j −1 .
(21.5) 182 Lecture 21: Discrete log-gases The validity of (21.5) can be checked directly using (21.1): LHS = ϕ−(m)w(m) ∏ j,k̸=i j<k (xj −xk)2∏ j̸=i (m−xj)2w(xj)m−x j −1 m−xj = ϕ+(m)w(m−1) ∏ j,k̸=i j<k (xj −xk)2∏ j̸=i (m−x j −1)2w(xj) m−xj m−xj −1 = RHS.
Remark 21.3. For the above cancellation argument we also should check the boundary cases when one of the configurations⃗ x,⃗ x−(i) is not well-defined. But the computation above shows that in such cases the corresponding term in res(i) m RN will vanish either due to ∏(xi −xj)2 in the law or due to the boundary condition ϕ+(M +1) = ϕ−(0) = 0.
Theorem 21.1 can be used to describe various probabilistic properties of log-gases. In particular, it can be applied to the asymptotic study of the uniformly random lozenge tilings of the hexagon. We are going to show two applications.
The first one produces another way (in addition to the approaches of Lectures 10 and 16) to compute the limit shape for the tilings and see the frozen bound-ary — inscribed ellipse. The second one (we only sketch the arguments there) leads to the Gaussian Fluctuations, which we already expect to be described by the Gaussian Free Field after Lectures 11 and 12.
21.2 Law of Large Numbers through loop equations We start by specializing Theorem 21.1 to the distribution of the lozenges on the section of hexagon as in (19.1), (21.1).
Note that for w(x) = (x+1)A−N(A+N −x)A−N we have w(x) w(x−1) = x+A−N x A+N −x 2A−x .
Hence, we can take ϕ+(z) = (z+A−N)(A+N −z), ϕ−(z) = z(2A−z), 183 and by Theorem 21.1 the function RN(Nz) N2 = z 2A N −z E N ∏ i=1 1− 1 N(z−xi N ) + z+ A−N N A+N N −z E N ∏ i=1 1+ 1 N(z−xi N −1 N ) !
is an entire function. For the limit shape of tilings, we want to consider N →∞ with A = ˜ aN for some fixed ˜ a. Note that for z away from [0, A+N N ] we have N ∏ i=1 1− 1 N(z−xi N ) = exp N ∑ i=1 ln 1− 1 N(z−xi N ) !
= exp −1 N ∑ i 1 z−xi N +O(1/N) !
→exp(−G(z)), where G(z) is the Cauchy–Stieltjes transform G(z) = ˆ ˜ a+1 0 1 z−yµ(y)dy, and µ is the limiting density at (1,y) of horizontal lozenges as N →∞, similarly to Section 10.3. Recall that the existence of the deterministic µ (or, more generally, the limit shape) was proved earlier using either a concentration of martingales or variational problem (see Lectures 5-7).
Repeating the same computation for the second term, for z away from [0, ˜ a+1] we have RN(Nz) N2 − − − → N→∞R∞(z) = z(2 ˜ a−z)exp(−G(z))+(z+ ˜ a−1)( ˜ a+1−z)exp(G(z)).
(21.6) Proposition 21.4. R∞(z) is an entire function.
Proof Note that outside of [0; ˜ a+1] the holomorphicity is immediate. More-over, outside of [0; ˜ a+1] entire functions RN(Nz) N2 converge to R∞(z).
Take a closed contour C around [0; ˜ a+1]. For any z inside C, Cauchy integral formula reads RN(Nz) N2 = 1 2πi ˛ C RN(Nv) N2(v−z)dv.
Taking N →∞we get a holomorphic inside C function 1 2πi ˛ C R∞(v) (v−z)dv, 184 Lecture 21: Discrete log-gases which coincides with R∞(z) outside of [0; ˜ a+1]. Hence R∞(z) is entire.
Corollary 21.5. R∞(z) is a degree two polynomial.
Proof R∞is an entire function growing as z2 when z →∞. Hence R∞is a degree two polynomial by Liouville’s theorem (by Cauchy’s integral formula the second derivative is bounded, hence constant).
Let us explicitly compute this degree two polynomial. Note that G(z) = 1 z + c z2 +..., when z →∞. Hence, using Taylor expansion for exp, for z →∞we have R∞(z) = z(2 ˜ a−z)(1−G(z)+ G(z)2 2 +o(z−2)) +(z+ ˜ a−1)( ˜ a+1−z)(1+G(z)+ G(z)2 2 +o(z−2)) = −z2 +(2 ˜ a+1)z+(c−2 ˜ a−1 2) + −z2 +(2−1)z+( ˜ a2 −1−c+2−1 2) +o(1) = −2z2 +2( ˜ a+1)z+ ˜ a2 −2 ˜ a+o(1).
But R∞(z) is a polynomial, hence o(1) term is equal to 0 and R∞(z) = −2z2 +2( ˜ a+1)z+ ˜ a2 −2 ˜ a.
Comparing the definition (21.6) of R∞with the expression above, we get a quadratic equation for exp(G(z)), namely z(2 ˜ a−z)exp(−G(z))+(z+ ˜ a−1)( ˜ a+1−z)exp(G(z)) = −2z2 +2( ˜ a+1)z+ ˜ a2 −2 ˜ a.
(21.7) Solving it, we can get an explicit formula for G(z), which is a Stieltjes trans-form of µ, hence, we can recover µ. Integrating it we get a formula for the limit shape (asymptotic height function of tilings). Therefore, Theorem 21.1 gives one more approach to computing the LLN, with others being the Kenyon-Okounkov theory and steepest descent analysis of the correlation kernel.
Exercise 21.6. For each fixed ˜ a the function G(z) found from (21.7) has two real branching points (of type √z; they correspond to the points where the density of µ reaches 0 or 1). If we start varying ˆ a, then these branching points form a curve in ( 1 ˜ a,z) plane. Find out how this curve corresponds to the circle inscribed into the unit hexagon.
Hint: Dealing with lozenges along the vertical line at distance 1 from the left boundary of ˜ a× ˜ a× ˜ a hexagon is essentially the same as dealing with lozenges along the vertical line at distance 1 ˜ a from the left boundary of 1×1×1 hexagon.
185 Exercise 21.7. Consider the distribution on 0 ≤x1 < x2 < ··· < xN ≤M with weight proportional to2 ∏ 1≤i 1.
21.3 Gaussian fluctuations through loop equations Our next step is to improve the arguments of the previous section by looking at lower–order terms of RN(Nz). Eventually this provides access to the global fluctuations in the system.
We only sketch the approach, for the detailed exposition see [BorodinGorinGuionnet15]. We have N ∏ i=1 1− 1 N(z−xi N ) = exp N ∑ i=1 ln 1− 1 N(z−xi N ) !
= exp −G(z)−∆GN(z)+ 1 2N2 N ∑ i=1 1 (z−xi N )2 +O(1/N2) !
, where ∆GN(z) is a random fluctuation of the (empirical) Stieltjes transform: ∆GN(z) = 1 N N ∑ i=1 1 z−xi N − ˆ 1 z−yµ(dy).
Note that as N →∞ 1 2N2 N ∑ i=1 1 (z−xi N )2 = −1 2N2 ∂ ∂z N ∑ i=1 1 z−xi N !
= −1 2N G′(z)+o(1).
Hence, N ∏ i=1 1− 1 N(z−xi N ) = exp −G(z)−∆GN(z)−1 2N G′(z)+O(1/N2) .
Making a similar computation for the second term of RN(Nz) we eventually 2 This distribution appears on sections of uniformly random domino tilings of the Aztec diamond as in Figure 1.10.
186 Lecture 21: Discrete log-gases get RN(Nz) N2 = ϕ− N (Nz)exp(−G(z))+ϕ+ N (Nz)exp(G(z)) +E ∆GN(z) −ϕ− N (Nz)exp(−G(z))+ϕ+ N (Nz)exp(G(z)) + 1 N (some explicit function of G(z))+o(1/N), (21.9) where ϕ+ N (Nz) = ϕ+(Nz) N2 = z+ A−N N A+N N −z , ϕ− N (Nz) = ϕ−(Nz) N2 = z 2A N −z .
Remark 21.8. We use the lower subscript N in ϕ± N (Nz) to emphasize that these functions might depend on N for general discrete log-gases. Note that in the hexagon case with scaling A = ˜ aN the functions ϕ± N (Nz) actually are N–independent.
Remark 21.9. The identity (21.9), or, more precisely, the o(1/N) part of it, requires an asymptotic estimate of E[∆GN(z)]. It turns out that E[∆GN(z)] = O(1/N), but the proof of this fact is a nontrivial argument based on “self-improving” estimates, see [BorodinGorinGuionnet15] for details.
The identity (21.9) has the form RN(Nz) N2 = R∞(z)+E ∆GN(z) Q∞(z)+E +S , (21.10) where E is an explicit expression in terms of the limiting G(z), S is a small term of order o(1/N) and Q∞(z) = −ϕ− N (Nz)exp(−G(z))+ϕ+ N (Nz)exp(G(z)).
Our next step is to find E[∆GN(z)] by solving (21.9) asymptotically as N →∞.
Let us emphasize that we have no explicit expression for RN(Nz), but, never-theless, the equation still can be solved by using a bit of complex analysis.
Note that R∞(z)2 −Q∞(z)2 = 4ϕ− N (Nz)ϕ+ N (Nz).
In the hexagon case, the right-hand side of the last identity is a degree four polynomial, hence Q∞(z)2 is a polynomial. But Q∞(z) grows linearly as z →∞.
Hence Q∞(z) = const · p (z−p)(z−q).
187 (one can show similarly to Exercise 21.6 that p,q above are the intersections of the vertical segment with the boundary of the liquid region).
Take a contour C encircling [0;1 + ˜ a] and let w be a point outside of C.
Integrating (21.10), we have const 2πi ˛ C E ∆GN(z) p (z−p)(z−q) z−w dz = 1 2πi ˛ C 1 N2 RN(Nz)−R∞(z)−E −S z−w dz.
(21.11) Note that 1 N2 RN(Nz)−R∞(z) /(z−w) is analytic inside C, hence its integral vanishes (removing the unknown and asymptotically dominating part of the equation). The function E[∆GN(z)] p (z−p)(z−q) z−w decays as O(z−2) when z →∞, hence, it has no poles outside of C. Therefore, the LHS of (21.11) is equal to minus the residue at w, so that (21.11) transforms into E ∆GN(w) p (w−p)(w−q) = (explicit)+(small), giving an explicit expression for the first moment of ∆GN(w) as N →∞.
Higher moments of ∆GN can be computed repeating the argument above for the deformed distribution (21.1) with weight ˜ w(x) = w(x) k ∏ a=1 1+ ta va −x N , where ta are parameters and va are complex numbers. Then, using the compu-tation above one can evaluate E˜ P[∆GN] for the deformed measure ˜ P. On the other hand: Exercise 21.10. Show that differentiating E˜ P[∆GN] by ∂k ∂t1...∂tk at t1 = ··· = tk = 0 one gets higher order mixed cumulants3 of ∆GN with respect to the original measure.
Further details on how this computation leads to the proof of asymptotic Gaussianity of fluctuations ∆GN (and explicit evaluation of the covariance of ∆GN(z1) and ∆GN(z2)) can be found in [BorodinGorinGuionnet15].
3 See Definition 23.15 and Exercise 23.16 for some further details.
188 Lecture 21: Discrete log-gases 21.4 Orthogonal polynomial ensembles The distributions on N–particle configurations of the form (21.1) are also known as orthogonal polynomial ensembles. The following statement explains the name.
Theorem 21.11. Consider the probability distribution on N–particle configu-rations x1 < x2 < ··· < xN, xi ∈Z of the form P(x1,...,xN) ∝ ∏ 1≤i N. So does the left-hand side.
For the case m = N the identity (21.14) relies on the observation: Exercise 21.12. Let P k(x) = xk + ... be any sequence of monic polynomials.
Then det P i−1(xj) N i, j=1 = ∏ 1≤i<j≤N (xj −xi).
(21.15) Using (21.15) we rewrite (21.12) as det "p w(xi) √ha−1 P a−1(xi) #N i,a=1 det "p w(xj) √ha−1 P a−1(xj) #N a, j=1 .
(21.16) Multiplying the matrices under determinants (summing over the a–indices), 189 we get (21.14) up to a constant normalization prefactor which is implicit in (21.12). For the prefactor verification we need to check that the sum of (21.16) over arbitrary choices of x1,x2,...,xN ∈Z (note that we do not impose any ordering here) is N!. This is done by expanding each determinant as a sum of N! terms, multiplying two sums and then summing each product over xi using (21.13).
For the case m = N −1, using the already proven m = N case, we need to compute the following sum (in which we set ui = xi, i = 1,...,N −1, to match (21.14)): ∑ xN∈Z det K(xi,xj) N i, j=1 (21.17) The definition of the kernel K(x,y) implies the reproducing properties:4 ∑ x∈Z K(x,x) = N, ∑ x∈Z K(y,x)K(x,z) = K(y,z).
(21.18) Expanding the determinant in (21.17) over the last row and column and using (21.18) to compute the sum over xN, we get N det K(xi,xj) N−1 i, j=1 + N−1 ∑ a=1 N−1 ∑ b=1 (−1)a+bK(xa,xB)det K(xi,xj) i̸=a; j̸=b.
The double sum is (minus) the sum of N −1 expansions of det K(xi,x j) N−1 i,j=1 over N −1 possible columns. Therefore, (21.17) equals the desired det K(xi,xj) N−1 i, j=1.
The general 1 < m < N case is similarly obtained by exploiting the orthog-onality relations between polynomials and we leave it to the reader as an ex-ercise. The details are, e.g., in [Mehta04, Section 5.8], [Deift00, Section 5.4], [BorodinGorin12, Section 3].
Let us remark that the particular choice of the lattice Z to which the points xi belong in Theorem 21.11 is not important and the statement remains valid for the orthogonal polynomials on any set. On the other hand, the fact that the interaction is ∏i< j |xi −xj|2 rather than ∏i 0 is crucial for the validity of this theorem.5 Theorem 21.11 reduces the asymptotic questions for β = 2 log-gases to 4 From the functional analysis point of view, these two properties arise because the operator with matrix K(x,y) is an orthogonal projector on the N–dimensional space spanned by the first N orthogonal polynomials. Hence, the trace of this operator is N and its square is equal to itself.
5 Similar, yet slightly more complicated formulas also exist for β = 1,4, cf. [Mehta04, Section 5.7].
190 Lecture 21: Discrete log-gases those for the orthogonal polynomials. Given the importance of this idea, the distribution (19.1) appearing in lozenge tilings of the hexagons is often called the Hahn ensemble, the distribution (21.8) appearing in domino tilings is the Krawtchouk ensemble, and the distribution (19.5) appearing in random Gaus-sian Hermitian matrices is the Hermite ensemble. Each name is derived from the name of the corresponding orthogonal polynomials, and in each case the polynomials are quite special: they belong to the families of hypergeometric or-thogonal polynomials, which can be expressed through hypergeometric func-tions and possess various nice properties, see [KoekoekLeskySwarttouw10] for the Askey scheme of such polynomials and their q–analogues. All the polyno-mials in the Askey scheme are related by various degenerations, which are reflected in limit relations between orthogonal polynomial ensembles. For in-stance, by taking a limit from the Hahn ensemble one can get the Krawtchouk ensemble, and the latter can be further degenerated into the Hermite ensemble (in Proposition 19.8 we observed a direct limit from Hahn to Hermite ensem-ble). [BorodinGorinRains09] found that the q–Racah ensemble (corresponding orthogonal polynomials sit on the very top of the Askey scheme and are related to the basic hypergeometic function 4φ3) can be still found in random tilings of the hexagon in the situation when we deal with more complicated measures than the uniform one.
The connections to classical orthogonal polynomials were exploited in [Johansson00, BaikKriecherbauerMcLaughlinMiller03, Gorin08, BorodinGorinRains09, BreuerDuits13] and applied to study various fea-tures of random lozenge tilings of hexagons. The periodically-weighted lozenge tilings (which we briefly mentioned at the end of Lecture 13) require much more sophisticated orthogonal polynomials, and this case is not so well developed, see [CharlierDuitsKuijlaarsLenells19] and references therein.
For general reviews of the appearances of orthogonal polynomial ensembles in probability we refer to [Deift00, Konig04, BaikDeiftSuidan16].
Lecture 22: Plane partitions and Schur functions Lectures 22 and 23 of the class were given by Andrew Ahn.
At the end of the last lecture we sketched an approach to analysis of the global fluctuations of random tilings based on the discrete loop equations.
When supplied with full details, the approach gives the convergence of fluctu-ations of the height function of random tilings along a vertical section in the hexagon towards a 1d section of the Gaussian Free Field.
The aim of the next two lectures, is to present another approach in a slightly different setting of qVolume–weighted plane partitions which can be viewed as random tilings of certain infinite domains. We are going to prove the full con-vergence of the centered height function to the Gaussian Free Field in this setting, thus checking one particular case of Conjecture 11.1.
22.1 Plane partitions Fix a B ×C rectangular table. A plane partition in B ×C rectangle is a fill-ing of the table with non-negative integers, such that the numbers are weakly decreasing along the rows and columns, as in Figure 22.1.
Observe that a plane partition may be viewed as a stepped surface consisting of unit cubes or as a tiling model with three types of lozenges formed by unions of adjacent equilateral triangles. The surface is given by stacking cubes on a plane to height given by entries in our B×C table. The lozenge tiling is given by projecting such an image (diagonally) onto a plane, as in Figure 22.1. If we know a priori that all the numbers in the plane partition are less or equal than A, then the resulting tiling is non-trivial only inside a A×B×C hexagon.
However, we are more interested in the case of unbounded entries, in which case the tiling might extend in a non-trivial way arbitrary far up. In any case, 191 192 Lecture 22: Plane partitions and Schur functions 0 A x y C B 0 0 1 1 1 2 2 3 3 C B Figure 22.1 Left panel: A plane partition in 3 × 3 rectangle. Right panel: corre-sponding stepped surface, projected onto a 2D plane.
far down the tiling consists solely of horizontal lozenges, far up-right one sees only lozenges and far up-left one sees only lozenges.
From now on in order to simplify the exposition we consider only the sym-metric case B = C and deal with plane partitions in N × N square (the same methods work for B ̸= C as well). On the plane we use the coordinate system shown in the right panel of Figure 22.1 with (0,0) placed at distance N in the positive vertical direction from the bottom tip of the tiling.
We define a modified height function h : Z×Z →Z≥0. We set h(x,−∞) = 0 and let h(x,y)−h(x,y−1) = 0 whenever the vector (x,y−1)−(x,y) crosses a horizontal lozenge and h(x,y)−h(x,y−1) = 1 if the vector (x,y−1)−(x,y) follows an edge of or lozenges. In other words, h(x,y) counts the total number of non-intersecting paths (following , lozenges, see Figures 2.2 and 18.1) below (x,y). This height function differs from the one of Lecture 1 by an affine transformation.
Exercise 22.1. Find explicitly the transformation which matches the modified height function of this section with the one given by (1.3) in Lecture 1.
Unlike tilings of a given hexagon, the total number of plane partitions in an N × N square is infinite. Thus there is no uniform probability measure on tilings. Instead, we study the following q–deformed version: 193 Definition 22.2. Let Volume(π) be the volume of a plane partition π, which is the sum of all N2 integers in the table. Given 0 < q < 1, define the qVolume probability measure on plane partition in N ×N square through PN,q(π) ∝qVolume(π).
If the parameter q is small, then a typical PN,q–random partition has just a few boxes. On the other hand, as q →1, the number of boxes grows to infinity and we can expect non-trivial asymptotic theory.
22.2 Schur polynomials Our treatment of random plane partitions uses Schur polynomials and their combinatorics. We refer to [Macdonald95, Chapter I] for the detailed treatment of symmetric polynomials and only review the necessary facts here.
Let GTN denote the set of all N–tuples of integers λ1 ≥λ2 ≥··· ≥λN. In representation-theoretic context, elements of GTN are often called signatures and the notation GT comes from the names of Gelfand and Tsetlin, who in-troduced an important basis (parameterized by a sequence of signatures) in an irreducible representation of unitary group U(N).
Definition 22.3. Given⃗ k = (k1,...,kN) ∈ZN, define a ⃗ k(x1,...,xN) = det xk1 1 ··· xkN 1 .
.
.
...
.
.
.
xk1 N ··· xkN N = ∑ σ∈SN sign(σ)x kσ(1) 1 ···x kσ(N) N , where the sum is taken over all permutations. Given an λ ∈GTN and setting δ = (N −1,N −2,...,0), define the Schur symmetric (Laurent) polynomials through sλ(x1,...,xN) = aλ+δ(x1,...,xN) aδ(x1,...,xN) .
Exercise 22.4. The denominator aδ is known as Vandermonde determinant.
Check that aδ(x1,...,xN) = ∏ 1≤i<j≤N (xi −xj).
In order to see that sλ(x1,...,xN) is, indeed, a (Laurent) polynomial, no-tice that a ⃗ k vanishes whenever xi = x j, hence, it must divisible by each factor (xi −xj), 1 ≤i < j ≤N, and, therefore, also by their product.
194 Lecture 22: Plane partitions and Schur functions Two signatures λ ∈GTN and µ ∈GTN−1 interlace denoted µ ≺λ if λ1 ≥µ1 ≥λ2 ≥µ2 ≥··· ≥µN−1 ≥λN.
Lemma 22.5 (Branching Rule). For any λ ∈GTN, we have sλ(x1,...,xN) = ∑ µ∈GTN−1,µ≺λ x|λ|−|µ| 1 sµ(x2,...,xN), (22.1) where |λ| = λ1 +···+λN and |µ| = µ1 +···+ µN.
Proof If⃗ ℓ= (λ1+N −1,...,λN) and ⃗ m = (µ1+N −2,...,µN−1), then µ ≺λ means that ℓ1 > m1 ≥ℓ2 > m2 ≥··· ≥ℓN.
Multiplying the right hand side of (22.1) by aδ(x1,...,xN) gives ℓ1−1 ∑ m1=ℓ2 ··· ℓN−1−1 ∑ mN−1=ℓN x|λ|−|µ| 1 ∑ τ∈SN−1 sign(τ)xm1 τ(2) ···xmN−1 τ(N) N ∏ i=2 (x1 −xi), where we consider the action of τ ∈SN−1 on {2,...,N}, so that 2 ≤τ(i) ≤N for i = 2,3,...,N. Writing |λ|−|µ| = (ℓ1 −m1 −1)+···(ℓN−1 −mN−1 −1)+ℓN, the summation becomes ∑ τ∈SN−1 sign(τ)xℓN 1 ℓ1−1 ∑ m1=ℓ2 xℓ1−m1−1 1 xm1 τ(2)(x1 −xτ(2)) ··· ℓN−1 ∑ mN−1=ℓN xℓN−1−mN−1−1 1 xmN τ(N)(x1 −xτ(N)).
Each interior summation telescopes, so we obtain ∑ τ∈SN−1 sign(τ)xℓN 1 (xℓ1−ℓ2 1 xℓ2 τ(2) −xℓ1 τ(2))···(xℓN−1−ℓN 1 xℓN τ(N) −xℓN−1 τ(N)).
The last sum is the determinant of the following N ×N matrix: 0 xℓ1 2 −xℓ1−ℓ2 1 xℓ2 2 ...
xℓ1 N −xℓ1−ℓ2 1 xℓ2 N .
.
.
0 xℓN−1 2 −xℓN−1−ℓN 1 xℓN 2 ...
xℓN−1 N −xℓN−1−ℓN 1 xℓN N xℓN 1 xℓN 2 ...
xℓN N We perform elementary manipulations preserving the determinant on the last matrix. Adding to the (N −1)st row the Nth row multiplied by xℓN−1−ℓN 1 195 we bring the (N −1)st row to the form xℓN−1 j . Further adding to (N −2)nd row a proper multiple of (N −1)st row we bring the (N −2)nd row to the form xℓN−2 j .
Further repeating this procedure, we bring the matrix to the form (xℓi j )N i, j=1, matching the left-hand side of (22.1).
Lemma 22.6 (Cauchy Identity). For |x1|,...,|xN|,|y1|,...,|yN| < 1, we have N ∏ i, j=1 1 1−xiyj = ∑ λ∈GTN,λN≥0 sλ(x1,...,xN)sλ(y1,...,yN).
Proof The Cauchy determinant formula of (20.7) reads det 1 ui −vj N i, j=1 = ∏ 1≤i<j≤N (ui −u j)(vj −vi) N ∏ i, j=1 (ui −vj) which implies det 1 1−xiyj N i, j=1 = ∏ 1≤i<j≤N (xi −xj)(yi −yj) N ∏ i, j=1 (1−xiyj) = aδ(⃗ x)aδ(⃗ y) N ∏ i, j=1 (1−xiyj)−1.
On the other hand, det 1 1−xiyj N i, j=1 = det " ∞ ∑ n=0 (xiy j)n #N i, j=1 = ∞ ∑ α1=0 ··· ∞ ∑ αN=0 det h x αj i y α j j iN i, j=1 = ∞ ∑ α1=0 ··· ∞ ∑ αN=0 aα(⃗ x)yα1 1 ···yαN N = ∑ λ∈GTN,λN≥0 aλ+δ(⃗ x)aλ+δ(⃗ y).
(22.2) Dividing by aδ(⃗ x), aδ(⃗ y) completes the proof.
22.3 Expectations of observables We do not have direct access to the distribution of the value of the height func-tion of random plane partitions at a particular point. However, we can produce contour integral formulas for the expectations of certain polynomial sums in these values, which we call observables. They are rich enough to fully charac-terize the distribution and therefore are useful for the asymptotic analysis.
The main purpose of this section is to obtain the following formula for the expectations of observables of the random height function distributed accord-ing to PN,q.
196 Lecture 22: Plane partitions and Schur functions Theorem 22.7. Let −N < x1 ≤··· ≤xk < N and 0 < t1,...,tk < 1. Then E " k ∏ i=1 ∑ y∈Z h(xi,y)ty i # = 1 (2πi)k ˛ ··· ˛ ∏ 1≤i<j≤k (zj −ti tj zi)(zj −zi) (zj −tizi)(zj −1 t j zi) k ∏ i=1 Gxi(zi;ti) dzi (t−1 i −1)(1−ti)zi , (22.3) where Gx(z;t) = tmin(0,x) ∏ i∈Z+ 1 2 −N<i<min(0,x) 1−t−1q−iz−1 1−q−iz−1 ∏ i∈Z+ 1 2 max(0,x)<i<N 1−qiz 1−tqiz.
(Note that the sum ∑y∈Z h(xi,y)ty i is infinite only in the positive y direction, as h(xi,y) vanishes for sufficiently small y.) The contours satisfy the following properties: • the zi-contour includes the poles at 0,qN−1 2 ,qN−3 2 ,...,q−min(0,xi)−1 2 , but does not contain any of the poles at t−1 i q−max(0,xi)−1 2 ,t−1 i q−max(0,xi)−3 2 ,...,t−1 i q−N+ 1 2 , • for 1 ≤i < j ≤k we have |zi| < |t jzj| (implying also |tizi| < |zj|) everywhere on the contours.
Exercise 22.8. Assume x1 = x2 = ··· = xk. Show that the residue of the inte-grand in (22.3) at zi = tjzj, 1 ≤i < j ≤k (but not for i > j) vanishes. In other words, it does not matter whether the integration contours are nested in such a way that |zi| < |tjzj| or |zi| > |tjzj|, 1 ≤i < j ≤k, — the integral is the same for both choices. (You should still have |tizi| < |zj|, though.) Remark 22.9. A skew plane partition is a generalization in which we place integers not in a rectangular table, but in a difference between a rectangle and a Young diagram. Theorem 22.7 has an extension to qVolume–weighted skew plane partitions, see [Ahn18] for the details.
In the rest of the lecture we outline the main ideas for the proof of the theo-rem.
We first observe that a plane partition π in the N × N square grid may be viewed as a sequence of interlacing signatures (/ 0) = π−N ≺π−N+1 ≺··· ≺π−1 ≺π0 ≻π1 ≻··· ≻πN−1 ≻πN = (/ 0). (22.4) In the notations of the left panel of Figure 22.1, each signature represents a vertical slice of the picture, so that π−N+1 ∈GT1,...,π0 ∈GTN,...,πN−1 ∈ 197 GT1. The slices are read from left to the right. In particular, the plane partition of Figure 22.1 corresponds to the sequence (/ 0) ≺(2) ≺(3,0) ≺(3,1,0) ≻(2,1) ≻(1) ≻(/ 0).
In what follows, we often treat a signature λ ∈GTN (with non-negative coor-dinates) as a signature µ ∈GTM with M > N by adding sufficiently many zero last coordinates. For instance, (2,1) ∈GT2 can be treated as (2,1,0) ∈GT3 or as (2,1,0,0) ∈GT4.
Lemma 22.10. Fix x ∈{−N +1,−N +2,...,N −1} and take arbitrary M ≥ N −|x|. Let λ ∈GTM be the signature encoding the vertical slice of the plane partition at abscissa x, i.e. λ is πx with M −N +|x| additional zeros. For any 0 < t < 1, we have ∑ y∈Z h(x,y)ty = tmin(0,x) 1−t M ∑ i=1 tλi−i+1 −t−M+1 1−t !
.
Proof The appearance of min(0,x) in the formulas is related to our particular choice of the coordinate system, as in the right panel of Figure 22.1. The ith from top horizontal lozenge in the vertical section of the tilings at abscissa x has top and bottom coordinates πx i −i + 1 + min(0,x) and πx i −i + min(0,x), respectively. With this in mind and using that h(x,y)−h(x,y−1) = 1 unless y corresponds to the top of a horizontal lozenge, we compute ∑ y∈Z h(x,y)ty = 1 1−t ∑ y∈Z h(x,y)(ty −ty+1) = 1 1−t ∑ y∈Z (h(x,y)−h(x,y−1))ty = 1 1−t ∑ y∈Z (1−1[y ∈{πx i −i+1+min(0,x)}∞ i=1])ty = 1 1−t ∑ y>−M+min(0,x) (1−1[y ∈{πx i −i+1+min(0,x)}M i=1])ty = 1 1−t − M ∑ i=1 tπx i −i+1+min(0,x) + t−M+1+min(0,x) 1−t !
, where in the third line πx is extended to have infinitely many coordinates by adding (infinitely many) zeros, in the fourth line πx is extended to an element of GTM (by adding M −N + |x| zeros), and the last line is obtained from the fourth by summing ones and indicators 1 separately.
198 Lecture 22: Plane partitions and Schur functions In view of the previous lemma, set ℘ t(λ) = M ∑ i=1 tλi−i+1 −t−M+1 1−t , λ ∈GTM, λM ≥0.
Note that if we increase M and simultaneously extend λ by adding zeros, then ℘ t(λ) is unchanged.
Our plan is to compute expectations of products of℘ t(πx), and then use the previous lemma to obtain Theorem 22.7. This is achieved using the following family of difference operators.
Definition 22.11. The first Macdonald operator D{x1,...,xn} t is defined by D{x1,...,xn} t = n ∑ i=1 " ∏ j̸=i xi −t−1xj xi −xj # Tt,xi −t−n+1 1−t where Tt,xi is the t–shift operator which maps F(x1,...,xi−1,xi,xi+1,...,xN) 7→F(x1,...,xi−1,txi,xi+1,...,xN).
Remark 22.12. The operators D{x1,...,xn} t are q = t versions of the difference op-erators used by Ian Macdonald in his study of the (q,t)–deformation of Schur polynomials bearing his name, see [Macdonald95, Chapter VI].
Lemma 22.13. The operator D{x1,...,xn} t is diagonalized by the Schur functions sλ(x1,...,xn), λ ∈GTn. We have D{x1,...,xn} t sλ(x1,...,xn) =℘ t(λ)sλ(x1,...,xn).
Proof Observe that D⃗ x t + t−n+1 1−t sλ(⃗ x) =t−n+1 n ∑ i=1 " ∏ j̸=i txi −xj xi −x j # Tt,xiaλ+δ(⃗ x) Tt,xiaδ(⃗ x) = n ∑ i=1 Tt,xiaλ+δ(⃗ x) aδ(⃗ x) where the second equality uses that the txi−x j xi−x j terms replace the t-shifted fac-tors in Tt,xiaδ(⃗ x) with the corresponding ordinary factors in aδ(⃗ x). Writing the alternant as aλ+δ(⃗ x) = ∑ σ∈Sn sgn(σ) n ∏ j=1 x λσ( j)+n−σ(j) j we have that n ∑ i=1 Tt,xiaλ+δ(⃗ x) = n ∑ i=1 tλi+n−i !
aλ+δ(⃗ x) which implies the statement of the lemma.
199 We are able to produce a contour integral representation for the action of these operators on multiplicative functions ∏n i=1 f(xi): Lemma 22.14 ([BorodinCorwin11]). Let f(z) be analytic and q ∈C such that f(qz) f(z) has no singularities in a neighborhood of {0}∪{xi}n i=1. Then D⃗ x t n ∏ i=1 f(xi) = n ∏ i=1 f(xi) ˛ n ∏ j=1 z−t−1xj z−x j · f(tz) f(z) dz (1−t−1)z, where the contour contains 0 and {xi}n i=1 and no other singularities of the integrand.
Proof This follows from the residue theorem and the definition of the opera-tor D⃗ x t .
The idea of the proof of Theorem 22.7 is to use the Dt operators to extract the volume information relying on the following lemma: Lemma 22.15. Let F denote a function of 2N variables a1,...,aN, b1,...bN given by F = N ∏ i, j=1 1 1−aibj .
(22.5) Then for qVolume random plane partitions identified with interlacing sequences of signatures through (22.4), any −N < x1 ≤··· ≤xk ≤0, N > x′ 1 ≥··· ≥x′ k′ ≥0, and any 0 < ti,t′ i < 1, we have E h ℘ t1(πx1)···℘ tk(πxk)·℘ t1(πx′ 1)···℘ t′ k(πx′ k′) i = F−1D {aN,...,aN+x1} t1 ···D {aN,...,aN+xk} tk D {bN,...,bx′ 1−1} t′ 1 ··· ···D {bN,...,bx′ k−1} t′ k F ai=bi=qi−1/2,i=1,...,N .
(22.6) Exercise 22.16. Given Lemma 22.15, prove Theorem 22.7 by a recursive ap-plication of Lemma 22.14 to convert the right-hand side (22.6) into contour integrals.
Let us present a historic overview for Lemma 22.15. Connection be-tween plane partitions and Schur polynomials was first exploited in prob-abilistic context in [OkounkovReshetikhin01]. The use of the differential operators for the case k′ = 0 and x1 = x2 = ··· = xk was introduced in [BorodinCorwin11]. The extension to arbitrary xi (still k′ = 0) was pre-sented in [BorodinCorwinGorinShakirov13, Section 4]. An observation that 200 Lecture 22: Plane partitions and Schur functions k′ > 0 is also possible can be found in [BufetovGorin17, Section 3]. Also [BorodinCorwinGorinShakirov13, Section 3] presents another way to reach Theorem 22.14 avoiding k′ > 0 case, and [Aggarwal14, Proposition 2.2.4] is a generalization of Theorem 22.14 obtained by this method. We also refer to [Ahn18] where another set of difference operators is used to extract observ-ables of plane partitions.
We only present the k = 1, k′ = 0 case of Lemma 22.15; general case is an extension of the arguments below and details can be found in the aforemen-tioned references.
Proof of k = 1, k′ = 0 case of Lemma 22.15 The first observation is that with the notation |πi| = πi 1 +πi 2 +..., we have qVolume(π) = q 1 2 (|π0|−|π1|)q 3 2 (|π1|−|π2|) ···q(N−1 2 )|πN−1| ×q 1 2 (|π0|−|π−1|)q 3 2 (|π−1|−|π−2|) ···q(N−1 2 )|π−N+1|.
Hence, if we take an expression ∑ λ∈GTN λN≥0 sλ qN−1 2 ,qN−3 2 ,...,q1/2 sλ qN−1 2 ,qN−3 2 ,...,q1/2 and expand both sλ factors into sums of monomials by recursively using the branching rule of Lemma 22.5, then each term in the expanded sum corre-sponds to a plane partition as in (22.4) and the value of such term is precisely qVolume(π).
We can also first expand ∑ λ∈GTN,λN≥0 sλ (a1,...,aN)sλ (b1,...,bN) (22.7) into monomials in ai, bi, getting a sum over plane partitions and substitute ai = bi = qi−1/2 later on. Note that (22.7) evaluates to the function F of (22.5) by Lemma 22.6. In particular, we get [F]ai=bi=qi−1/1 = ∑ π qVolume(π).
(22.8) Assume without loss of generality that x1 ≤0. Let us expand (22.7) into monomials corresponding to plane partitions π, then fix πx1,πx1+1,...,πN−1 and make a summation over π1−N,π2−N,...,πx1−1. We claim that the result has the form sπx1(aN,aN−1,...,a1−x1)×(terms depending on a−x1,...,a1, b1,...,bN).
(22.9) Indeed, the summation goes over interlacing signatures, hence, we can again 201 use Lemma 22.5 to turn the result into the Schur function. We can now act with D {aN,aN−1,...,a1−x1} t on F, by using (22.9) and eigenrelation of Lemma 22.13. The factor ℘ t(πx1) pops out, and afterwards we can again expand sπx1(aN,aN−1,...,a1−x1) into monomials. The conclusion is that h D {aN,aN−1,...,a1−x1} t F i ai=bi=qi−1/2 = ∑ π ℘ t(πx1)qVolume(π).
Dividing by F and using (22.8) we get the desired evaluation of E℘ t(πx1).
Exercise 22.17. Using (22.8) compute the partition function of qVolume– weighted plane partitions, that is, show that1 ∑ plane partitions qVolume = ∞ ∏ n=1 (1−qn)−n, (22.10) where the sum is taken over all plane partitions, which is the N = ∞version of the ones in Definition 22.2.
1 Similarly to (1.1), this formula also bears the name of MacMahon and has been known for more than 100 years, although the first proofs were not using Schur polynomials.
Lecture 23: Limit shape and fluctuations for plane partitions In this lecture we use Theorem 22.7 to analyze global asymptotics (LLN and CLT) for random qVolume–weighted plane partitions.
In what follows we fix ˆ N > 0 and set q = e−ε, N = ⌊ˆ N/ε⌋for a small pa-rameter ε > 0.
Our goal is to show that for the random height function h(x,y) of plane partitions in N × N rectangle and distributed according toPN,q of Definition 22.2, we have as ε →0 εh ˆ x ε , ˆ y ε →h∞(ˆ x, ˆ y) h ˆ x ε , ˆ y ε −Eh ˆ x ε , ˆ y ε →GFF where h∞is a deterministic limit shape, and GFF is the Gaussian Free Field governing fluctuations, as in Lectures 11 and 12.
23.1 Law of Large Numbers Our first result is the limit shape theorem for the height function of the qVolume– weighted random plane partitions in N ×N rectangle.
Theorem 23.1. For any ˆ x ∈[−ˆ N, ˆ N] and c > 0, we have lim ε→0E " ∑ ˆ y∈εZ ε2h ˆ x ε , ˆ y ε e−cˆ y # = ˆ h∞(ˆ x, ˆ y)e−cˆ y d ˆ y, (23.1) where h∞is an explicit function described in Theorem 23.5 below.
202 203 Remark 23.2. In the next section we show that the variance of the height func-tion decays, which automatically upgrades the convergence of expectations in (23.1) to convergence in probability.
Remark 23.3. Formally, the height function h(x,y) is defined only at integer points (x,y), and therefore, we should write h ˆ x ε , j ˆ y ε k , where ⌊·⌋is the integer part. However, we are going to omit the integer part in order to simplify the notations.
The proof of Theorem 23.1 starts by expressing (23.1) as a contour integral.
Lemma 23.4. For any c > 0 we have lim ε→0E " ∑ ˆ y∈εZ ε2h ˆ x ε , ˆ y ε e−cˆ y # = 1 2πi ˛ 0 Gˆ x(z)c dz c2z, (23.2) where Gˆ x(z) = 1−z e−ˆ N −z · e−ˆ x −z−1 e−ˆ N −z−1 , (23.3) the branch for raising to the cth power is the one giving positive real values to z = 0 and z = ∞, and integration goes over a positively oriented contour enclosing 0 and [e−ˆ N,emin(0,ˆ x)], but not [emax(0,ˆ x),e ˆ N].
Proof The k = 1 case of Theorem 22.7 reads E " ∑ y∈Z h(x,y)ty # = 1 2πi ˛ Gx(z;i) dz (t−1 −1)(1−t)z, (23.4) where Gx(z;t) = tmin(0,x) ∏ i∈Z+ 1 2 −N<i<min(0,x) 1−t−1q−iz−1 1−q−iz−1 ∏ i∈Z+ 1 2 max(0,x)<i<N 1−qiz 1−tqiz, and the integration contour includes the poles at 0,qN−1 2 ,qN−3 2 ,...,qmin(0,x)−1 2 .
We set t = exp(−cε), q = exp(−ε), x = ˆ x ε , y = ˆ y ε , N = ˆ N ε and send ε →0. Let us analyze the asymptotic behavior of Gx(z;t). Clearly, tmin(0,x) →e−cmin(0,ˆ x). Using the asymptotic expansion ln(1 + u) = u + o(u), 204 Lecture 23: Limit shape and fluctuations for plane partitions the logarithm of the first product in Gx(z,t) behaves as: ∑ i∈Z+ 1 2 −ˆ N/ε<i<min(0,ˆ x/ε) ln 1−t−1q−iz−1 1−q−iz−1 = ∑ i∈Z+ 1 2 −ˆ N/ε<i<min(0,ˆ x/ε) (1−t−1) q−iz−1 1−q−iz−1 +o(1) = 1−t−1 1−q ∑ i∈Z+ 1 2 −ˆ N/ε −ˆ N. Then for the first term in (23.9) we notice that e ˆ N w = e ˆ N+ˆ y > 1.
Hence, imaginary part of the logarithm makes a jump by 2π.
For the second term in (23.9), note that upon substitution w = e−ˆ y and def-inition of Gˆ x(z) from (23.3), we transform (23.10) into a quadratic equation in z with real coefficients. It either has real roots (in which case, there is no interesting jump for ln(z0)), or complex conjugate roots. One can check that the jump of ln(z0) comes precisely from the choice of the root in the upper half-plane changing to the choice of the one in the lower half-plane. Hence, the jump of the imaginary part of ln(z0) is twice the argument of z0. Let us record our findings in the theorem.
1 For simplicity, assume that u is a point of continuity of µˆ x. Then this fact is proven by direct computation of the imaginary part as we approach u from the upper and lower halfplanes.
207 Theorem 23.5. Let ζ(˜ x, ˜ y) be a root of the quadratic equation 1−z e−ˆ N −z · e−ˆ x −z−1 e−ˆ N −z−1 = e−ˆ y (23.11) chosen so that ζ(ˆ x, ˆ y) is continuous in (ˆ x, ˆ y) and ζ is in the upper half-plane H, if (23.11) has two complex conjugate roots. Then ∂ˆ yh∞(ˆ x, ˆ y) = 1−argζ(ˆ x, ˆ y) π .
Remark 23.6. Notice that if ζ(˜ x, ˜ y) is real, then ∂ˆ yh∞(ˆ x, ˆ y) = 0 or 1. This cor-responds to the frozen region, either beneath the plane partition (where h is not changing) or above the plane partition (where the sufrace is vertical, and h increases with ˆ y). For (ˆ x, ˆ y) such that ζ(ˆ x, ˆ y) ∈H, we have 0 < ∂ˆ yh∞(ˆ x, ˆ y) < 1.
This corresponds exactly to the liquid region D. Therefore, the liquid region may be alternatively defined as the set of (˜ x, ˜ y) ∈R2 such that (23.11) has a pair of complex conjugate roots.
Remark 23.7. Let us match the result of Theorem 23.5 with Kenyon-Okounkov theory developed in Lectures 9 and 10. We define a complex number z, so that ζ(ˆ x, ˆ y) = eˆ x(1−z(ˆ x, ˆ y)).
Note that since ζ is in the upper half-plane, so is z. Conjugating the equation (23.11) and rewriting it in terms of z, we get Q(zey,(1−z)ex) = 0, Q(u,v) = u(1−v)−(e−ˆ N −v)(1−ve−ˆ N).
This is precisely the equation of the form appearing in Theorems 9.8, 10.1 and giving solutions to the complex Burgers equation for the limit shape. Indeed, the parameter c appearing in those theorems is −1 in our setting, see Corollary 9.6 and identify ε with 1/L. Note also that arg(ζ) is computing the density of horizontal . Hence, this density is also computed by −arg(1 −z), which matches Figure 9.1. With a bit more care, one can show that z(ˆ x, ˆ y) is precisely Kenyon–Okounkov’s complex slope.
Exercise 23.8. Extend Theorems 23.1 and 23.5 to qVolume–weighted plane par-titions in N ×⌊αN⌋rectangle, α > 0, as N →∞.
Exercise 23.9. Sending ˆ N →∞in the result of Theorem 23.5 we get the limit shape for the plane partitions without bounding rectangle. Find explicitly the formula for the boundary of the frozen region in this situation. The answer (drawn on the plane of the right panel of Figure 22.1) should be invariant under rotations by 120 degrees.
208 Lecture 23: Limit shape and fluctuations for plane partitions The limit shape theorem for the qVolume–weighted plane partitions (with-out bounding rectangle) was addressed by several authors some of whom did not know about the existence of others. We refer to [BloteHilhorst82], [NienhuisHilhorstBlote84], [Vershik97] for the first appearances and to [CerfKenyon01], [OkounkovReshetikhin01] for later treatments in the math-ematical literature.
23.2 Central Limit Theorem Our next aim is to show that the centered height function √π h ˆ x ε , ˆ y ε −Eh ˆ x ε , ˆ y ε converges to the pullback of the Gaussian free field (GFF) under the map ζ(ˆ x, ˆ y) from Theorem 23.5.
We briefly recall the definition of the GFF, see Lecture 11 for more details.
Definition 23.10. The Gaussian Free Field on the upper half-plane H = {z ∈ C | ℑ(z) > 0} is a random generalized centered Gaussian function GFF on H with covariance EGFF(z)GFF(w) = −1 2π ln z−w z−¯ w .
Given a bijection ζ : D →H, where D ⊂R2, we define the GFF pullback.
Definition 23.11. The ζ-pullback of the GFF is a random distribution GFF ◦ζ such that E[GFF ◦ζ(x1,y1)·GFF ◦ζ(x2,y2)] = −1 2π ln ζ(x1,y1)−ζ(x2,y2) ζ(x1,y1)−¯ ζ(x2,y2) .
Theorem 23.12. For the height functions of qVolume–weighted plane partitions, we have ( √π ∑ y∈εZ ε h ˆ x ε , ˆ y ε −Eh ˆ x ε , ˆ y ε e−cˆ y ) c>0,−ˆ N≤ˆ x≤ˆ N → ˆ Dˆ x H ◦ζ(ˆ x, ˆ y)e−cy c>0,−N≤x≤N (23.12) in the sense of convergence of finite dimensional distributions as ε →0. Here ζ(ˆ x, ˆ y) is as in Theorem 23.5, D is the liquid region (which can be defined as 209 those (ˆ x, ˆ y), for which ζ is non-real), and Dˆ x is the section of D by the vertical line with abscissa ˆ x.
Remark 23.13. By Remark 23.7, ζ = ex(1 −z), where z is the Kenyon– Okounkov’s complex slope. Because Q(eyz,ex(1 −z)) = 0 for an analytic Q, the complex structure of ζ is the same as the one of eyz. Noting also that the Laplacian in the definition of the GFF is unchanged under the complex con-jugation of the complex structure, we conclude that Theorem 23.12 proves a particular case of Conjecture 11.1.
Exercise 23.14. Show that ζ : D →H is a homeomorphism from D to H, which maps sets of the form {(x,y) ∈D : y ∈R} for fixed x to half circles in H with real center points.
The convergence in the sense of finite dimensional distributions in Theorem 23.12 means that if we take any finite collection c1,...,ck, −ˆ N ≤ˆ x1,..., ˆ xk ≤ˆ N, then we have the convergence in distribution for random vectors lim ε→0 ∑ ˆ y∈εZ ε h ˆ xi ε , ˆ y ε −Eh ˆ xi ε , ˆ y ε e−ci ˆ y !k i=1 = ˆ Dˆ xi H ◦ζ(ˆ x, ˆ y)e−ci ˆ yd ˆ y !k i=1 .
(23.13) We can prove such a statement by showing that the random vectors above converge to a Gaussian limit and then matching their covariance. We outline some of the ideas of the proof here and refer to [Ahn18] for more details.
Definition 23.15. Given random variables X1,...,Xn, their joint mixed cumu-lant κ(X1,...,Xn) is defined as κ(X1,...,Xn) = ∂n ∂t1 ...∂tn lnE " exp i N ∑ i=1 tiXi !# t1=t2=···=tn=0 .
If we fix random variablesY1,...,Yk, then its family of cumulants are defined as all possible κ(X1,...,Xn), n = 1,2,..., where Xi coincide with various Yj with possible repetitions. n is then the order of the cumulant. For instance, if we have one random variable Y1 = Y, then its first cumulant is κ(Y) = EY, and the second cumulant κ(Y,Y) = EY 2 −(EY)2. For two random variables the second order cumulant is the covariance, κ(X1,X2) = cov(X1,X2).
210 Lecture 23: Limit shape and fluctuations for plane partitions For a Gaussian vector (Y1,...,Yk), the logarithm of its characteristic function is a second degree polynomial. This leads to the following statement: Exercise 23.16. A random vector (Y1,...,Yk) has Gaussian distribution if and only if all joint cumulants of its coordinates of order ≥3 vanish.
In principle, one can express joint cumulants through joint moments. In this way, vanishing of the higher-order cumulants is equivalent to the Wick’s for-mula for the computation of joint moments for Gaussian vector as a combina-torial sum involving perfect matchings.
Hence, in order to prove Theorem 23.12 we need two steps: 1 Compute the covariance for random variables ∑y∈εZ ε h ˆ xi ε , ˆ y ε e−ci ˆ y us-ing Theorem 22.7 and show that it converges to the covariance of the vectors in the right-hand side of (23.13).
2 Compute higher order cumulants for the same random variables and show that they converge to 0.
We only outline the first step. The asymptotic vanishing of the higher or-der cumulants for the second step holds in much greater generality for random variables, whose joint moments are given by contour integrals as in Theorem 22.7. We refer to [BorodinGorin13, Section 4.3], [GorinZhang16, Section 6.2], [Ahn18, Section 5.3] for such arguments. For the identification of the covari-ance structure, take c1,c2 > 0 and −ˆ N ≤ˆ x1, ˆ x2 ≤ˆ N. Then Cov ∑ ˆ y∈εZ ε h ˆ x1 ε , ˆ y ε −Eh ˆ x1 ε , ˆ y ε e−c1 ˆ y, ∑ ˆ y∈εZ ε h ˆ x2 ε , ˆ y ε −Eh ˆ x2 ε , ˆ y ε e−c2 ˆ y !
= E " 2 ∏ i=1 ∑ ˆ y∈εZ εh ˆ xi ε , ˆ y ε e−ci ˆ y # − 2 ∏ i=1 E " ε ∑ ˆ y∈εZ h ˆ xi ε , ˆ y ε e−ci ˆ y # .
By Theorem 22.7, the above is 1 (2πi)2 ˛ ˛ (z2 −t1 t2 z1)(z2 −z1) (z2 −t1z1)(z2 −1 t2 z1) −1 !
2 ∏ i=1 " Gxi(zi;ti) εdzi (t−1 i −1)(1−ti)zi # , where ti = e−ciε, xi = ˜ xi ε . Since (z2 −t1 t2 z1)(z2 −z1) (z2 −t1z1)(z2 −1 t2 z1) −1 = (1−t1)( 1 t2 −1)z1z2 (z2 −t1z1)(z2 −1 t2 z1), 211 the ε factors balance out with 1 −ti to give us an overall 1/(c1c2) factor. By (23.5), we obtain 1 c1c2 · 1 (2πi)2 ˛ ˛ Gˆ x1(z1)c1Gˆ x2(z2)c2 (z2 −z1)2 dz1 dz2.
(23.14) It remains to show that (23.14) is the covariance of the GFF paired with expo-nential test functions on the sections.
Let Ci be a closed contour symmetric with respect to the x-axis, and the part of Ci in H is the half circle which is ζ–image of the vertical section of the liquid region with abscissa ˆ xi, i.e., the set {ζ(ˆ xi, ˆ y) | y ∈Dˆ xi}. We deform the integration contours in (23.5) to Ci and then split Ci = C + i ∪C − i , where C + i = Ci ∩H. Changing variables via ζ(x,y) 7→(x,y) and using (23.11), we rewrite (23.14) as − 1 4π2c1c2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1e−c2 ˆ y2 (ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y2))2 ∂ζ ∂ˆ y1 (ˆ x1, ˆ y1) ∂ζ ∂ˆ y2 (ˆ x2, ˆ y2)d ˆ y1 d ˆ y2 + 1 4π2c1c2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1e−c2 ˆ y2 (ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y2))2 ∂ζ ∂ˆ y1 (ˆ x1, ˆ y1) ∂ζ ∂ˆ y2 (ˆ x2, ˆ y2)d ˆ y1 d ˆ y2 + 1 4π2c1c2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1e−c2 ˆ y2 (ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y2))2 ∂ζ ∂ˆ y1 (ˆ x1, ˆ y1) ∂ζ ∂ˆ y2 (ˆ x2, ˆ y2)d ˆ y1 d ˆ y2 − 1 4π2c1c2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1e−c2 ˆ y2 (ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y2))2 ∂ζ ∂ˆ y1 (ˆ x1, ˆ y1) ∂ζ ∂ˆ y2 (ˆ x2, ˆ y2)d ˆ y1 d ˆ y2.
(23.15) Integrate by parts on ˆ y1 and ˆ y2 for each summand, observing that the boundary terms cancel since the value of ζ(x,·) at the end points of Dx is real, to obtain −1 4π2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1−c2 ˆ y1 ln ζ(x1,y1)−ζ(x2,y1) −ln ζ(x1,y1)−ζ(x2,y1) −ln ζ(x1,y1)−ζ(x2,y1) +ln ζ(x1,y1)−ζ(x2,y1) dy1 dy1 = −1 2π2 ˆ Dˆ x2 ˆ Dˆ x1 e−c1 ˆ y1e−c2 ˆ y1 ln ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y1) ζ(ˆ x1, ˆ y1)−ζ(ˆ x2, ˆ y1) d ˆ y1 d ˆ y2.
This is precisely the desired formula for the covariance of the integrals of the GFF against exponential test functions. The prefactor 1 2π2 matches the one in Conjecture 11.1; it also mathes the ratio of 1 2π in Definition 23.11 and (√π)2 coming from the prefactor in (23.12).
Lecture 24: Discrete Gaussian compo-nent in fluctuations 24.1 Random heights of holes There are two frameworks for picking a uniformly random tiling: 1 Fix a domain, pick a tiling of this domain uniformly at random.
2 Fix a domain and a height function on the boundary, pick a height func-tion inside the domain extending the boundary height function uniformly at random.
For simply connected domains, the two points of view are equivalent. For do-mains with holes, the two might not be equivalent, as the height of holes are fixed in (2), but there can be multiple possibilities for the heights of the holes in (1). So in (1), the height of a hole is in general a nonconstant random variable.
In expectation, the height of a hole grows linearly with the size of the domain.
In this lecture, we are interested in determining the order of fluctuations of the height around its expectation. It turns out that the perspective that leads to the correct answer is thinking of these as global fluctuations, and expecting the fluctuations to have finite order, as the GFF heuristic predicts (cf. Lectures 11,12, 21, and 23). The limit is a discrete random variable, and we will present two approaches to finding it. The first is entirely heuristic, but applies to very general of domains. The second approach can be carried out rigorously (al-though our exposition here will still involve some heuristic steps), but only applies to a specific class of domains.
212 213 Figure 24.1 A hexagon with a rhombic hole.
24.2 Discrete fluctuations of heights through GFF heuristics.
In this section, we are going to work with an arbitrary domain with a single hole (which can be of any shape), and a reader can take a hexagon with a rhombic hole in Figures 24.1, 24.2 as a running example. Let h be the centered height function on the domain. By this we mean that at each point, h is the difference between the height function and the expected height at that point. Throughout this section we work with a modified version of the height function of (9.3), (9.4). This means that h(x,y) can be thought of as the (centered) total number of horizontal lozenges in the tiling situated on the same vertical line as (x,y) below (x,y).
We start with the GFF heuristic – see Lectures 11-12 for the derivation. For large domains, the distribution of the centered height function (or rather its density) can be approximated as P(h) ≈exp −π 2 ¨ ∥∇h∥2 .
The above integral is taken in the complex coordinates related to the Gaussian Free Field asymptotics; the integration domain is in bijection with the liquid region.
Outside the liquid region, the lozenges are deterministic with overwhelming probability. This means that for a typical tiling, the height increment between any two points in a frozen region is deterministic and equal to the increment 214 Lecture 24: Discrete Gaussian component in fluctuations Figure 24.2 A typical tiling of a hexagon with a hole. Simulation by Leonid Petrov.
in expectation. Hence, the centered height function is the same at all points of a frozen region. Thus, picking a point on the outer boundary of the domain to have zero centered height, we have that the centered height function on the outer frozen region is zero, and the centered height on the frozen region around the hole is δ, which is the same as the fluctuation h(B)1 of the height of the hole. Using our GFF heuristic, P(h(B) = δ) ≈ ˆ Ωδ exp −π 2 ¨ ∥∇h∥2 , (24.1) 1 h(B) stays for “Height of Boundary”.
215 where Ωδ is the subset of height functions for which the height of the inner frozen region is δ.
We would like to understand how (24.1) changes as we vary δ. For that we take a (unique) harmonic function gδ on the liquid region that satisfies gδ = 0 boundary condition on the outer boundary and gδ = δ boundary condition on the internal boundary surrounding the hole. The harmonicity is with respect to the local coordinates of the GFF complex structure. Then for h ∈Ωδ we can define ˜ h = h−gδ, and notice that ˜ h is 0 on all boundaries of the liquid region.
This takes us a step closer to evaluating the integral: h = ˜ h+gδ ¨ ∥∇h∥2 = ¨ ∥∇˜ h+∇gδ∥2 = ¨ ∥∇˜ h∥2 + ¨ ∥∇gδ∥2 +2 ¨ ⟨∇˜ h,∇gδ⟩ We can integrate the last summand by parts: ¨ ⟨∇˜ h,∇gδ⟩= boundary term− ¨ ˜ h∇2gδ.
The boundary term is 0 since ˜ h is 0 on the boundary, and the other term is 0 since ∇2gδ = 0 as gδ is harmonic.
Hence, we can write the probability we are interested in as P(h(B) = δ) ≈ ˆ Ωδ exp −π 2 ¨ ∥∇˜ h∥2 + ¨ ∥∇gδ∥2 .
Notice that for any δ, δg1 is harmonic and satisfies the same boundary condi-tions as gδ, so gδ = δg1. Hence, we can write P(h(B) = δ) ≈ ˆ Ωδ exp −π 2 ¨ ∥∇˜ h∥2 ·exp −π 2 δ 2 ¨ ∥∇g1∥2 = exp(−Cδ 2) ˆ Ωδ exp −π 2 ¨ ∥∇˜ h∥2 .
Note that ˜ h in the last integral is an element of Ω0 and therefore is independent of δ. Hence, the integral evaluates to a δ–independent constant. Therefore, we have provided a heuristic argument for the following conjecture.
Conjecture 24.1. As the linear size of the domain L →∞, the height H of a hole is an integer random variable that becomes arbitrarily close in distribu-tion to the discrete Gaussian Distribution const·exp(−C(H −m)2), (24.2) 216 Lecture 24: Discrete Gaussian component in fluctuations Figure 24.3 A domain with multiple holes where m is a certain (unknown at this point) shift and the scaling factor C can be computed as the Dirichlet energy C = π 2 ¨ ∥∇g1∥2, and g1 is the harmonic function in the liquid region (with respect to the Kenyon-Okounkov’s complex structure) with g1 = 0 on the external boundary and g1 = 1 on the internal boundary.
The same heuristic argument can also be given for domains with multiple holes, as in Figure 24.3, leading us to the following conjecture, which the au-thor learned from Slava Rychkov.
Conjecture 24.2. For a domain with an arbitrary collection of K holes, the asymptotic distribution of the hole heights is a K-dimensional discrete Gaus-sian vector with scale matrix given by the quadratic form ˜ ⟨∇f,∇g⟩of har-monic functions in the liquid region with Kenyon-Okounkov complex structure and with prescribed boundary conditions.
We are not aware of any nice formula for the shifts m and this is an interest-ing open question.
217 N1 N2 Figure 24.4 A hexagon with a symmetric rhombic hole, with the vertical line and some horizontal lozenges.
24.3 Approach through log-gases In this section we consider the case of a regular hexagon with side length N +D with a rhombic hole with side length D in the middle. We will make some heuristic steps, although the arguments can be carried out rigorously with more efforts.
The height of the hole is determined by the lozenges on the vertical line that goes through the middle of the hexagon. Specifically, the height is determined by the number of horizontal lozenges that lie on the vertical line below the hole. Let this number be N1, and the number of horizontal lozenges on this line above the hole be N2. Note that the geometry of the hexagon (the height function on the boundary) dictates that the total number of horizontal lozenges on this vertical line is N. So N1 +N2 = N. The fluctuation of the height of the hole is the same as the fluctuation of N1.
As we are looking for the distribution of N1, it would be great to know the probability distribution for the positions of these horizontal lozenges. Fortu-nately, this was found in Lecture 19 for the hexagon, and essentially the same argument (which is omitted) also works for the hexagon with a hole. With x1 < x2 < ... < xN being the vertical coordinates of the N horizontal lozenges, P(x1,x2,...,xN) ∝∏ i<j (xi −xj)2 N ∏ i=1 w(xi).
(24.3) 218 Lecture 24: Discrete Gaussian component in fluctuations In our case, w is the product of several Pochhammer symbols, cf.
[BorodinGorinGuionnet15, Section 9.2].
Exercise 24.3. Extending the argument which produced (19.1), compute w explicitly.
Now, let us define a ’potential’ V by w(x) = exp −NV x N .
Given a position vector x x x = (x1,x2,...,xN), we also define the following prob-ability measure: µN = 1 N N ∑ i=1 δxi/N.
(24.4) As x x x is a random variable, µN is a random distribution. In the limit N →∞, µN becomes a continuous probability measure (note that since adjacent xi are separated at least by distance 1, the density of the limiting measure is upper-bounded by 1). With this notation, we can rewrite P(x x x) ∝exp N2I[µN] , where I[µN] = ¨ x̸=y ln|x−y|µN(dx)µN(dy)− ˆ V(x)µN(dx).
As N →∞, the measure µN concentrates near the maximizer of I[·] — this is another face of the limit shape theorem for random tilings, cf.
[BorodinGorinGuionnet15].
Letting µ be the probability measure of density at most 1 that maximizes I[·], we can consider a deviation 1 N gN from the maximum (here, 1 N gN is the difference of two probability measures, i.e. has total mass zero), and expand: I[µN] = I µ + 1 N gN = I[µ] + 1 N 2 ¨ ln|x−y|µ(dy)gN(dx)− ˆ V(x)gN(dx) + 1 N2 ln|x−y|gN(dx)gN(dy).
Since µ is a maximizer and the middle term is a derivative of I[·] at µ (in the gN direction), it is zero. So plugging back into the equation for P(x x x), we get the following proposition.
219 Proposition 24.4. The law of fluctuations gN as N →∞becomes P(gN) ∝exp ¨ ln|x−y|gN(dx)gN(dy) .
This is an exponential of a quadratic integral, so we get a Gaussian law.
The above proposition can be proved rigorously using loop (Nekrasov) equations of Lecture 21, see [BorodinGorinGuionnet15] and [BorotGuionnet11], [Johansson98] for a continuous space version.
We now understand the distribution of the fluctuations gN. Note that the support of gN is the part of the vertical line in the liquid regions, that is, a pair of intervals [a,b] ∪[c,d], as in Figure 24.5. We would like to also understand the fluctuation of the height of the hole, which is equal to the fluctuation of gN[a,b] = −gN[c,d]. The distribution of the latter is P(gN[a,b] = −gN[c,d] = δ) ∝ ˆ Ωδ exp ¨ ln|x−y|gN(dx)gN(dy) .
(24.5) Here, Ωδ is the subset of all gN with gN[a,b] = −gN[c,d] = δ.
Lemma 24.5. The expression in (24.5) is the same as P(gN[a,b] = −gN[c,d] = δ) ∝exp max gN such that gN[a,b]=δ ¨ ln|x−y|gN(dx)gN(dy) = exp ¨ ln|x−y|gδ(dx)gδ(dy) , where gδ is defined to be the maximizer.
Proof This is a general statement about Gaussian vectors, which we leave as an exercise: the density of projection of a large–dimensional vector onto a smaller subspace can be found through the maximization procedure.
In analogy with the argument from before, we again have gδ = δg1, so we conclude with the following result.
Conjecture 24.6. The height of a hole is asymptotically a discrete Gaussian (integer-valued) random variable, i.e. it has distribution const·exp(−C(H −m)2), where m is a certain shift and the scale parameter C is given by the extremum 220 Lecture 24: Discrete Gaussian component in fluctuations a b c d Figure 24.5 The dotted lines are boundaries of frozen regions, and the dashed segments are the support of gN of the logarithmic energy: C = −max ¨ ln|x−y|gN(dx)gN(dy), with the maximum taken over all gN on [a,b] ∪[c,d], such that gN[a,b] = 1, gN[c,d] = −1.
Conjecture 24.6 can be proven rigorously, although with significant tech-nical effort, see [BorotGorinGuionnet20+]. The same argument can also be carried out for multiple symmetric holes on a vertical line.
24.4 2d Dirichlet energy and 1d logarithmic energy The material of this subsection is based on private communications of the au-thor with Alexei Borodin, Slava Rychkov, and Sylvia Serfaty.
We have seen in the previous two sections two approaches which both gave discrete Gaussian distributions for the height function of the hole. However, the constants C in the exponents in Conjectures 24.1 and 24.6 have different expressions. For the former, C is given as the Dirichlet energy of a harmonic function with prescribed boundary conditions, which is equal to the minimal Dirichlet energy with these boundary conditions. For the latter, C is given by a log energy minimization on the line. The two visually different expressions 221 have to be equivalent, and in this section we provide a mathematical explana-tion for this.
We start by repeating and extending the statements of Conjectures 24.1, 24.6 for the setting of k ≥1 holes in the domain.
Recall that we would like to understand the discrete component of the global fluctuations in uniformly random tilings of the domains with k holes. Figure 24.2 shows a sample of lozenge tilings in the one-hole situation and Figure 24.6 shows a sample of domino tiling in the two-holes situation. Asymptot-ically the discrete component should be given by discrete Gaussian random variables, which are random variables on Zk−1 depending on two sets of pa-rameters: shift and scale matrix (“covariance”, although the exact covariance of the coordinates is slightly different). In the case of one hole this random variable is one dimensional, has the weight 1 Z exp −(x−m)2 2s2 , x ∈Z.
(24.6) We would like to compare two different ways of computing s2 (and more gen-erally, the scale matrix): the first one proceeds through the 2d variational prin-ciple and its approximation by the Dirichlet energy in the appropriate coor-dinate system; the second one relies on the identification of a section of the tilings with 1d–log gas and variational principle for the latter. The two ways generalize those in the previous subsections.
Let us start from the first approach.
We know from the variational principle that the number of tilings with a height approximating a height profile h(x,y) can be approximately computed as exp L2 ˆ ˆ S(∇h)dxdy+o(L2) , (24.7) where L is the linear size of the system and the integration goes over the tiled domain (rescaled to be finite). As in Lecture 12, (24.7) can be used to give a quadratic approximation of the energy near the maximizer of (24.7) known as the limit shape. Denoting ˜ h = h−Eh, we get the law exp −π 2 ˆ ˆ ∇Ω(˜ h) 2dz .
(24.8) The integral in (24.8) goes over the liquid region in (x,y)–coordinates, as there are no fluctuations outside. The map Ωtransforms the liquid region into a certain Riemann surface of the same topology; in the applications this surface can be taken to be a domain in C through the Riemann Uniformization, as in Theorem 11.13. The meaning of (24.8) is that we need to introduce a map or 222 Lecture 24: Discrete Gaussian component in fluctuations change of coordinates (x,y) 7→Ω(x,y) in order to turn the quadratic variation of (24.7) near its maximum into the Dirichlet energy. For our purposes, the map Ω(x,y) is defined only up to compositions with conformal functions, as the latter leaves the Dirichlet energy invariant.
The discrete Gaussian component arises as the integral of (24.8) over vary-ing boundary conditions. A basic property of the Dirichlet energy in a domain is that for the sum of two functions f = u+v, such that v is harmonic in the do-main and u vanishes on the boundary of the domain, the Dirichlet energy of f is the sum of Dirichlet energies of u and v. Hence, we can identify the integrals of (24.8) with different boundary conditions. Therefore, the scale of the Gaussian component can be now reconstructed through the computation of the Dirichlet energy of the harmonic functions (in the liquid region, in Ω–coordinate) with appropriate boundary conditions reflecting the height differences on different connected components of the boundary. E.g. for the holey hexagon of Figure 24.4 there are two connected components of the boundary, and therefore, there are two heights. By recentering, we can assume that the height of the outer component is zero and, therefore, only one variable remains.
It is shown in [BufetovGorin17] that for a class of domains (gluing of trape-zoids in the lozenge case and gluings of Aztec rectangles in the domino case) the map Ω(x,y) can be chosen so that it maps the liquid region into the com-plex plane with cuts. For instance, in the holey hexagon case of Figure 24.4, the cuts can be identified with intervals on the vertical symmetry axis on the picture; their endpoints coincide with intersection of the frozen boundary of the liquid region with the symmetry axis, these are [a,b] and [c,d] in Figure 24.5. In the Aztec diamond with two holes of Figure 24.6 the cuts are on the horizontal axis going through the centers of the holes; again end-points of the cuts lie on the frozen boundary.
We thus are led to the problem of finding the Dirichlet energy of harmonic functions in a domain with cut. Let us now formulate the setting of the last problem in a self-contained form.
Take 2k real numbers a1 < b1 < a2 < ··· < ak < bk. We refer to intervals [ak,bk] as cuts. Let H denote the (open) upper half-plane and ¯ H denote the lower half-plane. Set D = H∪¯ H∪k i=1 (ai,bi).
Fix k real numbers n1,...,nk subject to the condition n1+···+nk = 0. These numbers are fluctuations of the height function in the setting of the previous text: they are increments of the height as we cross a connected component of the liquid region (cf. N1 and N2 in Figure 24.4) with subtracted expectations.
The zero sum condition corresponds to the deterministic full increment of the 223 Figure 24.6 Domino tiling of Aztec diamond with two holes; there are two types of horizontal and vertical lozenges (due to checkerboard coloring), hence 4 colors on the picture. Drawing by Sevak Mkrtchyan height function between the points of the outer boundary. In particular, k = 2 in Figures 24.2 and 24.4, and k = 3 in Figure 24.6.
We consider continuous functions h(z), z = x+iy, in C, differentiable in D, and subject to the following boundary conditions on the real axis: h(−∞) = h(+∞) = 0, (24.9) ∂ ∂xh(x) = 0 on [−∞,a1]∪[b1,a2]∪···∪[bk−1,ak]∪[bk,+∞], (24.10) h(bi)−h(ai) = ni, i = 1,...,k. (24.11) 224 Lecture 24: Discrete Gaussian component in fluctuations We are interested in the Dirichlet energy of such functions, E (h) = ¨ D |∇h|2dxdy = ¨ D [(hx)2 +(hy)2]dxdy.
We further define a quadratic form C through C : (n1,...,nk) →π 2 min h E (h), subject to (24.9),(24.10),(24.11).
(24.12) Note that by the Dirichlet principle, the minimizer in (24.12) is a harmonic function in D.
Claim 24.7. The quadratic form C is precisely the quadratic form for the dis-crete Gaussian component of the fluctuations of the height function in random tilings in Conjecture 24.1.
From the second perspective, we saw in the previous section that the sec-tion of a random tiling along a singled out axis gives rise to a discrete log-gas. This axis is the vertical symmetry axis in Figures 24.2, 24.4 and hor-izontal axis going through the centers of the holes in Figure 24.6. Other examples can be found in [BorodinGorinGuionnet15], [BufetovGorin17], [BorotGorinGuionnet20+]. Clearly, it is enough to study what is happening on such axis in order to find the desired discrete Gaussian component of the fluctuation.
In more details, in Figure 24.4 the distribution of the horizontal lozenges on the vertical section of the tiling by the symmetry axis of the domain has the form of the log-gas. The particles xi in (24.3) belong to two segments – below and above the hole. We are, thus, interested in fluctuations of the number of particles in one of the segments. For a large class of tilings we will have a generalisation of (24.3) with particles now confined to k segments instead of 2.
If we denote through µN the empirical measure of the particles as in (24.4), then (24.3) can rewritten (ignoring the diagonal) as exp N2 ¨ ln|x−y|µN(dx)µN(dy)+ ˆ lnw(Nx)µ(dx) (24.13) The maximizer of the functional in the exponent of (24.13) is the equilib-rium measure. The fluctuations of the numbers of particles (called filling fractions in this context) can be thus obtained by varying this functional near its extremum. Rigorous justification of this procedure is the subject or [BorotGorinGuionnet20+]. In the continuous setting (when xi in (24.3) are real numbers rather than integers), the detailed rigorous analysis for arbitrary num-ber of cuts k was done in [BorotGuionnet11].
225 Let us describe how the approximation of the functional near the equilibrium measure looks like. We assume that the bands of the equilibrium measure are k intervals [a1,b1],..., [ak,bk]. Recall that bands are regions where the density is not zero and not one as in the liquid region for tilings (i.e., not frozen/saturated, the latter corresponding to the areas in Figures 24.2 and 24.4 densely packed with particles).
Given n1,...,nk with n1 + ··· + nk = 0, we are led to consider signed mea-sures ν on the union of the intervals [ai,bi] subject to filling fractions: ν([ai,bi]) = ni, i = 1,...,k.
(24.14) We consider the logarithmic energy of such measure given by ˜ E = − ˆ ˆ ln|x−y|ν(dx)ν(dy).
(24.15) Define a quadratic functional ˜ C through ˜ C : (n1,...,nk) →min ν ˜ E (ν) subject to (24.14).
(24.16) Claim 24.8. The quadratic form ˜ C is precisely the quadratic form for the dis-crete Gaussian component of the fluctuations of the height function in random tilings, as in Conjecture 24.6.
Of course, both Claim 24.7 and Claim 24.8 describe the same object.
The advantage of Claim 24.8 is that we are actually able to prove it in [BorotGorinGuionnet20+]. Nevertheless, the answers in these claims also look visually different. Hence, we are led to proving the following statement.
Proposition 24.9. Quadratic forms (24.12) and (24.16) are the same.
Proof The key idea is to split the minimization procedure in (24.12) into two steps. First, we will fix the values of h on the real axis in arbitrary way subject to (24.9), (24.10), (24.11) and minimize over all such functions. The minimizer is then a harmonic function separately in the upper half–plane and in the lower half–plane; by symmetry, the values differ by conjugation of the arguments.
By the Schwarz integral formula (i.e. Poisson kernel), this harmonic function can be expressed as an integral over the real axis h(z) = ℜ 1 πi ˆ ∞ −∞ h(ζ) ζ −zdζ , z ∈H, (24.17) and by a similar expression with changed sign in the lower halfplane ¯ H.
In the second step we minimize the Dirichlet energy over the choices of the values of h on the real axis. We claim that the latter optimization is the same as the log-energy optimization of (24.16). Let us explain this.
226 Lecture 24: Discrete Gaussian component in fluctuations We identify a function on the real axis h(x) with a signed measure ν through h(y) −h(x) = ν([x,y]). Although, a priori the functions do not have to be differentiable and the signed measures do not have to be absolutely continu-ous, however, the minimum in (24.12) and (24.16) is attained on differentiable functions and measures with densities, respectively. Thus, we can restrict our-selves to the latter, in which case the derivative hx(x) becomes the density of ν.
Clearly, under such identification, conditions on h (24.9)-(24.11) become con-ditions on ν (24.14). It thus remains to show that the functional (24.12) turns into (24.16).
Let us compute the Dirichlet energy of h(x + iy) given by (24.17) in the domain y ≥ε > 0. Integrating by parts, we get ¨ y≥ε (hx)2 +(hy)2 dxdy = ˆ +∞ ε dy (hhx)(+∞+iy)−(hhx)(−∞+iy)− ˆ ∞ −∞ hhxxdx + ˆ ∞ −∞ dx (hhy)(x+i∞)−(hhy)(x+iε)− ˆ ∞ ε hhyydy .
(24.18) Further, note that since h(x) has a compact support on the real axis, h(z) given by (24.17) decays as O(1/|z|) when z →∞and its derivatives decay as O(1/|z|2). Hence, all the boundary terms at the infinity in (24.18) vanish.
Moreover, since h is harmonic, hhxx+hhyy = 0 and the double integral vanishes as well. We conclude that (24.18) is − ˆ ∞ −∞ (hhy)(x+iε)dx = − ˆ ∞ −∞ dxh(x+iε)ℜ 1 π ˆ ∞ −∞ h(ζ) −(ζ −(x+iε))2 dζ We integrate by parts both in x and in ζ to get −1 π ˆ ∞ −∞ dxhx(x+iε)ℜ ˆ ∞ −∞ hx(ζ)ln(ζ −(x+iε))dζ .
Note that we need to choose some branch of the logarithm here, but it does not matter, because we will get the logarithm of the absolute value when comput-ing the real part anyway.
At this point it remains to send ε →0 (the integrand is singular and we also have hx rather than h itself; hence, the limit needs some justifications, which we omit) and to add the same contribution from the lower half-plane to get the desired ¨ D |∇h|2dxdy = −2 π ˆ ∞ −∞ ˆ ∞ −∞ hx(x)hx(ζ)ln|ζ −x|dxdζ, which matches (24.16).
227 Remark 24.10. Although we were only concentrating on the discrete compo-nent, but essentially the same arguments explain the link between covariance structures for the entire field of fluctuations: Gaussian Free Field in the 2d pic-ture and universal covariance of random matrix theory and log-gases on the 1d section. This covariance coincidence is also discussed (through other tools) in [BufetovGorin17], [BorodinGorinGuionnet15].
Remark 24.11. It is interesting to also try to do similar identification for the shifts — m in (24.6). This involves computing the second order expansion of the logarithm of the partition function — we need O(N) rather than O(N2) terms. On the side of log-gases there is a certain understanding of how this can be done (at least for continuous log–gases). On the other hand, the author is not aware of approaches to the computation of this second term in the 2d setting of random tilings.
24.5 Discrete component in tilings on Riemann surfaces There is another natural situation in which the height function of lozenge tilings asymptotically develops a discrete Gaussian component.
Let us give an example by looking at lozenge tilings of a torus, as we did in Lectures 3, 4, and 6. The height function does not make sense on torus as a single-valued function: we define the heights by local rules and typically these rules result in a non-trivial increment as we loop around the torus and come back to the same point.
There are two closely related ways to deal with this difficulty. We could either split the height function into two components: affine multivalued part (sometimes called the instanton part) keeping track of the height change as we loop around the torus, and a scalar single-valued part. Alternatively, we can define the height function as a 1–form, so that it is a not function of the point, but rather of a path on the torus, and therefore is allowed to have different values on paths representing different homotopy classes. In both ways, one can single out the component which asymptotically converges to a discrete Gaussian random variable: in the first approach this is the instanton part, and in the second approach these are the values of the 1–form on closed loops.
More generally, a similar component can be singled out whenever we deal with tilings of domains embedded into Riemann surfaces of higher genus. We deal with random functions of homotopy classes of closed loops and we can ask about the asymptotic of such functions as the mesh size goes to zero.
In several situations the convergence of the discrete component of the 228 Lecture 24: Discrete Gaussian component in fluctuations tilings on Riemann surfaces to discrete Gaussian random variables was rig-orously proved, see [BoutillierDeTiliere06, Dubedat11, KenyonSunWilson13, DubedatGheissari14, BerestyckiLaslierRay19].
Lecture 25: Sampling random tilings In this lecture we are interested in algorithms for sampling a random lozenge tiling. Fix a tileable domain R. We consider the cases in which the probability of a particular tiling of R is either uniform or proportional to qVolume, where Volume is the number of cubes which one needs to add to the minimal tiling of R in order to obtain a given tiling.
We are interested in reasonably fast algorithms for sampling a tiling. The trivial algorithm – generating the set of all tilings of R and picking a tiling from this set according to some probability measure – is in general very slow in the size of R. Indeed, we saw in previous lectures that for a domain of linear size L, the number of tilings is exp(L2 ...), cf. Theorem 5.15. Fortunately, there are other sampling algorithms that do not require generating a list of all tilings.
We will discuss a few algorithms that are based on Markov chains.
One practical reason for wanting fast ways to sample tilings from a distri-bution sufficiently close to the desired distribution is so that one can guess the peculiar features of the tiling model (L1/3 behaviour, Tracy-Widom distribu-tions, Gaussian Free Field, etc.) from looking at some sampled tilings.
25.1 Markov Chain Monte-Carlo We start from a general theorem, see, e.g., [Shiryaev16, Theorem 1 in §1.12].
Theorem 25.1. Take a (time-homogeneous discrete time) Markov chain with finite state space [n] and transition matrix P. That is, the matrix entry P ij is the probability of transitioning from state i to state j in one time step. Suppose that there exists an integer k, such that for all pairs of states (i, j), (Pk)ij > 0.
Then P has a unique invariant distribution π π π (i.e. a distribution vector π π π for which π π πP = π π π). Furthermore, for any distribution vector x x x, lim t→∞x x xPt = π π π.
229 230 Lecture 25: Sampling random tilings add cube remove cube m m Figure 25.1 Adding/removing a cube at the vertex m This theorem motivates the following general idea for a sampling algorithm.
Design a Markov chain such that the assumptions of the above theorem are satisfied. Pick your favorite start state (the corresponding distribution vector x x x has a 1 in the corresponding location and 0’s elsewhere), and perform random transitions on it for long enough (so as to make the distribution of the end state as close to uniform as desired). Output the end state.
All Markov chains we consider in this lecture are going to have state space equal to the set of all tilings of our domain R. On each step, the Markov chain will either add a cube to the tiling or remove a cube from the tiling. For our Markov chains, we want the assumptions of Theorem 25.1 to be satisfied. In particular, we want any tiling to be reachable from any other tiling by a se-quence of cube addition/removal moves. This condition is guaranteed to hold when the domain R is a simply connected region of the plane, but it might fail to hold for domains with holes, since adding or removing cubes does not change the height function on the hole boundaries, while in general there are multiple possibilities for the height of a hole. However, for fixed height functions of the holes, any tiling is reachable from any other by a sequence of such moves, and the rest of the discussion in this lecture notes applies to that case (with appro-priate modifications). So for domains which are subsets of the plane that are not simply connected but for which the distribution of the height function on the holes is known, one can still sample a random tiling by first sampling the heights of the holes from the known distribution, and then sampling a random tiling, given these heights for the holes. But for clarity, we will from now on only consider the case in which R is a simply connected region of the plane.
Consider the Markov chain given by the following step.
Chain 25.2. Choose a vertex m of R uniformly at random, and flip a coin. With probability 1/2, add a cube at m (if possible; otherwise do nothing), and with probability 1/2, remove a cube at m (if possible; otherwise do nothing). The possible moves are shown in Figure 25.1.
When running this chain in practice, one can of course restrict to only pick-ing vertices m at which it is possible to either add or remove a cube. However, 231 the above formulation is what we will use formally, because it has a number of nice properties. Note that Chain 25.2 satisfies the assumption of Theorem 25.1, as any tiling can be reached from any other and there is always a nonzero probability of not changing the tiling, from which the assumption of the theo-rem follows. Hence, this chain has a unique invariant distribution, and we have the following proposition.
Proposition 25.3. The uniform distribution is the invariant distribution of Chain 25.2.
Proof For a general Markov chain, to show that π π π is an invariant distribution, it suffices to show reversibility, that is πiP ij = πjPji. Indeed, if this holds, then (π π πP)j = ∑ i πiP i j = ∑ i π jPji = πj∑ i Pji = πj = ⇒π π πP = π π π.
So it remains to show that for our cube switching Markov chain, π π π = the uni-form distribution satisfies πiP i j = πjPji. This is just a case check: 1 If i = j, then this is trivially true.
2 If tilings i and j are different, but do not differ by adding/removing a single cube, then P i j = Pji = 0, so we are done.
3 If tilings i and j differ by adding/removing a single cube, then both P ij and Pji are 1 2 · 1 number of vertices, so we are done since for the uniform distribution πi = πj. Note that it is crucial here that our Markov chain step involves picking a random vertex regardless of whether a cube flip can be performed there.
As for the qVolume distribution, consider the following Markov chain that is the same as Chain 25.2, except with different probabilities for adding/removing cubes.
Chain 25.4. Choose a vertex m of R uniformly at random, and flip a biased coin. With probability q 1+q, add a cube at m (if possible; otherwise do noth-ing), and with probability 1 1+q, remove a cube at m (if possible; otherwise do nothing).
As argued before, Chain 25.4 also has a unique invariant distribution. We have the following proposition.
Proposition 25.5. The qVolume distribution is the invariant distribution of Chain 25.4.
Proof As in the proof of Proposition 25.3, it suffices to show reversibility, i.e.
that πiP i j = π jPji. This is just a case check: 232 Lecture 25: Sampling random tilings 1 If i = j, then this is trivially true.
2 If tilings i and j are different but do not differ by adding/removing a single cube, then P i j = Pji = 0, so we are done.
3 If tilings i and j differ by adding/removing a single cube, with WLOG tiling j having 1 cube more than tiling i, then for the qVolume distribution π π π, πj πi = q, and with our transition probabilities, P ij Pji = q 1+q 1 1+q = q. The desired equality πiP i j = π jPji follows.
So we have the following algorithm for sampling a random tiling.
Algorithm 25.6. Pick your favorite tiling. Run either Chain 25.2 (for the uni-form distribution) or Chain 25.4 (for the qVolume distribution) long enough, and output the result.
Of course, this does not give exactly the distribution we are interested in for any finite time, but by Theorem 25.1, as the time goes to infinity, it gives an arbitrarily good approximation. One may ask what the runtime should be for the approximation to be ’sufficiently good’. To be more rigorous, define the total variation distance between two distributions µ1,µ2 on a finite space X to be ∑x∈X |µ1(x) −µ2(x)|. Define the mixing time of this Markov chain to be the minimal amount of time t the above algorithm needs to run (from any start state), so that for all t′ ≥t, the variation distance between the in-duced distribution and the desired distribution is at most some fixed constant ε. In [RandallTetali00] an upper bound O(L8 lnL) for the mixing time in the domain of linear size L was obtained. For a slightly different Markov chain (which allows addition/removal of 1×1×k stacks of cubes; it was introduced in [LubyRandallSinclair95]), the mixing time was shown to be O(L4 lnL) in [Wilson01] and for hexagons this bound is tight up to a constant factor.
[Wilson01, Section 5.6] discusses that the mixing time for our chain (adding only one cube at a time) is expected to be of the same order for “good” do-mains. This prediction (which also agrees with expectations from the the-oretical physics literature) matches more recent results on the mixing time for our chain in specific classes of domains [CaputoMartinelliToninelli11], [LaslierToninelli13].
A general behaviour (cf. [LevinPeresWilmer17]) is that before some time the chain is not mixed at all, and after that time it is mixed very well. To get an idea of what the above estimates mean in practice, for instance, when our domain is a hexagon of side length 1000, one should run the Markov chain on a personal computer for about an entire day to get a good approximation.
233 25.2 Coupling from the past [ProppWilson96] In any finite amount of time, the previous algorithm only gives a sample from an approximation of the desired distribution. This is often inconvenient, as we would like to look at very delicate features of the tilings, and therefore, knowing that our sample is exact rather than approximate becomes important.
With an upgrade known as coupling from the past, we can sample from the desired distribution without any error.
Let us first discuss some ideas that will eventually lead us to the coupling from the past algorithm. Suppose we can run our Markov chain in a coupled way from all states, i.e. picking a sequence of transitions (which we think of as transitions from any state via some canonical identification, and not just transitions from some fixed state) and applying this same sequence to all start states. Further suppose that at some time t, we notice that we have reached the same final state from all start states. Suppose we output this final state. Since this state is independent from the initial configuration, one might expect that it is a sample from the invariant distribution, as we can imagine starting from the invariant distribution originally. However, there are a few issues here.
First, with the algorithm as stated, we still run into our original problem – the total number of states is huge, so it would take a huge amount of time to simul-taneously run this procedure from all start states. It turns out that this problem can be fixed, due to the fact that transitions preserve the height poset structure on tilings. This will be explained in more detail in the next few paragraphs.
The second issue is that there is no way to choose the desired time t deter-ministically, while if t is random (e.g. we choose it as the first time when all initial states lead to the same current state), then it becomes unclear whether the distribution of the output of such an algorithm is the invariant distribution of the Markov chain. Unfortunately, this turns out to be false in general, as the following counterexample shows. Suppose we have a Markov chain that satisfies the assumptions of Theorem 25.1 with more than one state, and with a state into which one can transition from exactly one state (also counting the state itself). It is not hard to see that such Markov chains exist. Then the prob-ability of outputting this state at first time t is clearly zero. However, (under assumptions of Theorem 25.1), the invariant distribution can not assign zero probability to some state. So the distribution of the output of our algorithm is not the invariant distribution.
Let us first explain how to work around the issue of not being able to run the Markov chain from all initial states.
Proposition 25.7. Suppose we have two tilings A0 and B0, such that at each 234 Lecture 25: Sampling random tilings m m A i-1 A i H H+1 H+2 H+2 H+2 H+2 H+2 H+1 H+1 H+1 H+1 H+1 H+2 H+3 Figure 25.2 The configuration and height function around m in Ai−1 and Ai grid vertex m, the height of A0 is at most the height of B0. Suppose we have a sequence of transitions T1,T2,...,Tt. That is, each transition is a specification of a vertex m at which to perform an operation and whether this operation is adding or removing a cube. Let At and Bt be the tilings we obtain after performing this sequence of operations on A0 and B0, respectively. Then at every grid point m, the height of At is at most the height of Bt.
Proof Say for a contradiction that there is a vertex at which the height of At is greater than the height of Bt. Then there must be a first transition after which there exists such a vertex, let it be Ti. This means that at all grid vertices, the height of the tiling Ai−1 was at most that of the tiling Bi−1, but there exists a vertex m at which the height of Ai is greater than the height of Bi. Our oper-ations only change the height at one vertex – the vertex at which we add or remove a cube. So in our situation, we must have added or removed a cube at the vertex m, and at all neighboring vertices, the height of Ai is still at most the height of Bi. Let us consider the case in which we added a cube to Ai−1 at m.
The case of removing a cube from Bi−1 at m is completely analogous. Since the height of Ai became greater than that of Bi at m, we must have not added a cube to Bi, so Bi = Bi−1. As we know the local configuration of the tilings Ai−1 and Ai around the vertex m and that the heights of other vertices are preserved, we can write out the height functions in terms of the height H of m in Ai−1.
These are shown in Figure 25.2.
Note that in Bi−1, m has height at most H + 2 by our assumption about the height function of Ai becoming larger than that of Bi. Also note that the vertices adjacent to m in the three negative coordinate directions have heights at least H +2 in Bi−1 (see Figure 25.2), so in particular they have heights greater than the height of m. Hence, the edges between m and these three vertices must be diagonals of lozenges in Bi−1. But this implies that we could have also added a cube to Bi−1, a contradiction.
Recall from the first lecture that for any simply connected domain R, there 235 MIN MAX Figure 25.3 MIN and MAX tilings for the hexagon is a minimal tiling (for the hexagon, this is the hollowed out cube as in Figure 25.3) which has height function at most that of any other tiling at all points.
Also notice that there is a maximal tiling (for the hexagon, this is the filled in cube as in Figure 25.3) which has height function at least that of any other tiling at all points. Let us call these tilings MIN and MAX. By Proposition 25.7, if a sequence of transitions leads to the same tiling X from both MIN and MAX, then this sequence must lead to X from all initial tilings, as the height function of any final state is both lower bounded and upper bounded by the height function of X at all points. So this solves the issue of having to run the Markov chain from all states – now we just need to run it from MIN and MAX to be able to tell when all branches coalesce.
We still have the problem that the above algorithm of applying successive transitions might give the wrong distribution because of the randomness of t.
It turns out that this is not the case if instead of applying successive transitions to the end state, one applies all transitions in the opposite order instead, i.e.
starting from the transition that was chosen last. We consider the following algorithm.
Algorithm 25.8. (Coupling from the Past) Successively generate random tran-sitions T−1,T−2,.... That is, each transition is a specification of a vertex m at which to perform an operation and whether this operation is adding or re-moving a cube, generated at random with the same probability distribution as before (in either Chain 25.2 or Chain 25.4). After generating each new transi-tion T−t, apply the same sequence of transitions T−t ◦T−(t−1) ◦...◦T−1 to both 236 Lecture 25: Sampling random tilings MIN and MAX. If the two tilings obtained from both start states are the same, output this tiling (and halt).
In practice, it is more efficient to only apply the transitions to MIN and MAX after generating t = 1,2,4,8,... of them. It is easy to see that this algorithm halts almost surely. Let us now prove that this algorithm generates samples from exactly the desired distribution.
Proposition 25.9. The distribution of the output of Algorithm 25.8 is precisely the invariant distribution of the corresponding tiling Markov chain (uniform or qVolume).
Proof If T−t ◦T−(t−1) ◦... ◦T−1 takes both MIN and MAX to the same end state i, then it also does so for any other state (by Proposition 25.7). The crucial observation is that if T−t ◦T−(t−1) ◦... ◦T−1 takes all initial states to i, then for t′ > t, T−t′ ◦T−(t′−1) ◦... ◦T−1 = T−t′ ◦T−(t′−1) ◦...◦T−(t+1) ◦ T−t ◦T−(t′−1) ◦...◦T−1 also takes any initial state to i. Starting from any start state (i.e. a distribution vector x x x with a 1 in the corresponding location and 0’s elsewhere), Theorem 25.1 gives that x x xPt approaches the invariant distribution as t →∞. Since for any particular sequence of transitions ...,T−2,T−1, there almost surely exists a time t after which the end state is unchanged by adding transitions to the beginning, the limiting distribution (formally, the limiting distribution starting from some fixed start state – although it does not matter, as the result is the same for all start states) is also equal to the distribution of the end states generated by our algorithm. Hence, the distribution of these end states is equal to the invariant distribution of the Markov chain. This is what we wanted to show.
To recap, one advantage Algorithm 25.8 has is that it allows sampling from the exact distribution we desire, whereas the simple Markov chain Monte-Carlo algorithm considered before only samples from a good approximation of this distribution. The runtime is unbounded, but the algorithm almost surely stops in finite time. For the hexagon, the expected runtime is of the same or-der as the mixing time. Perfect sampling of random lozenge tilings was one of the important initial motivations for developing coupling from the past in [ProppWilson96], yet is also useful for sampling from other large systems, e.g.
the six-vertex model.
Although coupling from the past is a powerful and simple to implement method, it is by no means a unique way to sample random tilings. In the next sections we briefly mention other approaches.
237 25.3 Sampling through counting [JerrumValiantVazirani86] popularized a powerful idea: counting and sam-pling are closely related to each other. In our context, imagine that we could quickly compute a probability that a given position in the domain is occupied by the lozenge of a fixed type. Then we could flip a three-sided coin to choose between , , and for this position, and continue afterwards for the smaller domain.
We learned in Lectures 2 and 3 that the required probabilities can be com-puted either by inverting the Kasteleyn matrix, or through evaluation of the determinants of matrices build out of the binomial coefficients. Thus, the sam-pling now reduces to the well-studied linear algebra problems of fast compu-tations for the determinants and inverse matrices. These ideas were turned into sampling algorithms in [ColbournMyrvoldNeufeld96] for spanning trees and in [Wilson97] for tilings.
A fast recursive algorithm for computing the same probabilities which works for very general weights is further discussed in [Propp01].
25.4 Sampling through bijections Here is another idea. In many situations the total number of tilings (or weighted sum over tilings) is given by a simple product formula as in Theorem 1.1 of Lecture 1 or (22.8) in Lecture 22. If we can turn the enumeration identity into a bijection between tilings and some simpler objects enumerated by the same product, then as soon as we manage to sample these objects, we get access to random tilings as well.
Perhaps, the most famous of such bijections is the Robinson–Schensted– Knuth (RSK) correspondence between rectangular matrices filled with in-tegers and pairs of semistandard Young tableaux, see e.g. [Romik15, Sec-tion 5.3], [BaikDeiftSuidan16, Section 10.2] for the reviews in probabilis-tic context and [Sagan01] for connections to representation theory and sym-metric functions. It is straightforward to sample a rectangular matrix with independent matrix elements, and using RSK we immediately get ran-dom pairs of Young tableaux, which, in turn, can be identified with plane partitions. This approach can be used for sampling of qVolume–weighted plane partitions of Lecture 22. For the Schur processes, which generalize plane partitions, the detailed exposition of the approach can be found in [BetaBoutillierBouttierChapuyCorteelVuletic14].
More delicate bijections were used in [BodiniFusiPivoteau07] for sampling 238 Lecture 25: Sampling random tilings random plane partitions with fixed volume and in [Krattenthaler98] for sam-pling random lozenge tilings of hexagons.
25.5 Sampling through transformations of domains There are specific domains, for which the number of tilings is small and sam-pling is easy. This could be either because the domain itself is very small, or because the height function is extremal along the boundary. For instance, the A×B×0 hexagon has a unique tiling.
A central idea for a class of algorithms is to introduce a Markov chain (with local transition rules), which takes a uniformly random tiling of a domain as an input and gives a uniformly random tiling of a slightly more complicated domain as an output. In this way, we can start from a domain with a single tiling and reach much more complicated regions in several steps. The main difficulty in this approach is to design such a Markov chain, as there is no universal recipe to produce it.
The first example of such a chain is known as the shuffling algorithm for the Aztec diamond [ElkiesKuperbergLarsenPropp92] . It works with tilings of rhombuses drawn on the square grid (“diamonds”) with 2 × 1 dominos, as in Figure 1.10 of Lecture 1. Starting from 2×2 square, which has two tilings, it grows the size of the domain, ultimately getting to huge rhombuses.
There are several different points of view on the shuffling algorithm which lead to different generalizations. The first point of view is that it provides a stochastic version of a deterministic operation which modifies a bipartite graph through a sequence of local moves (known as urban renewal or spider move) with explicitly controlled change of the (weighted) count of all perfect match-ings. It is explained in [Propp01] how this can be used for sampling random domino tilings with quite generic weights. Since the admissible weights in this procedure are essentially arbitrary, it is is very flexible: by various degener-ations one can get all kinds of complicated domains, tilings on other lattices (including lozenge tilings), etc. Extending beyond the probabilistic context, urban renewal is studied as an abstract algebraic operation on weighted bi-partite graphs in [GoncharovKenyon11]; it was further linked to the Miquel move on cycle patterns (arrangements of intersecting circles on the plane) in [Affolter18, KenyonLamRamassamyRusskikh18].
Another point of view exploits the connections of the domino tilings to Schur polynomials (which is similar to the lozenge tilings case we saw in Proposition 19.3 and Lecture 22, see also [BufetovKnizel16]). Along these 239 3×6×0 3×5×1 3×4×2 3×3×3 3×2×4 3×1×5 3×0×6 empty Figure 25.4 Steps of the shuffling algorithm for lozenge tilings of hexagons.
lines [Borodin10] constructed an efficient sampler for qVolume-weighted plane partitions which we discussed in Lectures 22-23 and their skew versions.
The third point of view treats domino tilings as orthogonal polyno-mial ensembles, cf. footnote to Exercise 21.7 and see [CohnElkiesPropp96, Johansson05]. Along these lines [BorodinGorin08], [BorodinGorinRains09] designed a Markov chain on random lozenge tilings of the hexagon, which changes the side lengths, thus allowing to reach A × B ×C hexagon starting from the degenerate A×(B+C)×0 hexagon, see Figure 25.4.
The above points of view and generalizations of the shuffling algorithm share one feature: if we restrict our attention to a subset of tiles, then the time evolution turns into a one-dimensional interacting particle system, which is a suitable discrete version of the Totally Asymmetric Simple Exclusion Process (TASEP). The TASEP itself is a prominent continuous time dynamics on par-ticle configurations on the integer lattice Z: each particle has an independent exponential clock (Poisson point process) and whenever the clock rings the particle jumps one step to the right unless that spot is occupied (“exclusion”); then the clock is restarted and the process continues in the same way.
The link between shuffling algorithm for the domino tilings of the Aztec diamond and a discrete time TASEP was noticed and exploited for the iden-tification of the boundary of the frozen region in [JockushProppShor95]. For lozenge tilings (of unbounded domains) a direct link to the continuous time TASEP was developed in [BorodinFerrari08]. These results connect the edge 240 Lecture 25: Sampling random tilings limit theorems for tilings of Lecture 18 to the appearance of the t1/3 scaling and Tracy-Widom distribution in the large time asymptotic of TASEP1.
The connection between statistical mechanics models (such as random tilings) and interacting particle systems is a fruitful topic and we refer to [BorodinGorin12, BorodinPetrov13, BorodinPetrov16] for further reviews.
1 The Tracy-Widom fluctuations for TASEP were first established by [Johansson00].
References [AdlerJohanssonMoerbeke16] M. Adler, K. Johansson and P. van Moerbeke, Tilings of nonconvex Polygons, skew-Young Tableaux and determinantal Processes, Com-munications in Mathematical Physics 364 (2018), 287-342. arXiv:1609.06995 [AdlerJohanssonMoerbeke17] M. Adler, K. Johansson and P. van Moerbeke, Lozenge tilings of hexagons with cuts and asymptotic fluctuations: a new universality class, Mathematical Physics, Analysis and Geometry 21 (2018) 1-53. arXiv:1706.01055 [Affolter18] N. C. Affolter, Miquel Dynamics, Clifford Lattices and the Dimer Model.
arXiv:1808.04227 [Aggarwal14] A. Aggarwal, Correlation functions of the Schur process through Mac-donald difference operators.
Journal of Combinatorial Theory, Series A 131 (2015), 88–118. arXiv:1401.6979 [Aggarwal19] A.
Aggarwal, Universality for lozenge tilings local statistics.
arXiv:1907.09991 [AggarwalGorin21] A. Aggarwal, V. Gorin, Gaussian Unitary Ensemble in random lozenge tilings. arXiv:2106.07589 [AissenSchoenbergWhitney52] M. Aissen, I. J. Schoenberg, and A. Whitney, On the generating functions of totally positive sequences I. Journal d’Analyse Math´ ematique 2 (1952), 93–103.
[Ahn18] A. Ahn, Global Universality of Macdonald Plane Partitions, to appear in An-nales de l’Institut Henri Poincar´ e. arXiv:1809.02698 [AkemannBaikFrancesco11] G. Akemann, J. Baik, P. Di Francesco (editors), The Ox-ford Handbook of Random Matrix Theory, Oxford University Press, 2011.
[AndersonGuionnetZeitouni10] G. W. Anderson, A. Guionnet, O. Zeitouni, An Intro-duction to Random Matrices, Cambridge University Press, 2010.
[ArdilaStanley05] F. Ardila, R. P. Stanley, Tilings. The Mathematical Intelligencer, 32 (2010), 32–43. arXiv:math/0501170 [AstalaDusePrauseZhong20] K. Astala, E. Duse, I. Prause, X. Zhong, Dimer Models and Conformal Structures, arXiv:2004.02599 [Azuma67] K. Azuma. Weighted sums of certain dependent random variables. Tohoku Mathematical Journal, Second Series, 19(3):357–367, 1967 [BaikKriecherbauerMcLaughlinMiller03] J.
Baik, T.
Kriecherbauer, K.
T.
-R. McLaughlin, P. D. Miller, Uniform Asymptotics for Polynomials Orthogonal With Respect to a General Class of Discrete Weights and Universality Results for Associated Ensembles, International Mathematics Research Notices 2003, no.
15. arXiv:math/0310278 241 242 References [BaikDeiftSuidan16] J. Baik, P. Deift, T. Suidan, Combinatorics and Random Matrix Theory. Graduate studies in Mathematics 172, AMS, Providence, RI, 2016.
[Baryshnikov01] Y. Baryshnikov, GUEs and queues, Probability Theory and Related Fields, 119, no. 2 (2001), 256–274.
[BasokChelkak18] M. Basok, D. Shelksk, Tau-functions a la Dub´ edat and probabilities of cylindrical events for double-dimers and CLE(4). arXiv:1809.00690 [Baxter82] R. Baxter, Exactly solved models in statistical mechanics, The Dover Edi-tion, Dover, 2007.
[BerggrenDuits19] T. Berggren, M. Duits, Correlation functions for determinantal pro-cesses defined by infinite block Toeplitz minors. arXiv:1901.10877 [BeffaraChhitaJohansson20] V. Beffara, S. Chhita, K. Johansson, Local Geometry of the rough-smooth interface in the two-periodic Aztec diamond. arXiv:2004.14068 [BerestyckiLaslierRay16] N. Berestycki, B. Laslier, G. Ray, Dimers and Imaginary geometry. Annals of Probability 48, no. 1 (2020), 1-52. arXiv:1603.09740 [BerestyckiLaslierRay19] N. Berestycki, B. Laslier, G. Ray, The dimer model on Rie-mann surfaces, I. arXiv:1908.00832 [BetaBoutillierBouttierChapuyCorteelVuletic14] D. Betea, C. Boutillier, J. Bout-tier, G. Chapuy, S. Corteel, M. Vuleti´ c, Perfect sampling algorithm for Schur processes, Markov Processes and Related Fields 24 (2018), 381–418.
arXiv:1407.3764 [BloteHilhorst82] H. W. J. Bl¨ ote, and H. J. Hilhorst, Roughening transitions and the zero-temperature triangular Ising antiferromagnet, Journal of Physics A: Mathe-matical and General 15 (1982), 631–637.
[BodineauSchonmannShlosman04] T. Bodineau, R. H. Schonmann, S. Shlosman, 3D Crystal: How Flat its Flat Facets Are?, Communications in Mathematical Physics, 255 (2005), 747-766. arXiv:math-ph/0401010 [BodiniFusiPivoteau07] O. Bodini, E. Fusy, C. Pivoteau, Random sampling of plane partitions. Combinatorics, Probability and Computing 19 (2010), no. 2, 201-226.
arXiv:0712.0111 [Borodin10] A. Borodin, Schur dynamics of the Schur processes, Advances in Mathe-matics, 228, no. 4 (2011), 2268–2291. arXiv:1001.3442 [Borodin11] A. Borodin, Determinantal point processes. In: Oxford Handbook of Ran-dom Matrix Theory, edited by G. Akermann, J. Baik, P. Di Franceco, Oxford Uni-versity Press, 2011.
[BorodinBorodinA] A. Borodin, M. Borodin, A 3d representation for lozenge tilings of a hexagon,
[BorodinBorodinB] A. Borodin, M. Borodin, A 3d representation for domino tilings of the Aztec diamond,
[BorodinCorwin11] A. Borodin, and I. Corwin, Macdonald processes. Probability Theory and Related Fields 158, 1-2 (2014), 225–400. arXiv:1111.4408 [BorodinCorwinGorinShakirov13] A. Borodin, I. Corwin, V. Gorin, S. Shakirov, Ob-servables of Macdonald processes, Transactions of American Mathematical Soci-ety. 368 (2016), 1517–1558. arxiv:1306.0659.
[BorodinFerrari08] A. Borodin, P. Ferrari, Anisotropic growth of random surfaces in 2 + 1 dimensions. Communications in Mathematical Physics, 325 no. 2 (2014), 603–684. arXiv:0804.3035.
243 [BorodinGorin08] A. Borodin, and V. Gorin. Shuffling algorithm for boxed plane par-titions. Advances in Mathematics 220, no. 6 (2009), 1739-1770. arXiv:0804.3071 [Borodin10] A. Borodin, Schur dynamics of the Schur processes, Advances in Mathe-matics, 228, no, 4 (2011), 2268–2291. arXiv:1001.3442 [BorodinGorin12] A. Borodin, and V. Gorin, Lectures on integrable probability. In: Probability and Statistical Physics in St. Petersburg, Proceedings of Symposia in Pure Mathematics, Vol. 91, 155–214. AMS 2016. arXiv:1212.3351 [BorodinGorin13] A. Borodin, V. Gorin, General beta Jacobi corners process and the Gaussian Free Field, Communications on Pure and Applied Mathematics, 68, no.
10, 1774–1844, (2015). arXiv:1305.3627.
[BorodinGorinGuionnet15] A. Borodin, V. Gorin, A. Guionnet, Gaussian asymptotics of discrete β–ensembles, Publications math´ ematiques de l’IH ´ ES 125, no. 1 (2017), 1-78. arXiv:1505.03760 [BorodinGorinRains09] A. Borodin, V. Gorin, and E. M. Rains. q-Distributions on boxed plane partitions. Selecta Mathematica 16, no. 4 (2010), 731-789.
arXiv:0905.0679 [BorodinKuan08] A. Borodin, J. Kuan, Asymptotics of Plancherel measures for the infinite-dimensional unitary group, Advances in Mathematics, 219, no. 3 (2008), 894–931. arXiv:0712.1848 [BorodinKuan10] A. Borodin, J. Kuan, Random surface growth with a wall and Plancherel measures for O(infinity), Communications on Pure and Applied Math-ematics, 63, no. 7 (2010), 831-894. arXiv:0904.2607 [BorodinOlshanski12] A. Borodin and G. Olshanski, The boundary of the Gelfand-Tsetlin graph: A new approach. Advances in Mathematics 230 (2012), 1738– 1779; arXiv:1109.1412.
[BorodinOlshanski13] A. Borodin and G. Olshanski, The Young bouquet and its boundary, Moscow Mathematical Journal 13, no. 2 (2013), 193-232, arXiv:1110.4458 [BorodinOlshanski16] A. Borodin and G. Olshanski, Representations of the infinite symmetric group. Cambridge University Press, 2016 [BorodinPetrov13] A. Borodin, and L. Petrov. Integrable probability: From represen-tation theory to Macdonald processes. Probability Surveys 11 (2014), pp. 1-58.
arXiv:1310.8007 [BorodinPetrov16] A. Borodin, and L. Petrov. Lectures on Integrable probability: Stochastic vertex models and symmetric functions. In Stochastic Processes and Random Matrices: Lecture Notes of the Les Houches Summer School, 104, (2015). Edited by Gregory Schehr, Alexander Altland, Yan V. Fyodorov, Neil O’Connell, and Leticia F. Cugliandolo. Oxford University Press, 2017.
arXiv:1605.01349 [BorodinShlosman08] A. Borodin, S. Shlosman, Gibbs Ensembles of Nonintersect-ing Paths, Communications in Mathematical Physics 293 (2010), 145–170.
arXiv:0804.0564 [BorodinToninelli18] A. Borodin, F. Toninelli, Two-dimensional Anisotropic KPZ growth and limit shapes, Journal of Statistical Mechanics: Theory and Experi-ment, 2018. arXiv:1806.10467 [BorotGorinGuionnet20+] G. Borot, V. Gorin, A. Guionnet, Fluctuations for multi-cut discrete β-ensembles and application to random tilings, in preparation.
244 References [BorotGuionnet11] G. Borot, A. Guionnet, Asymptotic Expansion of β Matrix Models in the One-cut Regime, Communications in Mathematical Physics (2013) 317: 447. arXiv:1107.1167 [BoutillierDeTiliere06] C. Boutillier and B. De Tili ere. Loop statistics in the toroidal honeycomb dimer model. The Annals of Probability, 37, no. 5 (2009), 1747–1777.
arXiv:math/0608600 [BreuerDuits13] J. Breuer and M. Duits, Central Limit Theorems for Biorthogonal En-sembles and Asymptotics of Recurrence Coefficients, Journal of American Math-ematical Society 30 (2017), 27–66. arXiv:1309.6224 [BrezinHikami96a] E. Br´ ezin and S. Hikami. Correlations of nearby levels induced by a random potential. Nuclear Physics B, 479(3):697–706, 1996. arXiv:cond-mat/9605046 [BrezinHikami96b] E. Br´ ezin and S. Hikami. Spectral form factor in a random matrix theory. Physical Review E, 55(4):4067, 1997. arXiv:cond-mat/9608116 [BufetovGorin13] A. Bufetov, V. Gorin, Representations of classical Lie groups and quantized free convolution, Geometric and Functional Analysis (GAFA), 25, no.
3 (2015), 763–814, arXiv:1311.5780 [BufetovGorin16] A. Bufetov, V. Gorin, Fluctuations of particle systems determined by Schur generating functions, Advances in Mathematics 338, no. 7 (2018), 702– 781. arXiv:1604.01110 [BufetovGorin17] A. Bufetov, V. Gorin, Fourier transform on high–dimensional uni-tary groups with applications to random tilings, Duke Mathematical Journal, 168, no. 13 (2019), 2559–2649. arXiv:1712.09925 [BufetovKnizel16] A. Bufetov, A. Knizel, Asymptotics of random domino tilings of rectangular Aztec diamonds, Annales Institut Henri Poincare: Probability and Statistics 54, no. 3 (2018), 1250–1290. arXiv:1604.01491.
[BykhovskayaGorin20] A. Bykhovskaya, V. Gorin, Cointegration in large VARs, Preprint, 2020. arXiv:2006.14179 [CaputoMartinelliToninelli11] P. Caputo, F. Martinelli, F. L. Toninelli, Mixing times of monotone surfaces and SOS interfaces: a mean curvature approach, Communi-cations in Mathematical Physics 311 (2012), 157–189. arXiv:1101.4190 [CerfKenyon01] R. Cerf, R. Kenyon, The Low-Temperature Expansion of the Wulff Crystal in the 3D Ising Model, Communications in Mathematical Physics, 222 no. 1 (2001), 147–179.
[Corwin11] I. Corwin. The Kardar-Parisi-Zhang equation and universality class, Ran-dom Matrices: Theory and Applications, 1, no. 1 (2012). arXiv:1106.1596 [CharlierDuitsKuijlaarsLenells19] C. Charlier, M. Duits, A.B.J. Kuijlaars, J. Lenells, A periodic hexagon tiling model and non-Hermitian orthogonal polyno-mials, Communications in Mathematical Physics 378 (2020), 401–466.
arXiv:1907.02460 [Charlier20] C. Charlier, Doubly periodic lozenge tilings of a hexagon and matrix val-ued orthogonal polynomials. arXiv:2001.11095 [ChelkakLaslierRusskikh20] D. Chelkak, B. Laslier, M. Russkikh, Dimer model and holomorphic functions on t-embeddings of planar graphs, arXiv:2001.11871 [ChelkakRamassamy20] D. Chelkak, S. Ramassamy, Fluctuations in the Aztec dia-monds via a Lorentz-minimal surface, arXiv:2002.07540.
245 [ChhitaYoung13] S. Chhita and B. Young. Coupling functions for domino tilings of Aztec diamonds.
Advcances in Mathematics, 259 (2014), 173–251.
arXiv:1302.0615 [ChhitaJohansson14] S. Chhita and K. Johansson. Domino statistics of the two-periodic Aztec diamond. Advances in Mathematics, 294, (2016) 37–149.
arXiv:1410.2385 [CimasoniReshetikhin06] D. Cimasoni, N. Reshetikhin, Dimers on Surface Graphs and Spin Structures. I. Communications in Mathematical Physics, 275 (2007), 187–208. arXiv:math-ph/0608070 [Ciucu97] M. Ciucu, Enumeration of perfect matchings in graphs with reflective sym-metry. Journal of Combinatorial Theory, Series A, 77, no. 1 (1997), 67–97.
[CiucuKrattenthaler01] M. Ciucu, C. Krattenthaler, Enumeration of Lozenge tilings of hexagons with cut-off corners. Journal of Combinatorial Theory, Series A 100, no. 2 (2002), 201–231. arXiv:math/0104058 [Ciucu08] M. Ciucu, Dimer packings with gaps and electrostatics. Proceding of the National Academy of Sciences of USA 105 no. 8 (2008), 2766–2772.
[CiucuFischer15] M. Ciucu, I. Fischer, Proof of two conjectures of Ciucu and Krattenthaler on the enumeration of lozenge tilings of hexagons with cut off corners. Journal of Combinatorial Theory, Series A 133 (2015), 228–250.
arXiv:1309.4640 [Ciucu15] M. Ciucu, A generalization of Kuo condensation. Journal of Combinatorial Theory, Series A, 134 (2015), 221–241. arXiv:1404.5003 [Ciucu18] M. Ciucu, Symmetries of shamrocks, Part I. Journal of Combinatorial The-ory, Series A, 155 (2018), 376-397.
[CohnElkiesPropp96] H. Cohn, N. Elkies, and J. Propp, Local statistics for random domino tilings of the Aztec diamond, Duke Mathematical Journal, 85 (1996).
math.CO/0008243, 117–166.
[CohnLarsenPropp98] H. Cohn, M. Larsen, and J. Propp, The Shape of a Typical Boxed Plane Partition, New York Journal of Mathematics, 4 (1998), 137–165.
arXiv:math/9801059 [CohnKenyonPropp00] H. Cohn, R. Kenyon, and J. Propp. A variational principle for domino tilings. Journal of the American Mathematical Society, 14(2):297–346, 2001. arXiv:math/0008220 [ColbournMyrvoldNeufeld96] C. J. Colbourn, W. J. Myrvold, and E. Neufeld. Two algorithms for unranking arborescences. Journal of Algorithms, 20 (1996), 268– 281.
[CorwinHammond11] I. Corwin, and A. Hammond. Brownian Gibbs property for Airy line ensembles. Inventiones mathematicae 195, no. 2 (2014), pp. 441-508.
arXiv:1108.2291 [CorwinSun14] I. Corwin, and X. Sun. Ergodicity of the Airy line ensemble. Electronic Communications in Probability 19 (2014). arXiv:1405.0464 [DaleyVere-Jones03] D. J. Daley, D. Vere-Jones, An Introduction to the Theory of Point Processes, Vol.1, 2nd ed, Springer (2003) [Deift00] P. Deift, Orthogonal Polynomials and Random Matrices: A Riemann-Hilbert Approach. Courant Lecture Notes, AMS, New York, 2000 246 References [DeSilvaSavin08] D. De Silva and O. Savin, Minimizers of convex functionals aris-ing in random surfaces, Duke Mathematical Journal 151, no. 3 (2010), 487–532.
arXiv:0809.3816 [DestainvilleMosseruBailly97] N. Destainville, R. Mosseri, and F. Bailly, Configura-tional entropy of co-dimension one tilings and directed membranes, Journal of Statistical Physics 87 (1997), 697–754.
[Dimitrov16] E. Dimitrov, Six-vertex Models and the GUE-corners Process, In-ternational Mathematics Research Notices, 2020, no. 6 (2020), 1794–1881.
arXiv:1610.06893 [Dubedat07] J. Dubedat, SLE and the free field: Partition functions and couplings, Journal of the American Mathematical Society, 22, no. 4 (2009), 995–1054.
arXiv:0712.3018.
[Dubedat11] J. Dubedat, Dubedat. Dimers and families of Cauchy Riemann opera-tors I. Journal of American Mathematical Society, 28, no. 4 (2015), 1063–1167.
arXiv:1110.2808 [Dubedat14] J. Dubedat, Double dimers, conformal loop ensembles and isomon-odromic deformations. Journal of the European Mathematical Society, 21, no.
1 (2019), 1–54. arXiv:1403.6076 [DubedatGheissari14] J. Dubedat and R. Gheissari. Asymptotics of height change on toroidal Temperleyan dimer models. Journal of Statistical Physics, 159, no, 1 (2015), 75–100. arXiv:1407.6227 [Duits15] M. Duits, On global fluctuations for non-colliding processes, Annals of Probability 46, no. 3 (2018), 1279-1350. arXiv:1510.08248 [DuitsKuijlaars17] M. Duits and A.B.J. Kuijlaars, The two periodic Aztec diamond and matrix orthogonal polynomials, to appear in Journal of European Mathemat-ical Society. arXiv:1712.05636.
[DuseMetcalfe14] E. Duse, A. Metcalfe, Asymptotic geometry of discrete interlaced patterns: Part I, International Journal of Mathematics, 26, no. 11, 1550093 (2015).
arXiv:1412.6653 [DuseMetcalfe17] E. Duse, A. Metcalfe, Universal edge fluctuations of discrete inter-laced particle systems, Annales Math´ ematiques Blaise Pascal 25, no. 1 (2018), 75–197. arXiv:1701.08535 [DuseJohanssonMetcalfe15] E. Duse, K. Johansson, A. Metcalfe, The Cusp-Airy Pro-cess, Electronic Journal of Probability 21 (2016), paper 57. arXiv:1510.02057 [Edrei53] A. Edrei, On the generating function of a doubly infinite, totally positive sequence. Transactions of American Mathematical Society 74 (1953), 367–383.
[ElkiesKuperbergLarsenPropp92] N. Elkies, G. Kuperberg, M. Larsen, J. Propp, James. Alternating-sign matrices and domino tilings. Journal of Algebraic Combi-natorics 1 (1992), no. 2, 111–132; Journal of Algebraic Combinatorics 1 (1992), no. 3, 219-234. arXiv:math/9201305 [ErdosYau17] L. Erdos and H.-T. Yau. A dynamical approach to random matrix the-ory, volume 28 of Courant Lecture Notes in Mathematics. Courant Institute of Mathematical Sciences, New York; American Mathematical Society, Providence, RI, 2017.
[EynardMehta97] B. Eynard, M. L. Mehta, Matrices coupled in a chain. I. Eigenvalue correlations, Journal of Physics A: Mathematical and General 31 (1998), 4449– 4456. arXiv:cond-mat/9710230 247 [FerrariSpohn02] P. L. Ferrari, H. Spohn, Step fluctuations for a faceted crystal, Jour-nal of Statistical Physics 113 (2003), 1–46. arXiv:cond-mat/0212456 [Forrester10] P. J. Forrester, Log-Gases and Random Matrices, Princeton University Press, 2010.
[GelfandNaimark50] I. M. Gel’fand, M. A. Naimark, Unitary representations of the classical groups, Trudy Matematicheskogo Instituta imeni V. A. Steklova, Leningrad, Moscow (1950) (in Russian). (German transl.: Academie-Verlag, Berlin, 1957.) [GelfandTsetlin50] I. M. Gel’fand, M. L. Tsetlin, Finite-dimensional representations of the group of unimodular matrices, Dokl. Akad. Nauk SSSR (N.S.) 71 (1950), 825–828, [GesselViennot85] I. Gessel and G. Viennot, Binomial determinants, paths, and hook length formulae, Advances in mathematics 58, no. 3 (1985), 300–321.
[Gilmore17] T. Gilmore, Holey matrimony: marrying two approaches to a tiling prob-lem. S´ eminaire Lotharingien de Combinatoire 78B (2017), Art. 26.
[GoncharovKenyon11] A. B. Goncharov and R. Kenyon, Dimers and cluster integrable systems, Annales scientifiques de l’ ´ Ecole Normale Sup´ erieure 46, no. 5 (2013), 747–813. arXiv:1107.5588 [Golomb95] S. W. Golomb, Polyominoes: Puzzles, Patterns, Problems, and Packings, true classic in the recreational literature (Second Ed.). Princeton University Press, 1995.
[Gorin08] V. E. Gorin, Non-intersecting paths and Hahn orthogonal polynomial en-semble, Functional Analysis and its Applications, 42 (2008), no. 3, 180–197.
arXiv: 0708.2349 [Gorin10] V. Gorin, The q-Gelfand-Tsetlin graph, Gibbs measures and q-Toeplitz ma-trices. Advances in Mathematics 229, no. 1 (2012), 201–266. arXiv:1011.1769 [Gorin13] V. Gorin, From Alternating Sign Matrices to the Gaussian Unitary En-semble, Communications in Mathematical Physics, 332, no. 1 (2014), 437–447, arXiv:1306.6347.
[Gorin16] V. Gorin. Bulk universality for random lozenge tilings near straight bound-aries and for tensor products, Communications in Mathematical Physics 354, no.
1 (2017), 317–344. arXiv:1603.02707 [GorinOlshanski15] V. Gorin and G. Olshanski. A quantization of the harmonic analy-sis on the infinite-dimensional unitary group. Journal of Functional Analysis, 270, no. 1 (2016), 375-418. arXiv:1504.06832 [GorinPanova13] V. Gorin, G. Panova, Asymptotics of symmetric polynomials with applications to statistical mechanics and representation theory, Annals of Proba-bility, 43, no. 6, (2015) 3052–3132. arXiv:1301.0634.
[GorinZhang16] V. Gorin, L. Zhang, Interlacing adjacent levels of β–Jacobi corners processes. Probability Theory and Related Fields 172, no. 3-4 (2018), 915–981.
arXiv:1612.02321 [GrunbaumShephard16] B. Gr¨ unbaum and G. Shephard. Tilings and patterns: Second edition. Dover Books on Mathematics, 2016.
[Guionnet02] A. Guionnet, First order asymptotics of matrix integrals ; a rigorous ap-proach towards the understanding of matrix models, Communications in Mathe-matical Physics 244 (2004), 527–569. arXiv:math/0211131 248 References [Guionnet19] A. Guionnet, Asymptotics of Random Matrices and Related Models: The Uses of Dyson-schwinger Equations (CBMS Regional Conference Series in Mathematics). AMS, 2019 [Hoffe97] M. H¨ offe, Zufallsparkettierungen und Dimermodelle, Thesis, Institut f¨ ur Theoretische Physik der Eberhard-Karls-Universit¨ at, T¨ ubingen, September 1997 [HuMillerPeres09] X. Hu, J. Miller and Y. Peres, Thick points of the Gaussian free field, Annals of Probability, 38, no. 2 (2010), 896–926. arXiv:0902.3842 [Izergin87] A. G. Izergin, Partition function of the six-vertex model in a finite volume, Soviet Physics Doklady 32 (1987) 878–879.
[JerrumValiantVazirani86] M. R. Jerrum, L. G. Valiant, and V. V. Vazirani, Random generation of combinatorial structures from a uniform distribution, Theoretical Computer Science, 43 (1986), 169-188.
[JockushProppShor95] W. Jocush, J. Propp, P. Shor, Random Domino Tilings and the Arctic Circle Theorem, Preprint, 1995. arXiv:math/9801068 [Johansson98] K. Johansson, On Fluctuations of eigenvalues of random Hermitian ma-trices, Duke Mathematical Journal, 91, no. 1 (1998).
[Johansson00] K. Johansson, Shape fluctuations and random matrices, Communica-tions in Mathematical Physics, 209 (2000), 437–476. arXiv:math/9903134.
[Johansson00] K. Johansson, Non-intersecting Paths, Random Tilings and Ran-dom Matrices, Probability Theory and Related Fields, 123 (2002), 225–280.
arXiv:math/0011250 [Johansson05] K. Johansson, The arctic circle boundary and the Airy process, Annals of Probability 33 (2005), 1–30. arXiv:math/0306216 [Johansson06] K. Johansson, Random matrices and determinantal processes. In: Mathematical Statistical Physics 1–55. Elsevier, Amsterdam, 2006, arXiv:math-ph/0510038.
[Johansson17] K. Johansson, Edge fluctuations of limit shapes. Current Developments in Mathematics, 2016. arXiv:1704.06035 [JohanssonNordenstam06] K. Johansson, E. Nordenstam, Eigenvalues of GUE minors, Electronic Journal of Probability, 11 (2006), paper no. 50, 1342–1371.
[Johnstone08] I. Johnstone, Multivariate analysis and Jacobi ensembles: largest eigen-value, Tracy-Widom limits and rates of convergence, Annals of statistics, 36, no.
6 (2008), 2638–2716. arXiv:0803.3408 [Kasteleyn61] P. W. Kasteleyn, The statistics of dimers on a lattice: I. the number of dimer arrangements on a quadratic lattice, Physica 27, no. 12 (1961), 1209–1225.
[Kasteleyn63] P. W. Kasteleyn, Dimer statistics and phase transitions, Journal of Math-ematical Physics 4 (1963), no. 2, 287–293.
[Kasteleyn67] P. W. Kasteleyn, Graph theory and crystal physics, Graph theory and theoretical physics (1967), 43–110.
[KarlinMcGregor59a] S. Karlin, J. McGregor, Coincidence properties of birth and death processes, Pacific journal of Mathematics, 9, no. 4 (1959), 1109–1140.
[KarlinMcGregor59b] S. Karlin, J. McGregor, Coincidence probabilities, Pacific Jour-nal of Mathematics, 9, no. 4 (1959), 1141–1164.
[Kenyon97] R. Kenyon, Local statistics of lattice dimers, Annales de l’Institut Henri Poincar´ e, Probability and Statistics 33 (1997), 591–618. arXiv:math/0105054 [Kenyon99] R. Kenyon, Conformal invariance of domino tiling, Annals of Probability 28 (2000), 759–795. arXiv:math-ph/9910002 249 [Kenyon00] R. Kenyon, Dominos and the Gaussian free field, Annals of Probability 29 (2001), 1128–1137. arXiv:math-ph/0002027 [Kenyon04] R. Kenyon, Height Fluctuations in the Honeycomb Dimer Model, Communications in Mathematical Physics 281 (2008) 675–709. arXiv:math-ph/0405052 [Kenyon09] R. Kenyon, Lectures on Dimers, arXiv:0910.3129.
[KenyonOkounkov03] R. Kenyon and A. Okounkov, Planar dimers and Har-nack curves, Duke Mathematical Journal 131, no. 3 (2006), 499–524.
arXiv:math/0311062 [KenyonOkounkov05] R. Kenyon and A. Okounkov, Limit shapes and the complex burgers equation. Acta Mathematica, 199, no. 2 (2007), 263-302. arXiv:math-ph/0507007 [KenyonOkounkovSheffield03] R. Kenyon, A. Okounkov, S. Sheffield, Dimers and Amoebae, Annals of Mathematics, 163 (2006), 1019–1056. arXiv:math-ph/0311005 [Kenyon11] R. Kenyon, Conformal invariance of loops in the double-dimer model, Communications in Mathematical Physics 326 (2014), 477–497. arXiv:1105.4158 [KenyonSunWilson13] R. W. Kenyon, N. Sun, D. B. Wilson, On the asymptotics of dimers on tori, Probability Theory and Related Fields 166 (2016), 971–1023.
arXiv:1310.2603 [KenyonLamRamassamyRusskikh18] R. Kenyon, W. Y. Lam, S. Ramassamy, M. Russkikh, Dimers and circle packings. arXiv:1810.05616 [Kerov03] S. Kerov: Asymptotic Representation Theory of the Symmetric Group and its Applications in Analysis, Amer. Math. Soc., Providence, RI, 2003.
[KoekoekLeskySwarttouw10] R. Koekoek, P.A. Lesky, and R.F. Swarttouw, Hyperge-ometric orthogonal polynomials and their q-analogues. Springer Verlag, Berlin 2010.
[Konig04] W. K¨ onig, Orthogonal polynomial ensembles in probability theory, Proba-bility Surveys 2 (2005), 385–447. arXiv:math/0403090 [Korepin82] V. E. Korepin, Calculation of norms of Bethe wave functions, Communi-cations in Mathematical Physics 86 (1982) 391–418 [Krattenthaler90] C. Krattenthaler, Generating functions for plane partitions of a given shape, Manuscripta Mathematica 69, no. 1 (1990), 173–201.
[Krattenthaler99] C. Krattenthaler, Advanced Determinant Calculus. In: Foata D., Han GN. (eds) The Andrews Festschrift. Springer, Berlin, Heidelberg.
arXiv:math/9902004.
[Krattenthaler98] C. Krattenthaler, Another involution principle-free bijective proof of Stanley’s hook-content formula. Journal of Combinatorial Theory: Series A 88, no. 1 (1999), 66–92. arXiv:math/9807068 [Krattenthaler01] C. Krattenthaler, A (conjectural) 1/3-phenomenon for the number of rhombus tilings of a hexagon which contain a fixed rhombus. in: Number Theory and Discrete Mathematics, A. K. Agarwal et al., eds., Trends in Mathematics.
Birkh¨ auser, Basel, pp. 13-30. arXiv:math/0101009 [Krattenthaler15] C. Krattenthaler, Plane partitions in the work of Richard Stanley and his school. In: The Mathematical Legacy of Richard P. Stanley; P. Hersh, T. Lam, P. Pylyavskyy and V. Reiner (eds.), Amer. Math. Soc., R.I., 2016, pp. 246-277.
arXiv:1503.05934 250 References [Kuijlaars11] A. B. J. Kuijlaars. Universality. In: The Oxford handbook of random matrix theory, 103–134. Oxford Univ. Press, Oxford, 2011.
[Lai15] T. Lai, A q-enumeration of lozenge tilings of a hexagon with three dents. Ad-vances in Applied Mathematics, 82 (2017), 23–57. arXiv:1502.05780 [LaslierToninelli13] B. Laslier, F. L. Toninelli, Lozenge Tilings, Glauber Dynamics and Macroscopic Shape, Communications in Mathematical Physics, 338, no. 3 (2015), 1287–1326. arXiv:1310.5844 [LevinPeresWilmer17] D. A. Levin, Y. Peres, E. L. Wilmer, Markov Chains and Mix-ing Times, second edition. AMS, Providence, 2017 [Lindstrom73] B. Lindstr¨ om, On the vector representations of induced matroids, Bul-letin of the London Mathematical Society 5, no. 1 (1973), 85–90.
[Lubinsky16] D. S. Lubinsky. An update on local universality limits for correlation functions generated by unitary ensembles. SIGMA Symmetry Integrability Geom.
Methods Appl., 12:Paper No. 078, 36, 2016. arXiv:1604.03133 [LubyRandallSinclair95] M. Luby, D. Randall, A. Sinclair, Markov chain algorithms for planar lattice structures. In 36th Annual Symposium on Foundations of Com-puter Science 150–159 (1995). [Expanded version SIAM Journal on Computing (2001) 31.] [MacMahon1896] P. A. MacMahon, Memoir on the theory of the partition of numbers.
part I, Philosophical Transactions of the Royal Society of London. Series A, Con-taining Papers of a Mathematical or Physical Character 187 (1896), 619–673.
[MacMahon1915] P. A. MacMahon, Combinatory analysis, vol. 1–2, Cambridge Uni-versity Press, 1915–1916, reprinted in one volume by Chelsea, New York, 1960.
[Macdonald95] I. G. Macdonald, Symmetric functions and Hall polynomials. Second edition. Oxford University Press, 1995.
[Matytsin93] A. Matytsin, On the large N limit of the Itzykson-Zuber integral. Nuclear Physics B 411, no. 2–3 (1994), 805-820. arXiv:hep-th/9306077 [McCoyWu73] B. M. McCoy, T. T. Wu, The Two-Dimensional Ising Model, Harvard University Press, 1973.
[Mehta04] M. L. Mehta, Random Matrices (3rd ed.), Amsterdam: Elsevier/Academic Press, 2004.
[Metcalfe11] A. Metcalfe, Universality properties of Gelfand-Tsetlin patterns, Proba-bility Theory and Related Fields, 155, no. 1–2 (2013), 303–346. arXiv:1105.1272 [Mktchyan19] S. Mkrtchyan, Turning point processes in plane partitions with pe-riodic weights of arbitrary period, to appear in Progress in Mathematics.
arXiv:1908.01246 [MontrollPottsWard63] E. W. Montroll, R. B. Potts, J. C. Ward, Correlations and spon-taneous magnetization of the two-dimensional Ising model. Journal of Mathemat-ical Physics 4 (1963):308-322.
[MoralesPakPanova17] A. H. Morales, I. Pak, G. Panova, Hook formulas for skew shapes III. Multivariate and product formulas. Algebraic Combinatorics, 2 no. 5 (2019), 815–861. arXiv:1707.00931 [Nekrasov15] N. Nekrasov, BPS/CFT correspondence: non-perturbative Dyson-Schwinger equations and qq-characters, Journal of High Energy Physics, 2016, no. 181 (2016), arXiv:1512.05388 251 [Neretin03] Yu.
A.
Neretin, Rayleigh triangles and non-matrix interpolation of matrix beta integrals, Sbornik: Mathematics (2003), 194(4), 515–540.
arXiv:math/0301070 [NicaSpeicher06] A. Nica and R. Speicher, Lectures on the Combinatorics of Free Probability, Cambridge University Press, 2006.
[NienhuisHilhorstBlote84] B. Nienhuis, H. J. Hilhorst, and H. W. J. Bl¨ ote, Triangular SOS models and cubic-crystal shapes, Journal of Physics A: Mathematical and General 17 (1984), 3559–2581 [Nordenstam09] E. Nordenstam, Interlaced particles in tilings and random matrices, PhD thesis, KTH, 2009 [NordenstamYoung11] E. Nordenstam, B. Young, Domino shuffling on Novak half-hexagons and Aztec half-diamonds. Electronic Journal of Combinatorics 18 (2011), # P181. arXiv:1103.5054 [OkounkovReshetikhin01] A. Okounkov and N. Reshetikhin.
Correlation function of schur process with application to local geometry of a random 3-dimensional Young diagram. Journal of the American Mathematical Society, 16(3):581–603, 2003. arXiv:math/0107056 [OkounkovReshetikhin05] A. Okounkov and N. Reshetikhin, Random skew plane par-titions and the Pearcey process, Communications in Mathematical Physics 269, no. 3 (2007), 571–609. arXiv:math/0503508v2 [OkounkovReshetikhin06] A. Yu. Okounkov, N. Yu. Reshetikhin, The birth of a ran-dom matrix, Moscow Mathematical Journal, 6, no. 3 (2006), 553–566 [OlshanskiVershik96] G. Olshanski, A. Vershik, Ergodic unitarily invariant measures on the space of infinite Hermitian matrices, In: Contemporary Mathematical Physics. F. A. Berezin’s memorial volume. Amer. Math. Transl. Ser. 2, vol. 175 (R. L. Dobrushin et al., eds), 1996, pp. 137–175. arXiv:math/9601215 [PakShefferTassy16] I. Pak, A. Sheffer, and M. Tassy, Fast domino tileability, Discrete & Computational Geometry, 56, no. 2 (2016), 377–394. arXiv:1507.00770 [Pickrell91] D. Pickrell, Mackey analysis of infinite classical motion groups, Pacific Journal of Mathematics 150 (1991), 139–166.
[Petrov12a] L. Petrov. Asymptotics of random lozenge tilings via Gelfand–Tsetlin schemes.
Probability theory and related fields, 160(3-4):429–487, 2014.
arXiv:1202.3901 [Petrov12b] L. Petrov, Asymptotics of Uniformly Random Lozenge Tilings of Poly-gons. Gaussian Free Field, Annals of Probability, 43, no. 1 (2015), 1–43.
arXiv:1206.5123 [Petrov12c] L. Petrov, The Boundary of the Gelfand-Tsetlin Graph: New Proof of Borodin– Olshanski’s Formula, and its q-Analogue. Moscow Mathematical Jour-nal 14 (2014), 121–160; arXiv:1208.3443.
[PrahoferSpohn01] M Pr¨ ahofer, H. Spohn, Scale Invariance of the PNG Droplet and the Airy Process, Journal of Statistical Physics, 108, no. 5–6 (2002), 1071–1106.
arXiv:math/0105240 [Prasolov94] V. Prasolov, Problems and Theorems in Linear Algebra, Translations of Mathematical Monographs, Vol. 134, American Mathematical Society, 1994.
[ProppWilson96] J. Propp, D. Wilson, Exact sampling with coupled Markov chains and applications to statistical mechanics. Random Structures and Algorithms 9, no. 1-2 (1996), 223–252.
252 References [Propp01] J. Propp, Generalized domino-shuffling, Theoretical Computer Science 303 (2003), 267–301. arXiv:math/0111034 [Propp14] J. Propp, Enumeration of Tilings. Chapter 9 in Handbook of Enumera-tive Combinatorics (Discrete Mathematics and Its Applications) edited by Miklos Bona. CRC Press, 2015.
[Purbhoo07] K. Purbhoo, Puzzles, tableaux, and mosaics, Journal of Algebraic Com-binatorics 28, no. 4 (2008), 461–480, arXiv:0705.1184 [Rosengren16] H. Rosengren, Selberg integrals, Askey-Wilson polynomials and lozenge tilings of a hexagon with a triangular hole. Journal of Combinatorial Theory, Series A 138 (2016), 29-59 1503.00971 [QuastelSpohn15] J. Quastel and H. Spohn. The one-dimensional KPZ equation and its universality class. Journal of Statistical Physics 160, no. 4 (2015), 965–984.
arXiv:1503.06185 [RandallTetali00] D. Randall, and P. Tetali, Analyzing Glauber dynamics by compari-son of Markov chains, Journal of Mathematical Physics 41, 1598 (2000) [Romik15] D. Romik. The Surprising Mathematics of Longest Increasing Subse-quences. Cambridge University Press, 2015.
[Russkikh16] M. Russkikh, Dimers in piecewise Temperleyan domains, Communica-tions in Mathematical Physics, 359, no. 1 (2018), 189–222. arXiv:1611.07884 [Russkikh18] M. Russkikh, Dominos in hedgehog domains. arXiv:1803.10012 [Sagan01] B. E. Sagan, The Symmetric group: Representations, Combinatorial Algo-rithms, and Symmetric Functions, Second Edition, Springer, 2001.
[Sato19] R. Sato, Inductive limits of compact quantum groups and their unitary repre-sentations, arXiv:1908.03988 [Sheffield03a] S.
Sheffield, Random Surfaces.
Asterisque 304 (2005).
arXiv:math/0304049 [Shiryaev16] A. N. Shiryaev, Probability-1, Third Edition, Graduate Texts in Mathe-matics 95, Springer-Verlag New York, 2016.
[Shlosman00] S. Shlosman, Wulff construction in statistical mechanics and in combi-natorics, Russian Mathematical Surveys, 56, no. 4 (2001), 709–738. arXiv:math-ph/0010039 [Sheffield03b] S. Sheffield. Gaussian free fields for mathematicians, Probability The-ory and Related Fields, 139 (2007), 521–541. arXiv:math/0312099 [TaoVu12] T. Tao, V. Vu, Random matrices: The Universality phenomenon for Wigner ensembles, arXiv:1202.0068 [TemperleyFisher61] H. N. V. Temperley and M. E. Fisher, Dimer problem in statis-tical mechanics-an exact result, Philosophical Magazine 6, no. 68 (1961), 1061– 1063.
[Thiant03] N. Thiant, An O (n log n)-algorithm for finding a domino tiling of a plane picture whose number of holes is bounded, Theoretical Computer Science 303, no. 2-3 (2003), 353–374.
[Thurston90] W. P. Thurston, Conway’s tiling groups, The American Mathematical Monthly 97, no. 8 (1990), 757–773.
[Toninelli19] F. L. Toninelli, Lecture notes on the dimer model,
univ-lyon1.fr/homes-www/toninelli/noteDimeri_latest.pdf, 2019.
[Vershik97] A. Vershik, talk at the 1997 conference on Formal Power Series and Al-gebraic Combinatorics, Vienna.
253 [VershikKerov82] A. M. Vershik and S. V. Kerov, Characters and factor representa-tions of the infinite unitary group. Doklady AN SSSR 267 no. (1982), 272–276 (Russian); English translation: Soviet Mathematics Doklady 26 (1982), 570–574.
[Voiculescu76] D. Voiculescu, Repr´ esentations factorielles de type II1 de U(∞). Jour-nal de Math´ ematiques Pures et Appliqu´ ees 55 (1976), 1–20.
[Voiculescu91] D. Voiculescu, Limit laws for random matrices and free products, In-ventiones Mathematicae, 104 (1991), 201–220 [VoiculescuDykemaNica92] D. Voiculescu, K. Dykema, and A. Nica, Free random variables, CRM Monograph Series, vol. 1, American Mathematical Society, Prov-idence, RI, 1992.
[WernerPowell20] W. Werner, E. Powell, Lecture notes on the Gaussian Free Field, arXiv:2004.04720 [Weyl39] H. Weyl, The Classical Groups: Their Invariants and Representations.
Princeton, University Press, 1939.
[Wilson97] D. Wilson, Determinant Algorithms for Random Planar Structures, Pro-ceedings of the Eighth Annual ACM-SIAM Symposium on Discrete Algorithms (1997), 258–267.
[Wilson01] D.
Wilson, Mixing times of Lozenge tiling and card shuffling Markov chains, Annals of Applied Probability, 14, no. 1 (2004), 274–325.
arXiv:math/0102193 [Young10] B. Young, Generating functions for colored 3D Young diagrams and the Donaldson-Thomas invariants of orbifolds (with appendix by Jim Bryan) Duke Mathematical Journal, 152, no. 1 (2010), 115-153.
[Zeilberger95] D. Zeilberger, Reverend Charles to the aid of Major Percy and Fields-Medalist Enrico, The American Mathematical Monthly,103, no. 6 (1996), 501– 502. arXiv:math/9507220 [Zinn-Justin08] P. Zinn-Justin, Littlewood–Richardson Coefficients and Integrable Tilings. Electronic Journal of Combinatorics 16 #R12. arXiv:0809.2392 Index Airy line ensemble, 147, 152, 154 Airy process, 150 Arctic boundary, 80, 85, 137, 145–152, 156–157 Arctic circle, see Arctic boundary bulk limit, 14, 110, 131, 138 Burgers equation, 75–86 Cauchy determinant, 167 Cauchy identity, 192 complementation principle, 147 correlation function, 33, 44, 106, 111, 112, 119, 131, 147, 151, 185 coupling from the past, 229–232 Cusp-Airy process, 156 determinantal point process, 33, 119, 147, 151, 185 domino tiling, 20, 25, 113, 152, 182, 187, 218, 234 edge limit, 15, 145–157 free energy, 40, 45, 56 free probability, 86–89 frozen boundary, see Arctic boundary and frozen phase frozen phase, 7, 12, 85, 90, 112, 137, 204 frozen region, see frozen phase gas phase, 112 Gaussian Free Field, 13, 90–99, 199, 205, 210, 224 Gaussian Unitary Ensemble, 164–171, 187 Gelfand-Tsetlin pattern, 119, 162, 171, 190 Gessel-Viennot lemma, 26 Gibbs measure, 107, 109, 111, 113, 154, 170, 171, 175 GUE, see Gaussian Unitary Ensemble GUE-corners process, 156, 164 Hahn ensemble, 159, 187 height function, 7, 15, 20 concentration, 48 fluctuations, 90, 199, 212, 216 for torus, 57, 224 hole, 209 limit shape, 51, 181, 199 monotonicity, 46, 229 perturbations, 48 Hermite ensemble, 162, 164, 187 Ising model, 9 Karlin-McGregor lemma, 26 Kasteleyn matrix, 22, 24, 31, 33, 37, 39, 117, 129 kernel Airy, 151 Christoffel-Darboux, 185 double contour integral, 119 Dyson Brownian Motion, 119 incomplete beta, 136 Kasteleyn, 33 sine, 112, 136 tilings of trapezoids, 117 Krawtchouk ensemble, 182, 187 Lindstrom lemma, 26 liquid phase, 12, 76, 84, 90, 97, 98, 112, 131, 138, 204 liquid region, see liquid phase log-gas, 159, 176–187 Macdonald operator, 195, 196 MacMahon formula, 11, 29, 159, 198 multiply connected domains, tilings of, 14, 30, 34–38, 209–224 non-intersecting paths, 25–28, 145, 152, 154 orthogonal polynomial ensemble, 185 particle-hole involution, 147 partition function, 21, 30, 33, 39, 198 254 255 Pearcey process, 156 perfect matching, 21–25, 31–38, 105 periodic weights, 112–115 plane partition, 188 random matrices, 88, 119, 153, 157, 164–171, 176, 187 rough phase, see liquid phase sampling of random tilings, 15, 47, 110, 225–235 sawtooth domain, see trapezoid Schur function, see Schur polynomial Schur polynomial, 158, 173, 174, 190–192, 195, 234 semiclassical limit, 157 shuffling algorithms, 234–235 signature, 116, 190 sine process, 112, 136 six-vertex model, 10, 47, 171, 232 slope, 44, 52–57, 60, 76, 109 smooth phase, see gas phase square ice model, 10 steepest descent method, 123–136 surface tension, 50, 54, 56, 60, 69, 76, 102 Tacnode process, 157 Tracy-Widom distribution, 15, 152, 157, 235 trapezoid, tilings of, 86, 116–122, 138, 158, 159, 219 unitary group, 88, 116, 162, 169, 172, 190 |
7794 | https://physics.nist.gov/cgi-bin/cuu/Value?bg | CODATA Value: Newtonian constant of gravitation
Constants Topics: Values Energy Equivalents Searchable Bibliography Background Constants Bibliography Constants, Units & Uncertainty home page Newtonian constant of gravitation Numerical value6.674 30 x 10-11 m 3 kg-1 s-2 Standard uncertainty0.000 15 x 10-11 m 3 kg-1 s-2 Relative standard uncertainty2.2 x 10-5 Concise form6.674 30(15)x 10-11 m 3 kg-1 s-2 Click here for correlation coefficient of this constant with other constants Source: 2022 CODATArecommended valuesDefinition of uncertaintyCorrelation coefficient with any other constant |
7795 | https://math.jhu.edu/~js/Math202/example1.pdf | Example (Distance between skew lines) Find the distance between the lines L1 : x + 2 2 = y −1 3 = z + 1 −1 and L2 : x −1 −1 = y + 1 2 = z −2 4 .
The direction of L1 is ⃗ v =< 2, 3, −1 > and it passes through P = (−2, 1, −1).
The direction of L2 is ⃗ w =< −1, 2, 4 > and it passes through Q = (1, −1, 2).
The main step is to find parallel planes containing the lines passing through P and Q respectively. To this end we compute ⃗ N = ⃗ v × ⃗ w = ˆ i ˆ j ˆ k 2 3 −1 −1 2 4 =< 14, −7, 7 > .
Think of the two lines as lying in the plane ⃗ X · ⃗ N = 0 passing through the origin. That is translate the lines in the ⃗ N direction until they lie in this plane. Note that the two lines intersect. Therefore the required distance between the lines is just the distance between the planes. This is obtained by taking ⃗ PQ =< 3, −2, 3 > and projecting it in the direction of ⃗ N =< 14, −7, 7 >, that is d = |Pr ⃗ N ⃗ PQ| = | ⃗ PQ · ⃗ N| | ⃗ N| = 77 7 √ 6 = 11 √ 6 .
Draw yourself a picture of the parallel planes (with normal direction ⃗ N) passing through P (containing L1) and Q (containing L2) respec-tively. Then draw the vector ⃗ PQ and its projection onto ⃗ N (i.e the line through ⃗ N).
1 |
7796 | https://inis.iaea.org/records/n1ggt-qw825 | M1 and E1 transitions of heavy-light quarkonia
Skip to main
...
All INIS – International Nuclear Information System
Communities
My dashboard
Log in
Published December 2021 | Version v1
Book Metadata-only
M1 and E1 transitions of heavy-light quarkonia
Creators
Bhatnagar, Shashank 1
Guleria, Vaishali 1
Gebrehana, Eshete 2
Show affiliations
Department of Physics, Chandigarh University, Mohali 140413 (India)
Department of Physics, Woldia University, Woldia (Ethiopia)
Description
The radiative transitions of heavy quarkonia are of considerable experimental, and theoretical interest, and provide an insight into the dynamics of quarkonium. Electric dipole E1 transitions are much stronger than the magnetic dipole, M1 transitions, though the M1 are more sensitive to relativistic effects. For details of M1 and E1 transitions. In this work, we focus on the radiative decays of the charmed and bottom vector mesons through the processes, V → Pγ, V → Sγ, and S → Vγ, and A- → Pγ, where V; P; S and A- refer to vector, pseudoscalar, scalar and axial quarkonia, 1+-. The aim of doing this study was to mainly test our analytic forms of wave functions obtained as solutions of mass spectral equations in an approximate harmonic oscillator basis obtained from 4 × 4 BSE as a starting point, that has so far given good predictions not only of the mass spectrum of heavy-light quarkonia, but also their various transitions, such as leptonic decays, two photon decays, and single photon radiative decays
Additional details
Publishing Information
Publisher Bhabha Atomic Research Centre Imprint Place Mumbai (India)Imprint Title Proceedings of the DAE-BRNS symposium on nuclear physics. V. 65 Imprint Pagination[977 p.]Journal Page Range[2 p.]
Conference
Title 65. DAE-BRNS symposium on nuclear physics Dates 1-5 Dec 2021 Place Mumbai (India)
INIS
Country of Publication India Country of Input or Organization India INIS RN 53031687 Subject category S72: PHYSICS OF ELEMENTARY PARTICLES AND FIELDS; Resource subtype / Literary indicator Conference Descriptors DEI BOUND STATE; ELECTRIC DIPOLES; LEPTONIC DECAY; QUARK MODEL; QUARKONIUM; VECTOR MESONS; WAVE FUNCTIONS Descriptors DEC BOSONS; COMPOSITE MODELS; DECAY; DIPOLES; ELEMENTARY PARTICLES; FUNCTIONS; FUNDAMENTAL INTERACTIONS; HADRONS; INTERACTIONS; MATHEMATICAL MODELS; MESONS; MULTIPOLES; PARTICLE DECAY; PARTICLE MODELS; WEAK INTERACTIONS; WEAK PARTICLE DECAY
Optional Information
Lead record 63z8c-hqn29Notes Article No. D11
Contact
Inquire about this record
Details
Resource type Book Publisher Bhabha Atomic Research Centre Published in[2 p.], 2021.Conference 65. DAE-BRNS symposium on nuclear physics , Mumbai (India), 1-5 Dec 2021 Languages English
Versions
Citation
[object Object]
Style
APA
APA
BibTeX
Chicago
Harvard
IEEE
MLA
Vancouver
Export
JSON
JSON
JSON-LD
CSL
DataCite JSON
DataCite XML
Dublin Core XML
MARCXML
BibTeX
GeoJSON
DCAT
Codemeta
Citation File Format
Export
Technical metadata
Created January 18, 2025
Modified January 29, 2025
Jump up
International Atomic Energy Agency (IAEA)
Vienna International Centre, PO Box 100, A-1400 Vienna, Austria
Telephone: (+431) 2600-0,Facsimile: (+431) 2600-7,Official Mail
Contact UsDisclaimer
Copyright © 2025 International Atomic Energy Agency. All rights reserved. |
7797 | https://medium.com/@rondotsch/degrees-of-freedom-tutorial-8d8e5c7be6ec | Sitemap
Open in app
Sign in
Sign in
Degrees of Freedom Tutorial
Ron Dotsch
10 min readJan 18, 2022
A lot of researchers seem to be struggling with their understanding of the statistical concept of degrees of freedom. Most do not really care about why degrees of freedom are important to statistical tests, but just want to know how to calculate and report them. This page will help. For those interested in learning more about degrees of freedom, take a look at the following resources:
This chapter in the little handbook of statistical practice
Walker, H. W. (1940). Degrees of Freedom. Journal of Educational Psychology, 31(4), 253–269.
I couldn’t find any resource on the web that explains calculating degrees of freedom in a simple and clear manner and believe this page will fill that void. It reflects my current understanding of degrees of freedom, based on what I read in textbooks and scattered sources on the web. Feel free to add or comment.
Conceptual Understanding
Let’s start with a simple explanation of degrees of freedom. I will describe how to calculate degrees of freedom in an F-test (ANOVA) without much statistical terminology. When reporting an ANOVA, between the brackets you write down degrees of freedom 1 (df1) and degrees of freedom 2 (df2), like this: “F(df1, df2) = …”. Df1 and df2 refer to different things, but can be understood the same following way.
Imagine a set of three numbers, pick any number you want. For instance, it could be the set [1, 6, 5]. Calculating the mean for those numbers is easy: (1 + 6 + 5) / 3 = 4.
Now, imagine a set of three numbers, whose mean is 3. There are lots of sets of three numbers with a mean of 3, but for any set the bottom line is this: you can freely pick the first two numbers, any number at all, but the third (last) number is out of your hands as soon as you picked the first two. Say our first two numbers are the same as in the previous set, 1 and 6, giving us a set of two freely picked numbers, and one number that we still need to choose, x: [1, 6, x]. For this set to have a mean of 3, we don’t have anything to choose about x. X has to be 2, because (1 + 6 + 2) / 3 is the only way to get to 3. So, the first two values were free for you to choose, the last value is set accordingly to get to a given mean. This set is said to have two degrees of freedom, corresponding with the number of values that you were free to choose (that is, that were allowed to vary freely).
This generalizes to a set of any given length. If I ask you to generate a set of 4, 10, or 1.000 numbers that average to 3, you can freely choose all numbers but the last one. In those sets the degrees of freedom are respectively, 3, 9, and 999. The general rule then for any set is that if n equals the number of values in the set, the degrees of freedom equals n — 1.
This is the basic method to calculate degrees of freedom, just n — 1. It is as simple as that. The thing that makes it seem more difficult, is the fact that in an ANOVA, you don’t have just one set of numbers, but there is a system (design) to the numbers. In the simplest form you test the mean of one set of numbers against the mean of another set of numbers (one-way ANOVA). In more complicated one-way designs, you test the means of three groups against each other. In a 2 x 2 design things seem even more complicated. Especially if there’s a within-subjects variable involved (Note: all examples on this page are between-subjects, but the reasoning mostly generalizes to within-subjects designs). However things are not as complicated as you might think. It’s all pretty much the same reasoning: how many values are free to vary to get to a given number?
Df1
Df1 is all about means and not about single observations. The value depends on the exact design of your test. Basically, the value represents the number of cell means that are free to vary to get to a given grand mean. The grand mean is just the mean across all groups and conditions of your entire sample. The cell means are nothing more than the means per group and condition. We’ll call the number of cells (or cell means) k.
Let’s start off with a one-way ANOVA. We have two groups that we want to compare, so we have two cells. If we know the mean of one of the cells and the grand mean, the other cell must have a specific value such that (cell mean 1 + cell mean 2) / 2 = grand mean (this example assumes equal cell sample sizes, but unequal cell sample sizes would not change the number of degrees of freedom). Conclusion: for a two-group design, df1 = 1.
Get Ron Dotsch’s stories in your inbox
Join Medium for free to get updates from this writer.
Sticking to the one-way ANOVA, but moving on to three groups. We now have three cells, so we have three means and a grand mean. Again, how many means are free to vary to get to the given grand mean? That’s right, 2. So df1 = 2. See the pattern? For one-way ANOVA’s df1 = k — 1.
Moving on to an ANOVA with four groups. We know the answer if this is a one-way ANOVA (that is, a 4 x 1 design): df1 = k — 1 = 4 -1 = 3. However, what if this is a two-way ANOVA (a 2 x 2 design)? We still have four means, so to get to a given grand mean, we can have three freely varying cell means, right? Although this is true, we have more to deal with than just the grand mean, namely the marginal means. The marginal means are the combined cell means of one variable, given a specific level of the other variable. Let’s say our 2 x 2 ANOVA follows a 2 (gender: male vs. female) x 2 (eye color: blue vs. brown) design. In that case, the grand mean is the average of all observations in all 4 cells. The marginal means are the average of all eye colors for male participants, the average of all eye colors for female participants, the average of all genders for blue-eyed participants, and the average of all genders for brown-eyed participants. The following table shows the same thing:
The reason that we are now dealing with marginal means is that we are interested in interactions. In a 4 x 1 one-way ANOVA, no interactions can be calculated. In our 2 x 2 two-way ANOVA, we can. For instance, we might be interested in whether females perform better than males depending on their eye color. Now, because we are interested in cell means differences in a specific way (i.e., we are not just interested in whether one cell mean deviates from the grand mean, but we are also interested in more complex patterns), we need to pay attention to the marginal means. As a consequence, we now have less freedom to vary our cell means, because we need to account for the marginal means (if you want to know how this all works, you should read up on how the sums of squares are partitioned in 2 x 2 ANOVA’s). It is also important to realize that if all marginal means are fixed, the grand mean is fixed too. In other words, we do not have to worry about the grand mean anymore for calculating our df1 in a two-way ANOVA, because we are already worrying about the marginal means. As a consequence, our df1 will not lose a degree of freedom because we do not want to get to a specific grand mean. Our df1 will only lose degrees of freedom to get to the specific marginal means.
Now, how many cell means are free to vary before we need to fill in the other cell means to get to the four marginal means in the 2 x 2 design? Let’s start with freely picking the cell mean for brown eyed males. We know the marginal mean for brown eyed males and blue eyed males together (it is given, all marginal means are), so I guess we can’t choose the blue eyed males cell mean freely. There goes one degree of freedom. We also know the marginal mean for brown eyed males and brown eyed females together. That means we can’t choose the brown eyed female cell mean freely either. And as we know the other two marginal means, we have no choice in what we put in the blue eyed females cell mean to get to the correct marginal means. So, we chose one cell mean, and the other three cell means had to be filled in as a consequence to get to the correct marginal means. You know what that means don’t you? We only have one degree of freedom in df1 for a 2 x 2 design. That’s different from the three degrees of freedom in a 4 x 1 design. The same number of groups and they might even contain the same observations, but we get a different number of degrees of freedom. So now you see that using the degrees of freedom, you can infer a lot about the design of the test.
You could do the same mental exercise for a 2 x 3 design, but it is tedious for me to write up, so I am going to give you the general rule. Every variable in your design has a certain number of levels. Variable 1 in the 2 x 3 design has 2 levels, variable 2 has 3 levels. You get df1 when you multiply the levels of all variables with each other, but with each variable, subtract one level. So in the 2 x 3 design, df1 would be (2–1) x (3–1) = 2 degrees of freedom. Back to the 2 x 2 design, df1 would be (2–1) x (2–1) = 1 degrees of freedom. Now let’s see what happens with a 2 x 2 x 2 design: (2–1) x (2–1) x (2–1) = still 1 degrees of freedom. A 3 x 3 x 4 design (I hope you’ll never have to analyze that one): (3–1) x ( 3–1) x (4 -1) = 2 x 2 x 3 = 12 degrees of freedom.
By now, you should be able to calculate df1 in F(df1, df2) with ease. By the way, most statistical programs give you this value for free. However, now you’ll be able to judge whether researchers have performed the right analyses in their papers to some extent based on their df1 value. Also, df1 is calculated the same way in a within-subjects design. Just treat the within-subjects variable as any other variable. Let’s move on to df2.
DF2
Whereas df1 was all about how the cell means relate to the grand mean or marginal means, df2 is about how the single observations in the cells relate to the cell means. Basically the df2 is the total number of observations in all cells (n) minus the degrees of freedoms lost because the cell means are set (that is, minus the number of cell means or groups/conditions: k). Df2 = n — k, that’s all folks! Say we have 150 participants across four conditions. That means we will have df2 = 150–4 = 146, regardless of whether the design is 2 x 2, or 4 x 1.
Most statistical packages give you df2 too. In SPSS, it’s called df error, in other packages it might be called df residuals.
For the case of within subjects-designs, things can become a bit more complicated. The following paragraphs are work in progress. The calculation of df2 for a repeated measures ANOVA with one within-subjects factor is as follows: df2 = df_total — df_subjects — df_factor, where df_total = number of observations (across all levels of the within-subjects factor, n) — 1, df_subjects = number of participants (N) — 1, and df_factor = number of levels (k) — 1. Basically, the take home message for repeated measures ANOVA is that you lose one additional degree of freedom for the subjects (if you’re interested: this is because the sum of squares representing individual subjects’ average deviation from the grand mean is partitioned separately, whereas in between-subjects designs, that’s not the case. To get to a specific subjects sum of squares, N — 1 subject means are free to vary, hence you lose one additional degree of freedom).
Conclusion
You should be able to calculate df1 and df2 with ease now (or identify it from the output of your statistical package like SPSS). Keep in mind that the degrees of freedom you specify are those of the design of the effect that you are describing. There is no such thing as one set of degrees of freedom that is appropriate for every effect of your design (although, in some cases, they might seem to have the same value for every effect).
Moreover, although we have been discussing means in this tutorial, for a complete understanding, you should learn about sums of squares, how those translate into variance, and how test statistics, such as F-ratio, work. This will make clear to you how degrees of freedom are used in statistical analysis. The short functional description is that, primarily, degrees of freedom affect which critical values are chosen for test statistics of interest, given a specific alpha level (remember those look-up tables in your early statistics classes?).
Why do we use n-1 degrees of freedom? Computing the mean uses all observations, divided by n. The mean is then used in computing sum of squares (the mean needs to be known, otherwise you can’t compute sum of squares). That fixes one number (the mean), and therefore you lose one degrees in freedom in computing sum of squares. If the mean is known, n-1 observations are free to vary. The last one no longer gets to be freely picked to get to a given mean.
Statistics
Statistical Analysis
Degrees Of Freedom
Tutorial
## Written by Ron Dotsch
22 followers
·27 following
Responses (1)
Write a response
What are your thoughts?
Alignednow
Oct 23, 2023
```
Greetings from Indonesia. Thank you for the information
```
More from Ron Dotsch
Ron Dotsch
## Reverse correlation image classification using R
Here I describe how to use the actively maintained (and relatively easy to use) R package rcicr that I wrote to generate reverse…
Jan 18, 2022
Ron Dotsch
## Importing Inquisit data files into SPSS
So you have your first Inquisit dataset. The next step is to preprocess the data so it can be analyzed with a statistical program. This…
Jan 22, 2022
7
See all from Ron Dotsch
Recommended from Medium
Pablo Alcain
## The unbearable effectiveness of the Central Limit Theorem
To view a better formatted version of this post, go to
Aug 10
151
In
Long. Sweet. Valuable.
by
Ossai Chinedum
## I’ll Instantly Know You Used Chat Gpt If I See This
Trust me you’re not as slick as you think
May 16
24K
1439
Stat Hacks
## Analysis of Variance (ANOVA) vs t-Test: Differences, Uses, and Examples
Note: this post is part of a series of posts about How to Choose an Appropriate Statistical Test
Aug 11
5
Damini Saini
## Hyperparameters 101: Your One-Stop Guide for All Deep Learning Models (Part 1)
An ultimate guide to deep learning hyperparameters to help you master the critical hyperparameters that power every deep learning model.
6d ago
3
1
In
An Injustice!
by
J. Henderson
## 10 Charlie Kirk Quotes, Ranked from Simply Bad to Utterly Horrible
Everyone has the right to free speech, but what happens when it turns into radicalism and extremism?
Sep 20
7.6K
337
Jordan Gibbs
## ChatGPT Is Poisoning Your Brain…
Here‘s How to Stop It Before It’s Too Late.
Apr 29
26K
1287
See more recommendations
Text to speech |
7798 | https://precalculusfinal.weebly.com/17-power-reducing-identities.html | Home
(1-13) Trigonometric Functions
(1) Degrees and Radians
(2) Unit Circle
(3) Periodic Functions
(4) Coterminal Angles
(5) Initial Side, Terminal Side
(6-13) Graphs >
(6) Sine Function
(7) Cosine Function
(8) Tangent Function
(9) Secant Function
(10) Cosecant Function
(11) Cotangent Function
(12) Inverse Trigonometric Function
(13) Sinusoids
(14-18) Identies
(14) Sum and Difference Identies
(15) Pythagorean Identities
(16) Double Angle Identities
(17) Power Reducing Identities
(18) Half Angle Identities
(19-20) Laws
(19) Sine
(20) Cosine
(21) Mathematical Induction
(22-23) Polar Graphs
(22) Curves and Roses
(23) How to Use an (r, theta) Table
(24) Basic Statistics
The Five Number Summary
Box Plots
Variance/Standard Deviation
Power reducing identities
power reducing identities/ Reducing formulas:
Power reduction formulas function a lot like double-angle and half-angle formulas do. They are used to simplify the calculations necessary to solve a given expression. Power reduction formulas can be derived through the use of double-angle and half-angle formulas, and the Pythagorean Identity (sin ^2 a + cos a = 1). In power reduction formulas, a trigonometric function is raised to a power (such as sin^2 a or cos ^2 a ). The use of a power reduction formula expresses the quantity without the exponent.
check out video for A step TO step process:
Power Reducing Formulas for SINE AND COSINE
Powered by Create your own unique website with customizable templates. |
7799 | https://link.springer.com/article/10.1007/s00454-025-00770-1 | Combinatorics of Generalized Parking-Function Polytopes | Discrete & Computational Geometry
Your privacy, your choice
We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media.
By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection.
See our privacy policy for more information on the use of your personal data.
Manage preferences for further information and to change your choices.
Accept all cookies
Skip to main content
Menu
Find a journalPublish with usTrack your research
Search
Cart
Search
Search by keyword or author
Search
Navigation
Find a journal
Publish with us
Track your research
Home
Discrete & Computational Geometry
Article
Combinatorics of Generalized Parking-Function Polytopes
Open access
Published: 12 September 2025
(2025)
Cite this article
Download PDF
You have full access to this open access article
Discrete & Computational GeometryAims and scopeSubmit manuscript
Combinatorics of Generalized Parking-Function Polytopes
Download PDF
Margaret M. Bayer1,
Steffen Borgwardt2,
Teressa Chambers3,
Spencer Daugherty4,
Aleyah Dawkins5,
Danai Deligeorgaki6,
Hsin-Chieh Liao7,
Tyrrell McAllister8,
Angela Morrison10,
Garrett Nelson9&
…
Andrés R. Vindas-MeléndezORCID: orcid.org/0000-0002-7437-374511
Show authors
839 Accesses
Explore all metrics
Abstract
For b=(b 1,…,b n)∈Z>0 n, a b -parking function is defined to be a sequence (β 1,…,β n) of positive integers whose nondecreasing rearrangement β 1′≤β 2′≤⋯≤β n′ satisfies β i′≤b 1+⋯+b i. The b-parking-function polytope X n(b) is the convex hull of all b-parking functions of length n in R n. Geometric properties of X n(b) were previously explored in the specific case where b=(a,b,b,…,b) and were shown to generalize those of the classical parking-function polytope. In this work, we study X n(b) in full generality. We present a minimal inequality and vertex description for X n(b), prove it is a generalized permutahedron, and study its h-polynomial. Furthermore, we investigate X n(b) through the perspectives of building sets and polymatroids, allowing us to identify its combinatorial types and obtain bounds on its combinatorial and circuit diameters.
Similar content being viewed by others
Generalized Parking Function Polytopes
Article Open access 06 November 2023
On expected face numbers of random beta and beta’ polytopes
Article Open access 11 February 2022
Base polyhedra and the linking property
Article 08 May 2017
Explore related subjects
Discover the latest articles, books and news in related subjects, suggested using machine learning.
Algebraic Geometry
Combinatorics
Combinatorial Geometry
Discrete Mathematics
Polytopes
Special Functions
Use our pre-submission checklist
Avoid common mistakes on your manuscript.
1 Introduction
A classical parking function of length n is a list (α 1,α 2,…,α n) of positive integers whose nondecreasing rearrangement α 1′≤α 2′≤⋯≤α n′ satisfies α i′≤i. It is well-known that the number of classical parking functions of length n is (n+1)n−1. This number surfaces in a variety of places; for example, it counts the number of labeled rooted forests on n vertices and the number of regions of a Shi arrangement in R n (see for further discussion). Let PF n denote the convex hull in R n of all classical parking functions of length n. In 2020, Stanley asked for the volume and the number of vertices, facets, and lattice points of PF n. These questions were first answered by Amanbayeva and Wang in [1 (2022), S2R10, 10")] and independently (regarding the volume, number of vertices, and number of facets) by Stong in [41, 286–289 (2022)")]. Recently, Behrend obtained a general formula for the Ehrhart polynomials of partial permutahedra [7,
arxiv:2403.06975
"), Theorem 1], the m=n−1, t=1 case of which yields an expression for the number of lattice points in PF n that is more concise and explict than the Amanbayeva–Wang formula. In this article, we present results on the combinatorial and geometric properties of the convex hull of a generalization of classical parking functions, namely b-parking functions.
Definition 1.1
Let b=(b 1,…,b n)∈Z>0 n. A b -parking function is a list (β 1,…,β n) of positive integers whose nondecreasing rearrangement β 1′≤β 2′≤⋯≤β n′ satisfies β i′≤b 1+⋯+b i.
This generalization of parking functions has been previously explored from an enumerative perspective by Yan [45,46,47] and Pitman and Stanley . In, Hanada, Lentfer, and Vindas-Meléndez generalized the classical parking-function polytope to the b-parking-function polytope X n(b), i.e., the convex hull of all b-parking functions of length n in R n for b=(b 1,…,b n)∈Z>0 n. The work of Hanada, Lentfer, and Vindas-Meléndez focused on the b-parking-function polytope in the special case when b=(a,b,b,…,b). In particular, they described the face structure and provided a closed volume formula for the b-parking-function polytope in that special case. These parking-function polytopes have surprisingly arisen in different guises and in connection to other polytopes and combinatorial objects, including partial permutahedra 83
"), [9, Art. 64, 12"), 27, 2863–2888 (2022)")], Pitman-Stanley polytopes [26, 575–613 (2024)"), 40, 603–634 (2002)")], polytopes of win vectors [3"), 6, 1–15 (1997)")], and stochastic sandpile models [37 (2024), 3.26, 29")].
In this paper, we explore b-parking-function polytopes for general b. Our contributions about b-parking-function polytopes lend themselves to fruitful exploration from different perspectives and propose connections to seemingly unrelated combinatorial objects. We present definitions, terminology, and references in each relevant section. The paper is structured as follows:
We begin Section 2 by studying the b-parking-function polytope from a polyhedral perspective. We show that X n(b) is (apart from a single degenerate case) an n-dimensional simple polytope (Proposition 2.2 and Theorem 2.3(a)), and we give explicit descriptions of the vertices, facets, and edges of X n(b) (Theorems 2.3(b) and (c) and Proposition 2.7, respectively). We then show that, up to an affine unimodular isomorphism, X n(b) is a generalized permutahedron (Theorem 2.12). We conclude the section by presenting the h-polynomial and f-vector of X n(b) (Theorem 2.16 and 2.17, respectively).
In Section 3 we take a building-set perspective to study X n(b), which allows us to completely characterize the face structure of X n(b) for any b∈Z>0 n (Theorem 3.12). As a consequence, we show that, up to combinatorial equivalence, every b-parking-function polytope is either the classical parking-function polytope PF n or the stellohedron St n (Corollary 3.14).
We identify X n(b) as a special polymatroid (Theorem 4.3) in Section 4 and use this perspective to derive linear bounds on its combinatorial and circuit diameters (Theorem 4.4 and Theorem 4.6, respectively).
In Section 5, we revisit the classical parking-function polytope PF n and summarize its known and new connections to other established families of polytopes. We conclude by adding a description of PF n as a projection of a relaxed Birkhoff polytope (Theorem 5.1), and X n(b) as a projection of a relaxed-partition polytope (Corollary 5.3).
2 Polyhedral Perspectives on X n(b)
Fix, throughout this work, a positive integer n and a vector b=(b 1,…,b n)∈Z>0 n. The following definition introduces our main object of study, i.e., a polytope associated to b-parking functions, which we denote X n(b). Standard references on polytopes include and .
Definition 2.1
The b -parking-function polytope X n(b) is the convex hull of all b-parking functions in R n.
Recall that the linear action of the symmetric group S n on R n is defined by π(∑i=1 n a i e i):=∑i=1 n a i e π(i) for all π∈S n, where e 1,…,e n is the standard basis of R n. Note that by definition, the polytope X n(b) is invariant under the action of S n. For convenience, we use the notation
S i:=∑j=1 i b j for 1≤i≤n
\; and \;v k:=(1,…,1,S k+1,S k+2,…,S n)∈R n for 0≤k≤n.
Further observe that the v k are b-parking functions, so v k∈X n(b) for all k. We will see below that the permutations π(v k) of the v k are precisely the vertices of X n(b) (Theorem 2.3(b)).
A theme that will appear early in our study of X n(b) is that the case in which b 1=1 behaves differently from when b 1≥2. Indeed, if b 1≥2, then the b-parking functions v 0,v 1,…,v n are distinct, and the index k in v k is precisely the number of entries in v k that equal 1 for all k≥0. However, if b 1=1, then v 0=v 1=(1,S 2,S 3,…,S n). This trivial coincidence has important consequences for the structure of the polytope X n(b). Figure 1 shows the b-parking-function polytopes X 3(1,2,3) (left) and X 3(2,3,4) (right). Note also that there may be lattice points in X n(b) that are not b-parking functions. For example, (2,2,2) is in X 3(1,2,3) but is not a (1,2,3)-parking function.
Fig. 1
The polytopes X 3(1,2,3) (left) and X 3(2,3,4) (right)
Full size image
We begin by proving that, with the exception of a single “degenerate” case, X n(b) is an n-dimensional polytope.
Proposition 2.2
If n=1=b 1, then dim(X n(b))=0. However, if n≥2, or if n=1 and b 1≥2, then dim(X n(b))=n.
Proof
In the degenerate case n=1=b 1, the polytope X n(b) contains only the point(1) in R 1. In all other cases, X n(b) is n-dimensional because, if b 1≥2, then {v k:0≤k≤n} is a set of n+1 affinely independent points in X n(b), and, if b 1=1, but n≥2, then {(S 2,1,S 3,…,S n)}∪{v k:1≤k≤n} is a set of n+1 affinely independent points in X n(b). ◻
In view of Proposition 2.2, we exclude the degenerate case n=1=b 1 from this point forward. That is, we proceed throughout the rest of this paper with the additional assumption that either n≥2 or, if n=1, then b 1≥2. Thus, dimX n(b)=n.
2.1 Vertices, Facets, and Edges of X n(b)
In this subsection, we describe the vertices, facets, and edges of the b-parking-function polytope X n(b). In particular, we find that X n(b) is a simple polytope, meaning that the number of edges of X n(b) that are incident to each vertex of X n(b) is n. Here and below, we use the standard notation [n]:={1,2,…,n}.
Theorem 2.3
(a)
The polytope X n(b) is simple.
(b)
The set of vertices of X n(b) is {π(v k):π∈S n and 0≤k≤n}.
(c)
The minimal inequality description of X n(b) is as the set of points (x 1,…,x n) in R n such that x i≥1 for 1≤i≤n and
∑i∈I x i≤∑i=n−|I|+1 n S i, (b-parking)
or, equivalently,
∑i∈I x i≤∑j=1 n min{j,|I|}b n−j+1,
for all nonempty I⊆[n] if b 1≥2, and for all nonempty I⊆[n] with |I|≠n−1 if b 1=1. In other words, these inequalities are precisely the facet-defining inequalities for X n(b).
Before proceeding to the proof of Theorem 2.3, we note that, even when b 1=1, every point in X n(b) satisfies the (b-parking)inequalities for all subsets I⊆[n], including those where |I|=n−1. However, when b 1 = 1, these “|I|=n−1” (b-parking)inequalities are redundant because they follow from the (b-parking)inequality for [n] together with the inequalities x i≥1. For, let m∈[n] and I:=[n]∖{m}. Then the (b-parking)inequality for [n] (with b 1=1) is
∑i∈[n]x i≤∑j=1 n−1 j b n−j+1+n.
Thus, subtracting the inequality x m≥1 yields
∑i∈I x i≤∑j=1 n−1 j b n−j+1+(n−1)=∑j=1 n min{j,|I|}b n−j+1,
which is the (b-parking)inequality for I. Nonetheless, none of the other inequalities can be eliminated, as the following proof demonstrates inter alia.
We also remark here that Theorem 2.3 can be proved using the established theory of polymatroids. See Section 4 for further discussion of this connection. However, for completeness, we provide a proof here of Theorem 2.3 that does not rely on that theory. In fact, it is the details of the proof of Theorem 2.3 that motivate the forthcoming connections with generalized permutahedra and polymatroids.
Proof of Theorem 2.3
We temporarily write P(b) for the polytope of points in R n satisfying the system of inequalities given in Theorem 2.3(c). The outline of the proof is as follows. We first prove that X n(b)⊆P(b). We then prove that the vertices of P(b) are all of the form π(v k) as in Theorem 2.3(b). Thus, the vertices of P(b) are b-parking functions, which proves the converse containment P(b)⊆X n(b), giving us X n(b)=P(b) and the “⊆” direction of Theorem 2.3(b). In the course of proving that every vertex of P(b) is of the form π(v k), we will have shown that each such vertex satisfies precisely n of the inequalities given in Theorem 2.3(c), which demonstrates that X n(b) is simple (Theorem 2.3(a)). We will then show that every point of the form π(v k) is in fact a vertex of X n(b), completing the proof of Theorem 2.3(b). Finally, we will see that each of the inequalities in Theorem 2.3(c) is satisfied with equality by some point in X n(b). This fact, together with the fact that every vertex of X n(b) satisfies exactly n such inequalities, implies that the system of inequalities in Theorem 2.3(c) is minimal and thus will complete the proof of Theorem 2.3.
To prove that X n(b)⊆P(b), it suffices to show that every vertex of X n(b) is in P(b). To this end, let v=(v 1,…,v n)∈R n be a vertex of X n(b), and let I⊆[n]. Write v′=(v 1′,…,v n′) for the weakly increasing reordering of v. Thus, the final |I| coordinates of v′, namely v n−|I|+1′,v n−|I|+2′,…,v n′, are the |I| weakly largest coordinates of v. Moreover, v is a b-parking function (because the vertices of the convex hull of a finite set are elements of that set), so 1≤v i′≤∑j=1 i b j for 1≤i≤n. Hence, v i≥1 for 1≤i≤n, and
∑i∈I v i≤∑i=n−|I|+1 n v i′≤∑i=n−|I|+1 n∑j=1 i b j=∑j=1 n∑i=max{j,n−|I|+1}n b j=∑j=1 n min{n−j+1,|I|}b j.
Reversing the order of summation yields that v satisfies (b-parking). Thus, X n(b)⊆P(b).
We now prove the converse containment. Let w=(w 1,…,w n) be a vertex of P(b). It suffices to show that w is a b-parking function. Since P(b) and the set of b-parking functions are both invariant under the action of S n on R n, we may suppose that w 1≤⋯≤w n. We will show that w=v k for some k, making w evidently a b-parking function.
Let I be the family of all nonempty subsets I⊆[n] such that
∑i∈I w i=∑i=n−|I|+1 n S i,
(1)
and furthermore such that |I|≠n−1 if b 1=1. Thus, I indexes the set of inequalities of the form(b-parking) that the vertex w satisfies with equality. Let k be the number of coordinates of w that equal 1. Since P(b) is a polytope in R n, and w is a vertex of this polytope, w satisfies with equality at least n of the inequalities that define P(b). Hence, |I|≥n−k.
For each I⊆[n], let I+:={n−|I|+1,n−|I|+2,…,n}, so that I+ comprises the last |I| elements of [n]. (We will see below that in fact I=I+ for all I∈I.) Since
∑i=n−|I|+1 n S i=∑i∈I w i≤∑i=n−|I|+1 n w i≤∑i=n−|I|+1 n S i for all I∈I
(where the middle inequality holds because w is weakly increasing, and the last inequality holds because w∈P(b)), we have that
∑i∈I w i=∑i=n−|I|+1 n w i for all I∈I
(2)
and, in particular, by Equation (1), that I+∈I for all I∈I. Moreover, note that Equation (2) is an equation of two sums, each of which is a sum of the same number |I| of terms, and that the ℓ th term on the left-hand side of Equation (2) is bounded above by the ℓ th term on the right-hand side for 1≤ℓ≤|I|. It follows that the ℓ th term on the left-hand side equals the ℓ th term on the right-hand side for 1≤ℓ≤|I|. In particular, the first terms on each side are equal:
w min(I)=w n−|I|+1 for all I∈I.
(3)
Since w∈P(b), we have in particular that
∑i=n−|I|+2 n w i≤∑i=n−|I|+2 n S i for all I⊆[n],
(4)
while, from Equation (1) and (2),
∑i=n−|I|+1 n w i=∑i=n−|I|+1 n S i for all I∈I.
(5)
Subtracting Inequality (4) from Equation (5) yields that
w n−|I|+1≥S n−|I|+1 for all I∈I.
(6)
Moreover, since w∈P(b),
∑i=n−|I|n w i≤∑i=n−|I|n S i for all I⊆[n]such that|I|≤n−1.
(7)
Subtracting Equation (5) from Inequality (7) yields
w n−|I|≤S n−|I|for all I∈I such that|I|≤n−1.
(8)
Since S n−|I|<S n−|I|+1, it follows from Inequality (6) and (8) that
w n−|I|<w n−|I|+1 for all I∈I such that|I|≤n−1.
(9)
We now get that, in fact, I=I+ for all I∈I, since otherwise we would have an I∈I with |I|≤n−1 and min(I)<min(I+)=n−|I|+1, and so, from Equation (3) and w min(I)≤w n−|I|≤w n−|I|+1, we would get that w n−|I|=w n−|I|+1, contradicting Inequality (9). Thus, every element of I is of the form I=I j:={j,j+1,…,n} for some 1≤j≤n.
We will now show that w=v k=(1,…,1,S k+1,S k+2,…,S n). However, we must separately consider two cases. For the first case, suppose that w j>1 for all j such that I j∈I. Since w 1=⋯=w k=1, it follows that j≥k+1 for all j such that I j∈I. In other words, every I∈I is of the form I j with k+1≤j≤n. But there are only n−k such subsets of[n], and |I|≥n−k, so we must have that I is precisely the set {I k+1,I k+2,…,I n}. Subtracting the (b-parking)equation for I j+1 from the equation for I j, we get that w j=S j for k+1≤j≤n−1, and the I n equation itself states that w n=S n. Thus, w=v k∈X n(b), and moreover w satisfies precisely k+|I|=n of the inequalities in Theorem2.3(c) with equality.
This leaves the case in which w j=1 for some j such that I j∈I. From Inequality (6), it follows that S j=1. Since 1≤S 1<⋯<S n, we get that j=1 and hence b 1=S 1=1=w 1 and [n]=I 1∈I. (Recall that we are also now assuming that n≥2 because we have excluded the degenerate case n=1=b 1 in the remarks following Proposition 2.2.) Furthermore, k=1 because, since [n]=I 1∈I, we may subtract the (b-parking) inequality for I k from the (b-parking) equation for[n] to get that
k=∑i=1 k w i≥∑i=1 k S i≥∑i=1 k i,
which, together with k≥1 (because w 1=1), implies that k=1. Thus, |I|≥n−1. Now, since b 1=1, the set I does not contain the subset I 2 because |I 2|=n−1, so I must contain all n−2 subsets I 3,…,I n.Footnote 1 Hence, I={[n],I 3,…,I n}. Subtracting the (b-parking)equation for I 3 from the (b-parking)equation for[n] and applying w 1=1=b 1 yields w 2=S 2. For 3≤j≤n−1, subtract the (b-parking)equation for I j+1 from the equation for I j to get that w j=S j. Finally, the I n equation itself states that w n=S n. Thus, we have again found that w=v k∈X n(b) and that w satisfies precisely k+|I|=1+1+(n−2)=n of the inequalities in Theorem 2.3(c) with equality.
Since, in both cases, w∈X n(b), we now have that P(b)=X n(b). We have also found that every vertex of P(b) equals π(v k) for some π∈S n and 0≤k≤n, so we have now proved that every vertex of X n(b) has this same form. In addition, we have found that the number of inequalities in Theorem 2.3(c) that the vertex w satisfies with equality is precisely n, so X n(b) is simple (proving Theorem 2.3(a)) and moreover no inequality given in Theorem 2.3(c) that supports X n(b) is redundant. In particular, since every point of the form v k with 0≤k≤n satisfies precisely n of the inequalities in Theorem 2.3(c) with equality, it follows that every such point is a vertex of X n(b). This completes the proof of Theorem 2.3(b).
The final claim is that this is a minimal system of inequalities. Since X n(b) is simple, it remains only to show that every inequality in this system supports X n(b). Of course, the inequalities of the form x i≥1 are all satisfied with equality by the point (1,…,1)∈X n(b). For the (b-parking)inequalities, let I⊆[n] be given, and let k:=n−|I|. Then v k satisfies the (b-parking)inequality for I+ with equality. Let π∈S n be any permutation under which the image of I+ is I. Then π(v k) is a point in X n(b) that satisfies the (b-parking)inequality for I itself with equality. ◻
As a corollary of Theorem 2.3(b), we enumerate the vertices of X n(b). When b 1=1 (and so v 0=v 1), the number of vertices is the same as in the case of the classical parking-function polytope PF n. However, when b 1≥2, an additional n! vertices come into play.
Corollary 2.4
If b 1=1, then the number of vertices of X n(b) is n!(1 1!+⋯+1 n!). Otherwise, if b 1≥2, then the number of vertices of X n(b) is n!(1 0!+1 1!+⋯+1 n!).
Proof
When b 1≥2 and k≥0, or when b 1=1 and k≥2, each vertex v of X n(b) of the form π(v k) (π∈S n) corresponds to the unique pair (I,σ) of the k-subset I⊆[n] of entries of v that equal 1 and the bijection σ:[n−k]→[n]∖I such that σ(j) is the position of S j in v for k+1≤j≤n. On the other hand, when b 1=1 and k∈{0,1}, so that v 0=v 1, each vertex of the form π(v k) corresponds to the unique permutation in S n (namely, π) such that π(j) is the position of S j in v for 1≤j≤n. Thus, X n(b) has ∑k=0 n(n k)(n−k)! vertices if b 1≥2, or n!+∑k=2 n(n k)(n−k)! vertices if b 1=1. ◻
As an immediate corollary of Theorem 2.3(c), we enumerate the facets of X n(b).
Corollary 2.5
If b 1=1, then the number of facets of X n(b) is 2 n−1. Otherwise, if b 1≥2, then the number of facets of X n(b) is 2 n+n−1.
See Corollary 2.17 below for a formula for the number of k-dimensional faces of X n(b) for all k=0,1,…,n.
From the characterization of the vertices and facets of X n(b) in Theorem 2.3, it is straightforward to describe the facets that contain a given nondecreasing vertex v k. To this end, define the following notation:
L j:={(x 1,…,x n)∈X n(b):x j=1}for 1≤j≤n,U j:={(x 1,…,x n)∈X n(b):∑i=j n x i=∑i=j n S i}for 1≤j≤n,with j≠2 i f b 1=1.
Thus, L j is the facet of X n(b) defined by the inequality x j≥1, and U j is the facet of X n(b) defined by the (b-parking)inequality for the subset I=I j:={j,j+1,…,n} (where j≠2 if b 1=1). In particular, it is routine to check that the facets that contain the vertex v k are as follows.
Lemma 2.6
Fix k∈Z with 0≤k≤n.
(a)
If b 1≥2, or if b 1=1 and k≥2, then the n distinct facets that contain v k are
L 1,L 2,…,L k,U k+1,U k+2,…,U n.
2. (b)
If b 1=1 and k∈{0,1} (i.e., v k=v 0=v 1), then the n distinct facets that contain v k are
L 1,U 1,U 3,U 4,…,U n.
Proof
Since X n(b) is simple and n-dimensional, it suffices to confirm that the n given facets contain v k. ◻
We will now use Lemma 2.6 to describe the precise conditions under which two vertices of X n(b) share an edge. An informal summary of the conditions is as follows: Two vertices v and w share an edge if and only if either
(1)
the vertices are the same, except that the minimum entry >1 in one vertex is a 1 in the other vertex; or
(2)
the vertices are the same, except that two entries of the form S j−1 and S j swap positions.
In addition, we find that the edge vectors are all parallel to roots in the root system B n, with lengths easily expressed in terms of the entries of the vector b defining X n(b) or by their partial sums.
Proposition 2.7
Let v and w be vertices of X n(b). Fix π∈S n and k∈Z with 0≤k≤n such that v=π(v k).
(a)
If b 1≥2, or if b 1=1 and k≥2, then v and w share an edge of X n(b) if and only if one of the following conditions holds:
(1)
w=(π∘(j,k))(v k−1) for some 1≤j≤k, in which case
w−v=(S k−1)e π(j)
(informally, the vertices are the same, except that one of the 1 entries in v is S k in w); or
2. (2)
w=(π∘(j−1,j))(v k) for some k+2≤j≤n, in which case
w−v=b j(e π(j−1)−e π(j))
(informally, the vertices are the same, except that the S j−1 and S j entries swap positions);Footnote 2 or
3. (3)
k≤n−1 and w=π(v k+1), in which case
w−v=−(S k+1−1)e π(k+1)
(informally, the vertices are the same, except that the S k+1 entry in v is a 1 in w).Footnote 3
(b)
If b 1=1 and k∈{0,1} (i.e., v k=v 0=v 1), then v and w share an edge of X n(b) if and only if one of the following conditions holds:
(1)
w=(π∘(j−1,j))(v 1) for some 2≤j≤n, in which case
w−v=b j(e π(j−1)−e π(j))
(informally, the vertices are the same, except that the S j−1 and S j entries swap positions); or
2. (2)
w=π(v 2), in which case
w−v=−b 2 e π(2)
(informally, the vertices are the same, except that the S 2 entry in v is a 1 in w).
Proof
For both parts of the proposition to be proved, note that we will be done once we have proved the “if” direction, since we will then have found n vertices that are adjacent to v, which must be all of the vertices that are adjacent to v, because X n(b) is simple and of dimension n by Theorem 2.3(a) and Proposition 2.2. Thus, the “only if” direction will follow immediately. Note also that the calculation of the difference vector w−v under each condition is straightforward.
To prove the “if” direction of Proposition 2.7(a), suppose that b 1≥2, or b 1=1 and k≥2, as in Lemma 2.6(a). Thus, the facets that contain v k are
L 1,L 2,…,L k,U k+1,U k+2,…,U n.
(10)
By the S n invariance of X n(b), the vertices v and w share an edge if and only if π−1(w) lies on all but one of these facets. Now, if w=(π∘(j,k))(v k−1) for some 1≤j≤k, then π−1(w)=(j,k)(v k−1) lies on all facets in(10) except L j. If w=(π∘(j−1,j))(v k) for some k+2≤j≤n, then π−1(w)=(j−1,j)(v k) lies on all facets in(10) except U j. Finally, if k≤n−1 and w=π(v k+1), then π−1(w)=v k+1 lies on all facets in(10) except U k+1.
To prove the “if” direction of Proposition 2.7(b), suppose that b 1=1 and k∈{0,1}, as in Lemma 2.6(b). Thus, the facets that contain v k=v 0=v 1 are
L 1,U 1,U 3,U 4,…,U n.
(11)
If w=(π∘(1,2))(v 0), then π−1(w)=(1,2)(v 0) lies on all facets in(11) except L 1. If w=(π∘(j−1,j))(v 0) for some 3≤j≤n, then π−1(w)=(j−1,j)(v 0) lies on all facets in(11) except U j. Finally, if w=π(v 2), then π−1(w)=v 2 lies on all facets in(11) except U 1. ◻
Recall that, given a vertex v of a polytope P, the tangent cone T v(P) at v is the cone generated by the edge vectors pointing from v to its neighbors in the edge graph of P. That is,
T v(P):={∑w∈V e r t(P):w∼v λ w(w−v):λ w≥0 for all vertices w∼v},
where we write w∼v when w and v are adjacent vertices of P. As a corollary of Proposition 2.7, we get an explicit description of the tangent cone at each vertex of X n(b).
Corollary 2.8
Let v be a vertex of X n(b). Fix π∈S n and k∈Z with 0≤k≤n such that v=π(v k). Let T v:=T v(X n(b)) be the tangent cone at v.
(a)
Suppose that b 1≥2, or that b 1=1 and k≥2. Then T v is generated by
{e π(j):1≤j≤k}∪{−e π(k+1)}∪{e π(j−1)−e π(j):k+2≤j≤n}
if k≤n−1, and by {e j:1≤j≤n} if k=n.
(b)
Suppose that b 1=1 and k∈{0,1}. Then T v is generated by
{e π(j−1)−e π(j):2≤j≤n}∪{−e π(2)}.
2.2 X n(b) is Equivalent to a Generalized Permutahedron
Generalized permutahedra (sometimes spelled permutohedra) form a widely studied class of polytopes with a rich theory and many interesting subclasses. One definition of a generalized permutahedron is that it is a polyhedron whose normal fan is a coarsening of the braid arrangement. For sources on generalized permutahedra, we recommend [21, 28, 34, 35]. We use an equivalent characterization. (See [28, Theorem 2.3].) Here and below, we write e 1,…,e n for the standard basis vectors in R n, and e^1,…,e^n,e^n+1 for the standard basis vectors in R n+1.
Definition 2.9
A polytope in R n is a generalized permutahedron if and only if every edge is parallel to e i−e j for some i,j∈[n] with i≠j.
We show that the lifting of X n(b) is a generalized permutahedron, from which it, and X n(b) itself, immediately inherit a variety of combinatorial and geometric results and properties. Let B:=∑k=1 n S k=∑k=1 n(n−k+1)b k, and let ℓ:R n→R n+1 be the affine map that “vertically projects” R n onto the hyperplane H={x∈R n+1:x 1+⋯+x n+x n+1=B} via (x 1,…,x n)↦(x 1,…,x n,B−∑i=1 n x i). Using this mapping, Proposition 2.11 below follows.
Definition 2.10
Let X¯n(b):={ℓ(x):x∈X n(b)} be the lifting of X n(b) to the hyperplane H. We call X¯n(b) the lifted b -parking-function polytope.
Proposition 2.11
The vertex set of X¯n(b) is {ℓ(v):v is a vertex of X n(b)}. Furthermore, the minimal inequality description of X¯n(b) is the minimal inequality description of X n(b) together with the additional equality x 1+⋯+x n+x n+1=B.
Building off this combinatorial structure, we have the main result of this subsection.
Theorem 2.12
The lifted b-parking-function polytope X¯n(b) is a generalized permutahedron.
Proof
Proposition2.11 implies that the edges of X n(b) lift to edges of X¯n(b) in the natural way: [v,w] is an edge of X n(b) if and only if [ℓ(v),ℓ(w)] is an edge of X¯n(b). If [v,w] is an edge of X n(b) with v−w=c(e i−e j), then ∑i=1 n v i=∑i=1 n w i. Thus ℓ(v)n+1=B−∑i=1 n v i=B−∑i=1 n w i, so ℓ(v)−ℓ(w)=c(e^i−e^j). If [v,w] is an edge of X n(b) with v−w=c e j, then ∑i=1 n v i−∑i=1 n w i=c. Thus, ℓ(v)n+1=B−∑i=1 n v i=B−∑i=1 n w i−c=ℓ(v)n+1−c, so ℓ(v)−ℓ(w)=c(e^j−e^n+1). By Proposition2.7, all edges of X n(b) are of this form. Hence, X¯n(b) is a generalized permutahedron. ◻
We may now apply established results about generalized permutahedra to X¯n(b). One such result is [28, Corollary 4.8], which provides a formula for the number of lattice points of a generalized permutahedron. Since lifting a polytope does not change the number of lattice points of a polytope, by applying the formula we obtain the number of lattice points of X n(b). Furthermore, we note that the enumeration of b-parking functions is not known in the form of a completely closed formula [45, 46], but by applying the formula of [28, Corollary 4.8], we obtain an upper-bound for the number of b-parking functions since all b-parking functions will be lattice points in X n(b).
Additionally, by a theorem of Backman and Liu ,Theorem 2.12 implies that X¯n(b), and hence X n(b) itself, admits a regular unimodular triangulation. For a definition of and further details about regular unimodular triangulations, see . We can also express X¯n(b) as a Minkowski sum of simplices. In order to do that, we begin by expressing X¯n(b) in the notation introduced by Postnikov for generalized permutahedra. Following from [35, Section 6], every generalized permutahedron can be expressed as
for some collection of real parameters {z I} over subsets I of [n] with z∅=0.
Proposition 2.13
The lifted b-parking-function polytope X¯n(b) is the generalized permutahedron P n+1 Z({z I}) in which the values of the parameters z I for ∅≠I⊆[n+1] are given by
Proof
From Proposition 2.11, consider the defining inequalities,
∑i∈I x i≤S n+S n−1+⋯+S n−|I|+1,
for any I with |I|=k, n+1∉I, and x 1+…+x n+1=∑i=1 n S i. Subtracting these two yields the inequality
∑i∈[n+1]∖I x i≥S 1+S 2+⋯+S|[n+1]∖I|−1.
This shows that X¯n(b) can be described in terms of the following inequalities:
{x i≥1,for 1≤i≤n,∑i∈I x i≥∑i=1|I|−1 S i,for all I⊂[n+1]with n+1∈I,∑i=1 n+1 x i=∑i=1 n S i,
since they recover the defining inequalities in Theorem 2.3. To match with Postnikov’s definition of P n+1 Z({z I}), we can add the following redundant inequalities obtained from x i≥1 which have no effect on the solutions of the above system of inequalities:
∑i∈I x i≥|I|for all I⊂[n+1]such that n+1∉I.
These inequalities give all the values of z I for all nonempty I⊂[n+1]. ◻
In , Ardila, Benedetti, and Doker prove that any generalized permutahedron can be written as a signed Minkowski sum of scaled standard simplices. Specifically,
P n Z({z I})=P n Y({y I}):=∑I⊆[n]y I△I,
where y I=∑J⊆I(−1)|I|−|J|z J and △I:=c o n v{e i:i∈I}, for each I⊆[n]. In the case that y I≥0 for all I, we call this a Y -generalized permutahedron, which enjoys a variety of special properties including, if it is an integral polytope, Ehrhart positivity and h∗-real-rootedness ; for more on these properties, see [17, 18, 32].
Proposition 2.14
The lifted b-parking-function polytope X¯n(b) is the signed Minkowski sum ∑I⊆[n+1]y I△I, where
and △I:=c o n v{e^i:i∈I} for all I⊆[n+1].
Proof
We write X¯n(b)=P n+1 Z({z I}), with z I as given in Proposition 2.13. By [21, Prop.2.2.4], this becomes P n+1 Z({z I})=∑I⊆[n+1]y I△I, where the y I s are determined by z I=∑J⊆I y J for all nonempty I⊆[n+1].
First note that
For I⊆[n+1] not containing n+1 with |I|≥2, we apply the principle of inclusion–exclusion to get
y I=∑J⊆I(−1)|I|−|J|z J=∑k=0|I|(−1)|I|−k(|I|k)k=0.
For I⊆[n+1] with n+1∈I, |I|≥2, then again by the principle of inclusion–exclusion,
y I=∑J⊆I(−1)|I|−|J|z J=∑J⊆I∖{n+1}(−1)|I|−|J|z J+∑J⊂I n+1∈J(−1)|I|−|J|z J=−y I∖{n+1}+∑k=1|I|−1(−1)|I|−k−1(|I|−1 k)∑ℓ=1 k S ℓ.
(12)
When |I|=2, y I=−1+b 1. For |I|≥3, the second term on the right-hand side of Equation (12) can be rewritten as
∑ℓ=1|I|−1 S ℓ(∑k=ℓ|I|−1(−1)|I|−1−k(|I|−1 k))=∑ℓ=1|I|−1 S ℓ(∑k=0|I|−1−ℓ(−1)k(|I|−1 k))=∑ℓ=1|I|−1(∑i=1 ℓ b i)(−1)|I|−1−ℓ(|I|−2|I|−1−ℓ)=∑i=1|I|−1 b i∑ℓ=i|I|−1(−1)|I|−1−ℓ(|I|−2|I|−1−ℓ)=∑i=1|I|−1 b i∑ℓ=0|I|−1−i(−1)ℓ(|I|−2 ℓ)=∑i=2|I|−1(−1)|I|−1−i(|I|−3|I|−1−i)b i.
Hence, for |I|≥3,
y I=∑i=2|I|−1(−1)|I|−1−i(|I|−3|I|−1−i)b i=(−1)|I|−1∑i=0|I|−3(−1)i(|I|−3 i)b i+2.
◻
Proposition 2.14 allows us to determine from b=(b 1,…,b n) whether X¯n(b) is a Y-generalized permutahedron. In particular, the liftings of PF n and of the specific b-parking-function polytope X n(a,b) studied in (where (a,b):=b=(a,b,b,…,b)) are both Y-generalized permutahedra.
2.3 h-Polynomial of X n(b)
We now consider the h-polynomial of the b-parking-function polytope. Given an n-dimensional polytope P and 0≤i≤n, let f i(P) denote the number of i-dimensional faces of P. The f-polynomial of P is then defined as f(P;t):=∑i=0 n f i(P)t i and the h-polynomial of P is the result of applying a change of basis given by h(P;t):=f(P;t−1).
The stellohedron St n is a simple generalized permutahedron first defined by Postnikov, Reiner, and Williams as the nestohedron for a certain building set (see Section 3 for more details), which is intimately connected to X n(b). The h-polynomial of St n is the binomial Eulerian polynomial
A~n(z)=1+z∑k=1 n(n k)A k(z),
where A k(z)=∑σ∈S k z des(σ) is the Eulerian polynomial .
From Theorem 2.3(a) and Theorem 2.12, we have that X¯n(b) is a simple generalized permutahedron. By [34, Theorem 4.2], the h-polynomial of a simple generalized permutahedron can be computed using its so-called vertex posets, which are defined for every vertex of the polytope as follows.
For a generalized permutahedron P of dimension n in R n+1, the normal cone of every vertex v of P is the intersection of the half spaces in R n+1/(1,…,1)R defined by a set of inequalities of the form x i≤x j, where (i,j)∈[n+1]2. One can associate to each vertex v a vertex poset Q v by setting i≤Q v j whenever x i≤x j in the normal cone of v. We denote covering relations by ⋖. The descent set Des(Q v) of Q v is the set of ordered pairs (i,j) such that i⋖Q v j but i>Z j. The number of descents of Q v is des(Q v):=|Des(Q v)|. When P is a simple generalized permutahedron, by [34, Theorem 4.2], its h-polynomial is given by
h(P;z)=∑v∈V e r t(P)z des(Q v).
Since the face numbers of X n(b) do not change after the lifting ℓ, we may use this equation to compute the h-polynomial of X n(b). We characterize the vertex poset for every vertex of X¯n(b).
Fig. 2
The vertex poset Q v¯ in Lemma 2.15(a)
Full size image
Lemma 2.15
Let v¯ be a vertex of X¯n(b). Fix π∈S n and k∈Z with 0≤k≤n such that v¯ is the image under the lifting ℓ of the vertex π(v k) of X n(b).
(a)
If b 1≥2, or if b 1=1 and k≥2, then the vertex poset Q v¯ has the covering relations
π(j)⋖n+1 for 1≤j≤k,n+1⋖π(k+1),π(j−1)⋖π(j)for k+2≤j≤n.
(See the Hasse diagrams in Figure 2.)
(b)
If b 1=1 and k∈{0,1} (in which case v k=v 0=v 1), then the vertex poset Q v¯ has the covering relations
π(1)⋖π(2)⋖⋯⋖π(n),n+1⋖π(2).
(See the Hasse diagram in Figure 3.)
Fig. 3
The vertex poset Q v¯ in Lemma 2.15(b)
Full size image
Proof
Let v:=π(v k), so that v¯=ℓ(v). The normal cone at v¯ is the polar of the tangent cone at v¯. From the formula for the affine projection ℓ, we have that ℓ(w)−ℓ(v)=L(w−v) for all w∈R n, where L:R n→R n+1 is the linear map (x 1,…,x n)↦(x 1,…,x n,−∑i=1 n x i). To compute the tangent cone of X¯n(b) at v¯, we apply L to the generators of T v(X n(b)) given in Corollary 2.8. Now,
L(e j)=e^j−e^n+1
for 1≤j≤n. Thus, the tangent cone T v¯ of X¯n(b) at v¯ is generated as follows:
(a)
If b 1≥2, or if b 1=1 and k≥2, then T v¯ is generated by
{e^π(j)−e^n+1:1≤j≤k}∪{−e^π(k+1)+e^n+1}∪{e^π(j−1)−e^π(j):k+2≤j≤n}
if k≤n−1, and by {e^j−e^n+1:1≤j≤n} if k=n.
(b)
If b 1=1 and k∈{0,1}, then T v¯ is generated by
{e^π(j−1)−e^π(j):2≤j≤n}∪{−e^π(2)+e^n+1}.
These generating vectors of the tangent cone are the facet-defining outer normals of the normal cone. Thus, for example, in the case where b 1≥2 and k≤n−1, the normal cone at v¯ is defined by the inequalities
x π(j)≤x n+1 for 1≤j≤k,x n+1≤x π(k+1),x π(j−1)≤x π(j)for k+2≤j≤n;
and the other cases are handled similarly. These inequalities determine the covering relations given in the statement of Lemma 2.15. ◻
Theorem 2.16
The h-polynomial of X n(b) is
When b 1=1 one can write h(X n(b);z)=1+z∑k=1,k≠n−1 n(n k)A k(z).
Proof
Consider first the case in which b 1≥2. From Figures 2 (corresponding to Lemma 2.15(a)), we see that Q π(v k)¯ contains 1+des(π(k+1),…,π(n)) descents if 0≤k≤n−1, and 0 descents if k=n. Thus, enumerating vertices as in the proof of Corollary 2.4,
∑v∈V e r t(X n(b))z des(Q v¯)=∑k=0 n−1∑I⊆[n]:|I|=n−k∑σ∈S n−k z 1+des(σ)+1=1+z∑k=1 n(n k)∑σ∈S k z des(σ)=A~n(z).
Now consider the case in which b 1=1. From Figure 2 and 3, we see that Q π(v k)¯ contains 1+des(π) descents if k=1, 1+des(π(k+1),…,π(n)) descents if 2≤k≤n−1, and 0 descents if k=n. Again enumerating vertices as in the proof of Corollary 2.4,
∑v∈V e r t(X n(b))z des(Q v¯)=∑π∈S n z 1+des(π)+∑k=2 n−1∑I⊆[n]:|I|=n−k∑σ∈S n−k z 1+des(σ)+1=1+z∑k=1 n−2(n k)∑σ∈S k z des(σ)+z∑π∈S n z des(π)=A~n(z)−n z A n−1(z).
◻
Since the h-polynomial of X n(b) depends only on whether b 1=1 or b 1≥2, comparing with results on faces numbers of X n(a,b,b,…,b) in [26, Prop. 3.14], we immediately have the following result giving all face numbers of X n(b) for all b.
Corollary 2.17
Let f k be the number of k-dimensional faces of X n(b) for k=0,1,…,n. Then,
where S(n,k) is the Stirling number of the second kind.
3 A Building Set Perspective
In this section, we completely characterize the face structure and identify the combinatorial types of all b-parking-function polytopes. In particular, we show that even though not every b-parking-function polytope is a nestohedron (see Definition 3.2), up to combinatorial equivalence, every b-parking-function polytope is either the classical parking-function polytope PF n (if b 1=1) or the stellohedron St n (if b 1≥2).
We consider the Y-generalized permutahedron associated with a certain class of collections B. We recall some background on building sets, nestohedra, and nested set complexes from [34, 35].
Definition 3.1
([35, Definition 7.1]) Let E be a finite set. A building set on E is a collection B of nonempty subsets of E that satisfies the conditions:
(B1)
If I,J∈B and I∩J≠∅, then I∪J∈B.
(B2)
The singleton {i} lies in B for all i∈E.
In this case, we write B max⊂B for the collection of maximal subsets in B under inclusion.
Definition 3.2
([34, Definition 6.3]) Let B be a building set in [n]. A nestohedron P B is a Y-generalized permutahedron with positive coefficients y I only for I∈B: P n Y({y I})=∑I∈B y I△I.
We will relate nestohedra to certain complexes formed from building sets.
Definition 3.3
([34, Definition 6.4]) Let B be a building set on a finite set E. A nested set of B is a subset N⊆B∖B max that satisfies the following conditions:
(N1)
If I,J∈N, then either I⊆J or I⊇J or I∩J=∅.
(N2)
If J 1,…,J k∈N are pairwise disjoint with k≥2, then J 1∪⋯∪J k∉B.
From this definition, it follows that a subset of a nested set is a nested set. Therefore, the collection of all nested sets of B forms an abstract simplicial complex with vertex set B, called the nested set complex△B of B. It turns out that the nestohedron P B is always a simple polytope, and the face lattice of P B does not depend on the values of the positive parameters y I, but only on B, and this face lattice is dual to the face lattice of the nested set complex △B; see [35, Theorem 7.4].
Example 3.4
The collection B={{1},{2},{3},{4},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}} is a building set on . Note that a nested set is a subset of B. To simplify the notation, in this example and in Figure 4 each set {i} is abbreviated as i, and each set {i,j,k} is abbreviated as ijk. The corresponding nested sets and their dimensions as faces in △B are as follows:
{dim=−1∅dim=0{1},{2},{3},{4},{124},{134},{234}dim=1{1,2},{1,3},{2,3},{1,4},{2,4},{3,4},{1,124},{1,134},{2,124},{2,234},{3,134},{3,234},{4,124},{4,134},{4,234}dim=2{1,2,3},{1,2,124},{1,3,134},{2,3,234},{1,4,124},{1,4,134},{2,4,124},{2,4,234},{3,4,134},{3,4,234}}.
This building set’s nestohedron is the simple polytope in Figure 4. Each of the vertices corresponds to a maximal nested set. In fact, each of the faces of the nestohedron corresponds to the nested set that is the intersection of the maximal nested sets that correspond to the vertices of the face.
Fig. 4
The nestohedron from Example 3.4
Full size image
There are two special families of vectors b for which X¯n(b) is a translation by e^n+1 of a nestohedron whose face structure can be determined without much more work.
Definition 3.5
([34, Section 10.4]) A stellohedron St n is a nestohedron for the building set
{{1},{2},…,{n}}∪{S∪{n+1}:S⊆[n]}.
Proposition 3.6
Let b∈Z>0 n. Let y I (I⊆[n+1]) be the associated coefficients given in Proposition 2.14. Assume that y I=∑j=0|I|−3(−1)|I|+j−1(|I|−3 j)b j+2>0 for all sets I for which n+1∈I and |I|≥3,
(a)
If b 1≥2, then the b-parking-function polytope X n(b) is combinatorially equivalent to the stellohedron St n.
(b)
If b 1=1, then X n(b) is combinatorially equivalent to the nestohedron for the building set
{{1},{2},…,{n}}∪{S∪{n+1}:S⊆[n],|S|≠1}.
Proof
From the assumption of Proposition 3.6, observe that X n(b) is combinatorially equivalent to the following Minkowski sum:
X n(b)≅X¯n(b)≅∑i=1 n△{i}+χ(b 1)∑i=1 n△{i,n+1}+∑S⊆[n],|S|≥2△S∪{n+1},
where χ:Z>0→{0,1} is the indicator function getting value 1 if b 1≥2 and getting value 0 if b 1=1. Note that if b 1≥2 the indices of the standard simplices in the above Minkowski sum range over the collection {{1},{2},…,{n}}∪{S∪{n+1}:∅≠S⊆[n]}. If we add {n+1} to this collection, the building set that results corresponds to the stellohedron St n. Adding △{n+1} to the Minkowski sum only translates the polytope by e^n+1 and does not change the face structure. Hence, we have
X n(b)≅∑i=1 n△{i}+∑S⊆[n],S≠∅△S∪{n+1}+△{n+1}≅St n.
If b 1=1, after adding △{n+1} to the Minkowski sum, we have
X n(b)≅∑i=1 n△{i}+∑S⊆[n],|S|≥2△S∪{n+1}+△{n+1}.
The indices of the standard simplices in the above Minkowski sum range over the collection
{{1},{2},…,{n}}∪{S∪{n+1}:S⊆[n],|S|≠1},
which is also a building set. Therefore, X n(b) is combinatorially equivalent to the nestohedron corresponding to this building set. ◻
We will see later in Corollary 3.14 that the two face structures in Proposition 3.6 are the only combinatorial types that can occur on X n(b) for any b∈Z>0 n, even if the corresponding y I do not satisfy the positivity conditions of Proposition 3.6. We introduce the notation
B St n:={{1},{2},…,{n}}∪{S∪{n+1}:S⊆[n]},B PF n:={{1},{2},…,{n}}∪{S∪{n+1}:S⊆[n],|S|≠1},
for the two building sets in Proposition 3.6. We also introduce some terminology from to characterize the nested sets. We say a subset I⊆[n] and a (nonempty) chain F=(A 1⊊⋯⊊A k) in the Boolean lattice (2[n],⊆) are compatible, denoted by I≤F, if I⊆min(F)=A 1. If F is the empty chain∅, then we consider every subset I to be compatible with ∅. The nested sets for B St n and B PF n are characterized in the next propositions.
Proposition 3.7
([317
"), Prop.5.3]) The nested sets for B St n have form {{i}}i∈I∪{A∪{n+1}}A∈F, where I⊆[n] and F⊂2[n]∖{[n]} is a chain such that I≤F is a compatible pair.
The building set and the collection of nested sets in Example 3.4 is the case of n=3 of the proposition below.
Proposition 3.8
The nested set for B PF n is of the form {{i}}i∈I∪{A}A∈F, where I⊆[n+1] and |I|≤2 if n+1∈I, and F⊂2[n+1]∖{[n+1]} is a chain such that |min(F)|≥3 and n+1∈min(F), and I≤F is a compatible pair.
Proof
Let N be a nested set for B PF n. Let I be the subset consisting of i∈[n+1] for any singleton {i}∈N and F be the collection of all subsets in N that are not a singleton. By the definition of nested set for B PF n, we must have n+1∈A and |A|≥3 for any A∈F. Pick any A 1,A 2∈F. Then we always have n+1∈A 1∩A 2≠∅. Hence one of A 1 or A 2 must be a subset of the other and therefore F is a chain. For any i∈I, we must have i∈A for all A∈F; otherwise, there exists A′∈F such that {i} and A′ are disjoint subsets in N with A′∪{i}∈B PF n, which violates the condition (N2) in the definition of nested sets (Definition 3.3). This implies I⊆min(F), so I≤F. If n+1∈I and |I|≥3, then the union of all singletons in I, which is I itself, is in B P F n, which again violates condition (N2). Thus, we have |I|≤2 if n+1∈I. Conversely, it is routine to check that such a collection {{i}}i∈I∪{A}A∈F satisfies conditions (N1) and (N2) of Definition 3.3.
◻
In order to understand the face lattice of X n(b) in terms of building sets, we now generalize the notation U i for certain facets of X n(b) introduced immediately before Lemma 2.6 by setting
U I:={(x 1,…,x n)∈X n(b):∑i∈I x i=∑i=n−|I|+1 n S i},
for all nonempty I⊆[n], except in the case |I|=n−1 if b 1=1. Thus, U I is the facet of X n(b) defined by the (b-parking)inequality for the subset I. In terms of this notation, the facets U j defined prior to Lemma 2.6 are given by U j=U I j, where I j:={j,j+1,…,n}. Comparing the supporting hyperplanes of the facets of X n(b) in Theorem 2.3(c) and the two building sets in Proposition 3.6, we immediately arrive at the following lemma.
Lemma 3.9
Let B:=B St n if b 1≥2, and let B:=B PF n if b 1=1. The facets of X n(b) are in one-to-one correspondence with the elements of B∖B m a x=B∖{[n+1]} via the following bijection:
L i⟼{i},U[n]∖S⟼S∪{n+1}
for all i∈[n], all proper subsets S⊊[n] if b 1≥2, and all proper subsets S⊊[n] with |S|≠1 if b 1=1.
For any B∈B∖{[n+1]}, we denote by F B the corresponding facet of X n(b) under the bijection in Lemma 3.9.
Lemma 3.10
Let B:=B PF n if b 1=1, and let B:=B St n if b 1≥2. Then the map
N⟼⋂B∈N F B
is a bijection between the maximal nested sets N (i.e., the facets of △B) and the vertices of X n(b).
Proof
Consider b 1=1. The maximal nested sets for B PF n are of the form {{i}}i∈I∪{A 1⊊A 2⊊…⊊A n−|I|} satisfying the condition of Proposition 3.8 and |A i+1|−|A i|=1 for all i. If n+1∉I, then the corresponding vertex can be obtained as a vector in R n by placing 1s in positions I, then placing S n in position [n+1]∖A n−|I|, placing S n−1 in position A n−|I|∖A n−|I|−1, and so on, and in the final step placing S|I|+1 in position A 2∖A 1. In this way, we fill out all n entries of the vector in R n, and the resulting vector is the permutation of v|I| which is a vertex by Theorem 2.3(b). If n+1∈I, then I={{i},{n+1}} for some i∈[n]. Then we obtain the vertex as a vector in R n by placing 1=b 1=S 1 in position i, placing S n in position [n+1]∖A n−2, placing S n−1 in position A n−2∖A n−3, etc., placing S 3 in position A 2∖A 1, and in the final step placing S 2 in position A 1∖{i,n+1}. In this case, the resulting vector is the permutation of v 1=v 0 which is also a vertex by Theorem 2.3(b). As a result, this gives a bijection between the maximal nested set for B PF n and the vertices of X n(b) for b 1=1 via the intersections of facets corresponding to the maximal nested sets. For the case of b 1≥2, the proof is similar. ◻
Remark 3.11
In both cases b 1=1 and b 1≥2, one can further consider the vertex posets Q v for every v in the set of vertices V e r t(X n(b)) (see Section 2.3) cooperating with the bijection in Lemma 3.10. It gives the following commutative diagram
For example, the following are the correspondences between the weakly increasing vertices of X n(b) for b 1=1 and their vertex posets and the maximal nested sets:
With the aid of Lemma 3.9 and Lemma 3.10, we can completely characterize the face structure of the b-parking-function polytope for any b as follows.
Theorem 3.12
Let B:=B PF n if b 1=1, and let B:=B St n if b 2≥1. Then the map
△B⟶{nonempty faces of X n(b)},where N⟼F N:=⋂B∈N F B
is an isomorphism between the dual of the face lattice of the nested set complex △B and the face lattice of X n(b).
Proof
We will give the proof in detail only for the case in which b 1=1. A similar approach works for the case in which b 1≥2.
Consider b=(b 1,…,b n) with b 1=1. We first prove that the map from the nested sets in △B PF n to the set of nonempty faces of X n(b) is well-defined. From Lemma 3.9, the singleton nested set {B}∈△B PF n corresponds to the facet F B of X n(b). Since any intersection of facets is a face of the polytope, we know F N is a face of X n(b) for any nested set N∈△B PF n. For any nested set N, the face F N is nonempty by Lemma 3.10 because F N contains those vertices that correspond to the maximal nested sets containing N. Hence, the map is well-defined.
Now we check the injectivity of the map. By Theorem 2.3(a), X n(b) is a simple polytope. In a simple polytope, every nonempty intersection of facets determines a unique face. Hence, the injectivity of the map follows.
We now show that the map is surjective. Let N′ be a subset of B PF n such that the face ⋂B∈N′F B is nonempty. Say N′={{i}}i∈I′∪{A}A∈F′ for some I′⊂[n+1] and F′ consisting of non-singleton subsets in B PF n, then
⋂B∈N′F B=(⋂i∈I′L i)∩(⋂A∈F′U[n]∖A).
We now show that N′ must be a nested set for B PF n (Recall Proposition 3.8).
By definition of B PF n, we have n+1∈A and |A|≥3 for any A∈F′ right away.
First, for any A∈F′ we must have I′⊂A; otherwise, there is j∈I′ but j∉A for some A∈F′. Then this face ⋂B∈N′F B is contained in the intersection of the facets F j∩U[n]∖A, which is an empty set since j∈[n]∖A. One can check the defining equations of the hyperplanes that there are no vertices of X n(b) in this intersection. This contradicts the assumption that the face is nonempty.
Second, we claim if n+1∈I′, then |I′|≤2. This is because the face ⋂B∈N′F B is contained in the facet U[n] whose supporting hyperplane is ∑i=1 n x i=S n+S n−1+…+S 1. Any vertex of X n(b) on U[n] is the permutation of (S 1,S 2,…,S n)=(1,S 2,…,S n) which has exactly one coordinate being 1. Thus, I′ does not contain more than one element that is not n+1.
In the final step, we show that F′ must be a chain. Let A 1,A 2∈F′ with |A 1|≤|A 2| then we must have A 1⊆A 2. If not, there exists j∈A 1, but j∉A 2. Notice that any vertex v=(v 1,…,v n) of X n(b) on the facet U[n]∖A 2 has coordinates satisfying
{v i:i∈[n]∖A 2}={S n,S n−1,…,S|A 2|+1};
while the vertex v on U[n]∖A 1 has coordinates satisfying
{v i:i∈[n]∖A 1}={S n,…,S|A 2|+1,…,S|A 1|+1}.
Then there is no vertex of X n(b) that lies on U[n]∖A 2 and U[n]∖A 1 simultaneously since all the vertices have at most one coordinate equal to any one of S n,S n−1,…,S|A 2|+1.
Therefore, by Proposition 3.8 the subset N′ is a nested set for B PF n.
Now the map is a bijection. By the definition of the map, it is clear that any two nested sets satisfy N 1⊆N 2 if and only if F N 1⊇F N 2. Therefore, the map is a poset isomorphism. ◻
Recall, as discussed immediately following Definition 3.3, that given a building set B⊂2[n], the face lattice of the nestohedron P B=∑I∈B y I△I (where y I>0) is completely determined by the building set B and is dual to the face lattice of the nested set complex △B. It is somewhat surprising that the face lattice of X n(b) is determined by the two building sets B P F n and B St n because from Proposition 2.14 we have seen that for most cases of b the polytope X¯n(b) is not a nestohedron (or a translation of one). It will be interesting to understand how the value of y I affects the face structure of ∑∅≠I⊆[n]y I△I. More concretely, can we characterize the parameters {y I} (with no positivity constraint on y I∈R) such that ∑∅≠I⊆[n]y I△I is combinatorially equivalent to P B for a given building set B?
From Theorem 3.12, we can explicitly tell which facets intersect at a given face and all the vertices on this face by observing the nested set corresponding to the face, as illustrated in the following example.
Example 3.13
For b=(1,b 2,…,b 6), consider the nested set N={{1},{3},{1,3,4,7},{1,2,3,4,7}}. Then the face F N is the intersection of the following facets of X 6(b):
L 1={x∈X 6(b):x 1=1},L 3={x∈X 6(b):x 3=1},
U{2,5,6}={x∈X 6(b):x 2+x 5+x 6=S 6+S 5+S 4},
U{5,6}={x∈X 6(b):x 5+x 6=S 6+S 5}.
The face F N is 2-dimensional and has 4 vertices:
(1,S 4,1,1,S 5,S 6),(1,S 4,1,S 3,S 5,S 6),(1,S 4,1,1,S 6,S 5),(1,S 4,1,S 3,S 6,S 5).
Corollary 3.14
For any b=(b 1,b 2,…,b n)∈Z>0 n, the b-parking-function polytope X n(b) is combinatorially equivalent to the classical parking-function polytope PF n if b 1=1 or is combinatorially equivalent to the stellohedron St n if b 1≥2.
Next, we provide a visualization of the two 3-dimensional cases of Corollary 3.14.
Example 3.15
The combinatorial type of X n(b) is as in Figure 4 when n=3 and b 1=1 and is as in Figure 5 when n=3 and b 1≥2. The coordinate of each vertex can be obtained by using the bijection described in the proof of Lemma 3.10.
Fig. 5
The combinatorial type of X n(b) when n=3 and b 1≥2
Full size image
Up to this point we have observed X n(b) as a generalized permutahedron and have characterized its combinatorial types through the use of building sets. In the next section, we exhibit X n(b) as a special polymatroid, which allows us to obtain further results.
4 A Polymatroid Perspective
Here we study the b-parking-function polytope through a polymatroid lens. Polymatroids are a class of polyhedra that were introduced by Edmonds in 1970 as polyhedral generalizations of matroids. We note that generalized permutahedra are equivalent (up to translation) to polymatroids. By exhibiting X n(b) as a polymatroid, we are able to draw conclusions about properties such as its combinatorial and circuit diameters. For background on combinatorial and circuit diameters, see [30, 48] and [11, 12, 15, 23], respectively. We begin by recalling a few helpful terms and definitions from matroid theory .
Definition 4.1
Let f:2[n]→R be a set function. The function f is said to be submodular if for every X,Y⊆[n] with X⊆Y and every z∈[n]∖Y, the following inequality holds: f(X∪{z})−f(X)≥f(Y∪{z})−f(Y). The polymatroid associated with such a function f is the polytope
P f:={x∈R n:x i≥0 for all i∈[n],∑i∈S x i≤f(S)for all S⊆[n]}.
A set function f:2[n]→R is said to be nondecreasing if, for every X,Y⊆[n] with X⊆Y, we have f(X)≤f(Y). Henceforth, let f b:2[n]→Z≥0 be the set function defined by setting f b(∅):=0 and
f b(I):=−|I|+∑j=1 n min{j,|I|}b n−j+1,for∅≠I⊆[n].
Proposition 4.2
The function f b is a nondecreasing submodular function.
Proof
By construction, f b is integer and nondecreasing. It remains to be shown that f b is submodular. Observe that f b(I)=−|I|+∑j=1 n min{j,|I|}b n−j+1 only depends on the cardinality |I| of I, so we write f b(|I|)=f b(I). It suffices to prove that I⊂J⊊[n]⇒f b(|I|+1)−f b(|I|)≥f b(|J|+1)−f b(|J|), or equivalently
i≤j<n⇒f b(i+1)−f b(i)≥f b(j+1)−f b(j).
The difference f b(i+1)−f b(i) is
f b(i+1)−f b(i)=−(i+1)−(−i)+∑k=1 n(min{i+1,k}−min{i,k})b n−k+1=−1+∑k=i+1 n b n−k+1.
As b=(b 1,…,b n)∈Z>0 n,
∑k=i+1 n b n−k+1≥∑k=j+1 n b n−k+1 for i≤j.
Thus, f b is submodular. ◻
We translate the inequality description of X n(b) (Theorem 2.3(c)) to a description of the form P f. Setting y i:=x i−1, one obtains
x i≥1⇔y i≥0 for all i∈[n],
and the remaining constraints in Theorem 2.3(c) take the form
∑i∈I x i≤∑j=1 n min{j,|I|}b n−j+1⇔∑i∈I y i≤−|I|+∑j=1 n min{j,|I|}b n−j+1.
Note that this transformation corresponds to the translation of X n(b) by −(1,…,1)∈R n. The resulting polymatroid is now described as
X n′(b):={y∈R n:y i≥0 for all i∈[n],∑i∈I y i≤f b(I)for all I⊆[n]},
which proves the following.
Theorem 4.3
The polytope X n′(b) is a polymatroid. In fact, X n′(b) is a special polymatroid for which upper bounds on the sum of entries in any set I depend only on the cardinality |I|.
The polytopes X n(b) and X n′(b) are combinatorially equivalent. Thus, we can leverage the theory of polymatroids to obtain results about X n(b). In particular, we note that known results about polymatroids allow for alternate directions to recover Theorem 2.3. Topkis studied extreme point adjacency in polymatroids [42, Corollary 5.4], and his work can be used to deduce Theorem 2.3(a). Shapley’s observation on the structure of the vertices of a polymatroid can be used to obtain Theorem 2.3(b). Edmonds showed that we can optimize linear functions over polymatroids in strongly polynomial time by certifying total dual integrality through the greedy algorithm. In turn, this yields another direction for proving the inequality description in Theorem 2.3(c).
Next, using Theorem 4.3 and Proposition 2.7, we can carry over known upper bounds on the combinatorial diameters of polymatroids to X n(b), of the form min{2 n,1 2 n(n−1)+1}. Recall that the combinatorial diameter of a polytope is the diameter of its edge graph, that is, the maximum length of a shortest edge-path between two vertices. We now provide a direct proof of these upper bounds, which allows us to improve them for the case b 1=1 and see tightness in all cases.
Theorem 4.4
If b 1>1, the combinatorial diameter of X n(b) is exactly min{2 n,1 2 n(n−1)+1}. If b 1=1, the combinatorial diameter is exactly min{2(n−1),1 2 n(n−1)}.
Proof
Note that 2 n≤1 2 n(n−1)+1 and 2(n−1)<1 2 n(n−1) if and only if n≥5. For n≤4, tightness of the bounds can be verified exhaustively. For example, for n=3, it is clear from Figure 1 that the combinatorial diameters of X 3(1,2,3) and X 3(2,3,4) are 3 and 4, respectively.
We provide an argument for the bounds 2 n and 2(n−1). Let w k=π(v k) be a vertex of X n(b) where k2.7, w k is adjacent to w k+1=π(v k+1), which corresponds to reducing the (k+1)th entry of v k to 1. Consider the path P k=(w k,w k+1,…,w n−1,1) on X n(b), where P k ends at v n=w n=1. Because P k begins at w k, the length of P k is at most _n_ when b 1>1. Concatenating paths when necessary, we have a path of length at most 2 n between any two vertices of X n(b) with b 1>1. A path of this length is realized when traveling between the vertices w 0=(b 1,S 2,…,S n) and π′(w 0)=(S n,S n−1,…,S 2,b 1). When b 1=1, w 0=w 1, so P 0 has length at most n−1. A path of length 2(n−1) is realized when travelling between vertices w 1=(1,S 2,…,S n) and π′(w 1)=(S n,S n−1,…,S 2,1).
It remains to show tightness. We exhibit that the paths constructed above are shortest paths between the given pairs of vertices. For b 1>1, the bound 2 n for a path between w 0 and π′(w 0) is achieved through a combination of n reductions and n increases of entries S i, and no swaps. A path that exclusively uses swaps requires at least 1 2 n(n−1) steps, and so is not better. Let S j be the largest value that is reduced during an edge walk. It can be seen that any walk where j=1 has to perform the same number of swaps as one exclusively using swaps; any walk where j=n−1 has to perform at least two further steps (in addition to n−1 reductions and n−1 increases). Let now 2≤j≤n−2. Then the number of swaps required to arrive at a vertex where the j largest values S n,…,S n−j+1 are in the target positions is bounded below by j(n−j). (For every 1≤i 1<i 2≤n, it takes i 2−i 1 swaps to move S i 2 to the position in which S i 1 lies; in a best case, the target positions for S n,…,S n−j+1 start these swaps with values S j,…,S 1.) Combined with the at least j reductions and j increases, the path has length at least 2 j+j(n−j)≥2 j+2(n−2)≥2 n. An analogous argument with n−1 in place of n shows tightness of 2(n−1) for b 1=1. ◻
Circuits and circuit diameters generalize the concepts of edges and combinatorial diameters. We refer the reader to 11, 12, 20, 23, [36")] for background. The set of _elementary vectors_ or _circuits_ C(P) of a polyhedron _P_ correspond to the inclusion-minimal dependence relations in the constraint matrices of a (minimal) representation. Note that the elementary vectors or circuits of a polymatroid are _not_ the circuits of the underlying matroid. The elementary vectors C(P) include all edge directions that appear in any polyhedron that can be described with the same constraints matrices as _P_ and any right-hand sides. Thus, C(X n(b)) contains a superset of the edge directions found in Proposition 2.7.
Lemma 4.5
The elementary vectors of X n(b) include ±e i and e i−e j for all i,j≤n with i≠j.
Proof
It suffices to exhibit that for any vector of type ±e i or e i−e j for all i,j≤n with i≠j, there exists an X n(b) in which that vector appears as an edge direction. In fact, they all appear in the same polymatroid: this is the case for a matroid polyhedron X n(b) for a uniform matroid with rank ≥2. A full characterization of these edges can be found in [42, Theorem 5.2]. ◻
The associated circuit walks begin and end at vertices, and take maximal steps along elementary vectors instead of edges, in particular through the interior of a polyhedron. The study of circuit diameters is motivated by the search for lower bounds on the combinatorial diameter and the efficiency of circuit augmentation schemes . The circuit diameter of a polytope is the maximum length of a shortest circuit walk between any two vertices. Of particular interest are polyhedral families where there is a significant gap between the combinatorial and circuit diameter [13, 29]. We will prove that b-parking-function polytopes are one of these families.
Unlike the more technical conditions for the adjacency of vertices found in Proposition 2.7, steps along the elementary vectors of Lemma 4.5 allow for more general changes to a b-parking function than classic matroid operations: any two entries can be swapped (a step in direction ±(e i−e j)), even for non-bases and even including an entry that is 1; this contrasts with the basis-exchange property of matroids where specific pairs of elements of two bases can be exchanged. Further, any entry can be decreased or increased (a step in direction ±e i), respectively. The steps along the elementary vectors have to be of maximal length while remaining feasible. Recall from the proof of Proposition 4.2 that X n(b) is a special polymatroid for which f b(I) only depends on the cardinality |I|. Thus, any maximal step in direction ±(e i−e j) indeed is a swap of the corresponding entries. A maximal step in direction −e i corresponds to a decrease of an entry to 1. Note that such a step starting at π(v k) arrives at a non-vertex when decreasing entry S i to 1 for i>k+1. A maximal step in direction e i corresponds to an increase of an entry to the largest S i that does not appear yet.
We are now ready to prove a bound on the circuit diameter of X n(b).
Theorem 4.6
If b 1=1, the circuit diameter of X n(b) is at most n. If b 1≥2, the circuit diameter is at most n−1. Moreover, these bounds are tight when restricted to the elementary vectors of Lemma 4.5.
Proof
Recall that each vertex of X n(b) is a permutation of a point of the form v k for 0≤k≤n. If b 1=1, then k≥1. Consider vertices π 1(v k 1) and π 2(v k 2) for two permutations π 1,π 2 and k 1≤k 2. Note that π 1(y k 1) and π 2(y k 2) have at least k 1 entries 1 and share the largest n−k 2 values of y k 2 (permuted to different entries). Circuit walks are not reversible, so we will explain how to walk from π 1(v k 1) to π 2(v k 2) and how to walk from π 2(v k 2) to π 1(v k 1) in at most the claimed number of steps.
For the walk from π 1(v k 1) to π 2(v k 2), order the entries in descending order by value in π 2(v k 2). In the first n−k 2 iterations, a swap of two entries can guarantee that the largest n−k 2 entries match. Then follow k 2−k 1 iterations in which entries are decreased to 1. The remaining k 1 entries are already identical to 1. We obtain a circuit walk of at most n−k 1 steps.
For the walk from π 2(v k 2) to π 1(v k 1), order the entries in descending order by value in π 1(v k 1). In the first n−k 2 iterations, a swap of two entries can guarantee that the largest n−k 2 entries match. Then follow k 2−k 1 iterations in which a 1-entry is increased to the next-largest entry in π 1(v k 1). The remaining k 1 entries are already identical to 1. Again, we obtain a circuit walk of at most n−k 1 steps.
It remains to show that the given bounds are tight if one is restricted to the elementary vectors in Lemma 4.5. To this end, observe that any elementary vector can only increase one entry. If b 1>1, the case where 0=k 1≤k 2=n is possible, and all n entries need to be increased in a circuit walk from π 1(y k 1) to π 2(y k 2). If b 1>1 then k 1≥1. In the case where 1=k 1≤k 2=n, n−1 entries need to be increased. This proves the claim. ◻
For all n≥2, the bounds of Theorem 4.6 are strictly lower than the bounds for the combinatorial diameter of Theorem 4.4. Note that we only worked with a subset of C(X n(b)). A complete characterization of the whole set C(X n(b)) remains open. Due to this restriction, we constructed an integral circuit walk , where each intermediate point is integral. We believe that our bounds are tight if restricted to integral walks and leave this as an open question.
5 Conclusion - Further Perspectives
Recall that the classical parking-function polytope PF n is the convex hull of all parking functions of length n in R n. As noted in [26, Remark 2.10] and [83
"), Remarks 4.8 and 4.9], there are other alternative perspectives on PF n. It can be viewed as:
a special case of the main object of study for this paper, namely the b-parking-function polytope when b=(1,1,1,…,1),
a particular partial permutahedron P(n,n−1) as defined in and further studied in 83
"), [9, Art. 64, 12")],
a specific polytope of win vectors as defined by Bartels et al. in for the case of the complete graph on n vertices (the work of Backman [3, Theorem 4.5 and Corollary 4.6] provides alternate routes for computing the volume and lattice-point enumerator of PF n), and
essentially as the stochastic sandpile model for the complete graph .
In addition to these interpretations, as direct corollaries to Theorem 2.12, Theorem 3.12, and Theorem 4.3, we can also add that (a lifted version of) PF n:
is a special generalized permutahedron,
can be associated with a building set, and
is a special type of polymatroid, respectively.
We conclude by adding an interpretation of the classical parking-function polytope as a projection of a relaxed Birkhoff or assignment polytope, and an interpretation of b-parking-function polytopes as projections of relaxed partition polytopes.
5.1 Relation of PF n to Birkhoff/Assignment Polytopes
Consider an assignment problem for n cars and n parking spots, represented with decision variables x i j∈{0,1} indicating car i being assigned to spot j. The linear relaxation of this problem yields the Birkhoff polytope, also called the assignment polytope or the polytope of doubly stochastic matrices:
∑j=1 n x i j=1 for all 1≤i≤n∑i=1 n x i j=1 for all 1≤j≤n x i j≥0 for all 1≤i≤n,1≤j≤n.
See [5, 10, 49, Example 0.12], and [19, Chapter 9], as well as references within, for the extensive literature on these classical polytopes. Due to the total unimodularity of the underlying constraint matrix and integrality of the right-hand sides, the vertices satisfy x i j∈{0,1} [5, 10, 44]. In fact, they are in one-to-one correspondence with all possible assignments of cars to parking spots. We will prove that a linear projection of a mild relaxation of this polytope gives PF n.
To this end, note that either of the two types of equality constraints in the description of the assignment polytope, combined with the nonnegativity constraints, implies an upper bound of x i j≤1. This allows an equivalent description of the form
∑j=1 n x i j=1 for all 1≤i≤n∑i=1 n x i j≥1 for all 1≤j≤n x i j≥0 for all 1≤i≤n,1≤j≤n.
Informally, each car is assigned to precisely one spot, and each parking spot requires at least one car assigned to it, which is only possible if one retains a one-to-one assignment.
We now perform a relaxation where we allow for the assignment of multiple cars to the same (preferred) parking spot, following the design of PF n: the i th car to arrive picks from the first i spots. To represent the corresponding conditions α i′≤i in a nondecreasing rearrangement of the entries, one can replace the set of constraints ∑i=1 n x i j≥1 by sums of the first k constraints for 1≤k≤n:
∑j=1 n x i j=1 for all 1≤i≤n∑j=1 k∑i=1 n x i j≥k for all 1≤k≤n x i j≥0 for all 1≤i≤n,1≤j≤n.
As ∑i=1 n x i j≥1 for all 1≤j≤n implies ∑j=1 k∑i=1 n x i j≥k for all 1≤k≤n, this feasible set, which we call BF n, is a relaxation of the assignment polytope. Note that the associated constraint matrix lost total unimodularity. As part of our arguments below, we will prove that BF n is integral, i.e., it has integral vertices.
We consider the linear projection ψ n:R n 2→R n, (x i j)1≤i≤n,1≤j≤n↦(∑j=1 n j x i j)1≤1≤n, i.e., where the i th entry is ∑j=1 n j x i j. Note that for integral (x i j) in BF n or PF n, x i j=1 holds for exactly one j, the parking spot j for car i. We show that the projection of BF n results in exactly PF n, i.e., ψ n(BF n)=PF n.
Theorem 5.1
The polytope PF n is the linear projection of the relaxed Birkhoff/assignment polytope BF n, under ψ n.
Proof
Recall that PF n is an integral polytope in R n. Thus, for a∈PF n, which may have fractional components, there exists a convex combination a=∑ℓ=1 n λ ℓ a ℓ of integral parking functions a ℓ. Let x ℓ=(x i j ℓ)∈{0,1}n 2 be defined by x i j ℓ=1 if and only if a i ℓ=j. Then x=∑ℓ=1 n λ ℓ x ℓ lies in BF n and projects onto a under ψ n.
Conversely, let now x∈BF n. We have to prove that its projection ψ n(x)=a lies in PF n. To this end, we first prove that BF n is integral. Let x be fractional and consider a bipartite graph G with V(G)=C∪P, |C|=n=|P|, and edges (i,j) from C to P if and only if 0<x i j<1; informally, it is a representation of only the fractional assignments of parking spots to cars. If there exists a cycle in the undirected graph underlying G, one can send flow along either orientation of that cycle and obtain a new feasible solution. In this case, x is not a vertex of BF n.
If there does not exist a cycle in G, then there exist at least two nodes with degree 1 in each connected component, as any tree has at least two leaves. By ∑j=1 n x i j=1 for all cars i, the degree of i∈C in any component is at least two. Thus, there are at least two parking spots j′,j″∈P with degree 1, as well as a path between j′ and j″. Let j′<j″ and note that the constraints ∑j=1 j′∑i=1 n x i j≥j′ and ∑j=1 j″∑i=1 n x i j≥j″ must be satisfied strictly. By sending (maximal) flow along the oriented path from j′ to j″, one obtains a new feasible solution with a strictly larger set of active constraints. This again implies that x is not a vertex of BF n.
We conclude that BF n is an integral polytope with all vertices in {0,1}n 2. By the same arguments as above, x=∑ℓ=1 t λ ℓ x ℓ is a convex combination with x ℓ∈{0,1}n 2. Let a ℓ=ψ n(x ℓ) be the projection of x ℓ under ψ n, and observe it is an integral parking function. Thus, x projects to a convex combination a=∑ℓ=1 n λ ℓ a ℓ of integral parking functions a ℓ. This proves the claim. ◻
Remark 5.2
As an anonymous reviewer noted, it is shown in [27, Theorems 5.27 & 5.28] that similar projections of other relaxations of the Birkhoff polytope give partial permutahedra, of which certain cases are (up to translation) b-parking-function polytopes. As an example, consider [27, Theorem 5.27] with n≥m−1 and z=(1,2,...,n). Then the projection of the polytope of m×n doubly substochastic matrices P Perm(m,n) under ϕ z:R m×n→R m, such that
(x i j)1≤i≤m,1≤j≤n↦(∑j=1 n j x i j)1≤i≤m,
is the partial permutahedron P z(m,n), which coincides (up to translation) with X m(b) where b=(n−m+2,1,…,1).
5.2 Relation of X n(b) to Partition Polytopes
Most of the proof for Theorem 5.1 transfers to b-parking-function polytopes X n(b). By using x i j∈{0,1} to indicate car i being assigned a number 1≤j≤S n, one can formulate the set of b-parking functions in the form
∑j=1 S n x i j=1 for all 1≤i≤n∑j=1 S k∑i=1 n x i j≥k for all 1≤k≤n x i j≥0 for all 1≤i≤n,1≤j≤S n,
and the projection (x i j)1≤i≤n,1≤j≤S n→(∑j=1 S n j x i j)1≤i≤n of an associated linear relaxation gives X n(b) by the same arguments.
A relation of this system to well-known polytopes is more involved than for PF n. The system can be seen as a relaxation of so-called (bounded-size) partition polytopes [14, 15]. Partition polytopes are special transportation polytopes and generalized Birkhoff or assignment polytopes. They are named for representing the (possibly fractional) assignment of n items to ℓ clusters, where one specifies the total weight κ j assigned to each cluster j. A bounded-size partition polytope specifies lower and/or upper bounds on this weight instead of an exact weight. For our purposes, we set ℓ=S n and only use lower bounds. Formally,
∑j=1 S n x i j=1 for all 1≤i≤n∑i=1 n x i j≥κ j for all 1≤j≤S n x i j≥0 for all 1≤i≤n,1≤j≤S n.
The first type of constraint guarantees that each item is assigned to a cluster (or partially to multiple clusters). The second type of constraint guarantees that each cluster receives a certain minimum weight. Note that S n>n for any X n(b) that is not also a classical parking function PF n; informally, there are more clusters than items. For the existence of a feasible solution, it is necessary that at least some of the κ j satisfy κ j<1.
Let S 0=0 and t∈[n], and set κ j=1 S t−S t−1 for S t−1<j≤S t. One can represent the conditions β i′≤S i in a nondecreasing rearrangement of the entries by replacing the set of constraints ∑i=1 n x i j≥κ j by sums of the first S k constraints for 1≤k≤n. This gives the above relaxed system.
Corollary 5.3
The polytope X n(b) is a linear projection of a relaxed bounded-size partition polytope.
We conclude by posing the following question to encourage further exploration of generalized parking-function polytopes in view of possible relations to other well-known classes of polytopes.
Question 5.4
What is the relationship (if any) between b-parking-function polytopes and partial permutahedra, polytopes of win vectors, or stochastic sandpile models for graphs?
An anonymous reviewer points out the following partial answer to Question 5.4 regarding the relationship between b-parking-function polytopes and partial permutahedra. This reviewer notes that the following remark can be regarded as a generalization of [26, Proposition 3.16].
Remark 5.5
For nonnegative real numbers z 1≥⋯≥z m, let Π(z 1,…,z m) denote the standard permutahedron c o n v{(z σ(1),…,z σ(m)):σ∈S m}, and let Π^(z 1,…,z m) denote the associated antiblocking permutahedron
{(x 1,…,x m)∈R m:t h e r e e x i s t s(y 1,…,y m)∈Π(z 1,…,z m)w i t h 0≤x i≤y i f o r 1≤i≤m}.
Corollary 5.8 in [83
")] shows that the partial permutahedron P(m,n) is related to antiblocking permutahedra by
(13)
The vertices of Π^(z 1,…,z m) are characterized in [83
"), Proposition 5.7], and comparing these with those of X m(b) (as in Theorem 2.3(b)), we see X m(b) is related to antiblocking permutahedra by
X m′(b)=Π^(b 1+⋯+b m−1,…,b 1+b 2−1,b 1−1),
(14)
where X m′(b) is the translation by −(1,...,1) of X m(b). Since (b 1+⋯+b m−1,…,b 1+b 2−1,b 1−1) in (14) is a strictly decreasing sequence, it can not give the sequence (n,n−1,…,1,0,…,0) in the n≤m−2 case of (13) (since this contains m−n≥2 zeros). However, (b 1+⋯+b n−1,…,b 1+b 2−1,b 1−1) gives the sequence (n,n−1,…,n−m+1) in the n≥m−1 case of (13) if and only if b 1=n−m+2 and b 2=⋯=b m=1. Therefore, a partial permutahedron P(m,n) and a (translated) b-parking-function polytope X m′(b) coincide if and only if n≥m−1 and b=(n−m+2,1,…,1), in which case P(m,n)=X m′(b). Outside this overlap, there are partial permutahedra P(m,n) which are not (translated) b-parking-function polytopes (i.e., when n≤m−2) and (translated) b-parking-function polytopes X m′(b) which are not partial permutahedra (i.e., when b 2,…,b m are not all 1).
6 Acknowledgement & Funding
This work was initiated at the 2023 Graduate Research Workshop in Combinatorics, which was supported in part by the National Science Foundation (NSF) under Award 1953985, and a generous award from the Combinatorics Foundation. The authors thank the workshop organizers and the University of Wyoming for fostering an excellent research atmosphere. We also thank Thomas Magnuson for conversations at the start of this project and Martha Yip for helpful directions. We also thank the anonymous reviewers for their comments which greatly influenced the exposition of our work. In particular, we express gratitude to the anonymous reviewer’s comments that led to Remarks 5.2 and 5.5.
Steffen Borgwardt was supported by the Air Force Office of Scientific Research, Complex Networks, under Award FA9550-21-1-0233.
Steffen Borgwardt and Angela Morrison were supported by the NSF, CCF, Algorithmic Foundations, under Award 2006183.
Danai Deligeorgaki is supported by the Wallenberg Autonomous Systems and Software Program; part of this research was performed while visiting the Institute for Mathematical and Statistical Innovation, which is supported by NSF Grant No.DMS-1929348.
Andrés R.Vindas-Meléndez was supported by the NSF under Award DMS-2102921.
Data Availability
The authors do not have any research data outside the submitted manuscript file.
Notes
The claim that there are exactly n−2 such subsets is the point at which this argument fails in the degenerate case n=1=b 1.
When b 1=1 and k∈{0,1} (so that v k=v 0=v 1), we must use k=0 for this equation to hold.
When b 1=1 and k∈{0,1} (so that v k=v 0=v 1), we must use k=1 for this equation to hold.
References
Amanbayeva, A., Wang, D.: The convex hull of parking functions of length n, Enumer. Comb. Appl. 2(2) (2022), S2R10, 10
Ardila, F., Benedetti, C., Doker, J.: Matroid polytopes and their volumes. Discrete Comput. Geom. 43(4), 841–854 (2010)
ArticleMathSciNetMATHGoogle Scholar
Backman, S.: Partial graph orientations and the Tutte polynomial. Adv. in Appl. Math. 94, 103–119 (2018)
ArticleMathSciNetMATHGoogle Scholar
Backman, S., Liu, G.: A regular unimodular triangulation of the matroid base polytope, Sém. Lothar. Combin. 91B (2024), Art. 92, 10
Balinski, M., Russakoff, A.: On the assignment polytope. SIAM Rev. 16(4), 516–525 (1974)
ArticleMathSciNetMATHGoogle Scholar
Bartels, J.E., Mount, J., Welsh, D.J.A.: The polytope of win vectors. Ann. Comb. 1(1), 1–15 (1997)
ArticleMathSciNetMATHGoogle Scholar
Behrend, R.E.: Ehrhart polynomials of partial permutohedra, (2024), arxiv:2403.06975
Behrend, RE., Castillo, F., Chavez, A.,Diaz-Lopez, A., Escobar, L., Harris, P., Insko, E.: Partial permutohedra, (2022), arXiv:2207.14253
Behrend, RE., Castillo, F.: Partial permutohedra, Sém. Lothar. Combin. 89B (2023), Art. 64, 12
Birkhoff, G.: Tres observaciones sobre el algebra lineal. Univ. Nac. Tucumán. Revista A. 5, 147–151 (1946)
MathSciNetGoogle Scholar
Borgwardt, S., De Loera, J.A., Finhold, E.: Edges versus circuits: a hierarchy of diameters in polyhedra. Adv. Geom. 16(4), 511–530 (2016)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Finhold, E., Hemmecke, R.: On the circuit diameter of dual transportation polyhedra. SIAM J. Discrete Math. 29(1), 113–121 (2015)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Finhold, E., Hemmecke, R.: Quadratic diameter bounds for dual network flow polyhedra. Math. Program. 159(1–2), 237–251 (2016)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Viss, C.: Constructing clustering transformations. SIAM J. Discrete Math. 35(1), 152–178 (2021)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Viss, C.: Circuit walks in integral polyhedra, Discrete Optim. 44 (2022), 100566
Braden, T., Huh, J., Matherne, JP.: Nicholas Proudfoot, and Botong Wang, A semi-small decomposition of the Chow ring of a matroid, Adv. Math. 409 (2022), 108646, 49
Brändén, P., Solus, L.: Symmetric decompositions and real-rootedness. Int. Math. Res. Not. IMRN 10, 7764–7798 (2021)
ArticleMathSciNetMATHGoogle Scholar
Braun, B.: Unimodality problems in Ehrhart theory, Recent trends in combinatorics, IMA Vol. Math. Appl., Springer, [Cham], (159) 687–711 (2016)
Richard, A.: Brualdi, Combinatorial matrix classes, Encyclopedia of Mathematics and its Applications, vol. 108. Cambridge University Press, Cambridge (2006)
Google Scholar
De Loera, J.A.: Hemmecke, R., Lee, J.: On augmentation algorithms for linear and integer-linear programming: from Edmonds-Karp to Bland and beyond. SIAM J. Optim. 25(4), 2494–2511 (2015)
Doker, JS.: Geometry of generalized permutohedra, University of California, Berkeley, (2011), PhD thesis
Edmonds, J.: Submodular functions, matroids, and certain polyhedra, Combinatorial optimization—Eureka, you shrink!, Lecture Notes in Comput. Sci.,, Springer, Berlin, (2570) 11–26 (2003)
Ekbatani, F., Natura, B., Végh, László A.: Circuit imbalance measures and linear programming, London Mathematical Society Lecture Note Series, Cambridge University Press, 64–114, (2022)
Grünbaum, B.: Convex polytopes, second ed., Graduate Texts in Mathematics, vol. 221, Springer-Verlag, New York, (2003), Prepared and with a preface by Volker Kaibel, Victor Klee and Günter M. Ziegler
Haase, C., Paffenholz, A., Piechnik, LC., Santos, F.: Existence of unimodular triangulations—positive results, Mem. Amer. Math. Soc. 270 (2021), 1321, v+83
Hanada, M., Lentfer, J., Vindas-Meléndez, A.R.: Generalized parking function polytopes. Ann. Comb. 28(2), 575–613 (2024)
ArticleMathSciNetMATHGoogle Scholar
Heuer, D., Striker, J.: Partial permutation and alternating sign matrix polytopes. SIAM J. Discrete Math. 36(4), 2863–2888 (2022)
ArticleMathSciNetMATHGoogle Scholar
Jochemko, K., Ravichandran, M.: Generalized permutahedra: Minkowski linear functionals and Ehrhart positivity. Mathematika 68(1), 217–236 (2022)
ArticleMathSciNetMATHGoogle Scholar
Kafer, S., Pashkovich, K., Sanità, L.: On the circuit diameter of some combinatorial polytopes. SIAM J. Discrete Math. 33(1), 1–25 (2019)
ArticleMathSciNetGoogle Scholar
Kim, E.D., Santos, F.: An update on the Hirsch conjecture. Jahresber. Deutsch. Math.-Verein. 112(2), 73–98 (2010)
ArticleMathSciNetMATHGoogle Scholar
Liao, HC.: Stembridge codes, permutahedral varieties, and their extensions, (2024), arXiv:2403.10577
Liu, F.: On positivity of Ehrhart polynomials, Recent trends in algebraic combinatorics, Assoc. Women Math. Ser., Springer, Cham, (16) 189–237 (2019)
Oxley, J.: Matroid theory, second ed., Oxford Graduate Texts in Mathematics, Oxford University Press, Oxford, 21 (2011)
Postnikov, A., Reiner, V., Williams, L.: Faces of generalized permutohedra. Doc. Math. 13, 207–273 (2008)
ArticleMathSciNetMATHGoogle Scholar
Postnikov, A.: Permutohedra, associahedra, and beyond. Int. Math. Res. Not. IMRN 6, 1026–1106 (2009)
ArticleMathSciNetMATHGoogle Scholar
Ralph Tyrrell Rockafellar: The elementary vectors of a subspace of R N, University of North Carolina Press, Chapel Hill, NC, Combinatorial Mathematics and its Applications 104–127, (1969)
Selig, T.: The stochastic sandpile model on complete graphs, Electron. J. Combin. 31(3) (2024), 3.26, 29
Lloyd, S.: Shapley. Cores of convex games, International journal of game theory 1(1), 11–26 (1971)
Google Scholar
Stanley, R.: Problem 12191, in Problems and Solutions. Amer. Math. Monthly 127(6), 563 (2020)
ArticleGoogle Scholar
Richard, P.: Stanley and Jim Pitman, A polytope related to empirical distributions, plane trees, parking functions, and the associahedron. Discrete Comput. Geom. 27(4), 603–634 (2002)
ArticleMathSciNetGoogle Scholar
Stong, R.: The Polytope of Parking Functions, Problems and Solutions. Amer. Math. Monthly 129(3), 286–289 (2022)
Google Scholar
Topkis, D.M.: Adjacency on polymatroids. Math. Program. 30(2), 229–237 (1984)
ArticleMathSciNetMATHGoogle Scholar
Topkis, D.M.: Paths on polymatroids. Math. Programming 54(3), 335–351 (1992)
ArticleMathSciNetMATHGoogle Scholar
Neumann, J.v.: A certain zero-sum two-person game equivalent to the optimal assignment problem, Contributions to the Theory of Games, Volume II, Annals of Mathematics Studies, Princeton University Press, 28 5–12, (1953)
Yan, CH.: On the enumeration of generalized parking functions, Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000), 147 201–209 (2000)
Yan, CH.: Generalized parking functions, tree inversions, and multicolored graphs, Adv. in Appl. Math. Special issue in honor of Dominique Foata’s 65th birthday (Philadelphia, PA, 2000) 27(2-3) 641–670, (2001)
Yan, CH.: Parking functions, Handbook of enumerative combinatorics, Discrete Math. Appl. (Boca Raton), CRC Press, Boca Raton, FL, 835–893 (2015)
Yemelichev, V.A., Kovalëv, M.M., Kravtsov, K.: Polytopes, graphs and optimisation. Translated from the Russian by G.H. Lawden, Cambridge University Press (1984)
Günter, M.: Ziegler, Lectures on polytopes, Graduate Texts in Mathematics, vol. 152. Springer-Verlag, New York (1995)
Google Scholar
Download references
Funding
Open access funding provided by SCELC, Statewide California Electronic Library Consortium
Author information
Authors and Affiliations
University of Kansas, Lawrence, KS, USA
Margaret M. Bayer
University of Colorado, Denver, CO, USA
Steffen Borgwardt
University of North Carolina - Chapel Hill, Chapel Hill, NC, USA
Teressa Chambers
University of Colorado, Boulder, Boulder, CO, USA
Spencer Daugherty
Carnegie Mellon University, Pittsburgh, PA, USA
Aleyah Dawkins
KTH Royal Institute of Technology, Stockholm, Sweden
Danai Deligeorgaki
Washington University in St. Louis, St. Louis, MO, USA
Hsin-Chieh Liao
University of Wyoming, Laramie, WY, USA
Tyrrell McAllister
Whitman College, Walla Walla, WA, USA
Garrett Nelson
University of Calgary, Calgary, AB, Canada
Angela Morrison
Harvey Mudd College, Claremont, CA, USA
Andrés R. Vindas-Meléndez
Authors
1. Margaret M. BayerView author publications Search author on:PubMedGoogle Scholar
2. Steffen BorgwardtView author publications Search author on:PubMedGoogle Scholar
3. Teressa ChambersView author publications Search author on:PubMedGoogle Scholar
4. Spencer DaughertyView author publications Search author on:PubMedGoogle Scholar
5. Aleyah DawkinsView author publications Search author on:PubMedGoogle Scholar
6. Danai DeligeorgakiView author publications Search author on:PubMedGoogle Scholar
7. Hsin-Chieh LiaoView author publications Search author on:PubMedGoogle Scholar
8. Tyrrell McAllisterView author publications Search author on:PubMedGoogle Scholar
9. Angela MorrisonView author publications Search author on:PubMedGoogle Scholar
10. Garrett NelsonView author publications Search author on:PubMedGoogle Scholar
11. Andrés R. Vindas-MeléndezView author publications Search author on:PubMedGoogle Scholar
Corresponding author
Correspondence to Andrés R. Vindas-Meléndez.
Ethics declarations
Conflict of Interest
The authors state that there is no conflict of interest.
Additional information
Editor in Charge: Csaba D. Tóth
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
Reprints and permissions
About this article
Cite this article
Bayer, M.M., Borgwardt, S., Chambers, T. et al. Combinatorics of Generalized Parking-Function Polytopes. Discrete Comput Geom (2025).
Download citation
Received: 26 March 2024
Revised: 24 July 2025
Accepted: 26 July 2025
Published: 12 September 2025
DOI:
Share this article
Anyone you share the following link with will be able to read this content:
Get shareable link
Sorry, a shareable link is not currently available for this article.
Copy shareable link to clipboard
Provided by the Springer Nature SharedIt content-sharing initiative
Keywords
Parking functions
Permutahedron
h-polynomial
Building set
Polymatroid
Mathematics Subject Classification
05A10
05A15
05A19
52B05
52B11
52B15
52B20
52B40
Use our pre-submission checklist
Avoid common mistakes on your manuscript.
Sections
Figures
References
Abstract
1 Introduction
2 Polyhedral Perspectives on X n(b)
3 A Building Set Perspective
4 A Polymatroid Perspective
5 Conclusion - Further Perspectives
6 Acknowledgement & Funding
Data Availability
Notes
References
Funding
Author information
Ethics declarations
Additional information
Rights and permissions
About this article
Advertisement
Fig. 1
View in articleFull size image
Fig. 2
View in articleFull size image
Fig. 3
View in articleFull size image
Fig. 4
View in articleFull size image
Fig. 5
View in articleFull size image
Amanbayeva, A., Wang, D.: The convex hull of parking functions of length n, Enumer. Comb. Appl. 2(2) (2022), S2R10, 10
Ardila, F., Benedetti, C., Doker, J.: Matroid polytopes and their volumes. Discrete Comput. Geom. 43(4), 841–854 (2010)
ArticleMathSciNetMATHGoogle Scholar
Backman, S.: Partial graph orientations and the Tutte polynomial. Adv. in Appl. Math. 94, 103–119 (2018)
ArticleMathSciNetMATHGoogle Scholar
Backman, S., Liu, G.: A regular unimodular triangulation of the matroid base polytope, Sém. Lothar. Combin. 91B (2024), Art. 92, 10
Balinski, M., Russakoff, A.: On the assignment polytope. SIAM Rev. 16(4), 516–525 (1974)
ArticleMathSciNetMATHGoogle Scholar
Bartels, J.E., Mount, J., Welsh, D.J.A.: The polytope of win vectors. Ann. Comb. 1(1), 1–15 (1997)
ArticleMathSciNetMATHGoogle Scholar
Behrend, R.E.: Ehrhart polynomials of partial permutohedra, (2024), arxiv:2403.06975
Behrend, RE., Castillo, F., Chavez, A.,Diaz-Lopez, A., Escobar, L., Harris, P., Insko, E.: Partial permutohedra, (2022), arXiv:2207.14253
Behrend, RE., Castillo, F.: Partial permutohedra, Sém. Lothar. Combin. 89B (2023), Art. 64, 12
Birkhoff, G.: Tres observaciones sobre el algebra lineal. Univ. Nac. Tucumán. Revista A. 5, 147–151 (1946)
MathSciNetGoogle Scholar
Borgwardt, S., De Loera, J.A., Finhold, E.: Edges versus circuits: a hierarchy of diameters in polyhedra. Adv. Geom. 16(4), 511–530 (2016)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Finhold, E., Hemmecke, R.: On the circuit diameter of dual transportation polyhedra. SIAM J. Discrete Math. 29(1), 113–121 (2015)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Finhold, E., Hemmecke, R.: Quadratic diameter bounds for dual network flow polyhedra. Math. Program. 159(1–2), 237–251 (2016)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Viss, C.: Constructing clustering transformations. SIAM J. Discrete Math. 35(1), 152–178 (2021)
ArticleMathSciNetMATHGoogle Scholar
Borgwardt, S., Viss, C.: Circuit walks in integral polyhedra, Discrete Optim. 44 (2022), 100566
Braden, T., Huh, J., Matherne, JP.: Nicholas Proudfoot, and Botong Wang, A semi-small decomposition of the Chow ring of a matroid, Adv. Math. 409 (2022), 108646, 49
Brändén, P., Solus, L.: Symmetric decompositions and real-rootedness. Int. Math. Res. Not. IMRN 10, 7764–7798 (2021)
ArticleMathSciNetMATHGoogle Scholar
Braun, B.: Unimodality problems in Ehrhart theory, Recent trends in combinatorics, IMA Vol. Math. Appl., Springer, [Cham], (159) 687–711 (2016)
Richard, A.: Brualdi, Combinatorial matrix classes, Encyclopedia of Mathematics and its Applications, vol. 108. Cambridge University Press, Cambridge (2006)
Google Scholar
De Loera, J.A.: Hemmecke, R., Lee, J.: On augmentation algorithms for linear and integer-linear programming: from Edmonds-Karp to Bland and beyond. SIAM J. Optim. 25(4), 2494–2511 (2015)
Doker, JS.: Geometry of generalized permutohedra, University of California, Berkeley, (2011), PhD thesis
Edmonds, J.: Submodular functions, matroids, and certain polyhedra, Combinatorial optimization—Eureka, you shrink!, Lecture Notes in Comput. Sci.,, Springer, Berlin, (2570) 11–26 (2003)
Ekbatani, F., Natura, B., Végh, László A.: Circuit imbalance measures and linear programming, London Mathematical Society Lecture Note Series, Cambridge University Press, 64–114, (2022)
Grünbaum, B.: Convex polytopes, second ed., Graduate Texts in Mathematics, vol. 221, Springer-Verlag, New York, (2003), Prepared and with a preface by Volker Kaibel, Victor Klee and Günter M. Ziegler
Haase, C., Paffenholz, A., Piechnik, LC., Santos, F.: Existence of unimodular triangulations—positive results, Mem. Amer. Math. Soc. 270 (2021), 1321, v+83
Hanada, M., Lentfer, J., Vindas-Meléndez, A.R.: Generalized parking function polytopes. Ann. Comb. 28(2), 575–613 (2024)
ArticleMathSciNetMATHGoogle Scholar
Heuer, D., Striker, J.: Partial permutation and alternating sign matrix polytopes. SIAM J. Discrete Math. 36(4), 2863–2888 (2022)
ArticleMathSciNetMATHGoogle Scholar
Jochemko, K., Ravichandran, M.: Generalized permutahedra: Minkowski linear functionals and Ehrhart positivity. Mathematika 68(1), 217–236 (2022)
ArticleMathSciNetMATHGoogle Scholar
Kafer, S., Pashkovich, K., Sanità, L.: On the circuit diameter of some combinatorial polytopes. SIAM J. Discrete Math. 33(1), 1–25 (2019)
ArticleMathSciNetGoogle Scholar
Kim, E.D., Santos, F.: An update on the Hirsch conjecture. Jahresber. Deutsch. Math.-Verein. 112(2), 73–98 (2010)
ArticleMathSciNetMATHGoogle Scholar
Liao, HC.: Stembridge codes, permutahedral varieties, and their extensions, (2024), arXiv:2403.10577
Liu, F.: On positivity of Ehrhart polynomials, Recent trends in algebraic combinatorics, Assoc. Women Math. Ser., Springer, Cham, (16) 189–237 (2019)
Oxley, J.: Matroid theory, second ed., Oxford Graduate Texts in Mathematics, Oxford University Press, Oxford, 21 (2011)
Postnikov, A., Reiner, V., Williams, L.: Faces of generalized permutohedra. Doc. Math. 13, 207–273 (2008)
ArticleMathSciNetMATHGoogle Scholar
Postnikov, A.: Permutohedra, associahedra, and beyond. Int. Math. Res. Not. IMRN 6, 1026–1106 (2009)
ArticleMathSciNetMATHGoogle Scholar
Ralph Tyrrell Rockafellar: The elementary vectors of a subspace of R N, University of North Carolina Press, Chapel Hill, NC, Combinatorial Mathematics and its Applications 104–127, (1969)
Selig, T.: The stochastic sandpile model on complete graphs, Electron. J. Combin. 31(3) (2024), 3.26, 29
Lloyd, S.: Shapley. Cores of convex games, International journal of game theory 1(1), 11–26 (1971)
Google Scholar
Stanley, R.: Problem 12191, in Problems and Solutions. Amer. Math. Monthly 127(6), 563 (2020)
ArticleGoogle Scholar
Richard, P.: Stanley and Jim Pitman, A polytope related to empirical distributions, plane trees, parking functions, and the associahedron. Discrete Comput. Geom. 27(4), 603–634 (2002)
ArticleMathSciNetGoogle Scholar
Stong, R.: The Polytope of Parking Functions, Problems and Solutions. Amer. Math. Monthly 129(3), 286–289 (2022)
Google Scholar
Topkis, D.M.: Adjacency on polymatroids. Math. Program. 30(2), 229–237 (1984)
ArticleMathSciNetMATHGoogle Scholar
Topkis, D.M.: Paths on polymatroids. Math. Programming 54(3), 335–351 (1992)
ArticleMathSciNetMATHGoogle Scholar
Neumann, J.v.: A certain zero-sum two-person game equivalent to the optimal assignment problem, Contributions to the Theory of Games, Volume II, Annals of Mathematics Studies, Princeton University Press, 28 5–12, (1953)
Yan, CH.: On the enumeration of generalized parking functions, Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000), 147 201–209 (2000)
Yan, CH.: Generalized parking functions, tree inversions, and multicolored graphs, Adv. in Appl. Math. Special issue in honor of Dominique Foata’s 65th birthday (Philadelphia, PA, 2000) 27(2-3) 641–670, (2001)
Yan, CH.: Parking functions, Handbook of enumerative combinatorics, Discrete Math. Appl. (Boca Raton), CRC Press, Boca Raton, FL, 835–893 (2015)
Yemelichev, V.A., Kovalëv, M.M., Kravtsov, K.: Polytopes, graphs and optimisation. Translated from the Russian by G.H. Lawden, Cambridge University Press (1984)
Günter, M.: Ziegler, Lectures on polytopes, Graduate Texts in Mathematics, vol. 152. Springer-Verlag, New York (1995)
Google Scholar
Discover content
Journals A-Z
Books A-Z
Publish with us
Journal finder
Publish your research
Language editing
Open access publishing
Products and services
Our products
Librarians
Societies
Partners and advertisers
Our brands
Springer
Nature Portfolio
BMC
Palgrave Macmillan
Apress
Discover
Your privacy choices/Manage cookies
Your US state privacy rights
Accessibility statement
Terms and conditions
Privacy policy
Help and support
Legal notice
Cancel contracts here
34.34.225.232
Not affiliated
© 2025 Springer Nature |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.