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4101 | https://pressbooks.bccampus.ca/technicalwriting/chapter/figurestables/ | Skip to content
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DOCUMENT DESIGN
3.4 Figures and Tables
Suzan Last
Visual elements such as graphs, charts, tables, photographs, diagrams, and maps capture your readers’ attention and help them to understand your ideas more fully. They are like the illustrations that help tell the story. These visuals help to augment your written ideas and simplify complicated textual descriptions. They can help the reader understand a complicated process or visualize trends in the data. The key concept to remember here is that visuals clarify, illustrate, and augment your written text; they are not a replacement for written text. The old adage, “a picture is worth a thousand words” may not always not hold true in technical writing, but adding visuals may save you a hundred words or so of additional explanation and clarification. If you have visual elements in your document, they must be based on and supplement your written content. Throwing in “gratuitous graphics” just to decorate or take up space can confuse your reader.
It is important to choose the right kind of visual to convey the story you want your reader to understand. If visuals are poorly chosen or poorly designed for the task, they can actually confuse the reader and have negative consequences. For example, it’s very likely that the first thing you noticed when you opened this page was the image above. Did you wonder why is it there? Has it distracted you?
Conventions for Integrating Visuals in your Document
Each style of visual has its own conventions that you will recognize after you have seen enough of them. In addition, different publications have different style guides that dictate the specifics of how to format and integrate visual elements. In general, however, whenever you integrate any kind of visual, you should adhere to five key rules.
Five Rules for Integrating Visual Elements into your Document
Give each visual a numbered caption that includes a clear descriptive title
Refer to the caption number within the body text and discuss its content
Label all units (x and y axes, legends, column box heads, parts of diagrams, etc)
Provide the source of the data and/or visual image if you did not create it yourself
Avoid distorting the data or image.
In addition, visual elements should also be surrounded with sufficient passive space to emphasize the image and enhance its readability. If copying and pasting an image, make sure all elements are clear and the print size is readable. A visual that has been shrunk down to an unreadable size does not help the reader understand your ideas. Whenever possible, try to orient the visual image in the same direction as the body text.
Examine Figure 3.4.1 below. Do you understand what information it conveys? What story it’s trying to tell? What is missing?
If you look carefully, you might be able to guess what story this graph is telling. However, the lack of a descriptive caption and labelling of axes makes it impossible to know for sure. Compare it to Figure 3.4.2 below.
Figure 3.4.2 has a numbered caption (which I have just referred to in my paragraph), a descriptive title, and it has properly labelled x and y axes and legends. With this added information, the story starts to take shape. The graph tells the story of the fluxuating water demand in Edmonton during the 2010 Olympic gold medal hockey game. If you add some context that flushing of toilets is one of the main causes of water demand, the story comes into focus. The figure also cites the source the graph was retrieved from in the caption using an in-text citation, which is linked to a full reference below. Therefore, if you want more information about this data, you can find it. The original image has not been distorted in any way. Thus, Figure 3.4.2 follows the five key rules listed above.
In addition to those five general rules, there are specific guidelines for implementing them. These are outlined in detail in the Faculty of Engineering Co-op Work Term Report Guideline (.pdf).
Specific Guidelines for Integrating Graphics
Terminology
Visual elements are referred to as either Tables or Figures. Tables are made up of rows and columns and the cells usually have numbers in them (but may also have words or images). Figures refer to any visual elements—graphs, charts, diagrams, photos, etc.—that are not Tables. They may be included in the main sections of the report, or if they contain supplemental material they may be contained in an appendix. Try to ensure that figures and tables are not broken over two pages. Tables that require a full page might be best put in an appendix.
Labelling Tables and Figures
Tables and figures must all be labelled with numbered captions that clearly identify and describe them. Figure captions are generally placed below the figures, while table captions must be placed above the tables. This is because we generally read tables from the top down, and therefore want to see the caption at the top. Figures are not always read top down. When you open a page and see a figure, the first thing you want to know is “what is that?” The caption below it should immediately identify what the figure represents for the reader. If you choose to place figure captions above the figures, do so consistently throughout your document.
Use the following conventions to assist the reader in understanding your graphics:
Numbering: Table and Figures are numbered sequentially, but separately
e.g. Table 1, Table 2, Figure 1, Figure 2, Table 3, etc.
Captioning: After the Figure or Table number, add a descriptive caption that clearly indicate what the figure or table illustrates without having to read anything else on the page.
There are two systems for numbering figures and tables within your document:
Simple Consecutive Numbering: All figures and tables are numbered consecutively (Figure 1, Figure 2, Figure 3, Table 1, Table 2, Table, 3 etc.) throughout the document regardless of which section they are in.
Section-based Numbering: Within each section, figures and tables may be numbered sequentially through each section (e.g. Table 1.1 refers to the first table in section 1, Table 2.4 refers to the fourth table in section 2).
If a large number of illustrations are presented, the latter option is the better choice. This can become confusing, however, when using sub-sections.
If the table or figure that you present in your report was not created by you, but comes from other sources, you must include a reference for the original source in your caption: e.g.: Figure 1. Network Design . You must ensure that all figures and tables represent data accurately and ethically, and that they do not distort data to create bias.
Using the Insert → Caption … function will allow Word to keep track of the Figure and Table numbering for you, and allow you to auto-create a List of Figures and Tables at the beginning of your document.
If you don’t use the Insert Caption function, then you should manually change the font of your captions to distinguish them from body text. Caption font is usually slightly smaller than body font and is often italicized. The numbered portion is often bolded in both the caption and in the in-paragraph reference to the figure or table for ease of cross-referencing.
Referring to Tables and Figures in your Text
Any figures or tables you use in your document must be discussed in your text. Use the following guidelines when discussing and referring to tables and figures:
Place the table/figure close to where it is first referred to in the text (preferably immediately below the paragraph in which it is first mentioned).
Refer to tables and figures in your text by their numbers, not their placement in the text. E.g, “See Figure 9 for a detailed schematic” (not “see the figure below”); “the test results are summarized in Table 1.”
When referring to a figure or table in your body text, it is helpful to place the reference in bold font.
Selecting the Right Visual
Table 3.1.1 lists common kinds of visual elements used in technical writing, along with their general purpose or description (for a more detailed discussion of how and when to use these kinds of visuals, see Graves and Graves. Notice the “box head” on the top and “stubs” on the left are bolded and centred to enhance readability. Tables that have text in the cells instead of numbers can also be referred to as figures. Thus, Table 3.1.1 could have been captioned as Figure 3.7 instead.
TABLE 3.1.1 Common types of illustrative graphics
| [Skip Table] | | |
| Type of Visual | | Description and Purpose |
| Tables | | Place detailed data/information in categories formatted into rows and columns for comparison; use when exact figures are important. Label column headings (box heads) and/or rows (stubs). |
| Graphs | Bar Graph | Compare and contrast two or more subjects at the same point in time, or compare change over time. |
| Column Graph | Reveal change in a subject at regular intervals of time. |
| Line Graph | Show the degree and direction of change relative to two variables; compare items over time, show frequency or distribution, or show correlations. |
| Charts | Pie Chart | Display the number and relative size of the divisions of a subject; shows relation of parts to a whole (parts must sum to 100% to make sense). |
| Org. Chart | Map the divisions and levels of responsibility or hierarchy within an organization. |
| Flow Chart | Show the sequence of steps in a process or procedure. |
| Gantt Chart | Indicates timelines for multi-stepped projects, especially used in proposals and progress reports. |
| Illustrations | Diagram | Identify the parts of a subject and their spatial or functional relationship; emphasize detail or show dimensions. |
| Photo | Show what a subject looks like in realistic detail or show it being used. |
| Animation | Simulate a process, operation, or incident. |
| Film clip | Depict a process, operation, or incident in realistic detail. |
For more information on how to format a long report, see the Faculty of Engineering’s “Co-op Work Term Report Guidelines.”
EXERCISE 3.7 Design a figure to match the data
Using what you have learned about figures and tables, create two different visual representations of the data described in the following paragraph, and caption them as Figure 1. Add your descriptive caption. Explain why you chose those methods and list the pros and cons of each:
We surveyed the students in 3 sections of ENGR 240 (total of 100 students) to gauge which aspect of the writing process they found most challenging: Pre-writing, Drafting, or Revising. The results among the 3 sections were consistent. Overall 50% of students said that they found the Pre-writing stage to be the most challenging, while 28% found the Drafting stage most difficult and 22% said the Revision stage was most challenging (see Figure 1). These results suggest that we should place more emphasis on teaching and practicing pre-writing strategies during the course.
[create a Figure 1 and a descriptive caption that illustrates the data above]
Additional Resources
For a look at how professionals can animate data, check out Hans Rosling’s “The Joy of Stats” on YouTube [Online].
Image descriptions
Figure 3.4.1 image description:
A graph with no figure number or caption and no x or y axis labels, so it is difficult to determine what point it is trying to make. It shows something rising and falling during a hockey game. This thing spikes at the end of each period and drops dramatically when Canada wins.
[Return to Figure 3.4.1]
Figure 3.4.2 image description:
A graph charting water consumption in Edmonton during the 2010 Gold Medal Hockey Game. The graphs shows spikes in water consumption at the end of each period, followed by very low usage periods, especially near the end of the 3rd period, and between the end of the game and the medal ceremony. It also has a line depicting water usage the previous day, which was fairly steady throughout the day.
[Return to Figure 3.4.2]
EPCOR, Edmonton’s Water Utility. “Water Consumption in Edmonton during 2010 Gold Medal hockey game,” Cited on Flowing Data[Online] Available: ↵
H. Graves and R. Graves, “Communicating through visuals,” in A Strategic Guide to Technical Communications, 2nd ed. Peterborough, ONT: Broadview Press, 2011, pp. 137-148. ↵
Engineering Co-op Work Term Report Guidelines (.pdf) ↵ |
4102 | https://storage.outreach.psu.edu/autism/64.%20and%2074.%20Handout.PDF | State Board of Education-Adopted Appendix F: Methods Used for Solving Single-Digit Addition and Subtraction Problems Page 1 of 4 Appendix F: Methods used for solving single-digit addition and subtraction problems Level 1. Direct Modeling by Counting All or Taking Away. Represent situation or numerical problem with groups of objects, a drawing, or fingers. Model the situation by composing two addend groups or decomposing a total group. Count the resulting total or addend. Adding (8 + 6 = ): Represent each addend by a group of objects. Put the two groups together. Count the total. Use this strategy for Add To/Result Unknown and Put Together/Total Unknown. Subtracting (14 – 8 = ): Represent the total by a group of objects. Take the known addend number of objects away. Count the resulting group of objects to find the unknown added. Use this strategy for Take From/Result Unknown. Levels 8 + 6 = 14 14 - 8 = 6 Level 1: Count all Count All a b 1 2 3 4 5 6 7 8 1 2 3 4 5 6 ο ο ο ο ο ο ο ο ο ο ο ο ο ο 1 2 3 4 5 6 7 8 9 10 11 12 13 14 c Take Away a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ο ο ο ο ο ο ο ο ο ο ο ο ο ο 1 2 3 4 5 6 7 8 1 2 3 4 5 6 b Level 2: Count on Count On 8 ο ο ο ο ο ο ο ο ο ο ο ο ο ο 9 10 11 12 13 14 8 To solve 14 – 8 I count on 8 + ? = 14 10 11 12 13 I took away 8 9 14 8 to 14 is 6 so 14 – 8 = 6 Level 3: Recompose Make a ten (general): one addend breaks apart to make 10 with the other addend Make a ten (from 5’s within each addend) Recompose: Make a Ten ο ο ο ο ο ο ο ο ο ο ο ο ο ο 10 + 4 ο ο ο ο ο ο ο ο ο ο ο ο ο ο 10 + 4 14 – 8: I make a ten for 8 + ? = 14 ο ο ο ο ο ο ο ο ο ο ο ο ο ο 8 + 2 + 4 6 8 + 6 = 14 Doubles = n 6 + 8 = 6 + 6 + 2 = 12 + 2 = 14 Note: Many children attempt to count down for subtraction, but counting down is difficult and error-prone. Children are much more successful with counting on; it makes subtraction as easy as addition. The Mathematics Framework was adopted by the California State Board of Education on November 6, 2013. The Mathematics Framework has not been edited for publication. State Board of Education-Adopted Appendix F: Methods Used for Solving Single-Digit Addition and Subtraction Problems Page 2 of 4 Level 2. Counting On. Embed an addend within the total (the addend is perceived simultaneously as an addend and as part of the total). Count this total but abbreviate the counting by omitting the count of this addend; instead, begin with the number word of this addend. Some method of keeping track (fingers, objects, mentally imagined objects, body motions, other count words) is used to monitor the count. For addition, the count is stopped when the amount of the remaining addend has been counted. The last number word is the total. For subtraction, the count is stopped when the total occurs in the count. The tracking method indicates the difference (seen as an unknown addend). Counting on can be used to find the total or to find an addend. These look the same to an observer. The difference is what is monitored: the total or the known addend. Some students count down to solve subtraction problems, but this method is less accurate and more difficult than counting on. Counting on is not a rote method. It requires several connections between cardinal and counting meanings of the number words and extended experience with Level 1 methods in kindergarten. Adding (e.g., 8 + 6 = ) uses counting on to find a total: One counts on from the first addend (or the larger number is taken as the first addend). Counting on is monitored so that it stops when the second addend has been counted on. The last number word is the total. Finding an unknown addend (e.g., 8 + = 14): One counts on from the known addend. The keeping track method is monitored so that counting on stops when the known total has been reached. The keeping track method tells the unknown addend. Subtracting (14 – 8 = ): One thinks of subtracting as finding the unknown addend, as 8 + = 14 and uses counting on to find an unknown addend (as above). The problems in Table 1 in the Glossary which can be solved by Level 1 methods in kindergarten can also be solved using Level 2 methods: counting on to find the total (adding) or counting on to find the unknown addend (subtracting). Level 2 and 3 methods are generally used in grades one and two. Finding an unknown addend (e.g., 8 + = 14) is used for Add To/Change Unknown, Put Together/Take Apart/Addend Unknown, and Compare/Difference Unknown. It is also used for Take From/Change Unknown (14 - = 8) after a student has decomposed the total into two addends, which means they can represent the situation as 14 – 8 = . Adding or subtracting by counting on is used by some students for each of the kinds of Compare problems (see the equations in Table 1 of the Glossary). Grade one students do not necessarily master the Compare Bigger Unknown or The Mathematics Framework was adopted by the California State Board of Education on November 6, 2013. The Mathematics Framework has not been edited for publication. State Board of Education-Adopted Appendix F: Methods Used for Solving Single-Digit Addition and Subtraction Problems Page 3 of 4 Smaller Unknown problems with the misleading language in the bottom row of Table 1 of the Glossary. Solving an equation such as 6 + 8 = by counting on from 8 relies on the understanding that 8 + 6 gives the same total, an implicit use of the commutative property without the accompanying written representation 6 + 8 = 8 + 6. Level 3. Convert to an Easier Equivalent Problem. Decompose an addend and compose a part with another addend. These methods can be used to add or to find an unknown addend (and thus to subtract). These methods implicitly use the associative property. Adding Make a ten. E.g., for 8 + 6 = , 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14, so 8 + 6 becomes 10 + 4. Doubles plus or minus 1. E.g., for 6 + 7 = , 6 + 7 = 6 + 6 + 1 = 12 + 1 = 13, so 6 + 7 becomes 12 + 1. Finding an unknown addend. Make a ten. E.g., for 8 + = 14, 8 + 2 = 10 and 4 more makes 14. 2 + 4 = 6. So 8 + = 14 is done as two steps: how many up to ten and how many over ten (which can be seen in the ones place of 14). Doubles plus or minus 1. E.g., for 6 + = 13, 6 + 6 + 1 = 12 + 1. 6 + 1 = 7. So 6 + = 13 is done as two steps: how many up to 12 (6 + 6) and how many from 12 to 13. The Mathematics Framework was adopted by the California State Board of Education on November 6, 2013. The Mathematics Framework has not been edited for publication. State Board of Education-Adopted Appendix F: Methods Used for Solving Single-Digit Addition and Subtraction Problems Page 4 of 4 Subtracting. Thinking of subtracting as finding an unknown addend. E.g., solve 14 – 8 = or 13 – 6 = as 8 + = 14 or 6 + = 13 by the above methods (make a ten or doubles plus or minus 1). Make a ten by going down over ten. E.g., 14 – 8 = can be done in two steps by going down over ten: 14 – 4 (to get to 10) – 4 = 6. The Level 1 and Level 2 problem types can be solved using these Level 3 methods. Level 3 problem types can be solved by representing the situation with an equation or drawing, then re-representing to create a situation solved by adding, subtracting, or finding an unknown addend as shown above by methods at any level, but usually at Level 2 or 3. Many students only show in their writing part of this multi-step process of re-representing the situation. Students re-represent Add To/Start Unknown + 6 = 14 situations as 6 + = 14 by using the commutative property (formally or informally). Students re-represent Take From/Start Unknown - 8 = 6 situations by reversing as 6 + 8 = , which may then be solved by counting on from 8 or using a Level 3 method. At Level 3, the Compare misleading language situations can be solved by representing the known quantities in a diagram that shows the bigger quantity in relation to the smaller quantity. The diagram allows the student to find a correct situation by representing the difference between quantities and seeing the relationship among the three quantities. Such diagrams are the same diagrams used for the other versions of compare situations; focusing on which quantity is bigger and which is smaller helps to overcome the misleading language. Some students may solve Level 3 problem types by doing the above re-representing but use Level 2 counting on. As students move through levels of solution methods, they increasingly use equations to represent problem situations as situation equations and then to re-represent the situation with a solution equation or a solution computation. They relate equations to diagrams, facilitating such re-representing. Labels on diagrams can help connect the parts of the diagram to the corresponding parts of the situation. But students may know and understand things that they may not use for a given solution of a problem as they increasingly do various representing and re-representing steps mentally. (Adapted from Progressions, K-5 CC and OA 2011) The Mathematics Framework was adopted by the California State Board of Education on November 6, 2013. The Mathematics Framework has not been edited for publication. |
4103 | https://www.teacherspayteachers.com/browse?search=dividing%20fractions%20puzzles | Dividing Fractions Puzzles | TPT
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TEKS 6.3E ✩ Multiply &Divide Fractions ✩ Google Sheets Puzzle Activity
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Looking for a SELF-CHECKING digital activity? Make online learning fun with these 3 Google Sheets activities + a Paper Version! Try a Sample ➯ ✩ FREE ✩ Rounding Decimals ✩ Mystery PictureTry a Sample ➯ ✩ FREE ✩ Fraction Arithmetic ✩ Digital PuzzleTry a Sample ➯ ✩ FREE ✩ Direct Variation ✩ Digital ColoringThese 3 puzzles are SELF-CHECKING. Each correct solution automatically puts together a bit of the puzzle. Students do NOT need to be familiar with Google Sheets and there is NO prep. A paper
6 th
Arithmetic, Math, Other (Math)
Also included in:6.3E ★ Activities BUNDLE ★ Multiply & Divide Rational Numbers ★ STAAR 2.0 Prep
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Multiplying and Dividing Fractions and Mixed Numbers | Crossword Puzzles
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Multiplication and Division of Fractions and Mixed Numbers Doing crosswords is fun and exciting, so students will be happy to do the task. Students have to put the numbers into empty cells based on a crossword puzzle. This product contains 5 crosswords with a different number of problems (from 16 to 30). There are more empty cells in the “Hard” crossword. Each page is presented both in color and in an ink-saving black and white version. Answers included. This resource is great for:homewo
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CCSS
4.NF.B.4
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Your students will love this self checking activity! Students will divide fractions and mixed numbers. Answers will include whole numbers, fractions and mixed numbers in simplest form. Answers will use a backslash (/) for a fraction bar, and an underscore to separate the whole number from the fraction. (example 3_1/2) Google Sheet will reveal pixels of a mystery picture as students answer questions correctly. Incorrect answers will highlight red, allowing students to self check their understandi
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Divide Fractions Challenge Puzzles are a fun way to differentiate for your students while they gain a deeper understanding of dividing with unit fractions and fraction division. These puzzles are available as a printable PDF and a paperless version made with Google Slides™ for distance learning. Three different types of puzzles are included: grid logic puzzles, determine the fractions puzzlers, and missing digits puzzles. These 14 different dividing fractions enrichment puzzles allow students t
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Multiplying and Dividing Fractions Digital Puzzle Activity| Google Classroom
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Students will match up strips by solving multiplication and division fraction equations. This is digital activity and also printable. Perfect for in-class or remote learning! Students can drag and drop the strips on the digital version or can match up the strips on the printable version. Perfect for small group, intervention, stations, for early finishers, homework, review etc. Puzzle 1: Multiplying Fractions Puzzle 2 Dividing Fractions Puzzle 3 + 4: Multiplying and Dividing Fractions This pur
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Your students will love these self checking activities! Students will multiply and divide fractions and mixed numbers. Answers will include whole numbers, fractions and mixed numbers in simplest form. Answers will use a backslash (/) for a fraction bar, and an underscore to separate the whole number from the fraction. (example 3_1/2) Google Sheet will reveal pixels of a mystery picture as students answer questions correctly. Incorrect answers will highlight red, allowing students to self check t
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4th Grade Math Puzzles Math Riddles BUNDLE Division Multiplication Fractions
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This is a bundle of 4th grade secret code math riddle activities. Each individual resource includes two standards-aligned activities that provide students with targeted skill practice. Includes: Multiplication (2-digt x 1-digit)Multiplication (3-digit x 1-digit)Division (dividends up to 4 digits)Equivalent FractionsMultiplying Fractions by Whole NumbersAddition up to 999,999Subtraction up to 999,999To complete the activities, place a set of cards around the room. When students find a card, they
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This puzzle was created by Pupsaroni Puzzlesfor use in classrooms studying fractions. Typically, this is in grades 5 thru Algebra 1. The focus of this activity is to practice dividing fractions. Cut out the cards on pages 2-5. Have students match up the problem with the answer to the problem. When the problems are completed, the student should have the puzzle completed. An an-swer key has been provided on page 6 for your convenience in grading. This set of cards has a focus of dividing fractio
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Dividing Fractions Ghost Puzzle | 6th Grade | Math | Halloween
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Middles and Mayhem
Dividing Fractions Ghost Puzzle This is a quick resource to help students review, or practice, dividing fractions! This is a PHYSICAL resource and requires you to print it out yourself. Students will match 33 dividing fractions problems with their answer and glue them down to make a ghost puzzle. This is a great activity to use as a review, group activity, or if you need a quick, no prep, activity at the end of a unit! The answers are NOT in simplest form, but you can have your students work
5 th - 8 th
Applied Math, Math
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Fractions – multiplying and dividing – Math puzzleI’ve included three different sizes of the same puzzle. The smaller size is only three pages and is great if you are going to print of individual copies for students to practice in class or at home. The larger size requires 8 pieces of paper and quite a bit of space to solve – fun for centers and group work. The extra-large size requires 24 pieces of paper. Cut out the puzzle pieces (or even better if your students do it themselves) and students
6 th - 11 th
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Jessica Ridgway
This puzzle is a fun way to review dividing fractions with your students. Includes dividing a whole number by a unit fraction and unit fraction by a whole number. Simply cut the puzzle up and give it your students to put together. Perfect activity for a math center or small group time!
5 th - 6 th
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Divide Whole Numbers by Fractions Mystery Pixel Puzzle Digital Activity
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Your students will love this self checking activity! Students will find the reciprocal of fractions. Students will divide whole numbers by fractions. Answers include whole numbers, fractions and mixed numbers. Answers will use a backslash (/) for a fraction bar, and an underscore to separate the whole number from the fraction. (example 3_1/2) Google Sheet will reveal pixels of a mystery picture as students answer questions correctly. Incorrect answers will highlight red, allowing students to sel
4 th - 8 th
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Included are 6 challenging square puzzles. Students are given the puzzle pieces and must solve the problems to match the pieces together. This makes for a great center activity! Simply cut and laminate the pieces for repeated use OR give the students the scrambled puzzle pages to cut out and assemble. The 6 puzzles are: 1. Multiplying Fractions with Whole Numbers 2. Multiplying Fractions by Fractions 3. Multiplying Mixed Numbers 4. Dividing Fractions with Whole Numbers 5. Dividing Fractions b
5 th - 6 th
Fractions, Math
CCSS
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Dividing Fractions Number Search: Keep Change Flip Puzzle: Paper + Digital
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This comprehensive dividing fractions number search is a captivating and effective way to help students master the concept of division with fractions. This ready-to-use worksheet turns a typical math assignment into an engaging puzzle activity by challenging students to find the answers in a number search grid. How It Works: To complete the activity, students first solve a division problem using the "Keep, Change, Flip" method and then search for the numerical answer within the grid. Inc
6 th - 9 th
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Multiplying &Dividing Fractions and Mixed Numbers Crossword Puzzle
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Maya Khalil
You could create a word bank for this puzzle since students are going to need help spelling out the fraction answers. Make sure "Actual Size" is checked in the print dialogue box when printing, otherwise the printer might shrink the worksheet.
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Writing Fractions as Division Problems Activity 5th Grade Fractions Math Puzzle
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Chloe Campbell
Are you looking for a fun and simple puzzle activity to help your 5th grade students practice fractions as division problems? This fractions math puzzle is the perfect solution! Use it as a low-prep math center, for practice after a whole group lesson, in small groups, or as a teacher table game. It's also great for spiral review throughout the year and makes a fun test prep activity too! Your students will love completing these math puzzles while mastering the content standards along the w
3 rd - 5 th
Algebra, Fractions, Math
CCSS
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Multiplying And Dividing Fractions Puzzle Summer Packet For 5th To 6th Grade
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Printables and Worksheets
These are fun worksheets on multiplying and dividing fractions by whole numbers. It's a Summer coloring activity that your students will love! Select the right set of problems for your students. To solve the mystery picture, they must first answer the questions correctly. After that, they can match their answers to the colors and then color the grid to reveal the hidden image. This packet is a subset of Summer Fractions and Decimals Mystery Pictures. (Click for more details)Product Contents
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4104 | https://openstax.org/books/university-physics-volume-2/pages/11-5-force-and-torque-on-a-current-loop | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
University Physics Volume 2
11.5 Force and Torque on a Current Loop
University Physics Volume 211.5 Force and Torque on a Current Loop
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Evaluate the net force on a current loop in an external magnetic field
Evaluate the net torque on a current loop in an external magnetic field
Define the magnetic dipole moment of a current loop
Motors are the most common application of magnetic force on current-carrying wires. Motors contain loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted into mechanical work in the process. Once the loop’s surface area is aligned with the magnetic field, the direction of current is reversed, so there is a continual torque on the loop (Figure 11.15). This reversal of the current is done with commutators and brushes. The commutator is set to reverse the current flow at set points to keep continual motion in the motor. A basic commutator has three contact areas to avoid and dead spots where the loop would have zero instantaneous torque at that point. The brushes press against the commutator, creating electrical contact between parts of the commutator during the spinning motion.
Figure 11.15 A simplified version of a dc electric motor. (a) The rectangular wire loop is placed in a magnetic field. The forces on the wires closest to the magnetic poles (N and S) are opposite in direction as determined by the right-hand rule-1. Therefore, the loop has a net torque and rotates to the position shown in (b). (b) The brushes now touch the commutator segments so that no current flows through the loop. No torque acts on the loop, but the loop continues to spin from the initial velocity given to it in part (a). By the time the loop flips over, current flows through the wires again but now in the opposite direction, and the process repeats as in part (a). This causes continual rotation of the loop.
In a uniform magnetic field, a current-carrying loop of wire, such as a loop in a motor, experiences both forces and torques on the loop. Figure 11.16 shows a rectangular loop of wire that carries a current I and has sides of lengths a and b. The loop is in a uniform magnetic field: The magnetic force on a straight current-carrying wire of length l is given by To find the net force on the loop, we have to apply this equation to each of the four sides. The force on side 1 is
11.14
where the direction has been determined with the RHR-1. The current in side 3 flows in the opposite direction to that of side 1, so
11.15
The currents in sides 2 and 4 are perpendicular to and the forces on these sides are
11.16
We can now find the net force on the loop:
11.17
Although this result has been obtained for a rectangular loop, it is far more general and holds for current-carrying loops of arbitrary shapes; that is, there is no net force on a current loop in a uniform magnetic field.
Figure 11.16 (a) A rectangular current loop in a uniform magnetic field is subjected to a net torque but not a net force. (b) A side view of the coil.
To find the net torque on the current loop shown in Figure 11.16, we first consider and Since they have the same line of action and are equal and opposite, the sum of their torques about any axis is zero (see Fixed-Axis Rotation). Thus, if there is any torque on the loop, it must be furnished by and Let’s calculate the torques around the axis that passes through point O of Figure 11.16 (a side view of the coil) and is perpendicular to the plane of the page. The point O is a distance x from side 2 and a distance from side 4 of the loop. The moment arms of and are and respectively, so the net torque on the loop is
11.18
This simplifies to
11.19
where is the area of the loop.
Notice that this torque is independent of x; it is therefore independent of where point O is located in the plane of the current loop. Consequently, the loop experiences the same torque from the magnetic field about any axis in the plane of the loop and parallel to the x-axis.
A closed-current loop is commonly referred to as a magnetic dipole and the term IA is known as its magnetic dipole moment Actually, the magnetic dipole moment is a vector that is defined as
11.20
where is a unit vector directed perpendicular to the plane of the loop (see Figure 11.16). The direction of is obtained with the RHR-2—if you curl the fingers of your right hand in the direction of current flow in the loop, then your thumb points along If the loop contains N turns of wire, then its magnetic dipole moment is given by
11.21
In terms of the magnetic dipole moment, the torque on a current loop due to a uniform magnetic field can be written simply as
11.22
This equation holds for a current loop in a two-dimensional plane of arbitrary shape.
Using a calculation analogous to that found in Capacitance for an electric dipole, the potential energy of a magnetic dipole is
11.23
Example 11.7
Forces and Torques on Current-Carrying Loops
A circular current loop of radius 2.0 cm carries a current of 2.0 mA. (a) What is the magnitude of its magnetic dipole moment? (b) If the dipole is oriented at 30 degrees to a uniform magnetic field of magnitude 0.50 T, what is the magnitude of the torque it experiences and what is its potential energy?
Strategy
The dipole moment is defined by the current times the area of the loop. The area of the loop can be calculated from the area of the circle. The torque on the loop and potential energy are calculated from identifying the magnetic moment, magnetic field, and angle oriented in the field.
Solution
The magnetic moment μ is calculated by the current times the area of the loop or
The torque and potential energy are calculated by identifying the magnetic moment, magnetic field, and the angle between these two vectors. The calculations of these quantities are:
Significance
The concept of magnetic moment at the atomic level is discussed in the next chapter. The concept of aligning the magnetic moment with the magnetic field is the functionality of devices like magnetic motors, whereby switching the external magnetic field results in a constant spinning of the loop as it tries to align with the field to minimize its potential energy.
Check Your Understanding 11.4
Check Your Understanding
In what orientation would a magnetic dipole have to be to produce (a) a maximum torque in a magnetic field? (b) A maximum energy of the dipole?
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Authors: Samuel J. Ling, William Moebs, Jeff Sanny
Publisher/website: OpenStax
Book title: University Physics Volume 2
Publication date: Oct 6, 2016
Location: Houston, Texas
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© Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. |
4105 | https://openstax.org/books/biology-2e/pages/11-1-the-process-of-meiosis | Skip to Content
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Biology 2e
11.1 The Process of Meiosis
Biology 2e
11.1
The Process of Meiosis
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Learning Objectives
By the end of this section, you will be able to do the following:
Describe the behavior of chromosomes during meiosis, and the differences between the first and second meiotic divisions
Describe the cellular events that take place during meiosis
Explain the differences between meiosis and mitosis
Explain the mechanisms within the meiotic process that produce genetic variation among the haploid gametes
Sexual reproduction requires the union of two specialized cells, called gametes, each of which contains one set of chromosomes. When gametes unite, they form a zygote, or fertilized egg that contains two sets of chromosomes. (Note: Cells that contain one set of chromosomes are called haploid; cells containing two sets of chromosomes are called diploid.) If the reproductive cycle is to continue for any sexually reproducing species, then the diploid cell must somehow reduce its number of chromosome sets to produce haploid gametes; otherwise, the number of chromosome sets will double with every future round of fertilization. Therefore, sexual reproduction requires a nuclear division that reduces the number of chromosome sets by half.
Most animals and plants and many unicellular organisms are diploid and therefore have two sets of chromosomes. In each somatic cell of the organism (all cells of a multicellular organism except the gametes or reproductive cells), the nucleus contains two copies of each chromosome, called homologous chromosomes. Homologous chromosomes are matched pairs containing the same genes in identical locations along their lengths. Diploid organisms inherit one copy of each homologous chromosome from each genetic contributor.
Meiosis is the nuclear division that forms haploid cells from diploid cells, and it employs many of the same cellular mechanisms as mitosis. However, as you have learned, mitosis produces daughter cells whose nuclei are genetically identical to the original parent nucleus. In mitosis, both the parent and the daughter nuclei are at the same “ploidy level”—diploid in the case of most multicellular animals. Plants use mitosis to grow as sporophytes, and to grow and produce eggs and sperm as gametophytes; so they use mitosis for both haploid and diploid cells (as well as for all other ploidies). In meiosis, the starting nucleus is always diploid and the daughter nuclei that result are haploid. To achieve this reduction in chromosome number, meiosis consists of one round of chromosome replication followed by two rounds of nuclear division. Because many events that occur during each of the division stages are analogous to the events of mitosis, the same stage names are assigned. However, because there are two rounds of division, the major process and the stages are designated with a “I” or a “II.” Thus, meiosis I is the first round of meiotic division and consists of prophase I, prometaphase I, and so on. Likewise, Meiosis II (during which the second round of meiotic division takes place) includes prophase II, prometaphase II, and so on.
Figure
11.2
Overview of Meiosis. The production of gametes is a crucial process for sexually reproducing organisms. Meiosis is the mechanism used to reduce diploid cells to haploid gametes while introducing genetic diversity. Prior to meiosis, chromosomes are replicated in S-phase to ensure proper number of chromosomes in the resulting gametes. During meiosis, two successive rounds of division reduces the number of chromosomes (ploidy) of the cell by half. going from diploid cells to haploid gametes. Credit: Rao, A., Tag, A, Fletcher, S., and Ryan, K. Department of Biology, Texas A&M University.
Meiosis I
Meiosis is preceded by an interphase consisting of G1, S, and G2 phases, which are nearly identical to the phases preceding mitosis. The G1 phase (the “first gap phase”) is focused on cell growth. During the S phase—the second phase of interphase—the cell copies or replicates the DNA of the chromosomes. Finally, in the G2 phase (the “second gap phase”) the cell undergoes the final preparations for meiosis.
During DNA duplication in the S phase, each chromosome is replicated to produce two identical copies—sister chromatids that are held together at the centromere by cohesin proteins, which hold the chromatids together until anaphase II. (Note: these chromosome copies are called sister chromatids regardless of whether they are in a female gamete or a male gamete.)
Prophase I
Early in prophase I, before the chromosomes can be seen clearly with a microscope, the homologous chromosomes are attached at their tips to the nuclear envelope by proteins. As the nuclear envelope begins to break down, the proteins associated with homologous chromosomes bring the pair closer together. Recall that in mitosis, homologous chromosomes do not pair together. The synaptonemal complex, a lattice of proteins between the homologous chromosomes, first forms at specific locations and then spreads outward to cover the entire length of the chromosomes. The tight pairing of the homologous chromosomes is called synapsis. In synapsis, the genes on the chromatids of the homologous chromosomes are aligned precisely with each other. The synaptonemal complex supports the exchange of chromosomal segments between homologous nonsister chromatids—a process called crossing over. Crossing over can be observed visually after the exchange as chiasmata (singular = chiasma) (Figure 11.3).
In humans, even though the X and Y sex chromosomes are not completely homologous (that is, most of their genes differ), there is a small region of homology that allows the X and Y chromosomes to pair up during prophase I. A partial synaptonemal complex develops only between the regions of homology.
Figure
11.3
Early in prophase I, homologous chromosomes come together to form a synapse. The chromosomes are bound tightly together and in perfect alignment by a protein lattice called a synaptonemal complex and by cohesin proteins at the centromere.
Located at intervals along the synaptonemal complex are large protein assemblies called recombination nodules. These assemblies mark the points of later chiasmata and mediate the multistep process of crossover—or genetic recombination—between the nonsister chromatids. Near the recombination nodule, the double-stranded DNA of each chromatid is cleaved, the cut ends are modified, and a new connection is made between the nonsister chromatids. As prophase I progresses, the synaptonemal complex begins to break down and the chromosomes begin to condense. When the synaptonemal complex is gone, the homologous chromosomes remain attached to each other at the centromere and at chiasmata. The chiasmata remain until anaphase I. The number of chiasmata varies according to the species and the length of the chromosome. There must be at least one chiasma per chromosome for proper separation of homologous chromosomes during meiosis I, but there may be as many as 25. Following crossover, the synaptonemal complex breaks down and the cohesin connection between homologous pairs is removed. At the end of prophase I, the pairs are held together only at the chiasmata (Figure 11.4). These pairs are called tetrads because a total of four sister chromatids of each pair of homologous chromosomes are now visible.
The crossover events are the first source of genetic variation in the nuclei produced by meiosis. A single crossover event between homologous nonsister chromatids leads to a reciprocal exchange of equivalent DNA between an egg-derived chromosome and a sperm-derived chromosome. When a recombinant sister chromatid is moved into a gamete cell it will carry a combination of maternal and paternal genes that did not exist before the crossover. Crossover events can occur almost anywhere along the length of the synapsed chromosomes. Different cells undergoing meiosis will therefore produce different recombinant chromatids, with varying combinations of maternal and parental genes. Multiple crossovers in an arm of the chromosome have the same effect, exchanging segments of DNA to produce genetically recombined chromosomes.
Figure
11.4
Crossover occurs between nonsister chromatids of homologous chromosomes. The result is an exchange of genetic material between homologous chromosomes (for clarity, only one recombination event between two of the four chromatids is represented).
Prometaphase I
The key event in prometaphase I is the attachment of the spindle fiber microtubules to the kinetochore proteins at the centromeres. Kinetochore proteins are multiprotein complexes that bind the centromeres of a chromosome to the microtubules of the mitotic spindle. Microtubules grow from microtubule-organizing centers (MTOCs). In animal cells, MTOCs are centrosomes located at opposite poles of the cell. The microtubules from each pole move toward the middle of the cell and attach to one of the kinetochores of the two fused homologous chromosomes. Each member of the homologous pair attaches to a microtubule extending from opposite poles of the cell so that in the next phase, the microtubules can pull the homologous pair apart. A spindle fiber that has attached to a kinetochore is called a kinetochore microtubule. At the end of prometaphase I, each tetrad is attached to microtubules from both poles, with one homologous chromosome facing each pole. The homologous chromosomes are still held together at the chiasmata. In addition, the nuclear membrane has broken down entirely.
Metaphase I
During metaphase I, the homologous chromosomes are arranged at the metaphase plate—roughly in the midline of the cell, with the kinetochores facing opposite poles. Each homologous pair is oriented randomly at the equator. For example, if the two homologous members of chromosome 1 are labeled a and b, then the chromosomes could line up a-b or b-a. This is important in determining the genes carried by a gamete, as each will only receive one of the two homologous chromosomes. (Recall that homologous chromosomes are not identical. They contain different versions of the same genes, and after recombination during crossing over, each gamete will have a unique genetic makeup that has never existed before.)
The randomness in the alignment of recombined chromosomes at the metaphase plate, coupled with the crossing over events between nonsister chromatids, are responsible for much of the genetic variation in the offspring. To clarify this further, remember that the homologous chromosomes of a sexually reproducing organism are originally inherited as two separate sets, one from each parent. Using humans as an example, one set of 23 chromosomes is present in the egg cell, often called maternal chromosomes because the genetic contributor is often the mother. The other set of 23 chromosomes is contained in the sperm, and the genetic contributor is called a father who provides the paternal chromosomes. Every cell of the multicellular offspring has copies of the original two sets of homologous chromosomes. When the offspring human creates their own gametes through meiosis, the two sets of chromosomes will be rearranged. In prophase I of meiosis, the homologous chromosomes form the tetrads. In metaphase I, these pairs line up at the midway point between the two poles of the cell to form the metaphase plate. Because there is an equal chance that a microtubule fiber will encounter a maternally or paternally inherited chromosome, the arrangement of the tetrads at the metaphase plate is random. Thus, any maternally inherited chromosome may face either pole. Likewise, any paternally inherited chromosome may also face either pole. The orientation of each tetrad is independent of the orientation of the other 22 tetrads.
This event—the random (or independent) assortment of homologous chromosomes at the metaphase plate—is the second mechanism that introduces variation into the gametes or spores. In each cell that undergoes meiosis, the arrangement of the tetrads is different. The number of variations is dependent on the number of chromosomes making up a set. There are two possibilities for orientation at the metaphase plate; the possible number of alignments therefore equals 2n in a diploid cell, where n is the number of chromosomes per haploid set. Humans have 23 chromosome pairs, which results in over eight million (223) possible genetically-distinct gametes just from the random alignment of chromosomes at the metaphase plate. This number does not include the variability that was previously produced by crossing over between the nonsister chromatids. Given these two mechanisms, it is highly unlikely that any two haploid cells resulting from meiosis will have the same genetic composition (Figure 11.5).
To summarize, meiosis I creates genetically diverse gametes in two ways. First, during prophase I, crossover events between the nonsister chromatids of each homologous pair of chromosomes generate recombinant chromatids with new combinations of maternal and paternal genes. Second, the random assortment of tetrads on the metaphase plate produces unique combinations of maternal and paternal chromosomes that will make their way into the gametes.
Figure
11.5
Random, independent assortment during metaphase I is demonstrated by considering a cell with a set of two chromosomes (n = 2).
There are two possible homologous chromosome arrangements at the equatorial plane in metaphase I, that are then separated during anaphase I.
The total possible number of different gametes is 2n, where n equals the number of chromosomes in a set. In this example, there are four possible
genetic combinations for the gametes. With n = 23 in human cells, there are over eight million possible combinations of paternal and maternal
chromosomes. Credit: Rao, A.and Fletcher, S. Department of Biology, Texas A&M University.
Anaphase I
In anaphase I, the microtubules pull the linked chromosomes apart. The sister chromatids remain tightly bound together at the centromere. The chiasmata are broken in anaphase I as the microtubules attached to the fused kinetochores pull the homologous chromosomes apart (Figure 11.6).
Telophase I and Cytokinesis
In telophase, the separated chromosomes arrive at opposite poles. The remainder of the typical telophase events may or may not occur, depending on the species. In some organisms, the chromosomes “decondense” and nuclear envelopes form around the separated sets of chromatids produced during telophase I. In other organisms, cytokinesis—the physical separation of the cytoplasmic components into two daughter cells—occurs without reformation of the nuclei. In nearly all species of animals and some fungi, cytokinesis separates the cell contents via a cleavage furrow (constriction of the actin ring that leads to cytoplasmic division). In plants, a cell plate is formed during cell cytokinesis by Golgi vesicles fusing at the metaphase plate. This cell plate will ultimately lead to the formation of cell walls that separate the two daughter cells.
Two haploid cells are the result of the first meiotic division of a diploid cell. The cells are haploid because at each pole, there is just one of each pair of the homologous chromosomes. Therefore, only one full set of the chromosomes is present. This is why the cells are considered haploid—there is only one chromosome set, even though each chromosome still consists of two sister chromatids. Recall that sister chromatids are merely duplicates of one of the two homologous chromosomes (except for changes that occurred during crossing over). In meiosis II, these two sister chromatids will separate, creating four haploid daughter cells.
Link to Learning
Watch this video about meiosis to review the process.
Meiosis II
In some species, cells enter a brief interphase, or interkinesis, before entering meiosis II. Interkinesis lacks an S phase, so chromosomes are not duplicated. The two cells produced in meiosis I go through the events of meiosis II in synchrony. During meiosis II, the sister chromatids within the two daughter cells separate, forming four new haploid gametes. The mechanics of meiosis II are similar to mitosis, except that each dividing cell has only one set of homologous chromosomes, each with two chromatids. Therefore, each cell has half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of chromosomal content, cells at the start of meiosis II are similar to haploid cells in G2, preparing to undergo mitosis.
Prophase II
If the chromosomes decondensed in telophase I, they condense again. If nuclear envelopes were formed, they fragment into vesicles. The MTOCs that were duplicated during interkinesis move away from each other toward opposite poles, and new spindles are formed.
Prometaphase II
The nuclear envelopes are completely broken down, and the spindle is fully formed. Each sister chromatid forms an individual kinetochore that attaches to microtubules from opposite poles.
Metaphase II
The sister chromatids are maximally condensed and aligned at the equator of the cell.
Anaphase II
The sister chromatids are pulled apart by the kinetochore microtubules and move toward opposite poles. Nonkinetochore microtubules elongate the cell.
Figure
11.6
The process of chromosome alignment differs between meiosis I and meiosis II. In prophase I, pairs of homologous chromosomes form
chiasmata which allow for crossing over events (genetic diversity). These pairs of homologous chromosomes arrange at the metaphase plate in metaphase I. In anaphase I, homologous chromosomes separate. Telophase I and cytokinesis result in haploid cells with 2 sister chromatids of each chromosome.
In prophase II, spindle microtubules form and elongate and any nuclear envelope disappears. Sister chromatids arrange at the midpoint of the cells in
metaphase II. In anaphase II, the sister chromatids separate. Telophase II and cytokinesis result in haploid cells with a single copy of each chromosome. Credit: Rao, A., Ryan, K., Fletcher, S. and Tag, A. Department of Biology, Texas A&M University.
Telophase II and Cytokinesis
The chromosomes arrive at opposite poles and begin to decondense. Nuclear envelopes form around the chromosomes. If the parent cell was diploid, as is most commonly the case, then cytokinesis now separates the two cells into four unique haploid cells. The cells produced are genetically unique because of the random assortment of paternal and maternal homologs and because of the recombination of maternal and paternal segments of chromosomes (with their sets of genes) that occurs during crossover. The entire process of meiosis is outlined in Figure 11.7.
Figure
11.7
An animal cell with a diploid number of four (2n = 4) proceeds through the stages of meiosis to form four haploid daughter cells.
Comparing Meiosis and Mitosis
Mitosis and meiosis are both forms of division of the nucleus in eukaryotic cells. They share some similarities, but also exhibit a number of distinct processes that lead to very different outcomes (Figure 11.8). Mitosis is a single nuclear division that results in two nuclei that are usually partitioned into two new cells. The nuclei resulting from a mitotic division are genetically identical to the original nucleus. They have the same number of sets of chromosomes: one set in the case of haploid cells and two sets in the case of diploid cells. In contrast, meiosis consists of two nuclear divisions resulting in four nuclei that are usually partitioned into four new, genetically distinct cells. The four nuclei produced during meiosis are not genetically identical, and they contain one chromosome set only. This is half the number of chromosome sets of the original cell, which is diploid.
The main differences between mitosis and meiosis occur in meiosis I, which is a very different nuclear division than mitosis. In meiosis I, the homologous chromosome pairs physically meet and are bound together with the synaptonemal complex. Following this, the chromosomes develop chiasmata and undergo crossover between nonsister chromatids. In the end, the chromosomes line up along the metaphase plate as tetrads—with kinetochore fibers from opposite spindle poles attached to each kinetochore of a homolog to form a tetrad. All of these events occur only in meiosis I.
When the chiasmata resolve and the tetrad is broken up with the homologous chromosomes moving to one pole or another, the ploidy level—the number of sets of chromosomes in each future nucleus—has been reduced from two to one. For this reason, meiosis I is referred to as a reductional division. There is no such reduction in ploidy level during mitosis.
Meiosis II is analogous to a mitotic division. In this case, the duplicated chromosomes (only one set of them) line up on the metaphase plate with divided kinetochores attached to kinetochore fibers from opposite poles. During anaphase II, as in mitotic anaphase, the kinetochores divide and one sister chromatid—now referred to as a chromosome—is pulled to one pole while the other sister chromatid is pulled to the other pole. If it were not for the fact that there had been crossover, the two products of each individual meiosis II division would be identical (as in mitosis). Instead, they are different because there has always been at least one crossover per chromosome. Meiosis II is not a reduction division because although there are fewer copies of the genome in the resulting cells, there is still one set of chromosomes, as there was at the end of meiosis I.
Figure
11.8
Meiosis and mitosis are both preceded by one cycle of DNA replication; however, meiosis includes two nuclear divisions. The four daughter cells resulting from meiosis are haploid and genetically distinct. The daughter cells resulting from mitosis are diploid and identical to the parent cell.
Evolution Connection
The Mystery of the Evolution of Meiosis
Some characteristics of organisms are so widespread and fundamental that it is sometimes difficult to remember that they evolved like other simple traits. Meiosis is such an extraordinarily complex series of cellular events that biologists have had trouble testing hypotheses concerning how it may have evolved. Although meiosis is inextricably entwined with sexual reproduction and its advantages and disadvantages, it is important to separate the questions of the evolution of meiosis and the evolution of sex, because early meiosis may have been advantageous for different reasons than it is now. Thinking outside the box and imagining what the early benefits from meiosis might have been is one approach to uncovering how it may have evolved.
Meiosis and mitosis share obvious cellular processes, and it makes sense that meiosis evolved from mitosis. The difficulty lies in the clear differences between meiosis I and mitosis. Adam Wilkins and Robin Holliday1 summarized the unique events that needed to occur for the evolution of meiosis from mitosis. These steps are homologous chromosome pairing and synapsis, crossover exchanges, sister chromatids remaining attached during anaphase, and suppression of DNA replication in interphase. They argue that the first step is the hardest and most important and that understanding how it evolved would make the evolutionary process clearer. They suggest genetic experiments that might shed light on the evolution of synapsis.
There are other approaches to understanding the evolution of meiosis in progress. Different forms of meiosis exist in single-celled protists. Some appear to be simpler or more “primitive” forms of meiosis. Comparing the meiotic divisions of different protists may shed light on the evolution of meiosis. Marilee Ramesh and colleagues2 compared the genes involved in meiosis in protists to understand when and where meiosis might have evolved. Although research is still ongoing, recent scholarship into meiosis in protists suggests that some aspects of meiosis may have evolved later than others. This kind of genetic comparison can tell us what aspects of meiosis are the oldest and what cellular processes they may have borrowed from in earlier cells.
Link to Learning
Click through the steps of this interactive animation to compare the meiotic process of cell division to that of mitosis at How Cells Divide.
Footnotes
1Adam S. Wilkins and Robin Holliday, “The Evolution of Meiosis from Mitosis,” Genetics 181 (2009): 3–12.
2Marilee A. Ramesh, Shehre-Banoo Malik and John M. Logsdon, Jr, “A Phylogenetic Inventory of Meiotic Genes: Evidence for Sex in Giardia and an Early Eukaryotic Origin of Meiosis,” Current Biology 15 (2005):185–91.
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GMAT Math: Weighted Averages
By
Mike MᶜGarry
— Last updated on
February 23, 2015
in
GMAT Word Problems
Problems range from easy to hard.
1) On a ferry, there are 50 cars and 10 trucks. The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg. What is the average mass of all 60 vehicles on the ferry?
(A) 1200 kg
(B) 1500 kg
(C) 1800 kg
(D) 2100 kg
(E) 2400 kg
2) In Plutarch Enterprises, 70% of the employees are marketers, 20% are engineers, and the rest are managers. Marketers make an average salary of $50,000 a year, and engineers make an average of $80,000. What is the average salary for managers if the average for all employees is also $80,000?
(A) $80,000
(B) $130,000
(C) $240,000
(D) $290,000
(E) $320,000
3) At Didymus Corporation, there are just two classes of employees: silver and gold. The average salary of gold employees is $56,000 higher than that of silver employees. If there are 120 silver employees and 160 gold employees, then the average salary for the company is how much higher than the average salary for the silver employees?
(A) $24,000
(B) $28,000
(C) $32,000
(D) $36,000
(E) $40,000
4) In a company of only 20 employees, 10 employees make $80,000/yr, 6 employees make $150,000/yr, and the 4 highest-paid employees all make the same amount. If the average annual salary for the 20 employees is $175,000/yr, then what is the annual salary of each highest-paid employee?
(A) $250,000
(B) $300,000
(C) $350,000
(D) $400,000
(E) $450,000
5) In a certain apartment building, there are one-bedroom and two-bedroom apartments. The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments. Let R be the average rental price for all apartments in the building. If R is $5,600 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $10,400 higher that R, then what percentage of apartments in the building are two-bedroom apartments?
(A) 26%
(B) 35%
(C) 39%
(D) 42%
(E) 52%
6) At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year?
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
Solutions will appear at the end of the article.
Weighted average
For the purposes of the GMAT, the weighted average situations occurs when we combine groups of different sizes and different group averages. For example, suppose in some parameter, suppose male employees in a company have one average score, the females have another average score, and gender non-binary/non-conforming employees (GNB/NC) have yet another average. If there were an equal number of men, women, and GNB/NC folks, we could just average the three separate gender averages: that would be ridiculously easy, which is precisely why the GMAT will never present you with groups of equal size in such a question. Instead, the number of male, female, and GNB/NC employees will be profoundly different, one significantly outnumbering the other, and then we will have to combine the individual gender averages to produce a total average for all employees. That’s a weighted average.
It is common for weighted average problems on the GMAT to have only two groups, but sometimes there are three (like the above problem), and conceivably, on a very hard problem, there could be four groups. Each group is a different size, each has its own average, and the job is to find the average of everyone all together. Or, perhaps the question will give us most of the info for the individual groups, and give us the total average for everyone, and then ask us to find the size or average of a particular group.
We have three basic approaches we can take to these question.
Approach I: averages & sums
Remember that, even with ordinary average questions, thinking in terms of the sum can often be helpful. We can’t add or subtract averages, but we can add or subtract sums! Right there is the key to one approach to the weighted average situation. If we calculate the sums for each separate group, we can simply add these sums to get the sum of the whole group. Alternately, if we know the size of the total group and the total average for everyone, we can figure out the total sum for everyone, and simply subtract the sums of the individual groups in order to find what we need.
Some weighted average problems give percents, not actual counts, of individual groups. In that case, we could simply pick a convenient number for the size of the population, and use the sums method from there. For example, if group A has 20% of the population, group B has 40% of the population, and group C also has 40% of the population, then we could just pretend that group A has one person, groups B & C each have two people, and total population has five people. From this, we could calculate all our sums.
This method always works, although is not always the most convenient in more advanced problems. This will be demonstrated in a few of the answers below.
Approach II: proportional & percentages
Sometimes the information about the sizes of individual groups is given, not in absolute counts of members, but simply in percents or proportions. In the problems above, question #2 simply gives percents, and question #5 asks for a percent. Yes, we could use Approach I, but there’s a faster way.
In Approach II, we simply multiply each group average by the percent of the population, expressed as a decimal, which that group occupies. When we add these products up, the sum is magically the total average for everyone. Suppose we have three groups, groups J and K and L which together constitute the entire population. Suppose this summarizes their separate information:
The percent have to be expressed in decimal form, so that:
Then, the total average is simply given by
This approach will be demonstrated in #2 below.
Approach III: proportional placement of total average
This final method can be hard to understand at first, but if you appreciate it, it is an incredibly fast time-saver. This approach only works if there are exactly two groups, no more.
Suppose there are two groups in a population, group 1 and group 2, and suppose that group 1 is bigger than group 2. Let’s also suppose that group 1 has a lower group average, and group 2 has the higher group average. Of course, the total combined average of the two groups together will be between the two individual group averages. In fact, because group 1 is bigger, the combined average will have to be closer to group 1’s average, and further away from group 2’s average.
Now, think about a number line, with the two individual group averages and the total averages indicated on the number line.
On that number line, I have labeled d1, the distance from the average of group #1 to the combined average, and d2, the distance from the combined average to the average of group 2. The ratio of these two distances is equal to a ratio of the size of the two individual groups. Let’s think about this. The bigger group, here group 1, will have more of an effect on the combined average and therefore will be closer to the combined average—a smaller distance. Therefore, the ratio of the distances must equal the reciprocal of the same ratio of the sizes of the groups:
Let’s say group 1 is 3 times larger than group 2. This means d2, the distance from group 2’s average to the combined average, would have to be 3 times bigger than d1, the distance from group 1’s average to the combined average. If the latter is x, then the former is 3x, and the total distance is 4x. If we know the averages of the two groups individually, we would simply have to divide the difference between those group averages by 4: the combined average would be one part away from group 1’s average, or three parts away from group 2’s average.
This approach is hard to explain clearly in words. You really have to see it demonstrated in the solutions below to understand it fully. Once you understand it, though, this is an extremely fast method to solve many problems.
Summary
If the above article gave you any insights, you may want to give the practice problems another look. Remember, in your practice problems on weighted averages, practice all three of these methods. The more ways you have to understand any problems, the more options you will have on test day!
Practice problem explanations
1) Method I: using sums
We will divide the two masses by 1000, 1.2 and 3 respectively, to simplify calculations. Note the use of the Doubling and Halving trick in the first multiplication.
sum for cars = 501.2 = 1000.6 = 60
sum for trucks = 103 = 30
total sum = 60 + 30 = 90
To find the total average, we need to divide this total sum by the total number of vehicles, 60.
Since we divided masses by 1000 earlier, we need to multiply by 1000 to get the answer. Total average = 1500 kg. Answer =(B)
Method II: proportional placement of the total average
Cars to trucks is 5:1, so if the distance between the car’s average and truck’s average were divided into 6 parts, the car’s average is 1 part away from the total average, and the truck’s average is 5 parts away.
Well, the difference in the two group averages is 3000 – 1200 = 1800 kg. Divide that by six: each “part” is 300 kg. Well, the total average must be 300 kg bigger than 1200 kg, or 5300 kg smaller than 3000 kg. Either way, that’s 1500 kg. Answer =(B)
2) We will approach this use the proportion & percents approach. Divide all dollar amounts by 1000 for smaller numbers. Multiply each group average by the percent expressed as a decimal:
marketers = 0.7050 = 35
engineers = 0.2080 = 16
managers = 0.10x = 0.1x
where x is the average salary for the managers. These three should add up to the average for all employees:
35 + 16 + 0.1x = 80
0.1x = 80 – 35 – 16
0.1x = 29
x = 290
Now, multiply the 1000 again, to get back to real dollar amounts. The average salary for managers is $290,000. Answer = (D)
3) Method I: using sums
We can use this if we pick a value for the average salary for the silver employees. It actually doesn’t matter what value we pick, because averages will fall in the same relative places regardless of whether all the individual values are slid up and down the number line. The easiest value by far to pick is zero. Let’s pretend that the silver folks make $0, and the gold folks make $56. (I divided dollars by 1000 for simplicity).
Now, we also have to simplify the numbers of employees. We could reduce the number of employees as long as we preserve the relative ratio.
silver : gold = 120:160 = 12:16 = 3:4
So everything would be the same if we just had 3 silver employees and 4 gold employees. OK, now find the sums.
silver = 30 = 0
gold = 456
I am not even going to bother to multiply that out, because we know that the next step is to divide by 7, the total number of employees.
total average = 456/7 = 48 = 32
The average salary is $32,000, which is $32,000 higher than the average for the silver employees. Answer = (C)
Method II: proportional placement of the total average
The ratio of silver employees to gold employees is
silver : gold = 120:160 = 12:16 = 3:4
If we divide the distance between the two averages by 7, then silver will be “four parts” away from the total average, and gold will be “three parts” away.
Well, the difference is $56,000, so that divided by 7 is $8000. That’s one part. Four parts would be $32,000, which has to be the distance from the silver average to the total average.
Answer = (C)
4) We will approach this using sums. The individual employee numbers are small. We will divide all dollar amounts by 1000, for easier calculations. Call the highest-paid employee salary x. Then the sums are
lowest = 1080 = 800
middle = 6150 = 3300 = 900
highest = 4x
Individual sums must add up to the total sum.
800 + 900 + 4x = 3500
4x = 1800
2x = 900
x = 450
The salary of each of those four highest paid employees is $450,000.
Answer = (E)
5) This question is designed for an analysis involving proportional placement of the mean. First, observe that R is much closer to the average for one-bedroom apartments, so there must be more one-bedroom apartments and fewer two-bedroom apartments.
The ratio of the distances to R is
5600:10400 = 56:104
Cancel a factor of 8 from both 56 and 104
56:104 = 7:13
One-bedroom apartments are “13 parts” of the building, and two-bedroom apartments are “7 parts.” That’s a total of 7 + 13 = 20 parts in the building. Two-bedroom apartments constitute 7/20 of the apartments in the building. Since 1/20 = 5%, 7/20 = 35%.
Answer = (B)
6) Method I: using sums
First, last year. Let x be the number of entrees. Then (15 – x) is the number of appetizers. The sums are:
entrees = 30x
appetizers = (15 – x)12 = 1215 – 12x = 630 – 12x = 180 – 12x
total = 1518 = 309 = 270
Notice the use of the Doubling and Halving trick in the second and third lines. The two individual sums should add up to the total sum.
30x + 180 – 12x = 270
18x = 90
x = 5
They start out with 5 entrees and 10 appetizers.
Let N be the number of appetizers added, so now there are 5 entrees and (10 + N) appetizers. We need to solve for N. Again, the sums:
entrees = 530 = 150
appetizers = (10 + N)12 = 120 + 12N
total = (15 + N)15 = 225 + 15N
Again, the two individual sums should add up to the total sum.
150 + 120 + 12N = 225 + 15N
270 = 225 + 3N
45 = 3N
15 = N
They added 15 more appetizers. Answer = (E)
Method I was do-able, but we had to solve for many values.
Method II: proportional placement of the total average
Originally, the entrée price was 30 – 18 = 12 from the total average, and the appetizer price was 18 – 12 = 6. This means there must have been twice as many appetizers as entrees. Therefore , with 15 items, there must have been 10 appetizers and 5 entrees.
The number of entrees doesn’t change. The average drops to $15, so the distance from the entrée price is now 30 – 15 = 15, and the distance from the appetizer price is now 15 – 12 = 3. That’s a 5-to-1 ratio, which means there must be 5x as many appetizers as entrees. Since there still are 5 entrees, there must now be 25 appetizers, so 15 have been added.
Answer = (E)
If you know how to employ this method, it is much more elegant.
Author
Mike MᶜGarry
Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as “member of the month” for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike’s Youtubevideo explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test.
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Tools & Reference>Clinical Procedures
Sentinel Lymph Node Biopsy in Patients With Melanoma
Updated: Oct 14, 2021
Author: Amy E Somerset, MD, ABOM; Chief Editor: Arlen D Meyers, MD, MBA more...;)
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Sentinel Lymph Node Biopsy in Patients With Melanoma
Sections Sentinel Lymph Node Biopsy in Patients With Melanoma
Overview
Practice Essentials
Background
Indications
Contraindications
Guidelines
Anatomic Considerations
Staging
Outcomes
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Periprocedural Care
Patient Education and Consent
Patient Preparation
Monitoring & Follow-up
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Technique
Overview
Post Procedure
Complications
Show All
Media Gallery;)
Tables)
References;)
Overview
Practice Essentials
The incidence of melanoma has risen in the last three decades, with an estimated 106,110 new diagnoses and 7,180 deaths in 2021. Survival rates are profoundly affected by the stage at time of diagnosis, which is determined by nodal involvement. Biopsy of sentinel lymph nodes (SLNs)the first lymph nodes involved in lymphatic spreadcan predict the potential for cancer in the local nodal basin or beyond. The histology of the sentinel node reflects the histology of the remaining lymphatic basin. Lymphatic mapping with SLN biopsy (SLNB) (see the image below) is associated with significantly fewer complications than regional lymphadenectomy and is recommended for patients at risk for metastasis. SLNB provides important information for both prognostic and treatment purposes.
Intraoperative left axillary sentinel lymph node seen after uptake with blue dye.
View Media Gallery)
Technical expertise is necessary for adequate identification of the sentinel node. Indications (see below) for SNLB should be evaluated individually, and a thorough discussion regarding the risks and benefits of performing SLNB should be undertaken with the patient.
Indications for SLNB
SLNB is designated for patients with significant risk of metastasis. Risk of metastasis is directly related to tumor thickness.
SLNB should not be offered for melanoma in situ, where cancer cells are confined to the epidermis.
SLNB should not be recommended routinely for thin melanomas that are T1a (nonulcerated lesions < 0.8 mm Breslow thickness), as the probability of positivity is less than 5%. However, if adverse features such as mitotic index greater than 2 mm2 (especially in younger patients) or lymphovascular invasion are present, the probability is increased to 5-10%, and SLNB may be considered.
SLNB should be considered for thin melanomas that are T1b (0.8 to 1 mm thickness or < 0.8 mm with ulceration).
SLNB should be recommended to patients with a clinically negative nodal basin and intermediate thickness primary melanomas (T2 or T3; >1 mm to 4 mm).
SLNB should be considered for thick melanomas (T4; >4 mm) and clinically negative nodes.
SLNB may be considered for melanoma that exhibits regression (controversial).
Potential contraindications for SLNB
SLNB is unnecessary if systemic disease is present.
SLNB may not be reliable if there has been previous extensive surgery in the targeted lymph node basin.
Fine-needle aspiration (FNA) or core needle biopsy of the node is preferable to SLNB as a first step when a clinically evident node is present.
There is debate about SLNB following a wide local excision; however, it may be acceptable if extensive reconstruction has not been performed.
Some suggest that repeat SLNB following a prior SLNB is inadvisable; however, no definitive recommendations exist in the melanoma literature. The authors do not consider prior SLNB a contraindication for repeat biopsy.
For patients who are deemed nonsurgical candidates owing to health conditions, or for patients who will not follow postbiopsy recommendations, an SNLB may be avoided.
Currently available genetic expression profiling testing should not be used to determine SLNB candidacy.
Staging
Clinical staging is based on physical examination and lesion biopsy. Clinical staging commonly involves the tumor, node, metastasis (TNM) staging system as developed by the American Joint Committee on Cancer (AJCC) in collaboration with the Union for International Cancer Control (UICC). This system incorporates the primary tumor (T), lymph nodes (N), and distant metastases (M). Tumor thickness (T) is recorded to the nearest 0.1 mm. Pathologic staging is based on pathologic examination of excised tissue. Tis refers to melanoma in situ.
Clinical staging is as follows:
Stage 0: TisN0M0
Stage IA: T1aN0M0
Stage IB: T1bN0M0; T2aN0M0
Stage IIA: T2bN0M0; T3aN0M0
Stage IIB: T3bN0M0; T4aN0M0
Stage IIC: T4bN0M0
Stage III: Any T, ¥N1, M0
Stage IV: Any T, Any N, M1
Pathologic staging is as follows:
Stage 0: TisN0M0
Stage IA: T1aN0M0
Stage IIA: T2bN0M0; T3aN0M0
Stage IB: T1bN0M0; T2aN0M0
Stage IIB: T3bN0M0; T4aN0M0
Stage IIC: T4bN0M0
Stage IIIA: T1a/b-T2a, N1a or N2a, M0
Stage IIIB: T0N1b or N1c, M0
Stage IIIB: T1a/b-T2a, N1b/c or N2b, M0
Stage IIIB: T2b/T3a, N1a-N2b, M0
Stage IIIC: T0, N2b, N2c, N3b or N3c
Stage IIIC: T1a-T3a, N2c or N3a/b/c, M0
Stage IIIC: T4b, N1a-N2c, M0
Stage IIID: T4b, N3a/b/c, M0
Stage IV: Any T, Any N, M1
Procedure
The main elements of SLNB are as follows:
Preoperative intradermal injection of a technetium Tc-99m radiocolloid around the primary melanoma followed by lymphoscintigraphy to image lymphatic drainage patterns
The injection may occur the day prior to the procedure, but generally requires a larger dose of radiocolloid
Transport to the operative suite and induction of anesthesia
Optional intradermal injection of approximately 1 mL of blue dye (isosulfan blue or methylene blue [preferred]) at the site of the primary melanoma followed by massage of the lesion for 4-5 minutes to enhance lymphatic drainage
Use of a handheld gamma probe to identify hot spots with high radiotracer activity (ie, sentinel nodes)
Placement of a small incision overlying the hot spot; incisions should be planned to allow for further dissection if this proves necessary
Use of a handheld gamma probe to identify hot nodes in the field and guide dissection; visual search for grossly blue nodes (guided by blue lymphatics) may assist if dye is injected
Removal of any nodes with significant radiotracer activity, followed by ex vivo measurement of their radioactivity
Collection of SLNs (defined as any nodes that are grossly suspicious, harbor blue dye, or have a radioactivity count >10% of the hottest node value) for histologic evaluation, including serial sectioning and appropriate staining
Continuation of dissection until the nodal bed count is 10% of that of the hottest node removed (awareness that after four negative nodes, subsequent nodes are very unlikely to be positive)
Wide local excision (WLE) of the primary melanoma; lymphatic mapping is generally performed prior to WLE, although it is acceptable to perform after
Postprocedure follow up is as follows:
All patients: At least annual skin examination for life; regular self-skin and lymph node examination; routine blood tests not recommended
Stage I-IIA: Physical examinations and medical history, with attention to skin and lymph nodes, every 6-12 months for 5 years, then annually; routine radiologic surveillance not recommended
Stage IIB-IIC: Physical examinations and medical history, with attention to skin and lymph nodes, every 3-6 months for 2 years, and then every 3-12 months for 3 years; consider imaging every 3-12 months for 2 years, then every 6-12 months for an additional 1-3 years
Stage IIB-IV: Examinations every 3-6 months for 2 years, then every 3-12 months for 3 years, then annually; consider imaging every 3-12 months to screen for recurrence or metastasis (not recommended after 3-5 years)
Stage III with positive sentinel node: Consider ultrasonography
Follow up is dictated by the stage of disease, as outlined by the National Comprehensive Cancer Network (NCCN) (Guidelines Version 2.2021).
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Background
Although it has been known since the time of Hippocrates, melanoma was not described in Western literature until recognized by John Hunter in 1787. It was not until 1968 that specimens were examined microscopically when Dr Bodenham confirmed that the black cancer described by Hunter was in fact melanoma. Melanoma is the most lethal type of skin cancer and continues to rise in incidence.
The American Cancer Societys estimates for melanoma include 106,110 new diagnoses in 2021, with 7,180 deaths. Melanoma is one of the fastest growing cancers and its incidence has continued to rise for the last few decades. It is also one of the most common malignancies in patients younger than 30 years, surpassing cervical cancer in young women.
Risk factors for development of melanoma include multiple clinically atypical moles or dysplastic nevi, family history, prior melanoma, inherited genetic mutations, and sun exposure. [3, 4] Although melanoma often arises in patients with fair skin and with significant sun exposure, it can affect any ethnicity and areas of the body with limited sun exposure.
The concept of lymphatic spread of cancer has been known for quite some time. Accordingly, it was reasoned that removing lymph nodes before the spread of melanoma could prevent systemic disease. In 1898, Snow published his work on the concept of anticipatory gland excision, or what would be later known as elective lymph node dissection (ELND). Although Snows reasoning was sound, multiple randomized controlled trials, short of subset analysis, showed no beneficial effect on survival or local control. [7, 8, 9] These trials were performed mainly on intermediate-depth melanomas (1-4 mm). In patients with thin melanomas (< 1 mm), the rate of nodal involvement seemed low enough that an ELND was not warranted. For thicker melanomas (>4 mm), the risk of systemic metastasis appeared to outweigh the potential benefit of ELND.
In 1955, Seaman and Powers used radiolabeled colloidal gold to describe the path of lymphatic channels followed by cancer cells. Subsequent studies by Gould et al described the first node involved in lymphatic spread, which was then termed the "sentinel lymph node" (SLN). Finally, in 1992, Morton et al showed that the SLN could accurately predict the potential for cancer in the associated nodal basin.
Although the complex anatomy and lymphatic drainage of the head and neck (see Anatomic Considerations) can make treatment of melanoma in this area quite challenging, SLN biopsy (SLNB) can be performed with acceptable morbidity. SLNB carries profound prognostic implications and dictates subsequent therapeutic decisions. The impact of further therapy in patients with a positive SLN (eg, completion lymph node dissection [CLND] and/or systemic therapy) remains an active area of investigation.
For more information, see Malignant Melanoma, Head and Neck Mucosal Melanomas, and Malignant Melanoma Staging.
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Indications
The National Comprehensive Cancer Network (NCCN) provides recommendations based on probability of sentinel lymph node (SLN) positivity. In patients with clinical stage I/II melanoma, SLN status is the most significant predictor of survival.
SNL biopsy (SLNB) is not recommended for patients with melanoma in situ (stage 0) or melanoma less than 0.8 mm in thickness in the absence of ulceration (T1a). Patients with melanomas less than or equal to 1 mm in thickness are at lower risk for nodal involvement (2-5%). For clinical stage IB, T1b melanoma (Breslow depth < 0.8 mm with ulceration or other adverse features including age < 40 years, lymphovascular invasion, significant vertical growth phase, increased mitotic rate), the probability of SLN positivity is increased to 5-10% . Thus, SNLB should be considered in these patients.
For patients with stage IB (T2a) or II (T2b and higher) melanoma, the probability of a positive SLN is generally greater than 10%. SLNB should be offered to these patients. For some patients (nonmitogenic, older patients) with stage IB (T2a) and stage II, the probability is actually < 10%. For these cases, SNLB may be considered but not necessarily offered.
SLNB should be offered to all patients with a clinically negative nodal basin and a primary melanoma greater than 1 mm in depth. As noted, evaluation of the nodal basins via elective lymph node dissection (ELND) was most commonly limited to melanoma with a depth of 1-4 mm. Biopsy of the sentinel node carries a lower risk of morbidity in comparison to a more extensive lymph node dissection. This has allowed for the expansion of the indications for evaluation of the nodal basins associated with melanoma, and SLNB should be routinely recommended for patients with 1-4 mm thick lesions (T2-T3).
SLNB may also be offered to patients with deep (>4 mm; T4) melanoma and clinically negative nodes, as it has proven to yield valuable prognostic information, with rates of positive SLNs ranging from 30-40%. [16, 17] The choice of adjuvant therapy or clinical trial enrollment for patients with very thick melanomas may be influenced by SLNB results.
Patients with melanoma exhibiting regression may also be considered for SLNB, although this is more controversial. Some studies have indicated that there may be no association between regression and positive SLN status. [18, 19, 20]
SLNB may still be performed following prior wide local excision, despite possible concern regarding accuracy.
For clinically or radiographically negative regional nodes and absence of distant metastases, SNLB may be considered for isolated in-transit metastasis or local recurrence of a primary melanoma.
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Contraindications
Sentinel lymph node biopsy (SLNB) is not always warranted. It may be deemed unnecessary when patients present with systemic disease, as it does not alter treatment and does not offer any survival advantage. If a patient presents with a clinically evident node, the first step should be evaluation of that node with fine-needle aspiration (FNA) or core needle biopsy as opposed to SLNB.
Whether SLNB should be performed after a wide local excision is somewhat controversial as there is concern regarding altered lymphatics. It appears, however, that if extensive reconstruction has not been performed, SLNB can be used in patients with a prior wide local excision. [21, 22]
For patients who have previously undergone SLNB, there is nothing in the melanoma literature to establish definitive recommendations. The breast cancer literature suggests that SLNB may be feasible, but whether such limited data can be applied to melanoma patients is a matter of debate. Nevertheless, given the reliability of lymphatic mapping for the identification of nodal pathways, the authors believe that prior SLNB is not a contraindication to repeat biopsy.
Lastly, SNLB should not be recommended to patients who are poor surgical candidates. For patients who will not follow postbiopsy imaging and treatment recommendations, it is reasonable to avoid SNLB.
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Guidelines
Guidelines from the American Society of Clinical Oncology and Society of Surgical Oncology are as follows :
Routine sentinel lymph node biopsy (SLNB) not recommended for T1a melanomas (nonulcerated lesions < 0.8 mm in Breslow thickness)
SLNB considered for T1b melanomas (0.8-1 mm Breslow thickness or < 0.8 mm Breslow thickness with ulceration) after a thorough discussion with patient about potential benefits and risks associated with the procedure
SLNB recommended for T2 and T3 melanomas (Breslow thickness of >1 mm to 4 mm)
SLNB may be recommended for T4 melanomas (>4 mm in Breslow thickness), after a discussion of potential benefits and risks
In the case of a positive SLNB, completion lymph node dissection (CLND) or careful observation are options for patients with low-risk micrometastatic disease, with due consideration of clinicopathologic factors; for higher-risk patients, careful observation may be considered only after a thorough discussion with patients about potential risks and benefits of foregoing CLND
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Anatomic Considerations
Anatomically following the lymphatic channels below the neck is relatively straightforward. The lymphatics follow the path toward either the ipsilateral axilla or groin in the upper and lower extremities, respectively. Clinicians should be aware that although the majority of sentinel lymph nodes (SLNs) are found in the major nodal basins, some can also be found in the epitrochlear nodal basin in the upper extremities and the popliteal nodal basin in the lower extremities.
In the truncal region, lymphatic pathways can travel to multiple nodal basins. This highlights the value of SLN mapping, which eliminates some of the guesswork for determination of melanoma spread. It is important, however, to keep in mind that nodes may be found anywhere along the line from the melanoma site to the nodal basin and that any of these nodes has the potential to harbor metastasis.
The lymphatic drainage of the head and neck is more extensive and variable than that of the trunk or extremities. The lymphatic drainage of the anterior neck is based on a system developed at Memorial Sloan-Kettering in the 1930s. Seven levels of lymph nodes exist (see the image below).
Lateral cervical lymph nodes I-VI. Level VII nodes are not shown.
View Media Gallery)
Levels I-V describe the lateral neck lymph node basins, while levels VI and VII describe the central lymph node basins, as follows:
Level I: Submandibular nodes, which lie between the mandible and the posterior and anterior bellies of the digastric muscle
Level II: Upper jugular nodes, which lie between the skull and the hyoid bone along the internal jugular vein
Level III: Middle jugular nodes, which lie between the hyoid bone and the omohyoid muscle along the internal jugular vein
Level IV: Lower jugular nodes, which lie between the omohyoid muscle and the clavicle along the internal jugular vein
Level V: Posterior triangle nodes, which lie between the trapezius and the sternocleidomastoid muscle above the clavicle
Level VI: Centrally located nodes between the carotid sheath and above the suprasternal notch
Level VII: Superior mediastinal nodes, which lie below the suprasternal notch
Traditionally, a line that bisected the ear caudad to cephalad marked the boundaries of the anterior and posterior drainage of the scalp (see the image below). Lesions anterior to this line would drain to the parotid, submandibular, and cervical lymph nodes (levels I-V). Lesions posterior to this line would drain to the posterior auricular, suboccipital, upper jugular, and posterior triangle lymph nodes. It was assumed that lymphatic drainage would not cross the midline.
Drainage patterns of the scalp.
View Media Gallery)
The advent of lymphoscintigraphy, however, demonstrated that these assumptions are not always accurate. In 1994, Wells et al found that lymphoscintigraphy was discordant with clinical predictions of nodal drainage 84% of the time. Even after revision of the clinical parameters, a study by OBrien et al the following year still found a discordance rate of 33%. Uren also reported discordance rates of 33% and demonstrated nodal basins draining across midline in 10% of cases.
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Staging
The staging of melanoma has undergone multiple changes. Common staging is according to the tumor, node, metastasis (TNM) system of the American Joint Committee on Cancer (AJCC) in collaboration with the Union for International Cancer Control (UICC). Changes were made to the 8th edition, which are included in the Table below. Stage groupings are as above.
Table. AJCC TNM Classification of Melanoma (Open Table in a new window))
| | | |
---
| T | Thickness, mm | Ulceration Status/Mitosis |
| Tis | n/a | n/a |
| T1 | < 0.8 0.8-1 | a: Without ulceration b: < 0.8 with ulceration or 0.8-1 with or without ulceration |
| T2 | >1-2 | a: Without ulceration b: With ulceration |
| T3 | >2-4 | a: Without ulceration b: With ulceration |
| T4 | >4 | a: Without ulceration b: With ulceration |
| N | No. of Metastatic Nodes or Microsatellite, Satellite, or In-transit Metastasis (MSI) | Nodal Metastatic Burden |
| NX N0 | Regional lymph nodes not assessed 0 | No lymph node spread |
| N1 | 1 | a: 1 clinically occult, no MSI b: 1 clinically detected, no MSI c: No regional lymph node disease, but MSI present |
| N2 | 2-3 | a: 2-3 clinically occult, no MSI b: 2-3 with at least 1 clinically detected c: 1 clinically occult or clinically detected and MSI present |
| N3 | ¥4 | a: ¥4 clinically occult, no MSI b: ¥4 at least 1 clinically detected or any number of nodes stuck together, no MSI c: ¥2 clinically occult or clinically detected and MSI present |
| M | Site | Lactate Dehydrogenase Level |
| M0 | No distant metastases | n/a |
| M1a | Distant skin, soft tissue, and/or nonregional lymph node | M1a(0) Not elevated M1a(1) Elevated |
| M1b | Lung metastasis with or without M1a sites of disease | M1b(0) Not elevated M1b(1) Elevated |
| M1c | Non-CNS visceral metastases with or without M1a or M1b sites of disease | M1c(0) Not elevated M1c(1) Elevated |
| M1d | Distant to CNS with or without M1a, M1b, or M1c sites of disease | M1d(0) Not elevated M1d(1) Elevated |
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Outcomes
A positive sentinel lymph node (SLN) carries extremely valuable prognostic information. Survival rates for patients with a negative SLN have been shown to be superior to those with a positive SLN. In addition, the method of detection has been shown to be a predictor of survival as well. SLNs that are positive only on polymerase chain reaction (PCR) are associated with a better survival rate than those detected with hematoxylin and eosin (H&E) staining (see the image below).
Survival rates of negative sentinel lymph node versus detection by polymerase chain reaction (PCR) and detection by PCR and H&E.
View Media Gallery)
In patients with a positive SLN, previous recommendations were to perform completion lymph node dissection (CLND) and treat with adjuvant interferonthis was the first adjuvant therapy to show benefit in patients with positive nodes. There have since been major advances in immunotherapy for melanoma, and monoclonal antibodies like ipilimumab have shown equivalent efficacy. More recent studies show protein kinase (PK) inhibitors may be superior as adjuvant therapy and for stage IV disease. There has been identification of several drug targets for the treatment of stage IV disease, with newer drugs inhibiting BRAF, MEK, NRAS, and KIT. Significant clinical benefit has been seen for patients with BRAF mutations, although the most superior systemic therapy has not been established.
There is no question that SLN biopsy (SLNB) provides tremendous prognostic information but the actual degree of sentinel node disease that would benefit from treatment is not known. Some studies have shown that a small volume of disease within the SLN is rarely associated with disease in other lymph nodes. Some suggest that even a small amount of disease is associated with recurrence and decreased survival compared with patients with no SLN disease. The MSLT-II trial confirmed prior studies showing that with current treatment paradigms, while immediate CLND reduces regional recurrence slightly (ie, improves regional control), it does not increase melanoma-specific survival. Thus, current NCCN Guidelines indicate both nodal basin surveillance and CLND as primary treatment options in the setting of a microscopically positive SLN, with adjuvant systemic therapy versus observation. The threshold of disease at which patients truly benefit from further surgical or immunologic treatment has not been determined.
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Periprocedure
References
American Cancer Society. Key Statistics for Melanoma Skin Cancer. Cancer.org. Available at January 12, 2021; Accessed: October 10, 2021.
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Ribuffo D, Gradilone A, Vonella M, Chiummariello S, Cigna E, Haliassos N. Prognostic significance of reverse transcriptase-polymerase chain reaction-negative sentinel nodes in malignant melanoma. Ann Surg Oncol. 2003 May. 10(4):396-402. [QxMD MEDLINE Link].
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Scolyer RA, Thompson JF, McCarthy SW, Gershenwald JE, Ross MI, Cochran AJ. Intraoperative frozen-section evaluation can reduce accuracy of pathologic assessment of sentinel nodes in melanoma patients. J Am Coll Surg. 2005 Nov. 201(5):821-3; author reply 823-4. [QxMD MEDLINE Link].
Koopal SA, Tiebosch AT, Albertus Piers D, Plukker JT, Schraffordt Koops H, Hoekstra HJ. Frozen section analysis of sentinel lymph nodes in melanoma patients. Cancer. 2000 Oct 15. 89(8):1720-5. [QxMD MEDLINE Link].
Tanis PJ, Boom RP, Koops HS, Faneyte IF, Peterse JL, Nieweg OE, et al. Frozen section investigation of the sentinel node in malignant melanoma and breast cancer. Ann Surg Oncol. 2001 Apr. 8(3):222-6. [QxMD MEDLINE Link].
Alkhatib W, Hertzenberg C, Jewell W, Al-Kasspooles MF, Damjanov I, Cohen MS. Utility of frozen-section analysis of sentinel lymph node biopsy specimens for melanoma in surgical decision making. Am J Surg. 2008 Dec. 196(6):827-32; discussion 832-3. [QxMD MEDLINE Link].[Full Text].
Ariyan S, Ariyan C, Farber LR, Fischer DS, Flynn SD, Truini C. Reliability of identification of 655 sentinel lymph nodes in 263 consecutive patients with malignant melanoma. J Am Coll Surg. 2004 Jun. 198(6):924-32. [QxMD MEDLINE Link].
Gipponi M, Solari N, Lionetto R, Di Somma C, Villa G, Schenone F. The prognostic role of the sentinel lymph node in clinically node-negative patients with cutaneous melanoma: experience of the Genoa group. Eur J Surg Oncol. 2005 Dec. 31(10):1191-7. [QxMD MEDLINE Link].
Eicher SA, Clayman GL, Myers JN, Gillenwater AM. A prospective study of intraoperative lymphatic mapping for head and neck cutaneous melanoma. Arch Otolaryngol Head Neck Surg. 2002 Mar. 128(3):241-6. [QxMD MEDLINE Link].
Schmalbach CE, Nussenbaum B, Rees RS, Schwartz J, Johnson TM, Bradford CR. Reliability of sentinel lymph node mapping with biopsy for head and neck cutaneous melanoma. Arch Otolaryngol Head Neck Surg. 2003 Jan. 129(1):61-5. [QxMD MEDLINE Link].
Wells KE, Rapaport DP, Cruse CW, Payne W, Albertini J, Berman C, et al. Sentinel lymph node biopsy in melanoma of the head and neck. Plast Reconstr Surg. 1997 Sep. 100(3):591-4. [QxMD MEDLINE Link].
Jacobs IA, Chevinsky AH, Swayne LC, Magidson JG, Britto EJ, Smith TJ. Gamma probe-directed lymphatic mapping and sentinel lymphadenectomy in primary melanoma: Reliability of the procedure and analysis of failures after long-term follow-up. J Surg Oncol. 2001 Jul. 77(3):157-64. [QxMD MEDLINE Link].
Carlson GW, Page AJ, Cohen C, Parker D, Yaar R, Li A, et al. Regional recurrence after negative sentinel lymph node biopsy for melanoma. Ann Surg. 2008 Sep. 248(3):378-86. [QxMD MEDLINE Link].
Media Gallery
Lateral cervical lymph nodes I-VI. Level VII nodes are not shown.
Drainage patterns of the scalp.
Lymphoscintigraphy of a forehead melanoma with uptake in the parotid and submandibular nodes.
Survival rates of negative sentinel lymph node versus detection by polymerase chain reaction (PCR) and detection by PCR and H&E.
Intraoperative left axillary sentinel lymph node seen after uptake with blue dye.
Confirmation of the sentinel node based on a high radioactive count.
of 6
Tables
Table. AJCC TNM Classification of Melanoma;)
Table. AJCC TNM Classification of Melanoma
| | | |
---
| T | Thickness, mm | Ulceration Status/Mitosis |
| Tis | n/a | n/a |
| T1 | < 0.8 0.8-1 | a: Without ulceration b: < 0.8 with ulceration or 0.8-1 with or without ulceration |
| T2 | >1-2 | a: Without ulceration b: With ulceration |
| T3 | >2-4 | a: Without ulceration b: With ulceration |
| T4 | >4 | a: Without ulceration b: With ulceration |
| N | No. of Metastatic Nodes or Microsatellite, Satellite, or In-transit Metastasis (MSI) | Nodal Metastatic Burden |
| NX N0 | Regional lymph nodes not assessed 0 | No lymph node spread |
| N1 | 1 | a: 1 clinically occult, no MSI b: 1 clinically detected, no MSI c: No regional lymph node disease, but MSI present |
| N2 | 2-3 | a: 2-3 clinically occult, no MSI b: 2-3 with at least 1 clinically detected c: 1 clinically occult or clinically detected and MSI present |
| N3 | ¥4 | a: ¥4 clinically occult, no MSI b: ¥4 at least 1 clinically detected or any number of nodes stuck together, no MSI c: ¥2 clinically occult or clinically detected and MSI present |
| M | Site | Lactate Dehydrogenase Level |
| M0 | No distant metastases | n/a |
| M1a | Distant skin, soft tissue, and/or nonregional lymph node | M1a(0) Not elevated M1a(1) Elevated |
| M1b | Lung metastasis with or without M1a sites of disease | M1b(0) Not elevated M1b(1) Elevated |
| M1c | Non-CNS visceral metastases with or without M1a or M1b sites of disease | M1c(0) Not elevated M1c(1) Elevated |
| M1d | Distant to CNS with or without M1a, M1b, or M1c sites of disease | M1d(0) Not elevated M1d(1) Elevated |
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Contributor Information and Disclosures
Author
Amy E Somerset, MD, ABOM General and Bariatric Surgeon, Harper Bariatric Medicine Institute, General and Bariatric Surgery of Grosse Pointe; Clinical Assistant Professor of Surgery, Wayne State University School of MedicineAmy E Somerset, MD, ABOM is a member of the following medical societies: American College of Surgeons, American Society for Metabolic and Bariatric Surgery, Society of American Gastrointestinal and Endoscopic SurgeonsDisclosure: Nothing to disclose.
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Mark J Jameson, MD, PhD, FACS Medical Director, Avera Medical Group ENT-Head and Neck Surgery, Avera McKennan Hospital and University Health CenterMark J Jameson, MD, PhD, FACS is a member of the following medical societies: American Academy of Otolaryngology-Head and Neck Surgery, American Association for Cancer Research, American College of Surgeons, American Head and Neck SocietyDisclosure: Serve(d) as a speaker or a member of a speakers bureau for: Castle Biosciences.
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Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment.
Nader Sadeghi, MD, FRCSC Professor and Chairman, Department of Otolaryngology-Head and Neck Surgery, McGill University Faculty of Medicine; Chief Otolaryngologist, MUHC; Director, McGill Head and Neck Cancer Program, Royal Victoria Hospital, CanadaNader Sadeghi, MD, FRCSC is a member of the following medical societies: American Academy of Otolaryngology-Head and Neck Surgery, American Head and Neck Society, American Thyroid Association, Royal College of Physicians and Surgeons of CanadaDisclosure: Nothing to disclose.
Chief Editor
Arlen D Meyers, MD, MBA Emeritus Professor of Otolaryngology, Dentistry, and Engineering, University of Colorado School of MedicineArlen D Meyers, MD, MBA is a member of the following medical societies: American Academy of Facial Plastic and Reconstructive Surgery, American Academy of Otolaryngology-Head and Neck Surgery, American Head and Neck SocietyDisclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: Cerescan; Neosoma; MI10;Invitrocaptal,Medtechsyndicates
Received income in an amount equal to or greater than $250 from: Neosoma; Cyberionix (CYBX);MI10;Invitrocaptal;MTS
Received ownership interest from Cerescan for consulting for: Neosoma, MI10 advisor.
Additional Contributors
Mimi S Kokoska, MD Physician, Department of Otolaryngology-Head and Neck Surgery, Aurora Health CareMimi S Kokoska, MD is a member of the following medical societies: American Academy of Facial Plastic and Reconstructive Surgery, American Academy of Otolaryngology-Head and Neck Surgery, American Association for Physician Leadership, American College of Surgeons, American Head and Neck SocietyDisclosure: Nothing to disclose.
Richard D Keidan, MD Attending Surgeon, Director, Melanoma Clinic, William Beaumont HospitalRichard D Keidan, MD is a member of the following medical societies: American College of Surgeons, Association for Academic Surgery, Society of Surgical OncologyDisclosure: Nothing to disclose.
Philip Fong, MD Resident Physician, Department of Surgery, William Beaumont HospitalPhilip Fong, MD is a member of the following medical societies: American College of SurgeonsDisclosure: Nothing to disclose.
William L Kestenberg, MD Surgeon, Kestenberg Surgical GroupWilliam L Kestenberg, MD is a member of the following medical societies: American College of SurgeonsDisclosure: Nothing to disclose.
Thabet Abbarah, MD, FACS Consulting Staff, Department of Otolaryngology, North Oakland Medical Centers
Thabet Abbarah, MD, FACS is a member of the following medical societies: American College of Surgeons
Disclosure: Nothing to disclose.
Dhananjay A Chitale, MBBS Senior Staff Pathologist, Division Head, Molecular Pathology and Genomic Medicine, Director, Tissue Biorepository, Henry Ford Hospital
Dhananjay A Chitale, MBBS is a member of the following medical societies: American Society for Clinical Pathology and College of American Pathologists
Disclosure: Nothing to disclose.
Amit Dwivedi, MD Assistant Professor of Surgery, University of Louisville School of Medicine
Amit Dwivedi, MD is a member of the following medical societies: American College of Surgeons and American Medical Association
Disclosure: Nothing to disclose.
Venkata Subramanian Kanthimathinathan, MD Staff Physician, Department of General Surgery, Loma Linda University Medical Center
Disclosure: Nothing to disclose.
Nafisa K Kuwajerwala, MD Staff Surgeon, Breast Care Center, William Beaumont Hospital
Nafisa K Kuwajerwala, MD is a member of the following medical societies: American College of Surgeons, American Society of Breast Disease, and American Society of Breast Surgeons
Disclosure: Nothing to disclose.
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4108 | https://www.ncbi.nlm.nih.gov/books/NBK606103/ | Urinary Crystals Identification and Analysis - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Urinary Crystals Identification and Analysis
Prathap Kumar Simhadri; Preeti Rout; Stephen W. Leslie.
Author Information and Affiliations
Authors
Prathap Kumar Simhadri 1; Preeti Rout 2; Stephen W. Leslie 3.
Affiliations
1 Advent Health
2 Wilson Case Western University
3 Creighton University School of Medicine
Last Update: July 27, 2024.
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Continuing Education Activity
Urinary crystal formation is a marker of urinary supersaturation with different substances and can occur secondary to inherited diseases, metabolic disorders, or drugs. Urine sediment examination by microscopy is an excellent tool for clinicians to understand underlying pathology,although it is often not performed.
Different crystals have recognizable appearances and patterns. The presence of crystals can be a clue to underlying disease pathologies. Commonly associated conditions include hyperoxaluria, hypercalciuria, hyperuricemia, metabolic syndrome, and drug-related crystalluria. Proper identification of urinary crystals can aid in selecting therapeutic interventions, from dietary changes to prescribing or discontinuing medications. Given the significant prevalence of nephrolithiasis, proper diagnosis and treatment of risk factors can decrease patient morbidity.
This activity reviews the methods used in urine sediment examination and the appropriate identification of various urinary crystals. Participants learn to recognize different types of urine crystals, understand their clinical significance, and apply this knowledge to improve patient outcomes. This activity also highlights the role of the interprofessional team in identifying and interpreting urine crystals, leading to more accurate diagnoses and targeted and effective treatment plans for patients with nephrolithiasis and other urinary crystal–associated diseases.
Objectives:
Differentiate the various types of urinary crystals observed in urine sediment and their characteristic appearances.
Screen patients for metabolic disorders and drug-related causes based on the presence and type of urinary crystals.
Apply knowledge of urinary crystal pathogenesis to inform and adjust therapeutic interventions.
Coordinate follow-up care and monitoring for patients diagnosed with conditions associated with urinary crystals to prevent recurrence and manage complications.
Access free multiple choice questions on this topic.
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Introduction
Urinary crystal formation is a marker of urinary supersaturation with various substances. Crystals can occur secondary to inherited diseases, metabolic disorders, and medications. The presence of crystals in the urine is not always pathognomic for an abnormal metabolic or renal condition, as uric acid, calcium oxalate, calcium phosphate (CaP), and drug-induced crystals can also be observed under normal physiological conditions. The diagnosis of various monogenetic diseases, such as cystinuria, primary hyperoxaluria, and adenine phosphoribosyl transferase deficiency, can be suggested by examining the urinary sediment and correct identification of the associated urinary crystals.
Urinary crystal formation always precedes renal stone formation but does not always lead to nephrolithiasis. When kidney stones are formed, they have the potential to cause acute kidney injury, obstructive uropathy, hydronephrosis, renal colic, pyonephrosis, and urosepsis. Urinary crystals can also cause renal crystal deposition disease, resulting in acute kidney injury, progressive chronic kidney disease, and end-stage kidney disease.
Urine sediment examination can be considered a liquid biopsy of the kidney because it provides valuable information about the events occurring in an individual's kidney. However, with the increased utilization of automated systems, urine sediment examination is becoming a lost art. Nonetheless, it remains an excellent source of clinical information to help appropriately diagnose and manage patients with various diseases associated with urinary crystal formation. Please see StatPearls' companion resource, "Calcium Deposition and Other Renal Crystal Diseases," for more information.
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Etiology and Epidemiology
Metabolic factors, inherited diseases, medications, dietary habits, dehydration, and anatomical abnormalities are the common etiologies contributing to urinary crystal formation. Kidney stone formation is a known risk factor for developing diabetes, cardiovascular disease, fractures, and chronic renal disease. Conversely, these conditions also increase the risk of nephrolithiasis.
Kidney stone disease is a polygenetic and multifactorial health problem worldwide that affects between 1% and 13% of the global population and almost 11% of the Western population.Over the past 30 years, the prevalence of kidney stones has been rising across all ages, genders, racial and ethnic groups globally. Approximately 10% of the United States population experience nephrolithiasis in their lifetime.
Estimating the overall prevalence of urinary crystalluria is challenging, as this condition is underdiagnosed and underreported.Although scant literature is available on the prevalence of urinary crystalluria, several studies showed crystalluria in approximately 8% of urine collections.Calcium oxalate, uric acid, and amorphous urate crystals accounted for most of the crystals encountered. Brushite, ammonium biurate, and cystine crystals were rare.
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Pathophysiology
Urine contains both promoters and inhibitors of crystal formation. Crystallization occurs when the concentration of crystal promoters in the glomerular filtrate exceeds the kidney's ability to keep them as soluble molecules. Stone inhibitors include citrate; uromodulin, also known as Tamm-Horsfall protein; magnesium; glycosaminoglycans; and pyrophosphate. Some chemical mediators decrease the urinary crystal's growth, aggregation, and adhesion to the tubular epithelium and the tubular cell; examples are osteopontin, matrix Gla protein, Tamm-Horsfall protein, and urinary fragment 1 of prothrombin.
Calcium, oxalate, urate, and phosphate ions are the leading promoters of crystal formation. Physiological factors influencing kidney stone formation include high urinary osmolarity or low urine volume, bacteria in the urinary tract, and urine pH. Structural factors influencing nephrolithiasis include urinary reflux, instrumentation, and genetic abnormalities of the urinary tract. Please see StatPearls' companion resource, "Renal Calculi, Nephrolithiasis," for more information.
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Specimen Requirements and Procedure
A first-morning urine sample (void after awakening) is ideal for crystal analysis as it tends to be the most concentrated for crystal formation. The urine sample should be brought to the lab and processed within 2 hours of collection.Uric acid and phosphorous crystallization can occur if the urine stays stagnant for longer than 2 hours. Refrigeration of the urine specimen is necessary after 2 hours, and examination within 8 hours of collection is recommended, even when refrigerated.
Urine analysis through automated and manual methods can provide significant clues to the type and characteristics of urinary crystals. When examining the sediment, urine pH, specific gravity, and a dipstick analysis should always be recorded. Of note, casts degrade quickly in alkaline urine. Red blood cells (RBCs) and white blood cells (WBCs) can swell in low osmolality and shrink in high osmolality.
Urine pH plays a key role in crystallization, as CaP, struvite, and ciprofloxacin crystals precipitate in alkaline urine. In contrast, uric acid, cystine, and other drug-induced crystals tend to precipitate in acidic urine. These characteristics assist clinicians in accurately identifying urinary crystals.
For microscopic examination, 10 mL of urine should be centrifuged at 1500 to 3000 revolutions per minute for at least 5 min. The sediment is obtained through a pipette after discarding the top 9.5 cc of supernatant and gentle manual agitation of the test tube. A single drop of urine sediment is placed on the glass slide and covered with a cover slip for direct observation.The sediment is examined using a standard brightfield or phase contrast microscope at low and high magnification. A minimum of 10 fields should be reviewed at each power.The specimen is then examined under polarization to determine birefringence and identify other detailed characteristics.
Automated urinary analyzers also often miss crystals when compared to manual microscopic evaluation.Off note, a recent study found significant interoperator variability identifying urinary sediment components, even among experienced nephrologists. Therefore, when possible, a personal examination of urine sediment is preferable.
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Diagnostic Tests
A comprehensive workup is essential for diagnosing the etiology of urinary crystals. This approach includes obtaining a detailed history and performing a thorough physical examination, and may also include x-ray KUB, renal or abdominal ultrasound, and plain computed tomography (CT) of the abdomen and pelvis to help localize the renal stones. Simple urinalysis and analysis of the urinary crystals can help understand a stone's chemical composition. A chemical analysis of a passed stone can give a definitive answer.
Crystalluria indicates supersaturation of the components in the urine, leading to precipitation of the crystals. However, supersaturation alone is insufficient to cause nephrolithiasis; other inciting factors are also typically present. The process can be related to metabolic disorders, inherited diseases, drug use, or toxins. The initial diagnostic test is generally a 24-hour urine collection evaluating the following parameters—urinary volume, pH, calcium, citrate, creatinine, magnesium, phosphate, uric acid, sodium, serum calcium, oxalate, and uric acid levels. The collection should ideally be performed in nonacute conditions at the patient's home with their normal diet and fluid intake for optimal information.Additional workup is typically guided by the type of crystal identified and patient history. Please see StatPearls' companion resources, "24-Hour Urine Collection," and "24-Hour Urine Testing for Nephrolithiasis: Interpretation and Treatment Guidelines,"for more information.
Further Diagnostic Testing Based on Stone Composition
Calcium oxalate:Calcium oxalate stones comprise up to 80% of all nephrolithiasis cases. Calcium oxalate and CaP often coexist in stones, but 1 component is typically primary. Patients with calcium oxalate crystals should be considered for primary and secondary causes of hyperoxaluria. Please see StatPearls' companion resource, "Calcium Deposition and Other Renal Crystal Diseases," for more information.Calcium oxalate dihydrate (COD) crystals are also highly associated with hypercalcemia, which should also be considered. Calcium oxalate crystalluria is not necessarily pathologic, as these crystals can sometimes be found in patients with high oxalate diets.
Hyperoxaluria of >75 mg/d in a 24-hour urine collection suggests possible hereditary hyperoxaluria (types I, II, or III).
Calcium phosphate:CaP nephrolithiasis is the second most common type, representing about 10% to 20% of kidney stones. The presence of these stones should prompt evaluation for conditions causing hypercalciuria, hyperphosphaturia, and alkaline urine formation. These crystals can also be found under normal physiological conditions.
Struvite:Struvite comprises ammonium, magnesium, and calcium binding with phosphate (triple phosphate). The presence of struvite stones, even in lower quantities, is almost always associated with infection by urease-producing microorganisms, such as Proteus mirabilis,Klebsiella pneumonia,Staphylococcus aureus,Pseudomonas aeruginosa, _Providencia_ stuartii,Serratia,and Morganella morganii.
Uric acid:Uric acid dihydrate crystals primarily depend on low urine pH, whereas the amorphous ones are mainly related to high urinary urate concentration.Uric acid crystals are only found when the urinary pH is acidic, as uric acid becomes much more soluble as the pH increases.The presence of uric acid dihydrate crystals should prompt the workup for conditions causing impaired renal ammoniogenesis, such as metabolic syndrome and type 2 diabetes.
Xanthine:Xanthine crystals are found in xanthinuria, a rare genetic disorder with autosomal recessive pattern inheritance caused by the deficiency of the enzyme xanthine dehydrogenase. Not all laboratories can test for xanthine and hypoxanthine, nor are these part of a routine urine collection. If this rare disorder is suspected, this test must be ordered specifically. Biopsy of the gastrointestinal tract to quantify xanthine dehydrogenase is rarely performed.An allopurinol loading test can also be used to detect a lack of xanthine oxidase, as allopurinol is abnormally metabolized in patients with hereditary xanthinuria; however, these measurements may not be available in all laboratories. Xanthinuria is also associated with hypouricemia and hypouricosuria.
Patients with 2,8-dihydroxyadenine (DHA) crystalluria, cystinuria, and suspected primary hyperoxaluria can also be genetically tested to confirm the diagnosis.
Please see StatPearls' companion resources, "Calcium Deposition and Other Renal Crystal Diseases," "Hyperoxaluria," "Uric Acid Nephrolithiasis,""Struvite and Triple Phosphate Renal Calculi,"and "Cystinuria,"for more information.
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Testing Procedures
Effective testing procedures for urine crystal identification and analysis are crucial for accurate diagnosis and patient care. Techniques commonly used for crystal analysis in clinical practice include the following:
Standard light microscopy
Optical polarization microscopy
Automated microscopy
Flow cytometry
Digitalized microscopy
Scanning electron microscopy
Spectroscopy
Standard Light Microscopy
Manual direct microscopic examination is the gold standard technique used in clinics to identify various urinary crystals based on their morphology.When sufficiently trained and equipped, nephrologists can acquire important diagnostic information through direct microscopy. Brightfield microscopy is more commonly used compared to phase contrast microscopy, but phase contrast microscopy may be more beneficial for analyzing urine sediment.
Optical Polarization Microscopy
Polarized light microscopy is used to observe the color, morphology, and birefringence of crystals for identification. The color, refraction of the light, and double refraction are the parameters used in mineral identification. However, contaminants can sometimes distort or interfere with the interpretations of results.
Automated Microscopy
Automated microscopy is a method used in laboratory practice that does not require urine centrifugation or standardized microscope measuring parameters. These machines are based on the Colter principle and became popular in hematology laboratory procedures in 1954 for the rapid and accurate enumeration of cells in the blood. They have been used for urine microscopy since the 1990s. Automated microscopy has been expanded to laser-based flow cytometry for better cell characterization.
This technique can utilize a large sample size, reducing the laboratories' workload. Automated microscopy can deliver a complete urine analysis result if urinary biochemical data are combined and integrated with the microscopy analysis. The practical approach is to use automated and manual machines to survey many specimens and expert microscopic examination for selected specimens. Automated systems have significant advantages over manual microscopy, as the specimens do not need centrifugation. They are also highly effective in diagnostic screening for blood, bacteria, and WBCs. However, automated systems significantly underperform in recognizing urinary casts and crystals compared to manual examinations performed by experienced lab technicians, pathologists, and nephrologists. The 2 major types of automated microscopy techniques are as follows:
Flow cytometry:
These machines use laser-based flow cytometry with fluorescent dyes. Flow cytometry can reduce the need for microscopy or culture and is more useful in screening bacteriuria, pyuria, and hematuria, although sometimes microscopy is still required.
This machine produces scattergrams, not actual images. Microscopy is required to differentiate similar particles, such as various types of crystals, dysmorphic or isomorphic RBCs, and different types of epithelial cells. The advantage of this machine is that minimum expertise is required to operate it, and approximately 100 samples can be tested in 1 hour.
Digitalized microscopy:
The digitalized microscopy technique uses computerized software to analyze digitalized monochrome images of urine particles. Two different instruments that supply digitized images are available. One is based on automated intelligent microscopy, which shows particles in the samples by categories such as RBCs and WBCs. The other uses cuvette-based microscopy, which displays particles in the same way they appear in the microscopic field during manual examination.
A significant advantage of this technique is that it can be used for many samples and selective individual specimens. The images can be stored on a computer and sent to different sites, including other laboratories and clinics, for review by pathologists or nephrologists. In addition, this method is relatively cheaper compared to flow cytometry.
Scanning Electron Microscopy
Scanning electron microscopy produces high-resolution images of the crystal surface, aiding in studying the morphology and texture of urinary calculi.This method can analyze the characteristics of stones ranging from 1 mm to 5 mm without changing the morphology of their components,and it does not destroy the stone sample. This method is also a proposed tool for diagnosing primary hyperoxaluria.
Spectroscopy
Spectroscopy identifies the molecular structure of a sample by investigating its interaction with electromagnetic radiation.In this technique, infrared light passes through the sample, and the detector analyzes the intensity of the transmitted light. Vibrational spectroscopy identifies a sample's composition and molecular structure through its vibrational characteristics when interacting with a beam source. This technique is analyzed based on the sample's vibrational characteristics when interacting with electromagnetic waves. Other spectroscopy methods, such as Raman and Fourier transform-infrared spectroscopy,are increasingly used in clinical practice.
These rapid, nondestructive, and relatively inexpensive methods have been increasingly used in recent years to identify the chemical nature of urinary crystals and detect organic matter.Infrared spectroscopy is extremely useful in identifying calcium oxalate crystals.
Overall, clinicians should be able to identify most crystals by checking urine pH and urinalysis, then examining the urine sediment under bright field, phase contrast, and polarizing light microscopy. Fourier transform-infrared and Raman spectroscopy and solubility testing can be additional resources for unclear crystal composition.
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Results, Reporting, and Critical Findings
The presence of urinary crystals associated with active urinary sediment, acute kidney injury, or kidney stones suggests a pathogenic process. Identifying and characterizing urinary crystals are essential for diagnosing the type of stone, especially when the actual stone is not available for analysis (see Table 1. Characteristics of Commonly Observed Urinary Crystals). The results of urinary crystal analysis are reported after identifying various characteristics of urinalysis and microscopic examination of the urine sediment.
Commonly Reported Urinary Crystals
Calcium oxalate crystals:There are 2 significant types of calcium oxalate crystals—monohydrate and dihydrate. The calcium oxalate monohydrate (COM) crystals are typically colorless on light microscopy. Their shape varies from oval, biconvex, and dumbbells to elongated rods (see Image.Calcium Oxalate Monohydrate Crystals). These crystals are strongly birefringent on polarized microscopy.
COD crystals are colorless on light microscopy and are bipyramidal or envelope-shaped, and their size is variable (see Image. Calcium Oxalate Dihydrate Crystals). These crystals are non-birefringent on polarization. Calcium oxalate crystallization is commonly observed in the urine pH range of 5.0 to 6.5 and is rarely observed when the urine pH increases to >7.0. A rare third form of calcium oxalate crystals, calcium oxalate trihydrate or caoxite, is hexagonal and seldom found in urine.
Calcium phosphate crystals:There are calcium orthophosphate crystals, mainly amorphous carbonated CaP and carbapatite, whereas brushite is dicalcium phosphate dihydrate.
CaP crystals are colorless on light microscopy. Calcium orthophosphate crystals are amorphous crystals that appear as small granulations with or without large plates and have no polarization (seeImage.Amorphous Phosphate and Triple Phosphate Crystals in the Urine). Brushite crystals have a broad spectrum of shapes,such as prisms, rosettes, starts, needle-like or splinter-like, or sticks or rods (see Image. Calcium Phosphate Crystals). These crystals are strongly birefringent on polarized microscopy and precipitate in alkaline urine.
Struvite or triple phosphate crystals:These are radioopaque magnesium ammonium phosphate hexahydrate or triple phosphate crystals. Struvite crystals are colorless under light microscopy. The most common shape observed is a coffin lid; the other shapes are feather-like structures, elongated prisms, and trapezoids. These crystals are weak to strongly birefringent on polarization and precipitate in alkaline urine (see Images. Triple Phosphate Crystals; and Amorphous Phosphate and Triple Phosphate Crystals in the Urine).
Cystine crystals:Cystine crystals are colorless on light microscopy and are either found alone or heaped onto each other,resembling perfect hexagonal plates. These crystals show weak birefringence on polarized microscopy and precipitate in urine with an acidic pH (see Image. Cystine Crystals in the Urinary Sediment).Please see StatPearls' companion resource,"Cystinuria,"for more information.
Uric acid and urate crystals:There are 4 types of uric acid crystals in urine—amorphous, anhydrous uric acid (uricite), monohydrate, and dihydrate crystals. Among these, uric acid dihydrate and amorphous uric acid crystals are most commonly observed.
Uric acid crystals are often amber-colored on light microscopy and have various shapes, such as rho boards, barrels, rosettes, 4-sided plates, needles, rounds, or parallelograms. These crystals show a very strong polychromatic birefringence on polarization (seeImages. Uric Acid Crystals, Barrel-Shaped; Uric Acid Crystals, Rosette-Shaped; and Uric Acid Crystals, Birefringent on Polarized Light).These crystals precipitate in urine with an acidic pH.
Xanthine crystals:Xanthine crystals are nonspecific in appearance and can be observed as polarized granules or sticks without a characteristic aspect. Crystal identification requires infrared spectroscopy, high-pressure liquid chromatography, or chemical analysis of a xanthine stone. These crystals precipitate in urine with an acidic pH.
Dihydroxyadenine crystals:DHA crystals are somewhat brown or reddish-brown with a dark perimeter and central spicules that are round in shape on light microscopy. These crystals turn yellow and may demonstrate a characteristic birefringent Maltese-cross pattern on polarized microscopy. In suspected 2,8-DHA crystalluria, it is mandatory to investigate the crystals using infrared spectroscopy on a centrifuged pellet. These crystals precipitate in urine with an acidic or alkaline pH (4.8 to 9.0) and are relatively insoluble at any urinary pH.
Table
Table 1. Characteristics of Commonly Observed Urinary Crystals.
Drug-Induced Crystals
Drug-induced urinary crystals are variable in size, shape, and other characteristics depending on the specific medication (seeTable 2. Characteristics of Various Drug-Induced Crystals). These rare crystals comprise only 1% to 2% of all nephrolithiasis (seeImages.Sulfamethoxazole Crystals;Sulfadiazine Crystals; and Acyclovir Crystals).
Table
Table 2. Characteristics of Various Drug-Induced Crystals.
Rare Crystals
Tyrosine crystals are colorless, thin, and needle-shaped, whereas leucine crystals are oval or round with circular striations. Both these crystals are found in acidic urine and are associated with aminoaciduria secondary to liver disease. Bilirubin crystals, which are clumped needles or yellow granules, are highly birefringent on polarization, and they are observed in conditions associated with hyperbilirubinemia. Cholesterol crystals are thin, transparent plates with cut edges. They are highly birefringent on polarization, typically observed in association with nephrotic syndrome.
Solubility Testing
The solubility test involves exposing the crystals to 10% KOH, 30% acetic acid, 30% HCL, and chloroform. The solubility of crystals in these fluids provides additional clues that aid in crystal identification (see Table 3. Solubility Characteristics of Crystals). This test is beneficial in identifying crystals with similar shapes.
Table
Table 3. Solubility Characteristics of Crystals.
Interfering Factors
Identifying urinary crystals can be challenging due to similarities in their shapes. Calcium sulfate, CaP, and sodium urate crystals can be needle-shaped or rectangle-shaped with blunt ends. COM, hippuric acid, uric acid, and ammonium biurate crystals can be rectangles with pointed ends. Amorphous urate and phosphate granules can have similar minor granular appearances. Solubility tests can help differentiate urinary crystals with similar shapes. Key considerations include:
Free hemosiderin granules are yellow-brown coarse granules that can be confused with amorphous phosphate and urate granules.
2,8-DHA crystals can be confused with ammonium hydrogen urate as both have a small spherical shape.
Starch granules are oval or round and highly refractive. Their irregular central indentation can resemble molar tooth grooves.
Under polarized light,starch granules can appear similar to a Maltese cross and be confused with 2,8-DHA crystals.
Cholesterol and leucine crystals can also exhibit a typical Maltese-cross appearance under polarized light.
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Clinical Significance
Renal stone formation is always preceded by crystalluria, although the presence of urinary crystals does not always result in nephrolithiasis. Investigating crystalluria is crucial for diagnosing and managing acute and chronic renal failure due to crystal precipitations associated with various inherited and acquired diseases. Identifying the type of urinary crystals is also helpful in offering specific targeted treatment for nephrolithiasis.
Calcium Oxalate Crystals
Calcium oxalate stones comprise up to 80% of all nephrolithiasis. Patients with calcium oxalate crystals should be considered for primary and secondary causes of hyperoxaluria. Please see StatPearls' companion resource, "Calcium Deposition and Other Renal Crystal Diseases," for more information.Calcium oxalate crystalluria is not necessarily pathologic, as these crystals can sometimes be found in those with high oxalate diets.As an overview, calcium monohydrate crystals are generally associated with high urinary oxalate levels, calcium dihydrate crystals are generally associated with high urinary calcium, and the rare dodecahedral calcium dihydrate crystals are associated with very high urinary calcium.
Secondary hyperoxaluria is much more common compared to primary hyperoxaluria and is often related to 1 of 3 mechanisms as follows:
Increased dietary intake of oxalate-containing foods, such as peanuts, cashews, almonds, spinach, rhubarb, star fruit, chaga mushroom powder, and black iced tea.
Increased intestinal absorption of oxalate in a setting of fat malabsorption (enteric hyperoxaluria secondary to intestinal bypass surgery, short gut syndrome, Crohn disease, celiac disease, chronic pancreatitis, and orlistat).
Increased intake of oxalate precursors, such as ethylene glycol and vitamin C, can develop increased gut oxalate absorption, leading to excess urinary oxalate excretion. These patients are also at high risk for developing calcium oxalate crystalluria, crystalline nephropathy, chronic kidney disease, and end-stage kidney disease. Ingestion of the toxic substance, ethylene glycol, results in severe hyperoxaluria, and a very peculiar kind of monohydrate crystals with elongated and narrow hexagonal shapes similar to diamonds are found in the urine.
COD crystals are typically associated with hypercalciuria, such as hyperparathyroidism; hyperthyroidism; multiple myeloma; paraneoplastic syndrome; bone metastasis; vitamin D toxicity; particular drug use, including loop diuretics and vitamin-D supplements; and milk-alkali syndrome.
According to Daudon et al, over 94% of urine specimens containing calcium oxalate crystals with a calcium-to-oxalate ratio of <5 show either monohydrate crystals alone or along with dihydrate crystals. In contrast, over 90% of urine specimens containing calcium oxalate crystals with a calcium-to-oxalate ratio >14 contain dihydrate crystals alone. Dodecahedral COD crystals typically indicate severe hypercalciuria, >7 mmol/L. All types of calcium-containing crystals are radioopaque.
Calcium Phosphate Crystals
Carbapatite crystals are typically observed in patients with hypercalciuria on a background urine pH of 5.8 to 6.5. In contrast, amorphous CaP crystals are typically observed in patients with near-normal calcium and phosphate concentrations with a pH >6.6.
Amorphous CaP crystals suggest a possible environment created by urinary tract infections or metabolic disorders keeping the urine alkaline. When associated with the calcium monohydrate crystals, they indicate the presence of medullary sponge kidneys and other causes related to urinary stasis. When carbapatite crystals are associated with struvite crystals, they indicate the presence of urea-splitting organisms causing urinary tract infections.
Brushite crystals are made of dicalcium phosphate dihydrate ions. These crystals are mainly observed in conditions associated with marked hypercalciuria and hyperphosphaturia and are generally favored by low concentrations of citrates. The presence of >500/mL brushite crystals should prompt an evaluation for primary hyperparathyroidism. CaP crystals are radioopaque.
Struvite Crystals
Struvite crystals are associated with high ammonium content and precipitate in alkaline urine. These crystals are always associated with urinary tract infections by urea-splitting bacteria, where the bacteria hydrolyze the urea, producing ammonia and leading to alkaline urine. These bacteria are Klebsiella, Pseudomonas, Proteus, Providencia, and some species of Staphylococcus. These are radioopaque, sometimes large enough to be called staghorn stones.
Cystine Crystals
Cystine crystals are associated with hereditary cystinuria,in which a proximal tubular dysfunction leads to increased excretion of dibasic amino acids such as cystine, ornithine, lysine, and arginine. Cystinuria is an autosomal recessive Mendelian disorder, and there are reports of acquired cystinuria post-renal transplant. Cystine forms radiolucent stones, which are not detected on x-rays but can be detected by ultrasound if the stones are large enough. CT scan is the gold standard for diagnosing renal stones and can detect cystine stones. Cystine becomes more soluble as the pH increases. The therapeutic urinary pH level to maximize solubility is ≥7.5, typically higher compared to uric acid.
Uric Acid Crystals
Uric acid crystals are associated with medical conditions leading to increased uric acid production or altered uric acid metabolism, such as gout, Lesch-Nyhan syndrome, tumor lysis, chemotherapy, or conditions associated with decreased urine pH, such as chronic diarrhea, metabolic syndrome, and diabetes mellitus. However, over half of patients with uric acid stones do not have hyperuricemia. These crystals form radiolucent renal stones that are not detectable on x-rays but can be visualized on CT and ultrasound.
Xanthine Stones
Xanthine crystals are found in xanthinuria, a rare genetic disorder with autosomal recessive pattern inheritance caused by the deficiency of the enzyme xanthine dehydrogenase. This disorder leads to xanthine accumulation and highly insoluble xanthine excretion. Xanthinuria is associated with hypouricemia and hypouricosuria.Rarely, xanthinuria can occur related to allopurinol use for Lesch-Nyan syndrome or tumor lysis syndrome.More recently, the xanthine oxidase inhibitor, febuxostat, has also been noted to cause xanthinuria when used for tumor lysis syndrome.
2,8-Dihydroxyadenine Crystals
2,8-DHA crystalluria is secondary to adenine phosphoribosyltransferase (APRT) deficiency, an autosomal recessive condition leading to excess DHA production, 2,8-DHA crystalluria, and kidney stones. The condition is underdiagnosed in the United States, with fewer than 300 cases reported in the medical literature. The worldwide prevalence is 1/5000 to 1/100,000 population. A significant number of patients remain asymptomatic, whereas 90% of patients develop urinary crystals and radiolucent renal stones. In addition, 10% of patients with 2,8-DHA crystalluria progress to end-stage renal failure. The diagnosis is confirmed by kidney stone analysis, histologic findings of crystal nephropathy, genetic testing, or lack of APRT activity in RBC lysate. Treatment is recommended with allopurinol or febuxostat, which blocks the conversion of adenosine to DHA. The dosage should be titrated to eliminate 2,8-DHA crystalluria.
Drug-Induced Crystals
Drug-induced crystals are formed in association with the use of various drugs, as previously described. These crystals precipitate in the urine when used at a high dose, for long durations, or both. They form radiolucent renal stones when they contain no calcium or other radioopaque constituents. Patients experiencing signs and symptoms of nephrolithiasis while using these drugs should prompt the search for renal stones by ultrasound or contrast-enhanced CT imaging for diagnosis. Analyzing these crystals is crucial for identifying at-risk patients and clarifying the drugs or medications involved.
Follow-UpMonitoring
The complete disappearance of crystalluria is the best indicator of improved lithogenic activity in CaP, uric acid, and other common forms of nephrolithiasis. However, high-risk patients, such as those with cystinuria and PH1, can have active urinary crystallization even when the disease is under reasonable control.
Global crystal volume (GCV) helps determine lithogenicity in patients with crystalluria. GCV is the product of the number of crystals/mm 3 multiplied by the average size of crystals (in μm) and by a numerical factor given for each crystal species (the numerical number is provided based on the geometrical form of a crystal). GCV of <500 μm 3/mm 3 is associated with the lowest risk for tubular obstruction and crystalline nephropathy in patients with PH1. In patients with cystinuria, a GCV of >3000 μm 3/mm 3 is associated with an increased risk for recurrent nephrolithiasis.
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Enhancing Healthcare Team Outcomes
Clinical outcomes can be significantly improved by emphasizing the importance of urine sediment examination and ensuring that medical students, residents, physician assistants, nurse practitioners, laboratory technicians, and other healthcare professionals receive comprehensive training on its routine application.
The preanalytical phase is the most vulnerable part of the evaluation, accounting for up to 75% of all laboratory errors. Outcomes can be improved by following appropriate protocol-based patient preparation, obtaining mid-stream urine samples, rapid sample transportation, and preparation techniques. Nursing, lab processing, and transporting staff are crucial in these collection steps. Physicians must often collect samples shortly before examination for the optimal specimens. Effective communication among team members is paramount. Conscientious laboratory staff are paramount in the accurate processing and calibration of microscopic and analytic equipment. Clear and open communication facilitates rapid diagnosis and treatment decisions, preventing errors and ensuring a coordinated interprofessional team response.
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Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
Figure
Calcium Oxalate Monohydrate Crystals. The calcium oxalate monohydrate crystals are typically colorless on light microscopy, with shapes that vary from oval and biconvex to dumbbell and elongated rod forms. Contributed by R Naik, MD, PhD
Figure
Calcium Oxalate Dihydrate Crystals. The calcium oxalate dihydrate crystals are colorless on light microscopy and have a bipyramidal or envelope shape. Contributed by R Naik, MD, PhD
Figure
Calcium Phosphate Crystals. Calcium phosphate crystals are colorless on light microscopy. Brushite crystals have a broad spectrum of shapes, such as prisms, rosettes, starts, needle-like or splinter-like, sticks, or rods. Contributed by R Naik, (more...)
Figure
Triple Phosphate Crystals. Struvite crystals are colorless under light microscopy and often appear in distinctive coffin-lid shapes. Other shapes, such as feather-like structures, elongated prisms, and trapezoids, can also be observed. Contributed by (more...)
Figure
Amorphous Phosphate and Triple Phosphate Crystals in the Urine. Calcium orthophosphate crystals are amorphous crystals that appear as small granulations with or without large plates and have no polarization. Contributed by M Zubair, MBBS, FCPS
Figure
Uric Acid Crystals, Barrel-Shaped. The image shows barrel-shaped uric acid crystals on light microscopy. Contributed by R Naik, MD, PhD
Figure
Uric Acid Crystals, Rosette-Shaped. The image shows rosette-shaped uric acid crystals on light microscopy. Iqbal Osman, Public Domain, via Wikimedia Commons
Figure
Uric Acid Crystals, Birefringent on Polarized Light. Uric acid crystals show a strong polychromatic birefringence on polarization. Contributed by S Bhimji, MD
Figure
Sulfamethoxazole Crystals. Urine sediment without staining ×400. Left: bright field light microscopy;right: polarized light microscopy. Seltzer J, Velez JC, Buchkremer F, Poloni JAT. Urine sediment of the month: drugs & crystalluria. (more...)
Figure
Sulfadiazine Crystals. Unstained urine sediment ×100. Left: bright field light microscopy;right: polarized light microscopy. Seltzer J, Velez JC, Buchkremer F, Poloni JAT. Urine sediment of the month: drugs & crystalluria . Renal (more...)
Figure
Acyclovir Crystals. Left: light microscopy unstained;right: polarized light with birefringence. Seltzer J, Velez JC, Buchkremer F, Poloni JAT. Urine sediment of the month: drugs & crystalluria. Renal Fellow Network Web site. (more...)
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Disclosure:Prathap Kumar Simhadri declares no relevant financial relationships with ineligible companies.
Disclosure:Preeti Rout declares no relevant financial relationships with ineligible companies.
Disclosure:Stephen Leslie declares no relevant financial relationships with ineligible companies.
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4109 | https://eprints.lse.ac.uk/84520/7/Swenson_Legal%20Pluralism_Published.pdf | Geoffrey Swenson Legal pluralism in theory and practice Article (Published version) (Refereed) Original citation: Swenson, Geoffrey (2018) Legal pluralism in theory and practice. International Studies Review, 20 (2). p. 342. ISSN 1521-9488 DOI: 10.1093/isr/vix060 © 2018 International Studies Association CC BY 4.0 This version available at: Available in LSE Research Online: July 2018 LSE has developed LSE Research Online so that users may access research output of the School. Copyright © and Moral Rights for the papers on this site are retained by the individual authors and/or other copyright owners. Users may download and/or print one copy of any article(s) in LSE Research Online to facilitate their private study or for non-commercial research. You may not engage in further distribution of the material or use it for any profit-making activities or any commercial gain. You may freely distribute the URL ( of the LSE Research Online website. International Studies Review (2018) 0, 1–25 ANALYTICAL Legal Pluralism in Theory and Practice GEOFFREY SWENSON London School of Economics and Political Science Legal pluralism has vast policy and governance implications. In develop-ing countries, for instance, non-state justice systems often handle most dis-putes and retain substantial autonomy and authority. Legal pluralism’s im-portance, however, is rarely recognized and dramatically under theorized.
This article advances scholarly understanding of legal pluralism both theo-retically and empirically. It proposes a new typological framework for con-ceptualizing legal pluralism through four distinct archetypes – combative, competitive, cooperative, and complementary – to help clarify the range of relationships between state and non-state actors. It posits five main strate-gies used by domestic and international actors in attempts to influence the relationship between state and non-state justice systems: bridging, har-monization, incorporation, subsidization, and repression. As post-conflict situations are fluid and can feature a wide range of relationships between state and non-state actors, they are particularly instructive for showing how legal pluralism archetypes can be shifted over time. Case studies from Timor-Leste and Afghanistan highlight that selecting an appropriate pol-icy is vital for achieving sustainable positive outcomes. Strategies that rely on large scale spending or even the use of substantial military force in isolation are unlikely to be successful. The most promising approaches are culturally intelligible and constructively engage non-state justice net-works of authority and legitimacy to collectively advance the judicial state-building process. While the case studies focus on post-conflict states, the theory presented can help understand and improve efforts to promote the rule of law as well as good governance and development more broadly in all legally pluralist settings.
Keywords: rule of law, post-conflict, legal pluralism Mainstream international relations theory tends to assume a unitary state with a monopoly on the use of legitimate violence domestically; however, the reality of po-litical and legal authority is far more complicated (Weber 1978).1 Legal pluralism whereby “two or more legal systems coexist in the same social field” is the domi-nant feature of most legal orders worldwide (Merry 1988, 870). All states feature legal pluralism, and only a limited number of high capacity states have nonstate jus-tice actors firmly under their control. Even in these states, legal pluralism thrives through alternative dispute resolution mechanisms, arbitration agreements, and 1As Lake highlights, “the state is central to the study of international relations and will remain so for the foreseeable future” because it is “fundamental” to neorealist, neoliberal institutionalism, and constructivist approaches, and usually plays a key role in “critical, post-modern, and feminist theories” (Lake 2008, 41).
Swenson, Geoffrey. (2018) Legal Pluralism in Theory and Practice. International Studies Review, doi: 10.1093/isr/vix060 © International Studies Association 2018. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.
Downloaded from by London School of Economics user on 09 July 2018 2 Legal Pluralism in Theory and Practice international treaty obligations. In the developing world, an estimated 80 to 90 per-cent of disputes are handled outside of the state justice system (Albrecht and Kyed 2010, 1). The role of legal pluralism is particularly vital in conflict and postconflict settings, as they tend to have weak state institutions and contested governing au-thority (Fearon and Laitin 2004). In states with lower levels of capacity, legitimacy, or both, seeking support from nonstate actors can serve as a conflict avoidance tactic or even a broader governance strategy that attempts to secure buy-in from powerful groups that may be skeptical of the state.2 While the prevalence and endurance of nonstate justice mechanisms could be seen as an indictment of the need for state justice to underpin the rule of law, non-state justice mechanisms often have significant negative externalities. Nonstate legal orders frequently reflect cultural or religious norms unconcerned with basic human rights. Women and other vulnerable groups are particularly at risk when nonstate legal systems embrace overtly patriarchal ideals. These systems can also reflect sig-nificant bias toward powerful individuals and families, and the legal processes often lack core protections, such as procedural and substantive due process norms. As Waldorf highlights, nonstate ““judicial” elites are neither independent nor impar-tial, and their discretionary rulings serve community harmony not individualized justice” (Waldorf 2006, 10). Furthermore, the relationship between state and non-state justice is often unclear, and cases may be resolved in different ways, encourag-ing forum shopping by parties, particularly those with more economic or political clout. The state system’s predominance in itself does not guarantee a just outcome at a systemic level, as it could be a means for more effective despotism (Krygier 2011, 19–21). Nevertheless, an effective state-operated justice system is a functional prerequisite for a state capable of fulfilling the rule-of-law’s requirements of being prospective, generalized, clear, fixed, and equally applied (Tamanaha 2007, 3).
While central governments prefer a monopoly on legal authority, state power is frequently contested even after violence ends. Domestically, the state’s ability to function justly and effectively is a life or death matter for millions of people world-wide and dramatically impacts the quality of life for millions more. Internationally, “poorly governed societies can generate conflicts that spill across international bor-ders” as well as facilitate criminal networks and transnational violent extremism (Krasner 2004, 86). Establishing a viable state justice sector is vital to the overall success or failure of state-building efforts (Paris 2004, 205–6).3 As nonstate justice is the dominant form of legal order, it is important to understand its implications.
State-building provides a powerful analytical lens to examine and understand the implications of legal pluralism, as situations tend to be fluid with a wide range of re-alistic outcomes. Little attention, however, has been paid to the unique challenges and opportunities for judicial state-building presented by legal pluralism. The rules that matter most may not stem from the state at all. Custom, tradition, religion, 2 These strategies are by no means limited to postconflict states. In Malaysia, for instance, the long-time ruling party has used state-sponsored Islamic Courts as an integral part of its efforts “to validate its Islamic credentials—relegitimize the party and the state—and thus co-opt, or at least undercut, both the Islamic resurgents and the opposition party” (Peletz 2002, 10).
3The record of international state-building efforts has been at best uneven and all too often profoundly discour-aging (for example, see Doyle and Sambanis 2006; Paris 2010). Past state-building efforts have often left much to be desired, and the endeavor involves a number of intrinsic contradictions and tensions (Paris and Sisk 2009). Some schol-ars go even further and argue that the international state-building enterprise itself is inherently illegitimate (Bain 2006, 537–38, Chandler 2006). Jahn, for instance, contends that state building and related activities such as democracy pro-motion are “counterproductive . . . ultimately producing enemies instead of allies and heightening insecurity instead of enhancing security” (Jahn 2007, 212). While these criticisms deserve consideration, as Paris notes, “for all the shortcom-ings of liberal peacebuilding—and there have been many—most host countries would probably be much worse off if not for the assistance they received,” and forsaking state building entirely would be akin to “abandoning tens of millions of people to lawlessness, predation, disease and fear” (Paris 2009, 108). Regardless of how one ultimately conceptualizes state building, the practice is here to stay for the foreseeable future. Thus, deepening scholarly understanding of what makes state building more or less successful in environments marked by a high degree of legal pluralism constitutes a worthwhile endeavor.
Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 3 family lineage, powers not sanctioned by law (such as those of criminal enterprises or powerful commercial interests), and a host of other sources may have equal or even greater influence than state law.
Robust legal pluralism challenges the state’s claim to a monopoly on legitimate resolution of legal disputes as well as the ideal of uniform application of the law.
It enables participants to select dispute resolution forums based on accessibility, efficiency, legitimacy, jurisdiction, and cost, as well as the state and nonstate systems’ respective abilities to make binding decisions and sanction individuals that choose other systems. This process leads to a sustained struggle between state and nonstate justice actors for legitimacy, resources, and authority. Apart from their adjudicatory role, nonstate justice authorities can even become “state-building spoilers” and help trigger a return to conflict (Menkhaus 2007, 76). Whether state officials can “gain[] the right and ability to make . . . rules” and whether nonstate actors can effectively resist the state has major implications for state-building (Migdal 1988, 30–31). On the other hand, seeking support from nonstate judicial actors can be an important strategy for maintaining stability and securing elite and popular support for the state.
In short, legal pluralism has major implications in all states, but the prevalence, autonomy, role, and authority of nonstate justice systems vary dramatically across contexts. It is not enough to merely recognize that legal pluralism exists; schol-ars and policymakers must understand how legal pluralism actually functions. The purpose of this article is to advance scholarly understanding of legal pluralism both theoretically and empirically. This is done by, first, articulating four new le-gal pluralism archetypes along with the major strategies for domestic and interna-tional policymakers dealing with legally pluralistic environments. Second, as these situations tend to be fluid with a wide range of realistic outcomes, this paper draws on postconflict state-building to provide an analytical lens to examine and understand legal pluralism’s implications. The case studies of Timor-Leste and Afghanistan highlight how both positive and negative change in the overarching relationship between state and nonstate justice can occur. They demonstrate that when promoting the rule of law, judicial state-builders could benefit greatly from understanding the legal pluralism archetype and its programmatic implications, along with developing a credible and appropriate strategy for transforming the cur-rent environment toward a more constructive situation that ideally secures buy-in from nonstate actors or at least mitigates their opposition.
Article Overview This article examines legal pluralism and its implications for policy through five sections. The first section demonstrates the need for new legal pluralism archetypes through a brief literature review. The second section proposes four dis-tinct archetypes: (1) combative, (2) competitive, (3) cooperative, and (4) comple-mentary. It likewise identifies commonly used strategies for interacting with non-state justice actors across different legally pluralistic environments: (1) bridging, (2) harmonization, (3) incorporation, (4) subsidization, and (5) repression. Sec-tions three and four present two divergent case studies of postconflict judicial state-building in legally pluralist settings, specifically Timor-Leste and Afghanistan. The case studies highlight how domestic and international policy decisions can help shift the overarching archetype. These decisions can help promote a change from competitive legal pluralism to cooperative as occurred in Timor-Leste or a deterio-ration of the situation into combative legal pluralism as happened in Afghanistan.
The final section offers overarching lessons from the experiences in Afghanistan and Timor-Leste.
Downloaded from by London School of Economics user on 09 July 2018 4 Legal Pluralism in Theory and Practice Current Understandings of Legal Pluralism Remain Insufficient While important scholarly work exists, current explanations remain inadequate for understanding the form and implications of legal pluralism in all settings but par-ticularly those in the wake of conflict. The nonstate system is usually seen as either an obstacle to progress (Farran 2006) or effective, efficient, and reflective of true community preferences (Harper 2011). For example, a major United Nations re-port defines nonstate justice systems as involving “a neutral third party not part of the judiciary” while also noting “custom-based systems appear to have the advan-tages of sustainability and legitimacy” (United Nations Development Program et al.
2012, 8). In both instances, the stark binary between the state and nonstate justice sectors is radically oversimplified.
Existing work has been largely historical or theoretical. Legal pluralism has a long historical pedigree. It is deeply embedded with the creation of the modern state. Fukuyama (2011, 245–75), for instance, contends that competition between legal systems has driven the formation of certain modern states. Legal pluralism also shaped interactions between different societies, and “dual legal systems were widespread in colonized parts of Africa, Asia, Latin America, and the Pacific” (Merry 1991, 890). Legal pluralism became a defining feature of colonial administrations that sought to harness local dispute resolution mechanisms to help legitimize and institutionalize their rule (Benton 2002). Likewise, multiethnic domains, such as the Ottoman Empire, embraced legal pluralism (Barkey 2013). Nor is the preva-lence of legal pluralism largely a historical phenomenon. Currently, legal pluralism exists everywhere in forms as varied as community dispute resolution to the interna-tional system where there has been a proliferation of treaties as well as transnational regimes with veto capacity or even legislative ability (Berman 2012). State and non-state legal systems can work together relatively smoothly or find themselves clashing frequently (Tamanaha 2008, 400).
Some noteworthy attempts have been made to understand the interactions be-tween the state and nonstate justice sectors. Connolly (2005) proposes the ap-proaches of abolition of nonstate systems or, alternatively, incorporation, partial incorporation, or no incorporation of nonstate mechanisms. Forsyth (2009, 202) conceptualizes the relationship between state and nonstate justice mechanisms as a seven-stage “spectrum of increasing state acceptance of the validity of adjudica-tive power by the non-state justice system.” While stopping short of a full typology, numerous scholars have addressed the interaction between state and nonstate jus-tice systems in postconflict settings (Mac Ginty 2008; Richmond 2009; Baker 2010).
Country-specific analyses have also recognized that legal pluralism has implications for state-building in places such as Afghanistan (Barfield 2008) and Timor-Leste (Nixon 2012), but these works do not sufficiently account for different types of re-lationships between state and nonstate justice and the fluidity between different legal pluralism archetypes. While these approaches can help illuminate state efforts to engage nonstate justice mechanisms, they offer limited insights for understand-ing efforts to engage in postconflict judicial state-building in highly legally pluralist states.
Direct application of existing models to judicial state-building efforts presents serious problems because they do not recognize postconflict settings’ systemic chal-lenges. Successfully asserting and maintaining authority domestically and interna-tionally invariably challenges even relatively high capacity states (Fukuyama 2014).
As Krasner and Risse (2014, 549) note, “while no state governs hierarchically all the time, consolidated states possess the ability to authoritatively make, implement, and enforce decisions for a collectivity.” In postconflict settings, state power is al-most always actively contested. The archetypes and strategies proposed below help contextualize all state-building efforts regardless of whether they are undertaken Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 5 by local elites under the auspices of the state, international state-builders, or some combination thereof.
Legal Pluralism Archetypes An alternative typology more attuned to legal pluralism’s structure and implications provides for a more robust understanding of its function in all states, as well as its relationship to the rule of law. Here the concept of the rule of law is used in a “thin” rather than “thick” sense (Peerenboom 2002). At minimum, a “thin” concept of the rule of law requires that “law must be set forth in advance (be prospective), be made public, be general, be clear, be stable and certain, and be applied to everyone” (Tamanaha 2007, 3). While admirable, thicker conceptualizations include extensive institutional, economic, cultural, and political requirements that are unrealistic for most postconflict states in the short or even medium term (West 2003). The cre-ation of a state and society with even a thin version of the rule of law after conflict is a daunting, time-consuming task that requires buy-in from both state officials and society at large. On the part of the state, “even the most formal, minimalist concep-tion of the rule of law requires a normative commitment to the project of the rule of law itself” by state officials (Stromseth, Wippman, and Brooks 2006, 76). At the same time, state action alone is not enough; establishing the rule of law requires formal and informal popular accountability mechanisms. Regardless of the partic-ular setting, “the rule of law is as much a culture as a set of institutions, as much a matter of the habits, commitments, and beliefs of ordinary people as legal codes” (Stromseth, Wippman, and Brooks 2006, 310). Respect for the law largely hinges on the broad social belief that the law, at its core, is basically fair and legitimate (Tyler 2006). Thus, establishing the rule of law is inherently a dynamic, contested process that includes both top-down and bottom-up elements. These dynamics are particularly acute in postconflict settings because public trust in state institutions has almost inevitably been undermined, often severely. In practical terms, establish-ing the rule of law requires consolidation of popular legitimacy of new legal norms and institutions just as much as the construction of courthouses or the passage of laws.
This section posits new theoretical archetypes for understanding the fluid rela-tionship between state and nonstate justice in a wide range of settings: (1) com-bative, (2) competitive, (3) cooperative, and (4) complementary. Furthermore, it proposes and examines five main strategies for understanding international and do-mestic judicial state-builders’ engagement techniques. The strategies of: (1) bridg-ing, (2) harmonization, (3) incorporation, (4) subsidization, and (5) repression.
These strategies illuminate the main domestic and international approaches to en-gaging the nonstate legal sector in legally pluralist societies.
The strategies discussed here are top down in the sense that they examine how domestic and international actors can attempt to proactively influence the overar-ching type of legal pluralism in a given setting. In practice, however, the overarching legal pluralism archetype invariably reflects both top-down and bottom-up factors.
Whether nonstate actors choose to engage the state and on what terms is decidedly bottom-up, as is to what extent and under what conditions people use the state and nonstate justice systems. As with other areas of state policy, popular mobilization and political advocacy is a potential tool to change the state’s approach to nonstate justice (Jordan and Van Tuijl 2000).4 Therefore, both the overarching archetype of legal pluralism present at a given moment and movement between different types of legal pluralism reflect a mixture of top down and bottom up elements.
4For instance, bottom-up activities could include popular efforts to end repression of nonstate justice actors and institutions or alternatively through political advocacy against recognizing legal dispute systems that are seen as violating human rights.
Downloaded from by London School of Economics user on 09 July 2018 6 Legal Pluralism in Theory and Practice In practical terms, the strategies help inform which activities can help foster change from one archetype to another, as well as how advances in judicial state-building could be achieved within a given archetype. Rather than viewing the pro-cess as linear, advancing the rule of law after conflict should be conceptualized as highly fluid with points of progress and regression. Even in favorable circumstances, establishing the rule of law and the inclusive institutions that underpin it is usually a prolonged, highly contingent process with both top-down and bottom-up elements (North, Wallis, and Weingast 2009; Acemoglu and Robinson 2012). Nevertheless, a well-informed understanding of the legal pluralism archetype that is present can significantly bolster domestic and international initiatives designed to influence the nonstate justice sector and minimize the risk of costly missteps.5 While invariably a simplification, archetypes help conceptualize the core features of the relationship between state and nonstate judicial actors. They can inform ju-dicial state-building efforts as well as other domestic and international postconflict policy initiatives that influence or can be influenced by nonstate actors, such as economic development initiatives or broader governance endeavors. Although the case studies presented in this article focus on postconflict settings, the archetypes can also help actors understand and improve their engagement with a wide vari-ety of legally pluralist states. Whatever the context, by understanding the dominant archetype, it is possible to constructively engage with the nonstate justice sector in a given area.6 Constructive engagement with the nonstate justice sector requires thinking critically about how to deal with current realities while simultaneously seek-ing to transform the environment to one more favorable to judicial state-building.
Combative Legal Pluralism In combative legal pluralism, state and nonstate systems are overtly hostile to one another. Where the normative underpinnings of the respective legal systems are not even tacitly accepted, the state and nonstate justice sectors seek explicitly to undermine, discredit, supplant, and—ideally—destroy the other. Combative legal pluralism can involve nonstate actors rejecting the state system’s ideological founda-tion in a largely nonviolent manner. For instance, in the fight against the apartheid state in South Africa, there were active efforts to establish “structures of justice and policing that contested the legitimacy of their equivalent in the apartheid institu-tion” (Nina 2000, 24). While compatible with nonviolence, combative legal plural-ism flourishes in countries facing an active insurgency or separatist movement. In many instances, nonstate justice forms a cornerstone of those movements (Mampilly 2011; Arjona, Kasfir, and Mampilly 2015). Unsurprisingly, combative legal pluralism prevails where postconflict state-building has failed or is clearly trending in a nega-tive direction. In many instances, this dynamic coincides with an active insurgency characterized by parallel antagonistic state-building enterprises (Wickham-Crowley 1992; Kalyvas 2006, 218–20). Thus, the scope for even limited collaboration tends to be minimal.
5These missteps can take a wide variety of forms and can involve both domestic and international actors. Isser highlights that state policies attempting to regulate the conduct and jurisdiction of nonstate systems in Liberia and Guatemala “have undermined the effectiveness of TAs [traditional authorities] without providing an alternative outlet” (Isser 2011, 333). In Afghanistan, US-funded rule of law programs sought to establish new nonstate dispute resolu-tion mechanisms in the hopes of rendering Taliban justice unnecessary. These new shuras, however, could be quite destabilizing to local communities by distributing large amounts of external funding as well as empowering individuals through military force who did not enjoy substantial popular support (Miakhel and Coburn 2010).
6The relationship may vary by geographic location as well as by religious, tribal, or ethnic group. The state may enjoy a cooperative relationship with nonstate actors in one part of the country, while facing a competitive or even combative relationship in other areas. For example, the Afghan state has a competitive relationship with many tribal and religious nonstate authorities but a combative relationship with the Taliban insurgents.
Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 7 Competitive Legal Pluralism Competitive legal pluralism, where the state’s overarching authority is not chal-lenged but nonstate actors retain substantial autonomy, characterizes many develop-ing countries and is extremely common after conflict. Indeed, it is the default con-dition in postconflict settings. While a postconflict political and legal settlement has been reached, it has not yet been consolidated or institutionalized. The prospect of a return to conflict is disturbingly common (Collier, Hoeffler, and Söderbom 2008).
The state invariably finds itself trying to assert a new order in places previously be-yond its control or in places where that control was contested.7 It is also common in many developing counties where societal actors retain autonomy and still exercise the right to maintain order in a way not dictated by state officials (Migdal 1988).
Competitive legal pluralism features significant, often deep, tensions between state and nonstate legal systems, especially where legal norms and procedure diverge significantly (Tamanaha 2008). Yet, in these situations, clashes rarely endanger the state’s formal judicial authority because the nonstate justice sector does not make a concerted effort to supplant state authority entirely. While the nonstate justice sector retains a sizable degree of autonomy, the state and nonstate systems respect each other’s right to exist in some form and are willing to engage with each other, at least tactically (Baker and Scheye 2007).
Competitive nonstate legal systems are most frequently rooted in religious be-liefs or shared culture, custom, or heritage. These legal systems often exist outside the state and do not necessarily share the state’s values. Competitive legal plural-ism also exists where criminal actors have established separate, parallel orders that rarely seek to take over the state but actively work to retain autonomy by circum-venting state law (Volkov 2000). There may be collusion between political elites and criminal actors to evade the law or profit illegally; however, the activities themselves remain fundamentally opposed to state law.
In postconflict settings, the level of competition tends to mirror the success of the state-building process. For example, postconflict state-building efforts in Afghanistan and Timor-Leste both started against a backdrop of competitive legal pluralism. In Afghanistan, where state-building trended in a decidedly negative di-rection that helped spur renewed violent conflict, competitive legal pluralism has slipped into combative legal pluralism. Alternatively, when the state gains legitimacy, authority, and capacity, the environment and the incentives change, as occurred in Timor-Leste. Nonstate actors increasingly favor collaboration because both the ad-vantages of partnership with the state and the disadvantages of opposing it increase.
Nonetheless, as long as nonstate actors retain a high degree of autonomy, the po-tential for setbacks is ever-present.
Competitive legal pluralism can be prolonged and endemic, particularly when nonstate dispute resolution is seen as legitimate and authoritative. In relatively sta-ble states, it can even be an explicit strategy.8 These difficulties are compounded in postconflict settings where the state’s legal authority and institutions have often been severely compromised.9 While states often seek to supplant nonstate competi-tors over time, in other circumstances, engagement with nonstate authorities forms the basis of the governance system. Domestic judicial state-builders often rely on 7This dynamic is by no means a purely modern phenomenon, as almost every successful judicial endeavor has historically required the state to suppress, outperform, or collaborate with nonstate rivals (Fukuyama 2011).
8For example, during the period of stability from 1923 to 1978, Afghan rulers used just such a strategy. The state demanded the allegiance of tribal leaders and the local population. Local politics and the resolution of local disputes, however, were left largely to local populations (Barfield 2010, 220).
9With regards to the state, legitimacy reflects “normative acceptance and expectation by a political community that the cluster of rules and institutions that compose the state ought to be obeyed” and the extent that “the state is seen as the natural provider of goods and services” (Call 2008, 14).
Downloaded from by London School of Economics user on 09 July 2018 8 Legal Pluralism in Theory and Practice powerful, established nonstate actors and structures to support their rule, includ-ing but not limited to tribal or clan groups and religious authorities (Migdal 2001).
Cooperative Legal Pluralism In a cooperative legal pluralist environment, nonstate justice authorities still retain significant autonomy and authority. Nonstate judicial actors, however, have by and large accepted the state’s normative legitimacy and are generally willing to work together toward shared goals. Major clashes are far less frequent and tend to fo-cus on social issues where values clash, such as women’s rights, rather than exis-tential issues of state judicial power. Cooperative legal pluralism tends to thrive in places where progress is being made toward consolidating legitimate state author-ity. This shift may coincide with meaningful advances toward the consolidation of democratic governance bound by the rule of law. However, the establishment of a high-capacity state that enjoys a cooperative relationship with nonstate authorities does not necessarily require democracy or the rule of law. This dynamic is reflected by postconflict transformations in places like Zimbabwe (Kriger 2006). Yet coopera-tion has its limits. In many postcolonial African states, nonstate justice actors largely backed the nascent state, but their support often waned “when the state-building project [became] too exclusivist or predatory” (Dorman 2006, 1087).
At the same time, it is important to stress that cooperative legal pluralism is co-operative only in terms of the relationship between the predominant forms of state and nonstate justice. Cooperative legal pluralism does not necessarily mean that, in terms of substance, the law is just. A legal order characterized by this archetype invariably still produces winners and losers. Under cooperative legal pluralism, the law, whether state or nonstate, can still be used to violate human rights, oppress citizens, or perpetrate systematic discrimination against certain groups.
Complementary Legal Pluralism Legal pluralism does not disappear in a state with a high-capacity, effective legal system, but it is complementary. In other words, nonstate is subordinated and struc-tured by the state because the state enjoys both the legitimacy to have its rule ac-cepted and the capacity to actually enforce its mandates (Ellickson 1991; Mac Ginty 2008, 142). The United States, Western Europe, and many other countries with high-capacity legal systems choose to allow private arbitration, mediation, and other forms of alternative dispute resolution (ADR). ADR takes a wide variety of forms, but “they share the feature that a third party is involved who offers an opinion or communicates information about the dispute to the disputants” (Shavell 1995, 1).
For civil disputes, state courts frequently mandate claimants’ attempts to settle their disputes outside of court before being allowed to access state courts (Stipanowich 2004). However, substantively and procedurally, state and nonstate law can still clash. Arbitration agreements facilitate the evasion of state law and legal process, but the extent of circumvention depends on the policy preferences of state offi-cials. In all instances, these processes are integrated into, and fall under the ulti-mate regulatory purview of, the state, exist at its pleasure, and largely depend on state courts for enforcement. ADR processes are allowed and often encouraged be-cause the state deems them useful for addressing real and perceived “inefficiencies and injustices of traditional court systems” (Edwards 1986, 668).
This form of legal pluralism is complementary from a governance perspective be-cause the state has effectively outsourced alternative forums, such as court-referred mediation, or at least tactically licensed dispute venues, such as binding arbitra-tion. Complementary legal pluralism features a similarly cooperative ethos, but the nonstate justice authorities operate under the umbrella of state authority and without substantial autonomy to reject state decisions. Only complementary legal Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 9 pluralism can truly uphold that oft-stated requirement for the rule of law: that the law is applied equally to all people (Carothers 1998). Complementary legal plural-ism is a worthwhile long-term goal, but it is important to have reasonable expecta-tions about what is feasible in the short to medium term after conflict. As with co-operative legal pluralism, complementary legal pluralism refers only to the nature of the relationship between state and nonstate justice. These types of legal orders can uphold the rule of law, but that does not mean that states with complementary legal pluralism necessarily do uphold the rule of law.
Strategies for Addressing the Nonstate Justice Sector While the predominant legal pluralism archetype is important, it is not immovable.
Domestic and international state-builders have a number of strategies that have at least the potential to promote significant change. Of course, the results of those efforts will reflect domestic or international actors’ ability to persuade, or in some case coerce, nonstate actors and society at large.10 Based on the range of available options, this section identifies five main strategies for how best to understand inter-actions between state and nonstate systems: (1) bridging, (2) harmonization, (3) in-corporation, (4) subsidization, and (5) repression. These strategies are by no means mutually exclusive or hermetically sealed, but they are, nevertheless, conceptually and functionally distinct. Success cannot be guaranteed, but certain strategies are better suited to certain environments. As will be demonstrated in the case study sec-tions, local and international state-builders often employ these strategies with little regard for the complex relationship between the state and nonstate justice systems, which can lead to decidedly suboptimal outcomes. This need not be the case. Savvy strategic planning and pragmatic adaptation, ideally combined with a bit of good fortune, could improve the relationship between the state and nonstate justice sec-tors regardless of the dynamics present.
Bridging With a bridging strategy, judicial state-builders attempt to ensure that cases are allo-cated between the state and nonstate justice systems as appropriate based on state law, participants’ preferences, and venue appropriateness. Almost invariably, bridg-ing asserts jurisdictional claims regarding both state and nonstate venues. Certain legal matters, most notably homicide and other serious crimes involving physical harms, must be resolved in state courts, while small, nonviolent claims are almost always left to nonstate venues. Public information campaigns are frequently under-taken to enhance understanding of the state legal system and how to access it. Sim-ilarly, training is often provided to local leaders and citizens about the state legal system and how to access it. Free or subsidized legal aid can give citizens an econom-ically viable choice. A bridging strategy can be paired with a formal incorporation approach that seeks to provide a state legislative framework for nonstate justice, but it can also be a stand-alone initiative that seeks to increase individuals’ choices with-out trying to resolve larger questions regarding the relationship between state and nonstate justice venues.
Bridging can work well where unmet demand exists for access to state courts or when increased awareness stimulates demand for state justice. A bridging strat-egy’s impact also hinges on nonstate authorities’ willingness to facilitate or accept 10As with any other area of policy, there is no guarantee that all organs of the state are going to pursue a consistent, coordinated approach to nonstate justice. While justice policy is often dealt with at the national level, in states with a degree of decentralization, local or regional government may pursue distinct policy approaches (Benjamin 2008). In practice, decentralized units’ ability to pursue distinct approaches will vary according to both their de facto and de jure level of authority. Nevertheless, state actors have the same set of strategies available whether or not they are actually able to pursue them.
Downloaded from by London School of Economics user on 09 July 2018 10 Legal Pluralism in Theory and Practice referrals to the state system, which by extension decrease their autonomy. Thus, bridging is frequently a useful approach in competitive and cooperative legal plu-ralistic environments while offering little utility for situations where the state has a combative relationship with nonstate authorities.
Harmonization Harmonization attempts to ensure that the outputs of the nonstate justice system are consistent with the state system’s core values. At the same time, the nonstate justice system is incorporated and legitimatized to some extent. To support harmo-nization, states and international donors often fund activities to encourage non-state justice practitioners to act in a manner consistent with state law in general.
However, there is often at least tacit recognition that nonstate actors retain a signif-icant degree of autonomy and independent legitimacy. Thus, there is a willingness to tolerate some normative differences in adjudication standards. As opposed to try-ing to get nonstate venues to act like state courts of first instance, there is a focus on changing the treatment of certain legal matters, for example, nonstate actors’ treatment of women (Chopra and Isser 2012). State judicial actors also frequently discriminate against women, but usually this is done in violation of state law rather than as a matter of accepted practice (Campbell and Swenson 2016). In general, the greater the state’s ability to offer a compelling and legitimate forum for dispute resolution worth emulating, the greater the prospects of successfully implement-ing a harmonization approach. Successful harmonization occurs most frequently in competitive—and especially cooperative—legal pluralism environments. Neverthe-less, as long as nonstate actors retain a significant degree of autonomy, meaningful divergence with state policy remains possible.
Incorporation Under incorporation, the distinction between state and nonstate justice is elimi-nated at least from the state’s perspective. Nonstate justice, in a formal sense, be-comes state justice. In practical terms, the nonstate justice systems’ decisions are endorsed but also regulated by state officials. Incorporation can mean the creation of explicitly religious or customary courts with state support and regulation, the labeling of nonstate justice actors as courts of first instance, or simply offering an avenue for appeal from nonstate venues to state courts. Decisions of the nonstate system could be subject to appeal or ratification by the state system. For example, a local council’s decision regarding a property dispute could be formalized through a district court or administrative entity. At its extreme, the entire nonstate justice system could be brought under the state justice system’s purview (Peters and Ubink 2015). While nonstate systems are allowed to continue and perhaps even grow, in practice incorporation is a bold move to assert practical and ideological authority over nonstate actors by limiting their independence. Judicial state-builders seek to both harness the authority of the state system and control it. Once incorporated, states may further seek to regulate “customary” nonstate law by codifying it, an in-herently subjective and selective process, or even by creating new nonstate law.
Incorporation strategies may also be less overt through professional regulation or the induction of elections to nonstate posts previously allocated through other means. There is no guarantee that nonstate actors will be willing or able to be incor-porated into the state justice system. The state may envision itself as the principal with the nonstate judicial actors as its agent. This approach is highly problematic, however, if the nonstate actors have notably different norms and values from the state and accountability mechanisms are weak. Thus, the prospects and effective-ness of incorporation strategies track the state’s ability to persuade or compel large numbers of nonstate judicial actors to engage with it. This enabling dynamic is most Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 11 likely in cooperative settings, possible but tentative in competitive scenarios, and very unlikely in combative environments.
Subsidization Subsidization of the state system is the most common strategy, and certain key subsi-dization techniques tend to recur across settings, notably legislative reform, capacity building and establishing physical infrastructure used by the justice sector, support-ing symbolic representation, and promoting public engagement. Subsidization has a vast number of potential targets. Unlike harmonization, bridging, and incorpora-tion, it does not require any meaningful acquiescence by nonstate judicial actors as a prerequisite. Consequently, under subsidization approaches, the nonstate system is largely left alone or at least is not the primary target of action. The state system receives assistance to increase its capacity, performance, and appeal relative to the nonstate system. This task sounds straightforward. However, as Fukuyama (2004, 59) notes, “establishing the rule of law involves extensive construction not just of laws, but also courts, judges, a bar, and enforcement mechanisms across the country,” thereby making it “one of the most complex administrative tasks that state-builders need to accomplish.” Subsidization can take an immense variety of forms and can be implemented re-gardless of whether the environment is marked by cooperative, combative, or com-petitive legal pluralism. Subsidization is by far the most common strategy in post-conflict settings, as well as for judicial state-building in general. It is also most likely to influence the overarching relationship between the state and nonstate justice systems because it is the only form of aid directed at improving the performance, legitimacy, and effectiveness of state justice.
Repression Repression strategies seek to fundamentally undermine, and ideally eliminate, the state’s nonstate rivals. Repression strategies are not concerned with persuading or incentivizing nonstate justice actors to work with state authorities. Nor does re-pression hinge on the state system persuading or influencing its nonstate rivals or encouraging citizens to use state courts. Repression can simply be a matter of outlawing nonstate justice forums, particularly in relatively peaceful places such as Botswana, and using the state’s power to enforce its mandate (Forsyth 2009). Where the state can effectively outlaw nonstate justice, however, the state is already predom-inant. Almost invariably repression involves significant violence, rather than merely the threat of sanction. As the state seeks to eliminate nonstate justice actors, repres-sion often results in reciprocal violence by nonstate actors. Alternatively, violence by nonstate actors can trigger state repression efforts.
Repression efforts are rarely paired with bridging, harmonization, and incorpora-tion strategies because those approaches depend on constructive engagement with nonstate justice systems. However, the state frequently subsidizes the state justice system in an attempt to increase its authority and effectiveness relative to nonstate justice systems, as well as to protect state judicial authorities from insurgent attacks.
While invariably unpleasant in practice, repression can be an important tool when the state faces an existential threat from nonstate justice actors, particularly when linked to an armed insurgency. Nevertheless, repression alone, even when backed by force, is unlikely to be sufficient for the state to consolidate a monopoly on legal authority since force must be paired with another form of legitimacy to be sustain-able over time (Beetham 2013).
Downloaded from by London School of Economics user on 09 July 2018 12 Legal Pluralism in Theory and Practice Legal Pluralism Archetypes in Practice This article presents case studies “as an in-depth study of a single unit (a rela-tively bounded phenomenon) where the scholar’s aim is to elucidate the features of a larger class of similar phenomena,” in this instance postconflict judicial state-building in legally pluralist environments (Gerring 2004, 341). This examination relies on a structured comparison of approaches to the state and nonstate justice sectors after conflict (George and Bennett 2005). It is important to stress that the goal of these case studies is not to provide a definitive, detailed account of state-building relating to the state and nonstate justice sectors. Rather they highlight key developments and illuminate how the archetypes and strategies discussed above can offer important insights into how relationships between state and nonstate actors can change for the better or worse.
As the default condition for postconflict states and the most common arrange-ment in developing countries, it worth examining competitive legally pluralist states in more detail. Thus, this article focuses on the most common setting for postcon-flict states rather than trying to highlight examples from each archetype. At the on-set of judicial state-building efforts after war, states are very rarely in a postconflict situation characterized by combative legal pluralism. After all, combative legal plu-ralism is usually a sign of continued or renewed conflict. In contrast, after conflict, tensions are still fresh and powerful nonstate actors often have found themselves on both sides of the conflict (Staniland 2012). Thus, it is unlikely that the nascent postconflict regime will initially enjoy a cooperative, let alone complementary, re-lationship with the major nonstate judicial actors. The rare exceptions tend to be places where a legitimate, high capacity state that featured complementary legal pluralism existed prior to the conflict, and one side of the conflict has experienced a clear victory that has been generally accepted by the former combatants. For ex-ample, post–World War II Germany and Japan would meet these criteria (Dobbins, et al. 2003).
Timor-Leste and Afghanistan are paradigmatic examples of postconflict, com-petitive legally pluralistic settings where domestic and international policy choices had major consequences. From the outset, policymakers themselves recognized that these cases would have broad significance and inform future endeavors (Chesterman 2002; Chopra 2002). Timor-Leste was explicitly envisioned as a model for postconflict Libya, though that option was ultimately rejected (Doyle 2016, 26), while policymakers are now fixated on trying to avoid replicating the state-building failures of Afghanistan elsewhere (Swenson 2017).
Other important similarities exist that further enhance the value of a compari-son between states. Afghanistan and Timor-Leste both established new regimes and new legal systems in the early 2000s in contexts marked by competitive legal plu-ralism where the state justice sector needed to demonstrate its value, effectiveness, and legitimacy. Both states saw their infrastructures devastated by conflict and high levels of poverty. They enjoyed a period of stability, but the prospect of renewed violence remained a very real possibility. While it is now regarded as a success, state-building efforts in Timor were “punctuated by initially sporadic, and then intense, episodes of violence” (Gledhill 2012, 48). The 2006 Crisis cast serious doubt on the state-building endeavor and even the viability of the state itself as many proclaimed Timor-Leste a “failed state” (Cotton 2007). Afghanistan likewise faced a prospect of renewed conflict, but it also had a meaningful opportunity for successful state-building efforts after the Taliban regime’s fall (Rashid 2008, Barfield 2010, 300–10).
The country’s history demonstrates that peace is possible. Prior to the 1978 Com-munist coup, Afghanistan had enjoyed decades of domestic tranquility, and the Tal-iban itself had previously established a monopoly on the use of force under even more chaotic conditions in the 1990s. Even since late 2001, Afghanistan enjoyed relative peace before the Taliban insurrection rapidly metastasized and reached the Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 13 level of “a full-blown insurgency by 2006” (Jones 2008, 7). Thus, the divergent shifts in the legal pluralist environment were not inevitable. As the subsequent section highlights, their decisions had significant consequences.
In both cases, local and international state-builders invested heavily in the state justice sector in an initially competitive legal pluralism environment where the vast majority of disputes were settled outside of state courts. The prevalence of the non-state justice sector, unsurprisingly, coincided with a history of very limited state ca-pacity and weak central rule. Nonstate justice systems faced few jurisdictional re-strictions or meaningful attempts by the state to regulate their conduct, particularly outside of urban centers.
Despite these shared circumstances, the outcomes in each country have been very different. Timor-Leste has successfully shifted to cooperative legal pluralism, while decisions in Afghanistan helped trigger a slide to combative legal pluralism.
Notwithstanding the upheaval surrounding the 2006 Crisis, judicial state-builders in Timor-Leste have made significant progress toward developing the rule of law by working to build an effective state justice sector and collaborating with local nonstate authorities. In contrast, the post-Taliban regime has helped trigger a slide from competitive legal pluralism to combative legal pluralism due to state officials’ lack of commitment to the rule of law and failure to engage key nonstate actors.
The case studies draw on a mixture of primary and secondary sources relating to state-building and legal pluralism generally and in Timor-Leste and Afghanistan specifically. I interviewed over 40 key stakeholders related to postconflict judicial state-building and legal pluralism in these countries in 2014 and 2015. The case studies also draw on personal insights from working on judicial state-building ini-tiatives in both countries between 2008 and 2012. These cases possess clear tem-poral and spatial boundaries. The temporal boundaries result from the extent of large-scale international involvement after conflict, particularly the presence of sig-nificant military forces. The Timor-Leste case covers from independence in May 2002 through to the exit of international forces in 2012. The Afghanistan case study covers from the start of post-Taliban transitional administration in late 2001 through to the election of Karzai’s successor in 2014 and the subsequent drawdown of inter-national forces.
Timor-Leste (2002–2012): Competition to Cooperation Legal order in East Timor has long been predicated on competitive legal plural-ism. For over four centuries in East Timor, legal order hinged on tactical alliances between Portuguese colonial officials and nonstate authorities who oversaw order at the local level (Robinson 2009, 23). After the collapse of Marcello Caetano’s authoritarian regime in Portugal in 1974, East Timor entered a rapid, haphazard decolonization process. Decolonization culminated in an East Timorese declara-tion of independence in November 1975. Indonesia invaded on December 7, 1975, and embarked on an intense twenty-five-year occupation. Despite Indonesia’s brutal occupation, the vast majority of disputes continued to be settled through nonstate mechanisms (Babo Soares 2003, 267). The resistance movement combined sup-port from nonstate authority to sustain the domestic resistance (McWilliam 2005), coupled with a compelling vision of an independent, democratic East Timor that upholds the rule of law to bolster international support (Strohmeyer 2000). In-donesian President B. J. Habibie agreed to a referendum on East Timor’s status that was held in August 1999. Despite extensive intimidation efforts by militias, vot-ers overwhelming supported independence. Shortly thereafter, pro-integrationist militias unleashed systemic destruction that ultimately led to the dispatch of in-ternational peacekeepers in mid-September 1999. East Timor was placed under United Nations’ trusteeship until mid-2002 when it became the independent Demo-cratic Republic of Timor-Leste (RDTL). During this time, nonstate authorities were Downloaded from by London School of Economics user on 09 July 2018 14 Legal Pluralism in Theory and Practice essential in maintaining order. Preexisting nonstate structures remained and be-came even more important under United Nations’ administration. Thus, at the onset, the state faced a competitive situation as the nonstate justice system was en-trenched, legitimate, and could not simply be compelled to the work with the state.
Since independence, the nascent state has faced immense challenges in the judi-cial state-building process. Effective postconflict state-building demands legitimacy and authority (Call 2008). The state has worked hard to enhance both. By investing in a modern state justice system and democratic elections, while drawing on the legacy of the independence struggle and a vision of a modern state committed to development, the state has harnessed nonstate authorities’ power, legitimacy, and capacity to an extent far greater than Portuguese or Indonesian authorities ever achieved. The state and international community have invested heavily in the state justice sector. While the state still had limited reach and capacity, it enjoyed sub-stantial legitimacy as the end goal of the independence struggle and due to its asso-ciation with prominent independence leaders. The state’s legitimacy helped trans-form nonstate judicial actors long skeptical of state authority into almost de facto state actors through elections and by offering a vision of the state that commanded widespread support. This section shows how the state, with significant international support, transformed a competitive legal pluralism environment into a cooperative one.
Slow but Steady Subsidization Building a state justice system proved challenging due to the inherent difficulties arising from minimal human resources, limited judicial infrastructure, and the pub-lic impression inherited from Indonesian rule that courts were instruments of state power rather than neutral arbitrators. Thus, domestic initiatives and international assistance emphasized the creation of basic state justice institutions. This approach reflected a straightforward rationale. The state justice sector was still in an embry-onic state. International aid emphasized building modern state institutions that acted in accordance with the rule of law. In other words, international assistance focused overwhelmingly on subsidization.
Since independence in 2002, procedural due process concerns were endemic along with substantial case backlogs and spotty opening hours (West 2007, 336– 338). Even prior to independence during the period of UN control from 1999 to 2002, there was a focus on empowering domestic legal personnel. Yet, this strategy faced systemic problems because local judicial actors were inexperienced and needed extensive training as Indonesia had prevented the development of a professional class in East Timor. In 2005, all UN appointed Timorese court ac-tors underwent evaluation. Subsequently, all Timorese personnel were disqualified (Jensen 2008, 133). This left the justice sector almost entirely dependent on in-ternational staff. While a difficult situation, international judicial actors mitigated the dismissal’s impact and allowed time for domestic capacity to bear fruit. The most important of these institutions was the Legal Training Centre. It was estab-lished to oversee the training and professional certification of all judges, prose-cutors, and public defenders. By 2012, the judicial system was staffed almost ex-clusively with Timorese judges, courts operated consistently nationwide, backlogs had decreased significantly, and popular faith in the justice system had increased (Marx 2013).
Enhanced reach and robustness from to 2002 to 2012 made the state justice sector an increasingly powerful force in dispute resolution. The nonstate justice sector re-mained dominant in 2012, but an ever-increasing number of citizens experienced a meaningful choice between the two systems. Not surprisingly, challenges remained.
The quality of justice was uneven. The legal system retained a myopic focus on Portuguese as the preferred legal language. However, most people, including those with university educations, cannot understand the language and the often wholesale Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 15 importation of Portuguese laws gives them little relevance to the local context. Case resolution was time consuming, and participants did not necessarily understand the proceedings, especially when conducted in Portuguese. Nevertheless, real progress was achieved against a difficult backdrop.
Stealth Incorporation Historically, the relationship between state and nonstate justice in Timor-Leste has been overwhelmingly competitive whereby there has been tactical engagement but also deep skepticism. One of the nascent state’s greatest achievements was to facil-itate a transformation into cooperative legal pluralism where state, international, and nonstate actors worked together as well as established a largely respected juris-dictional divide between courts and local (suco) councils. The state handled major issues, particularly violent crimes, while most civil matters and petty crimes were left to local dispute resolution mechanisms. Nonstate authorities continued to resolve most disputes in Timor-Leste. Nevertheless, nonstate authorities viewed the state and the state-building endeavor as legitimate and were largely open to constructive engagement because state leaders were recognized as key actors in the indepen-dence struggle.
The state has achieved this feat through a remarkably simple yet powerful mech-anism: competitive local elections. Elections for suco chiefs were codified under state legislation on community authorities in 2004, with additional reforms in 2009 (RDTL, 2004a, b, 2009a). Suco elections were held during 2004 to 2005, and a sub-sequent set of elections was undertaken in October 2009. The legal framework is largely procedural and jurisdictional. However, its goal is transformational. The law roots the legitimacy of nonstate judicial actors in modern democratic ideas, most notably through recurring, state-administered democratic elections and a workable jurisdictional divide. Less than a decade after independence, state authorities were already able to establish democratic elections as the primary source of legitimacy— a view that has been internalized by the suco chiefs themselves (RDTL Ministry of State Administration and The Asia Foundation 2013). The state administered suco election system forms the backbone of state influence at the local level.
The suco reforms enacted in 2009 subtly changed the suco chief’s role by more tightly linking it with state development through program planning and monitor-ing, creating an annual development plan, and submitting an annual report (RDTL 2009a). These responsibilities were paired with major state community develop-ment initiatives. Suco chiefs were thus expected to bring state developmental funds to their communities. Consequently, not only were suco chiefs active state-building agents; their constituents demanded it.
The Limits of Cooperation However, the situation always fell short of complementary legal pluralism. There have been major harmonization efforts in the area of gender based-violence, which were backed by both state officials and international actors. Domestic violence was unequivocally a public crime under the Penal Code (RDTL 2009b: Articles 146, 154) and the Law Against Domestic Violence (RDTL 2010). However, the nonstate justice system continued to resolve most cases of domestic violence, including in many instances where the victim preferred to use state courts (Wigglesworth 2013).
Nonstate authorities still retained an effective veto over state law in their jurisdic-tion. Thus, their willingness to be directed by the state in most matters reflected their acceptance of the state’s legitimacy and persuasive authority rather than its ability to impose its will. At the end of 2012, Timor-Leste still faced a host of system-atic problems that could arrest or even retract advances toward the consolidation of the rule of law in the future. Nevertheless, its accomplishments were significant and offer insights into how to transform a situation marked by competitive legal pluralism into cooperative legal pluralism.
Downloaded from by London School of Economics user on 09 July 2018 16 Legal Pluralism in Theory and Practice Afghanistan (2001–2014): Competition to Combat Afghanistan from 2001 to 2014 demonstrates how domestic and international state-building missteps can help transform an environment from competitive into com-bative legal pluralism. Competitive legal pluralism has long defined Afghanistan.
Since the start of the constitutional era in 1923, the Afghan state apparatus has long sought sources of legitimacy that were less dependent on tribal support, ex-ternal religious sanction, and individual charismatic leadership (Poullada 1973).
Nevertheless, state power largely rested on its relationships with religious and tribal authority (Rubin 2002). The most effective form of legal order was not state law.
Rather it was Pashtunwali, a nonstate legal code that served as both “an ideology and a body of common law which has evolved its own sanctions and institutions,” as implemented through local dispute resolutions: jirgas for Pashtuns and shuras for non-Pashtuns (Roy 1990, 35). For nearly a hundred years, all legitimate state-sponsored legal orders in Afghanistan were grounded in a combination of state performance, Islam, and tribal approval. The system broke down, however, when Communists toppled the regime in 1978 and plunged the country into decades of civil strife.
Eventually the Taliban seized control. The Taliban imposed a harsh but effective state legal order in the mid-1990s based on religious authority in tacit agreement with prominent forms of tribal justice. Pashtunwali and other nonstate adjudication systems were largely tolerated and not seen as inconsistent with Taliban understand-ings of Islam. The Taliban regime was sympathetic to Al Qaeda’s radical brand of Islam and harbored the perpetrators of the September 11, 2011, terrorist attacks.11 Shortly thereafter, a major international effort was undertaken to dislodge the Taliban.
Subsidization Without Strategy After the Taliban’s defeat, postconflict judicial state-building efforts in Afghanistan started optimistically. The new, multiethnic state under President Hamid Karzai had the opportunity to prove itself a valid governing entity committed to democracy and the rule of law (Rashid 2008; Barfield 2010). While it enjoyed overwhelming inter-national support, the new regime eschewed the traditional pillars of legitimate au-thority domestically. Religion was acknowledged but the state was not clothed in re-ligious legitimacy. Karzai lacked robust religious credentials and, having dislodged the Taliban, his international backers were wary of any government that seemed too Islamic. Moreover, the international community’s emphasis on human rights clashed with Pashtunwali and Islamic law as frequently understood in Afghanistan and gave the Taliban grounds to criticize the regime as unrepresentative and un-Islamic. The state’s most plausible path to legitimacy was through elections and the provision of public goods, most notably a more just legal order, but it failed to de-liver.
From 2001 to 2014, Karzai, with international support, worked ceaselessly to cen-tralize authority and undermine constitutional checks and balances. While nomi-nally tolerant of democratic competition, the regime never displayed a normative commitment to democracy and the rule of law. The state built a judiciary that had the outside appearance of a modern state legal system but instead focused on rent extraction. At the same time, international subsidization occurred on a staggering scale. The legal system and human resource base had been devastated by decades of conflict, so the needs were certainly daunting (Swenson and Sugerman 2011).
During Karzai’s time in office, Afghanistan’s justice sector received over USD$1 bil-lion in aid from the United States alone, yet the Office of Special Inspector General for Afghanistan Reconstruction (SIGAR) determined that this assistance generated 11Sinno (2008: 244) goes so far as to posit that the reason the Taliban refused to turn over Bin Laden was that it would violate Pashtunwali’s guest hospitality requirements.
Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 17 no notable improvements (SIGAR 2015). The ability of international subsidization to advance the rule of law was quite limited absent an ideological commitment by the state to those ideals. There was no progress toward the rule of law, and there were few significant gains in the reach, effectiveness, and legitimacy of the state justice system by the end of Karzai’s presidency in 2014 (Singh 2015). Instead, the Karzai regime was phenomenally corrupt. Rather than seeking to promote the rule of law, it consistently functioned as a “vertically integrated criminal organization .
. . whose core activity was not in fact exercising the functions of a state but rather extracting resources for personal gain” (Chayes 2015, 62). International assistance unintentionally institutionalized a rentier state. On occasions when international actors tried to “investigat[e] corruption, they were rebuked by Karzai’s officials for misunderstanding the nature of patronage networks that served to support the gov-ernment” (Chaudhuri and Farrell 2011, 285).
State Attempts at Incorporation, Bridging, and Harmonization Tribal dispute resolution mechanisms continued to be the forum of choice for many Afghans, particularly in Pashtun tribal areas (Wardak and Braithwaite 2013). Non-state justice mechanisms remained the most prominent form of dispute resolution, settling 80 to 90 percent of disputes (Barfield, Nojumi and Thier 2006, 9). Nev-ertheless, engaging nonstate legal providers was not a state priority. Only in 2009 did Afghan state officials express meaningful interest in the nonstate justice sector through the “Draft National Policy on Relations between the Formal Justice System and Dispute Resolution Councils” (Ministry of Justice 2009). State policy, with inter-national support, envisioned simultaneous harmonization, bridging, and incorpora-tion strategies in relation to nonstate justice. At no point, however, was there serious outreach to tribal or religious authorities. The state policy bluntly proclaimed non-state legal decisions must be consistent with Sharia, the Constitution, other Afghan laws, and international human rights standards. The policy envisioned voluntary jir-gas and shuras, and their jurisdiction was limited to certain civil matters. Ultimately, the official policy consensus came at the expense of feasibility, as the final report was drafted in a manner designed to appease all state parties but was disconnected from reality.
As a state policy has no independent legal bearing, a law was required to opera-tionalize it. The “Draft Law on Dispute Resolution: Shuras and Jirgas” was produced in September 2010 (Ministry of Justice 2010). The legislation formally incorporated shuras and jirgas into the state system and asserted the state’s authority to regulate all aspects of nonstate dispute resolution. The law even imposed criminal liability by stipulating that jirga participants “and parties of dispute shall be duty bound to observe provisions of this law” or face potential criminal charges. The draft law only authorized shuras and jirgas to hear civil disputes and petty juvenile crimes on refer-ral from state authorities. The Ministry of Women’s Affairs and the Human Rights Commission strongly opposed the law because they believed it endorsed dispute resolution mechanisms that violated human rights standards by legitimizing and in-stitutionalizing jirgas and shuras (International Rule of Law Professional 2014). As the tenuous alliance among state agencies broke down, the legislation did not move forward.
Combative Legal Pluralism and the Failure of Repression The Taliban justice system constituted the state’s fiercest rival. Taliban justice ex-ploited the widespread view that state courts were corrupt, ineffective, and cultur-ally unintelligible in an attempt to displace state justice itself. Effective legal order rooted in religious beliefs constituted the core of the Taliban’s political program. It also underpinned their claim to be Afghanistan’s legitimate rulers and highlighted the state justice system’s failures. By the time its insurgency had become full blown in 2006, the Taliban had established a lean, but sophisticated network of parallel Downloaded from by London School of Economics user on 09 July 2018 18 Legal Pluralism in Theory and Practice governance structures (Jones 2010). The Taliban’s legal system tapped into the well-spring of legitimacy offered by religion, culture, and working constructively with tribal leaders. The Taliban justice system claimed to adjudicate based on Sharia law “strengthens their legitimacy in a deeply religious population, particularly when the codes of law used by the state are little known, misunderstood, and sometimes re-sented” (Giustozzi and Baczko 2014, 219).
By drawing a contrast with the highly corrupt state courts, Taliban insurgents ac-tively contended with the state system, especially outside the capital, by offering inexpensive, expedient, and relatively fair dispute resolution. The Taliban operated “a parallel legal system that is acknowledged by local communities as being legiti-mate, fair, free of bribery, swift, and enduring” if brutal, and their system was “easily one of the most popular and respected elements of the Taliban insurgency by lo-cal communities, especially in southern Afghanistan” (Johnson 2013, 9). Unlike in the state system, decisions were enforced, and addressing corruption was taken seri-ously (Kilcullen 2011). In short, the Taliban justice system sought to provide exactly what the state justice system did not: predictable, effective, legitimate, and accessi-ble dispute resolution.
Recognizing the profound threat to the state’s authority, the Afghan state and international forces undertook a robust repression campaign against the Taliban justice system (US Mission Afghanistan 2010). It achieved little, however, as state justice remained highly inefficient, ineffective, and corrupt. Unsurprisingly, the Tal-iban legal order steadily gained ground (Ahmed 2015). The campaign was mutual as the Taliban sought to disrupt state court operations, targeted judges and other state officials for assassination, and denied the courts’ legitimacy as a legal author-ity. Both sides refused to recognize the other’s right to exist, let alone promulgate binding legal decisions.
The Karzai regime lacked the capacity to defeat its rivals militarily or embody a cause worth fighting for. Nowhere is this failure more prevalent than in the jus-tice sector, where the state failed to promote, let alone provide, a just legal order.
The Taliban’s success at judicial state-building dispels the notion that establishing legal order was impossible in post-2001 Afghanistan. The Taliban’s justice system had major shortcomings, and its human rights record was appalling, but the Tal-iban possessed a comprehensive legal strategy with a clear long-term vision rooted in local values and beliefs. The state judicial system decidedly did not. State legal authority could have increased dramatically if the Afghan state improved its perfor-mance and committed to fostering the rule of law. Taliban justice thrived in large part due to the Afghan state’s abysmal performance.
Timor-Leste and Afghanistan in Comparison Since the early 2000s, Timor-Leste and Afghanistan have taken sharply divergent paths as noted in Table 1. Despite initial optimism, Afghanistan has seen little progress. The Afghan legal system, as with the state more broadly, became known primarily for corruption and predation. As the case study shows, the country was once again subsumed in widespread civil conflict. In contrast, Timor-Leste saw po-litical violence and major upheaval in 2006 but has since enjoyed domestic stability and made significant progress toward consolidating democratic governance and the rule of law.
Judicial state-building in Timor-Leste made progress from 2002 to 2012 because there was a credible and sustained effort to develop democratic institutions bound by the rule of law in a manner intelligible and compelling to nonstate actors. Dur-ing the decades-long independence struggle, political leaders offered a persua-sive vision of a democratic state committed to the rule of law. Even more impor-tantly, since achieving independence, policymakers in Timor-Leste sought to cre-ate an effective, just legal order and to constructively engage with nonstate actors Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 19 Table 1. Legal pluralism archetypes Archetype Key Features Examples Combative The state and nonstate justice sectors do not recognize each other’s right to exist and actively seek to destroy each other.
Afghan state since 2004 with Taliban justice system; Iraq since 2003.
Competitive Deep tensions exist between the state and nonstate justice sectors and there are frequent clashes between systems; however, the state’s formal juridical authority is not challenged.
While the nonstate justice sector retains autonomy, the state and nonstate systems respect each other’s right to exist in some form.
Afghanistan from 2001 to 2003; Afghan state with tribal authorities from 2004; East Timor from 1998 to 2002.
Cooperative The nonstate justice sector retains a significant degree of authority and autonomy; however, state and nonstate legal authorities are generally willing to work together towards shared goals.
Timor-Leste after independence, particularly since 2006; Zimbabwe after 1980.
Complementary Both state and nonstate justice exist, but nonstate justice mechanisms operate under the umbrella of state authority.
United States, United Kingdom, Germany, Japan.
nationwide. They worked to establish an independent judiciary and inclusive gov-ernance institutions underpinned by competitive, free, and fair national elections in 2002, 2007, and 2012. Both the ability to change the government through free elections and the creation of inclusive institutions are essential for a democratic state underpinned by the rule of law (Huntington 1993, 266–67). With regards to the legal system, state officials worked to develop a sensible legal framework for nonstate actors that granted them discretion over small matters but funneled more serious crimes to the state courts. Moreover, the foundation of nonstate actors’ legit-imacy was transformed through regular, competitive local elections for suco coun-cils in 2004/2005 and 2009. Local elections grounded the legitimacy of traditional nonstate judicial actors in modern democratic ideas of popular sovereignty and ac-countability rather than lineage or custom.
The international community’s subsidization of the nascent state justice sector reinforced positive domestic trends in Timor-Leste. While not without flaws, inter-national assistance improved state justice institutions and their auxiliaries (Marriott 2009). Furthermore, in the wake of the 2006 Crisis, international aid helped facil-itate the transition to cooperative legal pluralism by offering aid to improve the performance of nonstate justice and build links between local suco councils and state courts. International efforts reinforced the domestic trends that ultimately led to a shift to cooperative legal pluralism. In short, Timor-Leste demonstrates the need for regulations that reflect shared values and institutions that make a good faith effort to translate those values into reality.
Afghanistan presents a stark contrast. Afghan policymakers offered neither a com-pelling ideological vision of the state committed to justice nor a serious attempt to construct legitimate, culturally intelligible institutions. The Karzai-led state squan-dered the opportunity to build a new, more inclusive and effective democratic state. Despite rhetoric to the contrary, Afghanistan under Karzai’s leadership never displayed a normative commitment to the rule of law. Worse, the regime systemati-cally undermined institutional, legal, and political checks on its authority, includ-ing suppressing political parties, manipulating elections, and undermining judi-cial independence and all institutional accountability mechanisms. Finally, Karzai’s regime never seriously engaged with key tribal and religious nonstate justice actors that historically constituted the building blocks of legitimate order in Afghanistan.
Downloaded from by London School of Economics user on 09 July 2018 20 Legal Pluralism in Theory and Practice Table 2. Nonstate justice sector strategies Strategy Key Features Examples Bridging Judicial state-builders seek to ensure that cases are allocated between the state and nonstate justice systems as appropriate based on state law, participants’ preferences, and venue appropriateness.
State-builders seek to ensure serious crimes cannot be resolved outside state courts regardless of the disputants’ preferences by using paralegals to direct cases to state courts or offering trainings on how to access state courts.
Alternatively, minor disputes may be sent to nonstate venues by state courts.
Harmonization Judicial state-builders seek to ensure that the nonstate justice systems’ outputs are consistent with the state system’s core values.
Laws to outlaw discriminatory practices in nonstate adjudication and training to end discriminatory practices.
Incorporation Judicial state-builders eliminate the distinction between state and nonstate justice. Nonstate justice, at least in a formal sense, becomes state justice.
Outcomes of the nonstate justice systems are endorsed but also regulated by the state system. In practice, incorporation could mean the creation of explicitly religious or customary courts with state support or the labeling of nonstate justice venues as state courts of first instance.
Subsidization Judicial state-builders support the state system to increase its capacity, performance, and appeal relative to the nonstate system.
Facilitating legislative reform, establishing physical infrastructure used by the justice sector, supporting symbolic representation, capacity building, and promoting public engagement.
Repression Judicial state-builders seek to fundamentally undermine and ideally eliminate the state’s nonstate rivals.
Outlawing nonstate justice forums or seeking to arrest or kill nonstate justice actors.
Instead, state actors simply sought to impose their will on tribal and religious au-thorities. This failure to engage meaningfully with tribal and religious authorities helps explain judicial state-building’s lack of progress in Afghanistan and the cor-responding slide from competitive legal pluralism into combative legal pluralism against an increasingly potent Taliban insurgency.
Despite their tremendous influence, international actors in Afghanistan achieved little in terms of advancing the rule of law or constructively engaging with nonstate actors (Swenson 2017). International backing secured the top state post for Karzai and encouraged the immense concentration of power in the executive. Through-out Karzai’s tenure, international actors retained enough sway to encourage or dis-courage state actions. That influence was never used effectively, however, as the Karzai regime was ultimately deemed too strategically important to fail. At the same time, the international community never recognized the legal pluralism archetype in place and, therefore, did not strategize accordingly. Massive, largely uncoordi-nated international subsidization efforts were continued, even when it was apparent that such programs were not improving the situation. This left the Taliban as the only major group with a credible strategy for engaging nonstate justice actors.
The case studies highlight that judicial state-builders, whether domestic or inter-national, need an informed strategy backed by institutions, regulations, and poli-cies that recognize nonstate legal authorities’ importance. A successful, sustainable strategy must be rooted in a deep understanding of how a country’s culture, poli-tics, and history can help underpin a legitimate legal order. Strategies, detailed in Table 2, must also recognize that advancing the rule of law depends primarily on Downloaded from by London School of Economics user on 09 July 2018 GEOFFREY SWENSON 21 domestic actors, and longer-term time horizons are essential, as developing the rule of law takes decades not years. While Timor-Leste has achieved admirable progress, it has yet to consolidate either democracy or the rule of law. At the same time, Afghanistan by all accounts faces a challenging future, but the fluidity of legal plu-ralism and Afghanistan’s own history suggest that combative legal pluralism is not inevitable. In both instances, the international community can still offer incentives (or disincentives) to influence the behavior of both state and nonstate justice actors.
Conclusion Understanding legal pluralism is important for any legal or policy intervention, in-cluding but by no means limited to state building. Without understanding legal pluralism’s dynamics in a given context, interventions are likely to be ineffective.
Even initiatives that enjoy short-term success are unlikely to be sustainable, as they reflect good fortune rather than an informed approach. Sound strategy requires un-derstanding how state and nonstate actors interact systematically. This article helps build that knowledge by presenting the four main types of relationships between state and nonstate actors through a typological framework that illuminates the dy-namics of legal pluralism across contexts. By understanding the archetype in which a policy is operating, an appropriate strategy or package of strategies for engag-ing with nonstate actors can be selected. By identifying the main strategies avail-able to policymakers, the article illustrates how each approach works within each archetype and which strategies might be appropriate to deploy depending on the context.
Against a backdrop of competitive legal pluralism, state-builders in both Timor-Leste and Afghanistan had the chance to promote a democratic state bound by the rule of law or spark a return to conflict. Collectively, the success of Timor-Leste and the failure of Afghanistan demonstrate two key propositions. First, they illuminate the importance of strategy selection by showing how policy choices can influence outcomes. Second, they show how strategy selection, while vital for long-term suc-cess, is largely irrelevant if the policy intervention in question is not culturally in-telligible and ultimately persuasive. Thus, the structure and implications of legal pluralism must be considered when creating and implementing policy.
Legal pluralism in Timor-Leste and Afghanistan ultimately offers both reasons for hope and trepidation. Legal pluralism can help form a vital foundation of state le-gitimacy. As the relationship between state and nonstate sectors is inherently fluid, the institutions and initiatives present can help or hinder the development of more constructive relationships between state and nonstate actors. Competitive legally pluralist relationships can be positively developed into cooperative ones. Relation-ships between state and nonstate justice authorities can also quickly sour as com-petitive relationships deteriorate into combative ones. Relationships are not static.
Even situations of combative legal pluralism that seem dire are not preordained to remain that way in perpetuity. Insights from Afghanistan and Timor-Leste into ju-dicial state-building within a competitive legal pluralist archetype offer lessons for future endeavors. However, as legal pluralism has major implications for institu-tional design and policy initiatives in areas such as governance and development, the theory presented here has broader applications. Future research into state and nonstate relationships in other settings, postconflict and otherwise, would be a valu-able next step in enabling better policy decisions through enhanced knowledge.
Acknowledgements I am grateful to Nancy Bermeo, E. A. Brett, Frances Z. Brown, Richard Ca-plan, Meghan Campbell, Lillian Dang, Richard Ponzio, Lyal Sunga, and Richard Downloaded from by London School of Economics user on 09 July 2018 22 Legal Pluralism in Theory and Practice Youngs, as well as to seminar participants at the University of Oxford and the London School of Economics for their feedback. The extremely helpful com-ments of the two anonymous reviewers and the editors is likewise much appre-ciated. This article was made possible by fieldwork support from Pembroke Col-lege and the Cyril Foster Fund of the Department of Politics and International Relations, University of Oxford.
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ResourcesAops Wiki 1962 IMO Problems/Problem 2
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1962 IMO Problems/Problem 2
Contents
1 Problem
2 Solution
3 Solution 2
4 See Also
Problem
Determine all real numbers which satisfy the inequality:
(Note that the problem as written in the official PDF does not have the outer )
Solution
Obviously we need for the outer square root to be defined, for the first inner square root to be defined, and for the second inner square root to be defined. Solving these we get that the left hand side is defined for .
Now obviously the function is continuous on , with and . Moreover, as is a decreasing and an increasing function, both and are decreasing functions, and hence is a decreasing function. Therefore there is exactly one solution to .
We can now find this solution:
(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)
Solving the quadratic equation for , we get
The reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than is the additional solution, and is the root we need.
Hence the solutions to the given inequality are precisely the reals in the interval ![Image 33: $\boxed{ \left ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }$.
Solution 2
We first do some substitution, Let , then , and the expression becomes: . Squaring both sides, we have:
Solving yields or
Substituting back for , or . This is two solutions, however, note that for the square root to be determinant, we need to have:
Obviously, conflicted with , and the lower bound for x is , so our final solution is:
![Image 50: [\boxed{x \in \left ~ -1,\quad 1-\dfrac{\sqrt{127}}{32} ~ \right) }]
~IDKHowtoaddsolution
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Problem 11•2•3•4•5•6Followed by
Problem 3
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4111 | https://www.wyzant.com/resources/answers/933830/write-the-equation-of-the-transformed-graph-of-tangent-with-period-6-that-h | Write the equation of the transformed graph of tangent with period 6 that has been shifted vertically up 2 units. (3 points) | Wyzant Ask An Expert
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Trigonometry
Joe B.
asked • 10/05/23
Write the equation of the transformed graph of tangent with period 6 that has been shifted vertically up 2 units. (3 points)
Write the equation of the transformed graph of tangent with period 6 that has been shifted vertically up 2 units. (3 points)
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Note that tan(x) has a period of π (not 2π). Below, this affects the equation for our b coefficient.
y = ±a tan (b x – c) + d
a = amplitude = 1
There is no flipping across the x-axis means start with a = 1
(so there is no a coefficient in our transformed equation)
period = 6 = π/b, so b = π/(period) = π/6
there’s no phase shift, so c = 0
(so there is no c coefficient in our transformed equation)
shifted vertically up 2 units means d = 2
So our required equation is
y = tan ((π/6)x) + 2
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4112 | https://pubmed.ncbi.nlm.nih.gov/11506097/ | A double-blind, randomized comparison of i.v. lorazepam versus midazolam for sedation of ICU patients via a pharmacologic model - PubMed
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. 2001 Aug;95(2):286-98.
doi: 10.1097/00000542-200108000-00007.
A double-blind, randomized comparison of i.v. lorazepam versus midazolam for sedation of ICU patients via a pharmacologic model
J Barr1,K Zomorodi,E J Bertaccini,S L Shafer,E Geller
Affiliations Expand
Affiliation
1 VA Palto Alto Health Care System, Stanford University School of Medicine, Anesthesiology Service, California 94304, USA. barrj@leland.stanford.edu
PMID: 11506097
DOI: 10.1097/00000542-200108000-00007
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Clinical Trial
A double-blind, randomized comparison of i.v. lorazepam versus midazolam for sedation of ICU patients via a pharmacologic model
J Barr et al. Anesthesiology.2001 Aug.
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. 2001 Aug;95(2):286-98.
doi: 10.1097/00000542-200108000-00007.
Authors
J Barr1,K Zomorodi,E J Bertaccini,S L Shafer,E Geller
Affiliation
1 VA Palto Alto Health Care System, Stanford University School of Medicine, Anesthesiology Service, California 94304, USA. barrj@leland.stanford.edu
PMID: 11506097
DOI: 10.1097/00000542-200108000-00007
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Background: Benzodiazepines, such as lorazepam and midazolam, are frequently administered to surgical intensive care unit (ICU) patients for postoperative sedation. To date, the pharmacology of lorazepam in critically ill patients has not been described. The aim of the current study was to characterize and compare the pharmacokinetics and pharmacodynamics of lorazepam and midazolam administered as continuous intravenous infusions for postoperative sedation of surgical ICU patients.
Methods: With Institutional Review Board approval, 24 consenting adult surgical patients were given either lorazepam or midazolam in a double-blind fashion (together with either intravenous fentanyl or epidural morphine for analgesia) through target-controlled intravenous infusions titrated to maintain a moderate level of sedation for 12-72 h postoperatively. Moderate sedation was defined as a Ramsay Sedation Scale score of 3 or 4. Sedation scores were measured, together with benzodiazepine plasma concentrations. Population pharmacokinetic and pharmacodynamic parameters were estimated using nonlinear mixed-effects modeling.
Results: A two-compartment model best described the pharmacokinetics of both lorazepam and midazolam. The pharmacodynamic model predicted depth of sedation for both midazolam and lorazepam with 76% accuracy. The estimated sedative potency of lorazepam was twice that of midazolam. The predicted C50,ss (plasma benzodiazepine concentrations where P(Sedation > or = ss) = 50%) values for midazolam (sedation score [SS] > or = n, where n = a Ramsay Sedation Score of 2, 3, ... 6) were 68, 101, 208, 304, and 375 ng/ml. The corresponding predicted C50,ss values for lorazepam were 34, 51, 104, 152, and 188 ng/ml, respectively. Age, fentanyl administration, and the resolving effects of surgery and anesthesia were significant covariates of benzodiazepine sedation. The relative amnestic potency of lorazepam to midazolam was 4 (observed). The predicted emergence times from sedation after a 72-h benzodiazepine infusion for light (SS = 3) and deep (SS = 5) sedation in a typical patient were 3.6 and 14.9 h for midazolam infusions and 11.9 and 31.1 h for lorazepam infusions, respectively.
Conclusions: The pharmacology of intravenous infusions of lorazepam differs significantly from that of midazolam in critically ill patients. This results in significant delays in emergence from sedation with lorazepam as compared with midazolam when administered for ICU sedation.
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Pharmacokinetic-pharmacodynamic modeling and ICU sedation: unexplored territories.Bailey JM.Bailey JM.Anesthesiology. 2001 Aug;95(2):281-2. doi: 10.1097/00000542-200108000-00005.Anesthesiology. 2001.PMID: 11506095 No abstract available.
Useful information about the pharmacokinetics and pharmacodynamics of midazolam and lorazepam.Deem S.Deem S.Anesthesiology. 2002 Aug;97(2):522; author reply 522-3. doi: 10.1097/00000542-200208000-00039.Anesthesiology. 2002.PMID: 12151951 No abstract available.
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¿Qué es el dequeísmo y el queísmo? Definición y ejemplos
¿Qué es el dequeísmo y el queísmo? Definición y ejemplos
Publicado el 15 de junio de 2025 por Ana Fernández, MA
¿Te preocupa “de que” cometas este error sin darte cuenta? El “dequeísmo” y el “queísmo” son usos inadecuados de las secuencias “de que” y “que”, respectivamente. Te explicamos cómo evitarlos.
Es difícil que algo así vuelva a suceder.
Ni me di cuenta de que había acabado la película.
Es difícil ~~de que~~ algo así vuelva a suceder.
Ni me di cuenta ~~que~~ que había acabado la película.
Tabla de contenidos
¿Qué es el “dequeísmo” y qué es el “queísmo”?
¿Qué es y por qué se produce el “dequeísmo”? Ejemplos ilustrativos
¿Qué es el “dequeísmo” y qué es el “queísmo”?
Es probable “de que” hayas cometido un error como este en alguna que otra ocasión. De manera oral es habitual escuchar en distintas conversaciones un “que” o un “de que”que se cuela en cierto tipo de oraciones. Pero ¿qué son y por qué ocurren estos fenómenos antinormativos?
Según la RAE, la Real Academia Española, el dequeísmo es:
El uso indebido de la preposición “de” delante de la conjunción “que” cuando la preposición no viene exigida por ninguna palabra del enunciado.
A su vez, la RAE define el queísmo como:
La supresión indebida de una preposición (generalmente “de”) delante de la conjunción “que”, cuando la preposición viene exigida por alguna palabra del enunciado.
Histórica y socialmente estos errores se perciben de manera bastante negativa. La presión por no caer en el dequeísmo hace que, en algunos casos, se caiga en el fenómeno opuesto, el queísmo. A continuación, vamos a darte todas las pistas necesarias para aliviar esa presión y aprender a detectar si estás utilizando incorrectamente “de” o “que” al escribir o al hablar.
¿Qué es y por qué se produce el “dequeísmo”? Ejemplos ilustrativos
Ahora que ya sabemos qué son estos fenómenos lingüísticos que conducen a un uso indebido de la conjunción “que” y de la preposición “de”, es importante saber por qué se producen y cómo evitar el queísmo y dequeísmo.
Empecemos abordando el dequeísmo. Existen diversos casos que dan lugar al dequeísmo y que deben tenerse en cuenta para tratar de evitar este fenómeno:
1) En primer lugar, cuando se antepone la preposición “de” a una oración sustantiva que hace la función de sujeto. Para no caer en el dequeísmo debemos recordar siempre que el sujeto de una oración nunca va precedido de preposición.
Es posible ~~de que~~ no llueva más en todo el verano.
Es posible que no llueva más en todo el verano.
Dicen los expertos: “Es posible que” no llueva más en todo el verano.
2) Por otro lado, al anteponerse la preposición “de” a una oración sustantiva que hace la función de complemento directo. Suele ocurrir con los verbos que se conocen como verbos de pensamiento.
Forman parte de estos verbos: pensar, opinar, creer, considerar, etc. También ocurre con los verbos de habla, que serían verbos como comunicar, exponer, etc.; con los verbos de temor (verbos como temer, etc.), y con los verbos de percepción (como ver, oír, etc.). Debes tener en cuenta que el complemento directo nunca va precedido de la preposición “de”.
Pienso ~~de que~~ es mejor no ir a la cena.
Pienso que es mejor no ir a la cena.
3) Cuando aparece la preposición “de” antes de una oración subordinada con funciones de atributo en oraciones copulativas con el verbo ser u otros verbos similares.
Nuestra intención es ~~de que~~ todos participemos en el proyecto.
Nuestra intención es que todos participemos en el proyecto.
Las oraciones copulativas se llaman así porque sus formas verbales no tienen de por sí ningún significado pleno y su función es únicamente la de unir el sujeto con el atributo o el predicado.
4) Cuando se añade la preposición “de” en locuciones conjuntivas que no la llevan como una vez que, a medida que o a no ser que.
A medida ~~de que~~ pasa el tiempo, lo aprecio más.
A medida que pasa el tiempo, lo aprecio más.
5) Cuando se usa la preposición “de” en vez de la que realmente pide el verbo.
Nos fijamos ~~de que~~ no viniste al concierto.
Nos fijamos en que no viniste al concierto.
Siguiendo estas cinco reglas, deberíamos evitar el dequeísmo y escribir frases correctas sin ese “de” innecesario.
Sin embargo, existen otros verbos que dificultan nuestra misión de impedir el dequeísmo. Algunos verbos pueden construirse con o sin el “de”, según el sentido que tenga la propia frase. Esto ocurre con verbos como:
_\Advertir:\__\_Advertir\ (algo) a alguien y \advertir\ \de\ algo (a alguien)._ _\Avisar:\__\Avisar\ (algo) a alguien y \avisar\ \de\ algo (a alguien)._ _\Cuidar:\__\Cuidar\ (algo o a alguien) y \cuidar\ \de\ algo o alguien._ _\Dudar:\__\Dudar\ (algo) y \dudar\ \de\ algo._ _\Informar:\__\Informar\ (algo) a alguien (este uso se da solo en Latinoamérica) e \informar\ \de\ algo (a alguien) (este uso se da solo en España).__
“A medida que” iba leyendo, se iba preocupando más por todo el trabajo que tenía.
Cómo evitar el “dequeísmo”
Ya conocemos la definición de dequeísmo y hemos visto algunos ejemplos de frases en las que se produce dicho fenómeno. Ahora nos falta averiguar cómo evitarlo.
El truco más sencillo que casi siempre funciona, para evitar caer en el dequeísmo y hacer un uso indebido de la preposición “de” es sustituir la oración subordinada que incluya la conjunción “que” por “eso”. Si se debe seguir usando esa preposición, significa que su uso es correcto. Veamos algunos ejemplos:
Pienso ~~de que~~ mañana va a llover → Pienso ~~de eso~~.
Pienso que mañana va a llover → Pienso eso.
Otro truco que a menudo funciona es convertir la oración sobre la que estamos dudando en una oración interrogativa. Si la oración interrogativa solo tiene sentido con el uso de la preposición “de”, significa que la oración original también debe incluirla. Por ejemplo:
¿De qué nos enteramos? Nos enteramos de que era imposible parar el calentamiento global.
¿~~Qué~~ nos enteramos? Nos enteramos ~~que~~ era imposible parar el calentamiento global.
¿“De qué” me están hablando con todo esto del “dequeísmo y queísmo”?
Ejemplos de frases en las que se produce “dequeísmo”
Estas son algunas oraciones ejemplo en las que se produce dequeísmo y cómo podemos solucionarlo.
Es posible ~~de que~~ llueva mañana.
Es posible que llueva mañana.
He oído ~~de que~~ has cambiado de trabajo. ¡Enhorabuena!
He oído que has cambiado de trabajo. ¡Enhorabuena!
Le preocupa ~~de que~~ llueva el día de su boda.
Le preocupa que llueva el día de su boda.
Mi idea es ~~de que~~ nos levantemos temprano y salgamos a las 7:00 am.
Mi idea es que nos levantemos temprano y salgamos a las 7:00 am.
¿Por qué se produce y cómo evitar el “queísmo”? Ejemplos
Como ya hemos visto,el queísmo o antidequeísmo es el fenómeno contrario al dequeísmo. Los principales casos en los que no se debe suprimir la preposición “de” son:
1) Con verbos pronominales que llevan un complemento de régimen, como acordarse de algo, alegrarse de algo, arrepentirse de algo, fijarse en algo, olvidarse de algo, etc.
Nos olvidamos ~~que~~ teníamos que llegar pronto.
Nos olvidamos de que teníamos que llegar pronto.
2) Con verbos no pronominales que llevan un complemento de régimen como convencer de algo, insistir en algo, tratar de algo, etc.
Le convencí ~~que~~ viniera a cenar.
Le convencí de que viniera a cenar.
3) Con sustantivos que llevan complementos preposicionales.
Me muero de ganas ~~que~~ lleguen mis vacaciones.
Me muero de ganas de que lleguen mis vacaciones.
4) Con adjetivos que necesitan complementos preposicionales.
Estamos seguros ~~que~~ aprobarás el examen
Estamos seguros de que aprobarás el examen.
5) En locuciones como:
_\A condición de\_
_\A fin de\_
_\A pesar de\_
_\En caso de\_
_\A pesar de quees jueves\__\, las terrazas están llenas.\_
6) En _locuciones verbales_ como:
_\Caer en la cuenta de\_
_\Darse cuenta de\_
_\Dudar de algo\_
_\Dudo de quese\__\_haya dado cuenta del error que ha cometido.\__
Para comprobar cuándo se debe emplear la secuencia “preposición + que” o simplemente “que” se pueden usar los mismos trucos que vimos para evitar el dequeísmo y, además, se deben tener en cuenta los casos que acabamos de analizar.
Además de experimentar con estos trucos y estudiar la gramática de manera exhaustiva, siempre puedes disponer de un asistente de escritura como LanguageTool que haga la corrección ortográfica, gramatical y de estilo por ti. ¿Te has dado cuenta ya de que te ahorrarás muchísimo tiempo, esfuerzo y, hasta casi, salud mental? ¡Es posible que pronto no puedas vivir sin él!
Resumamos:
El dequeísmo es el uso incorrecto de la preposición “de” delante de la conjunción “que” cuando la preposición no viene exigida por ninguna palabra del enunciado.
El queísmo o antidequeísmo es el fenómeno contrario al dequeísmo. Consiste en la supresión incorrecta de la preposición “de” delante de la conjunción “que” cuando la preposición viene exigida por alguna palabra de la oración.
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Ana Fernández, MA
Ana estudió Traducción e Interpretación en la Universidad de Granada (España) y tiene un máster en Traducción Audiovisual de la University of Roehampton, en Londres (Reino Unido). Tiene experiencia como traductora, editora y como Content Manager.
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4115 | https://fiveable.me/key-terms/intermediate-algebra/slope-intercept-form | Slope-Intercept Form - (Intermediate Algebra) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Slope-Intercept Form
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key term - Slope-Intercept Form
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Definition
The slope-intercept form is a way to represent a linear equation in two variables, $y$ and $x$, in the form $y = mx + b$, where $m$ represents the slope of the line and $b$ represents the $y$-intercept, the point where the line crosses the $y$-axis. This form provides a straightforward method for graphing linear equations and understanding their key features.
5 Must Know Facts For Your Next Test
The slope-intercept form, $y = mx + b$, is a widely used representation of linear equations that provides a clear and intuitive way to understand the behavior of a line.
The slope, $m$, in the slope-intercept form determines the direction and steepness of the line, with positive slopes indicating an upward trend and negative slopes indicating a downward trend.
The $y$-intercept, $b$, in the slope-intercept form represents the point where the line crosses the $y$-axis, providing information about the starting value of the line.
The slope-intercept form is particularly useful for graphing linear equations, as the slope and $y$-intercept can be used to easily plot the line on a coordinate plane.
Understanding the slope-intercept form is crucial for solving linear equations, as it allows for the identification of key features of the line, such as its direction, steepness, and starting point.
Review Questions
Explain how the slope-intercept form, $y = mx + b$, can be used to solve linear equations in the context of 2.1 Use a General Strategy to Solve Linear Equations.
The slope-intercept form, $y = mx + b$, is a valuable tool for solving linear equations in the context of 2.1 Use a General Strategy to Solve Linear Equations. By rearranging the equation into the slope-intercept form, you can isolate the variable $y$ and solve for the unknown variable, whether it is $x$ or $b$. The slope $m$ and $y$-intercept $b$ provide key information about the structure of the linear equation, allowing you to determine the solution and understand the relationship between the variables.
Describe how the slope-intercept form can be used to graph linear equations in the context of 3.1 Graph Linear Equations in Two Variables.
In the context of 3.1 Graph Linear Equations in Two Variables, the slope-intercept form, $y = mx + b$, is essential for graphing linear equations. The slope $m$ determines the direction and steepness of the line, while the $y$-intercept $b$ provides the starting point on the $y$-axis. By using the values of $m$ and $b$, you can easily plot the line on a coordinate plane and visualize the relationship between the variables $x$ and $y$. This understanding of the slope-intercept form is crucial for accurately graphing and interpreting linear equations.
Analyze how the slope-intercept form, $y = mx + b$, can be used to find the equation of a line in the context of 3.3 Find the Equation of a Line.
In the context of 3.3 Find the Equation of a Line, the slope-intercept form, $y = mx + b$, provides a powerful tool for determining the equation of a line. Given the slope $m$ and a point $(x_1, y_1)$ on the line, you can use the slope-intercept form to solve for the $y$-intercept $b$, which, along with the slope $m$, fully defines the equation of the line. Alternatively, if you know two points on the line, you can use the slope-intercept form to calculate the slope $m$ and then solve for the $y$-intercept $b$, again yielding the complete equation of the line. This understanding of the slope-intercept form is essential for finding the equation of a line in various contexts.
Related terms
Slope:
The slope of a line is a measure of its steepness, indicating the rate of change between the $y$-values and $x$-values. It is represented by the variable $m$ in the slope-intercept form.
y-Intercept:The $y$-intercept is the point where the line crosses the $y$-axis, represented by the variable $b$ in the slope-intercept form. It indicates the starting value of the line on the $y$-axis.
Linear Equation:
A linear equation is an equation that represents a straight line in the coordinate plane, where the variables $x$ and $y$ are related by a constant rate of change, or slope.
"Slope-Intercept Form" also found in:
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Calculus I
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4116 | https://electronics.stackexchange.com/questions/408205/confusion-with-output-impedance-using-h-parameters | Stack Exchange Network
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Confusion with output impedance using h parameters
Regarding the below liner circuit model, as far as I understand to find the output impedance the input current Ib is set to zero and the output resistance is calculated as Vc/Ic. I think this way of modeling and calculating the output resistance comes from the Thevenin model. So the definition output impedance will be Vc/Ic when input port is open circuit (?).
So if I set Ib to zero, hfr Ib becomes zero and the only thing left is hoe where Ic = Vc hoe so can we say that the output impedance is 1/hoe ?
I'm not sure above is correct but in some texts they say the output impedance is infinity theoretically? How can that be estimated? If wrong how is the output impedance is derived following the circuit above?
What is meant by the following? I had to take photo to add the original text:
1 Answer 1
can we say that the output impedance is 1/hoe ?
The output admittance is simply \$h_{OE}\$. That's what \$h_{OE}\$ is by definition. So yes output impedance is the reciprocal.
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4118 | https://www.youtube.com/watch?v=HrD_vpFzrBk | Complement of a Set
Math and Stats Help
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This video defines the complement of a set. I also show all of the different symbols used by different authors of textbooks. I demonstrate a few examples of finding the complement of a set in a couple of different ways that could show up in a course.
If you want to view all of my videos in a nicely organized way, please visit . It is free for anyone to use. Please share with others you know who are struggling with math topics or statistics.
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Transcript:
hi for this video what i want to do is talk to you about a complement of a set so the complement of a set contains all the elements that are not in the given set okay um so different textbooks have different notations my current textbook uses this notation so all of the examples that i'm using i'm using that notation just to make it easier for my students but any one of these are acceptable notations for the complement of a set the first one is a to the c power a prime and tilde a but no matter what symbol is used we just read this as not in set a so it's not in a okay you don't always even have to say the set um so whenever you see these symbols you're looking for everything that is not in the given set so if i just looked at set a one three five seven and nine there are a lot of numbers in our real number system that are not in this set so it could be a very large amount of numbers or you can narrow it down to where you're dealing with multiple sets so i have set a as one three five seven and nine b is one four six seven and nine and d is two three four five and six i chose not to use c um as a set name even though you can just because of the symbol that i'm using but it doesn't matter what your sets are called okay so for this first one we are looking for everything that is not in b and in d so remember that this symbol means and so basically what we're looking for here is everything that is in set d that is not also in b so i have the numbers 2 3 4 5 and 6. we can see that 4 and 6 are both in b so those we would want to eliminate so our answer to this the not in b and in d we would just say are the numbers two three and five okay so moving to the next one we're looking for everything that is not in a but it is in b so this time we're going to look at b and i'm going to eliminate anything that i see in a so we know that a has 1 in common i tried to switch colors all right so we're looking at b and we know that one is in there four six seven and nine so the only two numbers that are left that we would not eliminate for this one would just be 4 and 6. so 4 and 6 are the numbers that are in b and not in a okay and then for the last one that we have we have um set d so it's not in set d and it is in a okay so this time we're gonna be looking at my elements in a and we're going to eliminate any elements that we see that are in d because we only want things that are in a but not d okay so i would eliminate the 3 and the 5 the 1 7 and 9 are not in d so i can just say that this is 1 7 [Music] and 9. okay so for this one we're looking for the elements that are in set a and not in d so we eliminated everything that they have in common a lot of times you won't see it like this in a textbook the textbook that i'm currently teaching from does do it this way most of the time you're going to see complement given this way where you're given a universal set and you're given values for two other sets so this is known as the universal set whenever a u is used and that's saying those are the only numbers that can be used for this particular problem so for this one our u is 1 2 3 4 5 6 7 8 9 and 10. and so for our first example we're looking for everything that is not in a so your textbook could have written this as find a prime or it could have said find tilde a same for this one it could have said find b prime or tilde b so all three of these mean the same thing as far as the notation goes so it's just a matter of preference of the author i personally prefer either this one or this one but like i said the book that i'm using right now uses this symbol okay so what we're going to do is we're looking at a so we're looking at all of the values that are in our universal set that are not in a okay so i can eliminate basically all of the even numbers for this one and so the complement of a would end up being 1 3 5 7 and 9. because in our universal set 1 3 5 7 and 9 are contained there but they are not in a okay um so for the last one for b um we're looking for all things that are in our universal set but not in b so we could say that b c or b prime or tilde b whichever one you want to use we can see that 1 and 2 are both there so if it helps you can always go through and cross out okay 1 and 2 are there 3 is not so that's the first thing that i would write down 4 is there so we would cancel that one out 5 is not in b 6 and 7 are both here so we would cancel those out so we would have 8 we would cancel out 9 and we would be left with 10. so these are all of the values that are in our universal set that are not in set b and for the previous one these were all the values that were in the universal set but not in set a as always thanks for watching if you have any questions please let me know if there are additional topics you would like me to cover please let me know that as well |
4119 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-6/section/3.1/primary/lesson/numbers-with-decimal-place-value-msm6/ | Skip to content
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3.1
Numbers with Decimal Place Value
Written by:Jen Kershaw, M.ed | Kimberly Hopkins
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Julie goes to the gas station with her father. She notices that the gas pump says 2.076 gallons, but she doesn't know how to say the number. What is the number written in words?
In this concept, you will learn about decimals.
Decimal and Place Values
There are a couple of ways to write a part of a whole. One of the ways is to write it is as a decimal. A decimal is a part of a whole.
An example of a decimal is 4.56. Notice that there is a point in the middle of the number. The point is called the decimal point. The decimal point divides the number between wholes and parts. To the right of the decimal point are the parts of the whole and to the left of the decimal point is the whole number.
Another example of a decimal is 0.43. This decimal has 0 whole parts and .43 parts of a whole.
This decimal has two decimal places. Each digit after the decimal is in a different place. These places are called place values.
To express a decimal using place value you need to use a place value chart.
Here is a place value chart.
| Tens | Ones | | Tenths | Hundredths | Thousandths | Ten Thousandths |
--- --- ---
| | | . | | | | |
Notice that the words after the decimal point each has a 'ths' in it. The 'ths' lets you know that you are working with a part of a whole.
You can take the decimal 0.43 and write it in the place value chart. Take the last example and write it in the place value chart. Above each number is a word. That word gives you the value of that digit according to its place in the chart. This number is forty-three hundredths. The three is the last number and is in the hundredths place so that lets you know to read the entire number as hundredths.
| Tens | Ones | | Tenths | Hundredths | Thousandths | Ten Thousandths |
--- --- ---
| | | . | 4 | 3 | | |
To better understand what whole the decimal is a part of, you can use a hundreds grid. Notice that the number in the last example was 0.43 or 43 hundredths. The hundredths tells you that this is “out of one hundred.” Here is a picture of a hundreds grid.
Now show 43 hundredths of the hundreds grid. To do that, shade 43 squares. Each square is one part of one hundred.
If you look at a place value chart, you can see that there are other decimal names besides hundredths. You can also have tenths. An example of tenths is 0.5.
Here is a number that is five-tenths. You can create a picture of five-tenths using a grid of ten units. If you want to show 0.5 in this box, shade five boxes of the ten.
You can make pictures of tenths, hundredths, thousandths and ten-thousandths.
Examples
Example 1
Earlier, you were given a problem about Julie and the number of gallons on the gas pump.
If there are 2.076 gallons, what is the number of gallons written in words?
First, place the number in a place value chart.
| Tens | Ones | | Tenths | Hundredths | Thousandths | Ten Thousandths |
--- --- ---
| | 2 | . | 0 | 7 | 6 |
Next, write the name of the place value above the 6.
thousandths
Then, write the number in words.
two and seventy-six thousandths
The answer is two and seventy-six thousandths.
Example 2
Write the number as a decimal using the grid.
First, identify the type of grid.
hundreds grid
Next, count the shaded squares.
57
Then, write the decimal.
0.57
The answer is 0.57.
Example 3
Write the number as a decimal using the grid.
Fist, identify the type of grid.
hundreds grid
Next, count the shaded squares.
12
Then, write the decimal.
0.12
The answer is 0.12.
Example 4
Write the number as a decimal using the grid.
First, identify the type of grid.
hundreds grid
Next, count the shaded squares.
25
Then, write the decimal.
0.25
The answer is 0.25.
Example 5
Write the number in words and as a decimal using the grid.
First, identify the type of grid.
hundreds grid
Next, count the shaded squares.
80
Then, write the decimal.
0.80
The answer is 0.80.
Review
Look at each hundreds grid and write a decimal to represent the shaded portion of the grid.
Identify the place value of each of the digits in the following decimals.
@$\begin{align}.32\end{align}@$
@$\begin{align}.43\end{align}@$
@$\begin{align}.125\end{align}@$
@$\begin{align}.6\end{align}@$
@$\begin{align}.789\end{align}@$
@$\begin{align}.209\end{align}@$
@$\begin{align}.1\end{align}@$
@$\begin{align}.009\end{align}@$
@$\begin{align}.08\end{align}@$
@$\begin{align}.003\end{align}@$
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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4120 | https://mathquestions.quora.com/If-a-2-b-2-c-2-d-2-ab-cd-how-can-you-prove-that-a-b-c-d | If a^2+b^2/c^2+d^2 = ab/cd, how can you prove that a:b=c:d? - Math Questions - Quora
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If a^2+b^2/c^2+d^2 = ab/cd, how can you prove that a:b=c:d?
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Enrico Gregorio
Associate professor in Algebra
·3y
If you set a=b x a=b x and c=d y c=d y, you get
a 2+b 2 c 2+d 2=b 2(x 2+1)d 2(y 2+1)a 2+b 2 c 2+d 2=b 2(x 2+1)d 2(y 2+1)
and
a b c d=b 2 x d 2 y a b c d=b 2 x d 2 y
Thus your problem becomes to prove that if
x 2+1 y 2+1=x y x 2+1 y 2+1=x y
then x=y x=y
Cross multiply in the given relation to obtain
x 2 y+y=x y 2+x x 2 y+y=x y 2+x
that can be rewritten as x 2 y−x y 2−x+y=0 x 2 y−x y 2−x+y=0 and so
x y(x−y)−(x−y)=0 x y(x−y)−(x−y)=0
hence
(x y−1)(x−y)=0(x y−1)(x−y)=0
Thus either x=y x=y or x y=1 x y=1.
No, you cannot prove that x=y x=y.
9 3
Lai Johnny
M. Phil in Mathematics Major, The Chinese University of Hong Kong (Graduated 1985)
·3y
We can’t prove the statement because it is not true!
For any a>b>0 a>b>0, then letting
a 2+b 2 c 2+d 2=a b c d=k a 2+b 2 c 2+d 2=a b c d=k
for some positive k k yields
a 2+b 2=k(c 2+d 2)and a b=k c d a 2+b 2+2 a b=k(c 2+d 2)+2 k c d(a+b)2=k(c+d)2 a 2+b 2=k(c 2+d 2)and a b=k c d a 2+b 2+2 a b=k(c 2+d 2)+2 k c d(a+b)2=k(c+d)2
a+b=√k(c+d)(1)(1)a+b=k(c+d)
If c<d c<d, then we similarly get
a−b=−√k(c−d)(2)(2)a−b=−k(c−d)
(1)+(2)(1)−(2)yields(1)+(2)(1)−(2)yields
a b=d c≠c d a b=d c≠c d
Continue Reading
We can’t prove the statement because it is not true!
For any a>b>0 a>b>0, then letting
a 2+b 2 c 2+d 2=a b c d=k a 2+b 2 c 2+d 2=a b c d=k
for some positive k k yields
a 2+b 2=k(c 2+d 2)and a b=k c d a 2+b 2+2 a b=k(c 2+d 2)+2 k c d(a+b)2=k(c+d)2 a 2+b 2=k(c 2+d 2)and a b=k c d a 2+b 2+2 a b=k(c 2+d 2)+2 k c d(a+b)2=k(c+d)2
a+b=√k(c+d)(1)(1)a+b=k(c+d)
If c<d c<d, then we similarly get
a−b=−√k(c−d)(2)(2)a−b=−k(c−d)
(1)+(2)(1)−(2)yields(1)+(2)(1)−(2)yields
a b=d c≠c d a b=d c≠c d
Remark:
By the way, if a>b>0 a>b>0 and c>d>0 c>d>0 satisfying the condition, then
a b=c d a b=c d
9 1
Kevin Solari
3y
Assuming you really mean (a^2+b^2)/(c^2+d^2) = ab/cd then the statement is incorrect. Eg put a=1,b=2,c=2 and d=1. The reason is that c and d are interchangeable so if (a^2+b^2)/(c^2+d^2) = ab/cd then either a:b=c:d or a:b=d:c
9 1
Sankaran Murthy
Retired Engineer, Teacher & Small Businessman
·3y
Cross multiply the denominators and solve it as a quadratic equation in a.
cd a^2 - b (c^2 + d^2) a + cdb^2 = 0
a = (b (c^2 + d^2) +or- sqrt(b^2 (c^4 + d^4 + 2c^2 d^2) - 4c^2 d^2 b^2)) /(2cd)
= (b (c^2 + d^2) +or- sqrt(b^2 (c^4 + d^4 - 2c^2 d^2))/(2cd)
= (b (c^2 + d^2) +or- b (c^2 - d^2))/(2cd)
= b...
Milana
11th Grader 2025-2026
·3y
If a≠b≠c≠d, then:
a^2cd + b^2cd = c^2ab + d^2ab
a^2cd + b^2cd - c^2ab - d^2ab = 0
ac((ad-bc) - bd(ad-bc) = 0
(ad-bc)(ac-bd) = 0
ad - bc = 0
ad = bc
a/b = c/d =>a:b = c:d PROVED
View 9 other answers on parent question
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4121 | https://www.youtube.com/watch?v=dUb2sT-oa2M | Simplifying Fractions | EasyTeaching
EasyTeaching
254000 subscribers
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Posted: 3 Aug 2021
Reducing fractions to their simplest form makes them easier to work with and helps with communication.
1) Find the greatest common factor (GCF) of both the numerator and the denominator.
2) Divide numerator and denominator by the GCF.
Find more resources at
64 comments
Transcript:
simplifying fractions we should always reduce fractions to their simplest form it makes fractions easier to work with and it also helps us to communicate more clearly let's take this example we can see from this fraction wall that one half is equivalent to 4 8. but what is the process of simplifying 4 8 so we end up with the simplest form one-half first we find the greatest common factor of both the top number and the bottom number the greatest common factor is the largest number that will divide evenly into both numbers 1 2 and 4 are all common factors the largest or greatest common factor is four then we divide both the numerator and the denominator by the greatest common factor 4 divided by 4 1. 8 divided by four two four eight simplified is one-half it's important to remember that we must always divide the numerator and the denominator by the same number whatever we do to the top number we must do to the bottom number and vice versa let's have a go together at simplifying the following two fractions we'll start with this one 12 16. step one find the greatest common factor the factors of 12 are 1 2 3 4 6 and 12. and here are the factors of 16. one two four eight and sixteen 12 and 16 have these factors in common the largest of the common factors is four step two divide the numerator and the denominator by the greatest common factor 12 divided by 4 and 16 divided by 4. 12 divided by 4 is 3. 16 divided by 4 is 4. 12 16 reduced to its simplest form is three quarters now let's simplify the second fraction 25 30 fifths step one find the greatest common factor the factors of 25 are 1 5 and 25. the factors of 35 are 1 5 7 and 35. 25 and 35 have these factors in common five is the largest common factor step two divide the numerator and the denominator by the greatest common factor 25 divided by 5 is 5. 35 divided by 5 is 7. 25 35 reduced to its simplest form is 5 7. if we are having trouble working out the greatest common factor we can choose any common factor divide both the numerator and the denominator by this factor and then repeat the steps until they can be reduced no further this can be particularly helpful when simplifying fractions with large numbers for example let's simplify this fraction I know a common factor of both numbers is 2 because both are even numbers I'll divide the numerator and the denominator by 2. 64 divided by 2 is 32. 112 divided by 2 is 56. we've reduced the original fraction but it's not yet in its simplest form we know this because we can still divide further both can still be divided by two 32 divided by 2 is 16. 56 divided by 2 is 28. we can divide further still we could divide by 2 again but now I can see a larger common factor can you see it [Music] 4. 4 is a factor of both the numerator and the denominator if we divide the numerator by 4 we get 4. and remember whatever we do to the top number we must do to the bottom number 28 divided by 4 is 7. 4 7. we know this fraction is now in its simplest form because there are no more common factors other than one now it's your turn reduce the following fractions to their simplest form pause here while you work let's see how you went the greatest common factor of 24 and 30 is 6. 24 divided by 6 is 4. 30 divided by 6 is 5. four fifths the greatest common factor of 10 and 15 is 5. 10 divided by 5 is 2. 15 divided by 5 is 3. two-thirds the greatest common factor of 28 and 32 is 4. 28 divided by 4 is 7. 32 divided by 4 is 8. 7 8. [Music] easyteaching.net |
4122 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6182472/ | Pediatric Dehydration Assessment at Triage: Prospective Study on Refilling Time - PMC
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Pediatric Dehydration Assessment at Triage: Prospective Study on Refilling Time
Samuele Caruggi
Samuele Caruggi
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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,✉, Martina Rossi
Martina Rossi
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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, Costantino De Giacomo
Costantino De Giacomo
Paediatrics Division, ASST Grande Ospedale Metropolitano Niguarda, Milan, Italy.
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, Chiara Luini
Chiara Luini
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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, Nicola Ruggiero
Nicola Ruggiero
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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, Alessandro Salvatoni
Alessandro Salvatoni
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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, Silvia Salvatore
Silvia Salvatore
Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
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Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Varese, Italy.
Paediatrics Division, ASST Grande Ospedale Metropolitano Niguarda, Milan, Italy.
✉
Corresponding author: Samuele Caruggi, Pediatric Department, Ospedale “F. Del Ponte”, University of Insubria, Via F. Del Ponte 19, 21100 Varese, Italy. Tel: +39-0332299390, Fax: +39-3460451714, s.caruggi@gmail.com
✉
Corresponding author.
Received 2017 Dec 18; Revised 2018 Apr 20; Accepted 2018 Apr 23; Issue date 2018 Oct.
Copyright © 2018 by The Korean Society of Pediatric Gastroenterology, Hepatology and Nutrition
This is an open-access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC6182472 PMID: 30345241
Abstract
Purpose
Dehydration is a paediatric medical emergency but there is no single standard parameter to evaluate it at the emergency department. Our aim was to evaluate the reliability and validity of capillary refilling time as a triage parameter to assess dehydration in children.
Methods
This was a prospective pilot cohort study of children who presented to two paediatric emergency departments in Italy, with symptoms of dehydration. Reliability was assessed by comparing the triage nurse's measurements with those obtained by the physician. Validity was demonstrated by using 6 parameters suggestive of dehydration. Comparison between refilling time (RT) and a validated Clinical Dehydration Score (CDS) was also considered. The scale's discriminative ability was evaluated for the outcome of starting intravenous rehydration therapy by using a receiver operating characteristic (ROC) curve.
Results
Participants were 242 children. All nurses found easy to elicit the RT after being trained. Interobserver reliability was fair, with a Cohen's kappa of 0.56 (95% confidence interval [CI], 0.41 to 0.70). There was a significant correlation between RT and weight loss percentage (r-squared=−0.27; 95% CI, −0.47 to −0.04). The scale's discriminative ability yielded an area under the ROC curve (AUC) of 0.65 (95% CI, 0.57 to 0.73). We found a similarity between RT AUC and CDS-scale AUC matching the two ROC curves.
Conclusion
The study showed that RT represents a fast and handy tool to recognize dehydrated children who need a prompt rehydration and may be introduced in the triage line-up.
Keywords: Dehydration, Triage, Child, Gastroenteritis, Diarrhea, Vomiting
INTRODUCTION
Dehydration is a significant depletion of body electrolytes and water, often secondary to acute gastroenteritis , or to other diseases that cause vomiting, diarrhea or polyuria . In the United States, acute diarrhea is responsible for around 1.5 million outpatient visits, 200,000 hospitalizations and 300 deaths annually . In Europe, according to an Italian study, it is estimated that rotavirus produces 3.6 million episodes of gastroenteritis [4,5].
Dehydration is a medical emergency, but there is no single standard parameter to evaluate it during the triage [6,7,8,9] and there are no systematic reviews regarding the usefulness of history, objective examination, and laboratory tests to assess dehydration . Weight loss recognized as an objective “gold standard” for dehydration rate often cannot be calculated because recent or “baseline” hydrated weights are rarely available on acute-care visits . Thus, several other related symptoms and signs should be considered including: urine output, sunken eyes, lack of tears, dry mucous membranes, heart rate (RR), respiratory rate (RR) and effort, capillary refilling time (RT), and skin turgor .
The European Society of Paediatric Gastroenterology Hepatology and Nutrition (ESPGHAN) guideline suggests different score systems to quantify dehydration, based on clinical signs and symptoms considered together (for example, RT, skin turgor, and urine output), called “dehydration scales” . The American Academy of Paediatrics (AAP) guidelines recommend making clinical decisions based on the degree of dehydration: mild (3–5%), moderate (6–9%), or severe (>10%) .
In 2008, the Working Group of ESPGHAN highlighted that none of the dehydration scales available at that time had a valid utility in individual patients and started a series of studies to validate the system introduced by Friedman et al. , the Clinical Dehydration Score (CDS). This scale consists in four clinical elements: general and eyes appearance, hydration of mucous and tears. Each item is rated from 0 to 2, and the total score is between 0 and 8. The three final categories are: no degree of dehydration (CDS score, 0), mild dehydration (CDS score, 1–4), and moderate/severe dehydration (CDS score, 5–8). In the last 15 years, the CDS has been validated by different studies [16,17,18,19]. In 2014, the Working Group of ESPGHAN published an update of the guidelines, which substantially confirm the results and conclusions obtained in 2008: the CDS scale had moderate interobserver reliability but should be used in combination with other criteria to guide the physician intervention .
The RT is a simple and quick parameter to be obtained; the examiner presses fingernail for 5 seconds, and estimates the time needed to return at normal color after releasing the finger pressure . The value for the non-dehydrated children is less than 2 seconds, whilst severity of dehydration increases with prolonged time up to 4 seconds which corresponds to severe shock .
Usefulness and validity of the RT in the assessment and severity of dehydration were attested in a systematic review included in the National Institute for Health and Care Excellence (NICE) guidelines of 2009 .
The aim of this study was to assess the reliability and validity of the RT in children who presented to the emergency department (ED) with symptoms suggestive of dehydration, such as diarrhea and vomiting. The second aim of the study was to demonstrate the feasibility of RT as a triage parameter.
MATERIALS AND METHODS
Setting and participants
This was a prospective comparative cohort pilot study. We collected data on children with symptoms at risk of dehydration, who presented to two Italian pediatric ED in Varese (between August 2015 and March 2017) and in Milano (between December 2016 and March 2017) whenever the physicians in charge of the study were on call. Informed consent was obtained from the guardians of all participants. The study was approved by the Institutional Review Board of Ospedale F. Del Ponte (No. 2015-0807).
Eligibility criteria included the presence of: acute diarrhea (at least three stools in liquid or semi-liquid consistency in the previous 24 hours) and/or vomiting (at least 3 episodes in the previous 24 hours). As exclusion criteria were considered: new born babies (correct age, <1 month), children on tube feeding, or with metabolic, or kidney or type 1 diabetes or mental retardation or severe neurological diseases, incomplete data collection form, poor Italian language comprehension and absence of informed consent.
Procedures
The study included two sequential parts, performed first by the nurse triage, and later by the attending clinician in the examining room. Eight nurses and four doctors participated in the study. We have included all nurses who were currently working in our accident and emergency (A&E) department as a fixed job and all doctors who were also working in the pediatric gastroenterology outpatient clinic. The decision to limit the number of health care professionals was based on personal willingness to participate to this pilot study and on frequent work overloading in A&E. Besides, we tried to have participants with both high motivation and similar experience to get better results.
Each nurse working in our ED was instructed about the study protocol and the correct way to elicit the RT prior to the study with a training session.
The nurses training on RT and dehydration assessment was based on a short frontal lecture on acute diarrhea and dehydration and explanation of the study performed by an expert physician (S.S.) followed by an interactive theorical and practical session during which doubts were clarified. The method of detecting the RT was first showed using slides, explained as reported in the literature and then practically performed and checked. Particularly it was highlighted that to obtain a correct measured the pressure should gradually be greater on the palmar surface of the distal fingertip, which must be released immediately after the capillary bed becomes pale, after which the elapsed time must be measured until the restoration of the normal color. A time less than 2 seconds means no dehydration; the longer the RT, the greater the dehydration. We then carried out practical tests on children admitted to A&E in the same day to make sure everyone has understood the method. All nurses found the explanation clear and the test simple to perform and the physicians were fully satisfied of the ability acquired by the nurses. We decided to repeat the training at the end of the first part of the study, to assess the nurses' performance and the re-evaluation was again excellent (Fig. 1).
Fig. 1. How to assess capillary refilling time at the triage.
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To all nurses the CDS scale appeared more subjective, difficult to score and time consuming than RT, so this scale was included only in the physician assessment. For data collection we used a standardized evaluation form which included items shown in Fig. 2.
Fig. 2. Data collection form.
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At triage the nurses enrolled all children reporting symptoms of acute diarrhea or vomiting and filled in the first part of the form recording the RT. According to the triage code determining the time priority, the doctor reassessed the child performed a detailed history and physical examination, assigned a CDS score and RT, completed the form and started treatment. Whenever the nurse noticed a state of dehydration based on an alteration of the capillary RT, the physician was informed about the general condition of the child and the nurse assessment and authorized the administration of oral rehydration therapy (ORS) even before the visit, unless a suspicion of severe infection, lethargy or acute abdomen was raised.
The partial results we obtained allowed us to create a new form (Fig. 3), which replaced the old one starting from July 2016. In this new form, we added the weight gain after rehydration, measured by a physician after the ORS rehydration therapy, to use it as an indicative weight of a state of good hydration and compare it with the weight measured at triage. At the beginning we not consider the patients' waiting time between nurse's and physician's RT assessment, so in the new form we introduce this measure and the quantity of ORS assumed by the dehydrated child.
Fig. 3. New data collection form.
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Patient assessment and data collection
The data collected by the triage nurse included basic demographics, presenting symptoms (diarrhea, vomiting), general appearance and state of consciousness (unconscious, lethargic, normal, irritable), RT at the arrival in ED and the providing of ORS (after triage rating). The physician completed the assessment and filled in the second part of the same standardized form, recording the general appearance and state of consciousness; HR and RR, temperature, current weight and the previous weight (if available); symptoms (presence and type of diarrhea or vomiting, timing of onset, number of episodes in the previous 24 hours); RT, skin turgor, urine output reported at home, intake of ORS or other drink at home and CDS scale. RR was measured for 60 seconds by observing chest wall movements with the child quiet and comfortable . We have defined tachycardia and tachypnea when HR and RR were above the 90th centiles, according to different age, as reported by a systematic review by Fleming et al. in 2011 .
Capillary RT was assessed at the fingertip by using a standardized technique . Skin turgor was assessed by pinching a small skin fold between the thumb and index finger on the lateral abdominal wall at the level of the umbilicus .
Type of rehydration needed and the pop up of episodes of vomiting during oral rehydration were also reported. Finally, we considered the patient evolution in terms of discharge timing or hospitalization. In the new form we considered weight gain after rehydration to overcome the lack of an available previous weight in most of cases.
Outcomes
The primary outcome of our project was reliability of RT; to demonstrate the equivalence between triage nurse and clinician assessments of the RT, by comparing results of the tests and finding out if they match for every child or if they were different. Our final purpose is to introduce this parameter in the triage line up for quantification of dehydration and assignment of the priority code.
The secondary outcome was to demonstrate the validity of RT for the rapid quantification of presence and degree of dehydration. We compared RT with parameters thought to be reflective of dehydration presence and severity; the CDS scale, the number of episodes of vomiting and diarrhea before presentation to the ED, RR, HR, length of hospital staying, percentage of weight loss . Discriminative validity was assessed by the need of intravenous rehydration therapy.
Analysis
Characteristics of the study sample were described with frequency counts and percentages for categorical variables. The triage nurse evaluation of general aspect and RT were compared with the valuation of attending physician by using chi-squre test of Pearson's and Cohen's kappa to demonstrate interobserver reliability. A p-value less than 0.05 was considered statistically significant. Results were interpreted according to the criteria of Landis and Koch , in which κ values of 0.21 to 0.40, 0.41 to 0.60, 0.61 to 0.80, and 0.80 signify fair, moderate, substantial, and almost perfect agreement, respectively. In estimating validity, correlations were quantified by using Pearson correlation coefficient (Pearson's ρ). We sought statistical associations between variables indicative of dehydration (weight loss, cardiac and respiratory frequency, the number of episodes of diarrhea, vomiting, combination of diarrhea and vomiting, and the discharge time) and the RT; we repeated the analysis also for the score obtained on the CDS scale. Finally, discriminative validity was assessed as the area under the receiver operating characteristic (ROC) curve. We calculated the area under the curve (AUC), to estimate the specificity and sensitivity of RT and CDS, using as discriminating positive for the state of dehydration the start of intravenous therapy, then measuring the difference between the two AUC. Sensitivity, specificity, positive likelihood ratio, positive predictive value, and negative predictive value were quantified by using the start of intravenous therapy as parameter reflective of dehydration severity.
RESULTS
We included 242 children aged 9 months to 17 years. In our population 133 (55.0%) children aged between 0 and 3 years, 92 (38.0%) children aged between 4 and 12 years and only 17 (7.0%) adolescents were between 13 and 17 years old. The presenting symptom was: diarrhea in 39 (16.1% of the population), vomiting in 112 (46.3%), association of vomiting and diarrhea in 89 (36.8%) children, whilst only one case (0.4%) complete refuse to feed and to drink was reported.
Only 57 (23.6%) patients had a documented previous weight assessed by a physician or, in the new form, a weight after the rehydration therapy (which was found in 24 children [9.9% of the total]). Among this group, 41 (71.9%) showed a <5% weight loss (respect to the previous evaluation) and 7 (12.3%) a weight loss between 5% and 10% of the usual weight. The others 9 children (15.8%) showed an increased weight: only one pointed out signs of dehydration (RT=2, CDS=1) (Table 1).
Table 1. Weight and CDS Scale Evaluation.
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CDS: Clinical Dehydration Score.
The new form allowed us to analyse patients' waiting time between nurse's and physician's RT assessment in 70 children: the average expectation between patients with altered RT at the triage was 24 minutes, with a median of 20 minutes, compared to an average of 41 minutes and a median of 30 minutes in children with a normal RT.
A post-rehydration weight was collected in 24 children: 41.7% showed a <1% weight gain; 33.3% a weight gain between 1% and 2%; 20.8% between 2% and 3%; only 4.2% a >3% weight gain.
The CDS at the visit was found equal to 0 in 151 (62.4%) cases and between 1 and 4 in 83 (34.3%) children. There were no cases with CDS above 4, as we can see in Table 1. The physician reported that the lack of crying limit the validity of assessment for the presence of tears in 27 cases (11.2%) and because of this uncertain parameter the score was not completed in 8 children (3.3%).
RT reliability
The RT evaluated by the triage nurse was <2 seconds in 196/242 children (81.0%) and ≥2 seconds in 46 cases (19.0%). The attending physician found the RT <2 seconds in 208/242 children (86.0%) and, ≥2 seconds in just 34 (14.0%) cases (Table 2).
Table 2. Refilling Time Elicited at Triage and at Visit.
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We compared the RT test results obtained by nurses and physicians. The evaluation was concordant in 207 cases (85.5%), discordant in 35 cases (14.5%): among these, in 26 children (74.3%, or 10.7% of the total) the triage nurse overestimated the dehydration compared to the physician's one. In 9 of these 26 children, the nurse gave the ORS to the children after the triage, so the rehydration therapy was started before the visit; this can explain the relative overrating of dehydration in these children. The correlation between the measurements obtained by the nurse with the physician's evaluations was rated as moderate in accordance with the valuation guidelines Cohen's kappa (κ=0.480; 95% confidence interval [CI], 0.073 to 0.534).
After that we ruled out the children whom a nurse gave ORS before the physician's visit and we calculated the Cohen's kappa with the new data: the result was similar, with a moderate Cohen's kappa value (κ=0.556; 95% CI, 0.411 to 0.701). Pearson's chi-square test showed a moderate contingency between the measurements (contingency coefficient, 0.42).
RT validity
Results of correlation between RT, CDS scale scores and variables are shown in Table 3.
Table 3. Statistical Analysis of the Correlation Coefficient of the Main Variables.
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The highest degree of correlation was found between RT and weight loss of percentage calculated using a previous weight reported by the parents, when available (ρ=−0.27; 95% CI, −0.47 to −0.04; p=0.019); it confirmed the statistical significance of the parameter. The correlation between length of stay and RT was lower but still significant (ρ=0.22; 95% CI, 0.09 to 0.35; p<0.01).
The number of diarrhea associated with vomiting episodes in 24 hours was correlated with the increase in RT (ρ=0.33; 95% CI, 0.12 to 0.51; p<0.01).
The correlation between CDS scale scores and the different parameters was similar: we found the maximum degree of correlation between CDS and percentage of weight loss (ρ=−0.37; 95% CI, −0.55 to −0.15; p<0.01), with statistical significance. A significant correlation was found between CDS and length of stay (ρ=0.15; 95% CI, 0.01 to 0.28; p=0.03) and between CDS and the number of episodes of diarrhea associated with vomiting during the previous 24 hours (ρ=0.37, 95% CI, 0.16 to 0.54; p<0.01).
For the outcome of intravenous rehydration therapy, the AUC was 0.65 for the CDS (95% CI, 0.57 to 0.73; p<0.01). Sensitivity and specificity were found to be respectively 51.2% and 72.5%. AUC for the RT score was 0.65 (95% CI, 0.57 to 0.72; p<0.01). The sensitivity and specificity were evaluated as 33.0% and 97.6%, respectively (Fig. 4).
Fig. 4. Receiver operating characteristic curve results of refilling time (A) and Clinical Dehydration Score (B).
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DISCUSSION
Proper assessment of dehydration is the critical step for a correct management and prompt rehydration treatment of children with acute diarrhea and vomiting. However, no specific parameter of dehydration is currently included in the triage line-up.
In our study, we evaluated the reliability of the RT performed by the triage nurses and compared the signs of dehydration, RT and CDS scale in a population of children who presented to our ED.
Our results confirmed that acute vomiting and diarrhea is a common cause of referring to ED and in most cases rehydration therapy with ORS is the only treatment needed and can be started immediately, even before the during the time the patient is waiting before the visit, as recommended by several authors [4,8,13] and international guidelines [11,13]. According to our results, the evaluation of RT in triage allowed nurses to give priority to patients who present dehydration and so to reduce the waiting time.
Our goal was twofold; first, to demonstrate the equivalence between RT measurement by triage nurse and attending physician and, then, to prove the reliability, validity and easiness of the RT in the evaluation of children at risk of dehydration because of diarrhea and vomit.
The Pearson's chi-square test evaluated the contingency between the two measurements (nurse and physician) as moderate, according to current statistical guidelines (ρ=0.42). The Cohen's kappa, however, defined a moderate correlation (κ=0.56): the result could have been higher, but due to there an imbalance in the appearance frequency of the system variables, with an excess of non-dehydrated children, the “kappa paradox” phenomenon occurred [25,26].
In most cases the triage nurse correctly assessed the hydration status of the children and ensured a prompt introduction of ORS and a speedy handling of the dehydrated patient who needs an earlier physician assessment. In 14.5% of our children we found a discrepancy in the evaluation of dehydration between nurses and physicians, mainly related to an overestimation by the nurse of the degree of dehydration (74.3%) compared to an underestimation (25.7%) of these cases. This discrepancy can be due to some limits of our study; the lack of data regarding the elapsed time between triage assessment and the physician visit and the number of inter-current episodes of vomiting or diarrhea could explain the nurses' underestimation. The children's grade of dehydration could get worse with a long time between triage evaluation and physician's visit. On the other hand, the assumption of ORS during the time spent in the waiting room could change in a better way the state of hydration of the children. The nurse's overestimation allows a good management of the patient; the prompt administration of ORS in case of not clinically detectable dehydration can be configured as a preventive treatment.
For the secondary outcome of our study we found that RT moderately correlated with other two parameters previously reported in the literature as indicative of dehydration, such as weight loss (ρ=−0.27) and the length of staying in ED, assumed as a direct indicator of the time required to complete rehydration of the patient (ρ=0.22) and showed a similar correlation compared to the CDS scale.
The ROC curve pointed out that there's no differences between the two AUC; it means that the two methods are comparable to assess dehydration in children. Current guidelines on gastroenteritis in children [11,13], consider the CDS scale as the best method to assess the presence and the severity of dehydration; the results obtained with the ROC curve in our population equate RT to CDS scale and demonstrate its validity and reliability.
RT has more specificity (97.6%) than CDS (72.5%) for the outcome of intravenous rehydration therapy need: children with a RT <2 seconds are certainly not dehydrated and a RT >2 seconds is indicative of rehydration therapy need.
It is noteworthy that weight loss estimation, considered as the ideal parameter to assess dehydration, is not frequently available in children and may limit the validity of our analysis.
Finally, all nurses of our ED learned to use RT in a few minutes, appreciated to have an easy, quick and reliable parameter for the dehydration assessment of the children at triage and reported that the CDS scale appeared more subjective than RT, difficult to score and time consuming.
The limits of our study consisted mainly by the small number of children referring to our A&E department with moderate or severe dehydration and by the lack of a recent weight data in most of our children.
In accordance with the literature [11,13], the utility of dehydration estimation by the triage nurses was limited by the low proportion of children with RT >2 seconds.
The low sensitivity shown by the RT and CDS could therefore be attributable to the large number of children who were not or slightly dehydrated (RT <2), who did not need this therapy.
Another limitation to our study was the impossibility to calculate the weight loss due to unavailable recent weight in many of our children. To minimize this effect, we introduced in the final form and analysis the post rehydration weight as a surrogate of the weight loss and we have considered the difference in weight as a mirror of the rehydration rate. We also analyzed, whenever reported, the need of intravenous rehydration that could be considered as an indirect indicator of dehydration. In our population the low sensitivity shown by the RT and CDS compared to this treatment could be due to the low sample size that was on intravenous treatment. Besides, the rate of intravenous rehydration could also be related to other factors not dependent on hydration state such as persistent vomit, parental or patient non-compliance with oral rehydration solution, suspicious of acute abdomen, electrolyte imbalance.
To improve the quality of the study we added other evaluation parameters to the study's form; the length of waiting time, the number of vomits or diarrhea episodes during that period and the quantity of ORS solution taken by children before visit. We reached some results about the evaluation of the length of the waiting time, but for the other aspects the results were accurate, but partial and so we decided not to include them in this study to not distort the structure of the original study. Our intention is to propose it as multicentric using the new form, to evaluate the replicability especially of the RT in different hospitals and triage and to increase the number of children with dehydration and allow a more accurate statistical analysis.
Our study showed that the RT is a quick and easily measurable parameter that can be introduced during the triage assessment and represents a reliable and valid method for the valuation of dehydration in children. Although most children referring to the ED are not dehydrated, the nurse fast recognition of dehydration can allow a prompt start of ORS rehydration and determine a different priority access to physician assessment compared to the symptom reported by parents or the general appearance in patients with acute vomiting or diarrhea.
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4123 | https://www.combinatorics.org/ojs/index.php/eljc/article/download/v28i1p35/pdf/ | Cycle decompositions of complete digraphs
A. C. Burgess ∗
Department of Mathematics and Statistics University of New Brunswick Saint John, NB E2L 4L5 Canada
P. Danziger †
Department of Mathematics Ryerson University Toronto, ON M5B 2K3 Canada
M. T. Javed
Department of Computer Science Ryerson University Toronto, ON M5B 2K3 Canada Submitted: Oct 3, 2018; Accepted: Oct 21, 2020; Published: Feb 12, 2021
© The authors. Released under the CC BY-ND license (International 4.0).
Abstract
In this paper, we consider the problem of decomposing the complete directed graph K∗
n
into cycles of given lengths. We consider general necessary conditions for a directed cycle decomposition of K∗
n
into t cycles of lengths m1, m 2, . . . , m t
to exist and provide a powerful construction for creating such decompositions in the case where there is one ‘large’ cycle. Finally, we give a complete solution in the case when there are exactly three cycles of lengths α, β, γ 6 = 2. Somewhat surprisingly, the general necessary conditions turn out not to be sufficient in this case. In particular, when γ = n, α + β > n + 2 and α + β ≡ n (mod 4), K∗
n
is not decomposable.
Mathematics Subject Classifications: 05C51, 05B30, 05C20, 05C38
1 Introduction
Let G be a graph and H = {H1, H 2, . . . , H r} be a collection of subgraphs of G. We say that H decomposes G if the edges of the graphs in H partition the edges of G. In this case, we write G = H1 ⊕ H2 ⊕ · · · ⊕ Hr. If H1 ∼= H2 ∼= · · · ∼= Hr ∼= H, we refer to an
∗
Supported by NSERC Discovery Grant RGPIN-2019-04328.
†
Supported by NSERC Discovery Grant RGPIN-2016-04178.
the electronic journal of combinatorics 28(1) (2021), #P1.35
H-decomposition of G; in this case, we call the decomposition uniform . An oft-studied case is that G is the complete graph Kn and each element of H is a cycle. It is easy to see that if n is even, then there can be no cycle decomposition of Kn. In this case, it is common instead to consider Kn − I the complete graph with the edges of a 1-factor I
removed. We will use Cm to denote a cycle of length m.The existence question for cycle decompositions of complete graphs has a history dating to the mid-1800s; among the first landmark results were Kirkman’s 1847 proof that Kn has a C3-decomposition if and only if n ≡ 1 or 3 (mod 6) and Walecki’s construction of Hamilton cycle decompositions of the complete graph . The existence problem for uniform cycle decomposition of Kn and Kn −I was only settled over a century later [2, 9].
Theorem 1 ([2, 9]) . Let n, k > 3 be integers. There exists a Cm-decomposition of Kn if and only if n is odd and k | n(n−1) 2 . There exists a Cm-decomposition of Kn − I if and only if n is even and m | n(n−2) 2 .
A more general question is the existence of possibly non-uniform cycle decompositions of Kn or Kn − I. It was conjectured by Alspach in 1981 that the obvious necessary conditions for the existence of such a decomposition were sufficient. Alspach’s conjecture was finally verified in a 2014 paper by Bryant, Horsley and Pettersson .
Theorem 2 () . Let n > 3 be an integer, and let Gn denote Kn if n is odd, and
Kn − I if n is even. Let H = {Cm1 , C m2 , . . . , C mr }. Then H decomposes Gn if and only if 3 6 mi 6 n for each i = 1 , 2, . . . , r and m1 + m2 + · · · + mr = nbn−12 c.
More recently, Bryant, Horsley, Maenhaut and Smith have extended this result, finding necessary and sufficient conditions for the existence of a cycle decomposition of the complete multigraph λK n.
Theorem 3 () . Let C = {Cm1 , C m2 , . . . , C mr }. Then C decomposes λK n if and only if the following conditions all hold: 1. λ(n − 1) is even; 2. 2 6 m1, m 2, . . . , m r 6 n;3. m1 + m2 + · · · + mr = λ(n
2
);4. max( m1, m 2, . . . , m r) + r − 2 6 λ
2
(n
2
) when λ is even; and 5. ∑
mi=2
mi 6 (λ − 1) (n
2
) when λ is odd. Also, C decomposes λK n − I if and only if the following conditions all hold: 1. λ(n − 1) is odd; 2. 2 6 m1, m 2, . . . , m r 6 n;
the electronic journal of combinatorics 28(1) (2021), #P1.35 2
m1 + m2 + · · · + mr = λ(n
2
) − n
2
; and 4. ∑
mi=2
mi 6 (λ − 1) (n
2
).
In this paper, we consider an alternative generalization of Alspach’s conjecture, to cycle decompositions of the complete symmetric digraph. For a graph G, we let G∗ denote the digraph formed from G by replacing each edge {x, y } with two arcs xy and yx . In particular, K∗
n
denotes the complete symmet-ric digraph on n vertices. We will use −→
P k to denote a directed path on k vertices, and −→
C k a directed cycle of length k. We use [ u1, u 2, . . . , u k] to denote a directed path with arcs u1u2, u 2u3, . . . , u k−1uk and ( u1, u 2, . . . , u k) to denote a directed cycle with arcs
u1u2, u 2u3, . . . , u k−1uk together with uku1.The case of uniform cycle decomposition of K∗
n
was settled by Alspach, Gavlas, ˇSajna and Verrall .
Theorem 4 () . Let n, m > 2 be integers. There is a −→
C m-decomposition of K∗
n
if and only if m | n(n − 1) and (n, m ) /∈ { (4 , 4) , (6 , 3) , (6 , 6) }.
Let C denote a collection of r directed cycles of lengths m1, m 2, . . . , m r. It is clear that by ignoring the direction on the arcs, any C-decomposition of K∗
n
yields a C-decomposition of 2 Kn. Thus, any such decomposition must satisfy the conditions of Theorem 3 with
λ = 2, giving the following necessary conditions:
Lemma 5. Let C = {−→
C m1 , −→
C m2 , . . . , −→
C mr }. If C decomposes K∗
n
, then: 1. 2 6 m1, m 2, . . . , m r 6 n;2. m1 + m2 + · · · + mr = n(n − 1) ; and 3. max( m1, m 2, . . . , m r) + r − 2 6 (n
2
)
In light of Theorem 3, one possible method of constructing directed cycle decomposi-tions of K∗
n
would be to orient the edges in a cycle decomposition of 2 Kn. However, it is worth noting that the methods of for 2 Kn do not in general produce decompositions whose cycles can be oriented to give a directed cycle decomposition of K∗
n
. In fact, as we shall see in Theorem 24, the necessary conditions stated in Lemma 5 are not also sufficient in all cases.
Definition 6. A nondecreasing list M = ( m1, m 2, . . . , m r) of nonnegative integers will be called n-admissible if it satisfies conditions 1–3 of Lemma 5. We refer to a directed cycle decomposition of K∗
n
corresponding to the admissible list M as an M -decomposition of
K∗
n
.Using the notation of and , for a given n-admissible list M , we let νi(M ) denote the number of instances of i in M . It is worth noting that one of the major differences in decomposing K∗
n
(or λK n) rather than Kn is the possibility of cycles of length 2. Further, if there are cycles of length 2 in a decomposition of K∗
n
, removing them corresponds to
the electronic journal of combinatorics 28(1) (2021), #P1.35 3
removing edges from the underlying undirected graph, Kn. Thus directed cycle decom-position of K∗
n
can be seen as a directed cycle decomposition of G∗ into cycles of lengths greater than 2, where G is an arbitrary graph of order n. This observation leads to a strategy for decomposing K∗
n
.
Definition 7. We say that a directed cycle −→
C decouples the arc xy if −→
C contains yx but not xy . If C = {−→
C m1 , −→
C m2 , . . . , −→
C mr } is a collection of pairwise edge-disjoint cycles, we say that an arc xy is decoupled by C if yx is contained in a cycle of C but xy is not. The following is an obvious consequence of Definition 7.
Lemma 8. Let C = {−→
C m1 , −→
C m2 , . . . , −→
C mr } be a set of pairwise edge-disjoint directed cycles in K∗
n
. There is a −→
C 2-decomposition of K∗
n
− C if and only if C leaves no arc decoupled.
For convenience of notation, we will henceforth let ν = ν2(M ) denote the number of cycles of length 2 in the list M . In light of Lemma 8, it will sometimes be convenient to list only the cycle lengths which are greater than 2, particularly when a decomposition is to contain many 2-cycles. Therefore, we make the following definition.
Definition 9. Let M be an n-admissible list. The associated canonical list is the non-decreasing list ˆM formed from M by removing all instances of 2 from M . Note that the sum of the entries in ˆM is n(n − 1) − 2ν.In this paper, we give further results on cycle decompositions of K∗
n
. In Section 2, we give general constructions and show sufficiency of the necessary conditions in certain cases. In Section 3, we determine necessary and sufficient conditions for the existence of an M -decomposition of K∗
n
when ν > |M | − 3, summarized in Theorem 24. As we will see, the necessary conditions of Lemma 5 turn out not to be sufficient in general. In particular, in Lemma 23 we exhibit a family of n-admissible lists for which K∗
n
admits no
M -decomposition.
2 General constructions
We begin by noting some easy consequences of conditions 2 and 3 of Lemma 5.
Lemma 10. Let M be an n-admissible list with associated canonical list ˆM = ( m1,. . . , m r). Then 1. m1 + · · · + mr−1 ≡ mr (mod 2) 2. m1 + · · · + mr−1 > mr + 2( r − 2) . Hence when r > 3, m1 + · · · + mr−1 > m r.Conversely, given a list ˆM = ( m1, . . . , m r) with 3 6 mi 6 n for each i ∈ { 1, . . . , r } and
m1 +· · · +mr 6 n(n−1) , if ˆM satisfies conditions 1 and 2 above, then ˆM is the associated canonical list of some n-admissible list M .
the electronic journal of combinatorics 28(1) (2021), #P1.35 4
Note that the case where equality holds in condition 2 corresponds with the case where equality holds in condition 3 of Lemma 5. We now give some constructions of cycle decompositions of K∗
n
from known decompo-sitions of complete graphs or complete symmetric digraphs.
Theorem 11. Let M = ( m1, m 2, . . . , m r) be an n-admissible list, where νk(M ) is even for each 3 6 k 6 n. If either 1. n is odd and either ν = 0 or ν > 3; or 2. n is even and either ν = n
2
or ν > n
2
3 ,then K∗
n
is M -decomposable. Proof. We form a new list M ′ which decomposes Kn. We then form the corresponding directed graph K∗
n
, using some of the cycles from Kn to form 2-cycles in K∗
n
. We consider two cases according to the parity of n.First, suppose that n is odd. Let s ∈ { 0, 4, 5} be such that ν ≡ s (mod 3). Let M ′ be the nondecreasing list satisfying that, for i ∈ { 3, 4, . . . , n },
νi(M ′) =
12
ν3(M ) + 13 (ν − s), if i = 3
12
νs(M ) + 1 , if i = s 6 = 0
12
νi(M ), otherwise Since m1 + m2 + · · · mr = n(n − 1), it follows that the sum of the entries of M ′ is (n
2
), and so by Theorem 2, Kn is M ′-decomposable. From such a decomposition, take
νi(M)2
cycles of length i for each 3 6 i 6 n, and orient them both ways to obtain νi(M )directed cycles of length i. Each edge in the remaining cycles can be directed both ways to form a directed 2-cycle, giving ν2(M ) in total. It is easy to see that we obtain an
M -decomposition of K∗
n
.The case where n is even is similar. Again, let s ∈ { 0, 4, 5} such that ν ≡ s + n
2
(mod 3). Let M ′ be the nondecreasing list satisfying that, for i ∈ { 3, 4, . . . , n }
νi(M ′) =
12
ν3(M ) + 13 (ν − s − n
2
), if i = 3
12
νs(M ) + 1 , if i = s 6 = 0
12
νi(M ), otherwise. In this case, we note that Kn − I is M ′-decomposable, and proceed as before, except that the edges of the 1-factor I are also directed both ways to form n
2
directed 2-cycles.
Lemma 12. Let M be an n-admissible list and M ′ an n′-admissible list, where n < n ′.If νk(M ) = νk(M ′) for each k > 3 and there exists an M -decomposition of K∗
n
, then there exists an M ′-decomposition of K∗
n′
.
the electronic journal of combinatorics 28(1) (2021), #P1.35 5
Proof. Decompose K∗
n′
= K∗
n
⊕ K∗
n′−n
⊕ K∗
n,n ′−n
. Decomposing K∗
n
into directed cycles of lengths given in M , it is easy to see that no arc of K∗
n′
is left decoupled. Since all remaining directed cycles to be formed are of length 2, the result follows by Lemma 8. In the remainder of this section, we will give constructions which give directed cycle decomposition in a wide array of cases. Before proceeding, we note the following result which settles existence in the case that the canonical list has size at most 2.
Theorem 13. Let M = ( m1, m 2, . . . , m r). Suppose M has at most two cycle lengths other than 2, i.e. ν > r − 2. There is an M -decomposition of K∗
n
if and only if M is
n-admissible. Proof. It is easy to see that if ν = r (i.e. M = (2 , 2, . . . , 2)), an M -decomposition of K∗
n
exists. Also, note that conditions 2 and 3 of Lemma 5 imply that no n-admissible list has
ν = r − 1. Finally, if M is an n-admissible list with ν = r − 2, its canonical list has the form ˆM = ( mr−1, m r). By condition 3 of Lemma 5 and since M is non-decreasing, mr−1+r−2 6
mr + r − 2 6 n(n−1) 2 . But condition 2 implies that 2( r − 2) + mr−1 + mr = n(n − 1), i.e. (mr−1 +r−2)+( mr +r−2) = n(n−1), so it must be that mr−1 +r−2 = mr +r−2 = n(n−1) 2 ,and hence mr−1 = mr. In this case, it is easy to see that an M -decomposition of K∗
n
exists. Recall that condition 3 of Lemma 5 states that in an n-admissible list M of size t
with maximum entry m, we have that m + t − 2 6 (n
2
). Written in terms of the canonical list ˆM = ( m1, m 2, . . . , m r), the total number of cycles is t = r + ν, so this condition becomes mr + r + ν − 2 6 (n
2
). The next result shows that if equality holds, then an
M -decomposition does indeed exist.
Lemma 14. Let M be an n-admissible list with associated canonical list ˆM = ( m1, m 2,. . . , m r). If
mr + r + ν − 2 =
(n
2
)
,
then K∗
n
is M -decomposable. Proof. First, suppose that mr−1 = mr. Since M is n-admissible and contains ν entries equal to 2, counting edges we have that 2ν + m1 + m2 + · · · + mr = 2 ν + m1 + m2 + · · · + mr−2 + 2 mr = n(n − 1) = 2( mr + r + ν − 2) .
Rearranging gives m1 + m2 + · · · + mr−2 = 2( r − 2), and since mi > 3 for each i ∈{1, 2, . . . , r − 2}, it follows that r = 2, which was dealt with in Theorem 13. Now suppose mr−1 < m r. Let the vertex set of K∗
n
be Zn. We form the cycles of lengths m1, m 2, . . . , m r. An illustration of this construction can be found in Figure 1.
the electronic journal of combinatorics 28(1) (2021), #P1.35 6
0 `0
`1
`2
`3
Figure 1: A (2 , . . . , 2, 4, 5, 7, 12)-decomposition of K∗
12
Let −→
C mr = ( mr − 1, m r − 2, . . . , 2, 1, 0) be the directed cycle of length mr. Define
0 = 1,r−1 = mr − 1, and for i = 1 , 2, . . . , r − 2, let `i =
(∑ij=1 mj
)
− (2 i − 1). Note that, since n(n − 1) = 2( mr + r + ν − 2) = 2 ν + m1 + m2 + · · · + mr,
`r−2 = (m1 + m2 + · · · + mr−2) − (2( r − 2) − 1) = (mr − mr−1 + 2 r − 4) − (2 r − 5) = mr − mr−1 + 1 .
Hence, i −i−1 = mi − 2 for each i = 1 , . . . , r − 1. For i = 1 , 2, . . . , r − 1, define the
mi-cycle −→
C mi = (0 , i−1, i−1 + 1 , i−1 + 2 , . . . , i).
It is easy to verify that −→
C mi has length i −i−1 + 2 = mi.Note that the cycles −→
C m1 , . . . , −→
C mr−1 , −→
C mr leave no decoupled arc, and the result follows by Lemma 8.
Theorem 15. Let M be an n-admissible list with associated canonical list ˆM =(m1, m 2, . . . , m r). Let S = ( s0, s 1, . . . , s r−2, s r−1, s r) be a sequence of non-negative in-tegers with s0 = sr−1 = sr = 0 , and let s be the sum of the entries of S. If M is an
n-admissible list such that: 1. mr + r + ν − 2 = (n
2
); and 2. mr + s 6 n,then K∗
n
is M ′-decomposable, where ˆM ′ = ( m′
σ(1)
, m ′
σ(2)
, . . . , m ′
σ(r)
) such that m′
i
= mi +
si−1 + si for each i ∈ { 1, . . . , r } and σ is a permutation which ensures that ˆM ′ is in nondecreasing order.
the electronic journal of combinatorics 28(1) (2021), #P1.35 7
Proof. We proceed in a similar manner to the proof of Lemma 14, except that we add s
vertices x1, . . . , x s to the cycles in ˆM , and modify the cycles −→
C mi , 1 6 i 6 r, from that lemma as follows. Let `i and −→
C mr be as defined as in the proof of Lemma 14. For i ∈ { 0, 1, . . . , r }, let
s′
i
= s0 + s1 + · · · + si. We define directed paths
Pi = [0 , x s′
i−1+1
, x s′
i−1+2
, . . . , x s′
i
, ` i].
Note that if si = 0, we take Pi = [0 , ` i] of length 1. We also define Qi to be the reversal of Pi, i.e. if si 6 = 0 then
Qi = [ `i, x s′
i
, x s′
i−1
, . . . , x s′
i−1+1
, 0] ,
and Qi = [ `i, 0] otherwise. Now, define −→
C m1 as the concatenation of the path [0 , 1, . . . , ` 1]and Q1. For i ∈ { 2, . . . , r − 2}, define −→
C mi as the concatenation of the paths Pi−1,[i−1, i−1 + 1 , . . . , ` i] and Qi. Finally, define −→
C mr−1 as the concatenation of Pr−2 and [r−2, r−2 + 1 , . . . , r−1, 0] (recall thatr−1 = mr − 1). See Figure 2. It is clear that these cycles leave no decoupled arcs, so the result follows by Lemma 8.
x3
x2
x4
x1
Figure 2: A (2 , . . . , 2, 5, 9, 10 , 12)-decomposition of K∗
16
formed using Theorem 15 with ˆM = (4 , 5, 7, 12), S = (0 , 1, 3, 0, 0). Note that although the statements of Lemma 14 and Theorem 15 are written so that the cycle lengths m1, . . . , m r are in non-decreasing order, this is not actually required in the proof, as long as mr > max 16i6r−1{mi}.
Theorem 16. Let H be an n-admissible list with associated canonical list ˆH = ( h1, . . . , h r)
such that r > 3, hr > 2( r − 1) , h1 + · · · + hr 6 2( n + r − 2) , and, if r is even, r 6 = 4 and
h1 6 r. Then K∗
n
is H-decomposable.
the electronic journal of combinatorics 28(1) (2021), #P1.35 8
Proof. We begin by defining a list K = ( k1, k 2, . . . , k r) by permuting the elements of ˆH
with the purpose of ensuring that K satisfies the inequalities (1) and (2) below, while retaining the property that kr > max 16i6r−1{ki}.(1) If r is odd, then k1 + k3 + · · · + kr−2 + kr > k2 + k4 + · · · + kr−1 + ( r + 1). (2) If r is even, then k2 + k4 + · · · + kr−2 + kr > k1 + k3 + · · · + kr−1 + ( r − 2). If r is odd, take kr = hr, and define k1, . . . , k r−1 by (k2, k 4, . . . , k r−1) = ( h1, h 2, . . . , h (r−1) /2)and (k1, k 3, . . . , k r−2) = ( h(r+1) /2, h (r+3) /2, . . . , h r−1).
This ensures that the elements of ( k1, . . . , k r) with odd index are greater than or equal to those with even index, i.e. k2i−1 > k2i for i ∈ { 1, . . . , (r − 1) /2}. Since kr > 2( r − 1), it is now easy to see that (1) is satisfied. If r is even, take kr = hr, kr−1 = h4, kr−2 = h3, kr−3 = h2 and k1 = h1, and define
k2, . . . , k r−4 so that (k3, k 5, . . . , k r−5) = ( h5, h 6 . . . , h (r+2) /2)and (k2, k 4, . . . , k r−4) = ( h(r+4) /2, h (r+6) /2, . . . , h r−1).
Thus, kr = max {k1, . . . , k r}, kr−2 > kr−3, kr−4 > kr−1, and k2i > k2i+1 for i ∈ { 1, . . . , (r −
6) /2}. Since the elements of H are written in nondecreasing order, using the assumptions that kr = hr > 2( r − 1) and k1 = h1 6 r, it is now easy to check that (2) is satisfied. We now define lists ( m1, . . . , m r) and ( s0, . . . , s r) of integers with mr > max
16i6r−1
{mi}
satisfying the conditions of Theorem 15, such that for each i ∈ { 1, . . . , r }, ki = mi +si−1 +
si. The inequalities (1) and (2) will ensure that mr−2 > 3. Note that by Lemma 10 and the assumptions that kr = hr > 2( r − 1) and r > 3, we have 3( r − 1) 6 kr + 2( r − 2) 6 k1 + · · · + kr−1.
Thus, we can find integers m1, . . . , m r such that 3 6 mi 6 ki for each i ∈ { 1, . . . , r } and
m1 + · · · + mr−1 = kr + 2( r − 2). Specifically, define
t = k1 + k2 + · · · + kr−1 − kr − 2( r − 2) 2 ,
and note that t is a non-negative integer by Lemma 10. We define a sequence of non-negative integers s0, s 1, s 2, . . . , s r by s0 = sr−1 = sr = 0, for i ∈ { 1, 2, . . . , r − 3}, set
si = min {ki − si−1 − 3, t − (s1 + · · · + si−1)},
and define sr−2 = t−(s1 +· · · +sr−3). It is evident from the definition that si−1 +si 6 ki −3for each i ∈ { 1, . . . , r − 3} and s1 + s2 + · · · + sr = t.
the electronic journal of combinatorics 28(1) (2021), #P1.35 9
We now show by induction that si > 0 for each 0 6 i 6 r. First note that s0 =0 > 0 and s1 = min {k1 − 3, t } > 0. Now suppose that i > 1 and assume as the inductive hypothesis that sj > 0 for each 0 6 j < i . Firstly, if si = ki − si−1 − 3 then
si−1 + si−2 6 ki−1 − 3, so si−1 6 ki−1 − 3, which implies that ki − si−1 − 3 > 0, since
ki > ki−1. Alternatively if si = t − (s1 + · · · + si−1), then by the definition of si−1,
si−1 6 t − (s1 + · · · + si−2) which implies t − (s1 + · · · + si−1) > 0. Hence, noting that
sr−1 = sr = 0, each si > 0. For each i ∈ { 1, . . . , r }, set mi = ki − si−1 − si. By choice of sr−2, it is easy to see that
s =
r−1
∑
i=1
si = t,
and hence
r−1
∑
i=1
mi =
r−1
∑
i=1
ki − 2
r−1
∑
i=1
si =
r−1
∑
i=1
ki − 2t = kr + 2( r − 2) = mr + 2( r − 2) .
It is now easy to check that ( m1, . . . , m r) satisfy conditions 1 and 2 of Lemma 10. Also, since mi 6 ki for each i ∈ { 1, . . . , r }, we have that each mi 6 ki 6 n, and m1 + · · · + mr 6
k1 + · · · + kr 6 n(n − 1). Thus, to show that ( m1, . . . , m r) can be viewed as the (suitably ordered) canonical list of an n-admissible list M , we need only to show that mi > 3 for each i ∈ { 1, . . . , r }. This is clear if 1 6 i 6 r − 3 since si 6 ki − si−1 − 3, and if i = r
since mr = kr.For i = r − 2, we have
mr−2 = kr−2 − sr−3 − sr−2.
If sr−2 = 0, then mr−2 = kr−2 − sr−3 > kr−3 − sr−3 > mr−3 > 3. Otherwise,
mr−2 = kr−2 − sr−3 − sr−2
= kr−2 − sr−3 − (t − (s1 + s2 + · · · + sr−3)) = kr−2 − t + s1 + s2 + · · · + sr−4.
To bound the value of mr−2, we rewrite this quantity in terms of k1, . . . , k r. Note that for each j ∈ { 1, . . . , r − 3}, if sj = t − (s1 + · · · + sj−1), then sj+1 = 0. Hence the condition
sr−2 6 = 0 implies that sj = kj − sj−1 − 3 for each j ∈ { 1, . . . , r − 3}. Thus
s1 + · · · + sr−4 =
r−4
∑
i=1
(ki − si−1 − 3) = ( k1 − 3) + [( k2 − 3) − (k1 − 3)] + [( k3 − 3) − (k2 − 3) + ( k1 − 3)] + · · ·
+[( kr−4 − 3) + · · · + ( −1) r−5(k1 − 3)] =
{ k1 + k3 + · · · + kr−4 − 3( r − 3) /2 if r is odd
k2 + k4 + · · · + kr−4 − 3( r − 4) /2 if r is even .
the electronic journal of combinatorics 28(1) (2021), #P1.35 10
If r is odd, it follows that
mr−2 = kr−2 − t + s1 + · · · + sr−4
= k1 + k3 + · · · + kr−4 + kr−2 − 3( r − 3) 2 − k1 + k2 + · · · + kr−1 − kr − 2( r − 2) 2= k1 + k3 + · · · + kr−2 + kr
2 − k2 + k4 + · · · + kr−1
2 − r − 52
3,
since k1 + k3 + · · · + kr−2 + kr > k2 + k4 + · · · + kr−1 + r + 1. If r is even, then
mr−2 = kr−2 − t + s1 + · · · + sr−4
= k2 + k4 + · · · + kr−4 + kr−2 − 3( r − 4) 2 − k1 + k2 + · · · + kr−1 − kr − 2( r − 2) 2= k2 + k4 + · · · + kr−2 + kr
2 − k1 + k3 + · · · + kr−1
2 − r − 82
3since k2 + k4 + · · · + kr−2 + kr > k1 + k3 + · · · + kr−1 + r − 2. Finally, we show that mr−1 > 3. Since sr−1 = 0 and by construction kr−1 > kr−2, we have that
mr−1 = kr−1 − sr−2 > kr−2 − sr−2 − sr−3 = mr−2 > 3.
We now show that the list satisfies Condition 2 of Theorem 15, i.e. that mr + s 6 n.Since
s =
r−1
∑
i=1
si =
r−1
∑
i=1
(ki − mi − si−1) =
(r−1∑
i=1
ki
)
− (kr + 2( r − 2)) − s,
we have 2s =
( r∑
i=1
ki
)
− 2kr − 2( r − 2) ,
giving
s = 12
( r∑
i=1
ki
)
− kr − r + 2 6 (n + r − 2) − kr − r + 2 = n − kr,
so that mr + s = kr + s 6 n.The result now follows by applying Theorem 15, taking ˆM ′ = K.We note that the conclusion of Theorem 16 holds whenever r > 3, hr > 2( r − 1),
h1 + · · · + hr 6 2( n + r − 2) and there is a reordering of the hi, 1 6 i < r , so that inequalities (1) and (2) are satisfied. In particular, when r is even, the conditions r 6 = 4 and h1 6 r may be dropped if (2) holds in the reordering of the hi.Recall that a directed cycle decomposition of K∗
n
can be considered as a decomposition of G∗ into directed cycles of lengths greater than 2 for some graph G of order n. Theo-rem 16 applies when the underlying (undirected) graph G formed by cycles not of length
the electronic journal of combinatorics 28(1) (2021), #P1.35 11
two is sparse. In particular, it solves all cases where G has one vertex of degree r > 5(r > 3 if r is odd) and all the rest of degree at most three, and which is a subdivision of a Hamiltonian graph. It thus gives a solution in all cases where G can be decomposed into subgraphs which either have this form or are 2-regular.
3 Admissible lists with three cycles of length greater than 2
Recall that Theorem 13 states that K∗
n
is M -decomposable whenever M is an admissible list containing at most two cycles of length greater than 2. In this section we give necessary and sufficient conditions for the existence of an M -decomposition of K∗
n
in the case that the associated canonical list ˆM has size 3. We begin by noting the following special case of Lemma 10.
Lemma 17. If M is an n-admissible list with associated canonical list ˆM = ( α, β, γ ),then α + β ≡ γ (mod 2) and α + β > γ .
For the case that α + β + γ 6 2( n + 1), sufficiency follows directly from Theorem 16. Specifically, we have the following lemma.
Lemma 18. Let M be an n-admissible list with associated canonical list ˆM = ( α, β, γ ).If α + β + γ 6 2( n + 1) , then K∗
n
is M -decomposable. Proof. Note that the conditions of Theorem 16 when r = 3 are that α + β + γ 6 2( n + 1) (as in the assumption) and γ > 4. However, if γ = 3, then α = β = γ = 3, so that
α + β 6 ≡ γ (mod 2), in contradiction to Lemma 17. If ˆM = ( α, β, γ ) and α + β + γ > 2( n + 1), the existence of an M -decomposition of
K∗
n
depends on the value of γ as well as the congruence classes of α + β and γ modulo 4.
Lemma 19. Let M be an n-admissible list with associated canonical list ˆM = ( α, β, γ ).If α + β + γ > 2( n + 1) and α + β 6 ≡ γ (mod 4) , then K∗
n
is M -decomposable. Proof. We construct directed cycles of lengths α, β and γ, leaving no decoupled arcs, so that the remaining arcs can be used to form directed 2-cycles by Lemma 8. By Lemma 12, it is sufficient to consider the case γ = n.Let −→
C n = (0 , n −1, n −2, . . . , 1). To construct −→
C α and −→
C β , we must use the decoupled arcs along the directed cycle −→
C = (0 , 1, . . . , n − 1), together with t = ( α + β − n)/2 further pairs of arcs. Note that t is odd and t > 1. We first form the cycle of length β. Let ` = n − β. If t = 3, let
−→
C β = (0 , + 3 , + 4 , + 1 , + 2 , + 5 , + 6 , ` + 7 , . . . , n − 1) .
Otherwise, we build −→
C β by concatenating directed paths. Let
P = [0 , + 3 , + 4 , + 1 , + 2] .
the electronic journal of combinatorics 28(1) (2021), #P1.35 12
For each i = 0 , 1, . . . , (t − 5) /2, let
Pi = [ + 4 i + 2 , + 4 i + 7 , + 4 i + 8 , + 4 i + 5 , ` + 4 i + 6] .
Note that:
• The initial vertex of P0 is ` + 2, the terminal vertex of P .
• For each i = 1 , 2, . . . , (t − 5) /2, the terminal vertex of Pi−1 and the initial vertex of Pi coincide. Moreover, the paths P, P 0, P 1, . . . , P (t−5) /2 are pairwise internally vertex-disjoint, so that the concatenation P P 0P1 · · · P(t−5) /2 is a path.
• Each Pi and P uses two arcs of −→
C and two other arcs. In total, these paths contain (t − 1) arcs of −→
C and decouple ( t − 1) further arcs. Next let Q = [ +2 t−4, +2 t−1]. Note that the initial vertex of Q is the terminal vertex of P(t−5) /2, and the terminal vertex of Q is +2 t−1 = α−1 6 n−1, which has not occurred in any of the paths so far. Finally, let Q′ = [+2 t−1, +2 t, . . . , +2 t−1+( n−2t−+1)]. Since the terminal vertex of Q′ is + 2 t − 1 + n − 2t − ` + 1 = n = 0, it is now easy to see that the concatenation
P P 0P1 · · · P(t−5) /2QQ ′
forms a directed cycle of length 4 + 4( t − 3) /2 + 1 + ( n − 2t − + 1) = n − = β.
0 00
Figure 3: An illustration of the construction of Lemma 19. Here α = 11, β = 15,
γ = 16 = n.It is not difficult to show that the remaining decoupled arcs form a cycle of length α;see Figure 3.
Lemma 20. Let M be an n-admissible list with associated canonical list ˆM = ( α, β, γ ).If α + β + γ > 2( n + 1) , α + β ≡ γ (mod 4) and γ < n , then K∗
n
is M -decomposable.
the electronic journal of combinatorics 28(1) (2021), #P1.35 13
Proof. First, note that α > 4, as otherwise 2( n+1) < α +β+γ 6 3+ β+( n−1) = β+n+2, implying β > n . Let M ′ be the ( n − 1)-admissible list with associated canonical list ˆM ′ = ( α − 1, β − 1, γ ). Since α + β + γ > 2( n + 1), we have that ( α − 1) + ( β − 1) + γ > 2n, and so by Lemma 19, K∗
n−1
is M ′-decomposable. Consider the cycles −→
C α−1, −→
C β−1 and −→
C γ in the decomposition constructed by Lemma 19. It is easy to see that there exist arcs xy and
yx with xy ∈ −→
C α−1 and yx ∈ −→
C β−1. Add a new vertex z, and replace arcs xy and yx
with directed paths xzy and yzx . We thus obtain three cycles of lengths α, β and γ in
K∗
n
which between them leave no arc decoupled. The result follows by Lemma 12. Note that the conditions of Lemma 20 require that the largest cycle be non-Hamiltonian. Indeed, if γ = n but all other conditions remain the same as those of Lemma 20, no M -decomposition exists. To prove this result, we will exploit a connection between an M -decomposition with ˆM = ( α, β, n ) and perfect 1-factorizations, which we now define.
Definition 21. A 1-factorization F of a graph G is a perfect 1-factorization if, for any two 1-factors F1 and F2 in F, F1 ∪ F2 is a Hamiltonian cycle. We will need the following result on perfect 1-factorizations of cubic bipartite graphs, due to Kotzig and Labelle .
Lemma 22 () . Let G be a cubic bipartite graph with bipartition (X, Y ), where |X| =
|Y | = t. If G admits a perfect 1-factorization, then t is odd.
Lemma 23. Let M be an n-admissible list with associated canonical list ˆM = ( α, β, n ).If α + β + n > 2( n + 1) (i.e. α + β > n + 2 ) and α + β ≡ n (mod 4) , then K∗
n
is not
M -decomposable. Proof. Suppose that such a decomposition exists, and let −→
C = ( n − 1, n − 2, . . . , 1, 0) be the directed n-cycle in the decomposition. Colour the edges of the directed α-cycle green and those of the directed β-cycle red, and without loss of generality, suppose that the arc 01 is green. Note that each arc of ←−
C = (0 , 1, . . . , n − 1), the reversal of −→
C , must be coloured green or red; in fact, ←−
C is partitioned into an equal number of green and red directed paths, say G0, G 1, . . . , G t−1 and R0, R 1, . . . , R t−1, where the green path Gi
has initial and terminal vertices xi and yi, and the red path Ri has initial and terminal vertices yi and xi+1 (where xt+1 = x0), and t = 12 (α + β − n) is the number of edges of
K∗
n
which are in the cycles of length α and β but not in ←−
C .We form a bipartite graph B with partite sets {x0, x 1, . . . , x t−1} and {y0, y 1, . . . , y t−1}
as follows. For i = 0 , 1, . . . , t − 1, we join xi to yi, forming a 1-factor F1 of B, and yi to
xi+1 (computing subscripts modulo t), forming a second 1-factor F2 of B. Also, join yi to
xj , where yixj is an arc of the green cycle; these edges form a third 1-factor F3 of B. Note that j 6 = i + 1, as otherwise the α-cycle in the decomposition must be ( yi, y i + 1 , . . . , y i+1 ), implying t = 1 and hence α + β − n = 2 ≡ 2 (mod 4). An example of the α- and β-cycles together with the associated cubic bipartite graph can be found in Figure 4.
the electronic journal of combinatorics 28(1) (2021), #P1.35 14
y0
R0
G0
x0
x1
G1
y1
R2
x2
G2
y2
R2
x3
G3
y3
R3
x3
G4
y4
R4
x4
x3
x2
x1
x0
y4
y3
y2
y1
y0
Figure 4: Directed α- and β-cycles together with the associated cubic bipartite graph It is clear that B is a cubic bipartite graph. We show that {F1, F 2, F 3} is a per-fect 1-factorization of B. First, notice that F1 ∪ F2 induces the Hamiltonian cycle (x0, y 0, x 1, y 1, . . . , x t−1, y t−1). Next, consider the green α-cycle in the decomposition. Re-placing the arcs of each Gi with a single arc xiyi and disregarding direction, we obtain a cycle whose edges correspond to those of F1 and F3, and so F1 ∪ F3 induces a Hamiltonian cycle in B. In a similar way, looking at the red β-cycle shows that F2 ∪ F3 induces a Hamiltonian cycle in B.By Lemma 22, the existence of a perfect 1-factorization of B implies that t is odd, contradicting the fact that α + β − n ≡ 0 (mod 4). Summarizing the results of this section, we have the following complete result for canonical lists of size at most three.
Theorem 24. Let M = (2 , 2, . . . , 2, α, β, γ ), where 2 < α 6 β 6 γ 6 n. The complete symmetric digraph K∗
n
is M -decomposable if and only if M is n-admissible and it is not the case that α + β > n + 2 , γ = n and α + β ≡ n (mod 4) .
One immediate consequence is the following result regarding cycle decomposition of certain 3-regular digraphs.
Corollary 25. Let α, β, γ > 3 be integers with α + β > γ + 2 and α + β ≡ γ (mod 4) . If
G is a 3-regular graph with γ vertices and (α + β + γ)/2 edges, then G∗ is not (α, β, γ )-decomposable.
4 Conclusion
In this paper, we have made progress on the problem of decomposing a complete symmetric digraph into cycles of given lengths. Theorem 16 shows that if the greatest cycle length
the electronic journal of combinatorics 28(1) (2021), #P1.35 15
and the number of 2-cycles in an n-admissible list M are both large enough, then K∗
n
admits an M -decomposition. In the case that there are at most three cycles of length greater than 2, we have given a complete solution (Theorem 24). Notably, there is a family of n-admissible lists M ,namely those with ˆM = ( α, β, n ) where α + β > n + 2 and α + β ≡ n (mod 4), for which 2 Kn is decomposable but K∗
n
is not. This result implies that no cubic graph G
of order n exists such that G∗ is decomposable into three cycles of lengths α, β and n,where α + β > n + 2 and α + β ≡ n (mod 4). In particular, G∗ has no Hamiltonian cycle decomposition if n ≡ 0 (mod 4). The method used in this paper to show non-existence by constructing an auxiliary bipartite graph with a perfect 1-factorization does not apply in general to canonical lists of size greater than three. It remains an interesting open question to determine if there are other infinite families of n-admissible lists M for which K∗
n
is not M -decomposable. We have checked all n-admissible lists with n 6 13 and have verified that the correspond-ing decompostions exist, except for those given by Lemma 23 and the decompositions corresponding to the canonical lists:
n = 4 : (4 , 4, 4);
n = 5 : (4 , 4, 5, 5) , (3 , 4, 4, 4, 5) , (3 , 3, 3, 3, 3, 5);
n = 6 : (3 , 3, 3, 3, 3, 3, 3, 3, 3, 3) , (3 , 3, 6, 6, 6, 6) , (4 , 6, 6, 6) , (6 , 6, 6, 6, 6) ,
which do not exist. We conjecture that these are the only exceptions.
References
B. Alspach. Research Problem 3. Discrete Math. 36 (1981), 333. B. Alspach and H. Gavlas. Cycle decompositions of Kn and Kn − I. J. Combin. Theory Ser. B 81 (2001), 77–99. B. Alspach, H. Gavlas, M. ˇSajna and H. Verrall. Cycle decompositions IV: complete directed graphs and fixed length directed cycles. J. Combin. Theory Ser. A 103
(2003), 165–208. D. Bryant, D. Horsley, B. Maenhaut and B.R. Smith. Decompositions of complete multigraphs into cycles of varying lengths. J. Combin. Theory Ser. B 129 (2018), 79–106. D. Bryant, D. Horsley and W. Pettersson. Cycle decompositions V: Complete graphs into cycles of arbitrary lengths. Proc. London Math. Soc. 108 (2014), 1153–1192. T.P. Kirkman. On a problem in combinations. Cambridge and Dublin Math J. 2
(1847), 191–204. A. Kotzig and J. Labelle. Strongly Hamiltonian graphs. Utilitas Math. 14 (1978), 99–116. E. Lucas. R´ ecr´ eations Math´ ematiques , Vol. II. Gauthier-Villars, Paris, 1892. M. ˇSajna. Cycle decompositions III: Complete graphs and fixed length cycles. J. Combin. Des. 10 (2002), 27–78. |
4124 | https://webstersdictionary1828.com/Dictionary/swing | Websters 1828 - Webster's Dictionary 1828 - Swing
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Swing
SWING, verb intransitive preterit tense and participle passive swung.
1. To move to and fro, as a body suspended in the air; to wave; to vibrate.
I tried if a pendulum would swing faster, or continue swinging longer in our receiver, if exhausted.
2. To practice swinging; as, a man swings for health or pleasure.
3. To move or float; also, to turn round an anchor; as, a ship swings with the tide.
SWING, verb transitive To make to play loosely; to cause to wave or vibrate; as a body suspended in the air.
1. To whirl round in the air.
--Swing thee in air, then dash thee down.
2. To wave; to move to and from; as, a man swings his arms when he walks.
He swings his tail, and swiftly turns him round.
3. To brandish; to flourish.
SWING, noun A waving or vibratory motion; oscillation; as the swing of a pendulum.
1. Motion from one side to the other. A haughty man struts or walks with a swing
2. A line, cord or other thing suspended and hanging loose; also, an apparatus suspended for persons to swing in.
3. Influence or power of a body put in motion.
The ram that batters down the wall,
For the great swing and rudeness of his poise--
4. Free course; unrestrained liberty or license.
Take thy swing
To prevent any thing which may prove an obstacle to the full swing of his genius.
5. The sweep or compass of a moving body.
6. Unrestrained tendency; as the prevailing swing of corrupt nature; the swing of propensities.
SWING'-BRIDGE, noun [swing and bridge.] A bridge that may be moved by swinging; used on canals.
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4125 | http://www.dfmf.uned.es/~aperea/material_docente/MyO/tema2_3bis.pdf | Problemas Resueltos 2.20 Un satélite describe una órbita circular en torno a la Tierra. Si se cambia de repente la dirección de su velocidad, pero no su módulo, estudiar el cambio en su órbita y en su período. • Al cambiar sólo la dirección de la velocidad, el momento angular varía pero la energía total no. La órbita nueva corresponderá a una trayectoria asociada a una energía negativa, que como hemos visto, puede ser una órbita circular o elíptica. El nuevo valor del momento angular es α α sin sin 0 L mVr L = = ′ siendo α el ángulo entre la velocidad y el radio vector en el instante del cambio, y 0 r el radio de la órbita circular, y la energía tiene el valor 0 2r GMm E E − = = ′ La fuerza central que actúa entre al satélite y la Tierra es i nversamente proporcional al cuadrado de la distancia entre ellos. Para dicha fuerza se satisfacen las leyes de Kepler. La tercera ley nos dice que el período de una órbita sólo está relacionado con la energía total, de la forma siguiente a GMm E 2 − = y 2 2 3 4π GM T a = donde a es el semieje mayor de la nueva órbita, que en este caso coincide con 0 r . Es decir, 3 2 2 2 4 − = E GMm GM T π Al no variar la energía durante el cambio de la velocidad, no varía el período orbital. • La segunda ley de Kepler es la ley de las áreas y establece que T S m L = 2 siendo S el área de la órbita y T el período orbital. Como el período no varía, T T ′ = , L S m L mS ′ ′ = 2 2 con lo cual, el área de la nueva órbita es α sin S S L L S = ′ = ′ Ya que la energía no varía, la nueva órbita es una elipse centrada en O, con semieje 0 r a = , y nos falta calcular su excentricidad. Tenemos 2 2 0 a r S π π = = y 2 2 1 e a ab S − = = ′ π π con lo cual 2 1 sin e S S − = ′ = α Es decir, la excentricidad de la nueva órbita viene dada por α cos = e 2.21 Un satélite de masa m describe una órbita circular a una distancia H de la superficie terrestre. Otra partícula de masa m/2 se mueve sobre la misma órbita pero en sentido contrario, de modo que choca con el satélite quedando unida a él. Calcular el apogeo y el perigeo de la órbita del cuerpo compuesto. • La velocidad de una masa m en una órbita circular de radio r puede obtenerse de la condición de equilibrio de las fuerzas en dirección radial r V m r GMm 2 2 = ó r GM V = La velocidad no depende de la masa en movimiento. Concluimos que la velocidad del satélite y de la partícula antes del choque tiene el mismo módulo y sentido contrario con el valor H R GM Vi + = siendo R el radio de la Tierra y M su masa. Durante el choque sólo actúan fuerzas interiores, y el momento angular respecto a O se conserva constante. Antes del choque ) ( 2 ) ( 2 ) ( H R GM m H R V m H R mV L i i + = + − + = Este es el valor del mom ento angular para el movimiento del cuerpo compuesto, después del choque. • Para determinar los puntos apsidales de la nueva órbita nos falta conocer la energía del cuerpo compuesto. Como la colisión no es elástica, la energía no se conserva en la colisión. Sin embargo, si se conserva la velocidad del centro de masas, que corresponde a la velocidad V del cuerpo compuesto después del choque. Tenemos i i V m mV V m m 2 2 − = + con lo cual i V V 3 1 = Justo después de la colisión, el cuerpo compuesto lleva una velocidad V y se encuentra a una distancia R+H del centro de la Tierra. Por tanto, su energía total tiene el valor + + − + = 2 2 2 1 2 m m H R GM V m m E Introduciendo el valor de la velocidad, llegamos a H R GMm H R GM m H R GM m E + − = + − + = 12 17 2 3 9 1 4 3 • Una vez conocidos los valores L y E, hallamos los parámetros de la órbita a y e, del cuerpo compuesto. Para el semieje mayor obtenemos el valor ) ( 17 9 2 2 3 12 17 H R a a GMm H R GMm + = − = + − La excentricidad de la órbita satisface la fórmula 3 2 2 2 2 3 2 1 + = m M G EL e Desarrollando el radicando 81 17 4 1 12 17 27 16 27 16 3 2 2 2 − = − = m M G EL encontramos 9 8 81 17 1 = − = e • La distancia de máximo acercamiento, perigeo, está dada por ( ) ) ( 17 1 ) ( 9 8 1 17 9 1 H R H R e a rP + = + − = − = y la distancia de máximo alejamiento, apogeo, es ( ) H R H R e a rA + = + + = + = ) ( 9 8 1 17 9 1 2.22 Un satélite artificial de masa m recorre una órbita elíptica, con período T. Las velocidades máxima y mínima en su órbita son min max ,V V respectivamente. Determinar los parámetros de la órbita. • La ecuación de la elipse es φ cos 1 ) 1 ( 2 e e a r + − = Las distancias al apogeo y perigeo son ) 1 ( ) 1 ( e a r e a r A P + = − = Por conservación del momento angular L en los puntos apsidales P A r mV r mV max min = con lo cual, de las ecuaciones anteriores, eliminando el semieje mayor a obtenemos la excentricidad de la órbita min max min max V V V V e + − = • Para hallar el semieje mayor, utilizamos la ley de las áreas. Si T es el período, ab π el área de la elipse, la ley de las áreas se escribe ) 1 ( 2 1 2 min e a V m L T ab + = = π Como 2 1 e a b − = , despejamos el valor del semieje mayor 2 min 1 1 2 e e T V a − + = π Tenemos min max 2 1 1 1 1 V V e e e e = − + = − + Por tanto, max min 2 V V T a π = • Finalmente la energía total E se encuentra a partir de la expresión a GMm E 2 − = obteniendo max minV V GMm E π − = 2.23 Un planeta de masa M tiene un satélite de masa m, describiendo en torno a él una trayectoria circular de radio R, con período T. Súbitamente el satélite se para. Determinar el tiempo de caída del satélite sobre el planeta. • La energía del satélite en su órbita circular es R GMm mV E − = 2 2 1 Además, hay equilibrio de las fuerzas en dirección radial. La fuerza gravitatoria se compensa con la fuerza centrífuga 2 2 R GMm R V m = con lo cual, la energía es R GMm E 2 − = • Cuando el satélite se frena, su energía cinética se hace cero, y su energía se reduce a la energía potencial, con lo cual E R GMm E 2 = − = ′ Según la tercera ley de Kepler, existe la relación entre el período de una órbita y la energía del sistema ( ) 3 2 2 2 3 2 2 m M G E T π = − El período de la nueva órbita satisface 2 2 ' 2 / 3 T T E E T = = ′ • Cuando el satélite se frena, su velocidad se hace cero, y así el momento angular de la nueva órbita es cero. Esto quiere decir que la órbita del satélite pasará por el centro del planeta. El punto de máximo alejamiento se produce en el momento inicial, R r = , y el punto de máximo acercamiento es 0 = r . Por tanto, el tiempo que tarda en caer es igual al tiempo que tarda en ir de R r = a 0 = r , es decir, el tiempo que tarda en ir del máximo alejamiento al máximo acercamiento, esto es, un semiperíodo. El tiempo de caída es, entonces 2 4 2 T T t = ′ = Problemas Propuestos 2.24 Un satélite artificial se lanza desde la superficie terrestre verticalmente hacia arriba con una velocidad inicial R GM U α = . En el momento en que se para, se le da una velocidad transversal R GM V β = . Hallar los parámetros de la órbita en función de α y β. Aquí, R es el radio terrestre. Solución: ( ) ( ) 2 2 2 2 2 2 2 2 2 4 1 2 α β β α β α − − − − = − − = e R a 2.25 Una nave espacial de masa m llega con una velocidad 0 V a las proximidades de la Luna siguiendo una trayectoria hiperbólica cuya asíntota está a una distancia b del centro de la Luna. Sea a la distancia de aproximación máxima de la n ave al centro de la Luna. Calcular la velocidad necesaria 0 V para que R a R b 2 , 3 10 = = , siendo R el radio de la Luna, y la velocidad en el punto de aproximación máxima en dicho caso. En el punto de máxima aproximación, la nave frena para describir una órbita circular de observación de radio a. Calcular la energía perdida por la nave. Solución: 2 2 2 2 0 2 4 5 4 3 a b a b a GMm E R GM V R GM V P − + = ∆ = = |
4126 | https://www.grc.nasa.gov/www/BGH/machrole.html | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- ---
| | | | --- | | + Text Only Site + Non-Flash Version + Contact Glenn | | | | | | | | | | --- --- --- | | | | | | | | | | | | | As an aircraft moves through the air, the air molecules near the aircraft are disturbed and move around the aircraft. If the aircraft passes at a low speed, typically less than 250 mph, it is observed that the density of the air remains constant. For higher speeds, some of the energy of the aircraft goes into compressing the air and locally changing the density of the air. This compressibility effect alters the amount of resulting force on the aircraft since the aerodynamic force depends on the air density. The effect becomes more important as speed increases. Near and beyond the speed of sound, about 330 m/s or 760 mph at sea level, small disturbances in the flow are transmitted to other locations isentropically or with constant entropy. Sharp disturbances generate shock waves that affect both the lift and drag of the aircraft, and the flow conditions downstream of the shock wave. On this slide, we will investigate the dependence of the density change on the Mach number of the flow. The Mach number is the ratio of the speed of the aircraft, or the speed of the gas, to the speed of sound in the gas. The speed of sound is equal to the speed of transmission of small, isentropic disturbances in the flow. To determine the role of the Mach number on compressibility effects. we begin with the conservation of momentum equation: rho V dV = - dp where rho is the fluid density, V is the velocity, and p is the pressure. dV and dp denote differential changes in the velocity and pressure. From our derivation of the conditions for isentropic flow, we know that: dp/p = gamma drho/rho dp = gamma p / rho drho where gamma is the specific heat ratio. We can use the ideal equation of state to simplify the expression on the right: p = rho R T dp = gamma R T drho where R is the specific gas constant and T is the absolute temperature. We recognize that: gamma R T = a^2 where a is equal to the speed of sound. So, dp = a^2 drho Substituting this expression for the change of pressure into the conservation of momentum equation gives: rho V dV = - a^2 drho - (V^2 / a^2) dV / V = drho / rho - M^2 dV / V = drho / rho where M is the Mach number. What does this expression tell us about the role of the Mach number in compressible flows? 1. For low speed, or subsonic conditions, the Mach number is less than one, M < 1 and the square of the Mach number is very small. Then the left hand side of the equation is very small, and the change in density is very small. For the low subsonic conditions, compressibility can be ignored. 2. As the speed of the object approaches the speed of sound, the flight Mach number is nearly equal to one, M = 1, and the flow is said to be transonic. If the Mach number is near one, the square of the Mach number is also nearly equal to one. For transonic flows, the change in density is nearly equal to the change in velocity, and compressibility effects can not be ignored. 3. As the speed increases beyond the speed of sound, the flight Mach number is greater than one M > 1 and the flow is said to be supersonic or hypersonic. For supersonic and hypersonic flows, the density changes faster than the velocity changes by a factor equl to the square of the Mach number. Compressibility effects become more important with higher Mach numbers. Activities: Guided Tours Navigation .. Beginner's Guide Home Page | | | | | | | | --- --- | | | + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification | | Editor: Tom Benson NASA Official: Tom Benson Last Updated: May 07 2021 + Contact Glenn | | | |
4127 | https://math.stackexchange.com/questions/2176767/prove-that-every-8-regular-graph-has-4-and-2-regular-spanning-subgraph | hamiltonian path - Prove that every 8-regular graph has 4-and 2-regular spanning subgraph! - Mathematics Stack Exchange
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Prove that every 8-regular graph has 4-and 2-regular spanning subgraph!
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Prove that every 8 8-regular graph has 4 4- and 2 2-regular spanning subgraphs.
Note: A graph is spanning subgraph, if it contains every vertex of the original graph. Furthermore this example's from Hamiltonian cycle/path topic.
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edited Mar 7, 2017 at 23:48
Botond KissBotond Kiss
asked Mar 7, 2017 at 23:36
Botond KissBotond Kiss
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The 2 2-regular claim is easy -- think about the components of the original 8 8-regular graph. Make a connection to the Hamiltonian cycle topic.hardmath –hardmath 2017-03-07 23:40:49 +00:00 Commented Mar 7, 2017 at 23:40
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This is part of a general result called Petersen's 2-factor theorem, which is not too difficult to prove as a whole.
Theorem (Petersen): Let G G be a 2 k 2 k-regular graph. Then G G has a decomposition into k k edge-disjoint 2 2-factors (where a 2 2-factor is a 2 2-regular spanning graph).
Proof: Without loss of generality, let G G be connected. Since G G has only vertices of even degree, it follows that G G has an Euler cycle. Take an Euler cycle of G G and define a corresponding orientation D(G)D(G) according to the cycle.
Now, split each vertex v v of G G into two distinct vertices colored red and blue, say v r v r and v b v b. Let us now define another (undirected) graph G′G′ with vertex set V r×V b V r×V b. For each directed arc (u,v)(u,v) in our digraph D(G)D(G), draw the edge (u r,v b)∈E(G′)(u r,v b)∈E(G′). This makes G′G′ into a k k-regular bipartite graph.
By Hall's marriage theorem, G′G′ has a decomposition into k k edge-disjoint perfect matchings. If we now identify v r v r and v b v b for each vertex, the previously defined perfect matchings gives us a decomposition of G G into k k edge-disjoint 2 2-factors, as required. □◻
Your result now follows easily from Petersen's theorem. Take any 2 2-factor to get your 2 2-regular spanning graph. Take any two 2 2-factors to get a 4 4-regular spanning graph.
In general, for any 2 k 2 k-regular graph G G, there exists 2 r 2 r-regular spanning graphs of G G for any 0<r≤k 0<r≤k.
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answered Mar 8, 2017 at 0:43
EuYuEuYu
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Thank you, I appreciate it, but as I have noticed this proof demands deeper knowledge of graph theory. (Hall's marriage theorem, perfect matching ect.) I should solve this problem without these. I will look for an alternative form of the Petersen theorem's proof!Botond Kiss –Botond Kiss 2017-03-08 15:27:24 +00:00 Commented Mar 8, 2017 at 15:27
@BotondKiss Did you ever find an alternative proof? Because I have the same problem now! These theorems are too advanced, yet I was supposed to solve this question on my exam.Sha Vuklia –Sha Vuklia 2017-05-09 16:30:59 +00:00 Commented May 9, 2017 at 16:30
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0Every nonhamiltonian 2-connected graph has a theta subgraph
3Spanning 2-regular subgraphs in even regular graphs.
2Using Dirac's theorem show if G G is a simple graph with |V(G)|=ν=2 k|V(G)|=ν=2 k and ν 2+1≤δ ν 2+1≤δ then G G has a 3-factor.
1What is a spanning path in graph theory?
1Prove that every simple graph G G contains a spanning subgraph, which is bipartite and deg(v′)≥1 2 deg(v)deg(v′)≥1 2 deg(v).
5Prove that if G G is a graph of order 101 101 and δ(G)=51 δ(G)=51, then every vertex of G G lies on a cycle of length 27 27
4Is there a graph with 2-factor that is not hamiltonian?
5Every connected graph has a spanning subtree
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4128 | https://warrenweckesser.github.io/papers/weckesser-scipy-linear-filters.pdf | DRAFT PyData CookBook 1 Signal Processing with SciPy: Linear Filters Warren Weckesser∗ !
Abstract—The SciPy library is one of the core packages of the PyData stack.
It includes modules for statistics, optimization, interpolation, integration, linear algebra, Fourier transforms, signal and image processing, ODE solvers, special functions, sparse matrices, and more. In this chapter, we demonstrate many of the tools provided by the signal subpackage of the SciPy library for the design and analysis of linear filters for discrete-time signals, including filter representation, frequency response computation and optimal FIR filter design.
Index Terms—algorithms, signal processing, IIR filter, FIR filter CONTENTS Introduction . . . . . . . . . . . . . . . . . . . .
1 IIR filters in scipy.signal . . . . . . . . . .
1 IIR filter representation . . . . . . . . .
2 Lowpass filter . . . . . . . . . . . . . .
3 Initializing a lowpass filter . . . . . . .
5 Bandpass filter . . . . . . . . . . . . . .
5 Filtering a long signal in batches . . . .
6 Solving linear recurrence relations . . .
6 FIR filters in scipy.signal . . . . . . . . . .
7 Apply a FIR filter . . . . . . . . . . . .
7 Specialized functions that are FIR filters 8 FIR filter frequency response . . . . . .
8 FIR filter design . . . . . . . . . . . . .
8 FIR filter design: the window method .
8 FIR filter design: least squares . . . . .
9 FIR filter design: Parks-McClellan . . .
10 FIR filter design: linear programming .
10 Determining the order of a FIR filter . .
13 Kaiser’s window method . . . . . . . .
13 Optimizing the FIR filter order . . . . .
13 References 14 Introduction The SciPy library, created in 2001, is one of the core pack-ages of the PyData stack. It includes modules for statistics, optimization, interpolation, integration, linear algebra, Fourier transforms, signal and image processing, ODE solvers, special functions, sparse matrices, and more.
The signal subpackage within the SciPy library includes tools for several areas of computation, including signal pro-cessing, interpolation, linear systems analysis and even some elementary image processing. In this Cookbook chapter, we ∗Corresponding author: warren.weckesser@gmail.com Copyright c ○2016 Warren Weckesser.
focus on a specific subset of the capabilities of of this sub-package: the design and analysis of linear filters for discrete-time signals. These filters are commonly used in digital signal processing (DSP) for audio signals and other time series data with a fixed sample rate.
The chapter is split into two main parts, covering the two broad categories of linear filters: infinite impulse response (IIR) filters and finite impulse response (FIR) filters.
Note. We will display some code samples as transcripts from the standard Python interactive shell. In each interactive Python session, we will have executed the following without showing it: >>> import numpy as np >>> import matplotlib.pyplot as plt >>> np.set_printoptions(precision=3, linewidth=50) Also, when we show the creation of plots in the transcripts, we won’t show all the other plot commands (setting labels, grid lines, etc.) that were actually used to create the corresponding figure in the paper. Complete scripts for all the figures in this paper are available in the papers/scipy directory at
IIR filters in scipy.signal An IIR filter can be written as a linear recurrence relation, in which the output yn is a linear combination of xn, the M previous values of x and the N previous values of y: a0yn = M ∑ i=0 bixn−i − N ∑ i=1 aiyn−N (1) (See, for example, Oppenheim and Schafer [OS], Chapter 6.
Note, however, that the sign convention for a, ..., a[N] in Eqn. (1) and used in SciPy is the opposite of that used in Oppenheim and Schafer.) By taking the z-transform of Eqn. (1), we can express the filter as Y(z) = H(z)X(z) (2) where H(z) = b0 +b1z−1 +···+bMz−M a0 +a1z−1 +···+aNz−N (3) is the transfer function associated with the filter. The functions in SciPy that create filters generally set a0 = 1.
Eqn. (1) is also known as an ARMA(N, M) process, where "ARMA" stands for Auto-Regressive Moving Average.
b holds the moving average coefficients, and a holds the auto-regressive coefficients.
DRAFT 2 PyData CookBook When a1 = a2 = ··· = aN = 0, the filter is a finite impulse response filter. We will discuss those later.
IIR filter representation In this section, we discuss three representations of a linear filter: • transfer function • zeros, poles, gain (ZPK) • second order sections (SOS) SciPy also provides a state space representation, but we won’t discuss that format here.
Transfer function. The transfer function representation of a filter in SciPy is the most direct representation of the data in Eqn. (1) or (3). It is two one-dimensional arrays, conventionally called b and a, that hold the coefficients of the polynomials in the numerator and denominator, respectively, of the transfer function H(z).
For example, we can use the function scipy.signal.butter to create a Butterworth lowpass filter of order 6 with a normalized cutoff frequency of 1/8 the Nyquist frequency. The default representation created by butter is the transfer function, so we can use butter(6, 0.125): >>> from scipy.signal import butter >>> b, a = butter(6, 0.125) >>> b array([ 2.883e-05, 1.730e-04, 4.324e-04, 5.765e-04, 4.324e-04, 1.730e-04, 2.883e-05]) >>> a array([ 1.
, -4.485, 8.529, -8.779, 5.148, -1.628, 0.217]) The representation of a filter as a transfer function with coef-ficients (b, a) is convenient and of theoretical importance, but with finite precision floating point, applying an IIR filter of even moderately large order using this format is susceptible to instability from numerical errors. Problems can arise when designing a filter of high order, or a filter with very narrow pass or stop bands.
ZPK. The ZPK representation consists of a tuple containing three items, (z, p, k). The first two items, z and p, are one-dimensional arrays containing the zeros and poles, respectively, of the transfer function. The third item, k, is a scalar that holds the overall gain of the filter.
We can tell butter to create a filter using the ZPK representation by using the argument output="zpk": >>> z, p, k = butter(6, 0.125, output=’zpk’) >>> z array([-1., -1., -1., -1., -1., -1.]) >>> p array([ 0.841+0.336j, 0.727+0.213j, 0.675+0.072j, 0.675-0.072j, 0.727-0.213j, 0.841-0.336j]) >>> k 2.8825891944002783e-05 A limitation of the ZPK representation of a filter is that SciPy does not provide functions that can directly apply the filter to a signal. The ZPK representation must be converted to either the SOS format or the transfer function format to actually filter a signal. If we are designing a filter using butter or one of the other filter design functions, we might as well create the filter in the transfer function or SOS format when the filter is created.
SOS. In the second order sections (SOS) representation, the filter is represented using one or more cascaded second order filters (also known as "biquads"). The SOS representation is implemented as an array with shape (n, 6), where each row holds the coefficients of a second order transfer function. The first three items in a row are the coefficients of the numerator of the biquad’s transfer function, and the second three items are the coefficients of the denominator.
The SOS format for an IIR filter is more numerically stable than the transfer function format, so it should be preferred when using filters with orders beyond, say, 7 or 8, or when the bandwidth of the passband of a filter is sufficiently small.
(Unfortunately, we don’t have a precise specification for what "sufficiently small" is.) A disadvantage of the SOS format is that the function sosfilt (at least at the time of this writing) applies an SOS filter by making multiple passes over the data, once for each second order section. Some tests with an order 8 filter show that sosfilt(sos, x) can require more than twice the time of lfilter(b, a, x).
Here we create a Butterworth filter using the SOS represen-tation: >>> sos = butter(6, 0.125, output="sos") >>> sos array() The array sos has shape (3, 6). Each row represents a biquad; for example, the transfer function of the biquad stored in the last row is H(z) = 1+2z−1 +z−2 1−1.681z−1 +0.8198z−2 Converting between representations. The signal module provides a collection of functions for converting one represen-tation to another: sos2tf, sos2zpk, ss2tf, ss2zpk, tf2sos, tf2zz, tf2zpk, zpk2sos, zpk2ss, zpk2tf For example, zpk2sos converts from the ZPK representation to the SOS representation. In the following, z, p and k have the values defined earlier: >>> from scipy.signal import zpk2sos >>> zpk2sos(z, p, k) array() Limitations of the transfer function representation. Earlier we said that the transfer function representation of moderate to large order IIR filters can result in numerical problems. Here we show an example.
DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 3 0 20 40 60 80 100 120 Sample number 6 4 2 0 Fig. 1: Incorrect step response of the Butterworth bandpass filter of order 10 created using the transfer function representation. Ap-parently the filter is unstable--something has gone wrong with this representation.
We consider the design of a Butterworth bandpass filter with order 10 with normalized pass band cutoff frequencies of 0.04 and 0.16.: >>> b, a = butter(10, [0.04, 0.16], btype="bandpass") We can compute the step response of this filter by applying it to an array of ones: >>> x = np.ones(125) >>> y = lfilter(b, a, x) >>> plt.plot(y) The plot is shown in Figure 1. Clearly something is going wrong.
We can try to determine the problem by checking the poles of the filter: >>> z, p, k = tf2zpk(b, a) >>> np.abs(p) array([ 0.955, 0.955, 1.093, 1.093, 1.101, 1.052, 1.052, 0.879, 0.879, 0.969, 0.969, 0.836, 0.836, 0.788, 0.788, 0.744, 0.744, 0.725, 0.725, 0.723]) The filter should have all poles inside the unit circle in the complex plane, but in this case five of the poles have mag-nitude greater than 1. This indicates a problem, which could be in the result returned by butter, or in the conversion done by tf2zpk. The plot shown in Figure 1 makes clear that something is wrong with the coefficients in b and a.
Let’s design the same 10th order Butterworth filter as above, but in the SOS format: >>> sos = butter(10, [0.04, 0.16], ...
btype="bandpass", output="sos") In this case, all the poles are within the unit circle: >>> z, p, k = sos2zpk(sos) >>> np.abs(p) array([ 0.788, 0.788, 0.8 , 0.8 , 0.818, 0.818, 0.854, 0.854, 0.877, 0.877, 0.903, 0.903, 0.936, 0.936, 0.955, 0.955, 0.964, 0.964, 0.988, 0.988]) We can check the frequency response using scipy.signal.sosfreqz: >>> w, h = sosfreqz(sos, worN=8000) >>> plt.plot(w/np.pi, np.abs(h)) [] 0.0 0.2 0.4 0.6 0.8 1.0 0.00 0.25 0.50 0.75 1.00 Gain 0.0 0.2 0.4 0.6 0.8 1.0 Normalized frequency /2 0 /2 Phase Fig. 2: Frequency response of the Butterworth bandpass filter with order 10 and normalized cutoff frequencies 0.04 and 0.16.
0 50 100 150 200 Sample number 0.2 0.0 0.2 Fig. 3: Step response of the Butterworth bandpass filter with order 10 and normalized cutoff frequencies 0.04 and 0.16.
The plot is shown in Figure 2.
As above, we compute the step response by filtering an array of ones: >>> x = np.ones(200) >>> y = sosfilt(sos, x) >>> plt.plot(y) The plot is shown in Figure 3. With the SOS representation, the filter behaves as expected.
In the remaining examples of IIR filtering, we will use only the SOS representation.
Lowpass filter Figure 4 shows a times series containing pressure measure-ments [SO]. At some point in the interval 20 < t < 22, an event occurs in which the pressure jumps and begins oscillating around a "center". The center of the oscillation decreases and appears to level off.
We are not interested in the oscillations, but we are interested in the mean value around which the signal is DRAFT 4 PyData CookBook 20 25 30 0 2 4 6 8 10 Pressure (MPa) 20 25 30 Time (ms) 0 5 10 15 20 25 Frequency (kHz) Fig. 4: Top: Pressure, for the interval 15 < t < 35 (milliseconds).
Bottom: Spectrogram of the pressure time series (generated using a window size of 1.6 milliseconds).
oscillating. To preserve the slowly varying behavior while eliminating the high frequency oscillations, we’ll apply a low-pass filter. To apply the filter, we can use either sosfilt or sosfiltfilt from scipy.signal. The function sosfiltfilt is a forward-backward filter--it applies the filter twice, once forward and once backward. This effectively doubles the order of the filter, and results in zero phase shift.
Because we are interesting in the "event" that occurs in 20 < t < 22, it is important to preserve the phase characteristics of the signal, so we use sosfiltfilt.
The following code snippet defines two convenience func-tions. These functions allow us to specify the sampling fre-quency and the lowpass cutoff frequency in whatever units are convenient. They take care of scaling the values to the units expected by scipy.signal.butter.
from scipy.signal import butter, sosfiltfilt def butter_lowpass(cutoff, fs, order): normal_cutoff = cutoff / (0.5fs) sos = butter(order, normal_cutoff, btype=’low’, output=’sos’) return sos def butter_lowpass_filtfilt(data, cutoff, fs, order): sos = butter_lowpass(cutoff, fs, order=order, output=’sos’) y = sosfiltfilt(sos, data) return y The results of filtering the data using sosfiltfilt are shown in Figure 5.
20 25 30 0 2 4 6 8 10 Pressure (MPa) 20 25 30 Time (ms) 0 5 10 15 20 25 Frequency (kHz) Fig. 5: Top: Filtered pressure, for the interval 15 < t < 35 (milliseconds). The light gray curve is the unfiltered data. Bottom: Spectrogram of the filtered time series (generated using a window size of 1.6 milliseconds). The dashed line is at 1250 Hz.
Comments on creating a spectrogram. A spectrogram is a plot of the power spectrum of a signal computed repeatedly over a sliding time window. The spectrograms in Figures 4 and 5 were created using spectrogram from scipy.signal and pcolormesh from matplotlib.pyplot. The func-tion spectrogram has several options that control how the spectrogram is computed. It is quite flexible, but obtaining a plot that effectively illustrates the time-varying spectrum of a signal sometimes requires experimentation with the parame-ters. In keeping with the "cookbook" theme of this book, we include here the details of how those plots were generated.
Here is the essential part of the code that computes the spectrograms. pressure is the one-dimensional array of measured data.
fs = 50000 nperseg = 80 noverlap = nperseg - 4 f, t, spec = spectrogram(pressure, fs=fs, nperseg=nperseg, noverlap=noverlap, window=’hann’) The spectrogram for the filtered signal is computed with the same arguments: f, t, filteredspec = spectrogram(pressure_filtered, ...) Notes: • fs is the sample rate, in Hz.
• spectrogram computes the spectrum over a sliding DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 5 segment of the input signal. nperseg specifies the number of time samples to include in each segment.
Here we use 80 time samples (1.6 milliseconds). This is smaller than the default of 256, but it provides sufficient resolution of the frequency axis for our plots.
• noverlap is the length (in samples) of the overlap of the segments over which the spectrum is computed. We use noverlap = nperseq - 4; in other words, the window segments slides only four time samples (0.08 milliseconds). This provides a fairly fine resolution of the time axis.
• The spectrum of each segment of the input is computed after multiplying it by a window function. We use the Hann window.
The function spectrogram computes the data to be plotted. Next, we show the code that plots the spectrograms shown in Figures 4 and 5. First we convert the data to decibels: spec_db = 10np.log10(spec) filteredspec_db = 10np.log10(filteredspec) Next we find the limits that we will use in the call to pcolormesh to ensure that the two spectrograms use the same color scale. vmax is the overall max, and vmin is set to 80 dB less than vmax. This will suppress the very low amplitude noise in the plots.
vmax = max(spec_db.max(), filteredspec_db.max()) vmin = vmax - 80.0 Finally, we plot the first spectrogram using pcolormesh(): cmap = plt.cm.coolwarm plt.pcolormesh(1000t, f/1000, spec_db, vmin=vmin, vmax=vmax, cmap=cmap, shading=’gouraud’) An identical call of pcolormesh with filteredspec_db generates the spectrogram in Figure 5.
Initializing a lowpass filter By default, the initial state of an IIR filter as implemented in lfilter or sosfilt is all zero. If the input signal does not start with values that are zero, there will be a transient during which the filter’s internal state "catches up" with the input signal.
Here is an example. The script generates the plot shown in Figure 6.
import numpy as np from scipy.signal import butter, sosfilt, sosfilt_zi import matplotlib.pyplot as plt n = 101 t = np.linspace(0, 1, n) np.random.seed(123) x = 0.45 + 0.1np.random.randn(n) sos = butter(8, 0.125, output=’sos’) # Filter using the default initial conditions.
y = sosfilt(sos, x) # Filter using the state for which the output # is the constant x[:4].mean() as the initial # condition.
0.0 0.2 0.4 0.6 0.8 1.0 t 0.0 0.2 0.4 0.6 Filter with different initial conditions x y (zero ICs) y2 (mean(x[:4]) ICs) Fig. 6: A demonstration of two different sets of initial conditions for a lowpass filter. The orange curve is the output of the filter with zero initial conditions. The green curve is the output of the filter initialized with a state associated with the mean of the first four values of the input x.
zi = x[:4].mean() sosfilt_zi(sos) y2, zo = sosfilt(sos, x, zi=zi) # Plot everything.
plt.plot(t, x, alpha=0.75, linewidth=1, label=’x’) plt.plot(t, y, label=’y (zero ICs)’) plt.plot(t, y2, label=’y2 (mean(x[:4]) ICs)’) plt.legend(framealpha=1, shadow=True) plt.grid(alpha=0.25) plt.xlabel(’t’) plt.title(’Filter with different ’ ’initial conditions’) plt.show() By setting zi=x[:4].mean() sosfilt_zi(sos), we are, in effect, making the filter start out as if it had been filtering the constant x[:4].mean() for a long time. There is still a transient associated with this assumption, but it is usually not as objectionable as the transient associated with zero initial conditions.
This initialization is usually not needed for a bandpass or highpass filter. Also, the forward-backward filters implemented in filtfilt and sosfiltfilt already have options for controlling the initial conditions of the forward and backward passes.
Bandpass filter In this example, we will use synthetic data to demonstrate a bandpass filter. We have 0.03 seconds of data sampled at 4800 Hz. We want to apply a bandpass filter to remove frequencies below 400 Hz or above 1200 Hz.
Just like we did for the lowpass filter, we define two functions that allow us to create and apply a Butterworth bandpass filter with the frequencies given in Hz (or any other units). The functions take care of scaling the values to the normalized range expected by scipy.signal.butter.
from scipy.signal import butter, sosfilt def butter_bandpass(lowcut, highcut, fs, order): DRAFT 6 PyData CookBook 0 500 1000 1500 2000 Frequency (Hz) 0.0 0.2 0.4 0.6 0.8 1.0 Gain lowcut: 400 Hz highcut: 1200 Hz Amplitude response for Butterworth bandpass filters order = 3 order = 6 order = 12 2/2 Fig. 7: Amplitude response for a Butterworth bandpass filter with several different orders.
nyq = 0.5 fs low = lowcut / nyq high = highcut / nyq sos = butter(order, [low, high], btype=’band’, output=’sos’) return sos def butter_bandpass_filt(data, lowcut, highcut, fs, order): sos = butter_bandpass(lowcut, highcut, fs, order) y = sosfilt(sos, data) return y First, we’ll take a look at the frequency response of the Butterworth bandpass filter with order 3, 6, and 12.
The code that generates Figure 7 demonstrates the use of scipy.signal.sosfreqz: for order in [3, 6, 12]: sos = butter_bandpass(lowcut, highcut, fs, order) w, h = sosfreqz(sos, worN=2000) plt.plot((fs0.5/np.pi)w, abs(h), ’k’, alpha=(order+1)/13, label="order = %d" % order) Figure 8 shows the input signal and the filtered signal. The order 12 bandpass Butterworth filter was used. The plot shows the input signal x; the filtered signal was generated with y = butter_bandpass_filt(x, lowcut, highcut, fs, order=12) where fs = 4800, lowcut = 400 and highcut = 1200.
Filtering a long signal in batches The function lfilter applies a filter to an array that is stored in memory. Sometimes, however, the complete signal 0.000 0.005 0.010 0.015 0.020 0.025 0.030 Time (seconds) 0.025 0.000 0.025 0.050 0.075 0.100 0.125 Noisy signal Filtered signal Fig. 8: Original noisy signal and the filtered signal. The order 12 Butterworth bandpass filter shown in Figure 7 was used.
to be filtered is not available all at once. It might not fit in memory, or it might be read from an instrument in small blocks and it is desired to output the filtered block before the next block is available. Such a signal can be filtered in batches, but the state of the filter at the end of one batch must be saved and then restored when lfilter is applied to the next batch. Here we show an example of how the zi argument of lfilter allows the state to be saved and restored. We will again use synthetic data generated by the same function used in the previous example, but for a longer time interval.
A pattern that can be used to filter an input signal x in batches is shown in the following code. The filtered signal is stored in y. The array sos contains the filter in SOS format, and is presumed to have already been created.
batch_size = N # Number of samples per batch # Array of initial conditions for the SOS filter.
z = np.zeros((sos.shape, 2)) # Preallocate space for the filtered signal.
y = np.empty_like(x) start = 0 while start < len(x): stop = min(start + batch_size, len(x)) y[start:stop], z = sosfilt(sos, x[start:stop], zi=z) start = stop In this code, the next batch of input is fetched by simply indexing x[start:stop], and the filtered batch is saved by assigning it to y[start:stop]. In a more realistic batch processing system, the input might be fetched from a file, or directly from an instrument, and the output might be written to another file, or handed off to another process as part of a batch processing pipeline.
Solving linear recurrence relations Variations of the question: How do I speed up the following calculation?
y[i+1] = alphay[i] + cx[i] DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 7 0.00 0.01 0.02 0.03 0.04 0.05 0.06 Time (seconds) 0.025 0.000 0.025 0.050 0.075 0.100 0.125 Noisy signal Filtered signal Fig. 9: Original noisy signal and the filtered signal. The order 12 Butterworth bandpass filter shown in Figure 7 was used. The signal was filtered in batches of size 72 samples (0.015 seconds). The alternating light and dark blue colors of the filtered signal indicate batches that were processed in separate calls to sosfilt.
often arise on mailing lists and online forums. Sometimes more terms such as betay[i-1] or dx[i-1] are in-cluded on the right. These recurrence relations show up in, for example, GARCH models and other linear stochastic models.
Such a calculation can be written in the form of Eqn. (1), so a solution can be computed using lfilter.
Here’s an example that is similar to several questions that have appeared on the programming Q&A website stackoverflow.com. The one-dimensional array h is an input, and alpha, beta and gamma are constants: y = np.empty(len(h)) y = alpha for i in np.arange(1, len(h)): y[i] = alpha + betay[i-1] + gammah[i-1] To use lfilter to solve the problem, we have to translate the linear recurrence: y[i] = alpha + betay[i-1] + gammah[i-1] into the form of Eqn. (1), which will give us the coefficients b and a of the transfer function. Define: x[i] = alpha + gammah[i] so the recurrence relation is: y[i] = x[i-1] + betay[i-1] Compare this to Eqn. (1); we see that a0 = 1, a1 = −beta, b0 = 0 and b1 = 1. So we have our transfer function coefficients: b = [0, 1] a = [1, -beta] We also have to ensure that the initial condition is set correctly to reproduce the desired calculation. We want the initial condition to be set as if we had values x[-1] and y[-1], and y is computed using the recurrence relation. Given the above recurrence relation, the formula for y is: y = x[-1] + betay[-1] We want y to be alpha, so we’ll set y[-1] = 0 and x[-1] = alpha. To create initial conditions for lfilter that will set up the filter to act like it had just operated on those previous values, we use scipy.signal.lfiltic: zi = lfiltic(b, a, y=, x=[alpha]) The y and x arguments are the "previous" values that will be used to set the initial conditions. In general, one sets y=[y[-1], y[-2], ..] and x=[x[-1], x[-2], ...], giving as many values as needed to deter-mine the initial condition for lfilter. In this example, we have just one previous value for y and x.
Putting it all together, here is the code using lfilter that replaces the for-loop shown above: b = [0, 1] a = [1, -beta] zi = lfiltic(b, a, y=, x=[alpha]) y, zo = lfilter(b, a, alpha + gammah, zi=zi) FIR filters in scipy.signal A finite impulse response filter is basically a weighted moving average. Given an input sequence xn and the M + 1 filter coefficients {b0,...,bM}, the filtered output yn is computed as the discrete convolution of x and b: yn = M ∑ i=0 bixn−i = (b∗x)n (4) where ∗is the convolution operator. M is the order of the filter; a filter with order M has M +1 coefficients. It is common to say that the filter has M +1 taps.
Apply a FIR filter To apply a FIR filter to a signal, we can use scipy.signal.lfilter with the denominator set to the scalar 1, or we can use one of the convolution functions available in NumPy or SciPy, such as scipy.signal.convolve. For a signal {x0,x1,...,xS−1} of finite length S, Eq. (4) doesn’t specify how to compute the result for n < M. The convolution functions in NumPy and SciPy have an option called mode for specifying how to handle this. For example, mode=’valid’ only computes output values for which all the values of xi in Eq. 4 are defined, and mode=’same’ in effect pads the input array x with zeros so that the output is the same length as the input. See the docstring of numpy.convolve or scipy.signal.convolve for more details.
For example, from scipy.signal import convolve # Make a signal to be filtered.
np.random.seed(123) x = np.random.randn(50) # taps is the array of FIR filter coefficients.
taps = np.array([ 0.0625, 0.25 , 0.375 , 0.25 , 0.0625]) # Filtered signal. y has the same length as x.
y = convolve(x, taps, mode=’same’) There are also convolution functions in scipy.ndimage.
The function scipy.ndimage.convolve1d provides an DRAFT 8 PyData CookBook axis argument, which allows all the signals stored in one axis of a multidimensional array to be filtered with one call.
For example, from scipy.ndimage import convolve1d # Make an 3-d array containing 1-d signals # to be filtered.
x = np.random.randn(3, 5, 50) # Apply the filter along the last dimension.
y = convolve1d(x, taps, axis=-1) Note that scipy.ndimage.convolve1d has a different set of options for its mode argument. Consult the docstring for details.
Specialized functions that are FIR filters The uniform filter and the Gaussian filter implemented in scipy.ndimage are FIR filters.
In the case of one-dimensional time series, the specific functions are uniform_filter1d and gaussian_filter1d.
The Savitzky-Golay filter [SavGol] is also a FIR filter. In the module scipy.signal, SciPy provides the function savgol_coeffs to create the coefficients of a Savitzy-Golay filter. The function savgol_filter applies the Savitzky-Golay filter to an input signal without returning the filter coefficients.
FIR filter frequency response The function scipy.signal.freqz computes the fre-quency response of a linear filter represented as a transfer function. This class of filters includes FIR filters, where the representation of the numerator of the transfer function is the array of taps and the denominator is the scalar a0 = 1.
As an example, we’ll compute the frequency response of a uniformly weighted moving average. For a moving average of length n, the coefficients in the FIR filter are simply 1/n.
Translated to NumPy code, we have taps = np.full(n, fill_value=1.0/n).
The response curves in Figure 10 were generated with this code: for n in [3, 7, 21]: taps = np.full(n, fill_value=1.0/n) w, h = freqz(taps, worN=2000) plt.plot(w, abs(h), label="n = %d" % n) The function freqz returns the frequencies in units of radians per sample, which is why the values on the abscissa in Figure 10 range from 0 to π. In calculations where we have a given sampling frequency fs, we usually convert the frequencies returned by freqz to dimensional units by multiplying by fs/(2π).
FIR filter design We’ll demonstrate how SciPy can be used to design a FIR filter using the following four methods.
• The window method. The filter is designed by computing the impulse response of the desired ideal filter and then multiplying the coefficients by a window function.
• Least squares design. The weighted integral of the squared frequency response error is minimized.
0 4 2 3 4 Frequency (radians/sample) 0.0 0.2 0.4 0.6 0.8 1.0 Gain Amplitude Response for Moving Average Filter n = 3 n = 7 n = 21 Fig. 10: Frequency response of a simple moving average. n is the number of taps (i.e. the length of the sliding window).
• Parks-McClellan equiripple design.
A "minimax" method, in which the maximum deviation from the desired response is minimized.
• Linear programming. The "minimax" design problem can be formulated as a linear programming problem.
In the following sections, we discuss each design method.
For this discussion, we define the following functions, where ω is the frequency in radians per sample: A(ω), the filter’s (real, signed) frequency response; D(ω), the desired frequency response of the filter; and W(ω), the weight assigned to the response error at ω (i.e. how "important" is the error A(ω)− D(ω)).
FIR filter design: the window method The window method for designing a FIR filter is to compute the filter coefficients as the impulse response of the desired ideal filter, and then multiply the coefficents by a window function to both truncate the set of coefficients (thus making a finite impulse response filter) and to shape the actual filter response. Most textbooks on digital signal processing include a discussion of the method; see, for example, Section 7.5 of Oppenheim and Schafer [OS].
Two functions in the module scipy.signal implement the window method, firwin and firwin2. Here we’ll show an example of firwin2. We’ll use firwin when we discuss the Kaiser window method.
We’ll design a filter with 185 taps for a signal that is sampled at 2000 Hz. The filter is to be lowpass, with a linear transition from the pass band to the stop band over the range 150 Hz to 175 Hz. We also want a notch in the pass band between 48 Hz and 72 Hz, with sloping sides, centered at 60 Hz where the desired gain is 0.1. The dashed line in Figure 12 shows the desired frequency response.
To use firwin2, we specify the desired response at the endpoints of a piecewise linear profile defined over the frequency range [0, 1000] (1000 Hz is the Nyquist frequency).
freqs = [0, 48, 60, 72, 150, 175, 1000] gains = [1, 1, 0.1, 1, 1, 0, 0] DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 9 0 92 184 Sample number 0.0 0.2 0.4 0.6 0.8 1.0 Window functions Hamming Kaiser, =2.70 rectangular Fig. 11: Window functions used in the firwin2 filter design example.
To illustrate the affect of the window on the filter, we’ll demonstrate the design using three different windows: the Hamming window, the Kaiser window with parameter β set to 2.70, and the rectangular or "boxcar" window (i.e. simple truncation without tapering).
The code to generate the FIR filters is fs = 2000 numtaps = 185 # window=None is equivalent to using the # rectangular window.
taps_none = firwin2(numtaps, freqs, gains, nyq=0.5fs, window=None) # The default window is Hamming.
taps_h = firwin2(numtaps, freqs, gains, nyq=0.5fs) beta = 2.70 taps_k = firwin2(numtaps, freqs, gains, nyq=0.5fs, window=(’kaiser’, beta)) Figure 12 shows the frequency response of the three filters.
FIR filter design: least squares The weighted least squares method creates a filter for which the expression Z π 0 W(ω)(A(ω)−D(ω))2 dω (5) is minimized. The function scipy.signal.firls im-plements this method for piecewise linear desired response D(ω) and piecewise constant weight function W(ω). Three arguments (one optional) define the shape of the desired response: bands, desired and (optionally) weights.
The argument bands is sequence of frequency values with an even length. Consecutive pairs of values define the bands on which the desired response is defined. The frequencies covered by bands does not have to include the entire spectrum from 0 to the Nyquist frequency. If there are gaps, the response in the gap is ignored (i.e. the gaps are "don’t care" regions).
The desired input array defines the amplitude of the desired frequency response at each point in bands.
0 50 100 150 200 Frequency (Hz) 0.0 0.2 0.4 0.6 0.8 1.0 Gain 185 taps Filters designed with the window method ideal Hamming Kaiser, =2.70 rectangular Fig. 12: Frequency response for a filter designed using firwin2 with several windows. The ideal frequency response is a lowpass filter with a ramped transition starting at 150 Hz. There is also a notch with ramped transitions centered at 60 Hz.
The weight input, if given, must be an array with half the length of bands. The values in weight define the weight of each band in the objective function. A weight of 0 means the band does not contribute to the result at all--it is equivalent to leaving a gap in bands.
As an example, we’ll design a filter for a signal sampled at 200 Hz. The filter is a lowpass filter, with pass band [0, 15] and stop band [30, 100], and we want the gain to vary linearly from 1 down to 0 in the transition band [15, 30]. We’ll design a FIR filter with 43 taps.
We create the arrays bands and desired as described above: bands = np.array([0, 15, 15, 30, 30, 100]) desired = np.array([1, 1, 1, 0, 0, 0]) Then we call firls: numtaps = 43 taps1 = firls(numtaps, bands, desired, nyq=100) The frequency response of this filter is the blue curve in Figure 13.
By default, the firls function weights the bands uni-formly (i.e. W(ω) ≡1 in Eqn. (5)). The weights argument can be used to control the weight W(ω) on each band. The argument must be a sequence that is half the length of bands.
That is, only piecewise constant weights are allowed.
Here we rerun firls, giving the most weight to the pass band and the least weight to the transition band: wts = [100, .01, 1] taps2 = firls(numtaps, bands, desired, nyq=100, weight=wts) The frequency response of this filter is the orange curve in Figure 13. As expected, the frequency response now deviates DRAFT 10 PyData CookBook 0 15 30 100 0.0 0.5 1.0 Gain Least Squares Filter Design uniform weight weight:[100, 0.01, 1] 0 15 0.99 1.00 1.01 Gain 30 100 Frequency (Hz) 0.00 0.01 0.02 Gain Fig. 13: Result of a least squares FIR filter design. The desired frequency response comprises three bands. On [0, 15], the desired gain is 1 (a pass band). On [15, 30], the desired gain decreases linearly from 1 to 0. The band [30, 100] is a stop band, where the desired gain is 0. The filters have 43 taps. The middle and bottom plots are details from the top plot.
more from the desired gain in the transition band, and the ripple in the pass band is significantly reduced. The rejection in the stop band is also improved.
Equivalence of least squares and the window method.
When uniform weights are used, and the desired result is specified for the complete interval [0,π], the least squares method is equivalent to the window method with no window function (i.e. the window is the "boxcar" function). To verify this numerically, it is necessary to use a sufficiently high value for the nfreqs argument of firwin2.
Here’s an example: >>> bands = np.array([0, 0.5, 0.5, 0.6, 0.6, 1]) >>> desired = np.array([1, 1, 1, 0.5, 0.5, 0]) >>> numtaps = 33 >>> taps_ls = firls(numtaps, bands, desired) >>> freqs = bands >>> gains = desired >>> taps_win = firwin2(numtaps, freqs, gains, ...
nfreqs=8193, window=None) >>> np.allclose(taps_ls, taps_win) True In general, the window method cannot be used as a replace-ment for the least squares method, because it does not provide an option for weighting distinct bands differently; in particular, it does not allow for "don’t care" frequency intervals (i.e.
intervals with weight 0).
FIR filter design: Parks-McClellan The Parks-McClellan algorithm [PM] is based on the Remez exchange algorithm [RemezAlg]. This is a "minimax" opti-mization; that is, it miminizes the maximum value of |E(ω)| over 0 ≤ω ≤π, where E(ω) is the (weighted) deviation of the actual frequency response from the desired frequency response: E(ω) = W(ω)(A(ω)−D(ω)), 0 ≤ω ≤π, (6) We won’t give a detailed description of the algorithm here; most texts on digital signal processing explain the algorithm (e.g. Section 7.7 of Oppenheim and Schafer [OS]). The method is implemented in scipy.signal by the function remez.
As an example, we’ll design a bandpass filter for a signal with a sampling rate of 2000 Hz using remez. For this filter, we want the stop bands to be [0, 250] and [700, 1000], and the pass band to be [350, 550]. We’ll leave the behavior outside these bands unspecified, and see what remez gives us. We’ll use 31 taps.
fs = 2000 bands = [0, 250, 350, 550, 700, 0.5fs] desired = [0, 1, 0] numtaps = 31 taps = remez(numtaps, bands, desired, fs=fs) The frequency response of this filter is the curve labeled (a) in Fig. 14.
To reduce the ripple in the pass band while using the same filter length, we’ll adjust the weights, as follows: weights = [1, 25, 1] taps2 = remez(numtaps, bands, desired, weights, fs=fs) The frequency response of this filter is the curve labeled (b) in Fig. 14.
It is recommended to always check the frequency response of a filter designed with remez. Figure 15 shows the fre-quency response of the filters when the number of taps is increased from 31 to 47. The ripple in the pass and stop bands is decreased, as expected, but the behavior of the filter in the interval [550, 700] might be unacceptable. This type of behavior is not unusual for filters designed with remez when there are intervals with unspecified desired behavior.
In some cases, the exchange algorithm implemented in remez can fail to converge. Failure is more likely when the number of taps is large (i.e. greater than 1000). It can also happen that remez converges, but the result does not have the expected equiripple behavior in each band. When a problem occurs, one can try increasing the maxiter argument, to allow the algorithm more iterations before it gives up, and one can try increasing grid_density to increase the resolution of the grid on which the algorithm seeks the maximum of the response errors.
FIR filter design: linear programming The design problem solved by the Parks-McClellan method can also be formulated as a linear programming problem ([Rabiner1972a], [Rabiner1972b]).
DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 11 0 200 400 600 800 1000 Frequency (Hz) 0.0 0.2 0.4 0.6 0.8 1.0 Gain 31 taps Bandpass filters designed with remez (a) (b) Fig. 14: Frequency response of bandpass filters designed using scipy.signal.remez. The stop bands are [0, 250] and [700, 1000], and the pass band is [350, 550]. The shaded regions are the "don’t care" intervals where the desired behavior of the filter is unspecified. The curve labeled (a) uses the default weights--each band is given the same weight. For the curve labeled (b), weight = [1, 25, 1] was used.
0 200 400 600 800 1000 Frequency (Hz) 0.0 0.2 0.4 0.6 0.8 1.0 Gain 47 taps Bandpass filters designed with remez (a) (b) Fig. 15: This plot shows the results of the same calculation that produced Figure 14, but the number of taps has been increased from 31 to 47. Note the possibly undesirable behavior of the filter in the transition interval [550, 700].
To implement this method, we’ll use the function linprog from scipy.optimize. In particular, we’ll use the interior point method that was added in SciPy 1.0. In the following, we first review the linear programming formulation, and then we discuss the implementation.
Formulating the design problem as a linear program.
Like the Parks-McClellan method, this approach is a "mini-max" optimization of Eq. (6). We’ll give the formulation for a Type I filter design (that is, an odd number of taps with even symmetry), but the same ideas can be applied to other FIR filter types.
For convenience, we’ll consider the FIR filter coefficients for a filter of length 2R+1 using centered indexing: b−R,b−R+1,...,b−1,b0,b1,...,bM−1,bR Consider a sinusoidal signal with frequency ω radians per sample. The frequency response can be written A(ω) = R ∑ i=−R bi cos(ωi) = b0 + R ∑ i=0 2bi cos(ωi) = R ∑ i=0 pi cos(ωi) where we define p0 = b0 and, for 1 ≤i ≤R, pi = 2bi. We’ve used the even symmetry of the cosine function and the of filter coefficients about the middle coefficient (b−i = bi).
The "minimax" problem is to minimize the maximum error.
That is, choose the filter coefficients such that |E(ω)| ≤ε for 0 ≤ω ≤π for the smallest possible value of ε. After substituting the expression for E(ω) from Eqn. (6), replacing the absolute value with two inequalities, and doing a little algebra, the problem can be written as minimize ε over p0, p1, ..., pR, ε subject to A(ω)− ε W(ω) ≤D(ω) and −A(ω)− ε W(ω) ≤−D(ω) ω is a continuous variable in the above formulation. To implement this as a linear programming problem, we use a suitably dense grid of L frequencies ω0,ω1,...,ωL−1 (not necessarily uniformly spaced). We define the L × (R + 1) matrix C as Cij = cos(ωi−1( j −1)), 1 ≤i ≤L and 1 ≤j ≤R+1 (7) Then the vector of frequency responses is the matrix product Cp, where p = [p0, p1,..., pR]T.
Let dk = D(ωk), and d = [d0,d1,...,dL−1]T. Similarly, de-fine v = [v0,v1,...,vL−1]T, where vk = 1/W(ωk). The linear programming problem is minimize ε over p0, p1, ..., pR, ε subject to C −v −C −v p ε ≤ d −d This is the formulation that can be used with, for example, scipy.optimize.linprog.
This formulation, however, provides no advantages over the solver provided by remez, and in fact it is generally much slower and less robust than remez. When designing a filter beyond a hundred or so taps, there is much more likely to be a convergence error in the linear programming method than in remez.
The advantage of the linear programming method is its ability to easily handle additional constraints. Any constraint, either equality or inequality, that can be written as a linear constraint can be added to the problem.
DRAFT 12 PyData CookBook We will demonstrate how to implement a lowpass filter design using linear programming with the constraint that the gain for a constant input is exactly 1. That is, A(0) = R ∑ i=0 pi = 1 which may be written Aeq p ε = 1, where Aeq = [1,1,...,1,0].
Implementing the linear program. Let’s look at the code required to set up a call to linprog to design a lowpass filter with a pass band of [0,ωp] and a stop band of [ωs,π], where the frequencies ωp and ωs are expressed in radians per sample, and 0 < ωp < ωs < π. We’ll also impose the constraint that A(0) = 1.
A choice for the density of the frequency samples on [0,π] that works well is 16N, where N is the number of taps (numtaps in the code). Then the number of samples in the pass band and the stop band can be computed as density = 16numtaps/np.pi numfreqs_pass = int(np.ceil(wpdensity)) numfreqs_stop = int(np.ceil((np.pi - ws)density)) The grids of frequencies on the pass and stop bands are then wpgrid = np.linspace(0, wp, numfreqs_pass) wsgrid = np.linspace(ws, np.pi, numfreqs_stop) We will impose an equality constraint on A(0), so we can can remove that frequency from wpgrid--there is no point in requiring both the equality and inequality constraints at ω = 0.
Then wpgrid and wsgrid are concatenated to form wgrid, the grid of all the frequency samples.
wpgrid = wpgrid[1:] wgrid = np.concatenate((wpgrid, wsgrid)) Let wtpass and wtstop be the constant weights that we will use in the pass and stop bands, respectively. We create the array of weights on the grid with weights = np.concatenate( (np.full_like(wpgrid, fill_value=wtpass), np.full_like(wsgrid, fill_value=wtstop))) The desired values of the frequency response are 1 in the pass band and 0 in the stop band. Evaluated on the grid, we have desired = np.concatenate((np.ones_like(wpgrid), np.zeros_like(wsgrid))) Now we implement Eq. (7) and create the L×(R+1) array of coefficients C that are used to compute the frequency response, where R = M/2: R = (numtaps - 1)//2 C = np.cos(wgrid[:, np.newaxis]np.arange(R+1)) The column vector of the reciprocals of the weights is V = 1/weights[:, np.newaxis] 0 0.16 Frequency (radians/sample) 0.999 1.000 1.001 1.002 Gain Pass band detail linear programming, H(0)=1 remez Fig. 16: Result of solving a lowpass FIR filter design problem by linear programming with the constraint A(0) = 1. The response without the extra constraint, solved using remez, is also plotted.
Next we assemble the pieces that define the inequality con-straints that are actually passed to linprog: A = np.block() b = np.block().T c = np.zeros(M+2) c[-1] = 1 In code, the arrays for the equality constraint needed to define A(0) = 1 are: A_eq = np.ones((1, R+2)) A_eq[:, -1] = 0 b_eq = np.array() Finally, we set up and call linprog: options = dict(maxiter=5000, tol=1e-6) sol = linprog(c, A, b, A_eq=A_eq, b_eq=b_eq, bounds=(None, None), method=’interior-point’, options=options) if sol.success: p = sol.x[:-1] taps = 0.5np.concatenate((p[:0:-1], [2p], p[1:])) Notes: • For different problems, the parameters defined in the dictionary options may have to be adjusted. See the documentation for linprog for more details.
• By default, linprog assumes that all the variables must be nonnegative. We use the bounds argument to override that behavior.
• We have had more success using the interior point method than the default simplex method.
See Figure 16 for a plot of the pass band response of the filter designed using linprog. The number of taps was N = 81, and the transition boundary frequencies, expressed in radians per sample, were ωp = 0.16π and ωs = 0.24π. For the weight in each band we used wtpass = 2 and wtstop = 1.
DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 13 Determining the order of a FIR filter Most of the filter design tools in SciPy require the number of taps as an input. Typically, however, a designer has require-ments on the pass band ripple and the stop band rejection, and wants the FIR filter with the minimum number of taps that satisfies these requirements. The diagram shown in Figure 17 illustrates the design parameters for a lowpass filter. The graph of the magnitude of the frequency response of the filter must not enter the shaded area. The parameter δp defines the allowed pass band ripple, and δs defines the required attenuation in the stop band. The maximum width of the transition from the pass band to stop band is ∆ω, and the cutoff frequency ωc is centered in the transition band.
In the next two sections, we’ll consider the following filter design problem. We need a lowpass filter for a signal that is sampled at 1000 Hz. The desired cutoff frequency is 180 Hz, and the transition from the pass band to the stop band must not exceed 30 Hz. In the pass band, the gain of the filter should deviate from 1 by no more than 0.005 (i.e. worst case ripple is 0.5%). In the stop band, the gain must be less than 0.002 (about 54 dB attenuation). In the next section, we’ll tackle the design using the Kaiser window method. After that, we’ll obtain an optimal design by using the Parks-McClellan method.
Kaiser’s window method The Kaiser window is a window function with a parameter β that controls the shape of the function. An example of a Kaiser window is plotted in Figure 11. Kaiser [Kaiser66], [Kaiser74] developed formulas that, for a given transition width ∆ω and error tolerance for the frequency response, determine the order M and the parameter β required to meet the requirements.
Summaries of the method can be found in many sources, including Sections 7.5.3 and 7.6 of the text by Oppenheim and Schafer [OS].
In Kaiser’s method, there is only one parameter that con-trols the passband ripple and the stopband rejection. That is, Kaiser’s method assumes δp = δs. Let δ be that common value.
The stop band rejection in dB is −20log10(δ). This value (in dB) is the first argument of the function kaiserord. One can interpret the argument ripple as the maximum deviation (expressed in dB) allowed in |A(ω) −D(ω)|, where A(ω) is the magnitude of the actual frequency response of the filter and D(ω) is the desired frequency response. (That is, in the pass band, D(ω) = 1, and in the stop band, D(ω) = 0.) In Figure 18, the bottom plot shows |A(ω)−D(ω)|.
The Kaiser window design method, then, is to de-termine the length of the filter and the Kaiser win-dow parameter β using Kaiser’s formula (implemented in scipy.signal.kaiserord), and then design the filter using the window method with a Kaiser window (using, for example, scipy.signal.firwin): numtaps, beta = kaiserord(ripple, width) taps = firwin(numtaps, cutoff, window=(’kaiser’, beta), [other args as needed]) For our lowpass filter design problem, we first define the input parameters: 0 c Frequency (radians per sample) 0 s 1 p 1 1 + p |H(ej )| Lowpass Filter Design Specifications Fig. 17: Lowpass filter design specifications. The magnitude of the frequency response of the filter should not enter the shaded regions.
Frequency values in Hz fs = 1000.0 cutoff = 180.0 width = 30.0 # Desired pass band ripple and stop band attenuation deltap = 0.005 deltas = 0.002 As already mentioned, the Kaiser method allows for only a single parameter to constrain the approximation error. To ensure we meet the design criteria in the pass and stop bands, we take the minimum of δp and δs: delta = min(deltap, deltas) The first argument of kaiserord must be expressed in dB, so we set: delta_db = -20np.log10(delta) Then we call kaiserord to determine the number of taps and the Kaiser window parameter β: numtaps, beta = kaiserord(delta_db, width/(0.5fs)) numtaps |= 1 # Must be odd for a Type I FIR filter.
For our lowpass filter design problem, we find numtaps is 109 and β is 4.990.
Finally, we use firwin to compute the filter coefficients: taps = firwin(numtaps, cutoff/(0.5fs), window=(’kaiser’, beta), scale=False) The results of the Kaiser method applied to our lowpass filter design problem are plotted in Figure 18. The tip of the right-most ripple in the pass band violates the δ-constraint by a very small amount; this is not unusual for the Kaiser method. In this case, it is not a problem, because the original requirement for the pass band is δp = 0.005, so the behavior in the pass band is overly conservative.
Optimizing the FIR filter order The Kaiser window method can be used to create a filter that meets (or at least is very close to meeting) the design requirements, but it will not be optimal. That is, generally there will exist FIR filters with fewer taps that also satisfy the design requirements. At the time this chapter is being written, SciPy does not provide a tool that automatically determines DRAFT 14 PyData CookBook 0 100 200 300 400 500 50 0 Gain (dB) Kaiser Window Filter Design 0 50 100 150 0.998 1.000 1.002 Gain 0 100 200 300 400 500 Frequency (Hz) 0.000 0.001 0.002 |A( ) - D( )| Fig. 18: Result of the Kaiser window filter design of a lowpass filter. The number of taps is 109. Top: Magnitude (in dB) of the frequency response. Middle: Detail of the frequency response in the pass band. Bottom: The deviation of the actual magnitude of the frequency response from that of the ideal lowpass filter.
the optimal number of taps given pass band ripple and stop band rejection requirements. It is not difficult, however, to use the existing tools to find an optimal filter in a few steps (at least if the filter order is not too large).
Here we show a method that works well, at least for the basic lowpass, highpass, bandpass and bandstop filters on which it has been tested. The idea: given the design requirements, first estimate the length of the filter. Create a filter of that length using remez, with 1/δp and 1/δs as the weights for the pass and stop bands, respectively. Check the frequency response of the filter. If the initial estimate of the length was good, the filter should be close to satisfying the design requirements. Based on the observed frequency response, adjust the number of taps, then create a new filter and reevaluate the frequency response. Iterate until the shortest filter that meets the design requirements is found. For moderate sized filters (up to 1000 or so taps), this simple iterative process can be automated. (For higher order filters, this method has at least two weaknesses: it might be difficult to get a reasonably accurate estimate of the filter length, and it is more likely that remez will fail to converge.) A useful formula for estimating the length of a FIR filter was given by Bellanger [Bellanger]: N ≈−2 3 log10 (10δpδs) fs ∆f (8) which has a straightforward Python implementation: def bellanger_estimate(deltap, deltas, width, fs): n = (-2/3)np.log10(10deltapdeltas)fs/width n = int(np.ceil(n)) return n We’ll apply this method to the lowpass filter design problem that was described in the previous section. As before, we define the input parameters: # Frequency values in Hz fs = 1000.0 cutoff = 180.0 width = 30.0 # Desired pass band ripple and stop band attenuation deltap = 0.005 deltas = 0.002 Then the code numtaps = bellanger_estimate(deltap, deltas, width, fs) numtaps |= 1 gives numtaps = 89. (Compare this to the result of the Kaiser method, where numtaps is 109.) Now we’ll use remez to design the filter.
trans_lo = cutoff - 0.5width trans_hi = cutoff + 0.5width taps = remez(numtaps, bands=[0, trans_lo, trans_hi, 0.5fs], desired=[1, 0], weight=[1/deltap, 1/deltas], fs=fs) The frequency response of the filter is shown in Figure 19.
We see that the filter meets the design specifications. If we decrease the number of taps to 87 and check the response, we find that the design specifications are no longer met, so we accept 89 taps as the optimum.
REFERENCES [Bellanger] M. Bellanger, Digital Processing of Signals: Theory and Practice (3rd Edition), Wiley, Hoboken, NJ, 2000.
[Kaiser66] J. F. Kaiser, Digital filters, in System Analysis by Digital Computer, Chapter 7, F. F. Kuo and J. F. Kaiser, eds., Wiley, New York, NY, 1966 [Kaiser74] J. F. Kaiser, Nonrecursive digital filter design using the I0-sinh window function, Proc. 1974 IEEE International Symp.
on Circuits and Systems, San Francisco, CA, 1974.
[Lyons] Richard G. Lyons. Understanding Digital Signal Processing (2nd ed.), Pearson Higher Education, Inc., Upper Saddle River, New Jersey (2004) [OS] Alan V. Oppenheim, Ronald W. Schafer. Discrete-Time Sig-nal Processing (3rd ed.), Pearson Higher Education, Inc., Upper Saddle River, New Jersey (2010) [PM] Parks-McClellan filter design algorithm.
Wikipedia, filter_design_algorithm [Rabiner1972a] L. R. Rabiner, The design of finite impulse response digital filters using linear programming techniques, The Bell System Technical Journal, Vol. 51, No. 6, July-August, 1972.
[Rabiner1972b] L. R. Rabiner, Linear program design of finite impulse response (FIR) digital filters, IEEE Trans. on Audio and Electroacoustics, Vol. AU-20, No. 4, Oct. 1972.
[RemezAlg] Remez algorithm.
Wikipedia, [SavGol] A. Savitzky, M. J. E. Golay. Smoothing and Differentiation of Data by Simplified Least Squares Procedures. Analytical Chemistry, 1964, 36 (8), pp 1627-1639.
DRAFT SIGNAL PROCESSING WITH SCIPY: LINEAR FILTERS 15 0 100 200 300 400 500 50 0 Gain (dB) 30 Hz Lowpass Filter Optimal Remez Design 0 50 100 150 0.995 1.000 1.005 Gain 0 100 200 300 400 500 Frequency (Hz) 0.000 0.002 0.004 |A( ) - D( )| Fig. 19: Optimal lowpass filter frequency response. The number of taps is 89.
[SO] Nimal Naser, How to filter/smooth with SciPy/Numpy?, |
4129 | https://thirdspacelearning.com/gcse-maths/algebra/algebraic-fractions/ | What are algebraic fractions?
How to solve equations including algebraic fractions
Algebraic fractions worksheet
Algebraic fractions examples
Example 1: Equation with one fraction Example 2: Equation with two fractions Example 3: Equation with x in the denominator Example 4: Equation with three fractions Example 5: Denominators are expressions in terms of x Example 6: Equation including a quadratic
Common misconceptions
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Algebraic fractions GCSE questions
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Introduction
What are algebraic fractions?
How to solve equations including algebraic fractions
Algebraic fractions worksheet
Algebraic fractions examples
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Example 1: Equation with one fraction Example 2: Equation with two fractions Example 3: Equation with x in the denominator Example 4: Equation with three fractions Example 5: Denominators are expressions in terms of x Example 6: Equation including a quadratic
Common misconceptions
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Algebraic fractions GCSE questions
Learning checklist
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GCSE Maths Algebra Algebraic Expressions Simplifying Expressions
Algebraic Fractions
Algebraic Fractions
Here we will learn about algebraic fractions, including operations with fractions, and solving linear and quadratic equations written in the form of algebraic fractions.
There are also algebraic fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
What are algebraic fractions?
Algebraic fractions are fractions that contain at least one variable.
The following algebraic expressions are examples of algebraic fractions:
x is the numerator: \quad \quad \quad \frac{x}{12} Both the numerator and the denominator contain an x term: \quad \quad \quad \frac{x+1}{2x}
An expression in terms of x is the denominator: \quad \quad \frac{3}{x+1}\ Both the numerator and the denominator contain an expression with x : \quad \quad \frac{3x+4}{2x-5}
The numerator is a multiple of x : \quad\quad \quad \quad \frac{2x}{15} The numerator and the denominator are quadratic expressions: \quad\quad \quad \quad \frac{(x+3)^{2}}{x^{2}-9}
The main aim of this lesson is to understand how to solve equations that include algebraic fractions.
All the examples above are expressions whereas the examples below are equations as we can find specific values for x for each example to solve the equation.
E.g.
One step equation: \quad \quad \frac{x}{12}=4 A separate constant term: \quad \frac{x+1}{2x}+4=x
A quadratic equation: \quad\frac{3}{x+1}=x+5 A linear equation: \quad \frac{3x+4}{2x-5}=6
A second fraction: \quad \frac{2x}{15}=\frac{5x}{2} Double brackets, difference of two squares and simultaneous equations: \quad \frac{(x+3)^{2}}{x^{2}-9}=2x-1
It is important to be able to simplify algebraic fractions into their simplest form. If you need to practice this or need a quick refresher, see the lesson on simplifying algebraic fractions for further information.
Step-by-step guide: Simplifying algebraic fractions
What are algebraic fractions?
How to solve equations including algebraic fractions
We need to be able to solve equations including algebraic fractions.
Let’s look at a simple example when \frac{8}{x}=2 .
Here, the denominator of the fraction contains the variable, so we first need to get the variable out of the denominator.
Rearranging the equation by multiplying both sides by x and then dividing by 2, we get the value of x=4.
We can substitute this into the original equation to prove that the answer is correct.
Here, \frac{8}{4}=2 so we have the correct answer.
We shall now consider more complicated cases when equations involve algebraic fractions.
In order to solve equations including algebraic fractions.
Convert each fraction so they all have a common denominator.
Multiply the equation throughout by the common denominator.
Solve the equation (linear or quadratic).
Explain how to solve equations including algebraic fractions
Algebraic fractions worksheet
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Algebraic fractions worksheet
Get your free algebraic fractions worksheet of 20+ questions and answers. Includes reasoning and applied questions.
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Algebraic fractions examples
Example 1: Equation with one fraction
Solve the equation
\frac{2x-1}{3}+x=3
Convert each fraction so they all have a common denominator.
Here, we only have one fraction and so we do not need to convert any other term into a fraction.
2Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by 3 (the denominator of the fractional term), we get
Make sure that you multiply every term in the equation by 3 .
3Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here,
Example 2: Equation with two fractions
Solve the equation
\frac{x+1}{2}+\frac{x+3}{5}=6
Convert each fraction so they all have a common denominator.
Here, we have the two fractions with the denominators of 2 and 5 . The lowest common multiple of 2 and 5 is 10 and so we can convert the two fractions so that they have the same denominator.
Remember to use brackets to ensure that you multiply the entire numerator by 5 .
Remember to use brackets to ensure that you multiply the entire numerator by 2 .
We now have the equation
Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by 10 (the denominator of the fractional terms), we get
Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here,
Example 3: Equation with x in the denominator
Solve the equation
3-\frac{5}{x+1}=1
Convert each fraction so they all have a common denominator.
Here, we have one fraction so we do not need to find a common denominator.
Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by x + 1 (the denominator of the fractional terms), we get
Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here,
Example 4: Equation with three fractions
Solve the equation
\frac{1}{x}+\frac{1}{2x}+\frac{1}{3x}=11
Convert each fraction so they all have a common denominator.
Here, we need to find the lowest common multiple of x, 2x, and 3x . As x is the highest common factor, x \times 1 \times 2 \times 3 is the lowest common multiple, which is equal to 6x. .
So by multiplying the numerator and denominator of each fraction by a constant, we can convert each fraction to have the common denominator of 6x:.
We now have an equation which we can immediately simplify.
Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by 6x (the denominator of the fractional terms), we get
Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here,
Example 5: Denominators are expressions in terms of x
Solve the equation
\frac{x+1}{x+2}+\frac{3}{x-4}=1
Convert each fraction so they all have a common denominator.
Here, we need to find a common denominator for (x + 2) and (x − 4) .
The easiest way to do this is to multiply the two expressions together.
Top tip: do not expand the brackets too soon as you may be able to simplify the fraction before solving the equation.
By multiplying each fraction by the denominator of the other fraction, we get
We now have an equation which we can immediately simplify.
Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by (x+2)(x-4) (the denominator of the fractional term), we get
Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here,
Example 6: Equation including a quadratic
Solve the equation
\frac{9x+4}{x}=-2x
Convert each fraction so they all have a common denominator.
Here, we have a single fraction and so we do not need to find a common denominator.
Multiply the equation throughout by the common denominator.
Multiplying the equation throughout by x (the denominator of the fractional terms), we get
Solve the equation (linear or quadratic).
You can check your solution by substituting the value for x into the original equation and evaluating it.
Here, we have two possible solutions for x so we can check both:
Common misconceptions
Multiplying the numerator by the denominator
Let us look at example 2 . When multiplying throughout by 10 to remove the denominator from each fraction, the numerator has also been multiplied by 10 .This means that the fractions have been multiplied by 100 , instead of 10 , leaving the next line of work to be incorrect.
Ignoring the denominators
When given an equation including an algebraic fraction, if the denominators are ignored the question will be answered incorrectly.
Not multiplying all terms by the denominator
When rearranging an equation, the denominator is moved to the other side of the equals sign, instead of each term being multiplied by it. You must remember to multiply throughout by any value, not just the opposite side of the equals sign.
Simplifying fractions incorrectly (1)
When adding two fractions, the denominator must be the same. A common misconception for adding two fractions is to add the numerators and the denominators together because this method is emphasised when looking at multiplying fractions.
Simplifying fractions incorrectly (2)
Seeing the same term on the numerator and denominator allow the misconception that they can both be cancelled.
Practice algebraic fractions questions
Solve the equation
\frac{x+5}{3}+x=7
x=4
x=1
x=8
x=3
Solve the equation
\frac{x+5}{2}+\frac{5+x}{2}=50
x=20
x=7.5
x=45
x=22.5
Solve the equation
7+\frac{6x+4}{8}=2x
x=6
x=1.1
x=-\frac{39}{46}
x=0
Solve the equation
\frac{5}{x}+\frac{2x}{5}=3
x=-1
x=2.5
x=\sqrt{11}
x=0
Solve the equation
\frac{6-x}{4}+\frac{12}{4-x}=4
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x=\frac{14}{3}
x=-5\pm\sqrt{31}
x=-2
Solve the equation
\frac{5}{8+x}+\frac{6-x}{9}=2
x=-3
x=\frac{23}{3}
x=-1\pm2\sqrt{23}
x=-\frac{23}{3}
Algebraic fractions GCSE questions
Arron earns £40 per hour. One day, he receives a bonus of £5 . He shares this day’s earnings equally with his brother and sister. If he gives away £190 , how many hours did Arron work that day?
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2(\frac{40x+5}{3})=190
(1)
x=7
(1)
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7x^{2}+19x=4x^{2}+20x+24
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(1)
x=\frac{1\pm{17}}{6}
(1)
x=3 or x=-\frac{8}{3}
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Two shapes given below have the same area. Calculate the value for x .
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Area of Triangle = \frac{2(x+8)}{2}=x+8
(1)
Area of Square = \left(\frac{3}{\sqrt{x}}\right)^{2}=\frac{9}{x}
(1)
x+8=\frac{9}{x}
(1)
x^2+8x-9=0
(1)
(x+9)(x-1)=0
(1)
x=1 or x=-9
(1)
Conclusion: x=1 only
(1)
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Solutions to sin x=sin y sinx=siny in 0≤x≤2 π 0≤x≤2 π
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How can I find the solutions to solutions to sin x=sin y sinx=siny in 0≤x≤2 π 0≤x≤2 π.
If (for instance) 0≤x≤π/2 0≤x≤π/2 then sin x=sin(x+π/2)sinx=sin(x+π/2) but thos doesnt work when π/2≤x≤π π/2≤x≤π in this case sin x=−sin(x+π/2)sinx=−sin(x+π/2)
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edited Mar 6, 2017 at 6:15
gymbvghjkgkjkhgfkl
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Hint. Draw the graph of the sin sin function. Find out where horizontal lines cross that graph.Ethan Bolker –Ethan Bolker 2017-03-06 00:19:24 +00:00 Commented Mar 6, 2017 at 0:19
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Consider the diagram below.
Two angles have the same sine if the y y-coordinates of the points where their terminal sides intersect the unit circle are equal. By symmetry, sin θ=sin φ sinθ=sinφ if φ=θ φ=θ or φ=π−θ φ=π−θ. Any angle coterminal with these angles will also have the same sine. In general, sin θ=sin φ sinθ=sinφ if
φ=θ+2 k π,k∈Z φ=θ+2 k π,k∈Z
or
φ=π−θ+2 m π,m∈Z φ=π−θ+2 m π,m∈Z
Observe that if 0≤θ≤π 0≤θ≤π, then 0≤π−θ≤π 0≤π−θ≤π. Thus, for 0≤θ≤π 0≤θ≤π, sin θ=sin φ⟹φ=θ sinθ=sinφ⟹φ=θ or φ=π−θ φ=π−θ.
If π<θ≤2 π π<θ≤2 π, 0>π−θ≥−π 0>π−θ≥−π, so we must add 2 π 2 π to π−θ π−θ to find an angle in [0,2 π][0,2 π] that has the same sine as θ θ. Hence, if π<θ<2 π π<θ<2 π, sin θ=sin φ⟹φ=θ sinθ=sinφ⟹φ=θ or φ=3 π−θ φ=3 π−θ.
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answered Mar 6, 2017 at 0:36
N. F. TaussigN. F. Taussig
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4131 | https://math.stackexchange.com/questions/1237256/find-n-such-that-x2-x-1-is-a-factor-of-x1n-xn-1 | polynomials - Find $n$ such that $x^2 + x + 1$ is a factor of $(x+1)^n - x^n - 1$. - Mathematics Stack Exchange
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Find n n such that x 2+x+1 x 2+x+1 is a factor of (x+1)n−x n−1(x+1)n−x n−1.
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I have to find the form of n i.e. whether n is even or odd and whether it is multiple of 2 or 3 such that:
x 2+x+1 x 2+x+1 is a factor of (x+1)n−x n−1(x+1)n−x n−1
What I tried:
x 2+x+1=(x+1)2−x x 2+x+1=(x+1)2−x
also,
(x+1)n−x n−1=(n o)x n+(n 1)x n−1+...+(n n−1)x+(n n).1−x n−1(x+1)n−x n−1=(n o)x n+(n 1)x n−1+...+(n n−1)x+(n n).1−x n−1
=(n 1)x n−1+...+(n n−1)x=(n 1)x n−1+...+(n n−1)x
after getting this far I tried with taking x and n common but still couldn't able to find the right direction.
I need extra hints to solve this.
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edited May 3, 2015 at 4:51
skdhfgeq2134skdhfgeq2134
asked Apr 16, 2015 at 10:30
skdhfgeq2134skdhfgeq2134
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The roots of x 2+x+1=0 x 2+x+1=0 are the cube roots of unity. Call them w,w 2 w,w 2. As x 2+x+1=0 x 2+x+1=0, we get w+1=−w 2 w+1=−w 2 and w 2+1=−w w 2+1=−w. Thus, the problem reduces to finding n n, such that (−1)n+1∗w 2 n+w n+1=0(−1)n+1∗w 2 n+w n+1=0. Thus the answer is, whenever n is not divisible by 3 and n is not divisible by 2, i.e., n≡1,5 mod 6 n≡1,5 mod 6.
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edited Apr 16, 2015 at 19:30
answered Apr 16, 2015 at 10:41
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1 I think it should be n≡1,5 mod 6 n≡1,5 mod 6.Ojas –Ojas 2015-04-16 10:55:58 +00:00 Commented Apr 16, 2015 at 10:55
Woah, thanks. The answer that I got was is n is even and it is a multiple of three. The final equation that I got is [(−1)n+1]∗[w 2 n−w n]=0[(−1)n+1]∗[w 2 n−w n]=0 which gives the required solution.skdhfgeq2134 –skdhfgeq2134 2015-04-16 11:37:46 +00:00 Commented Apr 16, 2015 at 11:37
@Bernard I still think, it shouldn't be 4 mod 6 4 mod 6. If n=6 k+4 n=6 k+4, then it becomes (1+w)4−w 4−1=w 8−w 4−1=2∗w 8(1+w)4−w 4−1=w 8−w 4−1=2∗w 8 which does not equal 0.Ojas –Ojas 2015-04-16 11:45:25 +00:00 Commented Apr 16, 2015 at 11:45
You're right. I'm sorry for such an error, I computed too hastily soon this morning and didn't take time to check I'll withdraw my comment. I think you should add the final form of the solutuion in terms of congruence mod 6 to you answer Bernard –Bernard 2015-04-16 18:23:23 +00:00 Commented Apr 16, 2015 at 18:23
@Bernard, done.Ojas –Ojas 2015-04-16 19:30:52 +00:00 Commented Apr 16, 2015 at 19:30
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0A divisibility question concerning polynomials.
2x 2+x+1 x 2+x+1 is a factor of (x+1)n−x n−1(x+1)n−x n−1, whenever
0How to find m m if (x 2+x+1)|((x+1)m−x m−1)(x 2+x+1)|((x+1)m−x m−1)
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Integrating a large product of sines
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Recently, I came across the following integral: ∫2 π 0 sin(x)sin(2 x)sin(3 x)sin(4 x)d x=π 4∫2 π 0 sin(x)sin(2 x)sin(3 x)sin(4 x)d x=π 4, which can be easily solved by some trigonometry.
But when trying to find a more general result: I n=∫0 2 π∏n k=1 sin(k x)d x I n=∫0 2 π∏n k=1 sin(k x)d x, I wasn't able to find a very general result for the problem. Is there any way to find I n I n as a "simple" formula?
My attempt: It's easy to see that I n=0 I n=0 for odd n. Using the sine product to sum formula for I 2 n I 2 n: I 2 n=(−1)n 4 n∫2 π 0∑e k∈{−1,1}(cos(e 1 x+2 e 2 x+3 e 3 x+⋯+2 n e 2 n x)e 1 e 2⋯e 2 n)d x I 2 n=(−1)n 4 n∫2 π 0∑e k∈{−1,1}(cos(e 1 x+2 e 2 x+3 e 3 x+⋯+2 n e 2 n x)e 1 e 2⋯e 2 n)d x.
Notice that every integral in the sum goes to zero, except when the expression inside the cosine is zero. We can rewrite I 2 n I 2 n as: I 2 n=2 π(−1)n(A 2 n−B 2 n)4 n I 2 n=2 π(−1)n(A 2 n−B 2 n)4 n, where A 2 n A 2 n and B 2 n B 2 n are related to the number of ways to zero this expression with an even and odd number of negative signs for e k e k. After that, I was not able to simplify the problem.
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edited Jul 10, 2024 at 19:07
amWhy
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asked Jun 19, 2024 at 2:12
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4 Playing around in Desmos, it seems that (for n n up to 100 100), I n≠0 I n≠0 only for the indices n∈{4,8,12,14,16,20,24,28,32,36,40,44,48,51,52,56,60,64,68,72,76,80,84,86,88,89,90,92,94,96,98,100}n∈{4,8,12,14,16,20,24,28,32,36,40,44,48,51,52,56,60,64,68,72,76,80,84,86,88,89,90,92,94,96,98,100} No clue what to make of this.PrincessEev –PrincessEev 2024-06-19 03:27:07 +00:00 Commented Jun 19, 2024 at 3:27
Playing around with Mathematica, I instead find that I n=0 I n=0 unless n n is a multiple of 4 4. In particular, I 14=0 I 14=0.Semiclassical –Semiclassical 2024-06-19 04:23:11 +00:00 Commented Jun 19, 2024 at 4:23
2 Related: math.stackexchange.com/q/3786668/42969Martin R –Martin R 2024-06-19 09:18:48 +00:00 Commented Jun 19, 2024 at 9:18
See also (2.10) in jstor.org/stable/26603380Travis Willse –Travis Willse 2024-07-10 19:46:29 +00:00 Commented Jul 10, 2024 at 19:46
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It can be expressed in the form π A n 2 n−1 π A n 2 n−1 as follows.
∫2 π 0 n∏k=1 sin(k x)d x=2 π∫0 n∏k=1 e i k x−e−i k x 2 i d x=1(2 i)n∫2 π 0 n∏k=1 1 e i k x⋅n∏k=1(e 2 i k x−1)d x=1(2 i)n∫2 π 0 1 e i N x⋅n∏k=1(e 2 i k x−1)d x where N=n∑k=1 k=n(n+1)2=1 i(2 i)n∫2 π 0 1 e i N x+i x⋅(n∏k=1(e 2 i k x−1))⋅i e i x d x Let z=e i x=1 i(2 i)n∫|z|=1 1 z N+1 n∏k=1(z 2 k−1)d z Contour integral=1 i(2 i)n∫|z|=1 f n(z)(z−0)N+1 d z where f n(z)=n∏k=1(z 2 k−1)=1 i(2 i)n⋅2 π i N!⋅f(N)n(0)Cauchy's integral formula∫2 π 0 n∏k=1 sin(k x)d x=π 2 n−1⋅f(N)n(0)i n N! where,
N=n(n+1)2
f(N)n(0) is the N th derivative of f n(z)=n∏k=1(z 2 k−1) at z=0.
i=√−1 is the imaginary unit, and i n=e i π n/2.
Intuition and Observation: Since i n is imaginary for n≡1,3(mod 4), f(N)n(0) and A n=f(N)n(0)i n N! must vanish as f(N)n(0) is real. Based on Mathematica, f(N)n(0)=0 and A n=0 for n≡2(mod 4). f(N)n(0) and A n are non-zero only if n is a multiple of 4. So, the integral I n=0 unless 4∣n.
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edited Jun 20, 2024 at 12:03
answered Jun 19, 2024 at 4:01
Pustam RautPustam Raut
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9 Based on Mathematica it seems that A 4 n=2,2,4,6,8,16,28,50,100,196,… for n≥1, with A n=0 otherwise. This matches OEIS sequence A269298 as far as I've been able to verify, and is moreover consistent with its link to sequence A231599.Semiclassical –Semiclassical 2024-06-19 04:34:31 +00:00 Commented Jun 19, 2024 at 4:34
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4133 | https://www.marchofdimes.org/find-support/topics/pregnancy/rubella-and-pregnancy | Find support
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Rubella and pregnancy
Key Points
Rubella is an infection that causes mild flu-like symptoms and a rash.
Vaccines usually given in childhood prevent you from getting the infection
Up to half of people infected with rubella don’t have any symptoms but still can spread it to others.
You can get rubella from an infected person who coughs or sneezes, or by sharing food or drinks with someone who is infected.
Rubella can cause serious problems for your baby if you get infected while pregnant, especially during the first three months of pregnancy.
Rubella, also called German measles, is an infection that causes mild flu-like symptoms and a rash. Up to half of people infected with rubella don’t have any symptoms and may not even know they’re infected but still spread it to other people.
If you get infected during pregnancy, rubella can cause serious problems for your baby. Infection causes the most severe damage when the pregnant person is infected early in their pregnancy, especially in the first 12 weeks (3 months).
Rubella has been nearly eliminated in the United States because of the routine vaccination of children. Vaccination protects a person against rubella for life. But people who were never vaccinated as children can get infected. You can be vaccinated as an adult when you are not pregnant.
Between 2005 and 2011, 67 cases of rubella were reported in this country. This is up from the previous period—2001 to 2004—when only 5 cases were reported. Vaccination is the best way to prevent infection. Today, less than 10 people in the United States are reported to have rubella and it is usually associated with travel and not being vaccinated.
Rubella is common in many other countries. Travelers can bring it into the United States, or you can get it when traveling outside the country.
It’s important to get vaccinated for rubella. Talk to your health care provider to make sure you’re protected against it. A blood test can provide you with information about your immunity to rubella
What are the signs and symptoms of rubella?
Rubella is usually mild, with flu-like symptoms followed by a pink or light red rash. The rash usually starts on the face, then spreads to the rest of the body. The rash lasts about 3 days. Other symptoms include:
Up to 7 in 10 (70 percent) teenage and adult women who get rubella develop arthritis. Arthritis is swelling and pain in your joints. Women who are infected with rubella may feel arthritis in their fingers, wrists and knees. This complication is rare in men or children who get rubella. In rare cases, rubella can cause serious problems, including brain infections and bleeding problems.
Some people have no symptoms so you may not know that you or they have rubella.
What causes rubella?
Rubella is caused by a virus (a tiny organism that can make you sick). It is not caused by the same virus that causes measles. It’s very contagious. It spreads through the air when an infected person coughs or sneezes, or by sharing food or drinks with someone who is infected. It also can be passed to a baby from a pregnant person who is infected.
A person with rubella may spread the disease to other people up to 1 week before they notice a rash. Infected people can continue to spread the disease up to 7 days after getting a rash.
People infected with rubella should tell their friends, family, and the people they work with if they have rubella. If your child has rubella, it’s important to tell their school or daycare provider. Telling others helps prevent the spread of rubella to even more people.
What problems can rubella cause during pregnancy?
Rubella can be serious for the baby, especially during the first 3 months. Having rubella during pregnancy increases the risk of:
Can you pass rubella to your baby during pregnancy?
Yes. The best way to protect your baby is to make sure you’re immune to rubella. Immune means being protected from an infection. If you’re immune to an infection, it means you can’t get the infection.
You are probably immune to rubella because you were vaccinated as a child or you had the illness during childhood. A blood test can tell if you’re immune to rubella. If you’re thinking about getting pregnant and aren’t sure if you’re immune, talk to your health care provider about getting a blood test. If you were not born in the United States or came as an adult you may not be vaccinated against rubella. Ask your health care provider to do a blood test to see if you are immune to rubella.
If you’re not immune to rubella, here’s what you can do to help protect your baby:
Before pregnancy. Get the measles, mumps and rubella (MMR) vaccine. Wait 1 month before trying to get pregnant after getting the shot.
During pregnancy. You can be tested at a prenatal visit to make sure you’re immune to rubella. If you’re not immune, the MMR vaccine isn’t recommended during pregnancy. But there are things you can do to help prevent getting infected with rubella:
After pregnancy. Get the MMR vaccination after you give birth. It is safe to get the MMR vaccine while nursing. Being protected from the infection means you can’t pass it to your baby before they get their own MMR vaccination at about 12 months. It also prevents you from passing rubella to your baby during a future pregnancy.
What are the chances of passing rubella to your baby during pregnancy?
You’re more likely to pass rubella to your baby the earlier you become infected during pregnancy. For example:
If you have rubella during pregnancy, your baby’s provider will carefully monitor your baby after birth to find any problems early.
How is rubella treated?
There is no specific medicine to treat rubella. In many cases, the symptoms of the disease are mild. Staying in bed (bed rest), drinking fluids and taking medicines to reduce fever can help.
Pregnant people with rubella may be treated with acetaminophen to relieve their symptoms. In more serious cases, treatment may include blood transfusion or steroids. A blood transfusion is when you have new blood put into your body. Steroids are synthetic hormones that treat inflammation.
Contact your health care provider if you are pregnant and think you have rubella or have been exposed to the disease.
Last reviewed: August 2021
See also: Your baby’s vaccinations, Vaccinations and pregnancy
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4134 | http://www.adichemistry.com/general/chemicalbond/vbt/valence-bond-theory-hybridization.html | VALENCE BOND THEORY (VBT) | HYBRIDIZATION | SP | SP2 | SP3 | SP3D | SP3D2 | SP3D3
Valence bond theory: Introduction
Hybridization
Types of hybridization
sp, sp 2, sp 3, sp 3 d, sp 3 d 2, sp 3 d 3
VALENCE BOND THEORY (VBT) & HYBRIDIZATION
The valence bond theory was proposed by Heitler and London to explain the formation of covalent bond quantitatively using quantum mechanics. Later on, Linus Pauling improved this theory by introducing the concept of hybridization.
The main postulates of this theory are as follows:
A covalent bond is formed by the overlapping of two half filled valence atomic orbitals of two different atoms.
The electrons in the overlapping orbitals get paired and confined between the nuclei of two atoms.
The electron density between two bonded atoms increases due to overlapping. This confers stability to the molecule.
Greater the extent of overlapping, stronger is the bond formed.
The direction of the covalent bond is along the region of overlapping of the atomic orbitals i.e., covalent bond is directional.
There are two types of covalent bonds based on the pattern of overlapping as follows:
(i) σ-bond: The covalent bond formed due to overlapping of atomic orbital along the inter nucleus axis is called σ-bond. It is a stronger bond and cylindrically symmetrical.
Depending on the types of orbitals overlapping, the σ-bond is divided into following types:
σ s-s bond:
σ p-p bond:
σ s-p bond:
(ii) π-bond: The covalent bond formed by sidewise overlapping of atomic orbitals is called π- bond. In this bond, the electron density is present above and below the inter nuclear axis. It is relatively a weaker bond since the electrons are not strongly attracted by the nuclei of bonding atoms.
Note: The 's' orbitals can only form σ-bonds, whereas the p, d & f orbitals can form both σ and π-bonds.
BONDING IN MOLECULES EXPLAINED BY VALENCE BOND THEORY
1) H 2 molecule:
The electronic configuration of hydrogen atom in the ground state is 1s 1.
In the formation of hydrogen molecule, two half filled 1s orbitals of hydrogen atoms overlap along the inter-nuclear axis and thus by forming a σ s-s bond.
2) Cl 2 molecule:
The electronic configuration of Cl atom in the ground state is [Ne]3s 2 3p x 2 3p y 2 3p z 1.
The two half filled 3p z atomic orbitals of two chlorine atoms overlap along the inter-nuclear axis and thus by forming a σ p-p bond.
3) HCl molecule:
In the ground state, the electronic configuration of hydrogen atom is 1s 1.
And the ground state electronic configuration of Cl atom is [Ne]3s 2 3p x 2 3p y 2 3p z 1.
The half filled 1s orbital of hydrogen overlap with the half filled 3p z atomic orbital of chlorine atom along the inter-nuclear axis to form a σ s-p bond.
4) O 2 molecule:
The electronic configuration of O in the ground state is [He] 2s 2 2p x 2 2p y 1 2p z 1.
The half filled 2p y orbitals of two oxygen atoms overlap along the inter-nuclear axis and form σ p-p bond.
The remaining half filled 2p z orbitals overlap laterally to form a π p-p bond.
Thus a double bond (one σ p-p and one π p-p) is formed between two oxygen atoms.
5) N 2 molecule:
The ground state electronic configuration of N is [He] 2s 2 2p x 1 2p y 1 2p z 1.
A σ p-p bond is formed between two nitrogen atoms due to overlapping of half filled 2p x atomic orbitals along the inter-nuclear axis.
The remaining half filled 2p y and 2p z orbitals form two π p-p bonds due to lateral overlapping. Thus a triple bond (one and two) is formed between two nitrogen atoms.
NEED FOR MODIFICATION OF VBT
However the old version of valence bond theory is limited to diatomic molecules only. It could not explain the structures and bond angles of molecules with more than three atoms.
E.g. It could not explain the structures and bond angles of H 2 O, NH 3 etc.,
However, in order to explain the structures and bond angles of molecules, Linus Pauling modified the valence bond theory using hybridization concept.
HYBRIDIZATION
What is hybridization?
The intermixing of two or more pure atomic orbitals of an atom with almost same energy to give same number of identical and degenerate new type of orbitals is known as hybridization.
The new orbitals formed are also known as hybrid orbitals.
What is intermixing?
The intermixing or hybridization of atomic orbitals is a mathematical concept based on quantum mechanics. During this process, the wavefunctions, Ψ of atomic orbitals of same atom are combined to give new wavefunctions corresponding to hybrid orbitals.
What are the requirements for atomic orbitals to undergo hybridization?
The atomic orbitals of same atom with almost same energy can only participate in the hybridization.
The full filled or half filled or even empty orbitals can undergo hybridization provided they have almost equal energy.
Do the orbitals of different atoms undergo hybridization?
No! The hybridization is the mixing of orbitals of same atom only. The combination of orbitals belonging to different atoms is called bonding.
What are hybrid orbitals? And what are its characteristics?
The new orbitals that are formed due to intermixing of atomic orbitals are also known as hybrid orbitals, which have mixed characteristics of atomic orbitals.
The shapes of hybrid orbitals are identical. Usually they have one big lobe associated with a small lobe on the other side.
The hybrid orbitals are degenerate i.e., they are associated with same energy.
How many hybrid orbitals are formed?
The number of hybrid orbitals formed is equal to the number of pure atomic orbitals undergoing hybridization.
E.g. If three atomic orbitals intermix with each other, the number of hybrid orbitals formed will be equal to 3.
How do the electrons are going to be filled in the hybrid orbitals?
The hybrid orbitals are filled with those electrons which were present in the pure atomic orbitals forming them.
The filling up of electrons in them follows Pauli's exclusion principle and Hund's rule.
What is the use of hybrid orbitals?
The hybrid orbitals participate in the σ bond formation with other atoms.
Why atomic orbitals in a given atom undergo hybridization?
The hybrid orbitals are oriented in space so as to minimize repulsions between them. This explains why the atomic orbitals undergo hybridization before bond formation.
The reason for hybridization is to minimize the repulsions between the bonds that are going to be formed by the atoms by using hybrid orbitals.
Remember that the hybridization is the process that occurs before bond formation.
And finally:
The bond angles in the molecule are equal to or almost equal to the angles between the hybrid orbitals forming the σ bonds. The shape of the molecule is determined by the type of hybridization, number of bonds formed by them and the number of lone pairs.
TYPES OF HYBRIDIZATION
During hybridization, the atomic orbitals with different characteristics are mixed with each other. Hence there is no meaning of hybridization between same type of orbitals i.e., mixing of two 's' orbitals or two 'p' orbitals is not called hybridization.
However orbital of 's' type can can mix with the orbitals of 'p' type or of 'd' type. Based on the type and number of orbitals, the hybridization can be subdivided into following types.
Note: Keep in mind that - only the orbitals of nearer energy values can participate in the hybridization.
sp HYBRIDIZATION
What is sp hybridization?
Intermixing of one 's' and one 'p' orbitals of almost equal energy to give two identical and degenerate hybrid orbitals is called 'sp' hybridization.
These sp-hybrid orbitals are arranged linearly at by making 180 o of angle.
They possess 50% 's' and 50% 'p' character.
JUMP TO EXAMPLES OF SP HYBRIDIZATION
sp 2 HYBRIDIZATION
What is sp 2 hybridization?
Intermixing of one 's' and two 'p' orbitals of almost equal energy to give three identical and degenerate hybrid orbitals is known as sp 2 hybridization.
The three sp 2 hybrid orbitals are oriented in trigonal planar symmetry at angles of 120 o to each other.
The sp 2 hybrid orbitals have 33.3% 's' character and 66.6% 'p' character.
JUMP TO EXAMPLES OF SP 2 HYBRIDIZATION
sp 3 HYBRIDIZATION
What is sp 3 hybridization?
In sp 3 hybridization, one 's' and three 'p' orbitals of almost equal energy intermix to give four identical and degenerate hybrid orbitals.
These four sp 3 hybrid orbitals are oriented in tetrahedral symmetry with 109 o 28' angle with each other.
The sp 3 hybrid orbitals have 25% ‘s’ character and 75% 'p' character.
JUMP TO EXAMPLES OF SP 3 HYBRIDIZATION
sp 3 d HYBRIDIZATION
What is sp 3 d hybridization?
In sp 3 d hybridization, one 's', three 'p' and one 'd' orbitals of almost equal energy intermix to give five identical and degenerate hybrid orbitals, which are arranged in trigonal bipyramidal symmetry.
Among them, three are arranged in trigonal plane and the remaining two orbitals are present above and below the trigonal plane at right angles.
The sp 3 d hybrid orbitals have 20% 's', 60% 'p' and 20% 'd' characters.
JUMP TO EXAMPLES OF SP 3 D HYBRIDIZATION
sp 3 d 2 HYBRIDIZATION
What is sp 3 d 2 hybridization?
Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by giving six identical and degenerate hybrid orbitals is called sp 3 d 2 hybridization.
These six sp 3 d 2 orbitals are arranged in octahedral symmetry by making 90 o angles to each other. This arrangement can be visualized as four orbitals arranged in a square plane and the remaining two are oriented above and below this plane perpendicularly.
JUMP TO EXAMPLES OF SP 3 D 2 HYBRIDIZATION
sp 3 d 3 HYBRIDIZATION
What is sp 3 d 3 hybridization?
In sp 3 d 3 hybridization, one 's', three 'p' and three 'd' orbitals of almost same energy intermix to give seven sp 3 d 3 hybrid orbitals, which are oriented in pentagonal bipyramidal symmetry.
Five among the sp 3 d 3 orbitals are arranged in a pentagonal plane by making 72 o of angles. The remaining are arranged perpendicularly above and below this pentagonal plane.
JUMP TO EXAMPLES OF SP 3 D 3 HYBRIDIZATION
< VSEPR TheoryChemical bonding: TOCIllustrations of Hybridization >
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4135 | https://delcode.delaware.gov/title8/c001/sc02/index.html | Delaware Code Online
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CHAPTER 1. General Corporation Law
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§ 121. General powers.
(a) In addition to the powers enumerated in § 122 of this title, every corporation, its officers, directors and stockholders shall possess and may exercise all the powers and privileges granted by this chapter or by any other law or by its certificate of incorporation, together with any powers incidental thereto, so far as such powers and privileges are necessary or convenient to the conduct, promotion or attainment of the business or purposes set forth in its certificate of incorporation.
(b) Every corporation shall be governed by the provisions and be subject to the restrictions and liabilities contained in this chapter.
8 Del. C. 1953, § 121; 56 Del. Laws, c. 50.;
§ 122. Specific powers [For application of this section, see 84 Del. Laws, c. 309, § 6].
Every corporation created under this chapter shall have power, whether or not so provided in the certificate of incorporation, to:
(1) Have perpetual succession by its corporate name, unless a limited period of duration is stated in its certificate of incorporation;
(2) Sue and be sued in all courts and participate, as a party or otherwise, in any judicial, administrative, arbitrative or other proceeding, in its corporate name;
(3) Have a corporate seal, which may be altered at pleasure, and use the same by causing it or a facsimile thereof, to be impressed or affixed or in any other manner reproduced;
(4) Purchase, receive, take by grant, gift, devise, bequest or otherwise, lease, or otherwise acquire, own, hold, improve, employ, use and otherwise deal in and with real or personal property, or any interest therein, wherever situated, and to sell, convey, lease, exchange, transfer or otherwise dispose of, or mortgage or pledge, all or any of its property and assets, or any interest therein, wherever situated;
(5) Appoint such officers and agents as the business of the corporation requires and to pay or otherwise provide for them suitable compensation; provided that any contract or other appointment or delegation of authority that empowers an officer or agent to act on behalf of the corporation shall be subject to § 141(a) of this title, to the extent it is applicable;
(6) Adopt, amend and repeal bylaws in accordance with § 109 of this title;
(7) Wind up and dissolve itself in the manner provided in this chapter;
(8) Conduct its business, carry on its operations and have offices and exercise its powers within or without this State;
(9) Make donations for the public welfare or for charitable, scientific or educational purposes, and in time of war or other national emergency in aid thereof;
(10) Be an incorporator, promoter or manager of other corporations of any type or kind;
(11) Participate with others in any corporation, partnership, limited partnership, joint venture or other association of any kind, or in any transaction, undertaking or arrangement which the participating corporation would have power to conduct by itself, whether or not such participation involves sharing or delegation of control with or to others;
(12) Transact any lawful business which the corporation’s board of directors shall find to be in aid of governmental authority;
(13) Make contracts, including contracts of guaranty and suretyship, incur liabilities, borrow money at such rates of interest as the corporation may determine, issue its notes, bonds and other obligations, and secure any of its obligations by mortgage, pledge or other encumbrance of all or any of its property, franchises and income, and make contracts of guaranty and suretyship which are necessary or convenient to the conduct, promotion or attainment of the business of (a) a corporation all of the outstanding stock of which is owned, directly or indirectly, by the contracting corporation, or (b) a corporation which owns, directly or indirectly, all of the outstanding stock of the contracting corporation, or (c) a corporation all of the outstanding stock of which is owned, directly or indirectly, by a corporation which owns, directly or indirectly, all of the outstanding stock of the contracting corporation, which contracts of guaranty and suretyship shall be deemed to be necessary or convenient to the conduct, promotion or attainment of the business of the contracting corporation, and make other contracts of guaranty and suretyship which are necessary or convenient to the conduct, promotion or attainment of the business of the contracting corporation;
(14) Lend money for its corporate purposes, invest and reinvest its funds, and take, hold and deal with real and personal property as security for the payment of funds so loaned or invested;
(15) Pay pensions and establish and carry out pension, profit sharing, stock option, stock purchase, stock bonus, retirement, benefit, incentive and compensation plans, trusts and provisions for any or all of its directors, officers and employees, and for any or all of the directors, officers and employees of its subsidiaries;
(16) Provide insurance for its benefit on the life of any of its directors, officers or employees, or on the life of any stockholder for the purpose of acquiring at such stockholder’s death shares of its stock owned by such stockholder;
(17) Renounce, in its certificate of incorporation or by action of its board of directors, any interest or expectancy of the corporation in, or in being offered an opportunity to participate in, specified business opportunities or specified classes or categories of business opportunities that are presented to the corporation or 1 or more of its officers, directors or stockholders; and
(18) Notwithstanding § 141(a) of this title, make contracts with 1 or more current or prospective stockholders (or 1 or more beneficial owners of stock), in its or their capacity as such, in exchange for such minimum consideration as determined by the board of directors (which may include inducing stockholders or beneficial owners of stock to take, or refrain from taking, 1 or more actions); provided that no provision of such contract shall be enforceable against the corporation to the extent such contract provision is contrary to the certificate of incorporation or would be contrary to the laws of this State (other than § 115 of this title) if included in the certificate of incorporation. Without limiting the provisions that may be included in any such contracts, the corporation may agree to: (a) restrict or prohibit itself from taking actions specified in the contract, (b) require the approval or consent of 1 or more persons or bodies before the corporation may take actions specified in the contract (which persons or bodies may include the board of directors or 1 or more current or future directors, stockholders or beneficial owners of stock of the corporation), and (c) covenant that the corporation or 1 or more persons or bodies will take, or refrain from taking, actions specified in the contract (which persons or bodies may include the board of directors or 1 or more current or future directors, stockholders or beneficial owners of stock of the corporation). Solely for purposes of applying the proviso in the first sentence of this subsection, a restriction, prohibition or covenant in any such contract that relates to any specified action shall not be deemed contrary to the laws of this State or the certificate of incorporation by reason of a provision of this title or the certificate of incorporation that authorizes or empowers the board of directors (or any 1 or more directors) to take such action. With respect to all contracts made under this paragraph (18), the corporation shall be subject to the remedies available under the law governing the contract, including for any failure to perform or comply with its agreements under such contract.
8 Del. C. 1953, § 122; 56 Del. Laws, c. 50; 57 Del. Laws, c. 148, § 3; 64 Del. Laws, c. 112, § 3; 65 Del. Laws, c. 127, § 2; 71 Del. Laws, c. 339, § 7; 72 Del. Laws, c. 343, § 3; 84 Del. Laws, c. 309, § 1;
§ 123. Powers respecting securities of other corporations or entities.
Any corporation organized under the laws of this State may guarantee, purchase, take, receive, subscribe for or otherwise acquire; own, hold, use or otherwise employ; sell, lease, exchange, transfer or otherwise dispose of; mortgage, lend, pledge or otherwise deal in and with, bonds and other obligations of, or shares or other securities or interests in, or issued by, any other domestic or foreign corporation, partnership, association or individual, or by any government or agency or instrumentality thereof. A corporation while owner of any such securities may exercise all the rights, powers and privileges of ownership, including the right to vote.
8 Del. C. 1953, § 123; 56 Del. Laws, c. 50.;
§ 124. Effect of lack of corporate capacity or power; ultra vires.
No act of a corporation and no conveyance or transfer of real or personal property to or by a corporation shall be invalid by reason of the fact that the corporation was without capacity or power to do such act or to make or receive such conveyance or transfer, but such lack of capacity or power may be asserted:
(1) In a proceeding by a stockholder against the corporation to enjoin the doing of any act or acts or the transfer of real or personal property by or to the corporation. If the unauthorized acts or transfer sought to be enjoined are being, or are to be, performed or made pursuant to any contract to which the corporation is a party, the court may, if all of the parties to the contract are parties to the proceeding and if it deems the same to be equitable, set aside and enjoin the performance of such contract, and in so doing may allow to the corporation or to the other parties to the contract, as the case may be, such compensation as may be equitable for the loss or damage sustained by any of them which may result from the action of the court in setting aside and enjoining the performance of such contract, but anticipated profits to be derived from the performance of the contract shall not be awarded by the court as a loss or damage sustained;
(2) In a proceeding by the corporation, whether acting directly or through a receiver, trustee or other legal representative, or through stockholders in a representative suit, against an incumbent or former officer or director of the corporation, for loss or damage due to such incumbent or former officer’s or director’s unauthorized act;
(3) In a proceeding by the Attorney General to dissolve the corporation, or to enjoin the corporation from the transaction of unauthorized business.
8 Del. C. 1953, § 124; 56 Del. Laws, c. 50; 71 Del. Laws, c. 339, § 8;
§ 125. Conferring academic or honorary degrees.
No corporation organized after April 18, 1945, shall have power to confer academic or honorary degrees unless the certificate of incorporation or an amendment thereof shall so provide and unless the certificate of incorporation or an amendment thereof prior to its being filed in the office of the Secretary of State shall have endorsed thereon the approval of the Department of Education of this State. No corporation organized before April 18, 1945, any provision in its certificate of incorporation to the contrary notwithstanding, shall possess the power aforesaid without first filing in the office of the Secretary of State a certificate of amendment so providing, the filing of which certificate of amendment in the office of the Secretary of State shall be subject to prior approval of the Department of Education, evidenced as hereinabove provided. Approval shall be granted only when it appears to the reasonable satisfaction of the Department of Education that the corporation is engaged in conducting a bona fide institution of higher learning, giving instructions in arts and letters, science or the professions, or that the corporation proposes, in good faith, to engage in that field and has or will have the resources, including personnel, requisite for the conduct of an institution of higher learning. Upon dissolution, all such corporations shall comply with § 8530 of Title 14. Notwithstanding any provision herein to the contrary, no corporation shall have the power to conduct a private business or trade school unless the certificate of incorporation or an amendment thereof, prior to its being filed in the office of the Secretary of State, shall have endorsed thereon the approval of the Department of Education pursuant to Chapter 85 of Title 14.
Notwithstanding the foregoing provisions, any corporation conducting a law school, which has its principal place of operation in Delaware, and which intends to meet the standards of approval of the American Bar Association, may, after it has been in actual operation for not less than 1 year, retain at its own expense a dean or dean emeritus of a law school fully approved by the American Bar Association to make an on-site inspection and report concerning the progress of the corporation toward meeting the standards for approval by the American Bar Association. Such dean or dean emeritus shall be chosen by the Attorney General from a panel of 3 deans whose names are presented to the Attorney General as being willing to serve. One such dean on this panel shall be nominated by the trustees of said law school corporation; another dean shall be nominated by a committee of the Student Bar Association of said law school; and the other dean shall be nominated by a committee of lawyers who are parents of students attending such law school. If any of the above-named groups cannot find a dean, it may substitute 2 full professors of accredited law schools for the dean it is entitled to nominate, and in such a case if the Attorney General chooses 1 of such professors, such professor shall serve the function of a dean as herein prescribed. If the dean so retained shall report in writing that, in such dean’s professional judgment, the corporation is attempting, in good faith, to comply with the standards for approval of the American Bar Association and is making reasonable progress toward meeting such standards, the corporation may file a copy of the report with the Secretary of Education and with the Attorney General. Any corporation which complies with these provisions by filing such report shall be deemed to have temporary approval from the State and shall be entitled to amend its certificate of incorporation to authorize the granting of standard academic law degrees. Thereafter, until the law school operated by the corporation is approved by the American Bar Association, the corporation shall file once during each academic year a new report, in the same manner as the first report. If, at any time, the corporation fails to file such a report, or if the dean retained to render such report states that, in such dean’s opinion, the corporation is not continuing to make reasonable progress toward accreditation, the Attorney General, at the request of the Secretary of Education, may file a complaint in the Court of Chancery to suspend said temporary approval and degree-granting power until a further report is filed by a dean or dean emeritus of an accredited law school that the school has resumed its progress towards meeting the standards for approval. Upon approval of the law school by the American Bar Association, temporary approval shall become final, and shall no longer be subject to suspension or vacation under this section.
8 Del. C. 1953, § 125; 56 Del. Laws, c. 50; 59 Del. Laws, c. 207, § 1; 59 Del. Laws, c. 377, § 1; 65 Del. Laws, c. 103, § 2; 71 Del. Laws, c. 339, § 9; 72 Del. Laws, c. 433, § 1; 73 Del. Laws, c. 65, § 7; 74 Del. Laws, c. 249, § 1;
§ 126. Banking power denied.
(a) No corporation organized under this chapter shall possess the power of issuing bills, notes, or other evidences of debt for circulation as money, or the power of carrying on the business of receiving deposits of money.
(b) Corporations organized under this chapter to buy, sell and otherwise deal in notes, open accounts and other similar evidences of debt, or to loan money and to take notes, open accounts and other similar evidences of debt as collateral security therefor, shall not be deemed to be engaging in the business of banking.
8 Del. C. 1953, § 126; 56 Del. Laws, c. 50; 57 Del. Laws, c. 148, § 4;
§ 127. Private foundation; powers and duties.
A corporation of this State which is a private foundation under the United States internal revenue laws and whose certificate of incorporation does not expressly provide that this section shall not apply to it is required to act or to refrain from acting so as not to subject itself to the taxes imposed by 26 U.S.C. § 4941 (relating to taxes on self-dealing), § 4942 (relating to taxes on failure to distribute income), § 4943 (relating to taxes on excess business holdings), § 4944 (relating to taxes on investments which jeopardize charitable purpose), or § 4945 (relating to taxable expenditures), or corresponding provisions of any subsequent United States internal revenue law.
8 Del. C. 1953, § 127; 58 Del. Laws, c. 87.;
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An Elegant New Proof of Herons Formula?
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Asked 4 years, 7 months ago
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A triangle with side lengths a,b,c a,b,c an altitude(h h), where the height(h a h a) intercepts the hypotenuse(a a) such that it is the sum of two side lengths, a=u+v a=u+v and height(h b h b) intercepts hypotenuse(b b) such that it is also the sum of two side lengths b=x+y b=x+y, we can find a simple proof of herons formula.
Image here:
(Note that h=a u−−√h=a u and h=b x−−√h=b x, giving a u=b x a u=b x)
First the equations we have are:
u+v=a u+v=a
x+y=b x+y=b
a u=b x a u=b x
a v−−√+b y−−√=c a v+b y=c
Substituting v v and y y with u u and x x for the 4 4 th equation:
a 2−a u−−−−−−√+b 2−b x−−−−−−√=c a 2−a u+b 2−b x=c
Then using a u=b x a u=b x we eliminate x x to find u u:
a 2−a u−−−−−−√+b 2−b(a u b)−−−−−−−−−−√=c a 2−a u+b 2−b(a u b)=c
u=1 4 a c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)u=1 4 a c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)
Finally, substituting this into h=a u−−√h=a u:
h=1 4 c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√h=1 4 c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)
Plugging this into the area formula (A=1 2 c h A=1 2 c h) gives:
A=1 2 c 1 4 c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√A=1 2 c 1 4 c 2(2 a 2 b 2+2 a 2 c 2−a 4−b 4+2 b 2 c 2−c 4)
A=1 16(c 2−(a−b)2)((a+b)2−c 2)−−−−−−−−−−−−−−−−−−−−−−−−−√A=1 16(c 2−(a−b)2)((a+b)2−c 2)
A=1 16(a+b−c)(a+b+c)(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√A=1 16(a+b−c)(a+b+c)(b+c−a)(a+c−b)
A=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√A=s(s−a)(s−b)(s−c)
Q.E.D.
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asked Feb 23, 2021 at 17:07
SpoonedBreadSpoonedBread
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This looks similar to this answer.robjohn –robjohn♦ 2021-02-23 17:12:30 +00:00 Commented Feb 23, 2021 at 17:12
1 That's a proof that uses Pythagorean theorem directly, mine avoids direct use.SpoonedBread –SpoonedBread 2021-02-23 17:14:17 +00:00 Commented Feb 23, 2021 at 17:14
1 What is the exact question please..?lone student –lone student 2021-02-23 17:16:02 +00:00 Commented Feb 23, 2021 at 17:16
2 @SpoonedBread: You want to make sure that the foot of altitude h h lies within side c c. (Otherwise, you need to accommodate the possibility that c c is not necessarily the sum of a v−−√a v and b y−−√b y.) Assuming that c c is the longest side will guarantee this. (I feel like I've offered this advice before. :)Blue –Blue 2021-02-23 17:38:54 +00:00 Commented Feb 23, 2021 at 17:38
3 It looks valid, however, it essentially proves, along the way, the part of the Pythagorean Theorem used. That seems to encumber the proof rather than making it elegant, but that is a matter of opinion.robjohn –robjohn♦ 2021-02-23 17:43:13 +00:00 Commented Feb 23, 2021 at 17:43
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It is certainly a correct proof of Heron's formula that I have not seen before. I don't believe that it is any simpler than the other proofs that I have seen, but I am still entertained by it. However, I do not know if this is an appropriate posting since you do not really have a question other than possibly 'is my proof correct?'. I believe that the stack exchange guidelines suggest we only post questions and answers, not discussions.
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answered Feb 23, 2021 at 17:17
Sohail FarhangiSohail Farhangi
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In my post, I had shown that
r 2(x+y+z)=x y z(result 1)(result 1)r 2(x+y+z)=x y z
and
K=r s(result 3)(result 3)K=r s
where s=x+y+z s=x+y+z is the semi-perimeter of the triangle ABC whose area is K K and sides a=y+z,b=z+x a=y+z,b=z+x and c=x+y c=x+y.
From (1), r 2 s=x y z r 2 s=x y z implies that
K 2=r 2 s 2=s x y z=s(s−a)(s−b)(s−c)K 2=r 2 s 2=s x y z=s(s−a)(s−b)(s−c)
Taking square root on both sides yields the Heron’s Formula
K=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√K=s(s−a)(s−b)(s−c)
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answered Jul 16, 2022 at 3:45
LaiLai
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That sure is a powerful formula!insipidintegrator –insipidintegrator 2022-07-16 03:48:38 +00:00 Commented Jul 16, 2022 at 3:48
See also Cut-the-Knot.com's page on this proof.Blue –Blue 2022-07-16 03:50:11 +00:00 Commented Jul 16, 2022 at 3:50
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4137 | https://projecteuclid.org/ebooks/books-by-independent-authors/Experiencing-Geometry/chapter/Chapter-4-Straightness-on-Cylinders-and-Cones/10.3792/euclid/9781429799850-8.pdf | Chapter 4 STRAIGHTNESS ON CYLINDERS AND CONES If a cut were made through a cone parallel to its base, how should we conceive of the two opposing surfaces which the cut has produced — as equal or as unequal? If they are unequal, that would imply that a cone is composed of many breaks and protrusions like steps. On the other hand, if they are equal, that would imply that two adjacent intersection planes are equal, which would mean that the cone, being made up of equal rather than unequal circles, must have the same appearance as a cylinder; which is utterly absurd. — Democritus of Abdera (~460 – ~380 B.C.) This quote shows that cylinders and cones were the subject of mathematical inquiry before Euclid (~365 – ~300 B.C.). In this chapter we investigate straightness on cones and cylinders. You should be comfortable with straightness as a local intrinsic notion — this is the bug’s view. This notion of straightness is also the basis for the notion of geodesics in differential geometry. Chapters 4 and 5 can be covered in either order, but we think that the experience with cylinders and cones in Problem 4.1 will help the reader to understand the hyperbolic plane in Problem 5.1. If the reader is comfortable with straightness as a local intrinsic notion, then it is also possible to skip Chapter 4 if Chapters 18 and 24 on geometric manifolds are not going to be covered. However, we suggest that you read the sections at the end of this chapter — Is “Shortest” Always “Straight”? and Relations to Differential Geometry — at least enough to find out what Euclid’s Fourth Postulate has to do with cones and cylinders. Chapter 4 Straightness on Cylinders and Cones 52 When looking at great circles on the surface of a sphere, we were able (except in the case of central symmetry) to see all the symmetries of straight lines from global extrinsic points of view. For example, a great circle extrinsically divides a sphere into two hemispheres that are mirror images of each other. Thus, on a sphere, it is a natural tendency to use the more familiar and comfortable extrinsic lens instead of taking the bug’s local and intrinsic point of view. However, on a cone and cylinder you must use the local, intrinsic point of view because there is no extrinsic view that will work except in special cases. PROBLEM 4.1 STRAIGHTNESS ON CYLINDERS AND CONES a. What lines are straight with respect to the surface of a cylinder or a cone? Why? Why not? b. Examine: Can geodesics intersect themselves on cylinders and cones? Can there be more than one geodesic joining two points on cylinders and cones? What happens on cones with varying cone angles, including cone angles greater than 360°? These are discussed starting in the next section. SUGGESTIONS Problem 4.1 is similar to Problem 2.1, but this time the surfaces are cylinders and cones. Make paper models but consider the cone or cylinder as continuing indefinitely with no top or bottom (except, of course, at the cone point). Again, imagine yourself as a bug whose whole universe is a cone or cylinder. As the bug crawls around on one of these surfaces, what will the bug experience as straight? As before, paths that are straight with respect to a surface are often called the “geodesics” for the surface. As you begin to explore these questions, it is likely that many other related geometric ideas will arise. Do not let seemingly irrelevant excess geometric baggage worry you. Often, you will find yourself getting lost in a tangential idea, and that’s understandable. Ultimately, however, the exploration of related ideas will give you a richer understanding of the scope and depth of the problem. In order to work through possible confusion on this problem, try some of the following suggestions others have found helpful. Each suggestion involves constructing or using models of cones and cylinders. You may find it helpful to explore cylinders first before beginning to explore cones. This problem has many aspects but focusing at first on the cylinder will simplify some things. If we make a cone or cylinder by rolling up a sheet of paper, will “straight” stay the same for the bug when we unroll it? Conversely, if we have a straight Chapter 4 Straightness on Cylinders and Cones 53 line drawn on a sheet of paper and roll it up, will it continue to be experienced as straight for the bug crawling on the paper? We are assuming here that the paper will not stretch, and its thickness is negligible. Lay a stiff ribbon or straight strip of paper on a cylinder or cone. Convince yourself that it will follow a straight line with respect to the surface. Also, convince yourself that straight lines on the cylinder or cone, when looked at locally and intrinsically, have the same symmetries as on the plane. If you intersect a cylinder by a flat plane and unroll it, what kind of curve do you get? Is it ever straight? (One way to see this curve is to dip a paper cylinder into water.) On a cylinder or cone, can a geodesic ever intersect itself? How many times? This question is explored in more detail in Problem 4.2, which the interested reader may turn to now. Can there be more than one geodesic joining two points on a cylinder or cone? How many? Is there always at least one? Again, this question is explored in more detail in Problem 4.2. There are several important things to keep in mind while working on this problem. First, you absolutely must make models. If you attempt to visualize lines on a cone or cylinder, you are bound to make claims that you would easily see are mistaken if you investigated them on an actual cone or cylinder. Many students find it helpful to make models using transparent material. Second, as with the sphere, you must think about lines and triangles on the cone and cylinder in an intrinsic way — always looking at things from a bug’s point of view. We are not interested in what’s happening in 3-space, only what you would see and experience if you were restricted to the surface of a cone or cylinder. And last, but certainly not least, you must look at cones of different shapes, that is, cones with varying cone angles. CONES WITH VARYING CONE ANGLES Geodesics behave differently on differently shaped cones. So an important variable is the cone angle. The cone angle is generally defined as the angle measured around the point of the cone on the surface. Notice that this is an intrinsic description of angle. The bug could measure a cone angle (in radians) by determining the circumference of an intrinsic circle with center at the cone point and then dividing that circumference by the radius of the circle. We can determine the cone angle extrinsically in the following way: Cut the cone along a generator (a line on the cone through the cone point) and flatten the cone. The measure of the cone angle is then the angle measure of the flattened planar sector. Chapter 4 Straightness on Cylinders and Cones 54 Figure 4.1 Making a 180° cone For example, if we take a piece of paper and bend it so that half of one side meets up with the other half of the same side, we will have a 180-degree cone (Figure 4.1). A 90º cone is also easy to make — just use the corner of a paper sheet and bring one side around to meet the adjacent side. Also be sure to look at larger cones. One convenient way to do this is to make a cone with a variable cone angle. This can be accomplished by taking a sheet of paper and cutting (or tearing) a slit from one edge to the center. (See Figure 4.2.) A rectangular sheet will work but a circular sheet is easier to picture. Note that it is not necessary that the slit be straight! Figure 4.2 A cone with variable cone angle (0 –360°) You are already familiar with a 360º cone — it’s just a plane. The cone angle can also be larger than 360º. A common larger cone is the 450º cone. You probably have a cone like this somewhere on the walls, floor, and ceiling of your room. You can easily make one by cutting a slit in a piece of paper and inserting a 90º slice (360º + 90º = 450º) as in Figure 4.3. Figure 4.3 How to make a 450º cone Chapter 4 Straightness on Cylinders and Cones 55 Two cone angles on a ceiling You may have trouble believing that this is a cone but remember that just because it cannot hold ice cream does not mean it is not a cone. If you will look around in the room you are, perhaps you can locate a corner where five right angles meet – that is 450º cone. It is important to realize that when you change the shape of the cone like this (that is, either it is with ruffles or straight lines), you are only changing its extrinsic appearance. Intrinsically (from the bug’s point of view) there is no difference. It may be helpful for you to discuss some definitions of a cone, such as the following: Take any simple (non-intersecting) closed curve a on a sphere and the center P of the sphere. A cone is the union of the rays that start at P and go through each point on a. The cone angle is then equal to (length of a)/ (radius of sphere), in radians. Do you see why?
You can also make a cone with variable angle of more than 180°: Take two sheets of paper and slit them together to their centers as in Figure 4.4. Tape the right side of the top slit to the left side of the bottom slit as pictured. Now slide the other sides of the slits. Try it! Figure 4.4 Variable cone angle larger than 360° Chapter 4 Straightness on Cylinders and Cones 56 Experiment by making paper examples of cones like those shown in Figure 4.4. What happens to the triangles and lines on a 450º cone? Is the shortest path always straight? Does every pair of points determine a straight line? Finally, also consider line symmetries on the cone and cylinder. Check to see if the symmetries you found on the plane will work on these surfaces and remember to think intrinsically and locally. A special class of geodesics on the cone and cylinder is the generators. These are the straight lines that go through the cone point on the cone or go parallel to the axis of the cylinder. These lines have some extrinsic symmetries (can you see which ones?), but in general, geodesics have only local, intrinsic symmetries. Also, on the cone, think about the region near the cone point — what is happening there that makes it different from the rest of the cone? Few more explorations of geodesics on the cone can be found It is best if you experiment with paper models to find out what geodesics look like on the cone and cylinder before reading further. GEODESICS ON CYLINDERS Let us first look at the three classes of straight lines on a cylinder. When walking on the surface of a cylinder, a bug might walk along a vertical generator. See Figure 4.5. Figure 4.5 Vertical generators are straight It might walk along an intersection of a horizontal plane with the cylinder, what we will call a great circle. See Figure 4.6 Figure 4.6 Great circles are intrinsically straight Chapter 4 Straightness on Cylinders and Cones 57 Or, the bug might walk along a spiral or helix of constant slope around the cylinder. See Figure 4.7 and the photo at the beginning of this chapter depicting lightening damage to the tree. Watch a squirrel running up the tree! Figure 4.7 Helixes are intrinsically straight Helixes can be seen on outside parking garages and in sculptures Why are these geodesics? How can you convince yourself? And why are these the only geodesics? GEODESICS ON CONES Now let us look at the classes of straight lines on a cone. Walking along a generator: When looking at straight paths on a cone, you will be forced to consider straightness at the cone point. You might decide that there is no way the bug can go straight once it reaches the cone point, and thus a straight path leading up to the cone point ends there. Or you might decide that the bug can find a continuing path that has at least some of the symmetries of a straight line. Do you see which path this is? Or you might decide that the straight continuing path(s?) is the limit of geodesics that just miss the cone point. See Figure 4.8. Figure 4.8 Bug walking straight over the cone point Help! Chapter 4 Straightness on Cylinders and Cones 58 Walking straight and around: If you use a ribbon on a 90º cone, then you can see that this cone has a geodesic like the one depicted in Figure 4.9. This particular geodesic intersects itself. However, check to see that this property depends on the cone angle. In particular, if the cone angle is more than 180°, then geodesics do not intersect themselves. And if the cone angle is less than 90°, then geodesics (except for generators) intersect at least two times. Try it out! Later, in Problem 4.2, we will describe a tool that will help you determine how the number of self- intersections depends on the cone angle. Figure 4.9 A geodesic intersecting itself on a 90° cone PROBLEM 4.2 GLOBAL PROPERTIES OF GEODESICS Now we will look more closely at long geodesics that wrap around on a cylinder or cone. Several questions have arisen. a. How do we determine the different geodesics connecting two points? How many are there? How does it depend on the cone angle? Is there always at least one geodesic joining each pair of points? How can we justify our conjectures? b. How many times can a geodesic on a cylinder or cone intersect itself? How are the self-intersections related to the cone angle? At what angle does the geodesic intersect itself? How can we justify these relationships? SUGGESTIONS Here we offer the tool of covering spaces, which may help you explore these questions. The method of coverings is so named because it utilizes layers (or sheets) that each cover the surface. We will first start with a cylinder because it is easier and then move on to a cone. n -SHEETED COVERINGS OF A CYLINDER To understand how the method of coverings works, imagine taking a paper cylinder and cutting it axially (along a vertical generator) so that it unrolls into a plane. This is probably the way you constructed cylinders to study this problem before. The unrolled sheet (a portion of the plane) is said to be a 1-sheeted covering of the cylinder. See Figure Chapter 4 Straightness on Cylinders and Cones 59 4.10. If you marked two points on the cylinder, A and B, as indicated in the figure, when the cylinder is cut and unrolled into the covering, these two points become two points on the covering (which are labeled by the same letters in the figure). The two points on the covering are said to be lifts of the points on the cylinder. Figure 4.10 A 1-sheeted covering of a cylinder Now imagine attaching several of these “sheets” together, end to end. When rolled up, each sheet will go around the cylinder exactly once — they will each cover the cylinder. (Rolls of toilet paper or paper towels give a rough idea of coverings of a cylinder.) Also, each sheet of the covering will have the points A and B in identical locations. You can see this (assuming the paper thickness is negligible) by rolling up the coverings and making points by sticking a sharp object through the cylinder. This means that all the A’s are coverings of the same point on the cylinder and all the B’s are coverings of the same point on the cylinder. We just have on the covering several representations, or lifts, of each point on the cylinder. Figure 4.11 depicts a 3-sheeted covering space for a cylinder and six geodesics joining A to B. (One of them is the most direct path from A to B and the others spiral once, twice, or three times around the cylinder in one of two directions.) cut Figure 4.11 A 3-sheeted covering space for a cylinder cut A B A B Chapter 4 Straightness on Cylinders and Cones 60 We could also have added more sheets to the covering on either the right or left side. You can now roll these sheets back into a cylinder and see what the geodesics look like. Remember to roll sheets up so that each sheet of the covering covers the cylinder exactly once — all of the vertical lines between the coverings should lie on the same generator of the cylinder. Note that if you do this with ordinary paper, part or all of some geodesics will be hidden, even though they are all there. It may be easier to see what’s happening if you use transparencies. This method works because straightness is a local intrinsic property. Thus, lines that are straight when the coverings are laid out in a plane will still be straight when rolled into a cylinder. Remember that bending the paper does not change the intrinsic nature of the surface. Bending only changes the curvature that we see extrinsically. It is important always to look at the geodesics from the bug’s point of view. The cylinder and its covering are locally isometric. Use coverings to investigate Problem 4.2 on the cylinder. The global behavior of straight lines may be easier to see on the covering. n-SHEETED (BRANCHED) COVERINGS OF A CONE Figure 4.12 1-sheeted covering of a 270° cone Figure 4.12 shows a 1-sheeted covering of a cone. The sheet of paper and the cone are locally isometric except at the cone point. The cone point is called a branch point of the covering. We talk about lifts of points on the cone in the same way as on the cylinder. In Figure 4.12 we depict a 1-sheeted covering of a 270° cone and label two points and their lifts. A 4-sheeted covering space for a cone is depicted in Figure 4.13. Each of the rays drawn from the center of the covering is a lift of a single ray on the cone. Similarly, the points marked on the covering are the lifts of the points A and B on the cone. In the covering there are four segments joining a lift of A to different lifts of B. Each of these segments is the lift of a different geodesic segment joining A to B. Chapter 4 Straightness on Cylinders and Cones 61 Figure 4.13 4-sheeted covering space for a 89° cone Think about ways that the bug can use coverings as a tool to expand its exploration of surface geodesics. Also, think about ways you can use coverings to justify your observations in an intrinsic way. It is important to be precise; you don’t want the bug to get lost! Count the number of ways in which you can connect two points with a straight line and relate those countings with the cone angle. Does the number of straight paths only depend on the cone angle? Look at the 450° cone and see if it is always possible to connect any two points with a straight line. Make paper models! It is not possible to get an equation that relates the cone angle to the number of geodesics joining every pair of points. However, it is possible to find a formula that works for most pairs. Make covering spaces for cones of different size angles and refine the guesses you have already made about the numbers of self-intersections. In studying the self-intersections of a geodesic l on a cone, it may be helpful for you to consider the ray R that is perpendicular to the line l. (See Figure 4.14.) Now study one lift of the geodesic l and its relationship to the lifts of the ray R. Note that the seams between individual wedges are lifts of R. Figure 4.14 Self-intersections on a cone with angle A recent tidbit about coverings: In 1914 Henri Lebesgue (French, 1875-1941) posed a question: What is the shape with the smallest area that can completely cover a host of other shapes (which all share a certain trait in common)? The shapes should be such that no two points are further than one unit apart. In 2014 retired software engineer Philip Gibbs ran computer simulations on 200 randomly generated shapes with diameter 1. He kept “trimming corners” of hexagon and found that to be smallest known covering. ( Chapter 4 Straightness on Cylinders and Cones 62 LOCALLY ISOMETRIC By now you should realize that when a piece of paper is rolled or bent into a cylinder or cone, the bug’s local and intrinsic experience of the surface does not change except at the cone point. Extrinsically, the piece of paper and the cone are different, but in terms of the local geometry intrinsic to the surface they differ only at the cone point. Two geometric spaces, G and H, are said to be locally isometric at the points G in G and H in H if the local intrinsic experience at G is the same as the experience at H. That is, there are neighborhoods of G and H that are identical in terms of their intrinsic geometric properties. A cylinder and the plane are locally isometric (at each point) and the plane and a cone are locally isometric except at the cone point. Two cones are locally isometric at the cone points only if their cone angles are the same. Because cones and cylinders are locally isometric with the plane, locally they have the same geometric properties. Later, we will show that a sphere is not locally isometric with the plane — be on the lookout for a result that will imply this. IS “SHORTEST” ALWAYS “STRAIGHT”? We are often told that “a straight line is the shortest distance between two points,” but is this really true? As we have already seen on a sphere, two points not opposite each other are connected by two straight paths (one going one way around a great circle and one going the other way). Only one of these paths is shortest. The other is also straight, but not the shortest straight path. Consider a model of a cone with angle 450°. Notice that such cones appear commonly in buildings as so-called “outside corners” (see Figure 4.3). It is best, however, to have a paper model that can be flattened. Figure 4.15 There is no straight (symmetric) path from A to B Use your model to investigate which points on the cone can be joined by straight lines (in the sense of having reflection-in-the-line symmetry). In particular, look at points such as those labeled A and B in Figure 4.15. Convince yourself that there is no path from A to B that is straight (in the sense of having reflection-in-the-line symmetry), and for these points the shortest path goes through the cone point and thus is not straight (in the sense of having symmetry). Chapter 4 Straightness on Cylinders and Cones 63 Figure 4.16 The shortest path is not straight (in the sense of symmetry) Here is another example: Think of a bug crawling on a plane with a tall box sitting on that plane (refer to Figure 4.16). This combination surface — the plane with the box sticking out of it — has eight cone points. The four at the top of the box have 270° cone angles, and the four at the bottom of the box have 450° cone angles (180° on the box and 270° on the plane). What is the shortest path between points X and Y, points on opposite sides of the box? Is the straight path the shortest? Is the shortest path straight? To check that the shortest path is not straight, try to see that at the bottom corners of the box the two sides of the path have different angular measures. (If X and Y are close to the box, then the angle on the box side of the path measures a little more than 180° and the angle on the other side measures almost 270°.) RELATIONS TO DIFFERENTIAL GEOMETRY We see that sometimes a straight path is not shortest, and the shortest path is not straight. Does it then make sense to say (as most books do) that in Euclidean geometry a straight line is the shortest distance between two points? In differential geometry, on “smooth” surfaces, “straight” and “shortest” are more nearly the same. A smooth surface is essentially what it sounds like. More precisely, a surface is smooth at a point if, when you zoom in on the point, the surface becomes indistinguishable from a flat plane. (For details of this definition, see Problem 4.1 in [DG: Henderson, See also the last section and especially the endnote in Chapter 1.) Note that a cone is not smooth at the cone point, but a sphere and a cylinder are both smooth at every point. The following is a theorem from differential geometry: THEOREM 4.1: If a surface is smooth, then an intrinsically straight line (geodesic) on the surface is always the shortest path between “nearby” points. If the surface is also complete (every geodesic on it can be extended indefinitely), then any two points can be joined by a geodesic that is the shortest path between them. See [DG: Henderson], Problems 7.4b and 7.4d. Consider a planar surface with a hole removed. Check that for points near opposite sides of the hole, the shortest path (on the planar surface with hole removed) is not straight Chapter 4 Straightness on Cylinders and Cones 64 because the shortest path must go around the hole. We encourage the reader to discuss how each of the previous examples and problems is in harmony with this theorem. Note that the statement “every geodesic on the surface can be extended indefinitely” is a reasonable interpretation of Euclid’s Second Postulate: Every limited straight line can be extended indefinitely to a (unique) straight line. Note that the Second Postulate does not hold on a cone unless you consider geodesics to continue through the cone point. Also, Euclid defines a right angle as follows: When a straight line intersects another straight line such that the adjacent angles are equal to one another, then the equal angles are called right angles. Note that if you consider geodesics to continue through the cone point, then right angles at a cone point are not equal to right angles at points where the cone is locally isometric to the plane. And Euclid goes on to state as his Fourth Postulate: All right angles are equal. Thus, Euclid’s Second Postulate or Fourth Postulate rules out cones and any surface with isolated cone points. What is further ruled out by Euclid’s Fourth Postulate would depend on formulating more precisely just what it says. It is not clear (at least to the authors!) whether there is something we would want to call a surface that could be said to satisfy Euclid’s Fourth Postulate and not be a smooth surface. However, we can see that Euclid’s postulate at least gives part of the meaning of “smooth surface,” because it rules out isolated cone points. When we were in high school geometry class, we were confused why Euclid would have made such a postulate as his Postulate 4 — how could they possibly not be equal? In this chapter we have discovered that on cones right angles are not all equal. |
4138 | https://en.wikipedia.org/wiki/Phasor | Jump to content
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Contents
(Top)
1 Notation
2 Definition
3 Arithmetic
3.1 Multiplication by a constant (scalar)
3.2 Addition
3.3 Differentiation and integration
3.4 Ratio of phasors
4 Applications
4.1 Circuit laws
4.2 Power engineering
4.3 Telecommunications: analog modulations
5 See also
6 Footnotes
7 References
8 Further reading
9 External links
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From Wikipedia, the free encyclopedia
Complex number representing a particular sine wave
For other uses, see Phasor (disambiguation).
Not to be confused with phaser.
"Complex amplitude" redirects here. For the quantum-mechanical concept, see Complex probability amplitude.
In physics and engineering, a phasor (a portmanteau of phase vector) is a complex number representing a sinusoidal function whose amplitude A and initial phase θ are time-invariant and whose angular frequency ω is fixed. It is related to a more general concept called analytic representation, which decomposes a sinusoid into the product of a complex constant and a factor depending on time and frequency. The complex constant, which depends on amplitude and phase, is known as a phasor, or complex amplitude, and (in older texts) sinor or even complexor.
A common application is in the steady-state analysis of an electrical network powered by time varying current where all signals are assumed to be sinusoidal with a common frequency. Phasor representation allows the analyst to represent the amplitude and phase of the signal using a single complex number. The only difference in their analytic representations is the complex amplitude (phasor). A linear combination of such functions can be represented as a linear combination of phasors (known as phasor arithmetic or phasor algebra: 53) and the time/frequency dependent factor that they all have in common.
The origin of the term phasor rightfully suggests that a (diagrammatic) calculus somewhat similar to that possible for vectors is possible for phasors as well. An important additional feature of the phasor transform is that differentiation and integration of sinusoidal signals (having constant amplitude, period and phase) corresponds to simple algebraic operations on the phasors; the phasor transform thus allows the analysis (calculation) of the AC steady state of RLC circuits by solving simple algebraic equations (albeit with complex coefficients) in the phasor domain instead of solving differential equations (with real coefficients) in the time domain.[a] The originator of the phasor transform was Charles Proteus Steinmetz working at General Electric in the late 19th century. He got his inspiration from Oliver Heaviside. Heaviside's operational calculus was modified so that the variable p becomes jω. The complex number j has simple meaning: phase shift.
Glossing over some mathematical details, the phasor transform can also be seen as a particular case of the Laplace transform (limited to a single frequency), which, in contrast to phasor representation, can be used to (simultaneously) derive the transient response of an RLC circuit. However, the Laplace transform is mathematically more difficult to apply and the effort may be unjustified if only steady state analysis is required.
Notation
[edit]
See also: Vector notation
Phasor notation (also known as angle notation) is a mathematical notation used in electronics engineering and electrical engineering. A vector whose polar coordinates are magnitude and angle is written can represent either the vector or the complex number , according to Euler's formula with , both of which have magnitudes of 1.
The angle may be stated in degrees with an implied conversion from degrees to radians. For example would be assumed to be which is the vector or the number
Multiplication and division of complex numbers become straight forward through the phasor notation. Given the vectors and , the following is true:
: ,
: .
Definition
[edit]
A real-valued sinusoid with constant amplitude, frequency, and phase has the form:
where only parameter is time-variant. The inclusion of an imaginary component:
gives it, in accordance with Euler's formula, the factoring property described in the lead paragraph:
whose real part is the original sinusoid. The benefit of the complex representation is that linear operations with other complex representations produces a complex result whose real part reflects the same linear operations with the real parts of the other complex sinusoids. Furthermore, all the mathematics can be done with just the phasors and the common factor is reinserted prior to the real part of the result.
The function is an analytic representation of Figure 2 depicts it as a rotating vector in the complex plane. It is sometimes convenient to refer to the entire function as a phasor, as we do in the next section.
Arithmetic
[edit]
See also: Complex number § Relations and operations
Multiplication by a constant (scalar)
[edit]
Multiplication of the phasor by a complex constant, , produces another phasor. That means its only effect is to change the amplitude and phase of the underlying sinusoid:
In electronics, would represent an impedance, which is independent of time. In particular it is not the shorthand notation for another phasor. Multiplying a phasor current by an impedance produces a phasor voltage. But the product of two phasors (or squaring a phasor) would represent the product of two sinusoids, which is a non-linear operation that produces new frequency components. Phasor notation can only represent systems with one frequency, such as a linear system stimulated by a sinusoid.
Addition
[edit]
The sum of multiple phasors produces another phasor. That is because the sum of sinusoids with the same frequency is also a sinusoid with that frequency: where:
and, if we take , then is:
if with the signum function;
if ;
if .
or, via the law of cosines on the complex plane (or the trigonometric identity for angle differences): where
A key point is that A3 and θ3 do not depend on ω or t, which is what makes phasor notation possible. The time and frequency dependence can be suppressed and re-inserted into the outcome as long as the only operations used in between are ones that produce another phasor. In angle notation, the operation shown above is written:
Another way to view addition is that two vectors with coordinates [A1 cos(ωt + θ1), A1 sin(ωt + θ1)] and [A2 cos(ωt + θ2), A2 sin(ωt + θ2)] are added vectorially to produce a resultant vector with coordinates [A3 cos(ωt + θ3), A3 sin(ωt + θ3)] (see animation).
In physics, this sort of addition occurs when sinusoids interfere with each other, constructively or destructively. The static vector concept provides useful insight into questions like this: "What phase difference would be required between three identical sinusoids for perfect cancellation?" In this case, simply imagine taking three vectors of equal length and placing them head to tail such that the last head matches up with the first tail. Clearly, the shape which satisfies these conditions is an equilateral triangle, so the angle between each phasor to the next is 120° (2π⁄3 radians), or one third of a wavelength λ⁄3. So the phase difference between each wave must also be 120°, as is the case in three-phase power.
In other words, what this shows is that:
In the example of three waves, the phase difference between the first and the last wave was 240°, while for two waves destructive interference happens at 180°. In the limit of many waves, the phasors must form a circle for destructive interference, so that the first phasor is nearly parallel with the last. This means that for many sources, destructive interference happens when the first and last wave differ by 360 degrees, a full wavelength . This is why in single slit diffraction, the minima occur when light from the far edge travels a full wavelength further than the light from the near edge.
As the single vector rotates in an anti-clockwise direction, its tip at point A will rotate one complete revolution of 360° or 2π radians representing one complete cycle. If the length of its moving tip is transferred at different angular intervals in time to a graph as shown above, a sinusoidal waveform would be drawn starting at the left with zero time. Each position along the horizontal axis indicates the time that has elapsed since zero time, t = 0. When the vector is horizontal the tip of the vector represents the angles at 0°, 180°, and at 360°.
Likewise, when the tip of the vector is vertical it represents the positive peak value, (+Amax) at 90° or π⁄2 and the negative peak value, (−Amax) at 270° or 3π⁄2. Then the time axis of the waveform represents the angle either in degrees or radians through which the phasor has moved. So we can say that a phasor represents a scaled voltage or current value of a rotating vector which is "frozen" at some point in time, (t) and in our example above, this is at an angle of 30°.
Sometimes when we are analysing alternating waveforms we may need to know the position of the phasor, representing the alternating quantity at some particular instant in time especially when we want to compare two different waveforms on the same axis. For example, voltage and current. We have assumed in the waveform above that the waveform starts at time t = 0 with a corresponding phase angle in either degrees or radians.
But if a second waveform starts to the left or to the right of this zero point, or if we want to represent in phasor notation the relationship between the two waveforms, then we will need to take into account this phase difference, Φ of the waveform. Consider the diagram below from the previous Phase Difference tutorial.
Differentiation and integration
[edit]
The time derivative or integral of a phasor produces another phasor.[b] For example:
Therefore, in phasor representation, the time derivative of a sinusoid becomes just multiplication by the constant .
Similarly, integrating a phasor corresponds to multiplication by The time-dependent factor, is unaffected.
When we solve a linear differential equation with phasor arithmetic, we are merely factoring out of all terms of the equation, and reinserting it into the answer. For example, consider the following differential equation for the voltage across the capacitor in an RC circuit:
When the voltage source in this circuit is sinusoidal:
we may substitute
where phasor and phasor is the unknown quantity to be determined.
In the phasor shorthand notation, the differential equation reduces to:
Derivation
| | |
--- |
| | Eq.1 |
Since this must hold for all , specifically: it follows that:
| | |
--- |
| | Eq.2 |
It is also readily seen that:
Substituting these into Eq.1 and Eq.2, multiplying Eq.2 by and adding both equations gives:
Solving for the phasor capacitor voltage gives:
As we have seen, the factor multiplying represents differences of the amplitude and phase of relative to and
In polar coordinate form, the first term of the last expression is: where .
Therefore:
Ratio of phasors
[edit]
A quantity called complex impedance is the ratio of two phasors, which is not a phasor, because it does not correspond to a sinusoidally varying function.
Applications
[edit]
Circuit laws
[edit]
With phasors, the techniques for solving DC circuits can be applied to solve linear AC circuits.[a]
Ohm's law for resistors
: A resistor has no time delays and therefore doesn't change the phase of a signal therefore V = IR remains valid.
Ohm's law for resistors, inductors, and capacitors
: V = IZ where Z is the complex impedance.
Kirchhoff's circuit laws
: Work with voltages and current as complex phasors.
In an AC circuit we have real power (P) which is a representation of the average power into the circuit and reactive power (Q) which indicates power flowing back and forth. We can also define the complex power S = P + jQ and the apparent power which is the magnitude of S. The power law for an AC circuit expressed in phasors is then S = VI (where I is the complex conjugate of I, and the magnitudes of the voltage and current phasors V and of I are the RMS values of the voltage and current, respectively).
Given this we can apply the techniques of analysis of resistive circuits with phasors to analyze single frequency linear AC circuits containing resistors, capacitors, and inductors. Multiple frequency linear AC circuits and AC circuits with different waveforms can be analyzed to find voltages and currents by transforming all waveforms to sine wave components (using Fourier series) with magnitude and phase then analyzing each frequency separately, as allowed by the superposition theorem. This solution method applies only to inputs that are sinusoidal and for solutions that are in steady state, i.e., after all transients have died out.
The concept is frequently involved in representing an electrical impedance. In this case, the phase angle is the phase difference between the voltage applied to the impedance and the current driven through it.
Power engineering
[edit]
Main article: Phasor measurement unit
In analysis of three-phase AC power systems, usually a set of phasors is defined as the three complex cube roots of unity, graphically represented as unit magnitudes at angles of 0, 120 and 240 degrees. By treating polyphase AC circuit quantities as phasors, balanced circuits can be simplified and unbalanced circuits can be treated as an algebraic combination of symmetrical components. This approach greatly simplifies the work required in electrical calculations of voltage drop, power flow, and short-circuit currents. In the context of power systems analysis, the phase angle is often given in degrees, and the magnitude in RMS value rather than the peak amplitude of the sinusoid.
The technique of synchrophasors uses digital instruments to measure the phasors representing transmission system voltages at widespread points in a transmission network. Differences among the phasors indicate power flow and system stability.
Telecommunications: analog modulations
[edit]
The rotating frame picture using phasor can be a powerful tool to understand analog modulations such as amplitude modulation (and its variants) and frequency modulation.
where the term in brackets is viewed as a rotating vector in the complex plane.
The phasor has length , rotates anti-clockwise at a rate of revolutions per second, and at time makes an angle of with respect to the positive real axis.
The waveform can then be viewed as a projection of this vector onto the real axis. A modulated waveform is represented by this phasor (the carrier) and two additional phasors (the modulation phasors). If the modulating signal is a single tone of the form , where is the modulation depth and is the frequency of the modulating signal, then for amplitude modulation the two modulation phasors are given by,
The two modulation phasors are phased such that their vector sum is always in phase with the carrier phasor. An alternative representation is two phasors counter rotating around the end of the carrier phasor at a rate relative to the carrier phasor. That is,
Frequency modulation is a similar representation except that the modulating phasors are not in phase with the carrier. In this case the vector sum of the modulating phasors is shifted 90° from the carrier phase. Strictly, frequency modulation representation requires additional small modulation phasors at etc, but for most practical purposes these are ignored because their effect is very small.
See also
[edit]
In-phase and quadrature components
Constellation diagram
Analytic signal, a generalization of phasors for time-variant amplitude, phase, and frequency.
Complex envelope
Phase factor, a phasor of unit magnitude
Footnotes
[edit]
^ a b Including analysis of the AC circuits.: 53
^ This results from which means that the complex exponential is the eigenfunction of the derivative operator.
References
[edit]
^ Huw Fox; William Bolton (2002). Mathematics for Engineers and Technologists. Butterworth-Heinemann. p. 30. ISBN 978-0-08-051119-1.
^ Clay Rawlins (2000). Basic AC Circuits (2nd ed.). Newnes. p. 124. ISBN 978-0-08-049398-5.
^ Bracewell, Ron. The Fourier Transform and Its Applications. McGraw-Hill, 1965. p269
^ K. S. Suresh Kumar (2008). Electric Circuits and Networks. Pearson Education India. p. 272. ISBN 978-81-317-1390-7.
^ Kequian Zhang; Dejie Li (2007). Electromagnetic Theory for Microwaves and Optoelectronics (2nd ed.). Springer Science & Business Media. p. 13. ISBN 978-3-540-74296-8.
^ a b c J. Hindmarsh (1984). Electrical Machines & their Applications (4th ed.). Elsevier. p. 58. ISBN 978-1-4832-9492-6.
^ a b Gross, Charles A. (2012). Fundamentals of electrical engineering. Thaddeus Adam Roppel. Boca Raton, FL: CRC Press. ISBN 978-1-4398-9807-9. OCLC 863646311.
^ William J. Eccles (2011). Pragmatic Electrical Engineering: Fundamentals. Morgan & Claypool Publishers. p. 51. ISBN 978-1-60845-668-0.
^ a b Richard C. Dorf; James A. Svoboda (2010). Introduction to Electric Circuits (8th ed.). John Wiley & Sons. p. 661. ISBN 978-0-470-52157-1.
^ Allan H. Robbins; Wilhelm Miller (2012). Circuit Analysis: Theory and Practice (5th ed.). Cengage Learning. p. 536. ISBN 978-1-285-40192-8.
^ a b c Won Y. Yang; Seung C. Lee (2008). Circuit Systems with MATLAB and PSpice. John Wiley & Sons. pp. 256–261. ISBN 978-0-470-82240-1.
^ Basil Mahon (2017). The Forgotten Genius of Oliver Heaviside (1st ed.). Prometheus Books Learning. p. 230. ISBN 978-1-63388-331-4.
^ Nilsson, James William; Riedel, Susan A. (2008). Electric circuits (8th ed.). Prentice Hall. p. 338. ISBN 978-0-13-198925-2., Chapter 9, page 338
^ Rawlins, John C. (2000). Basic AC Circuits (Second ed.). Newnes. pp. 427–452. ISBN 9780750671736.
^ Singh, Ravish R (2009). "Section 4.5: Phasor Representation of Alternating Quantities". Electrical Networks. Mcgraw Hill Higher Education. p. 4.13. ISBN 978-0070260962.
^ Clayton, Paul (2008). Introduction to electromagnetic compatibility. Wiley. p. 861. ISBN 978-81-265-2875-2.
^ de Oliveira, H.M. and Nunes, F.D. About the Phasor Pathways in Analogical Amplitude Modulations. International Journal of Research in Engineering and Science (IJRES) Vol.2, N.1, Jan., pp.11-18, 2014. ISSN 2320-9364
Further reading
[edit]
Douglas C. Giancoli (1989). Physics for Scientists and Engineers. Prentice Hall. ISBN 0-13-666322-2.
Dorf, Richard C.; Tallarida, Ronald J. (1993-07-15). Pocket Book of Electrical Engineering Formulas (1 ed.). Boca Raton, FL: CRC Press. pp. 152–155. ISBN 0849344735.
External links
[edit]
Wikimedia Commons has media related to Phasors.
Wikiversity has a lesson on Phasor algebra
Phasor Phactory
Visual Representation of Phasors
Polar and Rectangular Notation
Phasor in Telecommunication
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Shortest Distance, Perp Bis, and Angle Bis Conjectures
Author:gmitesser, canhalt_1
Topic:Geometry
Shortest Distance Conjecture Investigation
Shortest Distance Conjecture
Move point C above until the length of CD is the shortest. Notice the angle measure that results. Then, complete the conjecture and add to your vocabulary list. Shortest Distance Conjecture: The shortest distance from a point to a line is measured along the _________ segment from the point to the line.
𝜋
Check my answer
Perpendicular and Angle Bisectors Investigations
Perpendicular Bisector Conjecture
Line CM the perpendicular bisector of segment AB. Move point C along the perpendicular bisector. Then, complete the following conjecture and add to your vocabulary list. Perpendicular Bisector Conjecture: If a point is on the perpendicular bisector of a segment, then it is _________ to the endpoints of the line segment.
𝜋
Check my answer
Angle Bisector Conjecture
Ray FK is the angle bisector of angle LFN. Segments KL and KN are the shortest distances from point K to the sides of the angle. Move point K around. Then, complete the following conjecture and put in your vocabulary. Angle Bisector Conjecture: If a point is on the angle bisector of a segment, then it is ____ to the sides of the angle.
𝜋
Check my answer
Driveway Problem
Driveway Problem
Point C represents your house on the road (line AB). Construct (with compass and straightedge) the shortest possible driveway from your house to the road. What conjecture applies to this problem?
𝜋
Check my answer
Roommates Problem
Roommates Problem
Cassidy and Chelsey are roommates and want to find a place to live that is equidistant to their respective jobs. Construct (with compass and straightedge) a line that shows all possible places they can live that would be the same distance to their workplaces. What conjecture applies here?
𝜋
Check my answer
Soccer Goalie Problem
Soccer Goalie Problem
Construct a ray that shows places the goalie should stand. What conjecture applies to this scenario?
𝜋
Check my answer
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The Organic Chem Lab Survival Manual: A Student's Guide to Techniques, 11th Edition
James W. Zubrick
ISBN: 978-1-119-60855-4
February 2020
288 pages
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The Organic Chem Lab Survival Manual: A Student's Guide to Techniques, 11th Edition
James W. Zubrick
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February 2020
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Description
Teaches students the basic techniques and equipment of the organic chemistry lab — the updated new edition of the popular hands-on guide.
The Organic Chem Lab Survival Manual helps students understand the basic techniques, essential safety protocols, and the standard instrumentation necessary for success in the laboratory. Author James W. Zubrick has been assisting students navigate organic chemistry labs for more than three decades, explaining how to set up the laboratory, make accurate measurements, and perform safe and meaningful experiments. This practical guide covers every essential area of lab knowledge, from keeping detailed notes and interpreting handbooks to using equipment for chromatography and infrared spectroscopy. Now in its eleventh edition, this guide has been thoroughly updated to cover current laboratory practices, instruments, and techniques. Focusing primarily on macroscale equipment and experiments, chapters cover microscale jointware, drying agents, recrystallization, distillation, nuclear magnetic resonance, and much more.
This popular textbook:
Familiarizes students with common lab instruments
Provides guidance on basic lab skills and procedures
Includes easy-to-follow diagrams and illustrations of lab experiments
Features practical exercises and activities at the end of each chapter
Provides real-world examples of lab notes and instrument manuals
The Organic Chem Lab Survival Manual: A Student’s Guide to Techniques, 11th Edition is an essential resource for students new to the laboratory environment, as well as those more experienced seeking to refresh their knowledge.
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Table of Contents
Chapter 1 Safety First, Last, and Always 1
Chapter 2 Keeping a Notebook 11
Chapter 3 Mining Your Own Data 25
Chapter 4 Jointware 30
Chapter 5 Microscale Jointware 41
Chapter 6 Other Interesting Equipment 52
Chapter 7 Pipet Tips 56
Chapter 8 Syringes, Needles, and Septa 63
Chapter 9 Clean and Dry 66
Chapter 10 Drying Agents 68
Chapter 11 On Products 72
Chapter 12 The Melting-Point Experiment 74
Chapter 13 Recrystallization 92
Chapter 14 Recrystallization: Microscale 108
Chapter 15 Extraction and Washing 114
Chapter 16 Extraction and Washing: Microscale 128
Chapter 17 Sources of Heat 131
Chapter 18 Clamps and Clamping 140
Chapter 20 Microscale Distillation 175
Chapter 21 The Rotary Evaporator 179
Chapter 22 Reflux and Addition 183
Chapter 23 Reflux: Microscale 190
Chapter 24 Sublimation 192
Chapter 25 Microscale Boiling Point 195
Chapter 26 Chromatography: Some Generalities 199
Chapter 27 Thin-Layer Chromatography: TLC 202
Chapter 28 Wet-Column Chromatography 214
Chapter 29 Refractometry 222
Chapter 30 Gas Chromatography 227
Chapter 31 HP Liquid Chromatography 236
Chapter 32 Infrared Spectroscopy (and a bit of uv-vis, too) 245
Chapter 32 On The Dual -Beam Infra-Red Instrument (ONLINE)1
Chapter 33 Nuclear Magnetic Resonance 272
Chapter 34 Theory of Distillation (Online) 1
Index 291
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The Organic Chem Lab Survival Manual: A Student's Guide to Techniques, 11th Edition -----------------------------------------------------------------------------------
James W. Zubrick
E-Book Rental (150 Days)
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$30.00
E-Book
978-1-119-60786-1
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$71.95
Paperback
978-1-119-60855-4
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Description
Teaches students the basic techniques and equipment of the organic chemistry lab — the updated new edition of the popular hands-on guide.
The Organic Chem Lab Survival Manual helps students understand the basic techniques, essential safety protocols, and the standard instrumentation necessary for success in the laboratory. Author James W. Zubrick has been assisting students navigate organic chemistry labs for more than three decades, explaining how to set up the laboratory, make accurate measurements, and perform safe and meaningful experiments. This practical guide covers every essential area of lab knowledge, from keeping detailed notes and interpreting handbooks to using equipment for chromatography and infrared spectroscopy. Now in its eleventh edition, this guide has been thoroughly updated to cover current laboratory practices, instruments, and techniques. Focusing primarily on macroscale equipment and experiments, chapters cover microscale jointware, drying agents, recrystallization, distillation, nuclear magnetic resonance, and much more.
This popular textbook:
Familiarizes students with common lab instruments
Provides guidance on basic lab skills and procedures
Includes easy-to-follow diagrams and illustrations of lab experiments
Features practical exercises and activities at the end of each chapter
Provides real-world examples of lab notes and instrument manuals
The Organic Chem Lab Survival Manual: A Student’s Guide to Techniques, 11th Edition is an essential resource for students new to the laboratory environment, as well as those more experienced seeking to refresh their knowledge.
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4141 | https://leannortiz.weebly.com/uploads/8/2/7/7/8277852/tempconvpptkfc.pdf | Temperature Notes • Kelvin, Celsius, Fahrenheit • Write notes, put in por7olio. • Everything Underlined – COPY IT! • Study. Read over notes daily. • Enjoy the learning process. • Temperature is how hot or cold something is. Temperature is really a measure of how fast the atoms and molecules that make up a substance are moving (this movement is sub-‐microscopic; you cannot see it). • A thermometer measures the temperature of something, showing how hot or cold it is. • There are many different types of thermometers that measure the temperature of different things. We use thermometers to measure the temperature of the air, our bodies, food, and many other things. There are analog and digital thermometers. An analog thermometer shows the temperature on a scale or dial; a digital thermometer shows the temperature as a number. • The higher the number above zero, the higher the temperature. • Temperature is measured in many different scales, including Fahrenheit, Celsius (or CenDgrade), and Kelvin scales. Copy this table down in your notes. • The units of the Celsius and Fahrenheit scales are called degrees. The symbol for degree is °. • The units of the Kelvin scale are called Kelvins. • A change of one degree of temperature Celsius is the same as a change of one Kelvin. • A change of one degree Fahrenheit is a smaller change in temperature. • Each temperature scale is useful for different things. • Fahrenheit is a measure of temperature that is abbreviated F. • Water freezes at 32°F and boils at 212°F. • This scale was developed by Gabriel Daniel Fahrenheit in 1714. • The Fahrenheit scale is used in the USA but not in many other countries. • The degrees in Fahrenheit are smaller than those in Celsius, making weather-‐related temperatures more easily wriSen (without having to use a decimal point). • Celsius (also called CenTgrade) is a measure of temperature that is abbreviated C. • Water freezes at 0°C and boils at 100°C. • This scale was developed by Anders Celsius in 1742. • Kelvin is a measure of temperature that is abbreviated K. • Kelvin is a temperature scale designed so that 0 (zero) K is defined as absolute zero, and the size of one unit is the same as the size of one degree Celsius. • (Absolute zero is a hypotheTcal temperature at which all molecular movement stops). • All actual temperatures are above absolute zero. • Absolute zero would occur at -‐273.16°C, -‐459.69°F, or 0 (zero) K.] • Water freezes at 273.16K; water boils at 373.16K. • Many scienTsts use this scale (because all the temperatures are posiTve, making calculaTons simpler) or the Celsius scale. • This temperature scale was designed by Lord Kelvin (William Thomson). Conversion Formulas If you know the temperature in one temperature scale you can use a formula to figure out what it is equivalent to in other scales. The conversion formulas are in the table that follows. COPY THIS CONVERSION FORMULA CHART IN YOUR NOTES Temperature Conversion Formulas: COPY THIS CONVERSION FORMULA CHART IN YOUR NOTES Examples: • This example problem illustrates the method to convert temperatures from Celsius to Fahrenheit. • Problem: What is the temperature in Fahrenheit of 20 °C? • SoluTon: • The conversion formula for °C to °F is • TF = 9/5(TC) + 32 • TF = 9/5(20) + 32 • TF = 36 + 32 • TF = 68 °F • Answer: The temperature in Fahrenheit of 20 °C is 68 °F. COPY THIS INFORMATION IN YOUR NOTES Examples: • Here is an example problem that explains how to convert a temperature from degrees on the Celsius scale to Kelvin. • Celsius to Kelvin Problem #1 • Convert 27° C to Kelvin. • Celsius to Kelvin SoluTon #1 • K = °C + 273 • K = 27 + 273 • K = 300 • 300 K • Note that the answer is 300 K. Kelvin is not expressed in degrees. COPY THIS INFORMATION IN YOUR NOTES Examples: • Celsius to Kelvin Problem #2 • Convert 77° C to Kelvin. • Celsius to Kelvin • SoluTon #2 • K = °C + 273 • K = 77 + 273 • K = 350 • 350 K COPY THIS INFORMATION IN YOUR NOTES Temperature Unit Conversion Problem Worked Problem ConverTng from Fahrenheit to Celsius and Kelvin • Problem: Express normal body temperature, 98.6°F, in °C and K. • SoluTon: The Fahrenheit to Celsius conversion equaTon may be expressed in this form: • F° = = 1.8(°C) + 32 • Enter 98.6 for the F° • 98.6 = 1.8(°C) + 32 • 1.8(°C) = 98.6 – 32 • 1.8(°C) = 66.6 • °C = 66.6/1.8°C = 37.0 • To solve for Kelvin: • K = °C +273K = 37.0 + 273 • K = 310 • Answer: 98.6 °F is equal to 37.0°C and 310 K COPY THIS INFORMATION IN YOUR NOTES |
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Discrete MathematicsAn Open Introduction
Oscar Levin
Section5.1Generating Functions
¶
There is an extremely powerful tool in discrete mathematics used to manipulate sequences called the generating function. The idea is this: instead of an infinite sequence (for example: (2, 3, 5, 8, 12, \ldots)) we look at a single function which encodes the sequence. But not a function which gives the (n)th term as output. Instead, a function whose power series (like from calculus) “displays” the terms of the sequence. So for example, we would look at the power series (2 + 3x + 5x^2 + 8x^3 + 12x^4 + \cdots) which displays the sequence (2, 3, 5, 8, 12, \ldots) as coefficients.
An infinite power series is simply an infinite sum of terms of the form (c_nx^n) were (c_n) is some constant. So we might write a power series like this:
\begin{equation} \sum_{k=0}^\infty c_k x^k. \end{equation}
or expanded like this
\begin{equation} c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5 + \cdots. \end{equation}
When viewed in the context of generating functions, we call such a power series a generating series. The generating series generates the sequence
\begin{equation} c_0, c_1, c_2, c_3, c_4, c_5, \ldots. \end{equation}
In other words, the sequence generated by a generating series is simply the sequence of coefficients of the infinite polynomial.
Example5.1.1
What sequence is represented by the generating series (3 + 8x^2 + x^3 + \frac{x^5}{7} + 100x^6 + \cdots\text{?})
Solution
We just read off the coefficients of each (x^n) term. So (a_0 = 3) since the coefficient of (x^0) is 3 ((x^0 = 1) so this is the constant term). What is (a_1\text{?}) It is NOT 8, since 8 is the coefficient of (x^2\text{,}) so 8 is the term (a_2) of the sequence. To find (a_1) we need to look for the coefficient of (x^1) which in this case is 0. So (a_1 = 0\text{.}) Continuing, we have (a_2 = 8\text{,}) (a_3 = 1\text{,}) (a_4 = 0\text{,}) and (a_5 = \frac{1}{7}\text{.}) So we have the sequence
\begin{equation} 3, 0, 8, 1, \frac{1}{7}, 100, \ldots \end{equation}
Note that when discussing generating functions, we always start our sequence with (a_0\text{.})
Now you might very naturally ask why we would do such a thing. One reason is that encoding a sequence with a power series helps us keep track of which term is which in the sequence. For example, if we write the sequence (1, 3, 4, 6, 9, \ldots, 24, 41,\ldots) it is impossible to determine which term (24) is (even if we agreed that the first term was supposed to be (a_0)). However, if we wrote the generating series instead, we would have (1 + 3x + 4x^2 + 6x^3 + 9x^4 + \cdots + 24 x^{17} + 41 x^{18} + \cdots\text{.}) Now it is clear that 24 is the 17th term of the sequence (that is, (a_{17} = 24)). Of course to get this benefit we could have displayed our sequence in any number of ways, perhaps (\fbox{1}_0 \fbox{3}_1 \fbox{4}_2 \fbox{6}_3 \fbox{9}_4 \cdots \fbox{24}_{17}\fbox{41}_{18}\cdots\text{,}) but we do not do this. The reason is that the generating series looks like an ordinary power series (although we are interpreting it differently) so we can do things with it that we ordinarily do with power series such as write down what it converges to.
For example, from calculus we know that the power series (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots + \frac{x^n}{n!} + \cdots) converges to the function (e^x\text{.}) So we can use (e^x) as a way of talking about the sequence of coefficients of the power series for (e^x\text{.}) When we write down a nice compact function which has an infinite power series that we view as a generating series, then we call that function a generating function. In this example, we would say
\begin{equation} 1, 1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, \ldots, \frac{1}{n!}, \ldots \mbox{ has generating function } e^x \end{equation}
SubsectionBuilding Generating Functions
¶
The (e^x) example is very specific. We have a rather odd sequence, and the only reason we know its generating function is because we happen to know the Taylor series for (e^x\text{.}) Our goal now is to gather some tools to build the generating function of a particular given sequence.
Let's see what the generating functions are for some very simple sequences. The simplest of all: 1, 1, 1, 1, 1, …. What does the generating series look like? It is simply (1 + x + x^2 + x^3 + x^4 + \cdots\text{.}) Now, can we find a closed formula for this power series? Yes! This particular series is really just a geometric series with common ratio (x\text{.}) So if we use our “multiply, shift and subtract” technique from Section 2.2, we have
\begin{align} S \amp = 1 + x + x^2 + x^3 + \cdots\ \underline{- xS} \amp \underline{\,\, = ~~~~~~ x + x^2 + x^3 + x^4 + \cdots}\ (1-x)S \amp = 1 \end{align}
Therefore we see that
\begin{equation} 1 + x + x^2 + x^3 \cdots = \dfrac{1}{1-x} \end{equation}
You might remember from calculus that this is only true on the interval of convergence for the power series, in this case when (|x| \lt 1\text{.}) That is true for us, but we don't care. We are never going to plug anything in for (x\text{,}) so as long as there is some value of (x) for which the generating function and generating series agree, we are happy. And in this case we are happy.
(1,1,1,\ldots)
The generating function for (1,1,1,1,1,1,\ldots) is (\dfrac{1}{1-x})
Let's use this basic generating function to find generating functions for more sequences. What if we replace (x) by (-x\text{.}) We get
\begin{equation} \frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots \mbox{ which generates } 1, -1, 1, -1, \ldots \end{equation}
If we replace (x) by (3x) we get
\begin{equation} \frac{1}{1-3x} = 1 + 3x + 9x^2 + 27x^3 + \cdots \mbox{ which generates } 1, 3, 9, 27, \ldots \end{equation}
By replacing the (x) in (\frac{1}{1-x}) we can get generating functions for a variety of sequences, but not all. For example, you cannot plug in anything for (x) to get the generating function for (2,2,2,2, \ldots\text{.}) However, we are not lost yet. Notice that each term of (2, 2, 2, 2, \ldots) is the result of multiplying the terms of (1, 1, 1, 1, \ldots) by the constant 2. So multiply the generating function by 2 as well.
\begin{equation} \frac{2}{1-x} = 2 + 2x + 2x^2 + 2x^3 + \cdots \mbox{ which generates } 2, 2, 2, 2, \ldots \end{equation}
Similarly, to find the generating function for the sequence (3, 9, 27, 81, \ldots\text{,}) we note that this sequence is the result of multiplying each term of (1, 3, 9, 27, \ldots) by 3. Since we have the generating function for (1, 3, 9, 27, \ldots) we can say
\begin{equation} \frac{3}{1-3x} = 3\cdot 1 + 3\cdot 3x + 3\cdot 9x^2 + 3\cdot 27x^3 + \cdots \mbox{ which generates } 3, 9, 27, 81, \ldots \end{equation}
What about the sequence (2, 4, 10, 28, 82, \ldots\text{?}) Here the terms are always 1 more than powers of 3. That is, we have added the sequences (1,1,1,1,\ldots) and (1,3,9, 27,\ldots) term by term. Therefore we can get a generating function by adding the respective generating functions:
\begin{align} 2 + 4x + 10x^2 + 28x^3 + \cdots \amp = (1 + 1) + (1 + 3)x + (1 + 9)x^2 + (1 + 27)x^3 + \cdots\ \amp = 1 + x + x^2 + x^3 + \cdots + 1 + 3x + 9x^2 + 27x^3 + \cdots\ \amp = \frac{1}{1-x} + \frac{1}{1-3x} \end{align}
The fun does not stop there: if we replace (x) in our original generating function by (x^2) we get
\begin{equation} \frac{1}{1-x^2} = 1 + x^2 + x^4 + x^6\cdots \mbox{ which generates } 1, 0, 1, 0, 1, 0, \ldots. \end{equation}
How could we get (0,1,0,1,0,1,\ldots\text{?}) Start with the previous sequence and shift it over by 1. But how do you do this? To see how shifting works, let's first try to get the generating function for the sequence (0, 1, 3, 9, 27, \ldots\text{.}) We know that (\frac{1}{1-3x} = 1 + 3x + 9x^2 + 27x^3 + \cdots\text{.}) To get the zero out front, we need the generating series to look like (x + 3x^2 + 9x^3 + 27x^4+ \cdots) (so there is no constant term). Multiplying by (x) has this effect. So the generating function for (0, 1, 3, 9, 27, \ldots) is (\frac{x}{1-3x}\text{.}) This will also work to get the generating function for (0,1,0,1,0,1,\ldots\text{:})
\begin{equation} \frac{x}{1-x^2} = x + x^3 + x^5 + \cdots \mbox{ which generates } 0, 1, 0, 1, 0 , 1 \ldots \end{equation}
What if we add the sequences (1,0,1,0,1,0,\ldots) and (0,1,0,1,0,1,\ldots) term by term? We should get (1,1,1,1,1,1\ldots\text{.}) What happens when we add the generating functions? It works (try it)!
\begin{equation} \frac{1}{1-x^2} + \frac{x}{1-x^2} = \frac{1}{1-x}. \end{equation}
Here's a sneaky one: what happens if you take the derivative of (\frac{1}{1-x}\text{?}) We get (\frac{1}{(1-x)^2}\text{.}) On the other hand, if we differentiate term by term in the power series, we get ((1 + x + x^2 + x^3 + \cdots)' = 1 + 2x + 3x^2 + 4x^3 + \cdots) which is the generating series for (1, 2, 3, 4, \ldots\text{.}) This says
(1,2,3,\ldots)
The generating function for (1, 2, 3, 4, 5, \ldots) is (\d\frac{1}{(1-x)^2}.)
Take a second derivative: (\frac{2}{(1-x)^3} = 2 + 6x + 12x^2 + 20x^3 + \cdots\text{.}) So (\frac{1}{(1-x)^3} = 1 + 3x + 6x^2 + 10x^3 + \cdots) is a generating function for the triangular numbers, (1,3,6,10\ldots) (although here we have (a_0 = 1) while (T_0 = 0) usually).
SubsectionDifferencing
¶
We have seen how to find generating functions from (\frac{1}{1-x}) using multiplication (by a constant or by (x)), substitution, addition, and differentiation. To use each of these, you must notice a way to transform the sequence (1,1,1,1,1\ldots) into your desired sequence. This is not always easy. It is also not really the way we have analyzed sequences. One thing we have considered often is the sequence of differences between terms of a sequence. This will turn out to be helpful in finding generating functions as well. The sequence of differences is often simpler than the original sequence. So if we know a generating function for the differences, we would like to use this to find a generating function for the original sequence.
For example, consider the sequence (2, 4, 10, 28, 82, \ldots\text{.}) How could we move to the sequence of first differences: (2, 6, 18, 54,\ldots\text{?}) We want to subtract 2 from the 4, 4 from the 10, 10 from the 28, and so on. So if we subtract (term by term) the sequence (0, 2, 4, 10, 28,\ldots) from (2, 4, 10, 28\ldots\text{,}) we will be set. We can get the generating function for (0,2,4,10,28,\ldots) from the generating function for (2,4,10,28\ldots) by multiplying by (x\text{.}) Use (A) to represent the generating function for (2, 4, 10, 28, 82, \ldots) Then:
\begin{align} A \amp = 2 + 4x + 10x^2 +28x^3 + 82x^4 + \cdots\ \underline{-xA} \amp \underline{\,\,= 0 + 2x + 4x^2 + 10x^3 + 28 x^4 + 82x^5 + \cdots}\ (1-x)A \amp = 2 + 2x + 6x^2 + 18x^3 + 54x^4 + \cdots \end{align}
While we don't get exactly the sequence of differences, we do get something close. In this particular case, we already know the generating function (A) (we found it in the previous section) but most of the time we will use this differencing technique to find (A\text{:}) if we have the generating function for the sequence of differences, we can then solve for (A\text{.})
Example5.1.2
Find a generating function for (1, 3, 5, 7, 9,\ldots\text{.})
Solution
Notice that the sequence of differences is constant. We know how to find the generating function for any constant sequence. So denote the generating function for (1, 3, 5, 7, 9, \ldots) by (A\text{.}) We have
\begin{align} A \amp = 1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots\ \underline{-xA} \amp \underline{\,\,= 0 + x + 3x^2 + 5x^3 + 7x^4 + 9x^5 + \cdots}\ (1-x)A \amp = 1 + 2x + 2x^2 + 2x^3 + 2x^4 + \cdots \end{align}
We know that (2x + 2x^2 + 2x^3 + 2x^4 + \cdots = \dfrac{2x}{1-x}\text{.}) Thus
\begin{equation} (1-x)A = 1 + \frac{2x}{1-x}. \end{equation}
Now solve for (A\text{:})
\begin{equation} A = \frac{1}{1-x} + \frac{2x}{(1-x)^2} = \frac{1+x}{(1-x)^2}. \end{equation}
Does this makes sense? Before we simplified the two fractions into one, we were adding the generating function for the sequence (1,1,1,1,\ldots) to the generating function for the sequence (0, 2, 4, 6, 8, 10, \ldots) (remember (\frac{1}{(1-x)^2}) generates (1,2,3,4,5, \ldots\text{,}) multiplying by (2x) shifts it over, putting the zero out front, and doubles each term). If we add these term by term, we get the correct sequence (1,3,5,7, 9, \ldots\text{.})
Now that we have a generating function for the odd numbers, we can use that to find the generating function for the squares:
Example5.1.3
Find the generating function for (1, 4, 9, 16, \ldots\text{.}) Note we take (1 = a_0\text{.})
Solution
Again we call the generating function for the sequence (A\text{.}) Using differencing:
\begin{align} A \amp = 1 + 4x + 9x^2 + 16x^3 + \cdots\ \underline{- xA} \amp \underline{\,\, = 0 + x + 4x^2 + 9x^3 + 16x^4 + \cdots}\ (1-x)A \amp = 1 + 3x + 5x^2 + 7x^3 + \cdots \end{align}
Since (1 + 3x + 5x^2 + 7x^3 + \cdots = \d\frac{1+x}{(1-x)^2}) we have (A = \d\frac{1+x}{(1-x)^3}\text{.})
In each of the examples above, we found the difference between consecutive terms which gave us a sequence of differences for which we knew a generating function. We can generalize this to more complicated relationships between terms of the sequence. For example, if we know that the sequence satisfies the recurrence relation (a_n = 3a_{n-1} - 2a_{n-2}\text{?}) In other words, if we take a term of the sequence and subtract 3 times the previous term and then add 2 times the term before that, we get 0 (since (a_n - 3a_{n-1} + 2a_{n-2} = 0)). That will hold for all but the first two terms of the sequence. So after the first two terms, the sequence of results of these calculations would be a sequence of 0's, for which we definitely know a generating function.
Example5.1.4
The sequence (1, 3, 7, 15, 31, 63, \ldots) satisfies the recurrence relation (a_n = 3a_{n-1} - 2a_{n-2}\text{.}) Find the generating function for the sequence.
Solution
Call the generating function for the sequence (A\text{.}) We have
\begin{align} A \amp = 1 + 3x + 7x^2 + 15x^3 + 31x^4 + \cdots + a_nx^n + \cdots\ -3xA \amp = 0 - 3x - 9x^2 - 21x^3 - 45x^4 - \cdots - 3a_{n-1}x^n - \cdots\ \underline{+~~~2x^2A_{~}^{~^{~}}} \amp \underline{\,\, = 0 + 0x + 2x^2 + 6x^3 + 14x^4 + \cdots + 2a_{n-2}x^n + \cdots}\ (1-3x+2x^2)A \amp = 1 \end{align}
We multiplied (A) by (-3x) which shifts every term over one spot and multiplies them by (-3\text{.}) On the third line, we multiplied (A) by (2x^2\text{,}) which shifted every term over two spots and multiplied them by 2. When we add up the corresponding terms, we are taking each term, subtracting 3 times the previous term, and adding 2 times the term before that. This will happen for each term after (a_1) because (a_n - 3a_{n-1} + 2a_{n-2} = 0\text{.}) In general, we might have two terms from the beginning of the generating series, although in this case the second term happens to be 0 as well.
Now we just need to solve for (A\text{:})
\begin{equation} A = \frac{1}{1 - 3x + 2x^2}. \end{equation}
SubsectionMultiplication and Partial Sums
¶
What happens to the sequences when you multiply two generating functions? Let's see: (A = a_0 + a_1x + a_2x^2 + \cdots) and (B = b_0 + b_1x + b_2x^2 + \cdots\text{.}) To multiply (A) and (B\text{,}) we need to do a lot of distributing (infinite FOIL?) but keep in mind we will group like terms and only need to write down the first few terms to see the pattern. The constant term is (a_0b_0\text{.}) The coefficient of (x) is (a_0b_1 + a_1b_0\text{.}) And so on. We get:
\begin{equation} AB = a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2 + a_1b_1 + a_2b_0)x^2 + (a_0b_3 + a_1b_2 + a_2b_1 + a_3b_0)x^3 + \cdots \end{equation}
Example5.1.5
“Multiply” the sequence (1, 2, 3, 4, \ldots) by the sequence (1, 2, 4, 8, 16, \ldots\text{.})
Solution
The new constant term is just (1 \cdot 1\text{.}) The next term will be (1\cdot 2 + 2 \cdot 1 = 4\text{.}) The next term: (1 \cdot 4 + 2 \cdot 2 + 3 \cdot 1 = 11\text{.}) One more: (1 \cdot 8 + 2 \cdot 4 + 3 \cdot 2 + 4 \cdot 1 = 28\text{.}) The resulting sequence is
\begin{equation} 1, 4, 11, 28, 57, \ldots \end{equation}
Since the generating function for (1,2,3,4, \ldots) is (\frac{1}{(1-x)^2}) and the generating function for (1,2,4,8, 16, \ldots) is (\frac{1}{1-2x}\text{,}) we have that the generating function for (1,4, 11, 28, 57, \ldots) is (\frac{1}{(1-x)^2(1-2x)})
Consider the special case when you multiply a sequence by (1, 1, 1, \ldots\text{.}) For example, multiply (1,1,1,\ldots) by (1, 2, 3, 4, 5\ldots\text{.}) The first term is (1\cdot 1 = 1\text{.}) Then (1\cdot 2 + 1 \cdot 1 = 3\text{.}) Then (1\cdot 3 + 1\cdot 2 + 1 \cdot 1 = 6\text{.}) The next term will be 10. We are getting the triangular numbers. More precisely, we get the sequence of partial sums of (1,2,3,4,5, \ldots\text{.}) In terms of generating functions, we take (\frac{1}{1-x}) (generating (1,1,1,1,1\ldots)) and multiply it by (\frac{1}{(1-x)^2}) (generating (1,2,3,4,5,\ldots)) and this give (\frac{1}{(1-x)^3}\text{.}) This should not be a surprise as we found the same generating function for the triangular numbers earlier.
The point is, if you need to find a generating function for the sum of the first (n) terms of a particular sequence, and you know the generating function for that sequence, you can multiply it by (\frac{1}{1-x}\text{.}) To go back from the sequence of partial sums to the original sequence, you look at the sequence of differences. When you get the sequence of differences you end up multiplying by (1-x\text{,}) or equivalently, dividing by (\frac{1}{1-x}\text{.}) Multiplying by (\frac{1}{1-x}) gives partial sums, dividing by (\frac{1}{1-x}) gives differences.
SubsectionSolving Recurrence Relations with Generating Functions
¶
We conclude with an example of one of the many reasons studying generating functions is helpful. We can use generating functions to solve recurrence relations.
Example5.1.6
Solve the recurrence relation (a_n = 3a_{n-1} - 2a_{n-2}) with initial conditions (a_0 = 1) and (a_1 = 3\text{.})
Solution
We saw in an example above that this recurrence relation gives the sequence (1, 3, 7, 15, 31, 63, \ldots) which has generating function (\dfrac{1}{1 - 3x + 2x^2}\text{.}) We did this by calling the generating function (A) and then computing (A - 3xA + 2x^2A) which was just 1, since every other term canceled out.
But how does knowing the generating function help us? First, break up the generating function into two simpler ones. For this, we can use partial fraction decomposition. Start by factoring the denominator:
\begin{equation} \frac{1}{1-3x + 2x^2} = \frac{1}{(1-x)(1-2x)}. \end{equation}
Partial fraction decomposition tells us that we can write this faction as the sum of two fractions (we decompose the given fraction):
\begin{equation} \frac{1}{(1-x)(1-2x)} = \frac{a}{1-x} + \frac{b}{1-2x} \text{ ~~ for some constants } a \text{ and } b. \end{equation}
To find (a) and (b) we add the two decomposed fractions using a common denominator. This gives
\begin{equation} \frac{1}{(1-x)(1-2x)} = \frac{a(1-2x) + b(1-x)}{(1-x)(1-2x)}. \end{equation}
so
\begin{equation} 1 = a(1-2x) + b(1-x). \end{equation}
This must be true for all values of (x\text{.}) If (x = 1\text{,}) then the equation becomes (1 = -a) so (a = -1\text{.}) When (x = \frac{1}{2}) we get (1 = b/2) so (b = 2\text{.}) This tells us that we can decompose the fraction like this:
\begin{equation} \frac{1}{(1-x)(1-2x)} = \frac{-1}{1-x} + \frac{2}{1-2x}. \end{equation}
This completes the partial fraction decomposition. Notice that these two fractions are generating functions we know. In fact, we should be able to expand each of them.
\begin{equation} \frac{-1}{1-x} = -1 - x - x^2 -x^3 - x^4 - \cdots \mbox{ which generates } -1, -1, -1, -1, -1, \ldots. \end{equation} \begin{equation} \frac{2}{1-2x} = 2 + 4x + 8x^2 + 16x^3 + 32x^4 + \cdots \mbox{ which generates } 2, 4, 8, 16, 32, \ldots. \end{equation}
We can give a closed formula for the (n)th term of each of these sequences. The first is just (a_n = -1\text{.}) The second is (a_n = 2^{n+1}\text{.}) The sequence we are interested in is just the sum of these, so the solution to the recurrence relation is
\begin{equation} a_n = 2^{n+1} - 1 \end{equation}
We can now add generating functions to our list of methods for solving recurrence relations.
SubsectionExercises
¶
1
Find the generating function for each of the following sequences by relating them back to a sequence with known generating function.
(4,4,4,4,4,\ldots\text{.})
(2, 4, 6, 8, 10, \ldots\text{.})
(0,0,0,2,4,6,8,10,\ldots\text{.})
(1, 5, 25, 125, \ldots\text{.})
(1, -3, 9, -27, 81, \ldots\text{.})
(1, 0, 5, 0, 25, 0, 125, 0, \ldots\text{.})
(0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0, 5, \ldots\text{.})
Solution
(\dfrac{4}{1-x}\text{.})
(\dfrac{2}{(1-x)^2}\text{.})
(\dfrac{2x^3}{(1-x}^2\text{.})
(\dfrac{1}{1-5x}\text{.})
(\dfrac{1}{1+3x}\text{.})
(\dfrac{1}{1-5x^2}\text{.})
(\dfrac{x}{(1-x^3)^2}\text{.})
2
Find the sequence generated by the following generating functions:
(\dfrac{4x}{1-x}\text{.})
(\dfrac{1}{1-4x}\text{.})
(\dfrac{x}{1+x}\text{.})
(\dfrac{3x}{(1+x)^2}\text{.})
(\dfrac{1+x+x^2}{(1-x)^2}) (Hint: multiplication).
Solution
(0, 4, 4, 4, 4, 4, \ldots\text{.})
(1, 4, 16, 64, 256, \ldots\text{.})
(0, 1, -1, 1, -1, 1, -1, \ldots\text{.})
(0, 3, -6, 9, -12, 15, -18, \ldots\text{.})
(1, 3, 6, 9, 12, 15, \ldots\text{.})
3
Show how you can get the generating function for the triangular numbers in three different ways:
Take two derivatives of the generating function for (1,1,1,1,1, \ldots)
Use differencing.
Multiply two known generating functions.
Solution
The second derivative of (\dfrac{1}{1-x}) is (\dfrac{2}{(1-x)^3}) which expands to (2 + 6x + 12x^2 + 20x^3 + 30x^4 + \cdots\text{.}) Dividing by 2 gives the generating function for the triangular numbers.
Compute (A - xA) and you get (1 + 2x + 3x^2 + 4x^3 + \cdots) which can be written as (\dfrac{1}{(1-x)^2}\text{.}) Solving for (A) gives the correct generating function.
The triangular numbers are the sum of the first (n) numbers (1,2,3,4, \ldots\text{.}) To get the sequence of partial sums, we multiply by (\frac{1}{1-x}\text{.}) So this gives the correct generating function again.
4
Use differencing to find the generating function for (4, 5, 7, 10, 14, 19, 25, \ldots\text{.})
Solution
Call the generating function (A\text{.}) Compute (A - xA = 4 + x + 2x^2 + 3x^3 + 4x^4 + \cdots\text{.}) Thus (A - xA = 4 + \dfrac{x}{(1-x)^2}\text{.}) Solving for (A) gives (\d\frac{4}{1-x} + \frac{x}{(1-x)^3}\text{.})
5
Find a generating function for the sequence with recurrence relation (a_n = 3a_{n-1} - a_{n-2}) with initial terms (a_0 = 1) and (a_1 = 5\text{.})
Solution
(\dfrac{1+2x}{1-3x + x^2}\text{.})
6
Use the recurrence relation for the Fibonacci numbers to find the generating function for the Fibonacci sequence.
Solution
Compute (A - xA - x^2A) and the solve for (A\text{.}) The generating function will be (\dfrac{x}{1-x-x^2}\text{.})
7
Use multiplication to find the generating function for the sequence of partial sums of Fibonacci numbers, (S_0, S_1, S_2, \ldots) where (S_0 = F_0\text{,}) (S_1 = F_0 + F_1\text{,}) (S_2 = F_0 + F_1 + F_2\text{,}) (S_3 = F_0 + F_1 + F_2 + F_3) and so on.
Solution
(\dfrac{x}{(1-x)(1-x-x^2)}\text{.})
8
Find the generating function for the sequence with closed formula (a_n = 2(5^n) + 7(-3)^n\text{.})
Solution
(\dfrac{2}{1-5x} + \dfrac{7}{1+3x}\text{.})
9
Find a closed formula for the (n)th term of the sequence with generating function (\dfrac{3x}{1-4x} + \dfrac{1}{1-x}\text{.})
Solution
(a_n = 3\cdot 4^{n-1} + 1\text{.})
10
Find (a_7) for the sequence with generating function (\dfrac{2}{(1-x)^2}\cdot\dfrac{x}{1-x-x^2}\text{.})
Solution
Hint: you should “multiply” the two sequences. Answer: 158.
11
Explain how we know that (\dfrac{1}{(1-x)^2}) is the generating function for (1, 2, 3, 4, \ldots\text{.})
Solution
Starting with (\frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots\text{,}) we can take derivatives of both sides, given (\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \cdots\text{.}) By the definition of generating functions, this says that (\frac{1}{(1-x)^2}) generates the sequence 1, 2, 3…. You can also find this using differencing or by multiplying.
12
Starting with the generating function for (1,2,3,4, \ldots\text{,}) find a generating function for each of the following sequences.
(1, 0, 2, 0, 3, 0, 4,\ldots\text{.})
(1, -2, 3, -4, 5, -6, \ldots\text{.})
(0, 3, 6, 9, 12, 15, 18, \ldots\text{.})
(0, 3, 9, 18, 30, 45, 63,\ldots\text{.}) (Hint: relate this sequence to the previous one.)
Solution
(\frac{1}{(1-x^2)^2}\text{.})
(\frac{1}{(1+x)^2}\text{.})
(\frac{3x}{(1-x)^2}\text{.})
(\frac{3x}{(1-x)^3}\text{.}) (partial sums).
13
You may assume that (1, 1, 2, 3, 5, 8,\ldots) has generating function (\dfrac{1}{1-x-x^2}) (because it does). Use this fact to find the sequence generated by each of the following generating functions.
(\frac{x^2}{1-x-x^2}\text{.})
(\frac{1}{1-x^2-x^4}\text{.})
(\frac{1}{1-3x-9x^2}\text{.})
(\frac{1}{(1-x-x^2)(1-x)}\text{.})
Solution
(0,0,1,1,2,3,5,8, \ldots\text{.})
(1, 0, 1, 0, 2, 0, 3, 0, 5, 0, 8, 0, \ldots\text{.})
(1, 3, 18, 81, 405, \ldots\text{.})
(1, 2, 4, 7, 12, 20, \ldots\text{.})
14
Find the generating function for the sequence (1, -2, 4, -8, 16, \ldots\text{.})
Solution
15
Find the generating function for the sequence (1, 1, 1, 2, 3, 4, 5, 6, \ldots\text{.})
Solution
(\frac{x^3}{(1-x)^2} + \frac{1}{1-x}\text{.})
16
Suppose (A) is the generating function for the sequence (3, 5, 9, 15, 23, 33, \ldots\text{.})
Find a generating function (in terms of (A)) for the sequence of differences between terms.
Write the sequence of differences between terms and find a generating function for it (without referencing (A)).
Use your answers to parts (a) and (b) to find the generating function for the original sequence.
Solution
((1-x)A = 3 + 2x + 4x^2 + 6x^3 + \cdots) which is almost right. We can fix it like this: (2 + 4x + 6x^2 + \cdots = \frac{(1-x)A - 3}{x}\text{.})
We know (2 + 4x + 6x^3 + \cdots = \frac{2}{(1-x)^2}\text{.})
(A = \frac{2x}{(1-x)^3} + \frac{3}{1-x} = \frac{3 -4x + 3x^2}{(1-x)^3}\text{.}) |
4143 | https://chemistry.stackexchange.com/questions/115016/how-to-calculate-the-equivalent-weight-of-kmno4 | equilibrium - How to calculate the equivalent weight of KMnO4? - Chemistry Stack Exchange
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How to calculate the equivalent weight of KMnO4?
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Let us consider the following reaction
2 K M n O X 4+16 H C l⟶2 K C l+2 M n C l X 2+8 H X 2 O+5 C l X 2 2 K M n O X 4+16 H C l⟶2 K C l+2 M n C l X 2+8 H X 2 O+5 C l X 2
Now, in order to calculate the equivalent mass of K M n O X 4 K M n O X 4, first I need to calculate it's n n-factor which turns out to be 5 5 because the oxidation state of M n M n in K M n O X 4 K M n O X 4 is +7+7 whereas in M n C l X 2 M n C l X 2 it is +2+2. And the n n-factor of H C l H C l is 1 1. So, the equivalent weight of K M n O X 4 K M n O X 4 is
molecular weight 5 molecular weight 5
So, ratio of moles of K M n O X 4 K M n O X 4 reacting with H C l H C l should be 1:5 1:5.
But from the balanced reaction we can see that this ratio is 1:8 1:8.
Where am I wrong? Is my understanding and hence calculation of n n-factor and equivalent weight is wrong?
(Please pardon this childish fashion of writing the question. If there is a problem anywhere, please do comment before flagging or downvoting.)
equilibrium
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edited May 9, 2019 at 7:29
andselisk♦
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asked May 8, 2019 at 17:20
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Hint - What are the two redox half-cell reactions?MaxW –MaxW 2019-05-08 17:29:14 +00:00 Commented May 8, 2019 at 17:29
@MaxW Ummm...I think 2 H X++2 C l X−⟶C l X 2+2 e X−+2 H X+2 H X++2 C l X−⟶C l X 2+2 e X−+2 H X+ and K X++M n O X 4 X−+5 e X−⟶M n X 2++K X++4 O X 2−K X++M n O X 4 X−+5 e X−⟶M n X 2++K X++4 O X 2− ami_ba –ami_ba 2019-05-08 17:31:13 +00:00 Commented May 8, 2019 at 17:31
You can leave off the 2 H X+2 H X+ and K=K= since they are on both sides of the equation.MaxW –MaxW 2019-05-08 17:33:26 +00:00 Commented May 8, 2019 at 17:33
@MaxW Are not these two the Oxidation and Reduction halves?ami_ba –ami_ba 2019-05-08 17:34:46 +00:00 Commented May 8, 2019 at 17:34
Yes, that is right. So per C l X−C l X− atom how many electrons are exchanged? Per M n M n atom how many electrons are exchanged?MaxW –MaxW 2019-05-08 17:36:04 +00:00 Commented May 8, 2019 at 17:36
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Refer to the definition of equivalent (IUPAC Gold Book):
equivalent entity
Entity corresponding to the transfer of a H X+H X+ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.
In other words, 1 e q u i v 1 e q u i v is the amount of substance reacting with 1 m o l 1 m o l of hydrogen atom. If hydrogen acts as a reducing or oxidizing agent, then either way 1 m o l 1 m o l of hydrogen atoms liberates or accepts 1 m o l 1 m o l of electrons:
0.5 H X 2 0.5 H X 2+e X−⟶H X++e X−⟶H X−0.5 H X 2⟶H X++e X−0.5 H X 2+e X−⟶H X−
That's why an equivalent of a redox agent is its amount which liberates or accepts 1 m o l 1 m o l of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.
Note that K M n O X 4 K M n O X 4 participates in redox reaction
2 K M n+7 O X 4+16 H C l−1=2 K C l+2 M n+2 C l X 2+8 H X 2 O+5 C l 0 X 2 2 K M n+7 O X 4+16 H C l−1=2 K C l+2 M n+2 C l X 2+8 H X 2 O+5 C l 0 X 2
and the half-reaction for the reduction of manganese is
M n+7 O X 4 X−+8 H X++5 e X−⟶M n+2 X 2++4 H X 2 O M n+7 O X 4 X−+8 H X++5 e X−⟶M n+2 X 2++4 H X 2 O
Also note that n n-factor of permangante here is not the number of protons H X+H X+, which are also used up in water formation; rather it's the number of transferred electrons, e.g. n=5 n=5 and
M e q u i v(K M n O X 4)=M(K M n O X 4)n=158.03 g m o l−1 5=31.61 g m o l−1 M e q u i v(K M n O X 4)=M(K M n O X 4)n=158.03 g m o l−1 5=31.61 g m o l−1
Obviously, the equivalent mass of permanganate isn't a constant and depends on the p H p H of the reaction. For example, in neutral medium half-reaction appears as
M n+7 O X 4 X−+2 H X 2 O+3 e X−⟶M n+4 O X 2+4 O H X−M n+7 O X 4 X−+2 H X 2 O+3 e X−⟶M n+4 O X 2+4 O H X−
and there is no explicitly shown protons to count at all! However, the n n-factor is 3 3, and the equivalent mass of permanganate would be a different value:
M e q u i v(K M n O X 4)=M(K M n O X 4)n=158.03 g m o l−1 3=52.68 g m o l−1 M e q u i v(K M n O X 4)=M(K M n O X 4)n=158.03 g m o l−1 3=52.68 g m o l−1
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edited Jun 10, 2020 at 14:04
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answered May 9, 2019 at 7:27
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Asymptotics
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Multiplicative Number Theory
Tech Notes
Symbolic Mathematics: Basic Operations
Power Series
Making Power Series Expansions
Operations on Power Series
The Representation of Power Series
Converting Power Series to Normal Expressions
Implementation notes: Algebra and Calculus
See Also
SeriesCoefficient
InverseSeries
ComposeSeries
Limit
FunctionAnalytic
Normal
InverseZTransform
RSolve
O
SeriesData
PadeApproximant
FourierSeries
AroundReplace
Asymptotic
AsymptoticDSolveValue
AsymptoticIntegrate
AsymptoticSum
AsymptoticRSolveValue
AsymptoticSolve
ResidueSum
Related Guides
Series Expansions
Calculus
Manipulating Equations
Scientific Models
Asymptotics
Analytic Number Theory
Multiplicative Number Theory
Tech Notes
Symbolic Mathematics: Basic Operations
Power Series
Making Power Series Expansions
Operations on Power Series
The Representation of Power Series
Converting Power Series to Normal Expressions
Implementation notes: Algebra and Calculus
Series[f,{x,x0,n}]
generates a power series expansion for f about the point x=x0 to order (x-x0)n, where n is an explicit integer.
Series[f,xx0]
generates the leading term of a power series expansion for f about the point x=x0.
Series[f,{x,x0,nx},{y,y0,ny},…]
successively finds series expansions with respect to x, then y, etc.
Details and Options
Examples
Basic Examples
Scope
Univariate Series
Generalizations & Extensions
Options
Analytic
Assumptions
Applications
Properties & Relations
Possible Issues
See Also
Tech Notes
Related Guides
Related Links
History
Cite this Page
Series Copy to clipboard. ✖ Series
Series[f,{x,x0,n}]
Copy to clipboard.
✖
Series[f,{x,x0,n}]
generates a power series expansion for f about the point x=x0 to order (x-x0)n, where n is an explicit integer.
Series[f,xx0]
Copy to clipboard.
✖
Series[f,xx0]
generates the leading term of a power series expansion for f about the point x=x0.
Series[f,{x,x0,nx},{y,y0,ny},…]
Copy to clipboard.
✖
Series[f,{x,x0,nx},{y,y0,ny},…]
successively finds series expansions with respect to x, then y, etc.
Details and Options
Series can construct standard Taylor series, as well as certain expansions involving negative powers, fractional powers, and logarithms.
Series detects certain essential singularities. On[Series::esss] makes Series generate a message in this case.
Series can expand about the point x=∞.
Series[f,{x,0,n}] constructs Taylor series for any function f according to the formula .
Series effectively evaluates partial derivatives using D. It assumes that different variables are independent.
The result of Series is usually a SeriesData object, which you can manipulate with other functions.
Normal[series] truncates a power series and converts it to a normal expression.
SeriesCoefficient[series,n] finds the coefficient of the n-order term.
The following options can be given:
| | Analytic | True | whether to treat unrecognized functions as analytic |
| | Assumptions | $Assumptions | assumptions to make about parameters |
| | SeriesTermGoal | Automatic | number of terms in the approximation |
Examples
open all
close all
Basic Examples (4)Summary of the most common use cases
Power series for the exponential function around :
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Convert to a normal expression:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Power series of an arbitrary function around :
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
In any operation on series, only appropriate terms are kept:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Find the leading term of a power series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Scope (10)Survey of the scope of standard use cases
Univariate Series (10)
Series can handle fractional powers and logarithms:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Symbolic parameters can often be used:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Laurent series with negative powers can be generated:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Truncate the series to the specified negative power:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Find power series for special functions:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Copy to clipboard.
In:=4
✖
Direct link to example
Out=4
Copy to clipboard.
In:=5
✖
Direct link to example
Out=5
Copy to clipboard.
In:=6
✖
Direct link to example
Out=6
Find the series for a function at a branch point:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
With x assumed to be to the left of the branch point, a simpler result is given:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Piecewise functions:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Power series at infinity:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Series can give asymptotic series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Series expansions of implicit solutions to equations:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Series expansions of unevaluated integrals:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Generalizations & Extensions (4)Generalized and extended use cases
Power series in two variables:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Series is threaded element-wise over lists:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Series generates SeriesData expressions:
Copy to clipboard.
In:=1
✖
Direct link to example
Series can work with approximate numbers:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Options (4)Common values & functionality for each option
Analytic (1)
Series by default assumes symbolic functions to be analytic:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Assumptions (3)
Use Assumptions to specify regions in the complex plane where expansions should apply:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Without assumptions, piecewise functions appear:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Get expansions in Stokes regions:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Applications (8)Sample problems that can be solved with this function
Plot successive series approximations to :
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Find a series expansion for a standard combinatorial problem:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Find Fibonacci numbers from a generating function:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Find Legendre polynomials by expanding a generating function:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Set up a generating function to enumerate ways to make change using U.S. coins:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
The number of ways to make change for $1:
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Find the lowest-order terms in a large polynomial:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Find higher-order terms in Newton's approximation for a root of f[x] near :
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Plot the complex zeros for a series approximation to Exp[x]:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Properties & Relations (10)Properties of the function, and connections to other functions
Series always only keeps terms up to the specified order:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Operations on series keep only the appropriate terms:
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Normal converts to an ordinary polynomial:
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Any mathematical function can be applied to a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Adding a series of lower order causes the higher-order terms to be dropped:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Differentiate a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Solve equations for series coefficients:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Find the list of coefficients in a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Use O[x] to force the construction of a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
ComposeSeries treats a series as a function to apply to another series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
InverseSeries does series reversion to find the series for the inverse function of a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Use FunctionAnalytic to test whether a function is analytic:
Copy to clipboard.
In:=1
✖
Direct link to example
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
An analytic function can be expressed as a Taylor series at each point of its domain:
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
The resulting polynomial approximates near 0:
Copy to clipboard.
In:=4
✖
Direct link to example
Out=4
Possible Issues (7)Common pitfalls and unexpected behavior
When there is an essential singularity, Series will attempt to factor it out:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Numeric values cannot be substituted directly for the expansion variable in a series:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Use Normal to get a normal expression in which the substitution can be done:
Copy to clipboard.
In:=3
✖
Direct link to example
Out=3
Series must be converted to normal expressions before being plotted:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Power series with different expansion points cannot be combined:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Not all series are represented by expressions with head SeriesData:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Some functions cannot be decomposed into series of power-like functions:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Series does not change expressions independent of the expansion variable:
Copy to clipboard.
In:=1
✖
Direct link to example
Out=1
Copy to clipboard.
In:=2
✖
Direct link to example
Out=2
Wolfram Research (1988), Series, Wolfram Language function, (updated 2020).
Copy to clipboard.
✖
Wolfram Research (1988), Series, Wolfram Language function, (updated 2020).
Text
Wolfram Research (1988), Series, Wolfram Language function, (updated 2020).
Copy to clipboard.
✖
Wolfram Research (1988), Series, Wolfram Language function, (updated 2020).
CMS
Wolfram Language. 1988. "Series." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2020.
Copy to clipboard.
✖
Wolfram Language. 1988. "Series." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2020.
APA
Wolfram Language. (1988). Series. Wolfram Language & System Documentation Center. Retrieved from
Copy to clipboard.
✖
Wolfram Language. (1988). Series. Wolfram Language & System Documentation Center. Retrieved from
BibTeX
@misc{reference.wolfram_2025_series, author="Wolfram Research", title="{Series}", year="2020", howpublished="\url{ note=[Accessed: 24-September-2025]}
Copy to clipboard.
✖
@misc{reference.wolfram_2025_series, author="Wolfram Research", title="{Series}", year="2020", howpublished="\url{ note=[Accessed: 24-September-2025]}
BibLaTeX
@online{reference.wolfram_2025_series, organization={Wolfram Research}, title={Series}, year={2020}, url={ note=[Accessed: 24-September-2025]}
Copy to clipboard.
✖
@online{reference.wolfram_2025_series, organization={Wolfram Research}, title={Series}, year={2020}, url={ note=[Accessed: 24-September-2025]}
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Interpolating polynomial with integer coefficients
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Asked 9 years, 11 months ago
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Let n 1<⋯<n k n 1<⋯<n k and m 1,…,m k m 1,…,m k be integers. Can we find conditions for the existence polynomial p p with integer coefficients such that p(n i)=m i p(n i)=m i for every i i?
This isn't always possible: for example, since p(i)≡p(0)mod i p(i)≡p(0)mod i, we cannot have p(0)=0 p(0)=0 and p(2)=1 p(2)=1.
By this answer, if such a polynomial exists then the Lagrange interpolating polynomial has integer coefficients, so the question may be restated as when does the Lagrange interpolating polynomial have integer coefficients.
elementary-number-theory
lagrange-interpolation
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edited Apr 13, 2017 at 12:58
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asked Oct 7, 2015 at 0:31
Oliver MillerOliver Miller
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The coefficients of the Lagrange polynomial that goes through (n 1,m 1),…,(n k,m k)(n 1,m 1),…,(n k,m k), can be found by solving
⎛⎝⎜⎜1⋮1 n 1 n k⋯⋯n k−1 1 n k−1 k⎞⎠⎟⎟⎛⎝⎜⎜a 0⋮a n−1⎞⎠⎟⎟=⎛⎝⎜⎜m 1⋮m k⎞⎠⎟⎟(1 n 1⋯n 1 k−1⋮1 n k⋯n k k−1)(a 0⋮a n−1)=(m 1⋮m k)
Hence the coefficients a i a i are integers precisely when the determinant of the Vandermonde matrix above divides its adjugate matrix applied to (m 1,…,m k)(m 1,…,m k).
For example, for n 1<n 2<n 3 n 1<n 2<n 3 and m 1,m 2,m 3 m 1,m 2,m 3, the condition becomes
(n 1−n 2)(n 1−n 3)(n 2−n 3)|⎛⎝⎜n 2(n 2−n 3)n 3 m 1+n 1 n 3(n 3−n 1)m 2+n 1(n 1−n 2)n 2 m 3 n 2 3(m 1−m 2)+n 2 1(m 2−m 3)+n 2 2(m 3−m 1)n 3(m 2−m 1)+n 2(m 1−m 3)+n 1(m 3−m 2)⎞⎠⎟(n 1−n 2)(n 1−n 3)(n 2−n 3)|(n 2(n 2−n 3)n 3 m 1+n 1 n 3(n 3−n 1)m 2+n 1(n 1−n 2)n 2 m 3 n 3 2(m 1−m 2)+n 1 2(m 2−m 3)+n 2 2(m 3−m 1)n 3(m 2−m 1)+n 2(m 1−m 3)+n 1(m 3−m 2))
(by which is meant that the left side divides each of the terms on the right).
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edited Jan 9, 2020 at 13:07
answered Jan 9, 2020 at 12:23
ChrystomathChrystomath
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I don’t have enough reputation to comment, this is not a complete answer.
I have a necessary and sufficient condition for consecutive ni, which is that the terms of the nth level difference are divisible by n! (We define the kth level difference to be the differences between consecutive terms of the (k-1)th level difference, with the 0th level difference being the sequence itself
Ex:
2nd level differences: 2 2
1st level differences: 2 4 6
The sequence itself: 0 2 6 12
This is proven by observing that any integer polynomial can be written as the sum of integer combinations of the polynomials x(x-1)(x-2)...(x-i)
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answered Dec 11, 2019 at 2:56
user105362user105362
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4147 | https://www.quora.com/What-is-the-identity-of-sin-2x-2sin-x-cos-x | Something went wrong. Wait a moment and try again.
Double Angle Formula
Cos Function
Mathematical Identities
Trigonometric Equations
Sin Function
Trigonometry Maths
Cos (mathematics)
5
What is the identity of sin(2x) = 2sin(x) cos(x)?
Jonathan Lewis
Masters in Mathematics & Theoretical Physics, University of Cambridge (Graduated 1988)
·
Author has 13.3K answers and 1.6M answer views
·
1y
Using polar coordinates,
eix=cosx+isinx
sinx=eix−e−ix2i
cosx=eix+e−ix2
2sinxcosx=2(eix−e−ix2i)(eix+e−ix2)=e2ix−e−2ix2i
=sin 2x
Related questions
Why is sin 2x ≠ 2sin x?
What are the non-permissible values for this identity cos(2x) = 2sin(x) cos(x)?
Can someone help verify the identity Sin x sin 2x + cos x cos 2x= cos x?
How can I verify this identity: sin (2x) - tan (x) = tan(x) cos(2x)?
How can I prove the identity, (sin x - cos x) ² = 1 - sin 2x?
KK
Works at Quora (product)
·
Upvoted by
Shubhankar Datta
, Master of Science Mathematics, Jadavpur University (2022) · Author has 145 answers and 516.5K answer views
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
You can prove it using Euler’s formula (eiθ=cosθ+isinθ) and the identity (a+b)(a−b)=a2−b2:
2sinθcosθ
=2eiθ−e−iθ2ieiθ+e−iθ2
=ei2θ−e−i2θ2i
=sin2θ
You can also prove it in an elegant way using plane geometry (with a unit circle):
we have use:
base angles of an isosceles triangle
exterior angle of a triangle equals the sum of two opposite angles
definitions of sine and cosine
You can prove it using Euler’s formula (eiθ=cosθ+isinθ) and the identity (a+b)(a−b)=a2−b2:
2sinθcosθ
=2eiθ−e−iθ2ieiθ+e−iθ2
=ei2θ−e−i2θ2i
=sin2θ
You can also prove it in an elegant way using plane geometry (with a unit circle):
we have use:
base angles of an isosceles triangle
exterior angle of a triangle equals the sum of two opposite angles
definitions of sine and cosine
Anushka Solanki
Studied 10th Class in India at Delhi Public School, Nagpur
·
2y
The identity is:
sin(2x) = 2sin(x)cos(x)
This is known as the double angle identity for sine, because it expresses the sine of twice an angle (2x) in terms of the sine and cosine of the angle (x).
Shivam Kadam
Former Intern at Tata Motors (company) (2018–2018)
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
Consider an isosceles triangle with equal sides 1 and 2x the angle between them. Find the area in two ways and equate them.
First way: drop a perpendicular from one of the equal angles to the opposite side of length 1. That perpendicular has length 1sin(2x) and is the altitude of the isosceles triangle. Its base is 1. Therefore its area is (1/2)sin(2x)1.
Second way: bisect the angle between the two sides of length 1, extending the angle bisector to meet the opposite side. This splits the triangle into two right triangles, each with base half the opposite side, that is 1sin(x) and altitude th
Consider an isosceles triangle with equal sides 1 and 2x the angle between them. Find the area in two ways and equate them.
First way: drop a perpendicular from one of the equal angles to the opposite side of length 1. That perpendicular has length 1sin(2x) and is the altitude of the isosceles triangle. Its base is 1. Therefore its area is (1/2)sin(2x)1.
Second way: bisect the angle between the two sides of length 1, extending the angle bisector to meet the opposite side. This splits the triangle into two right triangles, each with base half the opposite side, that is 1sin(x) and altitude the length of the angle bisector, 1cos(x). Those right triangles therefore each have area (1/2)sin(x)cos(x) so adding the areas together gives area of the isosceles triangle as sin(x)cos(x).
Equate the areas:
(1/2)sin(2x)1 = sin(x)cos(x), multiply by 2:
sin(2x) = 2sin(x)cos(x).
Alternative method,
sin(A + B) = sin A cos B + cos A sin B
sin(2x) = sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x
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Related questions
Can you explain the difference between sin(2x) and 2sin(x) cos(x)?
What is the answer for sin(2x)-cos(2x)-sin(x) +cos(x) =0?
How do you prove 2sin(x) cos(x) =sin(2x)?
Can someone prove the identity, (sin t - cos t) ² = 1 - sin 2t?
How do I prove
sin
2
x
=
2
sin
x
cos
x
?
Kyle Weatherly
Still in High school, so I have no credentials yet, but math is one hobby I have
·
Author has 110 answers and 166.9K answer views
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
Say that we have cos(a-b), c=a-b, and the distance between (cos(c),sin(c)) and (1,0) is the same as the distance between (cos(a),sin(a)) and (cos(b),sin(b)). Then, if we plug these both into the distance formula, we get sqrt((cos(c)-1)^2+(sin(c))^2)=sqrt((cos(a)-cos(b))^2+(sin(a)+sin(b))^2). We can then square both sides to simplify the expression (we can do this because we are adding two positive numbers together in both radicals because we have an real number squared plus a real number squared) and expand the squared expressions and rearrange the expressions to get: cos^2(c)+sin^2(c)-2cos(c)
Say that we have cos(a-b), c=a-b, and the distance between (cos(c),sin(c)) and (1,0) is the same as the distance between (cos(a),sin(a)) and (cos(b),sin(b)). Then, if we plug these both into the distance formula, we get sqrt((cos(c)-1)^2+(sin(c))^2)=sqrt((cos(a)-cos(b))^2+(sin(a)+sin(b))^2). We can then square both sides to simplify the expression (we can do this because we are adding two positive numbers together in both radicals because we have an real number squared plus a real number squared) and expand the squared expressions and rearrange the expressions to get: cos^2(c)+sin^2(c)-2cos(c)+1= cos^2(a)+sin^2(a)+cos^2(b)+sin^2(b)-2cos(a)cos(b)-2sin(a)sin(b). Because sin^2(x)+cos^2(x)=1, we simplify this to: 1–2cos(c)+1=1+1-2cos(a)cos(b)-2sin(a)sin(b). Then, some easy algebra will tell us cos(c)=cos(a)cos(b)+sin(a)sin(b), but we defined c as a-b, so cos(a-b)=cos(a)cos(b)+sin(a)sin(b). Finally, because cos((pi/2)-x)=sin(x), we can write the equation sin(f+g)=cos((pi/2)-(f+g)), which can be rewritten as sin(f+g)=cos(((pi/2)-f)-g), and we already have a way to simplify cos(a-b), so we set a=(pi/2)-f and b=g to get sin(f+g)=cos(((pi/2)-f)-g)=cos((pi/2)-f)cos(g)+sin((pi/2)-f)sin(g) which again simplifies to sin(f+g)=cos(((pi/2)-f)-g)=sin(f)cos(g)+cos(f)sin(g), so sin(f+g)=sin(f)cos(g)+cos(f)sin(g). Given this, if we set both f and g equal to x, we get sin(x+x)=sin(x)cos(x)+cos(x)sin(x), which is the same as saying sin(2x)=2sin(x)cos(x).
Varsha KV
Student at RV College of Architecture (2018–present)
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
w.k.t.sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
Let y=x then,
sin(x+x)=sin(x)cos(x)+cos(x)sin(x)
sin(2x)=2sin(x)cos(x)
Hence it is true
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Akhil Dubey
MBA from Indian Institute of Management Bangalore (Graduated 2024)
·
Author has 108 answers and 815.5K answer views
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
By starting from R . H . S .
sin(2x) = sin( x + x )
now as we know sin( a + b ) = sin(a)cos(b) + cos(a)sin(b)
=> sin(2x) = sin(x)cos(x) + sin(x)cos(x)
=> sin(2x) = 2sin(x)cos(x)
Siddharth Bhandarkar
Former Junior Engineer
·
Author has 214 answers and 77.9K answer views
·
6y
Originally Answered: Why does 2sin(x) cos(x) = sin(2x)?
·
It is from the formula of sin(A+B)=sinA.cosB+cosA.sinB . In this case both angles being the same gives us that as a result
Promoted by Almedia
Charlee Anthony
Go-to Resource for Realistic, Side Hustle Ideas
·
Sep 22
How can I make an extra $200 a week online?
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How I consistently make $200 a week online
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signed up for free - got a $5 welcome bonus instantly just right after I started my first offer.
played games and completed offers
This one good method to make extra income brought me over $3,000, and I still make $150–$200 every single week from Freecash. As a gig hunter, this app came natural to me.
I hacked this Freecash platform so you don't have to. These GPT platforms can be hell to learn, so learn my top methods, rent free.👇
How I consistently make $200 a week online
Here’s exactly how I made $200 a week online with zero upfront investment. And yeah, it may have involved 10 hours a week 😂
signed up for free - got a $5 welcome bonus instantly just right after I started my first offer.
played games and completed offers - chose the highest-paying tasks first like monopoly GO and raid shadow legends.
made an in-app purchase - spent $3 on a power-up in one mobile game, which unlocked a $25 reward.
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Ajesh K C
Studied at Mar Athanasius College of Engineering, Kothamangalam
·
Author has 1.3K answers and 3.3M answer views
·
6y
Originally Answered: Why does 2sin(x) cos(x) = sin(2x)?
·
Sin (x + y) = Sin (x) cos(y) + cos(x) sin(y)
Sin (2x) =Sin (x +x)
= Sin (x) cos(x) + cos(x) sin(x)
= 2 Sin (x) cos(x)
Gregory Schoenmakers
I won't crack any jokes about knowing the angles.
·
Author has 4.4K answers and 8.5M answer views
·
8y
Originally Answered: Why is 2\sin(x)\cos(x)=\sin(2x) true?
·
It is a simple matter to show on a unit circle that:
cos2A+sin2A=1
cos(−A)=cosA
sin(−A)=−sinA
cos(A−B)=cosAcosB+sinAsinB
From these 4 basic relationships, all other compound angle formulas can be derived.
PD TechMath
Studied Pure Mathematics & Applied Mathematics ( Major) (Graduated 2017)
·
Author has 140 answers and 24.6K answer views
·
2y
We know that sin(A+B)=sinAcosB+cosAsinB
Here A=x and B=x
MATHEMATICS BADSHAH
Class 1, 2, 3, 4, 5, 6,7, 8, 9, 10, 11, 12, BSc. 1
·
Author has 79 answers and 22.3K answer views
·
2y
Sin2x= sin(x+x)
= Sinx.cosx + cosx. Sinx
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
8y
What is the proof that cos (x + y) = cos (x) cos (y) − sin (x) sin (y)?
This proof is so lovely and simple that I thought I must share it.
We need to prove sin(A...
Related questions
Why is sin 2x ≠ 2sin x?
What are the non-permissible values for this identity cos(2x) = 2sin(x) cos(x)?
Can someone help verify the identity Sin x sin 2x + cos x cos 2x= cos x?
How can I verify this identity: sin (2x) - tan (x) = tan(x) cos(2x)?
How can I prove the identity, (sin x - cos x) ² = 1 - sin 2x?
Can you explain the difference between sin(2x) and 2sin(x) cos(x)?
What is the answer for sin(2x)-cos(2x)-sin(x) +cos(x) =0?
How do you prove 2sin(x) cos(x) =sin(2x)?
Can someone prove the identity, (sin t - cos t) ² = 1 - sin 2t?
How do I prove
sin
2
x
=
2
sin
x
cos
x
?
How can I verify the identity of sin (x-3pi/2) + sin (pi/2+x) =2 cos x?
How does one prove
sin
(
2
x
)
sin
(
x
)
−
cos
(
2
x
)
cos
(
x
)
=
sec
(
x
)
?
What is the identity sin^2(x) + cos^2 (x) = 1 for all values of x?
How do I prove the identity sin x tan x = sec x - cos x?
How do you prove that sin(x) +sin(2x) +sin(3x) = -4sin(x) cos(x/2) sin (3x/2)?
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4148 | https://www.bbc.co.uk/learningenglish/features/prepositions_with_georgie/ep-240503 | BBC Learning English - Prepositions with Georgie / Prepositions of time
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Prepositions with Georgie
Intermediate level
Prepositions of time
Episode 240503 / 03 May 2024
Free worksheet
Watch all episodes
Prepositions of time
To talk about times of the day, the most common prepositions are in, on and at.
‘At’ is for exact times, like:
Clock times
I'll see youat7pm.
Exact times of the day (e.g. sunset, sunrise, dawn, dusk, bedtime, dinnertime, midnight)
The birds singatsunrise.
Exception: night (this is more of a duration than one exact time)
The bats come out at night.
‘In’ is for longer periods, like:
Months
My birthday isinNovember.
Years
I graduatedin2017.
Seasons
It usually snows hereinwinter.
Centuries
My parents were borninthe 60s.
‘In’ is also for sections of the day (exclude night), like:
The morning
It's too earlyinthe morning for talking.
The afternoon
We can go shopping tomorrowinthe afternoon.
The evening
Mum's coming to visitinthe evening.
‘On’ is for shorter periods, like:
Days
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4149 | https://aspe.org/pipeline/hazen-williams-and-darcy-weisbach-equations-similarities-and-differences/ | Hazen-Williams and Darcy-Weisbach Equations: Similarities and Differences - ASPE Pipeline
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Hazen-Williams and Darcy-Weisbach Equations: Similarities and Differences
September 5, 2023
Which equation is better? Here’s a brief comparison of their use in sprinkler piping sizing.
by Yasser Amer Ahmed, CPD, LEED AP BD+C, CFPS, MIFireE, PMSFPE
The Hazen-Williams and Darcy-Weisbach equations are both used to calculate head loss due to friction in fluid flow. The Hazen-Williams equation is commonly used for water flow through pipes, while the Darcy-Weisbach equation is a more versatile and accurate method for calculating head loss due to friction in fluid flow.
The Hazen-Williams equation is easier to use than the Darcy-Weisbach equation, as it doesn’t require iterative calculations. It estimates friction loss in a pipeline with just a few parameters. However, this equation is only applicable to water flow and cannot be used for other fluids. It also is less accurate for larger pipes and higher flow velocities.
On the other hand, the Darcy-Weisbach equation can be used for various fluids, not just water. It provides more accurate results for a wide range of pipe sizes and flow velocities. However, the equation is more complex, requiring iterative calculations and additional parameters.
Values of the friction loss coefficient, C, used in the Hazen-Williams equation are accurate only if the flow velocity is close to that at which the value of C was measured.
It is a matter of judgment as to what velocity is “too high” for the Hazen-Williams equation. American Water Works Association (AWWA) data list C factors measured at a velocity of 3 feet per second (fps) (0.9 m/s), yet it is standard practice in sprinkler calculations to accept velocities in sprinkler piping between 10 and 20 fps (3.05 and 6.1 m/s). Similarly, the tables of equivalent lengths for fittings and valves used by sprinkler system designers are based on fittings and valve types typical of sprinkler systems. Water mist systems may incorporate different types of fittings and valves for which the equivalent length values based on Hazen-Williams commonly accepted for sprinkler piping will be incorrect.
The Hazen-Williams (H-W) equation contains no terms that allow you to change the temperature, density, or viscosity of the liquid. It assumes that the water contains no additives and is close to 60°F (15.6°C). The Darcy-Weisbach (D-W) equation does allow for variation in fluid properties, and for that reason, it is widely used by mechanical engineers designing fluid systems.
NFPA 750: Standard on Water Mist Fire Protection Systems requires the use of the D-W equation if fluid velocity exceeds 25 fps (7.6 m/s). The limiting velocity was selected after a review of calculations comparing the agreement between H-W and DW calculations and measured head losses in 1- to 3-inch (25- to 80-mm) diameter black steel pipe over a range of velocities.
Finally, the Hazen-Williams formula comes within 10 percent accuracy of the value obtained using the Darcy-Weisbach equation, with significantly less effort. This is considered to be within the range of acceptable values in an engineering exercise involving fire protection system piping.
About the Author
Yasser Amer Ahmed is an accomplished Egyptian Fire Protection and Plumbing Specialist with 30+ years of international experience at Dar Al-Handasah Consultants, an internationally ranked consulting firm providing planning, design, management, and consultancy. As a Fire Protection & Plumbing Specialist, he excels in managing diverse projects and teams, with expertise in strategic leadership, project management, and engineering, and has strong and diversified experience in the design and coordination of plumbing and fire protection systems for mixed-use complexes, airports, educational, healthcare, residential, commercial, high-rise, and governmental buildings. Yasser graduated from Ain Shams University with a bachelor’s degree in Civil Engineering, Public Works Dept., in 1991, and he received a post-graduate diploma in Sanitary Engineering from Ain Shams University in 1996 and an MBA from Brookline Business School (New York) in August 2020. Yasser is a Certified Fire Protection Specialist (CFPS) since December 2016, Certified in Plumbing Design (CPD) since April 2016, and a LEED AP BD+C since April 2014. He is a member of the National Fire Protection Association (NFPA), ASPE, the Society of Fire Protection Engineers (SFPE), the Institution of Fire Engineers (MIFireE), the Egyptian Engineers Syndicate, and the Egyptian Society of Civil Engineers, and he is authorized as a Fire Protection Consultant Engineer from Egyptian Engineers Syndicate.
The opinions expressed in this article are those of the author and not the American Society of Plumbing Engineers.
References: SFPE Handbook of Fire Protection, 5th edition, and Fire Protection Handbook, 20th edition
plumbing design
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4150 | https://math.stackexchange.com/questions/2446475/splitting-a-summation-into-even-and-odd-components | Splitting a summation into even and odd components - Mathematics Stack Exchange
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Splitting a summation into even and odd components
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I'm having trouble understanding how the the second line of the splitting of this summation works. I don't understand how the sums of each individual parts can still run from n = 0 to infinity without over-counting and making the sum larger.
Definition of the matrix's properties
Splitting of summation into even/odd parts (trouble understanding bridge from line 1 to 2)
Thanks for any help.
summation
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asked Sep 26, 2017 at 19:35
falsovuotofalsovuoto
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If I write 2 2 n+2 2 n+1 2 2 n+2 2 n+1 for n=0…4 n=0…4 you have 2 0+2 1+2 2+2 3+2 4+2 5+2 6+2 7+2 8+2 9 2 0+2 1+2 2+2 3+2 4+2 5+2 6+2 7+2 8+2 9 No overcounting as you can see Raffaele –Raffaele 2017-09-26 19:42:36 +00:00 Commented Sep 26, 2017 at 19:42
Ah, but of course – thank you!falsovuoto –falsovuoto 2017-09-26 19:52:05 +00:00 Commented Sep 26, 2017 at 19:52
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The reason you are not overcounting is that the exponents in each of the split sums are advancing by 2 2 for increase in n n or 1 1. So if you went up to any particular (even) upper limit of the exponent, you would have the same number of terms in the sum of the split sums and the original sum.
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answered Sep 26, 2017 at 19:39
Mark FischlerMark Fischler
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4151 | https://ocw.mit.edu/courses/6-451-principles-of-digital-communication-ii-spring-2005/resources/lecture-5-introduction-to-binary-block-codes/ | Lecture 5: Introduction to Binary Block Codes | Principles of Digital Communication II | Electrical Engineering and Computer Science | MIT OpenCourseWare
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0:00 PROFESSOR: Chapter six, problem set three, problem set
0:04 two solutions.
3:27 OK.
3:28 Good morning.
3:29 Nice to see you all again on this mild day.
3:32 I want to start off by thanking Ashish greatly for
3:37 giving the last three lectures.
3:39 I have very high confidence in Ashish, and I'm sure he give
3:42 you excellent lectures and was able to answer all your
3:45 questions, right?
3:48 But if you feel dissatisfied at any point, you'll have an
3:52 opportunity to ask the question again today.
3:57 We will continue now in chapter six.
3:59 I understand Ashish got up to section six point one already,
4:04 so that makes my life a little easier.
4:07 And we'll just continue to plow ahead through the notes.
4:09 You have as handouts today chapter six, the outgoing
4:16 problem set, the solutions to the incoming problem set.
4:21 So does anyone have any comments or questions or
4:26 suggestions as we go forward at this point?
4:30 Let's say I'll do a very quick review of everything so far,
4:34 at least what we need to proceed, so you'll have a
4:37 chance to ask questions along there if you like.
4:40 Anyone have anything to say?
4:43 Anyone want to say thanks to Ashish?
4:48 All right.
4:49 Great.
4:51 OK.
4:52 So a very quick review.
4:57 We first said we were going to work with a continuous time
5:02 Additive White Gaussian noise channel, but we quickly
5:05 reduced it to a discrete time additive white
5:08 Gaussian noise model.
5:10 Vector model, Y equals X plus N, where the vectors may go on
5:15 as long as you like.
5:16 And implicit in here is we have some kind of power
5:19 limitation expressed by various --
5:21 I know I've probably introduced too many parameters
5:24 in the early part of the course.
5:27 But you'll see them all in the literature, and I thought you
5:31 might as well see the relation between them all at once.
5:36 The noise.
5:36 The main thing about this is it's an independent,
5:40 identically distributed noise, also independent of X, and
5:44 it's characterized simply by its variance per symbol, or
5:49 average power.
5:49 Again, there are various notations that we
5:54 can use for its power.
5:57 OK.
5:58 However you express the power of the signal of the noise,
6:03 you're going to get some signal-to-noise ratio out of
6:05 that, as long as you use consistent the same units for
6:09 signal and noise.
6:10 And so the key parameters of this channel turn
6:13 out to be just two.
6:16 One is the signal-to-noise ratio.
6:19 And the other is something which in the discrete time
6:23 domain we might think of as a data rate, rho, the number of
6:26 bits per two dimensions.
6:28 But we were at pains to show --
6:32 and the reason we measure it in bits per two dimensions is
6:35 that it converts directly to the spectral efficiency, or we
6:40 sometimes say, the nominal spectral efficiency, or the
6:42 Nyquist spectral efficiency.
6:45 Basically, this is the spectral efficiency if you
6:48 assume that you only use the nominal Nyquist bandwidth.
6:52 And both in 450 and here it was shown that you can really
6:57 get as close to the nominal bandwidth as you like, its
7:03 sharp or roll off as you like, by using filters.
7:08 Very sharp filters or very sharp digital filters that
7:15 today we can really realistically get effectively
7:20 to the Nyquist bandwidth.
7:23 So this is a good measure of nominal spectral efficiency,
7:27 which is an extremely important parameter, the
7:29 spectral efficiency in continuous time.
7:33 And then the Shannon limit is expressed in this very simple
7:36 form that we can never do better in data rate or nominal
7:44 spectral efficiency than log of 1 plus SNR.
7:48 And the units are bits per two dimensions.
7:51 So there's an awful lot about parameters and units in the
7:55 first part of the course, and sometimes it's a little
7:57 overwhelming and confusing.
8:01 But in my experience in engineering is, it's extremely
8:05 important to get the units right.
8:08 It helps you in thinking clearly.
8:11 It helps you to focus on the right things.
8:14 And so in the first couple of chapters, there's a lot of
8:16 emphasis on getting the units right.
8:18 For instance, bits per two dimensions.
8:21 I could simply assert to you that things are always to come
8:25 out better if you do things per two dimensions.
8:28 This basically has to do with Gaussian distributions being
8:34 simpler in two dimensions than they are in one dimension.
8:37 So you know you can get closed form integrals in two and four
8:41 and so forth dimensions that you can't get in one and three
8:44 and so forth dimensions.
8:45 So you know, the geometrical formulas for the volume of the
8:50 sphere and so forth are much simpler in even dimensions
8:53 than they are in odd dimensions.
8:55 This all, I think, has to do with the fact that really a
9:00 one-dimensional complex Gaussian variable, or in
9:03 general, complex Gaussian variables, are in some sense
9:06 mathematically simpler than real Gaussian variables.
9:10 For instance, the Q function is a very ugly thing.
9:14 In two dimensions, you can closed-form integrals with
9:16 probability of error.
9:18 So there's a lot of clues that we really want to think in
9:22 terms of pairs of real dimensions.
9:29 All right.
9:29 And then we introduced a couple more parameters.
9:32 And I understand Ashish got the questions
9:35 that I get every year.
9:36 Why do we introduce both of these things?
9:39 SNR norm, which is SNR normalized by 2 to the rho
9:43 minus 1, that's motivated directly by the
9:45 Shannon limit formula.
9:47 Or Eb over N_0, which is SNR divided by rho.
9:52 You know, these are both more or less
9:54 equivalent to each other.
9:55 From this Eb over N_0 is just 2 to the rho minus 1 over rho
10:00 times SNR norm, so why do we introduce both of them?
10:05 And there's not an intellectually very strong
10:11 argument to introduce both of them.
10:15 Because one is easily translated into the other.
10:20 For fixed nominal spectral efficiency rho, there clearly
10:26 is just a one-to-one translation.
10:28 We could use either one.
10:29 If you graph something versus Eb over N_0 or versus SNR
10:33 norm, it's just a matter of shifting it by this factor.
10:38 They're going to be exactly the same graph with the 0 dB
10:42 point in a different place, according to where 0 dB is.
10:47 Philosophically, of course, SNR norm, which is sometimes
10:52 called the gap to capacity, is exactly a measure of how many
10:58 dBs away are we from the Shannon limit.
11:02 Measuring things on a log scale in dB.
11:06 And so 0 dB is always the Shannon limit point with SNR
11:10 norm, because this statement translates into SNR norm is
11:15 greater than 1, which is 0 dB.
11:18 So here the focus is always, how far are you from capacity?
11:23 How far are you from the Shannon limit?
11:25 And that really is the modern view.
11:29 In the early days of coding, the view was, well, how much
11:32 coding gain can we get over no coding?
11:35 And as we'll see, Eb over N_0 is often a very convenient
11:42 parameter to use when we're focusing on coding gain.
11:46 In fact, for binary linear block codes, which is what
11:49 we're talking about in chapter six, we get an extremely
11:51 simple expression that Eb over N_0 it's just kd over N, where
11:57 N, k, d are the basic parameters of a block code.
12:02 If you don't know those yet, you will by
12:04 the end of this lecture.
12:06 And Eb over N_0 is what was put forward in the early days,
12:14 when the principal coding application was coding for
12:17 deep space communications.
12:19 It makes sense in the power-limited regime.
12:22 In the power-limited regime, we showed that this
12:27 essentially goes -- this is rho log2 over rho, so up to a
12:31 factor of natural logarithm of two, these are the same,
12:36 almost independent of the spectral efficiency, as well.
12:39 And so this is a very natural thing to use, especially in
12:44 the power-limited regime.
12:46 In the bandwidth-limited regime, as rho gets large,
12:50 then these things become very different.
12:52 SNR norm always keeps us in the neighborhood of 0 dB.
12:56 This thing goes up to 10 dB, 20 dB, 30 dB, and so forth.
13:01 Nonetheless, you'll see in some literature, people
13:03 continue to measure against Eb over N_0 maybe I would say,
13:09 just because they don't know any better.
13:11 Anyway, since I started writing these notes about
13:16 eight or nine years ago, I've been advocating SNR norm.
13:19 It's more and more widely used in our business,
13:23 in the coding business.
13:24 Or equivalently, one always shows nowadays how far are you
13:30 from capacity, and that's what SNR norm is about.
13:35 And you can always translate this into this and this into
13:38 this by this simple formula.
13:40 So that's why we introduce them both.
13:42 Eb over N_0 is traditional.
13:44 SNR norm is more of the modern gap to capacity viewpoint.
13:49 Any questions about that?
13:51 Because I understood that Ashish got a number of
13:54 questions about that.
13:56 Yes?
13:56 AUDIENCE: And the probability of error you mentioned
13:59[UNINTELLIGIBLE] is natural with SNR model.
14:03 PROFESSOR: Well, this is a slightly
14:04 different point, actually.
14:08 So we go on from this to talk about the power-limited
14:11 regime, which we defined more or less arbitrarily as the
14:14 regime where rho is less than or equal to two bits per two
14:17 dimensions, and the bandwidth-limited regime,
14:20 which is where rho is larger.
14:23 And at this point, I simply assert that it's better to do
14:30 everything per two dimensions in the bandwidth-limited
14:32 regime and to do everything per bit in the
14:39 power-limited regime.
14:41 And the reason for this is basically long practice and
14:47 intuition and experience that things do work out better, and
14:51 this is the proper normalization.
14:53 But I think at this point in the course, with your
14:55 background, this is only an assertion, all right?
15:00 So I simply say, this is the way we're going to do things
15:05 in bandwidth-limited and power-limited.
15:08 For most of the rest of the course, we're going to be in
15:13 the power-limited regime.
15:14 Then we'll come back to the bandwidth-limited very late in
15:18 the course.
15:19 So you can now forget this assertion for a while.
15:26 Yes?
15:26 AUDIENCE: [INAUDIBLE]
15:31 PROFESSOR: Well, I'm trying to focus on that here.
15:33 That if two systems have different rhos, then you
15:38 should take that into account in their comparison.
15:42 And when you see charts of a rate 7/8 system, or a rho 7/4
15:59 system versus one that's rate 1/8, those are very different
16:03 regimes, and it probably isn't fair to compare two systems,
16:10 just in terms of their probability of error versus Eb
16:13 over N_0 at two different spectral efficiencies.
16:16 What is more fair is to compare them in terms of their
16:21 gap to capacity.
16:27 You can get more powerful codes with more
16:30 error-correcting capability the lower in rate that you go.
16:34 And so you should start from the point of view that two
16:39 systems with different rho are incomparable, and then using
16:45 this you can say, but, you know, if we can get within 1
16:48 dB of the Shannon limit with both of them, then they're
16:51 both approximately equally powerful.
16:53 That would be the modern point of view.
16:56 But it's really apples and oranges if they have different
17:00 rates, different rhos.
17:03 Yes?
17:03 AUDIENCE: [INAUDIBLE]
17:07 If [UNINTELLIGIBLE]
17:10 systems modulation [INAUDIBLE].
17:13 Then is it fair --
17:16 because if one of them has coding --
17:21 PROFESSOR: If none of them have coding, well --
17:24 AUDIENCE: No, just modulation.
17:28 PROFESSOR: So what shall I say to that.
17:31 In the power-limited regime, an uncoded system is simply
17:35 binary modulation or 4-QAM, and there aren't different
17:41 systems to compare, really.
17:43 There's only one baseline system.
17:46 As you go up into the bandwidth-limited regime, then
17:51 it is fair to compare systems of the same
17:55 class, like M by M QAM.
17:58 That's a simple uncoded system.
18:01 Now there, rho keeps changing.
18:04 rho equals 2, rho equals 4, rho equals 6, rho equals 8.
18:08 Or you can easily find intermediate
18:11 rates that are uncoded.
18:13 And there you find that this normalization makes all these
18:17 systems comparable.
18:19 In fact, we saw that we got a universal baseline curve for
18:24 the bandwidth-limited regime, which was independent of rho,
18:28 for, say, the class of 4 by 4, by the class of M by M
18:33 bandwidth-limited QAM systems.
18:37 I'm sorry.
18:38 I got off a plane at nine o'clock last night, so I may
18:41 not be totally coherent today.
18:45 But all of these systems have exactly the same baseline
18:49 curve up to the approximations we've made that their exactly
18:55 four nearest neighbors and so forth.
18:57 But on a curve of probability of error per two dimensions,
19:02 per QAM symbol, versus SNR norm, we
19:06 get a universal curve.
19:08 So that indicates they really are comparable.
19:11 They form a class where the performance of the class is
19:15 independent of rho.
19:16 And that's sort of typical of the bandwidth-limited regime.
19:20 Two systems that differ, that basically use the same code
19:24 with smaller constellation or a larger-based uncoded
19:28 constellation are going to be directly comparable, and are
19:30 probably going to have the same gap to capacity.
19:35 Going to be just as far away from the Shannon limit.
19:37 But we don't have that phenomenon in the
19:39 power-limited case.
19:45 Other questions, or shall I proceed?
19:49 OK.
19:52 Again continuing what I hope will be a quick review, but I
19:58 don't want to go any faster than you're comfortable with.
20:05 We get this baseline curve for 2-PAM or 2
20:12 by 2 QAM, same thing.
20:18 So this is the baseline of Pb of E on a log scale, where we
20:26 typically go down to ten to the minus six, ten to the
20:29 minus five.
20:31 Here's 0 dB.
20:33 This goes down through about 9.6 dB here or
20:37 about 10.5 dB here.
20:40 And the ultimate Shannon limit is over here.
20:47 The ultimate Shannon limit for very low rho, as rho goes to
20:52 0, is about minus 1.6 dB.
20:55 And we get expressions.
20:57 0 dB is the Shannon limit at rho equals 1.
21:00 At rho equals 2, it's up about 1.8 dB.
21:05 Shannon limit for rho equals 2, and so forth.
21:08 And anyway, so we see as a function of rho, we can
21:12 measure the gap to capacity at different probabilities of
21:16 error and see how much coding gain is,
21:19 in principle, possible.
21:20 And then for a coded system, we can put that on here.
21:26 The effective coding gain is the distance
21:29 between here and here.
21:32 At some target probability of error.
21:36 It's going to differ according to probably of error.
21:39 You found a way of getting a good, rough-cut estimate, at
21:46 least for not very complicated codes called the union bound
21:50 estimate for any signal constellation in any number of
21:57 dimensions.
21:58 So if we have a signal constellation A, which
22:04 consists of M points in N dimensions, so forth,
22:10 basically found that we could, in the power-limited regime,
22:16 get an approximate expression from considering pairwise
22:22 error probabilities that's very simple.
22:25 And I used this notation.
22:27 Q of the square root of some coding gain of the
22:34 constellation times 2 Eb over N_0.
22:40 If it's just 2-PAM then the coding gain becomes 1.
22:45 This is the average number of nearest neighbors per
22:52 transmitted bit.
22:55 And so the whole thing reduces to this expression in terms of
23:00 couple of parameters.
23:01 the principal parameter is the nominal coding gain, which is
23:06 the minimum squared distance of A over 4 times the energy
23:13 per bit of this constellation A. So we really only need to
23:19 know this kind of normalized measure of goodness of the
23:24 constellation.
23:27 And K_b of A Is the average number of nearest neighbors.
23:35 I forget what we call it.
23:38 What is the numerator here?
23:43 K_min of A, which itself is an average, over the number of
23:50 bits that we're actually sending, which is log2 of the
23:56 size of A.
23:59 So basically we only need to know a couple of parameters of
24:03 this signal constellation.
24:04 Its minimum square distance is very important.
24:08 Energy parameter, which we choose to make the energy per
24:11 bit, or so that we get this expression.
24:15 And the average number of nearest neighbors per bit that
24:19 we transmit.
24:19 OK.
24:23 And our best example so far is the tetrahedral constellation,
24:33 where we basically pick every other
24:37 point from the 4 simplex.
24:48 Maybe I should do that in a different color.
24:50 I don't think that's a different color.
24:52 Nope.
24:54 Anyway.
24:55 You know this quite well by now.
24:57 And if we do that, we find that normalizing everything,
25:01 this is 4/3 or 1.25 dB.
25:05 And every point has three nearest neighbors, and we're
25:10 sending 2 bits, so this is 3/2.
25:15 So then we get this handy dandy engineering rule to make
25:20 a quick plot of the union bound estimate.
25:25 Given that we decided to put this on a log-log scale to
25:27 start with, all we have to do to plot this expression --
25:33 if we're given, as we always are, the baseline curve, which
25:42 is simply Q to the square root of 2 Eb over N_0.
25:49 How do you convert this curve to this curve
25:52 on a log-log plot?
25:53 Well, you simply move it to the left by the coding gain.
25:57 And you move it up by the log of whatever K_b is.
26:03 So if the dominant effect is moving it left, in this case,
26:08 by 1.25 dB --
26:12 I'm going to wind up getting about this curve, so I won't
26:15 draw it again --
26:15 then we move it up by a factor of 3/2, we developed a rule of
26:22 thumb that said a factor of two is basically going to cost
26:26 you about 0.2 dB.
26:30 Around ten to the minus five.
26:31 This is just based on the slope of this baseline curve,
26:35 as long as we're somewhat in that region.
26:38 So this will cost us about 1/10 of a dB, so we'll get an
26:43 effective coding gain of about 1.15 dB.
26:49 So graphically, we take this curve bodily.
26:52 We move it over 1.25 dB, and we move it up
26:58 by a factor of 3/2.
27:00 And what we'll find is the effective coding gain is
27:02 thereby reduced to, I estimate, about 1.15 dB.
27:07 And this is all just engineering rules of thumb.
27:11 Nice template.
27:12 I sometimes say you should fill out a copy of this
27:15 baseline curve, cut it out, put it in your wallet for the
27:18 duration of this course, because you'll have the
27:20 opportunity to make this kind of
27:22 calculation again and again.
27:25 All right.
27:25 So that's union bound SNR.
27:28 For simple constellations, the union bound
27:31 estimate is very accurate.
27:34 So this is a very good way to proceed, from an engineering
27:37 point of view.
27:38 We want to write a paper?
27:41 You know, it's a little late to write a paper on the
27:43 performance of the four simplex signal constellations,
27:46 but if you wanted to, and you wanted to have one graph in
27:49 that paper, you would have the graph of --
27:51 well, you would actually probably compute the exact
27:57 error of probability, either by analysis or by Monte Carlo
28:01 simulation, you would put that in there.
28:03 But in your first draft of the paper, you would put in union
28:05 bound estimate and you would find that wasn't far off.
28:11 OK.
28:12 Any questions on that?
28:13 This has basically got us up to chapter six.
28:26 So in chapter six now, which Ashish got into last time.
28:35 This is basically about binary linear block codes.
28:44 And well, we first just start talking about
28:48 binary block codes.
28:51 That's what Ashish did last time.
28:58 We basically take 0 and 1 as our binary alphabet.
29:03 We take a blocks of length n.
29:08 Sorry.
29:08 Are we using little n or big N?
29:12 Little n, where n is called the block plank.
29:15 So we take the set of all binary symbols of length n,
29:23 and we're going to convert this to real n space.
29:31 How do we convert it?
29:33 Component-wise, by the standard 2-PAM map.
29:36 We map 0 into plus alpha, 1 into minus alpha.
29:41 So this maps into plus or minus alpha to the n.
29:46 So the entire universe of possible code words that we
29:50 have is this set of 2 to the n real n-tuples, which are
29:58 simply the vertices of an n cube of size 2 alpha,
30:01 obviously, right?
30:02 Just like this.
30:04 We have eight possible vertices in 3-space.
30:10 They obviously all have the same power.
30:12 They all lie on the surface of a sphere of squared radius n
30:17 alpha squared.
30:20 So they're not only the vertices of a cube, they're
30:22 vertices that are spread on an equal energy sphere.
30:26 And the whole idea is that our code is going to be some
30:32 subset of this, and the code will map under the same map,
30:36 which we call S, into some subset of the
30:43 vertices of the n-cube.
30:45 OK?
30:45 And again, our favorite example is the tetrahedron.
30:50 For instance, if we take the code as being 0, 0, 0, 0, 1,
30:57 1, 1, 0, 1 1, 1, 0, those for binary three-tuples, this maps
31:06 into T, the tetrahedron.
31:10 OK?
31:11 Well, we get this signal structure, which we've already
31:15 found the coding gain.
31:16 It has a little bit of coding gain.
31:18 We actually accomplished something.
31:22 All right.
31:24 So that's the basic idea that we go through in 6.1.
31:30 That's our favorite example.
31:33 And Ashish also shows you that, you know, you might
31:37 think this is an awfully restricted way of designing
31:41 constellations.
31:43 But when we're talking about low spectral efficiencies, by
31:47 going through the capacity calculation, we can assure
31:50 ourselves that in principle, this is not going to be very
31:55 sub-optimal for it.
31:57 Capacity is about the regime where n gets very large for
32:00 very long block codes.
32:02 We don't lose much in potential coding gain, or in
32:06 the Shannon limit, equivalently, by restricting
32:10 the input output input alphabet to be just these two
32:15 numbers, plus or minus alpha, I should say, rather than
32:18 using the whole real line as an input, as long as the
32:24 nominal spectral efficiency is less than or equal to one bit
32:35 per two dimensions.
32:37 And the exact numbers --
32:41 at rho equals 1, you lose about 0.2 dB, in principle,
32:47 using the Shannon limit as your guide.
32:51 And for lower rho, it just goes to 0.
32:56 And you did this last time.
32:59 OK.
32:59 So it's a very reasonable, and obviously attractive from an
33:04 implementation point of view, way of designing signal
33:08 constellations.
33:09 And again, this is basically all we're going to be talking
33:12 about for the majority of the course, is signal
33:15 constellations like this.
33:16 But of course as we get to codes that are thousands of
33:19 bits long, or perhaps even infinitely long, as in the
33:22 case of convolutional codes and trellis codes, sometimes
33:28 you forget that we're really talking about just putting
33:31 points in Euclidian signal space.
33:35 Because we're going to be very abstracted back into the
33:39 binary domain.
33:42 But in this course, we're always thinking of codes as
33:46 means of designing signal constellations.
33:53 Of course, codes are used far more widely than just for
33:57 signaling over the additive white Gaussian noise channel.
34:01 I've sort of packaged this course as a search to get to
34:08 capacity, we have added the additive white
34:10 Gaussian noise channel.
34:11 First of all because this corresponds
34:13 very closely to history.
34:16 Within this package, we can talk about all of the
34:19 principal classes of codes that have been developed to
34:22 date, and we can compare them in some performance terms.
34:29 How close do they get to capacity on the additive white
34:31 Gaussian noise channel.
34:33 So you get most of what you would get if you took a
34:38 communications-free view of coding theory.
34:42 You know, people are interested in codes for
34:44 computer memories, for God knows what,
34:50 lots of other things.
34:51 And for many of these other applications, you are going to
34:57 be interested in the same class of codes.
34:59 Basically, you're going to want to maximize the distance
35:03 between code words for a given rate, which has to do with the
35:07 size of the code.
35:09 And so you're going to be interested in the same codes.
35:14 Putting it in the framework of getting to capacity on the
35:16 additive white Gaussian noise channel gives a motivation,
35:21 gives a very nice story, because over 50 years, we were
35:24 able to get to capacity, and gives it a real
35:28 communications flavor.
35:30 Some of you, I'm sure, are here without any interest in
35:33 communications whatsoever.
35:34 You simply want to know about coding.
35:36 You will still get that story, but you'll get it
35:38 in this nice package.
35:41 That's why I do it this way.
35:44 OK.
35:45 I'm about to get into new stuff.
35:47 End of review.
35:49 Any questions?
35:52 So you must have done a great job.
35:56 Everyone understands perfectly.
35:59 All right.
35:59 So now let's talk about, as I said, binary
36:03 linear block codes.
36:06 When you see the word "linear," that's a signal that
36:11 there's some algebra ahead.
36:12 And so this is the very first point at which we get into
36:18 what's called algebraic coding theory.
36:22 The algebra will be extremely simple at this
36:24 point, not to worry.
36:28 And what are we doing here?
36:36 First thing we do is to identify 0 and 1 with the
36:40 binary field, which I'm always going to write as F2.
36:45 The older way of writing this is GF(2), for Galois field
36:51 with two elements.
36:52 They mean exactly the same thing.
36:54 Nowadays most people write just F2.
36:59 Now we have [UNINTELLIGIBLE], we can write this blackboard
37:03 F.
37:06 And OK.
37:09 Step one.
37:09 We've identified our alphabet with a field.
37:16 So algebraically, this is a field.
37:20 Some of you know exactly what I mean by that.
37:23 Others don't.
37:24 We'll come back to this again.
37:26 Informally, I would say a field is simply something
37:29 where you can add, subtract, multiply, or divide.
37:32 Our best examples of that before now have been the real
37:36 and complex fields.
37:39 There is a more formal definition of that.
37:42 In this case, we have only two elements.
37:45 So let me just write down the tables, which you all know,
37:52 regardless of your background.
37:55 What's the addition table for this field?
37:57 Well, 0 is the additive identity.
38:01 So we know that 0 added to anything gives itself, so that
38:09 gives us three of the entries of this table.
38:14 What do we put down here?
38:16 There's really only one choice to satisfy one of the axioms
38:20 of the field, which is that under addition, the field must
38:27 form a group.
38:29 In particular, that means that all elements are invertible.
38:32 That means that each row or column has to be a permutation
38:36 of the group elements.
38:37 And so the only possibility here is to put in a 0.
38:40 And well, we are forced, if we're going to make 0 and 1
38:46 into a field to find an addition table which is the
38:50 table of mod2 addition.
38:52 You've had this stated as an axiom before.
38:55 You can derive it just from the fact that 0 needs to be
39:00 the additive identity, and then we need to put a 0 in
39:05 here in order to get invertibility.
39:12 Under multiplication.
39:14 What is the multiplication table?
39:16 Well, the additive identity is also a nullifier under
39:21 multiplication in any field.
39:22 So 0 times anything is equal to 0.
39:26 1 times anything is equal to itself.
39:29 So that completely determines the multiplication table.
39:34 Again, just the table of mod2 multiplication.
39:37 So if I'd done this axiomatically, I would have
39:41 given you the axioms of field.
39:43 Then I would show that under this binary addition
39:47 operation, this multiplication operation,
39:50 we satisfy the axioms.
39:52 I'll do that as we get into chapter seven.
39:57 OK.
40:00 So now we have 0 1 to the n, n-tuples, binary n-tuples, we
40:12 will now regard as n-tuples of field elements, F2.
40:20 And these will be regarded as vectors in a vector space.
40:29 Well, I'll just say that F2 to the n is a vector space, which
40:35 clearly has 2 to the n elements.
40:37 Now again, informally.
40:39 What's a vector space?
40:41 A vector space is always over a given field.
40:46 In this case, it's going to be over the
40:47 binary field, of course.
40:49 F2.
40:51 The given field is called a scalar.
40:54 Just like vector space over the reals, the scalars are
40:57 real numbers.
40:58 Here the scalars are elements of F2.
41:02 And what do we have to have to make something algebraically a
41:07 vector space?
41:08 We have to have that the addition of two vectors is
41:12 well-defined and gives another vector, and that the
41:19 multiplication of a vector by a scalar is well-defined and
41:23 gives another vector in our vector space.
41:27 All right?
41:30 OK.
41:31 So how are we going to define that to addition?
41:35 If we want to add two n-tuples --
41:38 again, you aren't going to see anything here that you haven't
41:40 already seen in some other context.
41:43 What do we do?
41:43 We add them component-wise, using the component-wise rules
41:49 of field addition.
41:51 OK.
41:51 So we just add two vector n-tuples,
41:55 component by component.
41:56 We obviously get some result, which is itself a binary
42:00 vector or an F2 vector.
42:03 No problem there.
42:05 Except that --
42:08 well, all right.
42:09 As long as we're talking about all possible n-tuples, the
42:11 result is certainly in the vector space, right?
42:14 We add two binary n-tuples, we get a binary n-tuple.
42:17 So the result is in F2 to the n.
42:20 This is going to be the key test when we get to subspaces
42:24 of F2 to the n.
42:27 OK.
42:28 And multiplication is even easier.
42:31 How do we define multiplication by scalars?
42:34 Well, we only have two scalars --
42:35 0 and 1.
42:37 So 0 times any vector is going to equal the all 0 vector.
42:45 Again, you can view that as just component-wise
42:47 multiplication of everything by 0, and since 0 times
42:51 anything equals 0, we're always going
42:53 to get the 0 vector.
42:55 All right?
42:57 So is the 0 vector in the vector space?
43:02 Well, yes.
43:02 If we're talking about several n-tuples, it of
43:05 course always is.
43:06 And one times anything, again, we can just do this
43:12 component-wise, and it just gives itself.
43:17 Trivially.
43:18 So since this was a vector in the vector space, this is
43:21 certainly a vector in the vector space.
43:25 OK.
43:26 This seems pretty trivial so far.
43:29 But what's a binary linear block code?
43:35 Again, focusing on the linear, is a --
43:40 now I'll give a formal definition.
43:43 it's a subspace of F2 to the n for some n
43:51 called the block length.
43:55 All right.
43:56 What do I mean when I say a subspace?
43:58 I mean subsets of the elements of a vector space that itself
44:03 forms a vector space.
44:05 So in this case when I say subset, I mean a
44:11 set of binary n-tuples.
44:14 OK?
44:16 That itself forms a vector space.
44:18 OK.
44:18 What are the components of that?
44:20 To check that it forms a vector space, let's see.
44:24 Multiplication by scalars is, again, easy to check.
44:30 If it's going to be a subspace, then the all 0 --
44:34 so it has to contain the all 0 vector in order that when I
44:48 multiply by the scalar 0, I get another
44:50 element of this subspace.
44:55 Multiplication by 1 is always trivially satisfied.
44:59 If I start off with a set, I multiply by 1, I'm going to
45:01 get the same set.
45:04 So let's check whether the elements of a subspace are
45:14 closed under vector addition.
45:17 What do I mean by that?
45:19 I mean if you add two elements of the of the subspace
45:24 together, you get another element of the subspace.
45:28 That's the key property to check.
45:36 Key property -- we can write that as closure
45:40 under vector addition.
45:47 Which is also called the group property.
45:52 It means that just under addition, under vector
45:56 addition, the set of elements that you have forms a group.
46:00 You add any two elements, you get another
46:01 element of the subset.
46:06 The example is our favorite example so far.
46:12 Let's take this little code, and I'm going to ask.
46:18 Is that a subspace of F2 to the three?
46:24 So does this equal subspace of the set of all binary
46:35 three-tuples?
46:40 Anyone care to hazard a guess whether it is or isn't?
46:46 AUDIENCE: It is.
46:47 PROFESSOR: It is?
46:48 Why?
46:53 AUDIENCE: It has the all 0 vector.
46:54 PROFESSOR: It has the all 0 vector, first of all.
46:58 Good.
47:00 AUDIENCE: [INAUDIBLE]
47:04 PROFESSOR: And it's closed under addition.
47:07 Now how might we see that?
47:11 You could, of course, just take all pair-wise --
47:14 you could form the addition table of these four elements,
47:17 and you would find that you always get another one of
47:20 these elements.
47:20 For instance, 0, 1, 1 plus 1, 0, 1, is equal to 1, 1, 0.
47:26 In fact, you easily see that you take any two of these, add
47:31 them together, you get the third one.
47:33 If you call this a, b, and c, a plus b plus c equals 0,
47:37 addition is the same as subtraction, because we're in
47:40 a binary field.
47:42 So that means that a plus b equals c, a equals b plus c, c
47:46 equals b plus a, whatever you like.
47:50 And of course, if you add 0 to anything, it's trivially
47:55 closed under that.
47:59 All right.
47:59 So it is.
48:01 It satisfies --
48:02 it's all you've got to check.
48:09 A more abstract proof of this would be, this is the set of
48:13 all even-weight three-tuples.
48:16 If I add even to even, I'm going to get even.
48:19 So of course my result is going to be another
48:21 even-weight three-tuple, therefore in the set.
48:24 That's the more algebraic proof.
48:29 All right.
48:31 Suppose I just add 0, 0, 1 to all of these things.
48:36 I'll get the set of all odd-weight n-tuples.
48:41 Add any odd-weight n-tuple to this.
48:43 Let me take the C prime, which is the set of
48:46 all odd-weight n-tuples.
48:50 Is that a vector space?
48:54 It doesn't have 0, and in fact, it's not even closed
48:57 under vector addition.
48:59 All right.
49:00 If I take in C double prime is equal to 0, 0, 0 0, 1, 1, 1,
49:06 0, 1 and I stop there, is that a subspace?
49:13 No.
49:13 Because?
49:15 Not closed.
49:17 For instance, if I add these two together, I would get
49:21 that, and that's missing.
49:23 OK.
49:24 So actually, everything is much simpler when we're
49:31 talking about finite fields.
49:32 All the finite dimensional vector spaces consist of a
49:35 finite number of elements.
49:38 It's easier than real and complex vector spaces.
49:42 There's no analysis involved.
49:46 So forth.
49:47 All right.
49:47 So a binary linear block code is in a subspace
49:52 of F2 to the n.
49:55 What are some of the key algebraic facts we know about
49:59 vector spaces from our study of linear algebra, which I
50:06 assume all of you have had in some form or another?
50:11 What's a key algebraic property of a vector space?
50:14 What's the very first property?
50:18 All vector spaces have a --
50:20 AUDIENCE: [INAUDIBLE]
50:23 PROFESSOR: Norm?
50:25 No.
50:26 Dimension.
50:27 Dimension, that's where I'm going.
50:30 What's the significance of dimension?
50:33 AUDIENCE: [INAUDIBLE]
50:36 PROFESSOR: The definition of dimension is the number of
50:40 generators in any basis.
50:42 So we're talking about generators of the vector
50:45 space, or a set of generators which form a basis.
50:49 If you think we have the same properties here in the case of
50:53 vector spaces over finite fields, well, we probably do.
51:01 But let's see how it works out.
51:02 So we're talking about things like
51:05 generators, basis, dimension.
51:10 These are all closely interlinked properties.
51:16 Again, the fact that everything is finite here
51:20 gives us very elementary ways of addressing
51:24 all of these concepts.
51:27 We don't have any geometry yet.
51:29 We don't have norms.
51:30 We don't have distances.
51:31 We don't have angles.
51:32 We'll talk about that in a minute.
51:34 Here we just have a set that basically has these two
51:38 properties.
51:39 It contains the all-0 vector, and it's closed
51:41 as the group property.
51:42 It's closed under vector addition.
51:46 All right.
51:46 So the first property is that the all-0 vector is always in
51:56 the subspace, which I'll represent by C, meaning code.
52:02 So when I talk about a code now, I'm talking about a
52:06 binary linear block code, which by definition is a
52:09 vector space, a subspace of F2 to the n, where
52:13 n is the code length.
52:16 All right.
52:16 Let me try to find a set of generators for the code.
52:22 All right?
52:22 If somebody gives me a code, they say, this is a binary
52:28 linear block code.
52:29 Let me see if I can find a set of generators for it.
52:33 If I find three generators, then I'll know the
52:37 dimension is three.
52:39 That's basically where I'm going.
52:40 I'll state this a little bit more formally as we go ahead.
52:44 So suppose I'm given a code.
52:46 I know it's a binary linear block code, so I know it has
52:49 the all-0 element.
52:52 So how might I go about finding a set of generators?
52:59 Let's just take a greedy algorithm.
53:01 All right?
53:05 Suppose the code contains only the all-0 vector.
53:08 Is that a subspace?
53:12 What's its dimension?
53:15 0.
53:15 All right?
53:18 So that's a very trivial vector space, but it's a
53:22 vector space.
53:23 Satisfies the axioms.
53:27 Suppose it is not the trivial code.
53:29 That means it has more than the all-0 vector.
53:32 So I take as my first generator any non-zero vector.
53:42 OK?
53:43 I can always do that.
53:46 Don't have the axiom of choice involved here, because
53:49 everything is finite.
53:51 So I'm going to take g1 to be any non-zero vector.
53:54 Now I've got a generator.
53:57 And how many code words does it generate?
54:10 I want to take the set of all binary linear combinations of
54:14 all the generators that I have so far.
54:16 At this point, the binary linear combinations are 0
54:19 times g1 and 1 times g1 And that just gives me this word
54:25 and this word.
54:26 So now I have counted for two words with one generator.
54:32 Could it be that that's the whole code?
54:36 Sure.
54:39 The all-0 vector and any non-zero vector together form
54:43 a one-dimensional subspace.
54:49 That's all you can get from one dimension.
54:51 So I could be finished here now.
54:53 But if I'm not finished, there's still more code words
54:55 that I haven't accounted for.
54:57 Then I greedily pick a second generator.
55:01 So this is now, let's say, any vector not generated by g1.
55:20 So I have a branch here.
55:22 Either I've finished or I can pick another generator g2.
55:27 Now with g1 and g-two, how many vectors can I generate?
55:33 Let me take all binary linear combinations.
55:41 So a binary linear combination is any vector of the form
55:47 alpha1 g1 plus alpha2 g2 where these are both scalars.
55:57 And therefore this can be 0 or 1, this could be 0 or 1.
56:02 So what have I got now?
56:03 I've got 0, g1 g2 and g1 plus g2 I've got four binary linear
56:09 combinations of two generators.
56:15 Could that be the whole code?
56:17 Certainly.
56:19 At this point, again, consider our standing example.
56:26 I start out what I take as my first generator.
56:29 Let me take g1 equal 1 0, 1, g2 equals whatever, 0, 1, 1.
56:38 Then if I take all binary linear combinations of these
56:42 two generators, I'm going to get the whole code, right?
56:45 These four code words.
56:46 They can all be expressed in this form.
56:51 Or I'm not done, and then I have to pick g3.
56:56 And how many binary linear combinations are there of g1,
56:59 g2, and g3?
57:03 Eight.
57:04 Are they all necessarily in my subspace?
57:09 Yes, by the fact that the subspace is closed under
57:14 scalar multiplication.
57:16 Alpha g1.
57:17 Alpha1 g1, alpha2 g2 alpha3 g3 are all in the subspace.
57:22 Any vector addition of any of these scalar multiples is in
57:29 the subspace.
57:30 So I get now eight possible elements of the subspace, and
57:36 I either may be done or not.
57:40 Continuing in this way, I get some number gk of generators,
57:48 just by picking greedily the next one until I'm done.
57:53 All right?
57:55 When I'm done --
57:57 so I have to stop.
57:58 Why do I have to stop?
58:02 All right.
58:03 Let's look at the size.
58:04 At each point here, this accounts for two code words,
58:07 this for four, this for eight, this for 2 to the k.
58:11 How many binary n-tuples are there?
58:15 2 to the n.
58:16 All right?
58:17 So I clearly can't find more than n generators.
58:21 More more than n independent generators, in the sense that
58:25 the set of all the binary linear
58:28 combinations are distinct.
58:32 All right.
58:32 So k is, at most, going to be n.
58:36 So I will stop in a finite number of steps.
58:39 I'll stop at some number k.
58:42 When I've stopped, that's because the code consists of
58:46 all binary linear combinations of these k generators, and
58:50 therefore has size 2 to the k.
58:54 So the only possible size of a subspace is 2 to the k for k
59:02 less than n.
59:04 A power of 2 where the power is, at most, n.
59:10 So any subspace besides the subspace --
59:13 I'm repeating myself.
59:17 OK.
59:18 So this means I found a basis.
59:25 Does this mean that all possible bases of any subspace
59:28 have the same size?
59:33 Well yeah, it must.
59:34 I mean, I've proved now that any subspace has to have this
59:40 size 2 to the k for some k.
59:45 So obviously, if I go through this process, no matter how I
59:47 choose my generators, I could choose any pair of these
59:54 non-zero n-tuples as my generators.
59:57 So that would be a legitimate basis.
59:59 Take any two out of these three.
1:00:01 But it's always going to take exactly two of them, right?
1:00:04 Why?
1:00:05 Because the size of this subspace is four.
1:00:08 Two to the two.
1:00:10 So if I go through this process, I'm always going to
1:00:13 come up with k generators, where k is the log to the base
1:00:17 2 of the size of this code that I was given, which I was
1:00:21 told was a linear code, meaning it's a subspace.
1:00:27 So I'm somewhat free to choose the generators, but I'm always
1:00:33 going to come up with k of them if the code has
1:00:35 size 2 to the k.
1:00:39 So a basis is a set of k linearly independent k-tuples,
1:00:50 where linear independence has the same meaning here as
1:00:52 you're accustomed to, meaning that all linear combinations
1:00:56 of the k generators are distinct.
1:01:00 So I get 2 to the k distinct elements of the code.
1:01:05 And I say the dimension of the code is k.
1:01:10 In this case, basically it's the size of the code is 2 to
1:01:13 the k, then its dimension is k.
1:01:16 Has to be.
1:01:18 And all basis have k generators in them.
1:01:24 And there are, in general, many ways to pick them.
1:01:27 All right?
1:01:29 So just by considering this greedy basis construction
1:01:33 algorithm, I basically find the size of any subspace is a
1:01:43 power of two, and the power is equal to the dimension.
1:01:46 And any basis is going to have that cardinality.
1:01:52 Are you with me?
1:01:53 This is a pretty simple proof.
1:01:57 And I call this an n, k binary linear block code.
1:02:02 n being the length.
1:02:04 That just means that every code word is an n-tuple over
1:02:09 the binary field.
1:02:11 And k is the dimension.
1:02:13 So an n, k binary linear block code has size 2 to the k.
1:02:24 That's 2 to the k distinct code words.
1:02:29 Main example is this guy.
1:02:35 Easy?
1:02:35 Any questions?
1:02:36 I think this is clear.
1:02:37 AUDIENCE: Can n use the number of code word, I take
1:02:41 like 2 to the k?
1:02:44 PROFESSOR: n is something I specify a priori as the length
1:02:48 of every vector in the code.
1:02:50 In other words, it has size 2 to the k, and it's a subset of
1:02:56 the set of all binary n-tuples, which
1:02:58 I write like that.
1:02:59 In other words, the elements of the
1:03:01 code are binary n-tuples.
1:03:03 If I write them out, each one has length n.
1:03:09 OK?
1:03:10 We're good?
1:03:12 What are some other examples?
1:03:18 The simplest codes you can think of is, first of
1:03:23 all, an n, 0 code.
1:03:26 That means a code with dimension zero, has size what?
1:03:32 1.
1:03:33 And what does it consist of?
1:03:35 AUDIENCE: [INAUDIBLE]
1:03:37 PROFESSOR: Yeah.
1:03:37 So this is the so-called trivial code, just containing
1:03:41 the all-0 word.
1:03:45 Has to mention 0.
1:03:46 It is a binary linear block code, but it's not much use
1:03:49 for communication.
1:03:52 So but nonetheless, we include that in this family.
1:03:58 Another trivial one is n, n.
1:04:01 What's that?
1:04:03 AUDIENCE: [INAUDIBLE]
1:04:06 PROFESSOR: F2 to the n, right.
1:04:07 What's its size?
1:04:08 2 to the n.
1:04:09 That means it has to contain all distinct binary n-tuples.
1:04:15 So this is called the trivial code.
1:04:18 This is called the universe code.
1:04:21 Contains the entire universe of binary n-tuples.
1:04:27 Let's get some slightly less trivial ones.
1:04:29 n, 1.
1:04:32 What would that be?
1:04:36 An n, 1 code.
1:04:38 What's its size?
1:04:40 2.
1:04:41 What does it consist of?
1:04:44 The 0 word, and?
1:04:48 Any other non-zero generator.
1:04:50 And that's correct.
1:04:52 This can be 0 and any generator, two words.
1:04:59 In communications, where we want to maximize the distance,
1:05:03 in some sense, between the two code words, what is g
1:05:10 always taken as?
1:05:12 0,1's, right.
1:05:14 So if it's in particular the all-0 and the all-1 word,
1:05:20 which I might write as a vector of 0's and a vector of
1:05:23 1's, this is called the repetition code.
1:05:28 The binary repetition code of length n.
1:05:32 It either gives me a 0 and I repeat it n times, or gives me
1:05:36 a 1 and I repeat it n times.
1:05:39 So whenever you see n,1, you can pretty well assume it's
1:05:41 the repetition code, though it might be any pair 0, g.
1:05:48 And n, minus 1.
1:05:49 This is an interesting one.
1:05:58 Again, while this could be a lot of things, in
1:06:07 communications, whenever you see this, this will always be
1:06:12 the set of all even-weight n-tuples.
1:06:22 In other words, the set of all n-tuples with even parity such
1:06:28 that if you sum up all of the components of any
1:06:31 vector, mod2 equals 0.
1:06:35 OK?
1:06:36 So this I will call the single parity check, or more briefly,
1:06:47 the SPC code, or the even-weight code.
1:06:51 That's equally good.
1:06:55 And here we maybe should do a little bit more work.
1:07:01 Say, is this in fact a subspace?
1:07:03 Does it include the all-zero code word?
1:07:06 Yes, all-zero has even weight.
1:07:08 The sum of any two even-weight code words, an even-weight
1:07:12 code word, an even-weight n-tuple.
1:07:15 Yes.
1:07:17 As here.
1:07:18 This is an example.
1:07:19 This is the three, two SPC code.
1:07:27 OK.
1:07:28 Why is this dimension n minus one?
1:07:32 AUDIENCE: [INAUDIBLE]
1:07:34 Is every code word orthogonal to the one-vector?
1:07:37 PROFESSOR: OK.
1:07:38 That's an excellent answer.
1:07:41 It's a little advanced for us right now.
1:07:43 I'm looking for an elementary argument.
1:07:45 AUDIENCE: [INAUDIBLE]
1:07:57 PROFESSOR: OK.
1:07:57 So you're saying we have a set of generators
1:08:00 that looks like this.
1:08:03 Is that what you're saying?
1:08:05 You are correct.
1:08:07 And how many such generators are there?
1:08:12 There are n minus 1 of them.
1:08:16 I always like to find the most elementary argument possible.
1:08:19 I think the most elementary argument here is that the
1:08:23 number of even-weight n-tuples is equal to the number of
1:08:26 odd-weight n-tuples, and together they form the set of
1:08:29 all n-tuples.
1:08:30 So exactly half of the n-tuples are even weight.
1:08:34 That means there are 2 to the n minus 1 of them, 2 to the n
1:08:38 over 2, and therefore, the dimension must be n minus 1.
1:08:42 But perhaps this is just as elementary a proof.
1:08:48 Well, however you do it, you'll find that there are 2
1:08:51 to the n minus 1 of them, or that here is clearly a set of
1:08:56 generators.
1:08:57 It might take a few more lines to show that every even-weight
1:09:00 code word is a linear combination of this particular
1:09:05 set of generators, but it's certainly true.
1:09:08 All right.
1:09:09 So these four classes of codes, these two entirely
1:09:13 trivial ones, these two which are actually somewhat more
1:09:17 interesting for coding purposes --
1:09:19 we've already seen, we can get a coding game with this length
1:09:23 three, dimension two code --
1:09:28 are basically the simplest codes we can think of.
1:09:33 The simplest binary linear block codes.
1:09:36 Now we'll see them again and again.
1:09:38 They turn up.
1:09:41 All right?
1:09:41 So the whole course is going to be about
1:09:43 finding 1's in between.
1:09:45 More complicated ones.
1:09:47 There's clearly more room to play.
1:09:49 For instance, if k is equal to half of n, which means that
1:09:55 rho is equal to one bit per two dimensions, there's a lot
1:09:59 of possibilities.
1:10:03 And so we're going to explore those possibilities.
1:10:12 AUDIENCE: [INAUDIBLE]
1:10:16[UNINTELLIGIBLE]
1:10:27 PROFESSOR: Sure.
1:10:31 Let's take the 6,5 code generated by these five
1:10:37 generators.
1:10:40 Well, it contains some
1:10:42 odd-weight code words, obviously.
1:10:46 But it's not as interesting from a coding point of view.
1:10:54 It's not the only one, but it's the only one
1:10:56 you'll ever see here.
1:11:01 All right.
1:11:01 Let's see.
1:11:07 One thing I didn't point out in the notes but probably
1:11:11 should have here is, what is rho for an n, k binary linear
1:11:15 block code?
1:11:19 How many bits can we send?
1:11:21 Suppose we take the Euclidean image of this.
1:11:24 It is going to have 2 to the k points, 2 to the k vertices of
1:11:29 the n-cube, and so what is rho?
1:11:35 What is the rate in bits per two dimensions?
1:11:37 AUDIENCE: [INAUDIBLE]
1:11:40 PROFESSOR: Right.
1:11:43 The rate is basically k over n over 2, if you like, or 2 k
1:11:49 over n bits per two dimensions.
1:11:54 You can send k bits in n dimensions, or 2 k over n bits
1:12:00 per two dimensions.
1:12:01 And since k can't be larger than n, this is going to be
1:12:08 less than or equal to two bits per two dimensions.
1:12:10 So again, we see we're definitely in the
1:12:14 power-limited regime, and that we can really get any nominal
1:12:20 spectral efficiency between 0 and 2 by choosing k and n
1:12:27 appropriately.
1:12:30 All right.
1:12:31 So n and k determine the rate, determine the nominal spectral
1:12:34 efficiency.
1:12:37 All right.
1:12:38 Let's now talk about things that I think are also very old
1:12:42 hat to you.
1:12:44 I mean weight and distance.
1:12:49 But we begin to get into areas that show us that these vector
1:12:57 spaces are very different from the real and complex vector
1:13:00 spaces that we're accustomed to.
1:13:03 So what are we doing now?
1:13:04 We're starting to get into the geometry of this vector space.
1:13:08 The geometry is not Euclidean geometry, but
1:13:11 it's Hamming geometry.
1:13:15 We define the Hamming weight of a vector as simply the
1:13:20 number of 1's in v. So the Hamming weight of the all-0
1:13:29 vector is 0, the Hamming weight of the all-1 vector is
1:13:31 n, and in general, the Hamming weight is somewhere
1:13:34 between 0 and n.
1:13:36 Just the number of 1's.
1:13:37 Pretty simple.
1:13:39 And given two vectors x and y, what is their distance?
1:13:53 The Hamming distance between x and y is equal to the Hamming
1:13:58 weight of x minus y.
1:14:00 This is the standard way of converting a weight metric
1:14:04 into a distance metric.
1:14:06 Or because addition is the same as subtraction in a
1:14:11 binary vector space, we might equally well write this is as
1:14:15 the Hamming weight of x plus y.
1:14:20 And more informally, this is simply the number of places in
1:14:28 which they differ.
1:14:37 OK.
1:14:37 So if x and y are identical, then x plus y is equal to 0
1:14:42 and the distance is 0.
1:14:44 If they are complementary, y is the complement of x, then
1:14:49 they differ in every place.
1:14:51 The sum will then be the all-1 vector, and the weight, the
1:14:57 Hamming distance, will be n.
1:14:59 And so again, the Hamming distance is somewhere between
1:15:02 0 and n, measures how different they are.
1:15:05 Clearly going to be important for coding.
1:15:07 It's going to translate directly into Euclidean
1:15:10 distance under this standard 2-PAM map.
1:15:13 OK.
1:15:17 Again, let's check that it satisfies the distance axioms.
1:15:24 I don't know how many of you have seen this, but let's see.
1:15:31 What are the distance axioms?
1:15:34 Strict non-negativity.
1:15:40 In other words, the Hamming distance between x and y --
1:15:46 that's a single comma --
1:15:48 is greater than or equal to 0, and equality if and
1:15:53 only if x equals y.
1:15:55 That's what strict means.
1:15:58 So if we find the Hamming distance is 0, we can assert
1:16:00 that x equals y.
1:16:04 We have, of course, symmetry.
1:16:09 The Hamming distance between x and y is the same as the
1:16:14 Hamming distance between y and x.
1:16:17 Order doesn't matter.
1:16:18 And finally we have the triangle inequality, that the
1:16:26 Hamming distance between x and z certainly can't be more than
1:16:31 the Hamming distance between x and y plus the Hamming
1:16:36 distance between y and z.
1:16:41 If x differs from y in only n1 places, and y differs from z
1:16:48 in only n2 places, then clearly z can't differ from x
1:16:52 in more than n1 plus n2 places.
1:16:57 So check, check, check.
1:16:59 This is a legitimate metric for defining a geometry on the
1:17:04 space, and this is the one that we use on the space of
1:17:10 all n-tuples.
1:17:12 But notice it's not all like the Euclidean Distance.
1:17:18 Now when we have a linear code --
1:17:22 let's combine these things.
1:17:23 When we have a linear code, we have a group property which
1:17:39 is, let me write it this way.
1:17:40 If we take any code word and add it to any other code word,
1:17:47 that's in the code.
1:17:50 And furthermore, c plus c prime is not equal to c prime
1:17:58 plus c single prime, c plus c single prime,
1:18:03 because we can --
1:18:04 well, I'll finish it.
1:18:06 Unless c prime equals c double prime.
1:18:13 Why is that?
1:18:14 We can do subtraction, cancellation.
1:18:17 Cancel c out from each side.
1:18:19 So if we add c to c double prime, we're going to get a
1:18:25 different result from adding c to c prime, if c prime and c
1:18:30 double prime are different.
1:18:33 So this implies that c plus C --
1:18:39 I write that as an abbreviation for the set of
1:18:43 all c plus c prime, as a c prime runs through the code C.
1:18:50 So this is the 2 to the k sums of the code plus any code word
1:18:59 in the code.
1:19:03 Sorry if I don't write down all steps.
1:19:05 What is that going to be?
1:19:08 C. How did we conclude that?
1:19:16 AUDIENCE: [INAUDIBLE]
1:19:19[UNINTELLIGIBLE]
1:19:24 PROFESSOR: Perfect.
1:19:25 Did you all hear that?
1:19:28 By the group property, each one of these
1:19:30 things is in the code.
1:19:32 By this argument, no two of them are the same.
1:19:36 That means I get 2 to the k distinct elements all in the
1:19:39 code, that's got to be the code, because the code only
1:19:42 has 2 to the k elements.
1:19:46 All right.
1:19:48 So if I add a code word --
1:19:51 in other words, if I write down the code --
1:19:54 0, 0, 0, 0, 1, 1, 1 0 1, 1 1 0- and I add any code word to
1:20:00 it -- say I add 0 1 1 to the code.
1:20:04 So let me just do one column of the addition table.
1:20:08 I get 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1.
1:20:14 I'll get the code itself again.
1:20:21 This has a very important property.
1:20:26 The set of Hamming distances C and c prime, as c prime runs
1:20:38 through c, from any given code word C, it is independent of
1:20:51 C. So I can start from any code word, measure the Hamming
1:20:59 distances, the 2 to the k Hamming distances to all other
1:21:02 code words, including C itself --
1:21:05 let c prime run through the entire code, I'm going to get
1:21:08 a set of 2 to the k distances called the distance
1:21:11 profile of the code.
1:21:16 And I claim that it doesn't matter which code
1:21:19 word I start from.
1:21:21 I'm going to get the same distance profile regardless of
1:21:24 where I start.
1:21:26 In other words, the set of all distances from the all-0 word
1:21:29 here, which is 0, 2, 2, 2, is the same as the set of all
1:21:35 distances from the 0, 1, 1, code word,
1:21:38 which is 2, 0, 2, 2.
1:21:43 And the proof is basically that this is simply the
1:21:47 Hamming weight of C plus c prime as c prime runs through
1:21:51 C. What is this equal to?
1:21:55 So the proof is that this is equal to the Hamming weight of
1:22:01 c prime as c prime runs through C.
1:22:07 So the distance profile from any code word is simply equal
1:22:10 to the weight profile of the code itself.
1:22:14 The weight profile of this code is 0, 2, 2, 2, Start from
1:22:18 any code word, measure the distances to other code words,
1:22:22 I'm always going to get 0, 2, 2, 2, 0, to
1:22:25 itself, and the others.
1:22:29 Sounds sort of like the tetrahedron, doesn't it?
1:22:32 It's zero distance from a code word to itself and equal
1:22:36 distance to all the other code words, in
1:22:39 that particular case.
1:22:41 OK.
1:22:41 So again, everything is very elementary here.
1:22:44 The distance profile is independent of C and equal to
1:22:50 the weight profile.
1:23:06 So this has an extremely important corollary.
1:23:10 What's the minimum Hamming distance of the code?
1:23:14 You might expect this is going to be an
1:23:16 important code parameter.
1:23:17 AUDIENCE: [INAUDIBLE]
1:23:23 PROFESSOR: The minimum Hamming distance between any two code
1:23:25 words is going to be equal to --
1:23:28 I think you said it.
1:23:29 AUDIENCE: The 0 [UNINTELLIGIBLE].
1:23:31 PROFESSOR: Non-zero is important here.
1:23:35 If the distance profile is equal to the weight profile,
1:23:37 one of the weights is always going to be zero.
1:23:40 And that corresponds to the distance between a code word
1:23:43 and itself.
1:23:45 All right.
1:23:46 If I go through all the other 2 to the k minus 1 distances,
1:23:50 they're going to be weights.
1:23:52 They're going to be the distances from a code to all
1:23:54 other code words.
1:23:56 And the minimum distance is simply going to be the minimum
1:24:01 non-zero weight of the code.
1:24:04 For example, in this code, the minimum Hamming distance
1:24:08 between any two distinct code words is going to be equal to
1:24:12 the minimum distance from the 0 code word --
1:24:16 that's another way of doing it.
1:24:17 Since it's independent of C, we may as well take the base
1:24:20 code word C to be zero.
1:24:23 And then what's the minimum distance to 0?
1:24:26 To the 0-code word?
1:24:27 It's the minimum weight.
1:24:29 From the 0 code word, distance equals weight.
1:24:33 So the minimum distance is the minimum weight of any non-zero
1:24:36 code word, which for this code is two.
1:24:43 Now here the weight profile is 0, 2, 2, 2, 0.
1:24:48 The distance profile from any code word to all the others is
1:24:52 0, 2, 2,2 2.
1:24:54 This is always the distance to itself.
1:24:56 So minimum distance to other code words is
1:24:59 always going to be two.
1:25:03 Furthermore, the number of nearest neighbors --
1:25:08 to go back and use chapter five terminology --
1:25:12 the number of nearest neighbors is going to be the
1:25:16 number of code words that have that minimum weight --
1:25:20 in this case, three.
1:25:22 Still sounding a lot like a tetrahedron, isn't it?
1:25:26 This easy map between Hamming distance and Euclidean
1:25:30 distance for this case and in general for all of our cases.
1:25:37 So corollary.
1:25:43 The minimum Hamming distance, which implicitly means between
1:25:48 two distinct code words of C, is equal to the minimum
1:25:55 non-zero weight of C, and the number of minimum weight code
1:26:06 words is independent --
1:26:11 I'm doing this backwards.
1:26:13 From any code word, the number of code words that distance,
1:26:18 let's call this d, is equal to the number of
1:26:25 weight d code words.
1:26:28 Sorry, you probably can't see that.
1:26:36 All right.
1:26:36 So we get this symmetry property for codes that
1:26:39 follows from the group property of the code that if
1:26:45 we stand on any code word and look out, we're always going
1:26:48 to see the same thing.
1:26:51 We have this constant.
1:26:53 It's actually easiest to see this when we make the map from
1:26:57 the code to the Euclidean image of the code.
1:27:08 So the Euclidean image S of C of the code word is going to
1:27:16 be some set of 2 to the k vertices of an
1:27:24 n-cube of side alpha.
1:27:35 Let's talk about the Euclidean image of these properties.
1:27:41 If the minimum Hamming distance of the code is d,
1:27:48 what's the minimum squared Euclidean distance between
1:27:52 elements of S of C going to be?
1:27:58 Well, let's do it coordinate by coordinates.
1:28:03 Let's take two code words, C and c prime, let's say.
1:28:11 And let's suppose we have some Hamming distance
1:28:14 between C and c prime.
1:28:18 That means that c and c prime differ in the Hamming
1:28:23 distance, number of places.
1:28:26 So if we map this into the corresponding pair of vertices
1:28:35 of the n-cube in Euclidean space, S of C and S of c
1:28:38 prime, how many coordinates are these going to differ in?
1:28:46 It's going to differ in same number of coordinates, D_h.
1:28:51 If they don't differ, what's Euclidean squared distance in
1:28:55 those coordinates?
1:28:56 0.
1:28:57 If they do differ, the Euclidean squared distance is
1:29:01 4 alpha squared.
1:29:03 So the Euclidean distance D_e between S of C and S of c
1:29:11 prime is simply going to be 4 alpha squared times the
1:29:15 Hamming distance between C and c prime, yes?
1:29:21 So I should say this is the squared Euclidean distance.
1:29:25 Why do we always talk about the
1:29:26 squared Euclidean distance?
1:29:27 Because it's additive, coordinate-wise.
1:29:30 And the Hamming distance is additive, coordinate-wise.
1:29:33 So there's a nice easy map here.
1:29:38 So what does this mean d_min squared is going to be?
1:29:44 d_min squared of, let's say, S of C. This constellation that
1:29:50 we've formed by taking the Euclidean image of c.
1:29:54 The minimum square distance between points in S of C is
1:30:00 just going to be 4 alpha squared times d, where I don't
1:30:05 think I ever --
1:30:06 d equals min Hamming distance.
1:30:13 And we're always going to talk about n, k, d as the three key
1:30:18 parameters of a binary linear block code.
1:30:22 n is the code length, F2 to the n, k is the dimension, d
1:30:26 is the minimum Hamming distance.
1:30:29 So by going into this Hamming geometry, we've got a third
1:30:32 key property of the code.
1:30:34 And we see it's key, because we can get the minimum squared
1:30:37 distance between this Euclidean image constellation,
1:30:40 just 4 alpha squared d.
1:30:43 AUDIENCE: nd makes a probability of
1:30:45[UNINTELLIGIBLE] that is dependent on [UNINTELLIGIBLE]
1:30:48 then.
1:30:49 PROFESSOR: Correct.
1:30:49 This is all we need to know to get the union bound estimate.
1:30:53 Well, a few more things.
1:30:55 We need to know what K_min average of S of C. And what is
1:31:05 that going to be?
1:31:08 This is simply going to be the number of words in the code.
1:31:13 To get this minimum squared distance, we need a Hamming
1:31:17 distance of d.
1:31:20 So the number of words in the code of distance d, which is
1:31:24 given by the parameter n sub d, is simply going to be the
1:31:29 number of nearest neighbors.
1:31:31 Not just the average distance, but I want to emphasize this
1:31:35 symmetry property.
1:31:36 If we stand on any point, on any vertex of this cube in
1:31:41 n-space, which is the code vertex, and we look at all the
1:31:44 other points in the constellation, no matter which
1:31:48 point we stand on, we will always see the
1:31:50 same profile of distances.
1:31:53 We'll see precisely nd code words at Euclidean distance 4
1:31:57 alpha squared d.
1:31:58 We'll see nd plus 1 at Euclidean squared distance 4
1:32:03 alpha squared d plus 1, and so forth, right up the profile.
1:32:08 So there's complete symmetry in the constellation.
1:32:12 In that universe, you don't know which code point you're
1:32:15 standing on just by looking out, because the world looks
1:32:17 the same to you.
1:32:20 Is that clear?
1:32:23 OK.
1:32:23 So from a communications point of view, this is important,
1:32:27 because it means it doesn't matter what code word we send.
1:32:30 The probability of error from any code word is going to be
1:32:34 the same as the probability of error from any other code
1:32:36 word, because the geometry is exactly the same.
1:32:40 The Voronoi regions are all the same shape.
1:32:43 So given the exact probability of error, not just the
1:32:46 union-bound estimate, is going to be independent of which
1:32:48 code word was sent.
1:32:50 This all follows from the fact that it's a linear code and
1:32:54 therefore has the group property, which translates
1:32:58 into this very strong geometrical uniformity
1:33:01 property in Euclidean space.
1:33:04 Or actually in Hamming space too, but it's more striking in
1:33:08 Euclidean space.
1:33:11 OK?
1:33:13 So we have everything we need to write down
1:33:17 the union bound estimate.
1:33:22 Union bound estimate was just the probably of error per bit
1:33:30 is well approximated by K_b of constellation, in this case, S
1:33:38 of C, times Q of the square root of the coding gain of the
1:33:48 constellation times 2 Eb over N_0.
Related Resources
Introduction to Binary Block Codes (PDF)
Course Info
Instructor
Prof. David Forney
Departments
Electrical Engineering and Computer Science
As Taught In
Spring 2005
Level
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Topics
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4153 | https://ditki.com/course/usmle-comlex-step-2/glossary/pathophysiologic-disorder/hyperparathyroidism-for-usmle-step-2 | USMLE / COMLEX - Step 2 Glossary: Hyperparathyroidism for USMLE Step 2 | ditki medical & biological sciences
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Hyperparathyroidism for the USMLE Step 2 Exam
Pathophysiology
Parathyroid hormone (PTH) plays a key role in regulating serum calcium by increasing bone resorption, enhancing renal calcium reabsorption, and stimulating vitamin D activation to promote gastrointestinal calcium absorption. PTH also promotes renal phosphate excretion.
Primary Hyperparathyroidism: Occurs when one or more of the parathyroid glands autonomously secrete excessive PTH, leading to hypercalcemia. Most cases are due to a parathyroid adenoma (85%), with fewer cases caused by parathyroid hyperplasia (10-15%) or, rarely, parathyroid carcinoma.
Secondary Hyperparathyroidism: Develops as a compensatory response to chronic hypocalcemia, commonly seen in chronic kidney disease (CKD) or vitamin D deficiency.
Tertiary Hyperparathyroidism: Occurs when prolonged secondary hyperparathyroidism leads to autonomous PTH secretion, often in patients with end-stage renal disease.
Etiology
Primary Hyperparathyroidism:
Parathyroid Adenoma: The most common cause.
Parathyroid Hyperplasia: Often associated with familial syndromes like MEN1 or MEN2A.
Parathyroid Carcinoma: Rare, but presents with severe hypercalcemia.
Secondary Hyperparathyroidism:
Chronic Kidney Disease (CKD): Impaired renal function leads to reduced vitamin D activation, hypocalcemia, and compensatory PTH secretion.
Vitamin D Deficiency: Causes hypocalcemia, prompting increased PTH secretion.
Clinical Features
Primary Hyperparathyroidism:
Often asymptomatic, diagnosed during routine screening for hypercalcemia.
Symptomatic Hypercalcemia:
Bone pain and fractures: Due to increased bone resorption (osteoporosis, osteitis fibrosa cystica).
Nephrolithiasis: Kidney stones due to hypercalciuria.
Gastrointestinal symptoms: Nausea, constipation, and abdominal pain.
Neuropsychiatric symptoms: Fatigue, depression, and cognitive impairment.
Secondary Hyperparathyroidism:
Bone pain and muscle weakness due to chronic hypocalcemia and high PTH levels.
Underlying conditions like CKD or vitamin D deficiency are common.
Diagnosis
Serum Calcium: Elevated in primary hyperparathyroidism, but low or normal in secondary.
Serum PTH: Elevated in both primary and secondary hyperparathyroidism.
Serum Phosphate: Low in primary hyperparathyroidism, but elevated in CKD-related secondary hyperparathyroidism.
Vitamin D Levels: Checked to rule out deficiency, especially in secondary hyperparathyroidism.
Imaging:
Sestamibi Scan: Localizes parathyroid adenomas for surgical planning.
Neck Ultrasound: Can help identify enlarged parathyroid glands.
Management
Primary Hyperparathyroidism:
Parathyroidectomy: The definitive treatment, especially indicated in symptomatic patients or those with complications (osteoporosis, nephrolithiasis, or significant hypercalcemia). It is also recommended for asymptomatic patients with calcium >1 mg/dL above the upper normal limit, osteoporosis, reduced kidney function, or age <50.
Medical Management:
Hydration: To prevent kidney stones.
Cinacalcet: A calcimimetic that reduces PTH secretion, used in patients who are not surgical candidates.
Bisphosphonates: Used to treat osteoporosis by reducing bone resorption.
Secondary Hyperparathyroidism:
Treat the underlying cause: Vitamin D supplementation and managing CKD.
Phosphate Binders: Lower serum phosphate in CKD.
Active Vitamin D Analogues (Calcitriol): Suppress PTH secretion in CKD.
Cinacalcet: Useful in reducing PTH levels in CKD patients who cannot undergo surgery.
Complications
Osteoporosis and Fractures: Due to excessive bone resorption.
Nephrolithiasis: Kidney stones from chronic hypercalcemia and hypercalciuria.
Hypercalcemia Crisis: Severe hypercalcemia leading to dehydration, confusion, and cardiac arrhythmias. Requires aggressive treatment with IV fluids, bisphosphonates, and calcitonin.
Renal Impairment: Hypercalcemia can cause nephrocalcinosis and renal insufficiency.
Key Points
Primary Hyperparathyroidism is most often caused by a parathyroid adenoma and leads to hypercalcemia. Common clinical features include bone pain, kidney stones, and gastrointestinal or neuropsychiatric symptoms.
Secondary Hyperparathyroidism arises due to chronic hypocalcemia, commonly seen in CKD or vitamin D deficiency. Treatment focuses on correcting the underlying cause with phosphate binders and vitamin D supplementation.
Parathyroidectomy is the definitive treatment for symptomatic primary hyperparathyroidism, while cinacalcet and bisphosphonates are used in patients who are not surgical candidates.
Complications include osteoporosis, nephrolithiasis, and, in severe cases, hypercalcemic crisis.
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Hyperparathyroidism |
4154 | https://planning.org/knowledgebase/resource/9182244 | Urban Expansion, Sprawl and Inequality
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Urban Expansion, Sprawl and Inequality
Landscape and Urban Planning, 177: 259-265, 2018
By: , Reid Ewing
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Benefits of Compact, Mixed Use Development
This article introduces a special issue of Landscape and Urban Planning on the topic of urban sprawl. It discusses the outcomes of sprawl in economic, social, and environmental dimensions and calls for more research on the relationships between sprawl, spatial inequality, and sustainable development.
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4155 | https://www.doubtnut.com/qna/644633694 | Let a line y=mx(m>0) intersect the parabola, y2=4x at a point P, other than the origin. Let the tangent to it at P meet the x-axis at the point Q. If area (ΔOPQ)=8 sq. units, then m is equal to _______ .
The correct Answer is:1
To solve the problem, we need to find the value of m such that the area of triangle OPQ is 8 square units, where O is the origin, P is a point on the parabola y2=4x, and Q is the point where the tangent at P meets the x-axis. 1. Identify the point P on the parabola: The parabola is given by the equation y2=4x. A point P on this parabola can be represented as P(t)=(t2,2t), where t is a parameter. 2. Equation of the tangent at point P: The equation of the tangent to the parabola at point P(t) is given by: 2ty=x+t2 Rearranging this gives: x=2ty−t2 3. Find the point Q where the tangent meets the x-axis: The x-axis is defined by y=0. Substituting y=0 into the tangent equation: x=2t(0)−t2=−t2 Thus, the coordinates of point Q are Q(−t2,0). 4. Coordinates of points O, P, and Q: - O(0,0) - P(t)=(t2,2t) - Q(−t2,0) 5. Area of triangle OPQ: The area A of triangle OPQ can be calculated using the formula: A=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)| Substituting the coordinates: A=12∣∣0(2t−0)+t2(0−0)+(−t2)(0−2t)∣∣ Simplifying this gives: A=12∣∣−t2(−2t)∣∣=12∣∣2t3∣∣=t3 We know from the problem statement that the area A=8: t3=8⟹t=2 6. Relate t to m: From the line equation y=mx, we also know that: 2t=m(t2) Substituting t=2: 2(2)=m(22)⟹4=4m⟹m=1 Final Answer: Thus, the value of m is: 1
To solve the problem, we need to find the value of m such that the area of triangle OPQ is 8 square units, where O is the origin, P is a point on the parabola y2=4x, and Q is the point where the tangent at P meets the x-axis.
Identify the point P on the parabola:
The parabola is given by the equation y2=4x. A point P on this parabola can be represented as P(t)=(t2,2t), where t is a parameter.
Equation of the tangent at point P:
The equation of the tangent to the parabola at point P(t) is given by:
2ty=x+t2
Rearranging this gives:
x=2ty−t2
Find the point Q where the tangent meets the x-axis:
The x-axis is defined by y=0. Substituting y=0 into the tangent equation:
x=2t(0)−t2=−t2
Thus, the coordinates of point Q are Q(−t2,0).
Coordinates of points O, P, and Q:
O(0,0)
P(t)=(t2,2t)
Q(−t2,0)
Area of triangle OPQ:
The area A of triangle OPQ can be calculated using the formula:
A=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|
Substituting the coordinates:
A=12∣∣0(2t−0)+t2(0−0)+(−t2)(0−2t)∣∣
Simplifying this gives:
A=12∣∣−t2(−2t)∣∣=12∣∣2t3∣∣=t3
We know from the problem statement that the area A=8:
t3=8⟹t=2
Relate t to m:
From the line equation y=mx, we also know that:
2t=m(t2)
Substituting t=2:
2(2)=m(22)⟹4=4m⟹m=1
Final Answer:
Thus, the value of m is:
1
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4156 | https://openorthopaedicsjournal.com/VOLUME/6/PAGE/108/ | Airway Compromise Due to Wound Hematoma Following Anterior Cervical Spine Surgery
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RESEARCH ARTICLE
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Airway Compromise Due to Wound Hematoma Following Anterior Cervical Spine Surgery
Mark A Palumbo, , 1
E-Mail Address of Dr. Mark A Palumbo
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mpalmd@aol.com
Jessica Pelow Aidlen 1 ##### E-Mail Address of Dr. Jessica Pelow Aidlen × mpalmd@aol.comAlan H Daniels 1 ##### E-Mail Address of Dr. Alan H Daniels × mpalmd@aol.comNikhil A Thakur 1 ##### E-Mail Address of Dr. Nikhil A Thakur × mpalmd@aol.comandJoseph Caiati 2 ##### E-Mail Address of Dr. Joseph Caiati × mpalmd@aol.comAuthors Info & Affiliations
The Open Orthopaedics Journal • 05 Mar 2012 • RESEARCH ARTICLE • DOI: 10.2174/1874325001206010108
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ABSTRACT
INTRODUCTION
CASE PRESENTATION
CLINICAL ANATOMY
ETIOLOGY
DIAGNOSTIC CONSIDERATIONS
PREVENTION
MANAGEMENT OF POSTOPERATIVE AIRWAY COMPROMISE
INITIAL ASSESSMENT
GENERAL GUIDELINES
CLINICAL SCENARIOS
CONCLUSION
CONFLICT OF INTEREST
REFERENCES
Abstract
One of the most serious adverse events associated with anterior cervical spine surgery is wound hematoma resulting in airway compromise. The reported incidence of this postoperative complication has varied from 0.2% to 1.9%. Obstruction of the airway secondary to bleeding presents a challenging clinical scenario given the rapidity of onset, distorted anatomy of the upper respiratory tract, urgent need to act and potential for catastrophic consequences. This high-risk, life-threatening clinical scenario requires specialized knowledge and a well-designed treatment protocol to achieve a positive outcome. In this review, we report a case of airway compromise secondary to wound hematoma following anterior cervical discectomy and fusion, followed by a review of relevant literature, anatomy, etiologic factors and diagnostic considerations. We also propose guidelines for the prevention and management of postoperative airway obstruction due to wound hematoma.
Keywords:Anterior cervical spine surgery, adverse event, airway obstruction, hematoma.
INTRODUCTION
Anterior cervical spine operations are commonly performed by orthopedic and neurological surgeons. These procedures generally involve decompression of the neural elements by discectomy or corpectomy in combination with reconstruction of the spinal column at one or more levels. Available reconstructive techniques include arthrodesis (with or without instrumentation) and disc arthroplasty. Reported rates of clinical success are high and complications are generally infrequent.
One of the most serious adverse events associated with anterior cervical spine surgery is postoperative airway obstruction due to wound hematoma. The reported incidence of this complication has varied from 0.2% to 1.9% [1-4]. To date, there have not been any Level 1, 2, or 3 studies published examining the topic of airway compromise due to wound hematoma following anterior cervical surgery. Current treatment protocols are based on Level 4 and 5 data derived from case reports, retrospective series and expert opinion.
Airway compromise caused by hemorrhage presents a challenging clinical scenario given the rapidity of onset, limited time to act, and potential for catastrophic consequences including cerebral ischemia and death. Further, standard airway control measures may prove difficult to employ due to anatomic distortion of the upper respiratory tract and ventral neck.
Specialized knowledge and advance planning is necessary to respond efficiently and effectively in the setting of this life-threatening complication. Due to the low incidence of the problem, however, it is not universal policy for the medical team to formulate a specific protocol for the prevention and management of airway compromise. The primary objective of this review is to describe a systematic approach to the patient with airway obstruction due to wound hematoma after anterior cervical spine surgery. Guidelines for prevention and management will be preceded by a review of clinically relevant anatomy, etiologic factors and diagnostic considerations.
CASE PRESENTATION
A 52 year old female presented with a two year history of neck and right upper extremity pain along with paresthesias involving the thumb and index finger. The pain was progressive in nature and had become disabling despite comprehensive nonoperative management. Magnetic Resonance Imaging (MRI) showed a moderate paracentral disc herniation to the right of midline at the C5/6 level with compromise of the exiting C6 nerve root.
Her past medical history was significant for emphysema, bipolar disorder and epilepsy. There was no history of prior neck surgery. Physical examination was significant for low grade weakness of the right wrist extensor muscles and a decreased right brachialradialis reflex.
Anterior cervical discectomy and fusion (ACDF) was performed at the C5-6 level through a transverse, left-sided incision utilizing iliac crest autograft and anterior plate fixation (Fig. 1). There were no intraoperative complications. At the time of wound closure, hemostasis was adequate. The patient was extubated and transferred to the recovery room in stable condition. Six hours following completion of the operation, the patient began complaining of difficulty swallowing and breathing. She was agitated and tachypneic; oxygen saturation was >95% on room air. Her neurologic exam was unchanged from her preoperative exam. Examination of her anterior neck revealed fullness and submandibular edema. The suture line was intact with no incisional drainage. The hemovac drain reservoir showed minimal bloody drainage and the tubing appeared to be clotted.
Fig. (1).
Intraoperative lateral cervical spine xray showing anterior cervical discectomy and fusion of C5/6.
OPEN IN VIEWER
Immediate consultation with anesthesia and an otolaryngologist was obtained. The patient was taken emergently to the operating room. Bag-mask assisted ventilation was instituted. An attempt at awake intubation was compromised by poor visualization of the upper airway landmarks and edema of the pharyngeal wall, thus bag mask assisted ventilation was continued with the plan to attempt endotracheal intubation after hematoma evacuation. Next, the incision was opened under local anesthesia and a large hematoma was evacuated from the space between the carotid sheath and the midline viscera. After hematoma evacuation, standard endotracheal intubation was still unsuccessful due to upper airway edema. With the patient becoming more hypercarbic and tachycardic, emergent open tracheostomy was performed by the otolaryngologist through the prior left-sided incision, which was extended across the midline. After securing the airway, formal wound exploration revealed no clear source of arterial or venous bleeding.
Postoperatively, she was transferred to the intensive care unit. Her tracheostomy was removed on postoperative day #4 and she was discharged home on the fifth postoperative day. The remainder of her postoperative course was unremarkable. She experienced major improvement in her preoperative pain symptoms and had normalization of strength in the right C6 innervated musculature.
CLINICAL ANATOMY
Effective management of airway compromise after cervical spine surgery requires an understanding of the surface anatomy of the anterior neck, the internal structure of the upper respiratory tract, and the surgical approach. The relevant external landmarks consist of the hyoid bone, the thyroid cartilage and the cricoid cartilage. Internally, the epiglottis separates the oropharynx from the laryngopharynx at approximately the level of the fourth cervical vertebra (Fig. 2). The larynx, extending from the lower pharynx to the trachea, is composed of a cartilaginous skeleton which includes the thyroid and cricoid cartilage along with the intervening cricothyroid membrane. Integration of ligaments and muscles with its cartilage components allows the larynx to serve as a soft tissue sphincter preventing passage of food and liquid into the tracheal lumen during swallowing.
Fig. (2).
Midsagittal section of the upper respiratory tract. OP: oropharnyx. LP: laryngopharnyx. TR: trachea. ES: esophagus. EP: epiglottis. TC: thyroid cartilage. CC: cricoid cartilage. Arrow A: path for direct laryngoscopic orotracheal intubation. Arrow B: path for surgical cricothyrotomy. Arrow C: path for high surgical traceostomy.
OPEN IN VIEWER
The surgical exposure for anterior cervical spine surgery can be performed from either side of the midline via a transverse or longitudinal incision. Access to the vertebral column is achieved by dissection in the plane between the carotid sheath and the midline viscera (Fig. 3) . During surgery, the upper airway structures and esophagus are retracted in a medial direction. The longus colli musculature is elevated off the ventral surface of the spine to expose the vertebral bodies and disc spaces. This exposure creates a potential space along one side and directly posterior to the larynx and trachea.
Fig. (3).
The anterior cervical approach to the spine (cross-sectional view): note the potential space created by dissection in the plane between the carotid sheath and the midline viscera. The carotid sheath structures are retracted laterally, and the esophagus and trachea are retracted medially to expose the ventral surface of the spine. Adapted with permission from: Albert T, Balderston R, Northrup B. Surgical Approaches to the Spine. Philadelphia: WB Saunders 1997; p. 10 .
OPEN IN VIEWER
ETIOLOGY
Hematoma following anterior cervical spine surgery may be the result of inadequate control of arterial or venous bleeding during the operation, and has been reported due to superior thyroid artery dissection . In other instances, a hematoma can form after surgery irrespective of adequate intraoperative hemostasis. Postoperative hemorrhage may occur secondary to coagulopathy, increased blood pressure during emergence from anesthesia, or elevated venous pressure due to the Valsalva effect of coughing at the time of extubation .
There are two potential pathophysiologic mechanisms by which hematoma can produce airway compromise. The first is direct mechanical compression leading to reduction in the cross-sectional area of the airway lumen. The second mechanism involves the development of intrinsic airway edema in response to the mass effect of collected blood within the surgical wound. In this situation, impaired venous drainage from the head leads to secondary swelling of the supraglottic structures, the epiglottis and the arytenoids. This mechanism of progressive airway obstruction has been promoted as more plausible than extrinsic compression given the low pressure required to occlude the venous and lymphatic systems in the neck and the relatively high pressure required to directly collapse the airway .
Cephalad to and including the glottic opening (at the C4 level), the airway can be deformed and obscured by both external compression and internal edema of the layngopharyngeal soft tissue. Caudal to the glottic opening, the airway is more resistant to external compression due to support provided by the laryngeal and tracheal cartilage. The subglottic region, however, is vulnerable to luminal narrowing when mucosal edema develops within the confines of the cartilaginous skeleton. The individual or combined effects of edema and mechanical compression can compromise both spontaneous respiration and the execution of laryngoscopy.
DIAGNOSTIC CONSIDERATIONS
The underlying cause of airway compromise is suggested by the elapsed time to onset after the spine operation. Upper airway compromise that occurs during the immediate postoperative period (within 12 hours of completion of surgery) is most commonly the result of an expanding wound hematoma. It is noteworthy that obstruction evident within the first hour after extubation may be related to residual neuromuscular blockade or hypoventilation from narcotic medication. Although delayed hematoma is possible beyond the first twelve hours, alternative causes of airway obstruction (e.g. pharyngeal/prevertebral edema, spinal construct failure, cerebrospinal fluid collection or retropharyngeal abscess) become more probable.
In general, airway compromise after spine surgery can produce a spectrum of clinical findings. The patient may progress at a variable rate from being asymptomatic to exhibiting symptoms and signs of partial occlusion and then complete obstruction (Table 1). In the early stages, complaints of difficulty talking and breathing predominate. Subtle changes in voice quality may be evident. Breathing problems may be exacerbated by the supine position. As the process evolves, the patient may become restless and agitated due to hypercarbia (with or without hypoxia). Dyspnea, inspiratory stridor and cyanosis develop during later stages of airway compromise with potential progression to respiratory failure/arrest.
Table 1.
Clinical Findings Associated with Postoperative Wound Hematoma
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OPEN IN VIEWER
Table 2.
Additional Risk Factors For Postoperative Airway Compromise
| Surgical | Patient | Anesthetic | Institutional |
--- --- |
| Exposure of > 3 vertebral bodies | Morbid Obesity | Suboptimal airway visualization | No 24 hour in-house anesthesia care |
| Exposure of C2-C4 levels | Obstructive Sleep Apnea | Multiple intubation attempts | No 24 hour in-house surgical staff |
| Blood loss > 300 ml | Pulmonary disease | Need for fiberoptic intubation |
| Operative time > 5 hours | Cervical Myelopathy |
| Dual approach operations | Prior anterior cervical surgery |
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OPEN IN VIEWER
Compared to other potential causative factors, wound hematoma tends to produce a more rapid rate of progression of respiratory compromise along with certain characteristic signs. Post-operative hemorrhage can be associated with significant drain output, suture line bleeding, progressive swelling of the anterior neck and submandibular region, facial edema/plethora and tracheal deviation [7,9-11]. Sudden hemodynamic instability has been observed in the setting of an expanding neck hematoma after anterior cervical surgery. In the report by Sethi et al., bradycardia and hypotension was attributed to mechanical distortion of the carotid sinus with activation of the baroreceptor reflex .
PREVENTION
Wound hematoma after anterior cervical spine surgery is difficult to predict and prevent. The complication often occurs in the absence of predisposing factors. Nonetheless, the probability of postoperative hemorrhage can theoretically be reduced by attention to certain basic surgical principles. Specifically, an effort should be made to control all soft tissue and osseous bleeding during the operation. Placement of a closed suction drain is also reasonable (though with recognition that the measure will not negate the potential for hematoma formation).
Postoperative measures to protect the upper respiratory tract can be employed in the patient at elevated risk of hemorrhage into the wound. The presence of a bleeding diathesis (e.g. anticoagulant therapy, severe liver disease, hereditary bleeding disorder) and/or excessive intraoperative blood loss should raise concern for bleeding after surgery. Monitoring in an intensive care unit for 24-72 hours (with or without delayed removal of the endotracheal tube) should be considered. The threshold for delaying extubation should be even lower in the presence of additional risk factors for postoperative airway compromise [12-14] (Table 2).
MANAGEMENT OF POSTOPERATIVE AIRWAY COMPROMISE
Postoperative airway compromise due to hematoma poses a challenging clinical scenario. This situation typically involves a patient who is extubated immediately after surgery and then develops respiratory distress after transfer to the post-anesthesia care unit or the surgical ward. The time to onset of this complication (< 12 hours after surgery) is such that the on-site night staff will often have responsibility for acute management of the patient [7,9,11].
An efficient and systematic approach to treatment is necessary to achieve a positive outcome. Early detection and intervention before the onset of occlusive laryngopharyngeal edema generally portends a positive outcome. The initial assessment functions to characterize the severity of airway obstruction. Based on this clinical information, the appropriate location to render care and the specific treatment methodology is established.
INITIAL ASSESSMENT
Following anterior cervical spine surgery, a patient with symptoms attributable to the airway must be evaluated immediately. If available in the hospital, the responsible spine surgeon and anesthesiologist should assess the situation. At this point, preparation to secure the airway is a priority; multispecialty assistance should be mobilized as rapidly as possible.
Upon reaching the bedside, the first responders must rapidly assess the clinical situation. A focused physical examination incorporates removal of the orthosis, visualization of the anterior neck tissues and identification of signs of airway obstruction. Identification of an expanding wound hematoma is based on exam findings and knowledge of the time elapsed since surgery. Any other immediately available data are taken into consideration (e.g. pulse oximetry, review of patient comorbidities, assessment of available hospital resources, etc).
The severity of airway compromise is characterized in order to determine whether the patient can be safely transported to the operating room. Non-critical airway compromise is associated with minimal/partial loss of airway patency. Clinically, the patient is not in extremis. Complaints of difficulty breathing and talking may be accompanied by dyspnea and subtle changes in voice quality. Characterization of the airway compromise as non-critical reflects a clinical judgment that the situation is not immediately life-threatening and that transfer of the patient to the operating room is reasonable.
Critical airway compromise is associated with near-total or completes loss of airway patency. Representative signs include a look of fear or panic associated with air hunger, excessive salivation, a rocking motion of the head and chest with the respiratory cycle, use of accessory muscles of respiration, inspiratory stridor and, eventually, central cyanosis. Characterization of the airway compromise as critical is indicative of a situation judged to be immediately life-threatening; management must be initiated at the bedside.
GENERAL GUIDELINES
The primary treatment objective is to establish and maintain patency of the airway by placement of an endotracheal tube. Intubation can be achieved by direct laryngoscopy, fiberoptic bronchoscopy, or by surgical means. In this regard, there are certain fundamental principles pertinent to the emergency management of airway compromise after anterior cervical spine surgery.
The optimal setting to secure the airway is in the operating room suite.
When supplemental oxygen is available, delivery of 100% oxygen via non-rebreather facemask should be the first step in treatment.
The use of intravenous sedatives should be minimized. Sedation can reduce voluntary efforts to maintain airway patency and risks converting partial occlusion to complete obstruction.
If direct laryngoscopy is the chosen method of intubation, it should be undertaken with full recognition that visualization of anatomic landmarks and endotracheal tube passage can be challenging due to edema of the pharyngeal wall, epiglottis and vocal cords. The first attempt provides the best opportunity for intubation and should be executed by the most qualified provider.
It is mandatory to proceed through the airway management protocol with maximal efficiency. Repetition of failed intubation attempts is to be avoided. Delay in the establishment of a surgical airway prolongs the period of hypoxia and can lead to cerebral ischemia.
If surgical airway access is deemed necessary, cricothyroidotomy provides a rapid means of securing the airway and is the procedure of choice.
CLINICAL SCENARIOS
Non-Critical Airway Compromise
For the most common scenario of non-critical airway compromise, the patient is efficiently transported to the operative suite. The surgical team should be assembled and ready with a tracheostomy tray available. Standard monitoring is initiated and the patient is hyper-oxygenated by face mask ventilation. Awake intubation utilizing a fiberoptic (or direct laryngoscopic) technique is attempted if there is imminent danger of losing the airway.
In the case of failed intubation and an obvious space-occupying hematoma, the incision is opened under local anesthetic while maintaining bag-mask ventilation. Separation of the subcutaneous tissue is followed by removal of all sutures reapproximating the platysma muscle. Blood clot is evacuated from alongside and posterior to the midline viscera. After decompression of the wound, a second trial of intubation is executed.
If intubation is unsuccessful, a surgical airway will need to be established by cricothyroidotomy. The spine surgeon, anesthesiologist, intensivists, and house staff must be competent in the execution of this procedure. The technique of cricothyroidotomy following anterior spine surgery incorporates extension of the original incision across the midline. The cricothyroid membrane is sharply divided and the opening dilated using a curved hemostat. A small endotracheal or tracheostomy tube (6-7 mm diameter) is inserted, and the cuff inflated.
In the presence of a space-occupying lesion, tracheal deviation and submandibular soft tissue swelling can substantially increase the technical difficulty of the surgical procedure. If time allows, recruitment of a specialist with expertise in surgical airway creation is beneficial. General anesthesia is instituted once the endotracheal tube is in place. The surgical team proceeds to formal wound exploration for complete evacuation of the hematoma and control of any ongoing hemorrhage.
Critical Airway Compromise
In the situation of critical airway compromise, the emergency response system is activated and treatment is initiated at the bedside. The first step is to administer 100% oxygen via non-rebreather facemask. If adequate bag-mask ventilation can be established, the patient is transported to the operative suite for definitive airway control. If ventilation is inadequate, awake intubation by fiberoptic airway management or direct laryngoscopy is attempted. In the case of successful endotracheal tube placement, transfer to the operating room is arranged for wound exploration.
If the initial intubation attempt fails (or airway management equipment is not immediately available), the surgical wound is opened and blood clot removed. Following wound decompression, another trial of face mask ventilation is initiated. If adequate ventilation is not possible and the clinical status allows, an additional attempt at intubation is performed.
If intubation is not rapidly achieved, a surgical airway is established by cricothyroidotomy. Once the endotracheal tube has been secured and ventilation restored, the patient is moved to the operative suite for formal wound exploration and control of any ongoing hemorrhage. Conversion to a tracheostomy can be performed during the same operation or at a later time.
CONCLUSION
Airway obstruction due to wound hematoma is a rare but potentially lethal complication of anterior cervical spine surgery. This clinical scenario is challenging given the often rapid onset of airway compromise, the limited time to act and the distortion of the upper respiratory tract and ventral neck anatomy. Furthermore, the clinicians of first response are often ill-prepared to deal with this unique type of airway emergency. We report a case of hematoma causing airway compromise following anterior cervical discectomy and fusion in which quick action to evacuate the hematoma and appropriate consultation with otolaryngology to secure the airway led to a successful outcome. There are no Level 1, 2 or 3 studies upon which to base guidelines for this challenging clinical scenario. To optimize outcome, specialized knowledge and a systematic approach to management is necessary. Progress through the airway management protocol must be maximally efficient. Delay in securing the airway prolongs the period of hypoxia and can lead to cerebral ischemia and death. As such, all medical personnel responsible for patient care after cervical spine surgery should be aware of this potential complication to respond effectively to this adverse event.
CONFLICT OF INTEREST
Declared none.
ACKNOWLEDGEMENT
We would like to acknowledge Michelle L. Daniels, DMD for assistance with illustrations.
REFERENCES
1
Tew JM, Mayfield FH. Complications of surgery of the anterior cervical spine Clin Neurosurg 1976; 23: 424-34.
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2
Bertalanffy H, Eggert HR. Complications of anterior cervical discectomy without fusion in 450 consecutive patients Acta Neurochir (Wien) 1989; 99: 41-50.
3
Emery SE, Bohlman HH, Bolesta MJ, Jones PK. Anterior cervical decompression and arthrodesis for the treatment of cervical spondylotic myelopathy: Two to seventeen-year follow-up J Bone Joint Surg Am 1998; 80A: 941-51.
4
Marotta N, Landi A, Tarantino R, Mancarella C, Ruggeri A, Delfini R. Five-year outcome of stand-alone fusion using carbon cages in cervical disc arthrosis Eur Spine J 2011; 20(Suppl 1): S8-12.
Go to reference
5
Albert T, Balderston R, Northrup B. Surgical approaches to the spine. Philadelphia: WB Saunders 1997; p. 10.
a [...] sheath and the midline viscera (Fig.
[b [...] Philadelphia: WB Saunders 1997; p. 10
6
Yu NH, Jahng TA, Kim CH, Chung CK. Life-threatening late hemorrhage due to superior thyroid artery dissection after anterior cervical discectomy and fusion Spine (Phila Pa 1976) 2010; 35(15): E739-42.
Go to reference
7
Sethi R, Tandon M, Ganjoo P. Neck hematoma causing acute airway and hemodynamic compromise after anterior cervical spine surgery J Neurosurg Anesthesiol 2008; 20(1): 69-70.
[a [...] of coughing at the time of extubation
[b [...] edema/plethora and tracheal deviation
[c [...] activation of the baroreceptor reflex
[d [...] for acute management of the patient
8
Carr EM, Benjamin E. In vitro study investigating post neck surgery haematoma airway obstruction J Laryngol Otol 2009; 122: 662-5.
Go to reference
9
Lee HS, Lee BJ, Kim SW, et al. Patterns of post-thyroidectomy hemorrhage Clin Exp Otorhinolaryngol 2009; 2(2): 72-.
[a [...] edema/plethora and tracheal deviation
[b [...] for acute management of the patient
10
Gwinnutt CL, Walsh GR, Kumar R. Airway obstruction after anterior cervical spine surgery J Neurosurg Anesthesiol 1992; 4(3): 99-202.
11
Shakespeare WA, Lanier WL, Perkins WJ, Pasternak JJ. Airway management in patients who develop neck hematomas after carotid endarterectomy Anesth Analg 2010; 110(2): 588-93.
[a [...] edema/plethora and tracheal deviation
[b [...] for acute management of the patient
12
Sagi HC, Beutler W, Carroll E, Connolly PJ. Airway complications associated with surgery on the anterior cervical spine Spine 2002; 27(9): 949-53.
Go to reference
13
Kwon B, Yoo JU, Furey CG, Rowbottom J, Emery SE. Risk factors for delayed extubation after single-stage, multi-level anterior cervical decompression and posterior fusion J Spinal Disord Tech 2006; (19): 389-93.
14
Terao Y, Matsumoto S, Yamashita K, et al. Increased incidence of emergency airway management after combined anterior-posterior cervical spine surgery J Neurosurg Anesthesiol 2004; (16): 282-6.
Go to reference
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Authors & Information
Authors Information
Authors
Affiliations Expand All
Mark A Palumbo, 1
Department of Orthopaedic Surgery, The Warren Alpert Medical School of Brown University, Providence, Rhode Island, USA
Jessica Pelow Aidlen1
Department of Orthopaedic Surgery, The Warren Alpert Medical School of Brown University, Providence, Rhode Island, USA
Alan H Daniels1
Department of Orthopaedic Surgery, The Warren Alpert Medical School of Brown University, Providence, Rhode Island, USA
Nikhil A Thakur1
Department of Orthopaedic Surgery, The Warren Alpert Medical School of Brown University, Providence, Rhode Island, USA
Joseph Caiati2
Department of Anesthesia, The Warren Alpert Medical School of Brown University, Providence, Rhode Island, USA
Information
Published In
The Open Orthopaedics Journal
RESEARCH ARTICLE
Article Information
Year: 2012
Volume: 6
E-location ID:TOORTHJ-6-108
DOI:10.2174/1874325001206010108
Cite As
Palumbo M, Aidlen J, Daniels A, Thakur N, Caiati J. Airway Compromise Due to Wound Hematoma Following Anterior Cervical Spine Surgery . Open Orthop J, 2012; 6: .
Article History
Received Date: 23 Dec 2011
Revision Received Date: 11 Jan 2012
Electronic Publication Date:05 Mar 2012
Copyright
© Palumbo et al.; Licensee Bentham Open.
open-access license: This is an open access article licensed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted, non-commercial use, distribution and reproduction in any medium, provided the work is properly cited.
ACKNOWLEDGEMENT
We would like to acknowledge Michelle L. Daniels, DMD for assistance with illustrations.
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Open-Access License: This is an open access article distributed under the terms of the Creative Commons Attribution 4.0 International Public License (CC-BY 4.0), a copy of which is available at: This license permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Address correspondence to this author at the Division of Spine Surgery, Department of Orthopaedic Surgery, Alpert Medical School of Brown University, 2 Dudley Street, Suite 200, Providence, Rhode Island 02905, USA; Tel: 401-457-5195; Fax: 401-457-2141; E-mail: mpalmd@aol.com
References
References
1
Tew JM, Mayfield FH. Complications of surgery of the anterior cervical spine Clin Neurosurg 1976; 23: 424-34.
Go to reference
2
Bertalanffy H, Eggert HR. Complications of anterior cervical discectomy without fusion in 450 consecutive patients Acta Neurochir (Wien) 1989; 99: 41-50.
3
Emery SE, Bohlman HH, Bolesta MJ, Jones PK. Anterior cervical decompression and arthrodesis for the treatment of cervical spondylotic myelopathy: Two to seventeen-year follow-up J Bone Joint Surg Am 1998; 80A: 941-51.
4
Marotta N, Landi A, Tarantino R, Mancarella C, Ruggeri A, Delfini R. Five-year outcome of stand-alone fusion using carbon cages in cervical disc arthrosis Eur Spine J 2011; 20(Suppl 1): S8-12.
Go to reference
5
Albert T, Balderston R, Northrup B. Surgical approaches to the spine. Philadelphia: WB Saunders 1997; p. 10.
a [...] sheath and the midline viscera (Fig.
[b [...] Philadelphia: WB Saunders 1997; p. 10
6
Yu NH, Jahng TA, Kim CH, Chung CK. Life-threatening late hemorrhage due to superior thyroid artery dissection after anterior cervical discectomy and fusion Spine (Phila Pa 1976) 2010; 35(15): E739-42.
Go to reference
7
Sethi R, Tandon M, Ganjoo P. Neck hematoma causing acute airway and hemodynamic compromise after anterior cervical spine surgery J Neurosurg Anesthesiol 2008; 20(1): 69-70.
[a [...] of coughing at the time of extubation
[b [...] edema/plethora and tracheal deviation
[c [...] activation of the baroreceptor reflex
[d [...] for acute management of the patient
8
Carr EM, Benjamin E. In vitro study investigating post neck surgery haematoma airway obstruction J Laryngol Otol 2009; 122: 662-5.
Go to reference
9
Lee HS, Lee BJ, Kim SW, et al. Patterns of post-thyroidectomy hemorrhage Clin Exp Otorhinolaryngol 2009; 2(2): 72-.
[a [...] edema/plethora and tracheal deviation
[b [...] for acute management of the patient
10
Gwinnutt CL, Walsh GR, Kumar R. Airway obstruction after anterior cervical spine surgery J Neurosurg Anesthesiol 1992; 4(3): 99-202.
11
Shakespeare WA, Lanier WL, Perkins WJ, Pasternak JJ. Airway management in patients who develop neck hematomas after carotid endarterectomy Anesth Analg 2010; 110(2): 588-93.
[a [...] edema/plethora and tracheal deviation
[b [...] for acute management of the patient
12
Sagi HC, Beutler W, Carroll E, Connolly PJ. Airway complications associated with surgery on the anterior cervical spine Spine 2002; 27(9): 949-53.
Go to reference
13
Kwon B, Yoo JU, Furey CG, Rowbottom J, Emery SE. Risk factors for delayed extubation after single-stage, multi-level anterior cervical decompression and posterior fusion J Spinal Disord Tech 2006; (19): 389-93.
14
Terao Y, Matsumoto S, Yamashita K, et al. Increased incidence of emergency airway management after combined anterior-posterior cervical spine surgery J Neurosurg Anesthesiol 2004; (16): 282-6.
Go to reference
Media
Figures
Figures
Fig. (1).
Intraoperative lateral cervical spine xray showing anterior cervical discectomy and fusion of C5/6.
Go to FigureOpen in Viewer
Fig. (2).
Midsagittal section of the upper respiratory tract. OP: oropharnyx. LP: laryngopharnyx. TR: trachea. ES: esophagus. EP: epiglottis. TC: thyroid cartilage. CC: cricoid cartilage. Arrow A: path for direct laryngoscopic orotracheal intubation. Arrow B: path for surgical cricothyrotomy. Arrow C: path for high surgical traceostomy.
Go to FigureOpen in Viewer
Fig. (3).
The anterior cervical approach to the spine (cross-sectional view): note the potential space created by dissection in the plane between the carotid sheath and the midline viscera. The carotid sheath structures are retracted laterally, and the esophagus and trachea are retracted medially to expose the ventral surface of the spine. Adapted with permission from: Albert T, Balderston R, Northrup B. Surgical Approaches to the Spine. Philadelphia: WB Saunders 1997; p. 10 .
Go to FigureOpen in Viewer
Tables
Clinical Findings Associated with Postoperative Wound Hematoma
Go to TableOpen in Viewer
Additional Risk Factors For Postoperative Airway Compromise
Go to TableOpen in Viewer
Share
Share
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figures
tables
Fig. (1).
Fig. (1).
Intraoperative lateral cervical spine xray showing anterior cervical discectomy and fusion of C5/6.
Fig. (2).
Fig. (2).
Midsagittal section of the upper respiratory tract. OP: oropharnyx. LP: laryngopharnyx. TR: trachea. ES: esophagus. EP: epiglottis. TC: thyroid cartilage. CC: cricoid cartilage. Arrow A: path for direct laryngoscopic orotracheal intubation. Arrow B: path for surgical cricothyrotomy. Arrow C: path for high surgical traceostomy.
Fig. (3).
Fig. (3).
The anterior cervical approach to the spine (cross-sectional view): note the potential space created by dissection in the plane between the carotid sheath and the midline viscera. The carotid sheath structures are retracted laterally, and the esophagus and trachea are retracted medially to expose the ventral surface of the spine. Adapted with permission from: Albert T, Balderston R, Northrup B. Surgical Approaches to the Spine. Philadelphia: WB Saunders 1997; p. 10 .
Clinical Findings Associated with Postoperative Wound Hematoma
Additional Risk Factors For Postoperative Airway Compromise
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Reference #1 |
4157 | https://www.youtube.com/watch?v=r_h0414ABSM | Image Under a Linear Transformation
Leah Howard
5230 subscribers
Description
14458 views
Posted: 10 Apr 2015
Transcript:
in this video we'll discuss the image of a vector under a linear transformation we're looking at the linear transformation T from R 2 to R 2 and we know how it acts on two specific vectors the image of two seven under the transformation T is negative one negative one the image of three four is the vector eight nine and we're trying to find the image of the vector 17 one meaning we're trying to figure out how t acts on this vector 17 1 well first of all just to keep our notation a little bit simpler let's and let's give some names to these vectors let's say 2 7 let's call that u and three four let's call that V the idea here is to see if we can write the vector 17 1 as a linear combination of U and V so let's say we figure out what that linear combination looks like that'll be your first step and then from there we've reduced the problem of finding T of 17 1 2 really finding T of C 1 u plus C 2 V so we'll be able to use the properties of a linear transformation to then find our answer so let's start with this first step here let's set up an Augmented matrix corresponding to this system that we're trying to solve so we're going to take C 1 times the vector U which is 2 7 and we're gonna add on C 2 tons the vector B which is 3 4 and we're going to make the vector 17 1 so let's do a little bit of gauss-jordan or Gaussian elimination see if we can try to figure out C 1 and C 2 we'll take Row 1 and divide it by 2 that gives us a 1 in the upper left-hand corner and we can then take Row 2 minus 7 times Row 1 so that's going to give us 0 minus 13 halves and over on the right side minus 117 house and the top row is not changing at this point so I'll just recopy that all right now we'd like to move down into the right this guy here is going to be our new pivot so let's try to make that equal to 1 we'll take Row 2 and we want to basically divide through by this fraction so in other words multiply by the reciprocal so you could say Row 2 times 2 over negative 13 I think that sounds kind of weird I'm going to call it Row 2 times negative 2 over 13 just bringing the negative up to the top here so that's going to give us 0 1 and over on the right side if we're careful with our signs we're gonna have a positive number it's going to be positive 9 and top row will not be changing all right the last thing we can do here is try to eliminate this three-halves so let's take Row 1 minus 3 halves of Row 2 that will give us 1 0 and negative 5 and bottom row won't be changing so what we were solving for here where C 1 and C 2 we can now read off the solution top row tells us C 1 is equal to negative 5 bottom row says C 2 equals 9 so let's go back to the problem at hand which was to find the image of the vector 17 1 under T let's write T of 17 1 is equal to T of negative 5 u + 9 V so that is thanks to the linear combination that we just found now let's have a quick recap of the two properties of a linear transformation first of all if you ply I'll apply a linear transformation to a sum then you're gonna get the sum of the images meaning you're gonna get T of vector a plus T vector B and if you apply a linear transformation to a scalar multiple of a vector then that scalar multiple can come outside and basically we're going to get the scalar multiple of the image so C times T of a so I think I'll put these in a box here I'm gonna use these two properties now to solve the problem so first of all we can express this as the sum of T of negative 5 u and T of negative 9 V and next we can bring those scalar multiples outside so we can say this is negative 5 times T of U plus 9 times T of V and we can refer back to the problem where we were given T of U and T of these so remember right back up at the top are told T of U is negative 1 negative 1 and we're also told that T of V is 8 9 and very last thing we can do is just a little bit of algebra so we take negative 5 times negative 1 plus 9 times 8 that gives us 77 and in the second component negative 5 times negative 1 plus 9 times 9 and I guess is 86 so the image of the vector 17 1 under the transformation T is this vector 77 86 |
4158 | https://www.onlinemath4all.com/simplifying-expressions-with-rational-exponents.html | Simplifying Expressions with Rational Exponents
SIMPLIFYING EXPRESSIONS WITH RATIONAL EXPONENTS
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The following properties of exponents can be used to simplify expressions with rational exponents.
x m⋅ x n = x m+n
x m÷x n = x m-n
(x m)n = x mn
(xy)m =x m⋅ y m
(x / y)m =x m/y m
x-m = 1 /x m
x m/n = y -----> x = y n/m
(x / y)-m = (y / x)m
√x = x 1/2
n√x = x 1/n
Example 1 :
Simplify :
y 2/3⋅y 7/3
Solution :
= y 2/3⋅y 7/3
Product of powers property.
= y 2/3 +7/3
Simplify exponents.
= y(2 + 7) / 3
= y 9/3
= y 3
Example 2 :
Simplify :
a 3/5⋅ a 7/5
Solution :
= a 3/5⋅ a 7/5
Product of powers property.
= a 3/5 +7/5
Simplify exponents.
= a(3 + 7) / 5
= a 10/5
= a 2
Example 3 :
Simplify :
(x 4 y 2)1/2
Solution :
=(x 4 y 2)1/2
Power of a Product Property.
=(x 4)1/2⋅(y 2)1/2
Power of a Power Property.
=(x 4⋅1/2)⋅(y 2⋅1/2)
Simplify exponents.
= (x 2)⋅(y 1)
= x 2 y
Example 4 :
Simplify :
(a 1/2 a 1/3)6
Solution :
=(a 1/2 a 1/3)6
Power of a Product Property.
=(a 1/2)6⋅(a 1/3)6
Power of a Power Property.
=(a 1/2⋅ 6)⋅(a 1/3⋅ 6)
Simplify exponents.
=a 3⋅a 2
Product of Powers Property.
=a 3 +2
= a 5
Example 5 :
Simplify :
(2a 1/2)(3a)
Solution :
=(2a 1/2)(3a)
=(3⋅ 2)⋅(a 1/2⋅ a)
Product of Powers Property.
= 6⋅a 1/2 + 1
Simplify exponents.
= 6⋅a 1/2 + 2/2
= 6⋅a(1 + 2)/2
= 6 a 3/2
Example 6 :
Simplify :
√(x 4 y 2)
Solution :
=√(x 4 y 2)
Write square root as exponent.
= (x 4 y 2)1/2
Power of a Product Property.
=(x 4)1/2⋅(y 2)1/2
Power of a Power Property.
=(x 4⋅ 1/2)⋅(y 2⋅ 1/2)
Simplify exponents.
=(x 2)⋅(y 1)
=x 2 y
Example 7 :
Simplify :
4√(x 4 y 12)
Solution :
=4√(x 4 y 12)
Write 4 th root as exponent.
= (x 4 y 12)1/4
Power of a Product Property.
=(x 4)1/4⋅(y 12)1/4
Power of a Power Property.
=(x 4⋅ 1/4)⋅(y 12⋅ 1/4)
Simplify exponents.
=(x 1)⋅(y 3)
=x y 3
Example 8 :
Simplify :
(x 1/3)6⋅4√y 4
Solution :
=(x 1/3)6⋅4√y 4
Write 4 th root as exponent.
=(x 1/3)6⋅ (y 4)1/4
Power of a Power Property.
=x 1/3⋅6⋅y 4⋅ 1/4
Simplify exponents.
=x 2⋅y 1
=x 2 y
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4159 | https://www.easycalculation.com/unit-conversion/curie-to-becquerel-conversion.php | Ci to Bq Calculator
Becquerel and Curie are the units of radioactivity. Here is the calculator to convert the values of Curie (Ci) to Becquerel (Bq). A Ci is a deprecated non-SI unit of radioactivity defined as 1 Ci = 3.7 × 1010 decays per second. A becquerel (Bq) is the SI derived unit of radiation activity. Enter the value of Curie and click calculate to convert Ci into Bq.
Curie to Becquerel
Becquerel to Curie
Curie to Becquerel Conversion
Becquerel and Curie are the units of radioactivity. Here is the calculator to convert the values of Curie (Ci) to Becquerel (Bq). A Ci is a deprecated non-SI unit of radioactivity defined as 1 Ci = 3.7 × 1010 decays per second. A becquerel (Bq) is the SI derived unit of radiation activity. Enter the value of Curie and click calculate to convert Ci into Bq.
One curie (Ci) is roughly the activity of 1 gram of the radium isotope 226Ra. The Becquerel (Bq) is defined as the activity of a quantity of radioactive material in which one nucleus decays per second, which is equivalent to inverse of a second, s-1.
Example
One Curie is equal to 370000000000 Bq and hence
0.0002 Ci is = 370000000000 x 0.0002 Bq
= 74000000 Bq
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4160 | https://thehacktivists.medium.com/off-by-one-error-cwe-193-the-hacktivists-7a62e9f0068f | Off-by-one Error [CWE-193] — The Hacktivists | by The Hacktivists | Medium
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Off-by-one Error [CWE-193] — The Hacktivists
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May 24, 2022
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Off-by-one error occurs when a program uses an improper maximum or minimum value that is one more or one less than the proper value.
Table of Content1.Description2.Potential impact3.Attack patterns4.Affected software5.Severity and CVSS Scoring
Description
………………………………
An off-by-one condition is a logic error in size calculation when working with strings and arrays. It usually produces a boundary condition, which may lead to memory corruption.
Off-by-one errors are often a result of incorrect null-termination of string sequence, which usually starts at zero rather than one. This scenario typically arises when software performs loop iteration a number of times that is greater or less than expected.
When an off-by-one condition occurs, the program is able to read or write beyond the bounds of allocated memory, which can result in data corruption, application crash, or even lead to code execution.
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The following example in C language uses a loop to read characters from an array:
include
define MAX_CHARS 19
int main ()
{
int x;
int ite_loop = 0;
char filename[MAX_CHARS] = "My mother loves me.";
printf("Length of filename array: %d\n",strlen(filename));
for (x = 0; x <= MAX_CHARS; x++) {
printf("%c",filename[x]);
ite_loop += 1;
}
printf("\nIterations: %d\n",ite_loop);
return 0;
}
However, the exit loop condition “x <= MAX_CHARS” is incorrectly defined; therefore, the loop reads one byte beyond the bounds of the array. Consider the following example:
include
define MAX_CHAR 19
define MAX_VALUE 30
int main ()
{
int x;
int ite_loop = 0;
char filename[MAX_CHAR] = "My mother loves me.";
printf("Length of filename array: %d\n",strlen(filename));
for (x = 0; x <= MAX_VALUE; x++) {
printf("%c",filename[x]);
ite_loop += 1;
}
printf("\nIterations: %d\n",ite_loop);
return 0;
}
In the previous code, the programmer has specified the wrong define directive variable in order to read the filename array. This mistake forces the code to read a few bytes beyond the bounds of the filename array; consequently, it prints random data to memory:
Press enter or click to view image in full size
An off-by-one error can also be introduced by improper usage of certain library functions. The following example uses the strncat function that always null-terminated its output string:
strcpy(buf, "buffer:");
strncat(buf, input, sizeof(buf)-strlen(buf));
The improper usage of the strncat function (third argument) produces an off-by-one condition, which depending on the application architecture, may lead to code execution.
Potential impact
………………………………
Off-by-one error leads to unpredictable behaviour of the application, depending on the nature of the vulnerability, and in most cases results in application crash or infinite loop. This weakness can also lead to buffer overflow and memory corruption. In cases of a heap-based buffer overflow, the most obvious result is the application crash. If an off-by-one error leads to a stack-based buffer overflow, successful code execution is more likely.
Attack patterns
………………………………
Software written in languages such as C and C++ that do not perform memory management is potentially vulnerable to this weakness.
Affected software
………………………………
Developers should pay extra attention to correct size parameter accounting for null terminator when copying character arrays or performing manipulations on arrays.
Severity and CVSS Scoring
……………………………………..
Off-by-one errors can be used to cause an application crash, data tampering or execution of arbitrary code. Depending on the software and vulnerable code, these weaknesses could be locally or remotely exploitable.
A common CVSSv3 score for locally exploitable vulnerability in the application would look like this:
3.3 [CVSS:3.0/AV:L/AC:L/PR:L/UI:N/S:U/C:N/I:N/A:L] — Low severity.
In cases of remote code execution, it is usually scored as:
10.0 [CVSS:3.0/AV:N/AC:L/PR:N/UI:N/S:C/C:H/I:H/A:H] — High severity.
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What Is Food Inventory Shrinkage & How Can You Prevent It? A Guide for Food Manufacturers
Rob Terry
Published On July 18, 2025
Table of Contents
Food manufacturers lose approximately $1.4 trillion annually to inventory shrinkage, with the food industry experiencing some of the highest shrink rates across all sectors. According to the National Retail Federation’s Security Survey, food businesses face shrinkage rates averaging 1.6% of total sales, nearly double the retail average.
This challenge extends far beyond simple accounting discrepancies. For food manufacturers, processors, and distributors, inventory shrinkage represents a complex operational issue that directly impacts profitability, regulatory compliance, and market competitiveness.
Unlike other industries, where shrinkage primarily stems from theft, food businesses must contend with unique factors, including spoilage, contamination risks, and stringent safety requirements, that compound traditional loss factors. TheFood and Agriculture Organisation reports that post-harvest losses in food supply chains can reach 30% in developing markets, while even advanced economies see 10-15% losses from farm to consumer.
Understanding and controlling these losses isn’t just about protecting margins, but it’s also about building sustainable, efficient operations that can adapt to market pressures while maintaining food safety standards.
What is Inventory Shrinkage in the Food Industry?
Inventory shrinkage refers to any unaccounted-for reduction in stock that cannot be attributed to documented sales, transfers, or approved waste disposal. In food businesses, this encompasses a wide range of losses, some of which are preventable, such as theft, administrative errors, and mishandling. In contrast, others are considered inevitable, like spoilage within accepted thresholds.
Unlike industries where product degradation is minimal, food businesses must carefully distinguish between acceptable spoilage and excessive losses. For example, a 2–5% spoilage rate may be normal for many perishable items, but higher rates can signal underlying operational inefficiencies.
Shrinkage challenges in the food sector are uniquely complex. Temperature fluctuations can spoil entire shipments, pest infestations can contaminate stock, and even if goods appear intact, they must be discarded if they have expired. These factors require dedicated processes for monitoring, identifying, and managing inventory loss.
How to Calculate Inventory Shrinkage
Inventory shrinkage is calculated by comparing recorded (book) inventory with actual physical stock. The basic formula is:
Shrinkage Rate = (Book Inventory – Physical Inventory) / Book Inventory × 100
For instance, if your system records $100,000 in inventory, but the physical count reveals only $96,000, the shrinkage rate would be 4%.
While this basic formula provides a general overview, food businesses often require more detailed calculations that break shrinkage down by product type, storage environment, period, and root cause. A dairy processor, for instance, may track separate shrinkage rates for refrigerated goods versus dry storage items, since spoilage and handling risks vary significantly.
To stay proactive, many companies conduct shrinkage analysis on a monthly or even weekly basis, especially in fast-moving categories like fresh produce, where short shelf lives leave little room for delay. Delayed identification of issues in these segments can lead to rapid inventory losses.
Modern inventory management systems can help automate this process by synchronising real-time inventory data with physical counts, identifying anomalies, and generating shrinkage reports across various locations, SKUs, or timeframes. These tools reduce manual errors and offer the visibility needed to identify loss patterns before they escalate.
What Are the Common Causes of Food Inventory Shrinkage?
Food inventory shrinkage stems from five primary sources, each requiring different food inventory management strategies and management approaches:
1. Theft:Internal and external theft are major contributors to shrinkage. Employees may manipulate inventory records or divert shipments, while external actors target easily resold food items. These actions often involve coordinated efforts and are challenging to detect without strong controls.
2. Spoilage:Improper temperature control, inadequate inventory rotation, and delays in usage lead to spoilage. Failures in cold storage systems and inadequate stock management lead to products expiring or degrading before they can be used.
3. Errors in Record-Keeping:Inventory discrepancies often result from administrative mistakes such as incorrect receiving, missed shipments, or inaccurate data entry. Scanning errors and mislabeling also contribute to inventory mismatches.
4. Damaged Goods:Products can become unsellable due to physical damage during handling, transport, or storage. Packaging failures, mishandling, and rough transportation conditions are common causes of such losses.
5. Administrative and Operational Inefficiencies:Lack of proper training, unclear procedures, and inefficient workflows increase the risk of shrinkage. These gaps in operations often lead to mistakes or create opportunities for inventory loss.
How Does Inventory Shrinkage Impact Food Businesses?
Food shrinkage has a significant impact on food businesses across financial, operational, and regulatory dimensions. While the immediate loss of products is apparent, the broader consequences can undermine profitability, disrupt operations, and jeopardise compliance.
Financial Losses
The most direct impact of shrinkage is financial. When inventory disappears due to theft, spoilage, damage, or errors, businesses lose not just the cost of the goods but also the potential revenue. Since many food businesses operate on narrow profit margins, even small percentages of shrinkage can erode a significant portion of net profits.
Additionally, the need to replace lost inventory often results in increased procurement costs, mainly when products must be sourced urgently. Administrative overhead also rises as companies invest more time and resources in documenting, auditing, and managing inventory discrepancies.
Operational Inefficiencies
Shrinkage distorts inventory records, making it difficult to plan production and manage supply chains effectively. When actual stock levels do not align with system data, businesses may encounter stockouts or overstocking.
Stockouts can lead to emergency purchasing, expedited shipping, and disrupted customer fulfilment, while excess inventory increases storage costs and poses a risk of spoilage. Furthermore, warehouse staff and managers spend additional time reconciling data, conducting audits, and investigating shrinkage-related issues, which reduces overall labour efficiency.
Pricing Pressure and Margin Compression
Persistent shrinkage puts pressure on pricing strategies. To maintain profitability, businesses may need to raise prices, which can reduce competitiveness in price-sensitive markets. Alternatively, they may absorb the losses, leading to margin compression. Both options can weaken market positioning and limit growth opportunities.
Customer Satisfaction and Trust
Inaccurate inventory records resulting from shrinkage can lead to fulfilment errors and missed deliveries, damaging customer trust. Repeated service failures can harm long-term relationships and lead to lost business, especially in industries where reliability and consistency are critical.
Regulatory and Compliance Challenges
Food shrinkage complicates compliance management with food safety regulations and traceability standards. Missing or undocumented inventory can hinder the ability to track product movement, which is essential for recalls and audits.
Regulatory bodies may view significant discrepancies as red flags, which could trigger investigations. High or unexplained shrinkage rates can also raise tax-related concerns if they fall outside accepted norms for business losses.
How Can Food Businesses Overcome Inventory Shrinkage?
Successfully reducing food shrinkage requires a systematic approach that addresses each major cause while building operational excellence throughout the organisation.
Implementing Accurate Inventory Tracking Systems
Modern inventory tracking systems help reduce shrinkage by providing real-time visibility into stock levels, movements, and storage conditions. Technologies such as barcoding or RFID eliminate many manual errors by automating data capture during receiving, storage, picking, and shipping.
When integrated with operational workflows, these systems ensure records are updated instantly, reducing lag time and minimising the risk of phantom inventory or unrecorded losses.
Regular Audits and Stocktaking
Frequent inventory audits help detect discrepancies before they escalate. Rather than relying solely on annual physical counts, many businesses adopt cycle counting, where a portion of inventory is checked on a rotating basis. This ensures ongoing accuracy without disrupting operations.
Using blind counts, where employees verify stock without viewing expected quantities, further improves count reliability. Statistical sampling can also be used to target high-risk categories, making audits more efficient and focused.
Improved Warehouse and Storage Management
Proper storage practices are essential to preventing shrinkage due to spoilage and physical damage. Temperature-controlled environments should be continuously monitored with alert systems to detect any deviations from the set temperature.
To minimise losses from expired goods, businesses should employ inventory rotation techniques, such as the First In, First Out (FIFO) method, which ensures that older stock is sold before newer stock. Investing in the proper handling equipment and maintaining orderly storage also reduces product damage and improves overall efficiency.
Employee Training and Security Measures
Educating employees on best practices for managing inventory, minimising waste, and understanding their role in preventing shrinkage is crucial. Well-trained staff are less likely to make procedural mistakes and more likely to identify and report potential issues.
Security measures such as CCTV monitoring, provided by areliable CCTV installation company, restricted access areas, and clear accountability protocols help deter theft and provide oversight of inventory handling. Additionally, anonymous reporting channels can encourage employees to report suspicious behavior without fear of retaliation.
Using Technology for Better Forecasting and Demand Planning
Accurate demand forecasting helps prevent overstocking, which can lead to spoilage, and understocking, which can cause stockouts. Predictive analytics tools can identify trends and seasonal patterns, enabling more informed purchasing and production decisions. These tools also help flag products at risk of becoming obsolete or expiring, allowing timely interventions such as promotions or redistribution before losses occur.
How Folio3 FoodTech Food Inventory Management Software Helps Combat Shrinkage?
Reducing shrinkage in food businesses often comes down to having better visibility, traceability, and control over inventory. A specialised food inventory management system, like the one developed by Folio3 FoodTech, is designed to address these challenges through integrated tools that support accuracy and operational efficiency. Here’s how Folio3 FoodTech food inventory management software can help:
Real-Time Inventory Tracking
Continuous tracking of raw materials, in-process items, and finished goods ensures stock levels are always up to date. This helps prevent data mismatches and makes it easier to spot irregularities early.
Multi-Location Inventory Management
When inventory is distributed across production sites, warehouses, or cold storage facilities, centralised oversight helps maintain consistency. It reduces miscounts, duplication, and blind spots in the inventory process.
Batch and Lot Tracking
Maintaining batch and lot histories supports better traceability. In the event of a quality issue or audit, businesses can quickly identify where the affected inventory originated and where it’s located, thereby reducing unnecessary waste.
Expiration and Shelf-Life Management
By tracking expiry dates and applying the First Expired, First Out (FEFO) principle, the system helps prioritise stock movement. This reduces the chance of products expiring before they’re used or sold.
Catch Weight Management
In operations involving variable-weight items, such as dairy, meat, or seafood, catch weight-based tracking helps ensure that inventory counts are accurate and consistent with what’s actually handled and sold.
Allergen Information Tracking
As part of allergen management, Tagging and separating allergen-related items minimises cross-contact risk and supports compliance. It also helps prevent unusable inventory shrinkage due to mishandling or misclassification.
Automated Replenishment
When stock levels drop below predefined thresholds, the system can automatically trigger reorders or production tasks. This reduces last-minute procurement and helps maintain optimal stock levels without overbuying.
Demand Forecasting
Analysing historical data and production trends enables more informed planning and decision-making. This helps avoid overstocking—often a significant cause of spoilage and ensures that procurement aligns more closely with actual demand.
Mobile & Barcode Scanning
On-the-go scanning capabilities reduce manual errors during receiving, picking, and stocktaking. Staff can update records in real time, improving accuracy and reducing data entry delays.
Conclusion
Inventory shrinkage remains a critical challenge for food businesses, but with the right systems and processes in place, it can be significantly reduced. A combination of operational discipline, employee training, and real-time inventory visibility is key to addressing both the causes and the costs of shrinkage.
For companies looking to strengthen their inventory control, a purpose-built solution like Folio3 FoodTech Food Inventory Management Software offers the tools needed to improve accuracy, minimise waste, and support long-term efficiency, making it a valuable component of any shrinkage reduction strategy.
FAQs
What Is Food Inventory Shrinkage?
Food inventory shrinkage refers to the discrepancy between recorded inventory levels and the actual physical stock on hand, encompassing losses due to theft, spoilage, damage, and administrative errors. It represents any unexplained reduction in inventory that can’t be attributed to legitimate sales or documented disposal.
How Can Folio3 FoodTech Food Inventory Management Software Help Reduce Inventory Shrinkage?
Folio3 FoodTech’s inventory software reduces shrinkage through real-time tracking, automated FIFO rotation, temperature monitoring, and comprehensive reporting capabilities. The system provides immediate visibility into discrepancies and helps identify root causes of losses, enabling proactive management rather than reactive responses.
What Causes Food Inventory Shrinkage?
The primary causes include spoilage (40-60% of losses), theft (30%), administrative errors, damaged goods, and operational inefficiencies. Food businesses face unique challenges, including temperature sensitivity, short shelf lives, and complex handling requirements, which differ from those in other industries.
How Can I Prevent Food Shrinkage In My Warehouse?
Prevention strategies include implementing automated tracking systems, conducting regular cycle counts, maintaining proper storage conditions, training employees on handling procedures, and using security measures to deter theft. Success requires addressing all significant causes of shrinkage simultaneously.
What Are The Financial Implications Of Food Inventory Shrinkage?
Financial impacts include direct product losses, increased procurement costs, administrative expenses, and margin compression. For food companies operating on 2-5% margins, shrinkage rates of 1-2% can eliminate 20-40% of potential profits, making effective shrinkage management critical for profitability.
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4162 | https://en.wikipedia.org/wiki/Factorization_of_polynomials | Jump to content
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Contents
(Top)
1 Formulation of the question
2 Primitive part–content factorization
3 Square-free factorization
4 Classical methods
4.1 Obtaining linear factors
4.2 Kronecker's method
5 Modern methods
5.1 Factoring over finite fields
5.2 Factoring univariate polynomials over the integers
5.3 Factoring over algebraic extensions (Trager's method)
6 Numerical factorization
7 See also
8 Bibliography
9 Further reading
Factorization of polynomials
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From Wikipedia, the free encyclopedia
Computational method
This article is about factorization algorithms. For paper-and-pencil methods, see Factorization § Polynomials.
In mathematics and computer algebra, factorization of polynomials or polynomial factorization expresses a polynomial with coefficients in a given field or in the integers as the product of irreducible factors with coefficients in the same domain. Polynomial factorization is one of the fundamental components of computer algebra systems.
The first polynomial factorization algorithm was published by Theodor von Schubert in 1793. Leopold Kronecker rediscovered Schubert's algorithm in 1882 and extended it to multivariate polynomials and coefficients in an algebraic extension. But most of the knowledge on this topic is not older than circa 1965 and the first computer algebra systems:
When the long-known finite step algorithms were first put on computers, they turned out to be highly inefficient. The fact that almost any uni- or multivariate polynomial of degree up to 100 and with coefficients of a moderate size (up to 100 bits) can be factored by modern algorithms in a few minutes of computer time indicates how successfully this problem has been attacked during the past fifteen years. (Erich Kaltofen, 1982)
Modern algorithms and computers can quickly factor univariate polynomials of degree more than 1000 having coefficients with thousands of digits. For this purpose, even for factoring over the rational numbers and number fields, a fundamental step is a factorization of a polynomial over a finite field.
Formulation of the question
[edit]
Polynomial rings over the integers or over a field are unique factorization domains. This means that every element of these rings is a product of a constant and a product of irreducible polynomials (those that are not the product of two non-constant polynomials). Moreover, this decomposition is unique up to multiplication of the factors by invertible constants.
Factorization depends on the base field. For example, the fundamental theorem of algebra, which states that every polynomial with complex coefficients has complex roots, implies that a polynomial with integer coefficients can be factored (with root-finding algorithms) into linear factors over the complex field C. Similarly, over the field of reals, the irreducible factors have degree at most two, while there are polynomials of any degree that are irreducible over the field of rationals Q.
The question of polynomial factorization makes sense only for coefficients in a computable field whose every element may be represented in a computer and for which there are algorithms for the arithmetic operations. However, this is not a sufficient condition: Fröhlich and Shepherdson give examples of such fields for which no factorization algorithm can exist.
The fields of coefficients for which factorization algorithms are known include prime fields (that is, the field of the rational numbers and the fields of the integers modulo a prime number) and their finitely generated field extensions. Integer coefficients are also tractable. Kronecker's classical method is interesting only from a historical point of view; modern algorithms proceed by a succession of:
Square-free factorization
Factorization over finite fields
and reductions:
From the multivariate case to the univariate case.
From coefficients in a purely transcendental extension to the multivariate case over the ground field (see below).
From coefficients in an algebraic extension to coefficients in the ground field (see below).
From rational coefficients to integer coefficients (see below).
From integer coefficients to coefficients in a prime field with p elements, for a well chosen p (see below).
Primitive part–content factorization
[edit]
See also: Content (algebra) and Gauss's lemma (polynomial)
In this section, we show that factoring over Q (the rational numbers) and over Z (the integers) is essentially the same problem.
The content of a polynomial p ∈ Z[X], denoted "cont(p)", is, up to its sign, the greatest common divisor of its coefficients. The primitive part of p is primpart(p) = p/cont(p), which is a primitive polynomial with integer coefficients. This defines a factorization of p into the product of an integer and a primitive polynomial. This factorization is unique up to the sign of the content. It is a usual convention to choose the sign of the content such that the leading coefficient of the primitive part is positive.
For example,
is a factorization into content and primitive part.
Every polynomial q with rational coefficients may be written
where p ∈ Z[X] and c ∈ Z: it suffices to take for c a multiple of all denominators of the coefficients of q (for example their product) and p = cq. The content of q is defined as:
and the primitive part of q is that of p. As for the polynomials with integer coefficients, this defines a factorization into a rational number and a primitive polynomial with integer coefficients. This factorization is also unique up to the choice of a sign.
For example,
is a factorization into content and primitive part.
Gauss proved that the product of two primitive polynomials is also primitive (Gauss's lemma). This implies that a primitive polynomial is irreducible over the rationals if and only if it is irreducible over the integers. This implies also that the factorization over the rationals of a polynomial with rational coefficients is the same as the factorization over the integers of its primitive part. Similarly, the factorization over the integers of a polynomial with integer coefficients is the product of the factorization of its primitive part by the factorization of its content.
In other words, an integer GCD computation reduces the factorization of a polynomial over the rationals to the factorization of a primitive polynomial with integer coefficients, and the factorization over the integers to the factorization of an integer and a primitive polynomial.
Everything that precedes remains true if Z is replaced by a polynomial ring over a field F and Q is replaced by a field of rational functions over F in the same variables, with the only difference that "up to a sign" must be replaced by "up to the multiplication by an invertible constant in F". This reduces the factorization over a purely transcendental field extension of F to the factorization of multivariate polynomials over F.
Square-free factorization
[edit]
Main article: Square-free polynomial
If two or more factors of a polynomial are identical, then the polynomial is a multiple of the square of this factor. The multiple factor is also a factor of the polynomial's derivative (with respect to any of the variables, if several).
For univariate polynomials, multiple factors are equivalent to multiple roots (over a suitable extension field). For univariate polynomials over the rationals (or more generally over a field of characteristic zero), Yun's algorithm exploits this to efficiently factorize the polynomial into square-free factors, that is, factors that are not a multiple of a square, performing a sequence of GCD computations starting with gcd(f(x), f '(x)). To factorize the initial polynomial, it suffices to factorize each square-free factor. Square-free factorization is therefore the first step in most polynomial factorization algorithms.
Yun's algorithm extends this to the multivariate case by considering a multivariate polynomial as a univariate polynomial over a polynomial ring.
In the case of a polynomial over a finite field, Yun's algorithm applies only if the degree is smaller than the characteristic, because, otherwise, the derivative of a non-zero polynomial may be zero (over the field with p elements, the derivative of a polynomial in xp is always zero). Nevertheless, a succession of GCD computations, starting from the polynomial and its derivative, allows one to compute the square-free decomposition; see Polynomial factorization over finite fields#Square-free factorization.
Classical methods
[edit]
This section describes textbook methods that can be convenient when computing by hand. These methods are not used for machine computations because they use integer factorization, which is currently slower than polynomial factorization.
The two methods that follow start from a univariate polynomial with integer coefficients for finding factors that are also polynomials with integer coefficients.
Obtaining linear factors
[edit]
All linear factors with rational coefficients can be found using the rational root test. If the polynomial to be factored is , then all possible linear factors are of the form , where is an integer factor of and is an integer factor of . All possible combinations of integer factors can be tested for validity, and each valid one can be factored out using polynomial long division. If the original polynomial is the product of factors at least two of which are of degree 2 or higher, this technique only provides a partial factorization; otherwise the factorization is complete. In particular, if there is exactly one non-linear factor, it will be the polynomial left after all linear factors have been factorized out. In the case of a cubic polynomial, if the cubic is factorizable at all, the rational root test gives a complete factorization, either into a linear factor and an irreducible quadratic factor, or into three linear factors.
Kronecker's method
[edit]
Kronecker's method is aimed to factor univariate polynomials with integer coefficients into polynomials with integer coefficients.
The method uses the fact that evaluating integer polynomials at integer values must produce integers. That is, if is a polynomial with integer coefficients, then is an integer as soon as a is an integer. There are only a finite number of possible integer values for a factor of a. So, if is a factor of the value of must be one of the factors of
If one searches for all factors of a given degree d, one can consider values, for a, which give a finite number of possibilities for the tuple Each has a finite number of divisors , and, each -tuple where the entry is a divisor of , that is, a tuple of the form , produces a unique polynomial of degree at most , which can be computed by polynomial interpolation. Each of these polynomials can be tested for being a factor by polynomial division. Since there were finitely many and each has finitely many divisors, there are finitely many such tuples. So, an exhaustive search allows finding all factors of degree at most d.
For example, consider
: .
If this polynomial factors over Z, then at least one of its factors must be of degree two or less, so is uniquely determined by three values. Thus, we compute three values , and . If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as
: 1×2, 2×1, (−1)×(−2), or (−2)×(−1).
Therefore, if a second degree integer polynomial factor exists, it must take one of the values
: p(0) = 1, 2, −1, or −2
and likewise for p(−1). There are eight factorizations of 6 (four each for 1×6 and 2×3), making a total of 4×4×8 = 128 possible triples (p(0), p(1), p(−1)), of which half can be discarded as the negatives of the other half. Thus, we must check 64 explicit integer polynomials as possible factors of . Testing them exhaustively reveals that
constructed from (g(0), g(1), g(−1)) = (1,3,1) factors .
Dividing f(x) by p(x) gives the other factor , so that . Now one can test recursively to find factors of p(x) and q(x), in this case using the rational root test. It turns out they are both irreducible, so the irreducible factorization of f(x) is:
Modern methods
[edit]
Factoring over finite fields
[edit]
Main articles: Factorization of polynomials over finite fields, Berlekamp's algorithm, and Cantor–Zassenhaus algorithm
Factoring univariate polynomials over the integers
[edit]
If is a univariate polynomial over the integers, assumed to be content-free and square-free, one starts by computing a bound such that any factor has coefficients of absolute value bounded by . This way, if is an integer larger than , and if is known modulo , then can be reconstructed from its image mod .
The Zassenhaus algorithm proceeds as follows. First, choose a prime number such that the image of remains square-free, and of the same degree as . A random choice will almost always satisfy these constraints, since only a finite number of prime numbers do not satify them, namely the prime divisors of the product of the discriminant and the leading coefficient of the polynomial. Then factor . This produces integer polynomials whose product matches . Next, apply Hensel lifting; this updates the in such a way that their product matches , where is large enough that exceeds : thus each corresponds to a well-defined integer polynomial. Modulo , the polynomial has factors (up to units): the products of all subsets of . These factors modulo need not correspond to "true" factors of in , but we can easily test them by division in . This way, all irreducible true factors can be found by checking at most cases, reduced to cases by skipping complements. If is reducible, the number of cases is reduced further by removing those that appear in an already found true factor. The Zassenhaus algorithm processes each case (each subset) quickly, however, in the worst case, it considers an exponential number of cases.
The first polynomial time algorithm for factoring rational polynomials was discovered by Lenstra, Lenstra and Lovász and is an application of the Lenstra–Lenstra–Lovász lattice basis reduction (LLL) algorithm (Lenstra, Lenstra & Lovász 1982). A simplified version of the LLL factorization algorithm is as follows: calculate a complex (or p-adic) root α of the polynomial to high precision, then use the Lenstra–Lenstra–Lovász lattice basis reduction algorithm to find an approximate linear relation between 1, α, α2, α3, . . . with integer coefficients, which might be an exact linear relation and a polynomial factor of . One can determine a bound for the precision that guarantees that this method produces either a factor, or an irreducibility proof. Although this method finishes in polynomial time, it is not used in practice because the lattice has high dimension and huge entries, which makes the computation slow.
The exponential complexity in the Zassenhaus algorithm comes from a combinatorial problem: how to select the right subsets of . State-of-the-art factoring implementations work in a manner similar to Zassenhaus, except that the combinatorial problem is translated to a lattice problem that is then solved by LLL. In this approach, LLL is not used to compute coefficients of factors, but rather to compute vectors with entries in {0,1} that encode the subsets of corresponding to the irreducible true factors.
Factoring over algebraic extensions (Trager's method)
[edit]
We can factor a polynomial , where the field is a finite extension of . First, using square-free factorization, we may suppose that the polynomial is square-free. Next we define the quotient ring of degree ; this is not a field unless is irreducible, but it is a reduced ring since is square-free. Indeed, if
is the desired factorization of p(x), the ring decomposes uniquely into fields as:
We will find this decomposition without knowing the factorization. First, we write L explicitly as an algebra over : we pick a random element , which generates over with high probability by the primitive element theorem. If this is the case, we can compute the minimal polynomial of over , by finding a -linear relation among 1, α, . . . , αn. Using a factoring algorithm for rational polyomials, we factor into irreducibles in :
Thus we have:
where corresponds to . This must be isomorphic to the previous decomposition of .
The generators of L are x along with the generators of over ; writing these as a polynomials in , we can determine the embeddings of and into each component . By finding the minimal polynomial of in , we compute , and thus factor over
Numerical factorization
[edit]
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"Numerical factorization" refers commonly to the factorization of polynomials with real or complex coefficients, whose coefficients are only approximately known, generally because they are represented as floating point numbers.
For univariate polynomials with complex coefficients, factorization can easily be reduced to numerical computation of polynomial roots and multiplicities.
In the multivariate case, a random infinitesimal perturbation of the coefficients produces with probability one an irreducible polynomial, even when starting from a polynomial with many factors. So, the very meaning of numerical factorization needs to be clarified precisely.
Let be a polynomial with complex coefficients with an irreducible factorization
where and the factors are irreducible polynomials with complex coefficients. Assume that is approximated through a polynomial whose coefficients are close to those of . The exact factorization of is pointless, since it is generally irreducible. There are several possible definitions of what can be called a numerical factorization of
If and 's are known, an approximate factorization consists of finding a polynomial close to that factors as above. If one does not know the factorization scheme, identifying becomes necessary. For example, the number of irreducible factors of a polynomial is the nullity of its Ruppert matrix. Thus the multiplicities can be identified by square-free factorization via numerical GCD computation and rank-revealing on Ruppert matrices.
Several algorithms have been developed and implemented for numerical factorization as an on-going subject of research.
See also
[edit]
Factorization § Polynomials, for elementary heuristic methods and explicit formulas
Swinnerton-Dyer polynomials, a family of polynomials having worst-case runtime for the Zassenhaus method
Bibliography
[edit]
^ FT Schubert: De Inventione Divisorum Nova Acta Academiae Scientiarum Petropolitanae v.11, pp. 172–182(1793)
^ Kaltofen (1982)
^ An example of degree 2401, taking 7.35 seconds, is found in Section 4 in: Hart, van Hoeij, Novocin: Practical Polynomial Factoring in Polynomial Time ISSAC'2011 Proceedings, pp. 163–170 (2011).
^ Fröhlich, A.; Shepherdson, J. C. (1955). "On the factorisation of polynomials in a finite number of steps". Mathematische Zeitschrift. 62 (1): 331–334. doi:10.1007/bf01180640. ISSN 0025-5874. S2CID 119955899.
^ Van der Waerden, Sections 5.4 and 5.6
^ M. van Hoeij: Factoring polynomials and the knapsack problem. Journal of Number Theory, 95, 167–189, (2002).
^ Ruppert, W. (1999). "Reducibility of polynomials f(x,y)". J. Number Theory. 77: 62–70. arXiv:math/9808021. doi:10.1006/jnth.1999.2381. S2CID 14316123.
Shaker, H. (2009). "Topology and factorization of polynomials". Math. Scand. 104: 51–59. arXiv:0704.3363. doi:10.7146/math.scand.a-15084. S2CID 14121840.
8. ^ For example: W. Wu and Z. Zeng (2017). "The numerical factorization of polynomials". Foundations of Computational Mathematics. 17: 259–286. arXiv:2103.04888. doi:10.1007/s10208-015-9289-1. S2CID 254171366.
9. ^ E. Kaltofen, J.P. May, Z. Yang and L. Zhi (2008). "Approximate factorization of multivariate polynomials using singular value decomposition". J. Symbolic Comput. 43 (5): 359–376. doi:10.1016/j.jsc.2007.11.005.{{cite journal}}: CS1 maint: multiple names: authors list (link)
Fröhlich, A.; Shepherson, J. C. (1955), "On the factorisation of polynomials in a finite number of steps", Mathematische Zeitschrift, 62 (1): 331–334, doi:10.1007/BF01180640, ISSN 0025-5874, S2CID 119955899
Trager, B.M. (1976). "Algebraic factoring and rational function integration". Proceedings of the third ACM symposium on Symbolic and algebraic computation - SYMSAC '76. pp. 219–226. doi:10.1145/800205.806338. ISBN 9781450377904. S2CID 16567619.
Bernard Beauzamy, Per Enflo, Paul Wang (October 1994). "Quantitative Estimates for Polynomials in One or Several Variables: From Analysis and Number Theory to Symbolic and Massively Parallel Computation". Mathematics Magazine. 67 (4): 243–257. doi:10.2307/2690843. JSTOR 2690843.{{cite journal}}: CS1 maint: multiple names: authors list (link) (accessible to readers with undergraduate mathematics)
Cohen, Henri (1993). A course in computational algebraic number theory. Graduate Texts in Mathematics. Vol. 138. Berlin, New York: Springer-Verlag. ISBN 978-3-540-55640-4. MR 1228206.
Kaltofen, Erich (1982), "Factorization of polynomials", in B. Buchberger; R. Loos; G. Collins (eds.), Computer Algebra, Springer Verlag, pp. 95–113, CiteSeerX 10.1.1.39.7916
Knuth, Donald E (1997). "4.6.2 Factorization of Polynomials". Seminumerical Algorithms. The Art of Computer Programming. Vol. 2 (Third ed.). Reading, Massachusetts: Addison-Wesley. pp. 439–461, 678–691. ISBN 978-0-201-89684-8.
Lenstra, A. K.; Lenstra, H. W.; Lovász, László (1982). "Factoring polynomials with rational coefficients". Mathematische Annalen. 261 (4): 515–534. CiteSeerX 10.1.1.310.318. doi:10.1007/BF01457454. ISSN 0025-5831. MR 0682664. S2CID 5701340.
Van der Waerden, Algebra (1970), trans. Blum and Schulenberger, Frederick Ungar.
Further reading
[edit]
Kaltofen, Erich (1990), "Polynomial Factorization 1982-1986", in D. V. Chudnovsky; R. D. Jenks (eds.), Computers in Mathematics, Lecture Notes in Pure and Applied Mathematics, vol. 125, Marcel Dekker, Inc., CiteSeerX 10.1.1.68.7461
Kaltofen, Erich (1992), "Polynomial Factorization 1987–1991" (PDF), Proceedings of Latin '92, Springer Lect. Notes Comput. Sci., vol. 583, Springer, retrieved October 14, 2012
Ivanyos, Gabor; Marek, Karpinski; Saxena, Nitin (2009). "Schemes for deterministic polynomial factoring". Proceedings of the 2009 international symposium on Symbolic and algebraic computation. pp. 191–198. arXiv:0804.1974. doi:10.1145/1576702.1576730. ISBN 9781605586090. S2CID 15895636.
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4163 | https://www.arxiv.org/pdf/2505.17672 | arXiv:2505.17672v1 [math.CO] 23 May 2025
Paired domination in trees: A linear algorithm and asymptotic normality
1,2
Michael A. Henning and 3Dimbinaina Ralaivaosaona
1
Department of Mathematics and Applied Mathematics University of Johannesburg Auckland Park, 2006, South Africa
Email: mahenning@uj.ac.za
2
DSTI-NRF Centre of Excellence in Mathematical and Statistical Sciences (CoE-MaSS), South Africa
3
Department of Mathematical Sciences Stellenbosch University Matieland, 7602, South Africa
Email: naina@sun.ac.za
Abstract
A set S of vertices in a graph G is a paired dominating set if every vertex of G is adjacent to a vertex in S and the subgraph induced by S contains a perfect matching (not necessarily as an induced subgraph). The paired domination number, γpr (G), of G is the minimum cardinality of a paired dominating set of G. We present a linear algorithm for computing the paired domination number of a tree. As an application of our algorithm, we prove that the paired domination number is asymptotically normal in a random rooted tree of order n
generated by a conditioned Galton–Watson process as n → ∞ . In particular, we have found that the paired domination number of a random Cayley tree of order n, where each tree is equally likely, is asymptotically normal with expectation approaching (0 .5177 . . . )n.
Keywords: Paired domination; Tree; Random trees; Linear algorithm, Conditioned Galton–Watson trees, Asymptotic normality
AMS subject classification: 05C69, 60C05
1 Introduction
A dominating set of a graph G is a set S ⊆ V (G) such that every vertex of V (G) \ S is adjacent to some vertex in S. A set S of vertices in G is a total dominating set , abbreviated TD-set, of G
if every vertex in G is adjacent to some other vertex in S. The domination number , γ(G), of G
is the minimum cardinality of a dominating set of G, and the total domination number , γt(G), of
G is the minimum cardinality of a TD-set of G. If A and B are two subsets of vertices in a graph
G, then the set A totally dominates the set B is every vertex in B has a neighbor in the set A,1where two vertices are neighbors if they are adjacent. For recent books on domination and total domination in graphs, we refer the reader to [6, 7, 8, 9]. A paired dominating set , abbreviated PD-set, of an isolate-free graph G is a dominating set
S of G with the additional property that the subgraph G[S] induced by S contains a perfect matching M (not necessarily induced). With respect to the matching M , two vertices joined by an edge of M are paired and are called partners in S. The paired domination number , γpr (G), of
G is the minimum cardinality of a PD-set of G. We call a PD-set of G of cardinality γpr (G) a
γpr -set of G. We note that the paired domination number γpr (G) is an even integer. For a recent survey on paired domination in graphs, we refer the reader to the book chapter . We denote the degree of a vertex v in a graph G by deg G(v). A vertex of degree 0 is called an isolated vertex , and a graph is isolate -free if it contains no isolated vertex. The maximum (minimum) degree among the vertices of G is denoted by ∆( G) ( δ(G), respectively). A leaf of a tree T is a vertex of degree 1 in T , and a support vertex of T is a vertex with a leaf neighbor. The distance d(u, v ) between two vertices u and v in a connected graph G, equals the minimum length of a ( u, v )-path in G from u to v.A rooted tree T distinguishes one vertex r called the root . For each vertex v̸ = r of T , the
parent of v is the neighbor of v on the unique ( r, v )-path, while a child of v is any other neighbor of v. A descendant of v is a vertex u̸ = v such that the unique ( r, u )-path contains v. We let
D(v) denote the set of descendants of v, and we define D[v] = D(v) ∪ { v}. The maximal subtree
at v is the subtree of T induced by D[v], and is denoted by Tv.The aim of this paper is twofold: First, we present a simple bottom-up linear-time algorithm that solves the minimum paired domination problem for trees and prove its correctness. Our algorithm has a similar flavour to that of Mitchell, Cockayne, and Hedetniemi for the dom-ination number problem. Second, we establish that the paired domination number in random trees of order n admits a nondegenerate Gaussian limit law as n → ∞ . The random trees con-sidered here are conditioned Galton–Watson trees, and our bottom-up algorithm plays a crucial role in proving the result on the limiting distribution. This paper is organised as follows: In Section 2, we focus on our new algorithm and prove its correctness. In Section 3, we analyse the limiting distribution of the paired domination number in conditioned Galton–Watson trees in general and provide explicit examples and simulations that are in support of our result.
2 A paired domination tree algorithm
In 1979 Mitchell, Cockayne, and Hedetniemi and in 1984 Laskar, Pfaff, Hedetniemi, and Hedetniemi presented linear algorithms for computing the domination number and, respec-tively, the total domination number, of a tree. Their algorithms consider an arbitrary tree, which is rooted, and they systematically consider the vertices of the tree, starting from the vertices at furthest distance from the root, and carefully select a minimum (total) dominating set S in such a way that the sum of the distances from the root to the vertices in S is a minimum. In 2003 Qiao, Kang, Cardei, and Du presented a linear-time algorithm for computing the paired domination number of a tree. The linear-time algorithm we present in this section has a similar flavour to the linear algorithms presented in [16, 17]. In order to formally state our 2algorithm for computing the paired domination number of a tree, we first introduce the necessary notation. Let T be a rooted tree. We commonly draw the root r of T at the top with the remaining vertices at the appropriate level below r depending on their distance from r. We label the root with label 1. We label all children of the root (at level 1 from the root) next with labels 2, . . . , deg(1) + 1. Next we label all children of the vertex 2 with labels deg(1) + 2 , . . . , deg(1) + deg(2)+1, and thereafter we label all children of the vertex 3, etc. Given such a rooted tree T with root vertex labelled 1 and with V (T ) = {1, 2, . . . , n }, we represent T by a data structure called a
Parent array in which the parent of a vertex labelled i is given by parent[ i] with parent = 0 (to indicate that the vertex labelled 1 has no parent). We assume that the vertices of T are labelled 1, 2, . . . , n so that for i < j , vertex i is at level less than or equal to that of vertex j (that is,
d(1 , i ) ≤ d(1 , j )). Figure 1 shows an example of a rooted tree T with its parent array.
1
2
5
10
6
11
3
7
12
15 14
8
13
4
9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Parent [ 0 1 1 1 2 3 3 3 4 5 6 7 8 12 12 ]Figure 1: A rooted tree T with its Parent array In Algorithm TREE PAIRED DOMINATION that follows, we will for each vertex, i, keep track of two boolean values, namely In PD Set[ i] and Dom by child[ i].
These values are initially set to false. As the algorithm progresses some of these values may change to true. Upon completion of the algorithm, all vertices, i, with In PD Set[ i] = true will be added to the PD-set. Further, all vertices with Dom by child[ i] = true will be dominated by one of their children. We define vertex 0 to be a dummy -vertex that does not exist and we define parent = parent = 0. We will never change In PD Set but may change Dom by child.
Algorithm TREE PAIRED DOMINATION :Input: A rooted tree T with V (T ) = {1, 2, . . . , n } rooted at 1 where n ≥ 2 (and where larger values are further from the root) and represented by an array parent[1 . . . n ]
Output: An array In PD Set[] that indicates which vertices belong to the PD-set
Code:
for i = 0 to n do {
Dom by child[ i] = false 3. In PD Set[ i] = false }
for i = n to 2 do
if (Dom by child[ i] = false )
and (In PD Set[parent[ i]] = false )
and (parent[ i]̸ = 1)
and (In PD Set[parent[parent[ i]]] = false )(we remark that in this case, no child of vertex i is in the PD-set, and neither is the parent of vertex i nor the parent of the parent of vertex i; further, the parent of vertex i is not the vertex 1)
then {
In PD Set[parent[ i]] = true 7. In PD Set[parent[parent[ i]]] = true 8. Dom by child[parent[parent[ i]]] = true 9. Dom by child[parent[parent[parent[ i]]]] = true (if the vertex exists) }
(we remark that in this case, we add the parent of vertex i and the parent of the parent of vertex i to the PD-set, and these two added vertices are paired in the PD-set. Further, we note that the parent of the parent of vertex i now is dominated by one of its children, and so we update Dom by child[parent[ i])
if (Dom by child[ i] = false )
and (In PD Set[parent[ i]] = false )
and (parent[ i] = 1) or (In PD Set[parent[parent[ i]]] = true )(we remark that in this case, the PD-set contains no child of vertex i and does not contain the parent of vertex i, but either the parent of vertex i is the vertex 1 or the PD-set contains the parent of the parent of vertex i.)
then {
In PD Set[parent[ i]] = true 12. and if i1 is the child of parent[ i] of largest label such that In PD Set[ i1] = false
then {
In PD Set[ i1] = true 14. Dom by child[parent[ i]] = true } }
(we remark that in this case, we add the parent of vertex i and a child of the parent of vertex i of the largest label to the PD-set, and these two added vertices are paired in the PD-set. Further, we note that the parent of vertex i now is dominated by one of its children, and so we update Dom by child[parent[ i]) 415. if (Dom by child = false )
then {
In PD Set = true 17. In PD Set = true 18. Dom by child = true }
(we remark that in this case, no child of the vertex 1 is in the PD-set, and so we add to the PD-set the vertices 1 and 2, and these two added vertices are paired in the PD-set )We prove next that algorithm TREE PAIRED DOMINATION produces a γpr -set of T .
Theorem 1. If T is tree of order n ≥ 2, then the set S produced by algorithm TREE PAIRED DOMINATION defined by
S = {i ∈ [n] : In PD Set[ i] = true }
is a γpr -set of T .Proof. Let S be the set produced by algorithm TREE PAIRED DOMINATION , and so S =
{i ∈ [n] : In PD Set[ i] = true }. By construction, S is a PD-set of T . Hence it remains to show that S is a γpr -set of T . Let S(i) denote all vertices, j, with In PD Set[ j] = true, immediately after performing line 14 in our tree algorithm with the value i. We show that property P (i)below holds for all i = n, n − 1, n − 2, . . . , 2, by induction.
Property P (i): The following properties hold. 1. The set S(i) totally dominates the set {i, i + 1 , i + 2 , . . . , n }.2. The set S(i) ⊆ Qi for some γpr -set Qi of T .We first consider Property P (n) (and so here i = n). If parent[ n] = 1, then T is a star rooted at its central vertex and S(n) = {1, 2} and Dom by child = true. If parent[ n]̸ = 1, then S(n) = {parent[ n], parent[parent[ n]] } and Dom by child[parent[parent[ n]]] = true. In both cases, the set S(n) contains the parent of vertex n, and therefore S(n) totally dominates the set
{n}. Let Qn be an arbitrary γpr -set of T . Moreover since the parent of vertex n is a support vertex (with leaf neighbor n), every γpr -set of T contains the parent of vertex n. In particular, parent[ n] ∈ Qn.Suppose that parent[ n] = 1. In this case, γpr (T ) = 2 and Qn contains a leaf neighbor of parent[ n]. If vertex 2 does not belong to Qn, then replacing the leaf neighbor of parent[ n] that belongs to Qn with the vertex 2 produces a γpr -set of T that is identically the set S(n). Suppose that parent[ n]̸ = 1. In this case, if parent[parent[ n]] does not belong to the set Qn, then replacing the leaf neighbor of parent[ n] that belongs to Qn with the vertex parent[parent[ n]] produces a new γpr -set of T that contains the set S(n). Furthermore, the array Dom by child is updated correctly noting that in both cases Dom by child[parent[ n]] = true upon completion of line 14 in our tree algorithm with the value n. Thus, property P (n) holds. This establishes the base case. For the inductive hypothesis, let 2 ≤ i < n and assume that property P (i + 1) holds. Thus, the set S(i + 1) totally dominates the set {i + 1 , i + 2 , . . . , n }. Moreover, S(i + 1) ⊆ Qi+1 for some 5γpr -set Qi+1 of T , and the array Dom by child is updated correctly. For notation convenience, let pi be the parent of vertex i and if pi̸ = 1, then let gi be the grandparent of vertex i. Thus,
pi = parent[ i] and gi = parent[parent[ i]]. We now show that property P (i) holds. Consider when the vertex i is first considered in the algorithm. Thus at this stage, the vertex i + 1 has already been analyzed by the algorithm and the next vertex to be considered is vertex i. If Dom by child[ i] = true or In PD Set[ pi] = true, then the vertex i is totally dominated by S(i + 1) and therefore S(i + 1) totally dominates
{i, i + 1 , i + 2 , . . . , n }. In this case, we let S(i) = S(i + 1) and Qi = Qi+1 and note that
S(i) ⊆ Qi+1 = Qi, which implies that S(i) is a subset of a γpr -set of T , namely Qi. Hence we may assume that Dom by child[ i] = false and In PD Set[ pi] = false, for otherwise Property P (i)holds as desired. We now consider the γpr -set, Qi+1 , of T that contains the set S(i + 1). We note that the set
Qi+1 exists by the inductive hypothesis. Some vertex, t, in Qi+1 must totally dominate vertex i.Thus either t is the parent of vertex i or a child of vertex i. Let t′ be the vertex paired with t in the set Qi+1 . We consider next the two possible outcomes of lines 5 and 10 when the vertex i is the current vertex considered.
Case 1. The statements in line 5 hold; that is,
• Dom by child[ i] = false,
• In PD Set[ pi] = false,
• pi̸ = 1, and In PD Set[ gi] = false. Thus the set S(i + 1) contains no child of vertex i, and neither pi nor gi belong to S(i + 1). According to lines 6 and 7, we have S(i) = S(i + 1) ∪ { pi, g i}.Suppose firstly that {pi, g i} ⊆ Qi+1 . In this case, we let Qi = Qi+1 and note that S(i) ⊆ Qi.Moreover since pi ∈ S(i), the set S(i) totally dominates the set {i, i + 1 , i + 2 , . . . , n }. Since Dom by child[ gi] = true in line 8, the array Dom by child is updated correctly. Hence, we may assume that pi /∈ Qi+1 or gi /∈ Qi+1 (or both pi /∈ Qi+1 and gi /∈ Qi+1 ). Suppose secondly that pi /∈ Qi+1 , implying that t is a child of i. Let Ti be the subtree of T
rooted at the vertex i. By the inductive hypothesis, the set S(i + 1) totally dominates the set
{i+1 , i +2 , . . . , n }. By supposition, Dom by child[ i] = false, and so no child of vertex i belongs to the set S(i+1). Thus the set S(i+1) totally dominates all vertices in the subtree Ti, except for the vertex i. By construction, the set S(i+1) is therefore a PD-set of Ti −i. Let S′
i
= V (Ti)∩S(i+1), and let Q′
i
= V (Ti) ∩ Qi+1 . Since S(i + 1) ⊂ Qi+1 and t ∈ Qi+1 \ S(i + 1), we note that S′
i
⊂ Q′
i
,implying that |S′
i
| ≤ | Q′
i
| − 2. If gi /∈ Qi+1 , then we let Qi = ( Qi+1 \ Q′
i
) ∪ (S′
i
∪ { pi, g i}). If
gi ∈ Qi+1 , then we let Qi = ( Qi+1 \ Q′
i
) ∪ (S′
i
∪ { pi, i }). In both cases, S(i) ⊆ Qi. Further, the set
Qi is a PD-set of cardinality |Qi| = |Qi+1 | − | Q′
i
| + ( |S′
i
| + 2) ≤ | Qi+1 | = γpr (T ), and is therefore a γpr -set of T . Hence if pi /∈ Qi+1 , then we have shown that the set S(i) totally dominates the set {i, i + 1 , i + 2 , . . . , n } and is contained in some γpr -set of T , namely the set Qi defined above. Suppose thirdly that pi ∈ Qi+1 and gi /∈ Qi+1 . In this case we can choose the vertex t to be the vertex pi, that is, t = pi. Let t′ be the vertex paired with t in the PD-set Qi+1 . Since gi /∈ Qi+1 ,the vertex t′ is a child of pi. Possibly, t′ = i. If t′ is a leaf, then we let Qi = ( Qi+1 { t′})∪{ gi} with
pi and gi paired in Qi, and with all other pairings in Qi the same as the pairings in Qi+1 \ { t, t ′}.As before, S(i) ⊆ Qi and the set S(i) totally dominates the set {i, i + 1 , i + 2 , . . . , n }. Hence, we may assume that t′ is not a leaf. Let T ′ be the subtree of T rooted at the vertex t′, and let j be 6an arbitrary child of t′. We note that j ≥ i + 1, and so by the inductive hypothesis the set S(j)totally dominates all vertices in the subtree Tj rooted at vertex j. Since S(j) ⊆ S(i + 1), we infer that all vertices in T ′ are totally dominated by S(i + 1), except possibly for the vertex t′. Hence as before, we let Qi = ( Qi+1 \ { t′}) ∪ { gi} with pi and gi paired in Qi, to produce the desired result.
Case 2. The statements in line 10 hold; that is,
• Dom by child[ i] = false,
• In PD Set[ pi] = false,
• pi = 1 or pi̸ = 1 and In PD Set[ gi] = true. Thus, the set S(i + 1) does not contain the parent pi of vertex i and contains no child of vertex i. However, either pi = 1 or pi̸ = 1 and In PD Set[ gi] = true. By the inductive hypothesis,
S(i + 1) ⊆ Qi+1 . In particular, if pi̸ = 1, then we infer that gi ∈ Qi+1 . According to lines 11–14, we have S(i) = S(i + 1) ∪ { pi, i 1} where i1 is the child of parent[ i] of largest label such that In PD Set[ i1] = false. Possibly, i1 = i.Suppose firstly that {pi, i 1} ⊆ Qi+1 . In this case, we let Qi = Qi+1 and note that S(i) ⊆ Qi.Moreover since pi ∈ S(i), the set S(i) totally dominates the set {i, i + 1 , i + 2 , . . . , n }. Since Dom by child[ pi] = true in line 14, the array Dom by child is updated correctly. Hence, we may assume that pi /∈ Qi+1 or i1 /∈ Qi+1 (or both pi /∈ Qi+1 and i1 /∈ Qi+1 ). Suppose secondly that pi /∈ Qi+1 , implying that t is a child of i. Let Ti be the subtree of T
rooted at the vertex i. By the inductive hypothesis, the set S(i + 1) totally dominates the set
{i + 1 , i + 2 , . . . , n }. By supposition, no child of vertex i belongs to the set S(i + 1). Thus the set
S(i + 1) totally dominates all vertices in the subtree Ti, except for the vertex i. We proceed now analogously as in Case 1. Adopting our notation in Case 1, let S′
i
= V (Ti) ∩ S(i + 1) and Q′
i
=
V (Ti)∩Qi+1 . In this case, |S′
i
| ≤ | Q′
i
|− 2. If i1 /∈ Qi+1 , then we let Qi = ( Qi+1 \Q′
i
)∪(S′
i
∪{ pi, i 1}). If i1 ∈ Qi+1 , then we let Qi = ( Qi+1 \ Q′
i
) ∪ (S′
i
∪ { pi, i }). In both cases, S(i) ⊆ Qi, where Qi is a γpr -set of T .Suppose thirdly that pi ∈ Qi+1 and i1 /∈ Qi+1 . In this case we choose the vertex t to be the vertex pi, that is, t = pi. Let t′ be the vertex paired with t in the PD-set Qi+1 .Suppose that the vertex t′ is a child of pi. Possibly, t′ = i. If t′ is a leaf, then we let
Qi = ( Qi+1 \ { t′}) ∪ { i1} with pi and i1 paired in Qi, and with all other pairings in Qi the same as the pairings in Qi+1 \ { t, t ′}. As before, S(i) ⊆ Qi and the set S(i) totally dominates the set
{i, i + 1 , i + 2 , . . . , n }. Hence, we may assume that t′ is not a leaf. Let T ′ be the subtree of T
rooted at the vertex t′, and let j be an arbitrary child of t′. We note that j ≥ i + 1, and so by the inductive hypothesis the set S(j) totally dominates all vertices in the subtree Tj rooted at j.Since S(j) ⊆ S(i + 1), we infer that all vertices in T ′ are totally dominated by S(i + 1), except possibly for the vertex t′. Hence as before, we let Qi = ( Qi+1 \ { t′}) ∪ { i1} with pi and i1 paired in Qi, to produce the desired result. Hence we may assume that the vertex t′ cannot be chosen to be a child of pi, implying that
t′ = gi. According to our algorithm, since gi ∈ Qi+1 and pi /∈ S(i + 1), the vertex gi was added to the set S(i + 1) when a grandchild j of gi was considered in line 5 of the algorithm for some
j ≥ i + 1. In this case, denoting the parent of vertex j by pj , and so pj = parent[ j], our tree algorithm added to the set S(j +1) the vertices pj and gi to produce the set S(j) upon completion 7of line 14 in our tree algorithm with the value j. By construction of the set S, we note that
S(j) ⊆ S(i + 1), and so {pj , g i} ⊆ Qi+1 .By assumption, the vertices t = pi and t′ = gi are paired in Qi+1 . Let p′
j
be the vertex paired with pj in the set Qi+1 . Since the parent, gi, of vertex pj is paired with vertex pi, the vertex p′
j
is a child of pj . Let Tj be the subtree of T rooted at the vertex j. By the inductive hypothesis, the set S(j), which is contained in the set S(i + 1), totally dominates all descendants of the vertex j,implying that the vertex p′
j
is not needed to totally dominate any descendant of the vertex j.Hence, the vertex p′
j
is needed only to partner the vertex pj . We let Qi = ( Qi+1 \ { gj }) ∪ { i1},where the vertices i1 and pi are paired and the vertices pj and gi are paired and where all other pairing of vertices in Qi+1 remain unchanged. The resulting set Qi is a PD-set of T satisfying
|Qi| = |Qi+1 |. Furthermore, S(i) ⊆ Qi. Thus, Qi is a γpr -set of T containing the set S(i). Thus, property P (i) also holds in Case 2. By Cases 1 and 2, property P (i) holds for all i = n, n − 1, n − 2, . . . , 2, by induction. In particular, property P (2) holds, and so the set S(2) totally dominates the set {2, 3, . . . , n }.Moreover, the set S(2) ⊆ Q2 for some γpr -set Q2 of T . Thus, S(2) is a subset of an optimal solution Q2, and S(2) totally dominates all vertices in G, except possibly for vertex 1. Therefore if vertex 1 is totally dominated by S(2), then S(2) = Q2 and S(2) is an optimal solution. In this case, we take Q1 = Q2 = S2, and the desired result follows. Hence we may assume that vertex 1 is not totally dominated by S(2). In this case, some vertex,
z, in Q2 must totally dominate vertex 1. Thus, z is a child of vertex 1. Since z / ∈ S(2) and both
|S(2) | and |Q2| are even, we infer that |S(2) | ≤ | Q2| − 2. We now let Q1 = S(2) ∪ { 1, 2} with the vertices 1 and 2 paired in Q1 and with all other pairing in S(2) unchanged. The resulting set Q1 is a PD-set of T . We infer that |Q2| = γpr (T ) ≤ | Q1| = |S(2) | + 2 ≤ | Q2| = γpr (T ), and so we must have equality throughout this inequality chain. In particular, |Q1| = γpr (T ), implying that Q1 is a γpr -set of T . Furthermore, S(1) ⊆ Q1. Thus, Q1 is a γpr -set of T containing the set
S(1). Thus, property P (1) holds. Since S(1) is a PD-set of T and |S(1) | ≤ | Q1| = γpr (T ), the set
S(1) is an optimal solution, that is, S(1) is a γpr -set of T . Since S = S(1), our algorithm TREE PAIRED DOMINATION does indeed find an optimal solution. This completes the proof of Theorem 1. We remark that the time complexity of the algorithm is O( n) for trees of order n. When we run our algorithm on the rooted tree T in Figure 1 the following steps will be performed, yielding the γpr -set {12 , 11 , 8, 7, 6, 5, 4, 3, 2, 1} as a minimum PD-set in T . In particular, we note that
γpr (T ) = 10.
3 Paired domination in random trees
In this section, we are interested in the limiting distribution of paired domination in random trees of order n as n → ∞ . The random tree model we have in mind is the so-called conditioned Galton–Watson tree , which forms a general model that includes many combinatorial models as special cases. We assume that all trees are rooted and ordered, i.e., the order of the root branches matters. Given a nonnegative integer-valued random variable ξ, the Galton–Watson tree T is a random 8i action i action 15 In PD Set = In PD Set = true 7 none 14 none 6 none 13 In PD Set = In PD Set = true 5 none 12 none 4 none 11 In PD Set = In PD Set = true 3 none 10 In PD Set = In PD Set = true 2 none 9 In PD Set = In PD Set = true 1 none 8 none Table 1: Algorithm TREE PAIRED DOMINATION implemented on the tree T in Figure 1 ordered rooted tree generated by the following process: Begin with a root vertex and create a certain number of children according to ξ. Each child then independently generates a number of children according to an independent copy of ξ. This process continues for each newly generated vertex until all vertices in the tree have been processed. Note that in general, such a process may not stop in finite time. Here, we consider only critical Galton–Watson trees where E(ξ) = 1. Furthermore, we assume throughout that
• Pr( ξ = 0) > 0, and
• gcd {i | Pr( ξ = i) > 0} = 1. Under these conditions, it is well known that T is finite almost surely and that Pr( |T | = n) > 0for sufficiently large n.The process above defines a probability distribution on the set of finite rooted ordered trees, where, for a given rooted tree T ,Pr( T = T ) = Y
j≥0
Pr( ξ = j)Dj (T ),
and Dj (T ) denotes the number of vertices in T with outdegree j. This induces a probability distribution on the set of rooted trees of order n by conditioning T to have exactly n vertices. The tree generated under this conditioning is called the conditioned Galton–Watson tree and is denoted by Tn. Many models of combinatorial trees can be obtained in this way.
• Plane trees : Among all rooted, unlabelled, plane (ordered) trees of order n, pick a random tree where each tree is equally likely. This model is equivalent to the conditioned Galton– Watson tree with offspring distribution ξ ∼ Geo( 12 ).
• Binary trees : A plane binary tree is a rooted, unlabelled plane tree where each vertex has outdegree at most two. The uniform binary tree model is equivalent to the conditioned Galton–Watson tree with offspring distribution ξ ∼ Bin(2 , 12 ).
• Rooted labelled trees : Pick one tree uniformly at random among all rooted unordered la-belled trees of order n. This model also produces a random uniform Cayley tree of order
n (by simply ignoring the root) and is equivalent to the conditioned Galton–Watson tree with offspring distribution ξ ∼ Pois(1). 9We are now ready to state the main result of this section.
Theorem 2. Let T be the Galton–Watson tree with offspring distribution ξ such that E(ξ) = 1
and 0 < Var( ξ) < ∞. Let Tn denote the corresponding conditioned Galton–Watson tree of order
n. Then, there exist constants μpr > 0 and σpr > 0 such that as n → ∞ ,
Eγpr (Tn) = μpr n + o(√n) and γpr (Tn) − μpr nσpr
√n
d
→ N (0 , 1) .
Table 2 shows the numerical values of the constant μpr for the three combinatorial tree models mentioned above.
ξ μpr
Binary trees Bin(2 , 1/2) 0.5255 Plane trees Geo(1 /2) 0.4747 Labelled trees Pois(1) 0.5177 Table 2: Numerical values for μpr .The proof of Theorem 2 is carried out in several steps, which make use of the algorithm described in the previous section. When we run the algorithm with an input tree T , we label each vertex as B, F , R, or P according to the following rules:
• A vertex i is labelled B if, when considering i (in the algorithm), both In PD Set[ i] and Dom by child[ i] are false.
• A vertex i is labelled F if, when considering i, In PD Set[ i] = false and Dom by child[ i] = true.
• A vertex i is labelled R if, when considering i, In PD Set[ i] = true and none of its children is labelled R.
• A vertex i is labelled P if, when considering i, In PD Set[ i] = true and at least one of its children is labelled R.We can also incorporate this vertex-labelling into the algorithm by inserting the following lines between Lines 4 and 5, and also between Lines 14 and 15, of Algorithm TREE PAIRED DOMI-NATION:
Additional Code: If ( In PD Set[ i] = false and Dom by child[ i] = false )
Then {ℓ[i] = B}
If ( In PD Set[ i] = false and Dom by child[ i] = true )
Then {ℓ[i] = F }
Elseif ( for every child j of i, ℓ(j)̸ = R )10 Then {ℓ[i] = R}
Else Then {ℓ[i] = P }
Since we have introduced new labels for the vertices, we shall refer to the numerical labels that we considered previously as ranks, e.g., the root has rank 1. At the end of the modified algorithm, each vertex of T is assigned a unique label: B, F , R, or P (see the example in Figure 2 for illustration). We also remark that no two vertices labelled R are adjacent to each other. The next lemma shows that the label of each vertex can be determined solely by the labels of its children.
B
F
F P F
P R P R P
R B R B
B B
B R
B
Figure 2: A tree with labelled vertices
Lemma 1. If the vertices of a tree T are labelled according to the rules above, then for every v
in T we have
• v is labelled B if and only if v is a leaf or every child of v is labelled F ,
• v is labelled F if and only if v has at least one child labelled P and all the other children of
v are labelled F ,
• v is labelled R if and only if v has at least one child labelled B and all the other children of
v are labelled P or F ,
• v is labelled P if and only if v has at least a child labelled R.Proof. For a given rooted tree T , we label the vertices of T in two ways: according to the original rules and according to the rules in the lemma. Assume, by contradiction, that these give two different labellings of T . Let v be the vertex with the highest rank (farthest from the root) such that the labelling given by the original definition differs from the labelling given in the statement of the lemma. Note that v cannot be a leaf, since leaves are labelled B in both labellings. Thus, we may assume from now on that v is an interior vertex. There are then different cases to consider:
• Assume that the statement of the lemma gives v the label B. This means that all children of v are labelled F . Since both statements agree on the labelling of the children of v, the 11 algorithm would have passed through each of these children without changing any of the vertex states. Additionally, no grandchildren of v are labelled B, since this would change the In PD Set state of a child of v to true. Thus, by the time v is considered by the algorithm, both Dom by child[ v] and In PD Set[ v] remain false. Therefore, the definition would also give v the label B.
• Assume that the statement of the lemma gives v the label F . This means that at least one of its children is labelled P , and the rest are labelled F (by both rules). This implies that when v is considered by the algorithm, Dom by child[ v] = true. It remains to show that In PD Set[ v] is false. Initially, In PD Set[ v] is set to false. If In PD Set[ v] has been changed to true before v has been considered by the algorithm, then v must have a grandchild labelled B. Let w be such a B-grandchild of v with the highest rank. Then, the parent of
w will be labelled R (by both rules). This is a contradiction since no child of v is labelled R.
• Assume that the statement of the lemma gives v the label R. Then, v has some children labelled B and the rest are labelled P or F (by both rules). It is clear that no child is labelled R. Let w be the B-child of v with the highest rank. By the time the algorithm considers w, both v and its parent have In PD Set and Dom by child set to false. So, after considering w, In PD Set[ v] = true. Hence, v will also be given the label R by the definition.
• Finally, assume that the statement of the lemma gives v the label P . Then, v must have a child labelled R, which in turn must have a child labelled B. Let w1 be the R-child of
v with the highest rank, and let w2 be the B-child of w1 with the highest rank. When
w2 is considered by the algorithm, both In PD Set[ w1] and In PD Set[ v] are set to true. Therefore, v is also labelled P by the definition. Hence, both the original rules and the rules in the lemma give the same label to v, which is a contradiction. It is useful to partition the class of rooted plane trees T into four subclasses: TB , TF , TR, and
TP , according to the label of the root. For instance, a tree in TB has its root vertex labelled B.Let Φ( T ) be the number of vertices labelled R in T after running the algorithm with the vertex labelling described above. By convention, we assume that a tree consisting of a single vertex belongs to TB ; hence, Φ( •) = 0.
Lemma 2. For every tree T of order at least 2, we have
Φ( T ) =
d
X
k=1
Φ( Tk) + φ(T ) (1)
where T1, T 2, . . . , T d are the root-branches of T , and
φ(T ) =
(
1 if T ∈ TR
0 otherwise .
Moreover, we have
γpr (T ) = 2Φ( T ) +
(
2 if T ∈ TB
0 otherwise .
12 Proof. Since Φ( T ) counts the number of R-vertices in T after running the algorithm, the first part of the lemma is obvious, as we only need to sum the number of R-vertices from all the branches and add the root of T if it is labelled R.The second part of the lemma is proved in a few steps. Let us first prove the following lower bound:
γpr (T ) ≥ 2Φ( T ). (2) To see this, notice that the γpr -set produced by the algorithm contains all the R-vertices. In addition, the R-set is independent by definition of the label R. Each of these R-vertices is paired with another vertex that is not labelled R in the γpr -set, so the claim is proved. Next, let us denote by S the set of vertices labelled R or P . We claim that S dominates T ,except possibly the root. To see this, assume that there is a vertex v different from the root that is not dominated by S. This implies that ℓ(v) /∈ { R, P } and no neighbor of v is labelled P or
R. In particular, the labels of the children of v are either B or F . Thus, we have the following possible cases:
• If v has a B-child, then In PD Set[ v] = true. So, by definition, ℓ(v) must either be R or P ,which is a contradiction.
• If all children of v are labelled F , then ℓ(v) = B. By Lemma 1, the parent of v cannot be labelled B or F . Therefore, the parent of v has to be labelled P or R, which is again a contradiction. We can then conclude that the set S of R and P vertices dominates T , except possibly the root. Let us now construct a PD-set from S. Consider the set S∗ in the following way: we go from the top to the bottom of T , and at each P -vertex, we pair it with one of its R-children, which we know exists by Lemma 1. Then, for the remaining R-vertices which have not yet been paired, pair each with one of its B-children (which we also know exists), and add this selected
B-child to S∗.We already know that S∗ dominates T , except possibly the root, which we will denote by r.So, we must separately consider the cases where the root is dominated or not. Observe first that the root is not dominated by S∗ if and only if the root is labelled B. To see this, if the root r
is not dominated by S∗, then none of its children is labelled R or P . Moreover, the root cannot have a child labelled B either, since otherwise, it would be labelled R according to Lemma 1. This means that all the children are labelled F , which implies that ℓ(r) = B. Conversely, if
ℓ(r) = B, then all its children are labelled F by Lemma 1, so they cannot be in S∗. Thus, r is not dominated by S∗. Now consider the following two cases:
• If the root r is dominated by S∗ (equivalently, ℓ(r)̸ = B), then S∗ is a PD-set of T .Therefore,
γpr (T ) ≤ | S∗| = 2Φ( T ).
This is in fact an equality by Equation (2). This proves the case T / ∈ TB .
• If the root is not dominated by S∗ (equivalently, ℓ(r) = B), then clearly none of the children of the root is in S∗. Therefore, we can add the root and one of its children to S∗ to obtain a PD-set of T . We obtain
γpr (T ) ≤ 2Φ( T ) + 2 .
13 We claim that we also have
γpr (T ) ≥ 2Φ( T ) + 2 .
Indeed, suppose to the contrary that this is not the case, i.e., γpr (T ) = 2Φ( T ). Then, let us take the γpr -set from the algorithm, with an appropriate pairing of the elements. In this case, since the R-vertices are independent, each pair of this γpr -set must have one R-vertex. Now, since it is a dominating set, it must contain a child of the root, say w. This means that ℓ(w) = R or w has a child labelled R. In either case, ℓ(w) cannot be F , which is in contradiction with ℓ(r) = B (according to Lemma1). This completes the proof of the lemma. Tree parameters that satisfy a recursion like the one in Equation (1) are called additive. In our case, the function φ is referred to as the associated toll function of the additive parame-ter Φ. There are many results in the literature concerning the limiting distributions of additive parameters of random rooted trees; see, for example, [13, 21, 24] and the references therein. The result relevant to our case can be found in [21, Theorem 1]. For completeness, we will state this theorem in full. But before we do so, let us recall some background and notation. We consider F to be an additive tree functional with a toll function f , which means that for every rooted tree T , we have
F (T ) =
d
X
k=1
F (Tk) + f (T ).
where T1, T 2, . . . , T d are the root-branches of T . We also assume that the toll function f is bounded. For a tree T and a positive integer M , let T (M ) denote the tree consisting of all vertices whose distance to the root is at most M . According to Kesten , we define the infinite size-biased Galton–Watson tree ˆT in the following way: Starting from an infinite path beginning at the root, we add to each vertex on the path and each side of the path a random number of additional branches, each of which is an independent copy of T . The random infinite tree ˆT plays an important role in the study of conditioned Galton–Watson trees because it is known that the truncated random tree T (M )
n
converges in distribution to ˆT (M ) for every fixed M , more details on this can be found in . The distribution of ˆT (M ) is given by the following formula: Pr( ˆT (M ) = T ) = wM (T ) Pr( T (M ) = T ),
where wM (T ) denotes the number of vertices at distance M from the root in T , see [12, Equa-tion (5.11)]. The above equation defines a probability distribution if E(ξ) = 1 since in this case we can show that E(wM (T )) = 1.
Theorem 3 () . Let Tn be a conditioned Galton–Watson tree of order n with offspring distri-bution ξ, where ξ satisfies E(ξ) = 1 and 0 < σ 2 := Var( ξ) < ∞. Consider an additive parameter
F with a bounded toll function f . Furthermore, assume that there exists a sequence (pM )M ≥1 of positive numbers with pM → 0 as M → ∞ such that
• for every M, N ∈ { 1, 2, . . . }, such that N ≥ M
E f ( ˆT (M )) − E
f ( ˆT (N )) | ˆT (M )
≤ pM , (3) 14 and
• there is a sequence of positive integers (Mn)n≥1 such that for large enough n,
E f (Tn) − f
T (Mn)
n
≤ pMn . (4)
If an := n−1/2(np Mn + M 2
n
) satisfies
lim
n→∞
an = 0 , and
∞
X
n=1
an
n < ∞, (5)
then F (Tn) − μn
√n
d
→ N (0 , γ 2) (6)
where μ = Ef (T ), for some 0 ≤ γ < ∞. Moreover, we have
E(F (Tn)) = μn + o(√n ), and Var F (Tn) = γ2n + o(n). (7) In order to prove Theorem 2, we need to show that the toll function φ associated with the functional Φ defined in Lemma 2 satisfies the conditions of Theorem 3. This is our main goal in the proof. The conditions of Theorem 3 essentially say that there are not “too many” trees T
such that f (T )̸ = f (T (M )). For our cases, this means that if we run our algorithm (twice), once with input T and once with T (M ), then most of the time the label of the root remains unchanged. To make this precise, we consider the following set of trees for M > 3:
BM =
n
T possibly infinite : ∃N > M, ℓ
root(T (N) )
̸
= ℓ
root(T (M) )
o
,
where ℓ(root(T)) indicates the label of the root of T , i.e., B, F , R or P , after running our algorithm with input T . Observe that there exists a tree τ0 of height 3 with the property that Pr( T = τ0) > 0 and that for every finite T such that T (3) = τ0,
ℓ(root(T)) = ℓ
root(T (3) )
.
The existence of a tree τ0 with the above property is crucial in the proof of Theorem 2, so let us explicitly construct an example of such a tree. We define
d0 := min {d ≥ 2 : Pr( ξ = d)̸ = 0 }, (8) whose existence follows from our assumptions on ξ. Then, the tree τ0 described in Figure 3, where each of the four internal vertices has outdegree d0, satisfies the desired properties. It is easy to see that if we run the algorithm on any extension of τ0, then the root will always be labelled P since the root of the left-most branch will be labelled R.
Lemma 3. There exists a constant c1 ∈ (0 , 1) such that Pr( T ∈ B M ) = O(cM
1
) as M → ∞ .
15 d0
d0 − 2
· · ·
· · ·
· · ·
· · ·
Figure 3: Fixed tree τ0 of height 3.
Proof. Observe, by definition of τ0, that for any tree T to be in BM , we must have T (3) ̸ = τ0.Hence, we can write Pr( T ∈ B M ) = X
T̸=τ0
Pr( T (3) = T ) Pr( T ∈ B M |T (3) = T ).
For any tree T , recall that w3(T ) denotes the number of vertices at distance 3 from the root. From the definition of T , given the event {T (3) = T }, the rest of T consists of w3(T ) independent copies of T . On the other hand, if we also have T ∈ B M , then at least one of those w3(T ) branches of T is an element of BM −3 (if M > 3). Thus, we have Pr( T ∈ B M |T (3) = T ) ≤ w3(T ) Pr( T ∈ B M −3).
Therefore, Pr( T ∈ B M ) ≤ Pr( T ∈ B M −3) X
T̸=τ0
Pr( T (3) = T )w3(T )= Pr( T ∈ B M −3)
E(w3(T )) − w3(τ0) Pr( T = τ0)
.
It is clear that the term in brackets in the last expression is independent of M and is nonnegative. Moreover, since we know that E(w3(T )) = 1 and w3(τ0) Pr( T = τ0) > 0, this term is also strictly less than 1. Thus, the lemma is proved by iteration.
Lemma 4. There exists a constant c2 ∈ (0 , 1) such that Pr( ˆT ∈ B M ) = O(cM
2
) as M → ∞ .Proof. The proof follows the same lines as the proof of the previous lemma. We have Pr( ˆT ∈ B M ) = X
T̸=τ0
Pr( ˆT (3) = T ) Pr( ˆT ∈ B M | ˆT (3) = T ).
Given the event { ˆT (3) = T }, the rest of ˆT consists of w3(T ) independent random trees, where one is a copy of ˆT and the rest are copies of T . Note that these branches are ordered, and the position of the copy of ˆT in this order is uniform. Hence, we have Pr( ˆT ∈ B M | ˆT (3) = T ) ≤ Pr( ˆT ∈ B M −3) + ( w3(T ) − 1) Pr( T ∈ B M −3).
16 Thus, we deduce that Pr( ˆT ∈ B M ) ≤
X
T̸=τ0
Pr( ˆT (3) = T )
Pr( ˆT ∈ B M −3) + E(w3( ˆT ) − 1) Pr( T ∈ B M −3).
We know that E(w3( ˆT ) − 1) is finite (see for example [11, Lemma 2.3]); hence, by Lemma 3, the second term on the right-hand side is bounded by O(cM
1
). On the other hand, 0 ≤ X
T̸=τ0
Pr( ˆT (3) = T ) = 1 − Pr( ˆT (3) = τ0) < 1.
Once again, the lemma is proved by iteration. We are now ready to prove our main theorem.
Proof of Theorem 2. Observe that if T / ∈ B M , then for every N ≥ M , φ(T (N )) = φ(T (M )). So, since the value of φ can only be 0 or 1, we have
φ( ˆT (M )) − E
φ( ˆT (N )) | ˆT (M )
≤ 1{ ˆT ∈B M }.
Using Lemma 4, we obtain
E φ( ˆT (M )) − E
φ( ˆT (N )) | ˆT (M )
≤ Pr( ˆT ∈ B M ) ≪ cM
2
. (9) By the same argument, we also have
E φ(Tn) − φ(T (M )
n
) ≤ Pr( Tn ∈ B M ) ≤ Pr( T ∈ B M )Pr( |T | = n) .
Based on our assumptions on ξ, it is well known that Pr( |T | = n) = Θ( n−3/2); see Janson [13, Equation (4.13)], which also refers to Kolchin . So, by Lemma 3, we have
E φ(Tn) − φ(T (M )
n
) = O(n3/2cM
1
). (10) If we choose Mn := (log n)2 and pM := cM
3
, where max {c1, c 2} < c 3 < 1, then it is not difficult to verify that all conditions of Theorem 3 are satisfied. Therefore, we obtain the asymptotic normality of Φ( Tn), with the estimates for its mean and variance exactly as stated in Theorem 3. Furthermore, the same result follows for γpr (Tn), since by Lemma 2, we have
|2Φ( Tn) − γpr (Tn)| ≤ 2.
Hence, the constant term μpr in Theorem 2 is equal to 2 E(φ(T )). On the other hand, we also have Var( γpr (Tn)) = σ2pr n + o(n).
It remains to show that the constant σ2pr is nonzero, to guarantee a nondegenerate Gaussian limit law for γpr (Tn). This follows from an argument already used in the proof of [21, Theorem 10]. By the same reasoning, it is enough to show that there exist two finite trees, τ1 and τ2, of the same size which satisfy the following properties: 17 • Pr( T = τi) > 0 for i ∈ { 1, 2},
• γpr (τ1)̸ = γpr (τ2), and
• ℓ(root( τ1)) = ℓ(root( τ2)) .
To briefly explain the idea, if we condition on the structure obtained by deleting every occur-rence of τ1 or τ2 as a fringe subtree of Tn, then γpr (Tn) is equal to a fixed term plus a constant (equal to |γpr (τ1)−γpr (τ2)|) times a sum of independent, nondegenerate Bernoulli variables, where the number of these Bernoulli variables is exactly the number of occurrences of τ1 and τ2 in Tn.Since, on average, the number of such occurrences in Tn is ≫ n (see, for example, [13, Theorem 1.3] which refers to among others), the law of total variance implies that Var( γpr (Tn)) ≫ n.Now, to construct such trees τ1 and τ2, let d0 be as defined in Equation (8). Then the two full
d0-ary trees shown in Figure 4 satisfy the second property. We denote these trees, from left to right in Figure 4, by t1 and t2, where the dots indicate leaves that need to be added to complete the trees as full d0-ary trees. It is clear that the paired domination numbers of t1 and t2 are 4 and 6, respectively, with the γpr -set consisting of the black-coloured vertices in the figure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 4: Two trees, t1 and t2, of the same size with different paired domination numbers. Finally, we obtain the trees τ1 and τ2 by attaching t1 and t2 (respectively) to the rightmost leaf at level 3 of the tree τ0 shown in Figure 3. It is straightforward to verify that the trees τ1
and τ2 constructed in this way satisfy all the desired properties. This completes the proof. We conclude this section with a method for computing the constant μpr and some further discussion. From the main theorem, we have μpr = 2 E(φ(T )), and since φ is an indicator function for TR, we have
μpr = 2 Pr( T ∈ TR). (11) It is convenient to define xX := Pr( T ∈ TX ) for each X ∈ { B, F, R, P }. Then, for each X, we can write
xX =
∞
X
k=0
Pr( ξ = k) Pr( T ∈ TX | deg T = k).
Applying the characterisations in Lemma 1 , we immediately deduce that Pr( T ∈ TB | deg T = k) = Pr( T ∈ TF )k,
Pr( T ∈ TF | deg T = k) = (Pr( T ∈ TP ) + Pr( T ∈ TF )) k − Pr( T ∈ TF )k,
Pr( T ∈ TR| deg T = k) = (Pr( T ∈ TB ) + Pr( T ∈ TP ) + Pr( T ∈ TF )) k − (Pr( T ∈ TP ) + Pr( T ∈ TF )) k , and Pr( T ∈ TP | deg T = k) = 1 − (Pr( T ∈ TB ) + Pr( T ∈ TP ) + Pr( T ∈ TF )) k .
18 Let us denote by g the probability generating function of ξ, i.e.,
g(x) :=
∞
X
k=0
Pr( ξ = k)xk,
which converges for every x in the interval [0 , 1] with g(1) = 1. Then, from the above equations, the constants xB , xF , xR, and xP satisfy the following system of equations:
xB = g(xF )
xF = g(xF + xP ) − g(xF )
xR = g(xB + xF + xP ) − g(xF + xP )
xP = 1 − g(xB + xF + xP ).
(12) Since Var( ξ) < ∞ implies that g(x) is at least twice differentiable on the interval (0 , 1), and also that g′(x) = E(ξ) = 1, we can use these facts to show that the above system of equations has a unique solution (e.g., by showing that the Jacobian determinant is nonzero). Then, such a solution can be numerically computed for specific examples, such as binary trees, plane trees, and labelled rooted trees, as shown in Table 3.
g(x) xB xF xR xP
Binary trees (1 + x)2/4 0.3347 0.1571 0.2627 0.2455 Plane trees (2 − x)−1 0.5145 0.0563 0.2374 0.1918 Labelled trees exp( x − 1) 0.4085 0.1046 0.2589 0.2281 Table 3: Numerical solutions of Equation (12). In the case of labelled trees, each Cayley tree on the vertex set {1, 2, . . . , n } corresponds to
n distinct labelled rooted trees, determined by the choice of the root. Therefore, choosing a random Cayley tree on the vertex set {1, 2, . . . , n } uniformly is equivalent to selecting a labelled rooted tree and then forgetting the root. As a result, the distribution of the paired domination number of a random Cayley tree of n is also asymptotically normal, with mean and variance as stated in Theorem 2. Figure 5 shows a histogram of paired domination numbers of 20000 randomly (and indepen-dently) sampled Cayley trees of size 10000. The simulation was run using SageMath . The horizontal axis represents the actual value of the paired domination number. It is, in fact, quite common to obtain central limit theorems for various graph-theoretic pa-rameters in random trees. In Banderier et al. , it is shown that the independence number, the path covering number, and the size of the kernel all satisfy a central limit theorem for simply generated trees. In Fuchs et al. , the same results were obtained for the independence number, the domination number, and the clique cover number in random recursive trees and binary search trees, which are other known models of random trees. Similar work on the independence number, the domination number, and the total domination number in conditioned Galton–Watson trees was conducted in the Master’s thesis of Rakotoniaina , supervised by the second author. It is well known that simply generated trees and conditional Galton–Watson trees are essen-tially equivalent, see [4, Section 1.2.7]. Hence, it might be possible to use the generating function 19 Figure 5: Histogram from 20000 random samples of Cayley trees of order n = 10000 approach as in Banderier et al. to prove Theorem 2. This is based on a method given in Dr-mota . However, to use such an analytic-based method, one needs to assume that all moments of ξ are finite (as implied by [2, Assumption 1]), and not just the second moment that we assumed here. One advantage of the analytic approach is that it can provide an explicit expression for the constant in the variance. In general, there is no guarantee that such an expression will be useful. For the parameters considered in , it is not even known if these constants are nonzero, see the remark on [2, Page 49]. Here, we chose to use a probabilistic approach. While we cannot provide an explicit formula for the constant in the variance, we are at least able to prove that, for our parameter, this constant is never zero, using an argument already employed in . Hence, we guarantee a nondegenerate Gaussian limit law. The same argument can certainly be used to address the problem stated in the remark on [2, Page 49]. For future work, it would also be interesting to explore other models of random trees. The results in Fuchs et al. were based on a general result in Holmgren and Janson . Since the toll function φ for the paired domination number is bounded, the same theorem applies in our case as well for the binary search trees and the recursive trees. However, more work needs to be done for other models of random trees, such as d-ary increasing trees or even P´ olya trees. The general results in [22, 24] for these models do not apply to the paired domination number or most of the graph-theoretic parameters mentioned above. The SageMath code related to this project is available in the GitHub repository .
4 Acknowledgements
The first author, Michael A. Henning, acknowledges support from the CoE-MaSS under graph number 2025-014-GRA-Structural Graph Theory. Opinions expressed and conclusions arrived at 20 by the first author are those of the author and not necessarily to be attributed to the CoE-MaSS.
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4164 | https://www.unipamplona.edu.co/unipamplona/portalIG/home_34/recursos/01general/21082014/unidad_2_termo_ii.pdf | TRANSIENT HEAT CONDUCTION Mehmet Kanoglu Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008SECOND UNIT SECOND UNIT 2 Objectives • Assess when the spatial variation of temperature is negligible, and temperature varies nearly uniformly with time, making the simplified lumped system analysis applicable • Obtain analytical solutions for transient one-dimensional conduction problems in rectangular, cylindrical, and spherical geometries using the method of separation of variables, and understand why a one-term solution is usually a reasonable approximation • Solve the transient conduction problem in large mediums using the similarity variable, and predict the variation of temperature with time and distance from the exposed surface • Construct solutions for multi-dimensional transient conduction problems using the product solution approach.
3 LUMPED SYSTEM ANALYSIS Interior temperature of some bodies remains essentially uniform at all times during a heat transfer process. The temperature of such bodies can be taken to be a function of time only, T(t). Heat transfer analysis that utilizes this idealization is known as lumped system analysis.
A small copper ball can be modeled as a lumped system, but a roast beef cannot.
4 Integrating with T = Ti at t = 0 T = T(t) at t = t The geometry and parameters involved in the lumped system analysis.
time constant 5 The temperature of a lumped system approaches the environment temperature as time gets larger.
• This equation enables us to determine the temperature T(t) of a body at time t, or alternatively, the time t required for the temperature to reach a specified value T(t).
• The temperature of a body approaches the ambient temperature Texponentially. • The temperature of the body changes rapidly at the beginning, but rather slowly later on. A large value of b indicates that the body approaches the environment temperature in a short time 6 Heat transfer to or from a body reaches its maximum value when the body reaches the environment temperature.
The rate of convection heat transfer between the body and its environment at time t The total amount of heat transfer between the body and the surrounding medium over the time interval t = 0 to t The maximum heat transfer between the body and its surroundings 7 Criteria for Lumped System Analysis Lumped system analysis is applicable if When Bi 0.1, the temperatures within the body relative to the surroundings (i.e., T −T) remain within 5 percent of each other.
Characteristic length Biot number 8 Small bodies with high thermal conductivities and low convection coefficients are most likely to satisfy the criterion for lumped system analysis.
Analogy between heat transfer to a solid and passenger traffic to an island.
When the convection coefficient h is high and k is low, large temperature differences occur between the inner and outer regions of a large solid.
9 Example: 10 Example: 11 Quiz: 12 TRANSIENT HEAT CONDUCTION IN LARGE PLANE WALLS, LONG CYLINDERS, AND SPHERES WITH SPATIAL EFFECTS We will consider the variation of temperature with time and position in one-dimensional problems such as those associated with a large plane wall, a long cylinder, and a sphere.
Schematic of the simple geometries in which heat transfer is one-dimensional.
Transient temperature profiles in a plane wall exposed to convection from its surfaces for Ti >T.
13 Nondimensionalized One-Dimensional Transient Conduction Problem 14 Nondimensionalization reduces the number of independent variables in one-dimensional transient conduction problems from 8 to 3, offering great convenience in the presentation of results.
15 Exact Solution of One-Dimensional Transient Conduction Problem 16 The analytical solutions of transient conduction problems typically involve infinite series, and thus the evaluation of an infinite number of terms to determine the temperature at a specified location and time.
17 Approximate Analytical and Graphical Solutions Solution with one-term approximation The terms in the series solutions converge rapidly with increasing time, and for > 0.2, keeping the first term and neglecting all the remaining terms in the series results in an error under 2 percent.
18 19 20 (a) Midplane temperature Transient temperature and heat transfer charts (Heisler and Grober charts) for a plane wall of thickness 2L initially at a uniform temperature Ti subjected to convection from both sides to an environment at temperature Twith a convection coefficient of h.
21 (b) Temperature distribution 22 (c) Heat transfer 23 24 25 The dimensionless temperatures anywhere in a plane wall, cylinder, and sphere are related to the center temperature by The specified surface temperature corresponds to the case of convection to an environment at Twith with a convection coefficient h that is infinite.
26 The fraction of total heat transfer Q/Qmax up to a specified time t is determined using the Gröber charts.
27 • The Fourier number is a measure of heat conducted through a body relative to heat stored.
• A large value of the Fourier number indicates faster propagation of heat through a body.
Fourier number at time t can be viewed as the ratio of the rate of heat conducted to the rate of heat stored at that time.
The physical significance of the Fourier number 28 Example: 29 Example: 30 Example: 31 TRANSIENT HEAT CONDUCTION IN SEMI-INFINITE SOLIDS Schematic of a semi-infinite body.
Semi-infinite solid: An idealized body that has a single plane surface and extends to infinity in all directions.
The earth can be considered to be a semi-infinite medium in determining the variation of temperature near its surface.
A thick wall can be modeled as a semi-infinite medium if all we are interested in is the variation of temperature in the region near one of the surfaces, and the other surface is too far to have any impact on the region of interest during the time of observation.
For short periods of time, most bodies can be modeled as semi-infinite solids since heat does not have sufficient time to penetrate deep into the body.
32 Transient temperature distributions in a semi-infinite solid for three surfac conditions: constant surface temperature, constant surface heat flux, and surface convection 33 Transformation of variables in the derivatives of the heat conduction equation by the use of chain rule.
Analytical solution for the case of constant temperature Ts on the surface error function complementary error function 34 Analytical solutions for different boundary conditions on the surface 35 Error function is a standard mathematical function, just like the sine and cosine functions, whose value varies between 0 and 1.
36 37 38 39 Variation of temperature with position and time in a semi-infinite solid initially at temperature Ti subjected to convection to an environment at T∞with a convection heat transfer coefficient of h.
40 Contact of Two Semi-Infinite Solids When two large bodies A and B, initially at uniform temperatures TA,i and TB,i are brought into contact, they instantly achieve temperature equality at the contact surface.
If the two bodies are of the same material, the contact surface temperature is the arithmetic average, Ts = (TA,i+TB,i)/2.
If the bodies are of different materials, the surface temperature Ts will be different than the arithmetic average.
Contact of two semi-infinite solids of different initial temperatures.
The interface temperature of two bodies brought into contact is dominated by the body with the larger kcp.
EXAMPLE: When a person with a skin temperature of 35C touches an aluminum block and then a wood block both at 15C, the contact surface temperature will be 15.9C in the case of aluminum and 30C in the case of wood.
41 Example: 42 Example: L=0.3 m Wood Slab Ti = 25C Hot gases T= 550C x 0 43 Example: Soil Ti =10C Winds T=-10C 44 TRANSIENT HEAT CONDUCTION IN MULTIDIMENSIONAL SYSTEMS • Using a superposition approach called the product solution, the transient temperature charts and solutions can be used to construct solutions for the two-dimensional and three-dimensional transient heat conduction problems encountered in geometries such as a short cylinder, a long rectangular bar, a rectangular prism or a semi-infinite rectangular bar, provided that all surfaces of the solid are subjected to convection to the same fluid at temperature T, with the same heat transfer coefficient h, and the body involves no heat generation.
• The solution in such multidimensional geometries can be expressed as the product of the solutions for the one-dimensional geometries whose intersection is the multidimensional geometry.
The temperature in a short cylinder exposed to convection from all surfaces varies in both the radial and axial directions, and thus heat is transferred in both directions.
45 A short cylinder of radius ro and height a is the intersection of a long cylinder of radius ro and a plane wall of thickness a.
The solution for a multidimensional geometry is the product of the solutions of the one-dimensional geometries whose intersection is the multidimensional body.
The solution for the two-dimensional short cylinder of height a and radius ro is equal to the product of the nondimensionalized solutions for the one-dimensional plane wall of thickness a and the long cylinder of radius ro.
46 A long solid bar of rectangular profile a b is the intersection of two plane walls of thicknesses a and b.
47 The transient heat transfer for a two-dimensional geometry formed by the intersection of two one-dimensional geometries 1 and 2 is Transient heat transfer for a three-dimensional body formed by the intersection of three one-dimensional bodies 1, 2, and 3 is 48 Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a uniform temperature Ti and exposed to convection from all surfaces to a medium at T 49 Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a uniform temperature Ti and exposed to convection from all surfaces to a medium at T 50 Summary • Lumped System Analysis Criteria for Lumped System Analysis • Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects Nondimensionalized One-Dimensional Transient Conduction Problem Exact Solution of One-Dimensional Transient Conduction Problem Approximate Analytical and Graphical Solutions • Transient Heat Conduction in Semi-Infinite Solids Contact of Two Semi-Infinite Solids • Transient Heat Conduction in Multidimensional Systems NUMERICAL METHOD IN TRANSIENT HEAT CONDUCTION Mehmet Kanoglu Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008 |
4165 | https://sites.lsa.umich.edu/idolga/wp-content/uploads/sites/1334/2024/08/autodelPezzoEven.pdf | AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 IGOR DOLGACHEV AND GEBHARD MARTIN ABSTRACT. We classify the automorphism groups of del Pezzo surfaces of degrees one and two over an algebraically closed field of characteristic two. This finishes the classification of automorphism groups of del Pezzo surfaces in all characteristics.
CONTENTS Introduction 1 1.
Notation 3 2.
Del Pezzo surfaces of degree ≥3 4 3.
Del Pezzo surfaces of degree 2 4 4.
Automorphism groups of del Pezzo surfaces of degree 2 12 5.
Del Pezzo surfaces of degree 1 20 6.
Automorphism groups of del Pezzo surfaces of degree 1 28 Appendix 43 References 45 INTRODUCTION This paper is a continuation of our paper , where we finished the classification of the auto-morphism groups of del Pezzo surfaces over an algebraically closed field of positive characteristic p ̸= 2. In this paper, we treat the remaining case when the characteristic equals two.
As we explained in the introduction to , the remaining part of the classification concerns del Pezzo surfaces of degrees one and two. The cases of odd and even positive characteristic are drastically different since, in the latter case, the anti-canonical map (resp. the anti-bicanonical map) is a separable Artin-Schreier cover of degree two but not a Kummer cover as in the cases of odd characteristic. So, no plane quartic curves (and no canonical genus four curves with vanishing theta characteristic) appear as branch curves.
Instead, in characteristic two, the branch curve B of the anti-canonical (resp. anti-bicanonical) map is not necessarily smooth plane conic (resp. a cubic in P3). The ramification curve R is a purely inseparable cover of B. Theorems 3.4 and 5.6 give normal forms for del Pezzo surfaces of degree two and one depending on the singularities of R and B.
Although plane quartics and canonical curves of genus four disappear in characteristic two, their familiar attributes, like 28 bitangent lines or 120 tritangent planes, persist. We call them fake bitagents and fake tritangent planes. They are defined to be lines in the plane (resp. planes in the three-dimensional space) that split under the anti-canonical (resp. anti-bicanonical) map.
It is well-known that the blow-up of the anti-canonical base point on a del Pezzo surface of degree one yields a rational elliptic surface with only irreducible fibers and, conversely, the contraction 1 2 IGOR DOLGACHEV AND GEBHARD MARTIN of a section of a rational elliptic surface with only irreducible fibers yields a del Pezzo surface of degree 1. Thus, the normal forms of Theorem 5.6 also give normal forms for all rational elliptic surfaces with only irreducible fibers.
Quite surprisingly, in characteristic 2, also every del Pezzo surface of degree 2 has a canonically associated rational elliptic surface. This surface is obtained by blowing up the base points of the preimage of the pencil of lines through the strange point of the branch locus B. We study the properties of this strange fibration in Section 5.4.
Using these geometric observations, we classify the automorphism groups of all del Pezzo surfaces of degree 2 and 1 in characteristic 2. The following result is proved in Theorems 4.3 and 6.8.
Theorem. A finite group G is realized as the automorphism group Aut(X) of a del Pezzo surface X of degree 2 (resp. 1) over an algebraically closed field k of characteristic char(k) = 2 if and only if G is listed in Table 3 (resp. Table 4) in the Appendix.
Table 3 (resp. Table 4) also gives the conjugacy classes in W(E7) (resp. W(E8)) of all elements of Aut(X) for all del Pezzo surfaces X of degree 2 (resp. degree 1). We refer to for a general discussion of the history of the problem and its relationship to the classification of conjugacy classes of finite subgroups of the planar Cremona group. Also, the reader finds there some general facts about del Pezzo surfaces, e.g. the relationship with the Weyl groups of roots systems and some classification results from group theory.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 3 1. NOTATION We recall the notations for some finite groups which we will encounter in this article. Throughout this article, p is a prime number, and q is a power of p. Unless stated otherwise, k denotes an algebraically closed field of characteristic 2.
• Cn is the cyclic group of order n.
• Sn and An are the symmetric and alternating groups on n letters.
• Q8 is the quaternion group of order 8.
• D2n is the dihedral group of order 2n.
• nk = (Z/nZ)k. In particular, n = n1 = Z/nZ.
• p1+2n ± , the extra special group. For odd p the sign + (−) defines a group of exponent p (p2).
For p = 2, the sign distinguishes the type of the quadratic forms on 22n = F2n 2 defined by the extension.
• GLn(q) = GL(n, Fq).
• PGLn(q) = GLn(q)/F∗ q. Its order is N = q 1 2 n(n−1)(qn −1) · · · (q2 −1).
• SLn(q) = {g ∈GLn(q) : det(g) = 1}. This is a subgroup of GLn(q) of index (q −1).
• Ln(q) = PSLn(q) is the image of SLn(q) in PGLn(q). Its order is N/(q −1, n).
• For odd n, On(q) is the subgroup of GLn(q) that preserves a non-degenerate quadratic form F.
• For even n, O+ n (q) (resp.
O− n (q)) is the subgroup of GLn(q) that preserves a non-degenerate quadratic form F of Witt defect 0 (resp. 1).
• SO± n (q) is the subgroup of O± n (q) of elements with determinant 1.
• PSO± n (q) is the quotient of SO± n (q) by its center.
• Sp2n(q) is the subgroup of SLq(2n) preserving the standard symplectic form on F2n q . Its order is qn2(q2n−1 −1) · · · (q2 −1).
• Sp2n(q) = Sp2n(q)/(±1).
• SUn(q2) is the subgroup of SLn(q2) of matrices preserving the hermitian form Pn i=1 xq+1 i .
Its order is q 1 2 n(n−1)(qn −(−1)n)(qn−1 −(−1)n−1) · · · (q3 + 1)(q2 −1). We have SU2(q2) = SL2(q).
• PSUn(q2) = SUn(q2)/C, where C is a cyclic group of order (q + 1, n) of diagonal Hermitian matrices. The simple group PSUn(q2) is denoted by Un(q) in .
• H3(3) is the Heisenberg group of 3 × 3 upper triangular matrices with entries in F3.
• A.B is a group that contains a normal subgroup A with quotient group B.
• A : B is the semi-direct product A ⋊B.
4 IGOR DOLGACHEV AND GEBHARD MARTIN 2. DEL PEZZO SURFACES OF DEGREE ≥3 For the convenience of the reader, we first recall the classification of automorphism groups of del Pezzo surfaces of degree at least 3.
2.1. Degree ≥5. For del Pezzo surfaces of degree at least 5, the description of Aut(X) is characteristic-free. We refer the reader to [12, Section 3] or for details.
2.2. Quartic del Pezzo surfaces. Starting from degree 4, the classification of automorphism groups depends on the characteristic. As in the other characteristics, a quartic del Pezzo surface X is a blow-up of five points in P2 no 3 of which are collinear. Moreover, the anti-canonical linear system | −KX| = |OP2(3) −p1 −p2 −p3 −p4 −p5| embeds X into P4 as a complete intersection of two quadrics.
Since p = 2, these quadrics cannot be diagonalized. Instead, as shown in , one can choose the normal forms (ab + b + 1)t2 2 + at2 3 + t2t3 + t3t4 = bt2 1 + (ab + a + 1)t2 2 + t1t3 + t2t4 = 0, (1) where a, b are parameters such that the binary form ∆= uv(u + v)(u + av)(bu + v) has five distinct roots.
As in the case p ̸= 2, the automorphism group Aut(X) contains a normal subgroup H isomor-phic to 24, and the quotient G = Aut(X)/H is isomorphic to a subgroup of S5. The classification is summarized in Table 1 in the Appendix. There, the first column refers to the values of the parameters a and b in Equation (1) above. The conjugacy classes of elements of Aut(X) can be obtained by combining [11, Table 2] and [6, Table 5].
2.3. Cubic surfaces. The classification of automorphism groups of cubic surfaces in characteristic 2 was achieved in [11, Table 7]. For the convenience of the reader, we recall the classification in Table 2 in the Appendix.
3. DEL PEZZO SURFACES OF DEGREE 2 3.1. The anti-canonical map. We start by describing the geometry of del Pezzo surfaces of degree d = 2 over an algebraically closed field k of characteristic p = 2. We refer to for the basic facts from the theory of del Pezzo surfaces over fields of any characteristic. It is known that the anti-canonical linear system | −KX| has no base points and defines a finite morphism f : X →P2 of degree 2.
If p ̸= 2, the map f is automatically separable and its branch curve is a smooth plane quartic. So any automorphism of X induces an automorphism of the quartic, and, conversely, any automorphism of the quartic can be lifted to two automorphisms of X that differ by the deck transformation, classically called the Geiser involution.
If p = 2, the structure of f, being a morphism of degree 2, is more complicated. Nevertheless, as a first step, we observe that f is still always separable.
Proposition 3.1. The anti-canonical linear system | −KX| defines a finite separable morphism f : X →P2 of degree 2.
Proof. Assume that f is not separable. Then, since deg(f) = 2, f is purely inseparable. Hence, f is a homeomorphism in the étale topology, which is absurd since H2 ét(X, Zℓ) has rank 8 (because X is the blow-up of 7 points in the plane), while H2 ét(P2, Zℓ) has rank 1.
□ AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 5 Let R(X, −KX) = ∞ M n=0 H0(X, OX(−nKX)) be the graded anti-canonical ring of X. By the Riemann-Roch Theorem, dimk R(X, −KX)1 = 3 and dimk R(X, −KX)2 = 7. One can show that R(X, −KX) is generated by R(X, −KX)1 and one element from R(X, −KX)2 that does not belong to the symmetric square of R(X, −KX)1. Let x, y, z be elements of R(X, −KX)1 and w ∈R(X, −KX)2 which together generate R(X, −KX).
Then, the relation between the generators is of the form w2 + A(x, y, z)w + B(x, y, z) = 0, (2) where A and B are homogeneous forms of degree 2 and 4, respectively. In particular, via equation (7), we can view X as a surface of degree 4 in the weighted projective space P(1, 1, 1, 2), and the anti-canonical map is the projection of this surface onto the x, y, z-coordinates.
If p ̸= 2, we can complete the square, get rid of A, and obtain the standard equation of a del Pezzo surface of degree 2. The curve V (B(x, y, z)) is the smooth plane quartic we mentioned in the introduction. The Geiser involution just negates w.
In our case, when p = 2, we cannot get rid of A, for otherwise the map would become inseparable.
Also, the coefficient B is not uniquely determined, since replacing w with w + Q for any quadratic form Q changes B to B +AQ+Q2, without changing the isomorphism class of the surface. Taking Q = A, we obtain the analog of the Geiser involution, so we keep the name for this involution.
The non-uniqueness of B becomes more natural if we take the following different point of view: By [13, Proposition 1.11], the double cover f is a torsor under a group scheme αL,s of order 2 over P2, defined by the exact sequence of fppf-sheaves 0 →αL,s →L φ →L⊗2 →0 for some line bundle L and a global section s. The homomorphism of sheaves φ is locally given by a 7→a2 U + aUsU, so s cuts out the branch locus of f. By [13, Proposition 1.7], we have ωX ∼ = f∗(OP2(−3) ⊗L−1), hence L ∼ = OP2(2) and s = A. The αL,s-torsor corresponding to f is defined by a cohomology class in H1 fppf(P2, αL,s). Since H1 fppf(P2, L) = H1(P2, L) = 0, we have H1 fppf(P2, αL,s) ∼ = H0(P2, L⊗2)/℘(H0(P2, L)), where ℘= H0(φ). The ternary form B is a representative of this space, and hence it is defined only up to a transformation of the form B 7→B + Q2 + AQ, where Q is a quadratic form in x, y, z.
By writing the equation of X locally as w2 U + aUwU + bU, and taking partial derivatives, we see that the differentials wUdaU + dbU restricted to V (A) glue together to define a global section α of Ω1 P2 ⊗L⊗2 ⊗OV (A). This section vanishes if and only if X is singular. So, in our case, when X is assumed to be smooth, we obtain the following.
Proposition 3.2. In (7), the equations A = 0, wAx + Bx = 0, wAy + By = 0, wAz + Bz = 0 have no common solutions.
6 IGOR DOLGACHEV AND GEBHARD MARTIN 3.2. Normal forms. Recall that X is given by an equation of the form w2 + Aw + B = 0, where A is a quadratic ternary form and B is quartic ternary form. We say that V (A) is the branch curve of the cover, and its preimage R = f−1(V (A)) under the anti-canonical map f : X →P2 will be called the ramification curve.
Remark 3.3. We use the notation A2n for singularities of curves whose formal completion is isomorphic to the unibranched singularity y2 + x2n+1 = 0. If n = 1, this is an ordinary cusp singularity. These are exactly the curve singularities that can occur on reduced purely inseparable double covers of smooth curves in characteristic 2. Indeed, after passing to formal completions, such a double cover is given by an equation of the form y2 + uxm, where u ∈k is a unit. Now, we can apply a substitution of the form y 7→y + f for a suitable power series f to assume that m is odd and then replace x by λx, where λ is an m-th root of u−1, which exists by Hensel’s lemma.
In other words, the singularity defined by y2 + uxm is of type A2n, where 2n + 1 is the smallest odd power of x that occurs in uxm.
The following theorem gives normal forms for the cover f : X →P2. In total, we obtain six normal forms, corresponding to the six possible combinations of singularities of V (A) and R.
Theorem 3.4. Every del Pezzo surface of degree 2 in characteristic 2 is isomorphic to a quartic surface in P(1, 1, 1, 2) given by an equation of the form w2 + A(x, y, z)w + B(x, y, z), where (A, B) is one of the following: Name A B B1 B0 #Parameters (1)(a) x2 + yz xB1 + B0 λyz(y + z) ay4 + by3z + cy2z2 + dyz3 + ez4 6 (1)(b) x2 + yz xB1 + B0 y2z ay4 + by3z + cy2z2 + dyz3 + ez4 5 (1)(c) x2 + yz xB1 + B0 y3 by3z + cy2z2 + dyz3 + ez4 4 (2)(a) xy B1 + B2 0 xz3 + yz3 ax2 + by2 + cz2 + dxz + eyz 5 (2)(b) xy B1 + B2 0 xz3 + y3z ax2 + cz2 + dxz + eyz 4 (3) x2 xB1 + B0 z3 + ayz2 y3z + by2z2 + cz4 3 The parameters satisfy the following conditions: (1) (a) λ ̸= 0, λ2 + a + b + c + d + e ̸= 0, b2 + a ̸= 0, d2 + e ̸= 0.
(1) (b) b2 + a ̸= 0, d2 + e ̸= 0.
(1) (c) d2 + e ̸= 0.
(2) (a) a ̸= 0, b ̸= 0.
(2) (b) a ̸= 0.
(3) None.
In terms of these normal forms, the singularities of the irreducible components of Rred are as follows: (1) (a) Three A2-singularities, over [0 : 1 : 0], [0 : 0 : 1] and [1 : 1 : 1].
(1) (b) An A4-singularity over [0 : 0 : 1] and an A2-singularity over [0 : 1 : 0].
(1) (c) An A6-singularity over [0 : 0 : 1].
(2) (a) Two A2-singularities, over [1 : 0 : 0] and [0 : 1 : 0].
(2) (b) Two A2-singularities, over [1 : 0 : 0] and [0 : 0 : 1].
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 7 (3) An A2-singularity over [0 : 0 : 1].
Proof. Since f : X →P2 is separable, A is non-zero. Hence, up to projective equivalence, there are three possibilities for A, corresponding to (1), (2), and (3). Now, we study those cases separately. The conditions on the parameters will follow from Proposition 3.2 by computing partial derivatives, a task which we will leave to the reader.
(1) A = x2 + yz Applying a substitution of the form w 7→w + Q for a suitable quadratic form Q, allows us to assume that B = xB1 + B0 for homogeneous forms B0 and B1 in y and z.
Let x = uv, y = u2, z = v2 define the Veronese isomorphism between V (A) and P1. Substitut-ing in B, we get that R is isomorphic to the double cover of P1 given by the equation w2 + uvB1(u2, v2) + B0(u2, v2) = 0.
By taking the partials, we find that R is singular exactly over the roots of B1.
After applying a suitable substitution that preserves A, we can move these roots to special positions. Note that the substitution w 7→w + Q of the first paragraph does not change the position of these singularities, so we can still assume that B = xB1 + B0.
If the roots are distinct, we get Case (a), if there are two distinct roots, we get Case (b), and if there is only a single root, we get Case (c). Note that in Cases (b) and (c), the substitution y 7→λy, z 7→λ−1z preserves the location of the roots and scales B1, which is why we can assume that xB1 occurs with coefficient 1. Finally, in Case (c), we can apply a substitution of the form z 7→z + λ2y, x 7→x + λy for a suitable λ to assume that B0(1, 0) = 0.
(2) A = xy After applying a substitution of the form w 7→w + Q for a suitable quadratic form Q, we may assume that B does not contain monomials divisible by xy. This allows us to write B = (a1x3 + a2y3)z + (a3x + a4y)z3 + B0(x, y, z)2.
Note that the preimages R1 and R2 of V (x) and V (y) on X are members of | −KX|, hence they must be reduced.
Restricted to V (x), the equation becomes w2 + a2y3z + a4yz3, so R1 is singular over [0 : √a4 : √a2]. Similarly, R2 is singular over [√a3 : 0 : √a1]. Note that these points must be distinct, for otherwise X is singular over [0 : 0 : 1] by Proposition 3.2.
If these two points are distinct and different from [0 : 0 : 1], we can apply a suitable substitution that preserves A to move them to [0 : 1 : 0] and [1 : 0 : 0]. Then, we can repeat the substitution of the first paragraph and, after rescaling, arrive at Case (a).
If the two points are distinct and one of them is [0 : 0 : 1], we can assume without loss of generality that the other one lies on V (y) and move it to [1 : 0 : 0]. After repeating the substitution of the first paragraph and rescaling, we may assume that B1 is as in Case (b). Finally, after applying a substitution of the form z 7→z + λy, w 7→w + λz2 + λ2yz + λ3y2 for a suitable λ, we may assume that B0(0, 1, 0) = 0.
(3) A = x2 Applying a substitution of the form w 7→w + Q for a suitable quadratic form Q, allows us to assume that B = xB1 + B0 for homogeneous forms B0 and B1 in y and z.
8 IGOR DOLGACHEV AND GEBHARD MARTIN Let R′ be the preimage of V (x). As in Case (2), since R′ ∈| −KX|, R′ must be reduced.
Restricted to V (x), the double cover becomes w2 + B0(y, z) = 0, hence R′ is singular over the common zero of B0,y and B0,z. We can assume that this zero lies at [0 : 0 : 1], that is, that yz3 does not occur in B0 and y3z occurs with non-zero coefficient. After rescaling, we may assume that y3z occurs with coefficient 1.
Applying a substitution of the form z 7→z + λ1x + λ2y, y 7→y + λ3x for suitable λi and repeating the substitution of the first paragraph, we can eliminate the monomials y3 and y2z in B1 and the monomial y4 in B0. Computing partials, we see that X is singular if and only if B1(0, 1) = 0. Hence, after rescaling, we may assume that B is as claimed.
□ 3.3. Fake bitangents and odd theta characteristics. It is known that a del Pezzo surface X of degree 2 contains 56 (−1)-curves (see [9, 8.7], where the proof is characteristic free). They come in pairs Ei + E′ i ∈| −KX| with Ei · E′ i = 2. The Geiser involution γ switches the two curves in a pair. The image of each pair under any birational morphism π : X →P2 given by the blow-up of seven points p1, . . . , p7 ∈P2 is either the union of a line through two points pi, pj and the conic through the remaining 5 points, or a cubic passing through p1, . . . , p7 with a double point at some pi (and one curve of the pair is contracted by π). The image of Ei + E′ i under the anti-canonical map f is a line ℓ.
If p ̸= 2, each of the resulting 28 lines is a bitangent line to the branch quartic curve and, conversely, every bitangent to the branch quartic gives rise to a pair of (−1)-curves. A bitangent line intersects the branch curve at two points, not necessarily distinct, whose sum is an odd theta characteristic of the curve. It is known that the number of odd theta characteristics on a smooth curve of genus 3 is equal to 28.
For arbitrary p, we still have the following.
Lemma 3.5. The preimage f−1(ℓ) of a line ℓis a sum of two (−1)-curves if and only if f−1(ℓ) is reducible.
Proof. Since f has degree 2 and ℓis integral, the curve f−1(ℓ) is reducible if and only if it has two irreducible components L1 and L2. These components satisfy L1 + L2 ∈| −KX|. Since Li maps birationally to ℓ, we have pa(Li) = 0, hence L1.L2 = 2 by adjunction. We have L2 1 = L2 2, because the two curves are interchanged by the covering involution, so the equality 2 = K2 X = (L1 + L2)2 = L2 1 + L2 2 + 2L1.L2 implies that Li is a (−1)-curve. The converse is clear.
□ So, even if p = 2, we have 28 splitting lines, which we call fake bitangent lines in analogy with the situation in the other characteristics. For the rest of this section, we assume p = 2. Since the anti-canonical map is étale outside the branch curve V (A), the intersection Ei ∩E′ i lies on the ramification curve R. Let L = OR(Ei) ∼ = OR(E′ i). It is an invertible sheaf on R of degree 2. We have L⊗2 ∼ = OR(Ei + E′ i) ∼ = OR(−KX).
Since B ∈|OP2(2)|, we have R ∈| −2KX|. By the adjunction formula ωR ∼ = OR(−2KX + KX) ∼ = L⊗2.
Invertible sheaves L on R that satisfy this property are called invertible theta characteristics.
They are called even, odd, or vanishing according to whether their space of global sections is AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 9 even-dimensional, odd-dimensional, or at least two-dimensional, respectively. We note that, on singular curves, there can be theta characteristics which are not invertible, see , . In the following, we will only discuss invertible theta characteristics, so we drop the “invertible” from the notation.
Let Θ(R) be the set of isomorphism classes of theta characteristics on R and let J(R) be the identity component of the Picard scheme of R, also called the generalized Jacobian of R.
Lemma 3.6. The generalized Jacobian J(R) of R is isomorphic to G3 a.
Proof. Since R is of arithmetic genus 3, J(R) is a commutative group scheme of dimension 3.
As Rred has only unibranched singularities, [5, Proposition 5, Proposition 9] shows that J(R) is unipotent. Finally, we have a factorization of the absolute Frobenius F : R →V (A) →R. Note that J(V (A)) is trivial, even if V (A) is non-reduced, since H1(V (A), OV (A)) = 0. Since F ∗is multiplication by p on J(R), we obtain that J(R) is p-torsion, hence isomorphic to G3 a.
□ In particular, J(R)(k) is an infinite 2-torsion group and it acts on Θ(R) via tensor products. It is easy to check that Θ(R) is a torsor under J(R)(k) via this action. This already shows that the problem of finding (fake) bitangents using theta characteristics on R in characteristic 2 is much more subtle than it is in the other characteristics. Let us give an example that further illustrates this point.
Example 3.7. Assume that V (A) is a smooth conic.
Consider π : R →V (A) ∼ →P1. We have π∗OP1(2) = (f|R)∗OV (A)(1) = (ωX)|R, so L := π∗OP1(1) is a theta characteristic on R. Moreover, we have h0(R, π∗OP1(1)) = 2, so L is a vanishing theta characteristic. In fact, this is the unique vanishing theta characteristic on R: Indeed, let L′ be another vanishing theta characteristic. Then, the Riemann–Roch formula yields h0(R, L ⊗L′) −h0(R, ωR ⊗L−1 ⊗L′−1) = 2.
Since h0(R, L) ≥2 and h0(R, L′) ≥2, we have h0(R, L⊗L′) ≥3, so h0(R, ωR⊗L−1⊗L′−1) ̸= 0. Since R is integral and ωR ⊗L−1 ⊗L′−1 has degree 0, this implies that L ∼ = L′.
Next, let ℓbe any line in P2 such that f−1(ℓ) meets R in two distinct smooth points. Then, f−1(ℓ∩V (A))red defines an effective theta characteristic L on R. By the previous paragraph, we have h0(R, L) = 1, hence all the infinitely many theta characteristics arising in this way are odd. It would be interesting to find an abstract characterization of the fake bitangent lines among the odd theta characteristics of R.
Nevertheless, we can find explicit equations of fake bitangent lines using the following result.
Lemma 3.8. Let C →P1 be an Artin–Schreier double cover given by an equation of the form w2 + f(u, v)w + g(u, v) = 0, where f and g are homogeneous polynomials of degree n and 2n, respectively, and f ̸= 0.
Then, C is reducible if and only if there exists a homogeneous polynomial h of degree n with g(u, v) = f(u, v)h(u, v) + h(u, v)2.
Proof. If there exists an h as in the assertion, then w2 + fw + g = (w + f + h)(w + h), so C is obviously reducible.
Conversely, assume that C is reducible. Then, C has exactly two irreducible components and these components are interchanged by the substitution w 7→w + f. In other words, we can write w2 + fw + g = h′(h′ + f), where h′ is a weighted homogeneous polynomial of degree n. This is 10 IGOR DOLGACHEV AND GEBHARD MARTIN only possible if h′ is of the form h′ = w + h for some h homogeneous of degree n in the variables u and v. Then, w2 + fw + g = (w + h)(w + h + f) = w2 + fw + h2 + fh, hence g = fh + h2, as claimed.
□ Finally, for later use, we record some simple restrictions on the possible positions of fake bitangent lines with respect to the singularities of R.
Proposition 3.9. Let ℓbe a fake bitangent line that passes through the image P of a singular point of an irreducible component of Rred. Then, V (A) is smooth and ℓis tangent to V (A) at P.
Proof. Write f−1(ℓ) = L1 + L2. Since Rred is singular at f−1(P), Li and R have intersection multiplicity at least 2 in f−1(P). Since R ∈| −2KX| and L1 + L2 ∈| −KX|, we have (L1 + L2).R = 2K2 X = 4, hence L1 + L2 and R meet only in f−1(P). Therefore, their images in P2 meet only in P. If V (A) is smooth, this implies that ℓis tangent to V (A) in P. If V (A) is the union of two lines, this implies that ℓpasses through their intersection. However, in this case, Li and R have intersection multiplicity at least 3 in f−1(P), which is absurd. Finally, if V (A) is a double line, then Rred ∈| −KX| and 2 = K2 X = (L1 + L2).Rred ≥4, a contradiction.
□ Remark 3.10. We note that there are del Pezzo surfaces for which fake bitangents satisfying the properties of Proposition 3.9 exist. See Proposition 5.10 for a classification in terms of the normal forms of Theorem 3.4.
3.4. Strange elliptic fibrations. To each del Pezzo surface X of degree 2 in characteristic 2 with branch locus V (A) of the anti-canonical map f : X →P2, there is a naturally associated point PX in P2: If V (A) is smooth, we let PX be the strange point of V (A), if V (A) is the union of two lines, we let PX be their intersection, and if V (A) is a double line, we let PX be the image of the singular point of f−1(V (A))red. We call PX the strange point of X and note that the action of Aut(X) fixes PX.
The pencil P of lines through PX is Aut(X)-invariant as well. Its preimage C in X is an Aut(X)-invariant pencil of curves of arithmetic genus 1 with two base points if V (A) is smooth and with one base point of multiplicity 2 if V (A) is singular. We let π : Y →X be the blow-up of the two (possibly infinitely near) base points of C. Then, C defines a relatively minimal genus one fibration φ : Y →P1. Since the map X →P2 is separable and a general line in the pencil is not contained in V (A), its pre-image on Y is a smooth elliptic curve. Thus, the genus one fibration is an elliptic fibration. We call it the strange elliptic fibration associated to X.
By construction, the group Aut(X) lifts to a subgroup of Aut(Y ) and we will use this in Proposition 4.2 to find restrictions on the possible structure of Aut(X). To make the most of this connection, we will now describe the singular fibers of the elliptic fibration φ : Y →P1. We employ Kodaira’s notation: we say that a fiber isomorphic to an irreducible cuspidal cubic curve is of type II, a fiber that consists of two smooth rational curves intersecting non-transversally at one point is of type III, and a fiber that consists of three smooth rational curves intersecting at one point is of type IV.
We use the normal forms of Theorem 3.4, so that A = x2 + yz, xy, or x2 and PX = [1 : 0 : 0] in the first case and PX = [0 : 0 : 1] in the other two cases. In the first case, we let ℓ[t0:t1] be the line V (t0y + t1z) and in the other two cases, we let ℓ[t0:t1] be the line V (t0x + t1y). The fiber of φ corresponding to ℓ[t0:t1] is denoted by F[t0:t1].
Proposition 3.11. The generic fiber of the strange elliptic fibration associated to X is a super-singular elliptic curve. Its singular fibers are of type II, III, or IV and its Mordell-Weil group is torsion-free. More precisely: AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 11 (1) If A = x2 + yz, then the following hold: • The fiber F[t0:t1] is smooth if and only if t0y + t1z ∤B1.
• The fiber F[t0:t1] is of type III if ℓ[t0:t1] is a fake bitangent and of type II otherwise.
• The line ℓ[1:0] is a fake bitangent if and only if e = 0.
• The line ℓ[0:1] is a fake bitangent if and only if a = 0.
• The line ℓ[1:1] is a fake bitangent if and only if a + b + c + d + e = 0.
(2) (a) If A = xy and B1 = xz3 + yz3, then the following hold: • The fiber F[t0:t1] is smooth if and only if [t0 : t1] ̸= [1 : 0], [0 : 1], [1 : 1].
• F[1:0] and F[0:1] are of type II.
• F[1:1] is of type IV if ℓ[1:1] is a fake bitangent and of type III otherwise.
• The curve ℓ[1:1] is a fake bitangent if and only if c = d2 + e2.
(2) (b) If A = xy and B1 = xz3 + y3z, then the following hold: • The fiber F[t0:t1] is smooth if and only if [t0 : t1] ̸= [1 : 0], [0 : 1].
• The curve F[0:1] is of type II.
• The curve F[1:0] is of type III.
(3) If A = x2, then the following hold: • The fiber F[t0:t1] is smooth if and only if [t0 : t1] ̸= [1 : 0].
• The curve F[1:0] is of type III.
Proof. We study each case separately.
(1) In this case, A = x2 + yz.
First, consider ℓ[1:t] = V (y + tz). Plugging y = tz into the equation of X, we obtain w2 + (x2 + tz2)w + xB1(tz, z) + (at4 + bt3 + ct2 + dt + e)z4 with B1(tz, z) ∈{λt(t + 1)z3, t2z3, t3z3}. If y + tz ∤B1, then B1(tz, z) ̸= 0, so taking partials with respect to x and w shows that a singular point must satisfy x = z = 0, which is absurd. If y + tz | B1, then B1(tz, z) = 0 and F[1:t] is singular over [t : t : 1]. Similarly, one checks that F[0:1] is singular.
The equation w2 + (x2 + tz2)w + xB1(tz, z) + (at4 + bt3 + ct2 + d + e)z4 shows that F[1:t] is a double cover of P1 branched over a single point. Hence, if F[1:t] is smooth, then it is supersingular, and if it is singular and irreducible, it is a cuspidal cubic.
Finally, consider the curve F[1:0] given by w2 + x2w + ez4.
By Lemma 3.8, it is clear that F[1:0] is reducible if and only if e = 0. The calculation for F[1:1] and F[0:1] is similar.
(2) (a) In this case, A = xy and B1 = xz3 + yz3.
First, consider ℓ[1:t] = V (x + ty) with t ̸= 0, 1. Plugging x = ty into the equation of X, we obtain w2 + ty2w + (t + 1)yz3 + B0(ty, y, z)2.
Then, taking partials shows that F[1:t] is smooth. Since it is a double cover of P1 branched over a single point, it is supersingular.
Next, consider F[1:1], whose image in X is given by w2 + y2w + ((a + b)y2 + (d + e)yz + cz2)2.
12 IGOR DOLGACHEV AND GEBHARD MARTIN This curve is singular over [0 : 0 : 1], so F[1:1] has one irreducible component contracted by Y →X. By Lemma 3.8, the image of F[1:1] in X is reducible if and only if c = d2 + e2.
Finally, the curves F[1:0] and F[0:1] are isomorphic to their images in X and these images are irreducible and cuspidal by Theorem 3.4.
(2) (b) In this case, A = xy and B1 = xz3 + y3z.
First, consider ℓ[1:t] = V (x + ty) with t ̸= 0. Plugging x = ty into the equation of X, we obtain w2 + ty2w + tyz3 + y3z + B0(ty, y, z)2.
As in the previous cases, taking partials shows that F[1:t] is smooth and supersingular.
The curve F[0:1] is isomorphic to its image in X, since f−1(ℓ[0:1]) is smooth over the point [0 : 0 : 1]. Hence, F[0:1] is of type II. On the other hand, the curve F[1:0] is of type III, since its image in X has multiplicity 2 over [0 : 0 : 1].
(3) In this case A = x2.
First, consider ℓ[t:1] = V (tx + y). Plugging y = tx into the equation of X, we obtain w2 + x2w + xz3 + (bt2 + at)x2z2 + cz4.
Then, taking partials shows that F[t:1] is smooth. Since it is a double cover of P1 branched over a single point, it is supersingular.
The curve F[1:0] is of type III, by the same argument as in the previous case.
The Mordell-Weil group of the fibration is torsion free by [17, Main Theorem], since the lattice spanned by fiber components is of rank at most 4 in each case.
□ Remark 3.12. The classification of singular fibers of rational elliptic surfaces with a section in characteristic 2 can be found in . Lang shows that in the cases where the general fiber is a supersingular elliptic curve, the number of singular fibers is at most 3, which agrees with what we observed in the case of strange elliptic fibrations. Proposition 3.11 shows that the singular fibers that occur on strange genus one fibrations are of type 9A, 9B, 10A, 10B, 10C or 11 in Lang’s terminology.
4. AUTOMORPHISM GROUPS OF DEL PEZZO SURFACES OF DEGREE 2 4.1. Preliminaries. Recall once more from Section 3.1 that a del Pezzo surface X of degree 2 is a surface of degree 4 in P(1, 1, 1, 2) given by an equation of the form w2 + A(x, y, z)w + B(x, y, z) = 0.
Since this is the anti-canonical model of X and ω−n X admits a natural Aut(X)-linearization for all n, we obtain that Aut(X) is isomorphic to the subgroup of Aut(P(1, 1, 1, 2)) of automorphisms that preserve X.
The structure of the group Aut(P(1, 1, 1, 2)) is well-known. The vector space k[x, y, z]2 of quadratic forms is a normal subgroup of Aut(P(1, 1, 1, 2)) that acts via (x, y, z, w) 7→(x, y, z, w + Q). The quotient by this subgroup is the group of transformations that change (x, y, z) linearly and multiply w by a scalar. Since the transformation (x, y, z, w) 7→(λx, λy, λz, λ2w) is the identity, this quotient is isomorphic to GL3(k)/µ2(k). Since we are in characteristic 2, the subgroup µ2(k) is trivial. This gives an isomorphism Aut(P(1, 1, 1, 2)) ∼ = k[x, y, z]2 : GL3(k).
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 13 We denote elements of this group by (Q, g) ∈k[x, y, z]2 × GL3(k) where the semi-direct product structure is (Q, g) ◦(Q′, g′) = (g∗(Q′) + Q, gg′).
Using this description of Aut(P(1, 1, 1, 2)), it is straightforward to calculate the subgroup of automorphisms preserving X. We obtain Aut(X) ∼ = {(Q, g) : g∗(A) = A, g∗(B) = B + AQ + Q2}.
The kernel of the homomorphism Aut(X) →GL3(k), (Q, g) 7→g is generated by the Geiser involution γ. We let G(X) be the image of Aut(X) in GL3(k).
Lemma 4.1. The homomorphism G(X) →GL3(k) →PGL3(k) is injective.
Proof. Let g ∈G(X) be in the kernel of this homomorphism. Then, g = λI3 for some λ ∈k×.
On the other hand, by definition of G(X), we have g∗(A) = A. Since A has degree 2, this implies λ2 = 1. Hence, λ = 1.
□ We recall from [12, §1] that a choice of a blow-up X →P2 of seven points defines an injective homomorphism ρ : Aut(X) →W(E7).
(3) The image of the Geiser involution is equal to −idE7. It is known that W(E7) = ⟨−idE7⟩×W(E7)+, where W(E7)+ ⊆W(E7) is the kernel of the determinant map.
In particular, to determine Aut(X), it suffices to determine G(X) and both groups are isomor-phic to subgroups of W(E7) via ρ. This puts severe restrictions on the possible structure of G(X).
Finally, we can use the strange genus one fibrations of the previous section to get information on G(X).
Proposition 4.2. Let φ : Y →P1 be the strange elliptic fibration associated to X. Choose an exceptional curve E of Y →X as the zero section of φ and let C be the second exceptional curve.
Then, there is a homomorphism ϕ : Aut(X) →Aut(Y ) that satisfies the following properties: (1) ϕ is injective.
(2) ϕ(γ) preserves every fiber of φ.
(3) If V (A) is smooth, then C is a section of φ. We have ϕ(γ) = tC ◦ι, where ι is the negation automorphism and tC is translation by C.
(4) If V (A) is singular, then C is a component of a reducible fiber of φ. We have ϕ(γ) = ι and ϕ factors through the stabilizer of the pair (E, C).
Proof. The surface Y is obtained by blowing up X at two points that are uniquely determined by V (A), hence stable under the action of Aut(X). This shows existence and injectivity of the homomorphism ϕ. The fibration φ is induced by the pencil of lines in P2 through the strange point of X. Since γ preserves these lines, it preserves the fibers of φ.
If V (A) is smooth, then E and C are interchanged by ϕ(γ). The automorphism t−C ◦ϕ(γ) ◦ι maps E to E and −C to −C. It is well-known that every fixed point of a non-trivial automorphism of an elliptic curve is a torsion point. On the other hand, by Proposition 3.11, φ has no torsion sections, so t−C ◦ϕ(γ) ◦ι = id, which yields Claim (3).
If V (A) is singular, then C is a (−2)-curve which meets E, hence it is the identity component of a reducible fiber of φ. Since ϕ(γ) is an involution that preserves E, we have ϕ(γ) = ι and we obtain Claim (4).
□ 14 IGOR DOLGACHEV AND GEBHARD MARTIN 4.2. Classification.
Theorem 4.3. Every del Pezzo surface of degree 2 in characteristic two such that, in the decompo-sition Aut(X) ∼ = 2 × G(X), the group G(X) is non-trivial is isomorphic to a surface of degree 4 in P(1, 1, 1, 2) given by an equation of the form w2 + Aw + B, where (A, B, G(X)) is one of the following: Name A B B1 B0 G(X) #Parameters (1)(a)(i) x2 + yz xB1 + B0 λyz(y + z) ay4 + by3z + cy2z2 + byz3 + az4 2 4 (1)(a)(ii) x2 + yz xB1 + B0 λyz(y + z) ay4 + λy3z + ay2z2 + λyz3 + az4 S3 2 (1)(c)(i) x2 + yz xB1 + B0 y3 by3z + cy2z2 + ez4 23 3 (2)(a)(i) xy B1 + B2 0 xz3 + yz3 ax2 + ay2 + cz2 + dxz + dyz 2 3 (2)(a)(ii) xy B1 + B2 0 xz3 + yz3 ax2 + by2 3 2 (2)(a)(iii) xy B1 + B2 0 xz3 + yz3 ax2 + ay2 6 1 (3)(i) x2 xB1 + B0 z3 y3z + cz4 3 1 (3)(ii) x2 xB1 + B0 z3 y3z 9 0 The parameters satisfy the following conditions: (1) (a) (i) λ ̸= 0, λ2 + c ̸= 0, b2 + a ̸= 0, (b, c) ̸= (λ, a).
(1) (a) (ii) λ ̸= 0, λ2 + a ̸= 0.
(1) (c) (i) e ̸= 0.
(2) (a) (i) a ̸= 0, (c, d) ̸= (0, 0).
(2) (a) (ii) a ̸= 0, b ̸= 0, a ̸= b.
(2) (a) (iii) a ̸= 0.
(3) (i) c ̸= 0.
(3) (ii) None.
Proof. We use the normal forms of Theorem 3.4 and the description of Aut(X) and G(X) given in the beginning of the current section. We go through the cases of Theorem 4.3.
(1) (a) Here, X is given by an equation of the form w2 + (x2 + yz)w + λxyz(y + z) + B0 with B0 = ay4 + by3z + cy2z2 + dyz3 + ez4 and the cusps lie over [0 : 1 : 0], [0 : 0 : 1], and [1 : 1 : 1]. Let (Q, g) ∈Aut(X) be an automorphism of X. Then, g preserves the three points lying under the cusps. Moreover, if g fixes the three cusps, then it fixes V (A) pointwise, hence g is trivial in PGL3(k), so, by Lemma 4.1, g is the identity and (Q, g) coincides with the Geiser involution. Hence, G(X) acts faithfully on {[0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1]}.
Note that G(X) contains an involution if and only if X admits an equation where this involution is given by y ↔z. This involution is in G(X) if and only if there exists a quadratic form Q such that Q2 + (x2 + yz)Q = g∗B0 + B0.
(4) Since Q2 + (x2 + yz)Q contains a non-zero monomial divisible by x2 as soon as it is non-zero and g∗B0 + B0 does not contain such a monomial, we must have Q ∈{0, x2 + yz} and Equation (4) holds if and only if a = e and b = d, as claimed.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 15 Next, note that G(X) contains an automorphism g of order 3 if and only if g is given by x 7→x + z, y 7→z, z 7→y + z and there exists a quadratic form Q such that Q2 + (x2 + yz)Q = λyz2(y + z) + g∗B0 + B0.
(5) By the same argument as in the previous paragraph, we have Q ∈{0, x2 + yz} and Equation (5) holds if and only if a = c = e and b = d = λ. In particular, note that these conditions imply the conditions of the previous paragraph in this case, hence G(X) = S3.
(1) (b) In this case, V (A) is smooth and R has two non-isomorphic singularities. Then, g ∈G(X) must fix the images of them on V (A). Since an automorphism of order 2 of P1 has only one fixed point, we may assume that the order of g is odd. By Proposition 3.9, the line ℓ through the images of the singularities is not a fake bitangent. Its preimage E in X is an integral curve of arithmetic genus one and the Geiser involution has two fixed points on E. Hence, either E is smooth and ordinary, or nodal. In both cases, there is no non-trivial automorphism of odd order that commutes with the involution, hence g fixes ℓpointwise.
Since g also fixes the strange point P on V (A) and the projection from P is inseparable, g fixes V (A) pointwise, hence g is the identity. We conclude that G(X) = {1}.
(1) (c) Here, X is given by an equation of the form w2 + (x2 + yz)w + xy3 + B0 with B0 = by3z + cy2z2 + dyz3 + ez4.
The singularity of R lies over [0 : 0 : 1]. An element g ∈G(X) of odd order has at least two fixed points on V (A) and then the same argument as in the previous case shows that g is the identity. Therefore, G(X) is a 2-group that acts on V (A) ∼ = P1 with a fixed point. In particular, G(X) is isomorphic to a subgroup of Ga(k), hence isomorphic to 2n for some n ≥0.
We may assume that g acts as x 7→x + αy, y 7→y, z 7→z + α2y. Then g lifts to Aut(X) if and only if there exists a quadratic form Q such that (x2 + yz)Q + Q2 = αy4 + g∗(B0) + B0.
Since the right-hand side does not contain a monomial divisible by x2, we get, as in the previous cases, Q = x2 + yz or Q = 0. Comparing coefficients yields the system of equations dα2 = 0 dα4 = 0 eα8 + dα6 + cα4 + bα2 + α = 0 So, if d ̸= 0, then α = 0 and G(X) is trivial. If d = 0, there are 8 possibilities for α, one for each root of ex8 + cx4 + bx2 + x. All the roots are distinct since the derivative of this polynomial is 1. Here, we also use that e ̸= 0 by Theorem 3.4. Thus G(X) ∼ = 23.
(2) (a) Here, X is given by an equation of the form w2 + xyw + xz3 + yz3 + B2 0 with B0 = ax2 + by2 + cz2 + dxz + eyz.
16 IGOR DOLGACHEV AND GEBHARD MARTIN The singularities of the irreducible components of R lie over [1 : 0 : 0] and [0 : 1 : 0]. Let (Q, g) ∈Aut(X). Then, g preserves these two points and the intersection of V (x) and V (y). Moreover, by Proposition 4.2 and Proposition 3.11, g preserves the line V (x + y).
Assume that g has odd order. Then, g preserves the three lines V (x), V (y), and V (x+y), hence it is of the form (x, y, z) 7→(x, y, αz). The quadratic form Q satisfies Q2 + xyQ = g∗(B1 + B2 0) + B1 + B2 0.
The right-hand side does not contain monomials divisible by xy, hence Q ∈{0, xy}. Now, g∗B1 + B1 = 0 implies that α3 = 1, and if α ̸= 1, then g∗B2 0 + B2 0 = 0 holds if and only if c = d = e = 0.
Assume that g has order a power of 2. If g does not swap the points [1 : 0 : 0] and [0 : 1 : 0], then it acts diagonally, hence it is the identity. Therefore, we may assume that g swaps these two points and g2 = id. Hence, g acts as x ↔y. The quadratic form Q satisfies Q2 + xyQ = g∗(B1 + B2 0) + B1 + B2 0, hence a = b and d = e.
(2) (b) Here, X is given by an equation of the form w2 + xyw + xz3 + y3z + B2 0 with B0 = ax2 + cz2 + dxz + eyz and the singularities of the irreducible components of R map to [1 : 0 : 0] and [0 : 0 : 1].
Let (Q, g) ∈Aut(X).
If g has odd order, then there is a g-invariant line ℓthrough [1 : 0 : 0] and we may assume that ℓ̸⊆V (A). By the same argument as in Case (1) (b), ℓis fixed pointwise.
Then, every line through [0 : 0 : 1] is g-invariant. Since g∗A = A, this means that g acts as (x, y, z) 7→(x, y, αz). An automorphism of this form satisfies g∗B1 = B1 if and only if α = 1, so g is trivial.
If g has order a power of 2, then by Proposition 3.11, ϕ((Q, g)) ∈Aut(Y ) preserves the two singular fibers of φ : Y →P1, hence ϕ((Q, g)) acts trivially on the base of φ. The 2-Sylow subgroup of automorphisms of the geometric generic fiber of φ is the quaternion group Q8 and ϕ((Q, g)) commutes with the unique involution ϕ(γ) in Q8. This implies that (Q, g) ∈⟨γ⟩, so g is trivial.
(3) If V (A) is a double line, then X is given by an equation of the form w2 + x2w + xB1 + B0 with B1 = z3 + ayz2 and B0 = y3z + by2z2 + cz4. The singularity of Rred lies over [0 : 0 : 1]. Let (Q, g) ∈Aut(X). Then, g is of the form (x, y, z) 7→(x, αx + βy, γx + δy + ϵz) with β, ϵ ̸= 0 and Q satisfies the equation Q2 + x2Q = x(g∗B1 + B1) + g∗B0 + B0.
(6) The monomials y3z, xz3, xyz2, xy2z and xy3 do not appear on the left-hand side, hence their coefficients on the right-hand side must be zero. This yields the following conditions: ϵ = β−3 AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 17 β9 = 1 δ = a(β + β6) α = a2(1 + β) γ = a3(1 + β2) So, the order of g is equal to the order of β in k×, hence it is equal to 1, 3 or 9. Now, we calculate that if β3 = 1, then g3 acts as (x, y, z) 7→(x, y, a3(β + β2)x + z).
Hence, if a ̸= 0, then g is the identity.
So, assume that a = 0, so that, in particular, α = δ = γ = 0. Equation (6) becomes Q2 + x2Q = (ϵ3 + 1)xz3 + (β3ϵ + 1)y3z + b(β2ϵ2 + 1)y2z2 + c(ϵ4 + 1)z4.
On the left-hand side, the coefficients of z4 and y2z2 are the squares of the coefficients of x2z2 and x2yz, respectively. Since the latter monomials do not appear on the right-hand side, the coefficients of the former monomials must vanish. Therefore, we get the four conditions: ϵ3 + 1 = 0 β3ϵ + 1 = 0 b(β2ϵ2 + 1) = 0 c(ϵ4 + 1) = 0 Hence, if b ̸= 0, then β = ϵ = 1, so G(X) is trivial. If b = 0 and c ̸= 0, then ϵ = 1 and β3 = 1, and so G(X) ∼ = C3. If b = c = 0, then ϵ = β−3 and β9 = 1, hence G(X) ∼ = C9.
□ Remark 4.4. With our choice of normal form in Theorem 4.3, the map g 7→(0, g) defines an explicit section of the surjection Aut(X) →G(X) in every case.
Remark 4.5. The group 24 that appears in Theorem 4.3 occurs as a group of automorphisms of a del Pezzo surface of degree 4 in all characteristics . In characteristic 0, there is a unique conjugacy class of subgroups isomorphic to 24 in the Cremona group. One can prove, using the theory of birational links, that in characteristic 2, the two subgroups of Crk(2) are not conjugate.
Remark 4.6. The fact that 2 and 3 are the only primes that divide the order of Aut(X) can be proven without the classification. It is known that 2, 3, 5, and 7 are the only primes that divide the order of W(E7). To exclude the primes 5 and 7, one can use the Lefschetz fixed-point formula and the known traces of elements of W(E7) acting on the the root lattice of type E7 to get a contradiction with the possible structure of the set of fixed points of an element of the group G(X).
4.3. Conjugacy classes and comparison with the classification in characteristic 0. In this section, we determine the conjugacy classes in W(E7) of the elements of the groups that occur in Theorem 4.3 and, whenever possible, compare the surfaces in Theorem 4.3 with their counterparts in characteristic 0 (see [9, Table 8.9]). To do this, we use the following result.
Lemma 4.7. Let X be a del Pezzo surface of degree 2 in characteristic 2. Let X′ be a geometric generic fiber of a lift of X to characteristic 0 and let sp : Aut(X′) →Aut(X) be the specialization map. Then, sp is injective and preserves conjugacy classes.
18 IGOR DOLGACHEV AND GEBHARD MARTIN Proof. Let X →S be a lift of X with geometric generic fiber X′. The map sp sends an automor-phism g ∈Aut(X′) to the special fiber of the closure of g considered as a point of the relative automorphism scheme AutX/S. To see that this is well-defined, we have to explain why AutX/S is proper over S.
By passing to the anti-canonical model of X in PS(1, 1, 1, 2), the scheme AutX/S is identified with the stabilizer of X under the action of AutPS(1,1,1,2) on the space HdP2,S of smooth quartic hypersurfaces in PS(1, 1, 1, 2). To check that this stabilizer is proper, it suffices to show that the shear map (G3 a : GL3) × HdP2 → HdP2 × HdP2 (Q, g, w2 + Aw + B) 7→ (w2 + Aw + B, w2 + (g∗A + 2Q)w + g∗B + g∗AQ + Q2) is a proper morphism of schemes over Spec Z. For this, it suffices to check that the individual shear maps for the G3 a-action and the GL3-action are proper. We check this using the valuative criterion.
So, let R be an arbitrary discrete valuation ring with field of fractions K and let F = w2 +Aw +B and F ′ = w2 +A′w+B′ be equations of smooth quartics in PR(1, 1, 1, 2) with A, A′ ∈R[x, y, z]2 and B, B′ ∈R[x, y, z]4.
Given Q ∈K[x, y, z]2 sending F to F ′, we have the two conditions A + 2Q −A′ = 0 B + AQ + Q2 −B′ = 0.
Comparing the valuations of the coefficients of Q2 and AQ + B −B′ in the second equation shows that Q ∈R[x, y, z]2. This proves the valuative criterion of properness for the shear map of the G3 a-action on HdP2.
Given g ∈GL3(K) sending F to F ′, we have the two conditions g∗A = A′ g∗B = B′.
Replacing g by its Smith normal form, we may assume that g acts as x 7→πexx y 7→πeyy z 7→πezz, where π is a uniformizer of R. Thus, if a monomial xiyjzk appears in A or B with unit coefficient, then iex + jey + kez = 0. We leave it to the reader to check that, because of the smoothness of F and F ′ modulo π, there are enough such monomials to check that ex = ey = ez, that is, that g ∈GL3(R). This is the valuative criterion of properness for the shear map of the GL3-action on HdP2.
Since H0(X, TX) = 0, the scheme AutX/S is discrete, hence the specialization map is injective.
As H1(X, OX) = H2(X, OX), the relative Picard scheme PicX/S is constant over S. Now, specialization of line bundles is sp-equivariant and compatible with the intersection pairing, hence sp preserves conjugacy classes.
□ Remark 4.8. It would be interesting to determine all integers a1, . . . , an and d such that the action of AutP(a1,...,an) on the space Hd of smooth hypersurfaces of degree d in P(a1, . . . , an) is proper.
This would be a generalization of [14, Proposition 11.8.2] to the weighted case.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 19 By Theorem 4.3, we have | Aut(X)| ≤18, so Types I,. . .,V and of [9, Table 8.9] do not have a reduction modulo 2 which is a del Pezzo surface. Similarly, Type VII of [9, Table 8.9] has no analogue in characteristic 2.
The surface of Type (3)(ii) of Theorem 4.3 is a reduction modulo 2 of the surface of Type VI in [9, Table 8.9]. Hence, we call this surface Type VI. Since the conjugacy classes of elements of Aut(X) are the same as the ones of the lift, the entry for Type VI in Table 3 is the same as the one in [12, Table 7].
The equations of the surfaces of Type (2)(a)(iii) of Theorem 4.3 define smooth surfaces in characteristic 0 and the automorphisms x ↔y and z 7→ζ3z make sense in characteristic 0. Hence, these surfaces lift to characteristic 0 as del Pezzo surfaces with an action of 2 × 6. As explained above, del Pezzo surfaces of degree 2 with an automorphism group of order bigger than 18 do not have a smooth reduction modulo 2, hence these lifts are of Type VIII [9, Table 8.9], so we also call the surfaces of Type (2)(a)(iii) Type VIII. As in the previous case, the conjugacy classes are the same as in [12, Table 7].
As for the surfaces of Type (1)(a)(ii), we rewrite their equations using the substitution x 7→ x + y + z as w2 + (x2 + y2 + z2 −yz)w + λxyz(z −y) + a(y2 + z2 −yz)2.
This equation defines a lift of X to characteristic 0 and the Aut(X)-action lifts as well, since it is generated by the Geiser involution γ : w 7→−w, the involution y ↔z and the automorphism g of order 3 given by x 7→ −x y 7→ z z 7→ z −y.
Hence, all surfaces of Type (1)(a)(ii) are reductions modulo 2 of surfaces of Type IX in [9, Table 8.9]. In particular, we can read off the conjugacy classes of elements of Aut(X) from [12, Table 7].
The surfaces of Type (1)(c)(i) are the characteristic two analogs of Type X from [9, Table 8.9].
We claim that every involution on a surface X of type (1)(c)(i) which is different from the Geiser involution is of conjugacy class 3A1/4A1. It suffices to check this for the surface given by w2 + (x2 + yz)w + xy3 + z4, where G(X) acts as gα : x 7→x + αy, z 7→α2x + z with α8 = α. After using the substitution z 7→αx + y + z, y 7→α6x + α6y, the equation of X becomes w2 + (x2 + xy + y2 + α6(y + x)z + α4(x2 + y2))w + α4(x3y + x2y2 + xy3) + α3(x4 + y4 + z4) and the involution gα acts as x ↔y. Then, the above equation makes sense in characteristic 0 and defines a lift of X together with the involution gα. In particular, by [12, Table 7], the conjugacy class of gα is 3A1 or 4A1.
The equations of Types (2)(a)(ii) and (3)(i) make sense in characteristic 0, where they define a lift of the surface together with the C3-action. These lifts must be of Type XI from [9, Table 8.9].
Similarly, the equations of Types (1)(a)(i) and (2)(a)(i) define lifts to characteristic 0 together with the C2-action. Hence, these lifts are of Type XII from [9, Table 8.9].
We summarize the classification of automorphism groups of del Pezzo surfaces of degree 2 in Table 3 in the Appendix. There, in the first column, we give the name of the corresponding family, 20 IGOR DOLGACHEV AND GEBHARD MARTIN both in the notation of Theorem 4.3 and in the notation of [9, Table 8.9]. The second and third columns give the group Aut(X) and its size. In the remaining columns, we list the number of elements of a given Carter conjugacy class in Aut(X).
5. DEL PEZZO SURFACES OF DEGREE 1 5.1. The anti-bicanonical map. As in the case of degree 2, we start by describing the geometry of del Pezzo surfaces of degree d = 1 and we refer to for characteristic-free facts on del Pezzo surfaces. Recall that the anti-bicanonical system | −2KX| defines a finite morphism f : X →Q onto a quadratic cone Q ⊆P3. As in degree 2, it turns out that this map is always separable, even in characteristic 2.
Proposition 5.1. The anti-bicanonical linear system | −2KX| defines a finite separable morphism f : X →Q of degree 2.
Proof. If f is not separable, then p = 2 and f is purely inseparable. But then f is a homeomorphism in the étale topology. This is impossible, since H2 ét(X, Zℓ) has rank 9 (because X is the blow-up of 8 points in the plane), while H2 ét(Q, Zℓ) has rank 1.
□ Let R(X, −KX) = ∞ M n=0 H0(X, OX(−nKX)) be the graded anti-canonical ring of X. By the Riemann–Roch-Theorem, we have dimk R(X, −KX)1 = 2 dimk R(X, −KX)2 = 4 dimk R(X, −KX)3 = 7.
Thus, we can choose u, v from R(X, −KX)1, x ∈R(X, −KX)2 \ S2(R(X, −KX)1), and y ∈ R(X, −KX)3 \ S3(R(X, −KX)1) + R(X, −KX)1 ⊗R(X, −KX)2 and obtain the following relation between the generators y2 + y(a1x + a3) + x3 + a2x2 + a4x + a6 = 0, (7) where ak denotes a binary form of degree k in u and v. In particular, via Equation (7), we can view X as a surface of degree 6 in the weighted projective space P(1, 1, 2, 3), the anti-canonical map is the projection of this surface onto the u, v-coordinates, and the anti-bicanonical map is the projection onto the u2, uv, v2, x-coordinates.
If p ̸= 2, we can replace y with y + 1 2(a1x+a3) to assume that a1 = a3 = 0. The surface X is a double cover of a quadratic cone Q ∼ = P(1, 1, 2). The branch curve B = V (x3 + a2x2 + a4x + a6) is a curve of degree 6 not passing through the vertex of Q. It is a smooth curve of genus 4 with a vanishing theta characteristic g1 3 defined by the ruling of Q. If we blow up the vertex of Q, we obtain a surface isomorphic to the rational minimal ruled surface F2. The preimage of the curve B is a curve in the linear system |6f + 3e|, where f and e are the standard generators of Pic(F2) with f2 = 0 and e2 = −2. The curve B is its canonical model in P3.
In our case, when the characteristic p = 2, the analog of B is the curve V (a1x + a3) in Q. In particular, Proposition 5.1 tells us that a1x + a3 ̸= 0 and there is no way of removing these terms.
Moreover, the curve B always passes through the vertex of Q and its strict transform on F2 is in |3f| if a1 = 0 and in |3f + e| if a1 ̸= 0. The analog of the involution y 7→−y, classically called the AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 21 Bertini involution, is the involution β defined by replacing y with y + a1x + a3. As in the classical case, we call this β Bertini involution.
By calculating the partial derivatives in Equation (7), the smoothness of X yields the following restrictions on the ai: Proposition 5.2. In (7), the smoothness of X is equivalent to the condition that the equations a1x + a3 = 0 x2 + a1y + a4 = 0 a1,uxy + a3,uy + a2,ux2 + a4,ux + a6,u = 0 a1,vxy + a3,vy + a2,vx2 + a4,vx + a6,v = 0 with ai,u := ∂ai ∂u and ai,v := ∂ai ∂v have no common solutions on X.
5.2. Normal forms. In this section, we find normal forms for del Pezzo surfaces of degree 1 in characteristic 2. In total, we will have 14 different normal forms, corresponding to the 14 possible combinations of singularities of the ramification curve R and the branch curve B. First, we simplify the equations of the branch curve.
Lemma 5.3. Let X be a del Pezzo surface of degree 1 given by Equation (7). Then, after a suitable change of coordinates, we may assume that the equation a1x + a3 of B is one of the following: (1) ux + v3 (2) ux (3) uv(u + v) (4) u2v (5) u3 Proof. If a1 ̸= 0, we may assume that a1 = u after applying a linear substitution in u and v. Then, a substitution of the form x 7→x + b2 for a suitable binary form b2 of degree 2 in u and v allows us to set a3 = λv3. Then, rescaling v, we can assume λ ∈{0, 1}.
If a1 = 0, we get three cases according to the number of distinct roots of a3. The equation can be normalized by applying a linear substitution in u and v to get Cases (3), (4), and (5).
□ If we consider P(1, 1, 2) as a quadratic cone Q in P3, these 5 normal forms for a1x + a3 correspond to the cases where B is a twisted cubic, a union of a line and a conic, a union of three lines, a union of a double line and a simple line, or a triple line, respectively. Later, we will use automorphisms of P(1, 1, 2) that preserve the equation of B and the form of Equation (7) in order to move the images of the singular points of R to special positions. In the following lemma, we describe this group of automorphisms.
Lemma 5.4. Let H ⊆Aut(k[u, v, x]) ⊆Aut(k[u, v, x, y]) be the subgroup of automorphisms that preserve a1x + a3, act on x as x 7→x + b2 for some binary quadratic form b2 in u and v, and that map Equation (7) to one of the same form, with possibly different a2, a4, and a6. Then, H consists of substitutions of the form u 7→ αu + βv v 7→ γu + δv x 7→ x + b2 where α, β, γ, δ ∈k such that αδ + βγ ̸= 0, and 22 IGOR DOLGACHEV AND GEBHARD MARTIN (1) if a1x + a3 = ux + v3, then α = 1, β = 0, δ3 = 1, b2 = γ3u2 + γ2δuv + γδ2v2.
In particular, H ∼ = k+ : 3.
(2) if a1x + a3 = ux, then α = 1, β = b2 = 0.
In particular, H ∼ = k+ : k×.
(3) if a1x+a3 = uv(u+v), then αγ(α+γ) = βδ(β +δ) = 0, α2δ +βγ2 = αδ2 +β2γ = 1.
In particular, H ∼ = k[u, v]2 : (3 × S3) (4) if a1x + a3 = u2v, then β = γ = 0, δ = α−2.
In particular, H ∼ = k[u, v]2 : k×.
(5) if a1x + a3 = u3, then β = 0, α3 = 1.
In particular, H ∼ = k[u, v]2 : (k : k× × 3).
For the convenience of the reader, we record the effect of a general substitution on the remaining ai in Equation (7). The proof is a straightforward calculation.
Lemma 5.5. A substitution of the form u 7→ αu + βv v 7→ γu + δv x 7→ x + b2 y 7→ y + b1x + b3 where α, β, γ, δ ∈k and bi ∈k[u, v]i such that αδ + βγ ̸= 0, changes the coefficients (a2, a4, a6) in Equation (7) as follows a2 7→ σ∗a2 + σ∗a1b1 + b2 1 + b2 a4 7→ σ∗a4 + σ∗a3b1 + σ∗a1b1b2 + σ∗a1b3 + b2 2 a6 7→ σ∗a6 + σ∗a4b2 + σ∗a3b3 + σ∗a2b2 2 + σ∗a1b2b3 + b2 3 + b3 2, where σ∗ai := ai(αu + βv, γu + δv).
Now, we are ready to describe the normal forms for del Pezzo surfaces of degree 1.
Theorem 5.6. Every del Pezzo surface of degree 1 in characteristic 2 is isomorphic to a surface of degree 6 in P(1, 1, 2, 3) given by an equation of the form y2 + y(a1(u, v)x + a3(u, v)) + x3 + a2(u, v)x2 + a4(u, v)x + a6(u, v) = 0, (8) where (a1, a2, a3, a4, a6) is one of the following: AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 23 Name a1x + a3 a2 a4 a6 #Parameters (1)(a) ux + v3 av2 bu4 + cu2v2 + dv4 eu6 + fu4v2 + gu2v4 + hv6 8 (1)(b) ux + v3 av2 cu2v2 + dv4 eu6 + fu4v2 + gu2v4 + hv6 7 (1)(c) ux + v3 av2 dv4 eu6 + fu4v2 + gu2v4 + hv6 6 (1)(d) ux + v3 av2 cu2v2 eu6 + fu4v2 + gu2v4 + hv6 6 (1)(e) ux + v3 av2 0 eu6 + fu4v2 + gu2v4 + hv6 5 (2)(a) ux av2 v4 bu6 + du4v2 + eu3v3 + fu2v4 + guv5 + hv6 7 (2)(b) ux av2 v4 bu6 + du4v2 + fu2v4 + guv5 + hv6 6 (2)(c) ux av2 v4 bu6 + du4v2 + eu3v3 + fu2v4 + hv6 6 (2)(d) ux av2 v4 cu5v + du4v2 + fu2v4 + hv6 5 (2)(e) ux av2 0 bu6 + du4v2 + eu3v3 + fu2v4 + euv5 + hv6 6 (2)(f) ux av2 0 bu6 + du4v2 + fu2v4 + uv5 + hv6 5 (3) uv(u + v) auv bu3v + (b + c)u2v2 + cuv3 du5v + eu3v3 + fuv5 6 (4) u2v 0 au3v + bu2v2 + cuv3 du5v + eu3v3 + uv5 5 (5) u3 0 au3v + bu2v2 + cuv3 uv5 + dv6 4 Moreover, the parameters satisfy the conditions summarized in the following table, where ∆:= a4 3 + a3 1a3 3 + a4 1(a2 4 + a1a3a4 + a2a2 3 + a2 1a6).
In this table, we also describe the singularities of the irreducible components of the reduction Rred of the ramification curve R.
Name Conditions on the parameters Singularities of the irreducible components of Rred (1)(a) ∆has only simple roots A2 over [1 : v : v3] with v8 + dv6 + cv4 + bv2 = 0 v8 + dv6 + cv4 + bv2 has four distinct roots (1)(b) ∆has only simple roots, c, d ̸= 0 A4 over [1 : 0 : 0] 2A2 over [1 : v : v3] with v4 + dv2 + c = 0 (1)(c) ∆has only simple roots, d ̸= 0 A6 over [1 : 0 : 0] A2 over [1 : d 1 2 : d 3 2 ] (1)(d) ∆has only simple roots, c ̸= 0 2A4 over [1 : 0 : 0] and [1 : c 1 4 : c 3 4 ] (1)(e) e ̸= 0 A8 over [1 : 0 : 0] (2)(a) u−4∆has only simple roots, e, g, (g2 + a + h) ̸= 0 3A2 over [0 : 1 : 1], [1 : 0 : 0] and [g 1 2 : e 1 2 : 0] (2)(b) b, g, (g2 + a + h) ̸= 0 A4 over [1 : 0 : 0] A2 over [0 : 1 : 1] (2)(c) b, e, (a + h) ̸= 0 3A2 over [0 : 1 : 1], [1 : 0 : 0] and [0 : 1 : 0] (2)(d) c, (a + h) ̸= 0 A4 over [0 : 1 : 0] A2 over [0 : 1 : 1] (2)(e) u−6∆has only simple roots, e ̸= 0 3A2 over [0 : 1 : 0], [1 : 0 : 0] and [1 : 1 : 0] (2)(f) u−6∆has only simple roots A4 over [1 : 0 : 0] A2 over [0 : 1 : 0] (3) d, f ̸= 0, (d + e + f) ̸∈{0, 1} 3A2 over [1 : 0 : 0], [0 : 1 : 0] and [1 : 1 : 0] (4) d ̸= 0 2A2 over [1.0 : 0] and [0 : 1 : 0] (5) − A2 over [0 : 1 : 0] Remark 5.7. The conditions on the parameters that guarantee the smoothness of X are equivalent to the conditions that Equation (8) is the Weierstrass equation of an elliptic fibration with only irreducible fibers. We will study this fibration later in Section 5.4. The homogeneous polynomial ∆ appearing in Theorem 5.6 is the discriminant of this fibration.
24 IGOR DOLGACHEV AND GEBHARD MARTIN Proof of Theorem 5.6. By Lemma 5.3, there are, up to choice of coordinates, 5 possible equations for B. We will now give normal forms in each case.
(1) a1x + a3 = ux + v3 Here, the ramification curve R is given by the two equations ux + v3 = 0 y2 + x3 + a2x2 + a4x + a6 = 0.
One checks that the curve R is smooth at the points with u = 0. On the affine chart u = 1, it is given in A2 by the single equation y2 + v9 + a2(1, v)v6 + a4(1, v)v3 + a6(1, v), so it has singularities over the roots of the derivative F ′ of F := v9 + a2(1, v)v6 + a4(1, v)v3 + a6(1, v). After applying an element of H in Lemma 5.4, we may assume that 0 is the root of highest multiplicity of F ′.
Now, substitutions as in Lemma 5.5 that fix u, v, and x do not change the location of the points that lie under singularities of R and thus, by Lemma 5.5, we can assume that a2 = av2, a4 = bu4 +cu2v2 +dv4, a6 = eu6 +fu4v2 +gu2v4 +hv6. With this notation, the polynomial F ′ becomes v8+dv6+cv4+bv2 and the conditions of Proposition 5.2 boil down to v8+dv6+cv4+bv2 and ∆(1, v) = v12 + v9 + (d2 + a)v8 + dv7 + hv6 + cv5 + (c2 + g)v4 + bv3 + fv2 + b2 + e not having a common solution. The former is the derivative of the latter, hence we want that the latter has only simple zeroes.
Now, if F ′ has four distinct roots, we are in Case (a). If F ′ has less than four distinct roots, we may assume b = 0. If F ′ has exactly three roots, then we are in Case (b). If b = 0, the polynomial F ′ has exactly two roots if and only if either c = 0 and d ̸= 0, which is Case (c), or d = 0 and c ̸= 0, which is Case (d). Finally, F ′ has a single root if and only if b = c = d = 0, which is Case (e).
(2) a1x + a3 = ux Here, the ramification curve has two components R1 and R2. The curve R1 is given by u = 0 y2 + x3 + a2x2 + a4x + a6 = 0.
This curve has a unique singularity, which is of type A2 and located over [0 : 1 : a4(0, 1) 1 2 ].
Rescaling v, we may assume that a4(0, 1) ∈{0, 1}.
The curve R2 is given by x = 0 y2 + a6 = 0.
This curve has singularities over the points [u : v : 0] where the derivatives of a6 by u and v both vanish.
First, assume that a4(0, 1) = 1 and one of the singularities of R2 does not lie over [0 : 1 : 0].
Then, using a substitution in v as in Lemma 5.4, we can assume that one of them lies over [1 : 0 : 0]. Substitutions as in Lemma 5.5 which fix u, v, and x do not change the location of these points and, after applying one of them, we may assume that a2 = av2, a4 = v4, and a6 = bu6 + du4v2 + eu3v3 + fu2v4 + guv5 + hv6. If e, g ̸= 0, this is Case (a), if e = 0 and g ̸= 0, this is Case (b), and if e ̸= 0 and g = 0, this is Case (c). The conditions of Proposition AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 25 5.2 boil down to ∆(1, v) = v8 + hv6 + gv5 + fv4 + ev3 + dv2 + b having only simple roots and g2 ̸= a + h. In particular, (e, g) ̸= (0, 0).
If a4(0, 1) = 1, R2 has a unique singularity, and this singularity lies over [0 : 1 : 0], then the only odd monomial in a6 is u5v. A substitution of the form v 7→v + µu and substitutions as in the previous paragraph allow us to assume that a2 = av2, a4 = v4, and a6 = cu5v + du4v2 + fu2v4 + hv6. The conditions of Proposition 5.2 become a + h ̸= 0 and c ̸= 0. This is Case (d).
If a4(0, 1) = 0, then Proposition 5.2 implies that R2 is smooth over [0 : 1 : 0]. Hence, we can assume that one of the singularities of R2 lies over [1 : 0 : 0]. Using a substitution as in Lemma 5.5 which fixes u, v and x, we may assume that a2 = av2, a4 = 0, and a6 = bu6 + du4v2 + eu3v3 + fu2v4 + guv5 + hv6. Since R2 is smooth over [0 : 1 : 0], we have g ̸= 0.
If e ̸= 0, we can scale v so that g = e. This is Case (e). If e = 0, we scale v so that g = 1. This is Case (f).
(3) a1x + a3 = uv(u + v) The curve B has the three irreducible components B1, B2, and B3, given by V (u), V (v), and V (u + v), respectively. The corresponding components R1, R2, and R3 of R are given by y2 + x3 + a2(0, v)x2 + a4(0, v)x + a6(0, v), y2 + x3 + a2(u, 0)x2 + a4(u, 0)x + a6(u, 0), and y2 + x3 + a2(u, u)x2 + a4(u, u)x + a6(u, u), respectively. The singular points of R1, R2, and R3 lie over [0 : 1 : a4(0, 1) 1 2 ], [1 : 0 : a4(1, 0) 1 2 ], and [1 : 1 : a4(1, 1) 1 2 ], respectively.
A substitution as in Lemma 5.4 which fixes u and v allows us to set a4(0, 1) = a4(1, 0) = a4(1, 1) = 0, that is, that a4 = bu3v + (b + c)u2v2 + cuv3 for some b, c ∈k. Then, a substitution as in Lemma 5.5 which fixes u, v, and x allows us to set a2 = auv and a6 = du5v +eu3v3 +fuv5.
The conditions of Proposition 5.2 become d ̸= 0, f ̸= 0 and d + e + f ̸∈{0, 1}.
(4) a1x + a3 = u2v The curve B has two irreducible components B1 and B2, given by V (u) and V (v), respectively.
The corresponding components R1 and R2 of R are given by y2 + x3 + a2(0, v)x2 + a4(0, v)x + a6(0, v), and y2 + x3 + a2(u, 0)x2 + a4(u, 0)x + a6(u, 0), respectively. The singular points of R1 and R2 lie over [0 : 1 : a4(0, 1) 1 2 ] and [1 : 0 : a4(1, 0) 1 2 ], respectively.
A substitution as in Lemma 5.5, which fixes u and v allows us to set a4(0, 1) = a4(1, 0) and that a2 is a square. Then, a substitution with b2 = b3 = 0 allows us to eliminate a2. Finally, a substitution with b1 = b2 = 0 allows us to assume that a6 contains no squares. If we write a6 = du5v + eu3v3 + fuv5, then the conditions of Proposition 5.2 becomes d ̸= 0 and f ̸= 0, and we can rescale f to 1.
(5) a1x + a3 = u3 The curve R is given by y2 + x3 + a2(0, v)x2 + a4(0, v)x + a6(0, v) and it is singular over [0 : 1 : a4(0, 1) 1 2 ].
26 IGOR DOLGACHEV AND GEBHARD MARTIN We apply the same substitutions as in the previous case to remove a2. Then, we apply a substitution as in Lemma 5.5 with b2 = b2 1 to remove the v4-term in a4. Next, using a substitution that fixes u, v, and x with b1 = 0, we eliminate the squares in a6, write a6 = du5v +eu3v3 +fuv5, and rescale f to 1. After that, a substitution of the form v 7→v + λu, and eliminating the square again, allows us to set d = 0. Next, a substitution as in Lemma 5.5 which fixes u and v, with b1 = λu, b2 = λ2u2, and b3 = µu3 for suitable λ and µ allows us to eliminate the u4-term in a4 without changing a6. Finally, we apply a substitution with b3 = ev3 and rename the parameters to assume that a6 = uv5 + dv6. The conditions of Proposition 5.2 are fulfilled for every choice of parameters.
□ 5.3. Fake tritangent planes and odd theta characteristics. It is known that a del Pezzo surface X of degree 1 contains 240 (−1)-curves (see [9, 8.7], where the proof is characteristic free). They come in pairs Ei + E′ i ∈| −2KX| with Ei · E′ i = 3. The Bertini involution β swaps the two curves in a pair. The image of Ei + E′ i under the anti-bicanonical map f is a plane section of Q not passing through the vertex.
If p ̸= 2, each of the resulting 120 planes is a tritangent plane to the branch sextic curve and, conversely, every tritangent plane to the branch sextic gives rise to a pair of (−1)-curves Ei + E′ i with Ei + E′ i ∈| −2KX|. A tritangent plane intersects the branch curve in twice a positive divisor of degree 3. This divisor is an odd theta characteristic of the curve. It is known that the number of odd theta characteristics on a smooth curve of genus 4 is equal to 120.
For arbitrary p, we still have the following.
Lemma 5.8. The preimage f−1(C) of an integral conic C = V (x+b2) is a sum of two (−1)-curves if and only if it is reducible.
Proof. Since f has degree 2 and C is integral, the curve f−1(C) is reducible if and only if it has two irreducible components L1 and L2. These components satisfy L1 + L2 ∈| −2KX|, L1 · L2 = 3, and L2 1 = L2 2. Via adjunction, this easily implies that L1 and L2 are (−1)-curves. The converse is clear.
□ So, even if p = 2, we have 120 splitting conics and we call the corresponding planes in P3 fake tritangent planes in analogy with the situation in the other characteristics. For the rest of this section, we assume p = 2.
Since the anti-bicanonical map is étale outside the branch curve V (A), the intersection Ei ∩E′ i lies on the ramification curve R. Let L = OR(Ei) ∼ = OR(E′ i). It is an invertible sheaf on C of degree 2. We have L⊗2 ∼ = OR(Ei + E′ i) ∼ = OR(−2KX).
Since B ∈|OP(1,1,2)(3)|, we have R ∈| −3KX|. By the adjunction formula, we have ωR ∼ = OR(−3KX + KX) ∼ = L⊗2.
As in the case of degree 2, invertible sheaves on R that satisfy this property are called invertible theta characteristics. Let Θ(R) be the set of isomorphism classes of such invertible theta characteristics on R and let J(R) be the generalized Jacobian of R. As in Lemma 3.6, one can prove that J(R) is a product of additive groups.
Lemma 5.9. The generalized Jacobian J(R) of R is isomorphic to G4 a.
Thus, as in degree two, finding fake tritangent planes using theta characteristics on R is subtle in characteristic two. We refer to Example 3.7 for an example in degree two that further illustrates this point and leave it to the reader to find a similar example in degree one.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 27 5.4. Rational elliptic surfaces. Equation (7) can also serve as the Weierstrass equation of the rational surface with a genus 1 fibration φ : Y →P1 obtained by blowing up the base point p0 of | −KX|. Since X is a del Pezzo surface, all members of | −KX| are irreducible, hence so are all fibers of φ. The discriminant of φ is ∆= a4 3 + a3 1a3 3 + a4 1(a2 4 + a1a3a4 + a2a2 3 + a2 1a6).
The singular fibers of φ lie over the zeroes of ∆. Moreover, the Bertini involution, which is given by β : y 7→y + (a1x + a3), induces the inversion on the group structure of each fiber. In particular, for [u0 : v0] ∈P1, if a1(u0, v0)x + a3(u0, v0) = 0, the corresponding fiber F of φ is cuspidal, if a1(u0, v0) = 0 and a3(u0, v0) ̸= 0, then F is smooth and supersingular, and in the other cases, F is either nodal, or smooth and ordinary, according to whether ∆(u0, v0) is zero or not. Applying these observations to the normal forms of Theorem 5.6, we obtain the following information on φ.
Proposition 5.10. Let X be a del Pezzo surface of degree 1 given by one of the normal forms in Theorem 5.6. Then, the associated genus 1 fibration φ is elliptic and all its fibers are irreducible.
The discriminant ∆and the singular fibers of φ are given in the following table.
Name ∆ Nodal fibers over the Cuspidal fibers over (1)(a) v12 + u3v9 + (d2 + a)u4v8 + du5v7 + hu6v6 + cu7v5 12 roots of ∆ – +(c2 + g)u8v4 + bu9v3 + fu10v2 + (b2 + e)u12 (1)(b) v12 + u3v9 + (d2 + a)u4v8 + du5v7 + hu6v6 + cu7v5 12 roots of ∆ – +(c2 + g)u8v4 + fu10v2 + eu12 (1)(c) v12 + u3v9 + (d2 + a)u4v8 + du5v7 + hu6v6 12 roots of ∆ – +gu8v4 + fu10v2 + eu12 (1)(d) v12 + u3v9 + au4v8 + hu6v6 + cu7v5 12 roots of ∆ – +(c2 + g)u8v4 + fu10v2 + eu12 (1)(e) v12 + u3v9 + au4v8 + hu6v6 12 roots of ∆ – +gu8v4 + fu10v2 + eu12 (2)(a) u4(v8 + u2(bu6 + du4v2 + eu3v3 + fu2v4 + gsv5 + hv6)) 8 roots of u−4∆ [0 : 1] (2)(b) u4(v8 + u2(bu6 + du4v2 + fu2v4 + gsv5 + hv6)) 8 roots of u−4∆ [0 : 1] (2)(c) u4(v8 + u2(bu6 + du4v2 + eu3v3 + fu2v4 + hv6)) 8 roots of u−4∆ [0 : 1] (2)(d) u4(v8 + u2(cu5v + du4v2 + fu2v4 + hv6)) 8 roots of u−4∆ [0 : 1] (2)(e) u6(bu6 + du4v2 + eu3v3 + fu2v4 + euv5 + hv6) If h ̸= 0 : 6 roots of u−6∆ [0 : 1] If h = 0 : 5 roots of u−7∆ (2)(f) u6(bu6 + du4v2 + fu2v4 + uv5 + hv6) If h ̸= 0 : 6 roots of u−6∆ [0 : 1] If h = 0 : 5 roots of u−7∆ (3) u4v4(u + v)4 – [1 : 0], [0 : 1], [1 : 1] (4) u8v4 – [1 : 0], [0 : 1] (5) u12 – [0 : 1] Remark 5.11. As in Remark 3.12, we point out the connection to Lang’s classification of singular fibers on rational elliptic surfaces: Our normal forms for del Pezzo surfaces of degree 1 yield normal forms for all rational elliptic surfaces with a section in characteristic 2 all of whose fibers are irreducible.
28 IGOR DOLGACHEV AND GEBHARD MARTIN 6. AUTOMORPHISM GROUPS OF DEL PEZZO SURFACES OF DEGREE 1 This section consists of three parts. In the first part, we collect various restrictions on the group G(X) = Aut(X)/⟨β⟩arising from the geometry of X. In the second part, we give an explicit description of Aut(X) in terms of Equation (7) and use it to classify all surfaces where G(X) is non-trivial and to determine the group Aut(X) in every case. In the third part, we compare our classification with the classification in characteristic 0 from [9, Table 8.14] and use this to determine the conjugacy classes of all elements in Aut(X) (see Table 4 in the Appendix). Throughout, we assume p = 2.
6.1. Restrictions on G(X). Since the elliptic fibration φ : Y →P1 associated to X is obtained by blowing up the base point of | −KX|, we can identify Aut(X) with the subgroup of Aut(Y ) preserving a chosen section. Let r : Aut(X) →Aut(P1) be the natural homomorphism defined by the action of Aut(X) on the coordinates [u : v] of the base of φ. Since φ is the unique relatively minimal smooth proper model of its generic fiber Fη, the kernel K = Ker(r) is isomorphic to the group of automorphisms of the elliptic curve Fη. In particular, K contains the Bertini involution β and it can contain more automorphisms only if the j-invariant of Fη is equal to 0 = 1728, in which case K is a subgroup of Q8 : 3 ∼ = SL2(F3).
Let P be the image of r. Evidently, P is a finite subgroup of Aut(P1) that leaves invariant the set S1 of points p = [ui : vi] corresponding to the singular fibers. It also leaves invariant the set S2 of the projections of singular points of the irreducible components of the ramification curve R.
The following proposition shows what kind of groups can be expected to occur for P. We use the known classification of finite subgroups of Aut(P1) ∼ = PGL2(k) ∼ = SL2(k) [12, Theorem 2.5].
Proposition 6.1. The group P is isomorphic to Gξ,A or D2n.
Proof. Since SL2(2) ∼ = S3 ∼ = D6, it suffices to show that SL2(Fq) ̸⊆P for q = 2m and m ≥2.
Since the set S2 has cardinality at most 4 and P preserves S2, every homogeneous polynomial F with simple roots along S2 is P-semi-invariant of degree at most 4. On the other hand, by [16, Theorem 6.1.8], the ring k[u, v]SL2(Fq) is generated over Fq by the Dickson polynomials L and d2,1 of degrees q + 1 and q2 −q, respectively. If SL2(q) ⊆P, then F is also a semi-invariant polynomial for SL2(q) and if q ̸= 2, then SL2(Fq) is simple, so F ∈k[u, v]SL2(q) = k[L, d2,1].
Hence, q = 2, as claimed.
□ We recall from [12, §1.3] that the image of the Bertini involution β under the injective homo-morphism ρ : Aut(X) →W(E8) is equal to −idE8. However, in contrast to the situation in degree 2, the extension W(E8) →W(E8)/(−idE8) ∼ = O+ 8 (2) does not split. The semi-direct product W(E8) = 2.GO+ 8 (2) corresponds to a non-trivial homomorphism O+ 8 (2) →C2, whose kernel is a simple group O8(2), where we use the ATLAS notation.
Therefore, in order to determine Aut(X), it is not enough to determine the image G(X) of the homomorphism Aut(X) →Aut(X)/⟨β⟩, and thus the calculation of Aut(X) is more complicated than in the case of del Pezzo surfaces of degree 2.
Let us summarize the restrictions on Aut(X) and G(X) that we have collected by now.
Theorem 6.2. Let X be a del Pezzo surface of degree 1 in characteristic 2. Let G(X) be the image of the homomorphism Aut(X) →Aut(P(1, 1, 2)), let K be the kernel of the homomorphism r : Aut(X) →Aut(P1), let P be the image of r, and let φ : Y →P1 be the elliptic fibration associated to X. Then, the following hold: AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 29 (i) Aut(X) is a central extension of G(X) by ⟨β⟩∼ = C2.
(ii) Aut(X) is an extension of P by K.
(iii) Aut(X) is a subgroup of W(E8).
(iv) G(X) is a subgroup of O+ 8 (2).
(v) K is the automorphism group of the generic fiber of φ.
(vi) P is isomorphic to Gξ,A or D2n.
(vii) P preserves the set S1 of points lying under singular fibers of φ. Moreover, it preserves the decomposition of S1 into subsets corresponding to isomorphic fibers.
(viii) P preserves the set S2 of points lying under the singularities of R. Moreover, it preserves the decomposition of S2 into subsets of isomorphic singularities.
(ix) The j-function of φ is P-invariant.
This yields the following preliminary restrictions on Aut(X) and G(X).
Corollary 6.3. Let X be a del Pezzo surface of degree 1 in characteristic 2 given by one of the normal forms in Theorem 5.6.
(i) In Case (1), G(X) is a subgroup of A4.
(ii) In Cases (2) (a), (2) (b), (2) (c), and (2) (d), G(X) is a subgroup of 23.
(iii) In Cases (2) (e) and (2) (f), G(X) is a subgroup of C5 or C2.
(iv) In Case (3), K is a subgroup of SL2(3) and P is a subgroup of S3.
(v) In Case (4), K is a subgroup of SL2(3) and P is cyclic of order 1, 3, 5, 7, 9, or 15.
(vi) In Case (5) K is a subgroup of SL2(3) and P ∼ = Gξ,A, where ξ is a primitive n-th root of unity with n ∈{1, 3, 5, 7, 9, 15}.
Proof. In Case (1), the generic fiber of φ is ordinary, hence K = ⟨β⟩and G(X) ∼ = P. The fibration φ has 12 nodal fibers, hence the j-function has 12 poles, so |P| | 12. Since P is isomorphic to Gξ,A or D2n with n odd, this implies that P is isomorphic to a subgroup of A4.
In Cases (2), we also have K = ⟨β⟩and G(X) ∼ = P. In Cases (2) (a), (2) (b), (2) (c), and (2) (d), the fibration φ has 8 nodal fibers, hence |P| | 8. This implies that P is elementary abelian of order 1, 2, 4 or 8. In Cases (2) (e) and (2) (f), the fibration φ has 5 or 6 nodal fibers. If it has 5 nodal fibers, then |P| | 5, hence P is a subgroup of C5. If it has 6 nodal fibers, then P is either a subgroup of C2 or isomorphic to the dihedral group D6. In the latter case, P acts without fixed point on P1, which is impossible, since φ admits a unique cuspidal fiber.
In Case (3), we have K ⊆SL2(3), since the generic fiber of φ is supersingular. Since φ has three singular fibers, P is isomorphic to a subgroup of S3.
In Case (4), we also have K ⊆SL2(3). Since one of the components of R is reduced and the other is not, P acts trivially on S2, hence with two fixed points on P1. So, P is cyclic of odd order.
Moreover, P is a subgroup of O+ 8 (2). In particular, P admits a faithful representation of dimension at most 8. Hence, if we denote Euler’s totient function by ϕ, then ϕ(|P|) ≤8. Thus, P is of order 1, 3, 5, 7, 9 or 15.
In Case (5), we have K ⊆SL2(3) and the action of P on P1 fixes the point lying under the unique singular fiber of φ, hence P ∼ = Gξ,A. The order of ξ can be bounded by the same argument as in the previous paragraph.
□ In particular, we get upper bounds on the size of Aut(X) in every case. Further information on the 2-groups that can occur in Case (5) can be obtained using the following remark.
Remark 6.4. Since the maximal powers of 2 that divide |W(E8)| and |W(D8)| are both 214, and since W(D8) is a subgroup of W(E8), the 2-Sylow subgroups P in W(E8) are isomorphic to 30 IGOR DOLGACHEV AND GEBHARD MARTIN the 2-Sylow subgroups in W(D8) = 27 : S8. Hence, P is isomorphic to 27 : (S8)2, where 27 acts on Z8 by an even number of sign changes and (S8)2 is a 2-Sylow subgroup of S8 acting as permutations on Z8. The group (S8)2 is isomorphic to the symmetry group of a binary tree of depth 3, considered as a subgroup of S8 via the permutation it induces on the leaves of the tree. An equivalent description is as the wreath product D8 ≀C2, where D8 × D8 is a subgroup of S4 × S4 ⊂S8. The Bertini involution β corresponds to the element (−1, id) that changes all signs. The 2-groups that can occur in Corollary 6.3 are isomorphic to subgroups of P.
In the following example, we apply this remark to give an explicit description of the group 21+6 + , which will occur in our classification.
Example 6.5. With notation as in the previous remark, let G ⊆P be a subgroup containing β such that G/⟨β⟩is an elementary abelian 2-group and such that β ∈Q8 ⊆G. Then, each element of G is of the form (σ, τ), where ord(τ) ≤2 and either τ preserves the set of coordinates whose sign is changed by σ and then (σ, τ) has order 1 or 2, or τ swaps this set with the set of coordinates whose sign is not changed and then (σ, τ) has order 4. In particular, in the latter case, τ has cycle type (2, 2, 2, 2). Since Q8 ⊆G, the image of G →(S8)2 contains a subgroup H of order 4 generated by involutions of cycle type (2, 2, 2, 2). The centralizer C of H is of order 8 and its non-trivial elements are involutions of cycle type (2, 2, 2, 2). The kernel of G →(S8)2 consists of sign changes σ that are compatible with all τ ∈H in the sense that (σ, τ)2 ∈⟨β⟩. One checks that the group N of all such compatible sign changes has order 16 and that all elements of N are also compatible with C. Then, G is a subgroup of the resulting extension M of C by N.
We have M/⟨β⟩= 26. This is a quadratic space over F2 with the quadratic form q : M/⟨β⟩→ ⟨β⟩defined as q(x) = ˜ x2, where ˜ x is a lift of x to M. The subspace N/⟨β⟩is totally isotropic of dimension 3 and the description of M in the previous paragraph shows that q is non-degenerate.
Hence, by [1, (23.10)], M is isomorphic to the extra-special 2-group of 21+6 + .
6.2. Classification. Recall that X is a hypersurface of degree 6 in P(1, 1, 2, 3) given by Equation (7). An automorphism of P(1, 1, 2, 3) is induced by a substitution of the form u 7→ αu + βv v 7→ γu + δv x 7→ εx + b2 y 7→ ζy + b1x + b3 where α, β, γ, δ, ε, ζ ∈k, bi ∈k[u, v]i, and αδ + βγ, ε, ζ ̸= 0. The substitutions that induce the identity on P(1, 1, 2, 3) are the ones with β, γ, b1, b2, b3 = 0 and γ = α, ε = α2, ζ = α3.
Since X is anti-canonically embedded into P(1, 1, 2, 3), all automorphisms of X are induced by the substitutions as above that map Equation (7) to a multiple of itself. Clearly, we can represent every such automorphism by a substitution with ζ = 1. Then, the substitution does not change the coefficient of y2 in Equation (7), hence ε3 = 1. Therefore, we may assume ε = 1 as well. In particular, using Lemma 5.5, we obtain the following description of Aut(X), where we write σ for the substitution u 7→ αu + βv v 7→ γu + δv and σ∗ai := ai(αu + βv, γu + δv).
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 31 Lemma 6.6. Let X be a del Pezzo surface of degree 1 given by Equation (7). Then, Aut(X) can be identified with the group of 4-tuples (b1, b2, b3, σ), where bi ∈k[u, v]i and σ ∈GL2(k) such that σ∗a1 + a1 = 0, σ∗a2 + a2 = a1b1 + b2 1 + b2, σ∗a3 + a3 = a1b2, σ∗a4 + a4 = a3b1 + a1b3 + b2 2, σ∗a6 + a6 = a4b2 + +a3(b3 + b1b2) + a2b2 2 + a1(b2b3 + b1b2 2) + b2 3 + b3 2 + b2 1b2 2 and where the composition is given by (b1, b2, b3, σ) ◦(b′ 1, b′ 2, b′ 3, σ′) = (σ′∗b1 + b′ 1, σ′∗b2 + b′ 2, σ′∗b3 + b′ 3 + σ′∗b1b′ 2, σ ◦σ′) In particular, there is a homomorphism Aut(X) →H ⊆Aut(P(1, 1, 2)), where H is the group from Lemma 5.4.
Lemma 6.7. The kernel of the homomorphism Aut(X) →H is generated by the Bertini involution.
Proof. Let (b1, b2, b3, σ) be in the kernel. Then, σ = id and b2 = 0. The conditions σ∗a2 = a2 + a1b1 + b2 1, σ∗a4 = a4 + a3b1 + a1b3, and σ∗a6 = a6 + a3b3 + b2 3 show that (b1, b3) ∈ {(0, 0), (a1, a3)}, so we recover our explicit description of the Bertini involution.
□ Now, we use the normal forms of Theorem 5.6 to classify all del Pezzo surfaces X of degree one with non-trivial G(X).
32 IGOR DOLGACHEV AND GEBHARD MARTIN Theorem 6.8. Every del Pezzo surface of degree one in characteristic two such that G(X) is non-trivial is isomorphic to a surface of degree 6 in P(1, 1, 2, 3) given by an equation of the form y2 + (a1x + a3)y + x3 + a2x2 + a4x + a6 where (a1, a2, a3, a4, a6, G(X), Aut(X)) is one of the following: Name a1x + a3 a2 a4 a6 G(X) Aut(X) #Parameters (1)(a)(i) ux + v3 av2 bu4 + (b + 1)u2v2 eu6 + fu4v2 + (a + b + b2 + f)u2v4 + bv6 2 4 4 (1)(a)(ii) ux + v3 0 bu4 eu6 + hv6 3 6 3 (1)(a)(iii) ux + v3 av2 u4 eu6 + au4v2 + v6 22 Q8 2 (1)(a)(iv) ux + v3 0 u4 eu6 + v6 A4 SL2(3) 1 (1)(d)(i) ux + v3 av2 u2v2 eu6 + fu4v2 + (a + f)u2v4 2 4 3 (1)(e)(i) ux + v3 0 0 eu6 + hv6 3 6 2 (2)(a)(i) ux av2 v4 bu6 + (efg−1 + e 3 2 g−1 2 + e3g−3)u4v2 + eu3v3 + fu2v4 + guv5 + e−1 2 g 3 2 v6 2 22 5 (2)(d)(i) ux av2 v4 cu5v + du4v2 + fu2v4 23 24 4 (2)(e)(i) ux av2 0 bu6 + (e + f)u4v2 + eu3v3 + fu2v4 + euv5 + ev6 2 22 4 (2)(f)(i) ux 0 0 bu6 + uv5 5 10 1 (3)(i) uv(u + v) auv bu3v + buv3 du5v + eu3v3 + duv5 2 22 4 (3)(ii) uv(u + v) auv a 1 2 u3v + a 1 2 uv3 (e + e 1 2 )u5v + eu3v3 + (e + e 1 2 )uv5 S3 2 × S3 2 (3)(iii) uv(u + v) 0 0 du5v + eu3v3 + duv5 6 2 × 6 2 (3)(iv) uv(u + v) 0 bu3v + ζ3bu2v2 + ζ2 3buv3 (e + e 1 2 )u5v + eu3v3 + (e + e 1 2 )uv5 3 6 2 (3)(v) uv(u + v) 0 0 (e + e 1 2 )u5v + eu3v3 + (e + e 1 2 )uv5 3 × S3 6 × S3 1 (4)(i) u2v 0 0 du5v + eu3v3 + uv5 3 6 2 (5)(i) u3 0 au3v + bu2v2 uv5 + dv6 26 21+6 + 3 (5)(ii) u3 0 0 uv5 + dv6 26 : 3 21+6 + : 3 1 (5)(iii) u3 0 0 uv5 26 : 15 21+6 + : 15 0 Here, S3, D8, Q8, and 21+6 + , denote the symmetric group on 3 letters, the dihedral group of order 8, the quaternion group, and the even extra-special group of order 128, respectively. In each case, the parameters have to satisfy the conditions of Theorem 5.6 and the obvious genericity conditions that keep them from specializing to other subcases.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 33 Proof. We use the normal forms of Theorem 5.6 and let H be the group of Lemma 5.4. By Lemma 6.6, we have G(X) ⊆H. We apply Lemma 6.6 to calculate Aut(X).
Case (1) (a) Let (b2, σ) ∈H. If (b2, σ) ∈G(X), then σ permutes the roots of the polynomial F ′ := v8 + dv6 + cv4 + bv2, since these are determined by the singularities of R. We have σ∗F ′ = δ2v8 + dv6 + δ(γ2d + c)v4 + δ2(γ4d + b)v2 + γ8 + γ6d + γ4c + γ2b.
If d ̸= 0, this is a multiple of F ′ if and only if δ = 1 and γ = 0, hence σ is the identity and G(X) is trivial. If d = 0, it is a multiple of F ′ if and only if γ8 + γ4c + γ2b = 0 (9) and δ = 1 or c = 0.
So, assume first that c ̸= 0 and δ = 1. If (b2, σ) ∈G(X), then there exist polynomials b1 and b3 such that σ∗a2 = a2 + a1b1 + b2 1 + b2 and σ∗a4 = a4 + a3b1 + a1b3 + b2 2. In our case, this means 0 = γ2au2 + ub1 + b2 1 + γ3u2 + γ2uv + γv2 0 = γ2cu4 + v3b1 + ub3 + γ6u4 + γ4u2v2 + γ2v4, hence b1 = λu + γ2t with λ2 + λ = γ2a + γ3 and γ4 = γ, and b3 = (γ2c + γ3)u3 + γuv2 + λv3.
If γ ̸= 0, then γ4 = γ implies γ3 = 1. Modifying the equation of X by an element of H, we may assume that γ = 1. Plugging this into Equation (9), we obtain c = b + 1. Hence, b1 = λu + v with λ2 + λ = a and b3 = bu3 + uv2 + v3. Plugging this into the equation for σ∗a6 and comparing coefficients in Lemma 6.6, we obtain the conditions h = b and g = a + b + b2 + f. Since γ is uniquely determined by Equation (9), we have G(X) ∼ = C2. The square of any lift of a non-trivial element of G(X) to Aut(X) is the Bertini involution, hence Aut(X) ∼ = C4.
Next, assume that c = 0. If (b2, σ) ∈G(X), then there exist polynomials b1 and b3 such that σ∗a2 = a2 + a1b1 + b2 1 + b2 and σ∗a4 = a4 + a3b1 + a1b3 + b2 2. In our case, this means 0 = γ2au2 + (1 + δ2)av2 + ub1 + b2 1 + γ3u2 + γ2δuv + γδ2v2 0 = v3b1 + ub3 + γ6u4 + γ4δ2u2v2 + γ2δ4v4, hence b1 = λu + γ2δv with λ2 + λ = γ2a + γ3 and γ4 + γ = (1 + δ)a, as well as b3 = γ6u3 + γ4δ2uv2 + λv3.
First, assume that δ ̸= 1. Then, σ has order 3, hence if (b2, σ) ∈G(X), then it fixes one of the four roots of F ′. After conjugating by a suitable element of H and repeating the substitutions we used in Theorem 5.6, we may assume that (b2, σ) fixes [1 : 0 : 0]. This implies that γ = 0, hence (1 + δ)a = 0 implies a = 0. Now, we plug everything into the equation for σ∗a6 and compare coefficients to obtain the conditions f = g = 0.
If δ = 1, then γ4 + γ = 0. Hence, if (b2, σ) is non-trivial, then γ3 = 1. Modifying the equation of X by an element of H, we may assume γ = 1, that is, that (b2, σ) maps [1 : 0 : 0] to [1 : 1 : 1].
Then, Equation (9) implies b = 1. Plugging into the equation for σ∗a6 and comparing coefficients yields g = f + a and h = 1. The square of both lifts of (b2, σ) to Aut(X) is the Bertini involution, hence the subgroup generated by these lifts is isomorphic to C4.
Suppose next that G(X) contains two distinct non-trivial automorphisms with δ = 1. Then, we can assume that one of them acts as in the previous paragraph, so b = h = 1 and g = f + a. The other one satisfies γ ̸= 1. Plugging this into the equation for σ∗a6 and comparing coefficents yields f = a. As in the previous paragraph, the square of all lifts of these automorphisms is the Bertini involution, hence they generate a subgroup isomorphic to the quaternion group Q8.
34 IGOR DOLGACHEV AND GEBHARD MARTIN Finally, Corollary 6.3 shows that G(X) acts on the four singular points of R through A4, so if G(X) contains a non-trivial automorphism with δ = 1 and a non-trivial automorphism with δ ̸= 1, then G(X) ∼ = A4. In particular, the previous two paragraphs show that b = h = 1 and g = 0 and f = a, while the above paragraph for δ ̸= 1 shows a = f = g = 0. In this case, Aut(X) ∼ = SL2(3).
Cases (1) (b) and (1) (c) In these cases, the singularity of R over [1 : 0 : 0] is not isomorphic to the other singularities of R, hence G(X) is a subgroup of C3 acting through the subgroup of H with γ = 0. In particular, G(X) fixes the points [1 : 0 : 0] and [0 : 1 : 0]. Since the number of singular points of R that lie over points different from [1 : 0 : 0] and [0 : 1 : 0] is not divisible by 3, G(X) fixes all of them, hence G(X) is trivial.
Case (1) (d) In this case, R has singularities over [1 : 0 : 0] and [1 : c 1 4 : c 3 4 ]. An element of H that fixes both of these points is trivial, and the unique one that swaps the two points is of the form (b2, σ) where σ acts as v 7→v + c 1 4 u and b2 = c 3 4 u2 + c 1 2 uv + c 1 4 v2. If such an element lies in G(X), then there exist polynomials b1 and b3 such that 0 = (ac 1 2 + c 3 4 )u2 + ub1 + b2 1 + c 1 2 uv + c 1 4 v2 0 = v3b1 + ub3 + cu2v2 + c 1 2 v4, hence b1 = λu + c 1 2 v with λ2 + λ = ac 1 2 + c 3 4 and c4 = c, and b3 = λv3 + cuv2. By Theorem 5.6 we have c ̸= 0, hence we can apply an element of H to assume that c = 1. Plugging this into the equation for σ∗a6 and comparing coefficients in Lemma 6.6, we obtain the conditions h = 0 and g = a + f. The square of this automorphism (b1, b2, b3, σ) is the Bertini involution, hence Aut(X) ∼ = C4 in this case.
Case (1) (e) In this case, we have G(X) ⊆C3, since G(X) fixes [1 : 0 : 0]. Non-trivial elements of H that fix [1 : 0 : 0] are of the form (0, σ), where σ acts as v 7→δv with δ3 = 1 and δ ̸= 1. Such an automorphism lifts to X if and only if there exist polynomials b1 and b3 such that (1 + δ2)av2 = ub1 + b2 1 0 = v3b1 + ub3 (1 + δ2)fu4v2 + (1 + δ)gu2v4 = v3b3 + b2 3.
The first equation implies a = 0 and b1 = λu with λ2 + λ = 0 and then the second equation implies that also b3 = λv3. Finally, the third equation shows f = g = 0.
Case (2) (a) Here, G(X) ⊆H fixes the point [0 : 1 : 1]. Moreover, if G(X) fixes the images of the other two singularities, then, by our description of H, G(X) is trivial. Hence, G(X) ⊆C2 with equality if and only if G(X) contains the involution (0, σ), where σ acts as v 7→v + e 1 2 g−1 2 u.
If this involution is in G(X), then there exist polynomials b1 and b3 such that aeg−1u2 = ub1 + b2 1 e2g−2u4 = ub3, AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 35 hence b1 = λu with λ2 + λ = aeg−1, and b3 = e2g−2u3. Plugging this into the equation for σ∗a6 and comparing coefficients in Lemma 6.6, we obtain the conditions 0 = e4 + he3g + fe2g2 + deg3 0 = e 1 2 (g 3 2 + he 1 2 ).
Since e ̸= 0 by Theorem 5.6, we have h = e−1 2 g 3 2 and d = efg−1 + e 3 2 g−1 2 + e3g−3. Note that both lifts of (0, σ) have order 2, hence Aut(X) ∼ = 22.
Case (2) (b) and (2) (c) Here, G(X) ⊆H fixes [0 : 1 : 1] and [1 : 0 : 0], since these are the points that lie under the singularities of the irreducible components of R, but not under the intersection of the two components R1 and R2. By our description of H in Lemma 5.4, this implies that G(X) is trivial.
Case (2) (d) In this case, G(X) fixes [0 : 1 : 1], but we get no other restrictions from the position of the singularities of R. Therefore, an element of G(X) ⊆H is of the form (0, σ) where σ acts as v 7→v + γu for some γ ∈k. Such an element is in G(X), if and only if there exist polynomials b1 and b3 such that aγ2u2 = ub1 + b2 1 γ4u4 = ub3 (cγ + dγ2 + fγ4 + hγ6)u6 + hγ4u4v2 + hγ2u2v4 = b2 3.
Such b1 and b3 exist if and only if h = 0 and γ8 + hγ6 + fγ4 + dγ2 + cγ = 0, and then b1 = λu with λ2 + λ = aγ2, and b3 = γ4u3. By Theorem 5.6, we have c ̸= 0, hence, as soon as h = 0, there are exactly 8 choices for γ. This shows G(X) ∼ = 23. Every lift of every non-trivial element in G(X) has order 2, hence Aut(X) ∼ = 24.
Case (2) (e) Here, the elements of G(X) ⊆H fix [0 : 1 : 0] and preserve the pair {[1 : 0 : 0], [1 : 1 : 0]}. Using our description of γ, it is clear that an element of H that fixes all of these three points is the identity.
An element that swaps [1 : 0 : 0] and [1 : 1 : 0] is of the form (0, σ), where σ acts as v 7→v + u.
Such an element is in G(X), if and only if there exist polynomials b1 and b3 such that au2 = ub1 + b2 1 0 = ub3 (d + f + h)u6 + (e + h)u4v2 + (e + h)u2v4 = b2 3, hence if and only if h = e and d = e + f, and then b1 = λu with λ2 + λ = a and b3 = 0. The square of the lift of this automorphism to Aut(X) is the identity, hence Aut(X) ∼ = 22.
Case (2) (f) In this case, G(X) ⊆H fixes [1 : 0 : 0] and [0 : 1 : 0]. Hence, by our description of H in Lemma 5.4, every element in G(X) is of the form (0, σ), where σ acts as v 7→δv for some δ ∈k×. A non-trivial element of this form is in G(X) if and only if there exist b1 and b3 such that a(1 + δ2)v2 = ub1 + b2 1 0 = ub3 d(1 + δ2)u4v2 + f(1 + δ4)u2v4 + (1 + δ5)uv5 + h(1 + δ6) = b2 3.
36 IGOR DOLGACHEV AND GEBHARD MARTIN Hence, we always have b1 = b3 = 0 and δ5 = 1. Since δ ̸= 1 by assumption, we deduce that (0, σ) lifts if and only if a = d = f = h = 0.
Case (3) Here, the group G(X) ⊆H fixes [1 : 0 : 0], [0 : 1 : 0], and [1 : 1 : 0]. Hence, every element of G(X) is of the form (0, σ) and σ satisfies the conditions of Lemma 5.4 (3).
First, assume that σ has even order and interchanges two components of B. Without loss of generality, we may assume that σ swaps u and v. Then, (0, σ) lifts to X if and only if there exist b1 and b3 such that 0 = b2 1 (b + c)(u3v + uv3) = uv(u + v)b1 (d + f)(u5v + uv5) = uv(u + v)b3 + b2 3.
This holds if and only if b = c, and then b1 = 0 as well as b3 = λuv(u + v) with λ2 + λ = 0 and d = f. The square of both lifts of (0, σ) is the identity, hence they generate a group isomorphic to 22.
Next, assume that σ is non-trivial and preserves the three components of B. Then, it acts as u 7→αu, v 7→αv, where α3 = 1, α ̸= 1. This automorphism lifts to X if and only if there exist polynomials b1 and b3 such that a(1 + α−1)uv = b2 1 (1 + α)(bu3v + (b + c)u2v2 + cuv3) = uv(u + v)b1 0 = uv(u + v)b3 + b2 3, hence if and only if a = b = c = 0.
Finally, assume that σ has odd order and interchanges components of B. Without loss of generality, we may assume that σ acts as u 7→βv, v 7→β(u + v) with β3 = 1. This lifts to X if and only if there exist b1 and b3 such that a(1 + β2)uv + aβ2v2 = b2 1 (b + βc)u3v + (b + c + βb)u2v2 + (c + β(b + c))uv3 = uv(u + v)b1 (d + f)u5v + fu4v2 + eu2v4 + (d + e)uv5 + (d + e + f)v6 = uv(u + v)b3 + b2 3.
The third equation implies f = d and d = e + e 1 2 and then b3 = λu2v + λuv2 + e 1 2 v3, where λ2 + λ = e + e 1 2 . If β = 1, the first equation implies b1 = a 1 2 v and the second equation implies b = c = a 1 2 . If β ̸= 1, the first equation implies b1 = a = 0 and the second equation implies b = βc.
Case (4) In this case, the group G(X) ⊆H fixes [1 : 0 : 0] and [0 : 1 : 0], hence every element of G(X) is of the form (b2, σ) with b2 = λuv for some λ ∈k and where σ acts as u 7→αu, v 7→α−2v with α ∈k×.
If such an automorphism lifts to X, then the condition σ∗a2 = a2 + b2 1 + b2 forces b2 = b2 1, hence b1 = b2 = 0. The other conditions of Lemma 6.6 become a(1 + α)u3v + b(1 + α−2)u2v2 + c(1 + α−5)uv3 = 0 d(1 + α3)u5v + e(1 + α−3)u3v3 + f(1 + α−9)uv5 = u2vb3 + b2 3.
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 37 Since d ̸= 0, the second equation implies α3 = 1. Hence, if σ is non-trivial, then (0, σ) lifts to X if and only if a = b = c = 0.
Case (5) Here, G(X) ⊆H fixes [0 : 1 : 0], hence every element of G(X) is of the form (b2, σ) with b2 = λu2 + µuv for some λ, µ ∈k and where σ acts as u 7→αu, v 7→γu + δv with α3 = 1, γ ∈ k, δ ∈k×.
If such an automorphism lifts to X, then there exists b1 with b2 1 + b2 = 0, hence µ = 0 and b1 = λ 1 2 u. Comparing coefficients in the equation for σ∗a4, we obtain λ2 + λ 1 2 + aγ + bα2γ2 + cαγ3 = 0 (10) a + aδ + cαδγ2 = 0 (11) b + bα2δ2 + cαδ2γ = 0 (12) c + cαδ3 = 0 (13) The automorphism lifts to X if and only if, additionally, there exists a b3 = λ0u3 + λ1u2v + λ2uv2 + λ3v3 satisfying the following conditions: λ2 0 + λ0 = λ3 + (aγ + bα2γ2 + cαγ3)λ + αγ5 + dγ6 λ1 = (aδ + cαδγ2)λ + αδγ4 λ2 1 + λ2 = (bα2δ2 + cαδ2γ)λ + dδ2γ4 λ3 = cαδ3λ (14) λ2 2 = αδ4γ + dδ4γ2 0 = 1 + αδ5 (15) λ2 3 = d + dδ6 (16) Equation (15) shows that α = δ−5. In particular, as α3 = 1, we have δ15 = 1.
First, assume that δ = 1, hence α = 1. Then, Equation (16) shows that λ3 = 0. Equation (14) shows cλ = 0 and Equation (11) shows cγ = 0. Hence, if c ̸= 0, then (b2, σ) is the identity, so we assume c = 0 in the following. Let Ga,b,d be the group of lifts of such automorphisms to X. By the description above, these Ga,b,d form a family Ga,b,d of finite group schemes over Spec k[a, b, d] cut out in Spec k[a, b, d, λ, λ0, γ] by the equations F1 := λ4 + λ + a2γ2 + b2γ4 = 0 (17) F2 := a4λ4 + b2λ2 + γ + dγ2 + d2γ8 + γ16 = 0 (18) F3 := λ2 0 + λ0 + λ3 + (aγ + bγ2)λ + γ5 + dγ6 = 0.
In the following, we show that all geometric fibers of Ga,b,d →Spec k[a, b, d] are reduced of length 128. In particular, Ga,b,d is étale over Spec k[a, b, d], hence all the Ga,b,d are isomorphic and we will show afterwards that Ga,b,d ∼ = 21+6 + .
• If a ̸= 0 and b ̸= a2, we argue as follows: The condition a8F 2 1 + F 2 2 + b4 a4 F2 = 0 yields the following expression for λ: (a12 + b6)λ2 = b4γ + (a4 + b4d)γ2 + (a4d2 + b4d2 + a16)γ4 +(a12b4 + b4d2)γ8 + (a4d4 + b4)γ16 + a4γ32 By our assumptions, we can divide by (a12 + b6) and we obtain an expression of λ2 in terms of γ. Plugging this back into Equation (18), we obtain a polynomial F in γ of the 38 IGOR DOLGACHEV AND GEBHARD MARTIN form F = P5 i=0 ciγ2i of degree 64 with c0 = 0 c1 = a12 a12 + b6 c5 = a8 (a12 + b6)2 .
Since a ̸= 0, both c1 and c5 are non-zero, so ∂γF = 1 and F has only simple roots. Hence, there are exactly 64 choices for γ such that (b2, σ) lifts and λ is uniquely determined by γ.
In particular, Ga,b,d has order 128 and it acts on the base of the associated elliptic fibration through 26.
• If a ̸= 0 and b = a2, we argue as follows: The condition a8F 2 1 + F 2 2 + a4F2 = 0 becomes 0 = a8γ + (a4 + a8d)γ2 + (a4d2 + a8d2 + a16)γ4 +(a20 + a8d2)γ8 + (a4d4 + a8)γ16 + a4γ32 =: F.
Note that, since a ̸= 0, F1 = F2 = 0 holds if and only if F2 = F = 0. There are 32 choices for γ with F(γ) = 0 and for each choice of γ, there are exactly 2 choices for λ such that F2(γ, λ) = 0. As in the previous case, Ga,b,d has order 128, but in this case, it acts on the base of the associated elliptic fibration through 25.
• Next, assume that a = 0 and b ̸= 0. We can immediately solve Equation (18) for λ and obtain b2λ2 = γ + dγ2 + d2γ8 + γ16.
Plugging this into the square of Equation (17), we obtain a polynomial F in γ of the form F = P5 i=0 ciγ2i of degree 64 with c0 = 0 c1 = b−2 c5 = b−8.
Hence, there are 64 choices for γ such that (b2, σ) lifts and λ is uniquely determined by γ. Therefore, Ga,b,d has order 128 and acts on the base of the associated elliptic fibration through 26.
• Now, assume that a = b = 0. The equations simplify to λ4 + λ = 0 λ2 0 + λ0 = λ3 + γ5 + dγ6 λ1 = γ4 λ2 = dγ4 + γ8 λ3 = 0 γ + dγ2 + d2γ8 + γ16 = 0.
Hence, there are 16 choices for γ and 4 choices for λ. Hence, Ga,b,d has order 128 and it acts on the base of the associated elliptic fibration through 24.
It remains to determine the group Ga,b,d. By the last bullet point, the subgroup of G0,0,0 of automorphisms that act trivially on the base of the associated elliptic fibration has order 8. Thus, by Corollary 6.3, it is isomorphic to Q8. Hence, every Ga,b,d contains a quaternion group Q8 with AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 39 β ∈Q8. On the other hand, in the cases where a ̸= 0, b ̸= a2, we have seen that Ga,b,d/⟨β⟩∼ = 26.
Hence, by Example 6.5, we have Ga,b,d ∼ = 21+6 + .
Next, assume that δ ̸= 1, δ3 = 1. Then, Equations (11), (12), and (13) show that a = b = c = 0.
The remaining equations become λ4 + λ = 0 λ2 0 + λ0 = λ3 + δγ5 + dγ6 λ1 = δ2γ4 λ2 = dδ2γ4 + δγ8 λ3 = 0 δ2γ16 + d2δγ8 + dδγ2 + δ2γ = 0.
We see that if γ = λ = 0, then (b2, σ) admits a lift to X as an automorphism g of order 3. For a fixed γ, there are at most 128 possible choices of (γ, λ). All of them are obtained by composing g with an element of G0,0,d, hence all choices are realized.
Finally, assume that δ ̸= 1, δ5 = α = 1. As in the previous paragraph, we have a = b = c = 0.
But in this case, Equation (16) yields the condition d = 0.
So, in summary, if c ̸= 0, then G(X) is trivial and if c = 0, then Aut(X) admits a unique 2-Sylow subgroup isomorphic to 21+6 + . If a, b, or c is non-zero, this is the full automorphism group. If a = b = c = 0 and d ̸= 0, then Aut(X)/21+6 + ∼ = C3 and if a = b = c = d = 0, then Aut(X)/21+6 + ∼ = C15.
□ Remark 6.9. The largest order of an automorphism group of a del Pezzo surface of degree one over the complex numbers is equal to 144 and the surface with such a group of automorphisms is unique . In our case, the maximal order is equal to 1920 = 27 · 15 and the surface with such an automorphism group is also unique. We also see the occurrence of the group G = 24 in Case (5). It is obtained as the pre-image in 21+6 + of a maximal isotropic subspace of F6 2. Since del Pezzo surfaces of degree one are superrigid (see [8, Definition 7.10, Corollary 7.11]) and the corresponding G-surface is minimal, this group is not conjugate in the Cremona group of P2 to the isomorphic subgroup of the group of automorphisms of del Pezzo surfaces of degree 4 or 2 that appeared in and .
6.3. Conjugacy classes and comparison with the classification in characteristic 0. In this section, we determine the conjugacy classes in W(E8) of the elements of the groups that occur in Theorem 6.8 and, whenever possible, compare the surfaces in Theorem 6.8 with their counterparts in characteristic 0 (see [9, Table 8.14]).
For a del Pezzo surface X of degree 1, we denoted by NX and PX the kernel and image of the morphism Aut(X) →Aut(P1) induced by the action of Aut(X) on the base of the associated elliptic pencil. First, we note the following: Lemma 6.10. Let g be a non-trivial element of NX. Then, the conjugacy class of g is either 8A1, 4A2, 2D4(a1), or E8(a8).
Proof. Since g acts trivially on the base of the pencil, it cannot preserve any (−1)-curve on X.
Then, the lemma follows from the classification of conjugacy classes in W(E8) (see e.g. [12, Table 3]), by checking which of them fix no (−1)-class in E8.
□ 40 IGOR DOLGACHEV AND GEBHARD MARTIN Corollary 6.11. Let X be a del Pezzo surface of degree 1 in characteristic 2. Let X′ be a geometric generic fiber of a lift of X to characteristic 0 and let sp : Aut(X′) →Aut(X) be the specialization map. Then, the following hold: (1) sp is injective and preserves conjugacy classes.
(2) sp induces morphisms NX′ →NX and PX′ →PX.
(3) The kernel H of PX′ →PX is an elementary 2-group and if g is an element of Aut(X′) that maps to a non-trivial element of H, then the conjugacy class of g is 2D4(a1).
Proof. The proof of Claim (1), including the existence of sp, is analogous to Lemma 4.7.
The existence of the morphisms in Claim (2) is clear, as, for a given lift X →S, the groups that appear can be defined as fibers of kernel and image of the homomorphism of S-group schemes AutX/S →AutP1 S/S that describes the action of AutX/S on the anti-canonical system.
For Claim (3), recall that sp preserves conjugacy classes by Claim (1). Therefore, by Lemma 6.10, all non-trivial elements of H are represented by elements g of Aut(X′) of conjugacy class 8A1, 4A2, 2D4(a1), or E8(a8). If g is of class 8A1, then it is the Bertini involution, hence g ∈NX′.
If g is of class 4A2, then it has negative trace on E8, so, by the Lefschetz fixed point formula, it must act trivially on the base of the elliptic pencil. Hence, g ∈NX′. If g is of class E8(a8), then, by what we just proved, g2 and g3 are in NX′, hence g ∈NX′. Thus, g must be of conjugacy class 2D4(a1). Then, g2 is the Bertini involution, so H is 2-elementary.
□ By Theorem 6.8, | Aut(X)| ≤36 or | Aut(X)| ∈{128, 384, 1920}, so Types I and II [9, Table 8.14] do not have a reduction modulo 2 which is a del Pezzo surface.
The surfaces of Type VI, VII, IX, XII, and XV from [9, Table 8.14] admit an automorphism of order 2n with n > 1 acting faithfully on P1, which is impossible in characteristic 2, so by Corollary 6.11 they do not have good reduction mod 2.
The equation of the surfaces of Type (3) (v) in Theorem 6.8 can be rewritten as y2 + uv(u −v)y + x3 + a(u2 −uv + v2)3 + bu2v2(u −v)2 for certain a, b ∈k. This equation makes sense in characteristic 0, and it is stable under S3-action generated by (u, v, x, y) 7→(v, u, x, −y) and (u, v, x, y) 7→(u −v, −u, x, −y) as well as the C3-action (u, v, x, y) 7→(u, v, ζ3x, y), where ζ3 is a primitive 3-rd root of unity. Hence, the automorphism group of has order at least 36, hence it is isomorphic to 6 × S3. Thus, surfaces of Type (3) (v) are reductions mod 2 of the surfaces of Type III from [9, Table 8.14]. In particular, we can read off the conjugacy classes from [12, Table 8].
The equation of Type (5) (iii) makes sense in characteristic 0, where it is isomorphic to y2 + x3 + u(u5 + v5), which is the equation of Type IV in [9, Table 8.14].
The equation of the surfaces of Type (3) (ii) in Theorem 6.8 can be rewritten as y2 + uv(u −v)y + x3 + c(u2 −uv + v2)x2 + a(u2 −uv + v2)3 + bu2v2(u −v)2 for certain a, b, c ∈k. Similar to the case of Type (3) (v) above, these equations are stable under a S3-action, both in characteristic 0 and in characteristic 2. In characteristic 0, these equations can be simplified to the normal forms of Type X from [9, Table 8.14].
The equation of the surfaces of Type (3) (iii) makes sense in characteristic 0, where it defines a lift of X together with the action of Aut(X). Both X and the lift admit an automorphism of order 6 that acts trivially on the base of the elliptic pencil. Hence, these surfaces are reductions mod 2 of the surfaces of Type XI from [9, Table 8.14].
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 41 The equations of the surfaces of Type (2) (f) (i) in Theorem 6.8 define a 1-dimensional family of surfaces in characteristic 0 with an action of C10. These lifts must be of Type XIII [9, Table 8.14].
The equations of the surfaces of Type (4) (i) in Theorem 6.8 define a 2-dimensional family of surfaces in characteristic 0 with an action of C6 that is trivial on the base of the elliptic pencil.
Hence, these lifts are of Type XVII [9, Table 8.14].
Next, consider the equations y2 + (aux + bu3 + cv3)y + x3 + (du4 + euv3)x + fu6 + gu3v3 + hv6, where a, b, c, d, e, f, g, h are parameters. In characteristic 0, we can simplify this equation to the normal form of Type XVIII from [9, Table 8.14]. In characteristic 2, these equations cover three of the families of Theorem 6.8: If a, c ̸= 0, we can simplify the equation to the normal form for Type (1) (a) (ii) which, in turn, specializes to Type (1) (e) (i) for special values of the parameters. If a = 0 but b, c ̸= 0, we can simplify the equation to y2 + (u3 + v3)y + x3 + euv3x + fu6.
This is an alternative normal form for our surfaces of Type (3) (iv).
Finally, consider the equations y2 + (a(u + v)x + b(u + v)3 + cuv(u + v))y + x3 + (d(u + v)4 + euv(u + v)2 + fu2v2)x + (g(u + v)6 + huv(u + v)4 + iu2v2(u + v)2 + ju3v3).
In characteristic 0, we can simplify this equation to the normal form of Type XX from [9, Table 8.14]. In characteristic 2, these equations cover four of the families of Theorem 6.8: If a ̸= 0, we can simplify the equation to y2 + (u + v)xy + x3 + cuvx2 + du2v2x + g(u + v)6 + huv(u + v)4 + iu2v2(u + v)2 + ju3v3.
If d, j ̸= 0, we can rescale one of them to 1 and obtain an alternative normal form for Type (2) (a) (i). If d ̸= 0 and j = 0, we obtain a normal form for Type (2) (e) (i). If j ̸= 0 and d = 0, we obtain an alternative normal form for Type (2) (d) (i). Note that d = j = 0 would lead to a singular surface. Since the family of Type (2) (d) (i) occurs as a reduction mod 2 of certain surfaces of Type XX from [9, Table 8.14], we call them Type XX’.
If a = 0 and c ̸= 0, we can simplify the equation to y2 + (b(u + v)3 + uv(u + v))y + x3 + (euv(u + v)2 + fu2v2)x + (g(u + v)6 + ju3v3).
This defines a 4-dimensional family of surfaces with 22-action (one parameter is redundant). By Theorem 6.8, the corresponding surfaces must be of Type (3) (i).
The surfaces in the families (1)(a)(i), (1)(a)(iii), (1)(a)(iv), (1)(d)(i), (5)(i), (5)(ii), and (5)(iii) admit an automorphism of order 4 and it turns out that writing down integral equations for such automorphisms similar to the ones above is hard. So, instead, to determine the conjugacy classes of the automorphisms of this family and to compare with the classification in characteristic 0, we will use the following observation.
Lemma 6.12. Let g be an automorphism of a del Pezzo surface of degree one. Let m := ord(g) and let n be the order of the induced automorphism of P1. Assume that m is even. Then, the conjugacy class Γ of g in W(E8) is one of the following: (1) If (m, n) = (2, 1), then Γ = 8A1.
(2) If (m, n) = (2, 2), then Γ = 4A1.
(3) If m = 4, then Γ = 2D4(a1).
(4) If (m, n) = (6, 1), then Γ = E8(a8).
42 IGOR DOLGACHEV AND GEBHARD MARTIN (5) If (m, n) = (6, 2), then Γ = E6(a2) + A2.
(6) If (m, n) = (6, 3) and g2 is of class 3A2, then Γ = E7(a4) + A1.
(7) If (m, n) = (6, 3) and g2 is of class 2A2, then Γ = 2D4.
(8) If m = 10, then Γ = E8(a6).
(9) If m = 12, then Γ = E8(a3).
(10) If m = 20, then Γ = E8(a2).
(11) If m = 30, then Γ = E8.
Proof. By Theorem 6.8, we know that the only possible values for m and n are the ones in the statement.
In Case (1), g is the Bertini involution, hence Γ = 8A1. In Case (2), we may assume that g acts as u ↔v. Then, we proved in this section that g lifts to characteristic 0, so by [12, Table 8], Γ = 4A1. In Case (3), g2 is the Bertini involution, because PGL2(k) does not contain elements of order 4, and it is known (see [12, Table 3]) that the only conjugacy class of automorphisms of order 4 whose square is the Bertini involution is 2D4(a1). In Cases (8) and (11), g m 2 is the Bertini involution and g2 lifts to characteristic 0, hence g lifts to characteristic 0 and we can read off the conjugacy class Γ from [12, Table 8]. Then, we deduce Case (10) from Case (8). In Case (9), g2 must be of type E8(a8), since PGL2(k) does not contain any elements of order 4 or 6. Then, from [12, Table 3], we see that Γ = E8(a3). Finally, Cases (4), (5), (6), and (7) follow from [12, Table 3] by comparing the conjugacy classes of g2 and g3.
□ Now, we can complete Table 4 by using the description of Aut(X) in Theorem 6.8. We observe that the conjugacy classes for Types (1)(a)(i) and (1)(d)(i) are the same as for Type XIX from [9, Table 8.14], the conjugacy classes for Type (1)(a)(iii) are the same as for Type XIV from [9, Table 8.14], and the conjugacy classes for Type (1)(a)(iv) are the same as for Type V from [9, Table 8.14]. The only groups in Theorem 6.8 that contain D8 are 21+6 + , 21+6 + : 3, and 21+6 + : 15, and the only group that contains an automorphism of order 20 is 21+6 + : 15. Hence, if the Types XVI, M, and VIII from [9, Table 8.14] and [12, Table 8] have good reduction modulo 2, then they must reduce to our Types (5)(i) and (5)(ii), respectively. In each of these cases, we determine the conjugacy classes using Lemma 6.12.
We summarize the classification of automorphism groups of del Pezzo surfaces of degree one in Table 3 in the Appendix. There, in the first column, we give the name of the corresponding family, both in the notation of Theorem 4.3 and in the notation of [9, Table 8.14]. The second and third columns give the group Aut(X) and its size. In the remaining columns, we list the number of elements of a given Carter conjugacy class in Aut(X).
AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 43 APPENDIX Name Aut(X) Order id 2A1 4A1 A2 A2 + 2A1 A3 A3 + A1 A4 D4 D4(a1) D5 (φ, φ) 24 : A5 960 1 70 5 80 80 120 384 160 60 (ζ3, ζ3) (same as (φ, φ)) (i, i) (does not exist) (a, a) 24 : 22 64 1 22 5 24 12 general 24 16 1 10 5 TABLE 1. Automorphism groups of quartic del Pezzo surfaces, cf. Section 2.2 Name Aut(X) Order id 2A1 4A1 A2 A2 + 2A1 2A2 3A2 A3 + A1 A4 A5 + A1 D4 D4(a1) D5 E6 E6(a1) E6(a2) I / 3C PSU4(2) 25920 1 270 45 240 2160 480 80 3240 5184 1440 1440 540 4320 5760 720 II / 5A (same as V) III / 12A (same as I) IV / 3A H3(3) : 2 54 1 9 24 2 18 V / 4B 23 : S4 192 1 30 13 32 72 32 12 VI / 6E (same as V) VII / 8A (does not exist) VIII / 3D S3 6 1 3 2 IX / 4A (same as V) X / 2B 24 16 1 10 5 XI / 2A 2 2 1 1 XII / 1A 1 1 1 TABLE 2. Automorphism groups of cubic del Pezzo surfaces, cf. Section 2.3 44 IGOR DOLGACHEV AND GEBHARD MARTIN Name Aut(X) Order id 3A1 4A1 7A1 2A2 3A2 2A3 2A3 + A1 A5 + A2 A6 D4(a1) D4(a1) + A1 D5 D5 + A1 D6(a2) + A1 E6 E6(a1) E6(a2) E7 E7(a1) E7(a2) E7(a4) I - V (do not exist) VI / (3)(ii) 18 18 1 1 2 6 6 2 VII (does not exist) VIII / (2)(a)(iii) 2 × 6 12 1 1 1 1 2 2 2 2 IX / (1)(a)(ii) 2 × S3 12 1 3 3 1 2 2 X / (1)(c)(i) 24 16 1 7 7 1 XI / (2)(a)(ii), (3)(i) 6 6 1 1 2 2 XII / (1)(a)(i), (2)(a)(i) 22 4 1 1 1 1 XIII 2 2 1 1 TABLE 3. Automorphism groups of del Pezzo surfaces of degree 2, cf. Section 4.3 Name Aut(X) Order id 4A1 8A1 2A2 3A2 4A2 2A3 + A1 2A4 A5 + A2 + A1 D4 + A2 2D4 D4(a1) + A1 2D4(a1) D8(a3) E6 + A1 E6(a2) E6(a2) + A2 E7(a2) E7(a4) + A1 E8 E8(a1) E8(a2) E8(a3) E8(a5) E8(a6) E8(a8) I - II (do not exist) III / (3)(v) 6 × D6 36 1 6 1 2 4 2 2 12 4 2 IV / (5)(iii) 21+6 + : 15 1920 1 70 1 8 64 56 80 512 384 160 512 64 8 M / (5)(ii) 21+6 + : 3 384 1 70 1 8 56 80 160 8 V / (1)(a)(iv) SL2(3) 24 1 1 8 6 8 VI - VII (do not exist) VIII (same as IV) IX (does not exist) X / (3)(ii) D12 12 1 6 1 2 2 XI / (3)(iii) 2 × 6 12 1 2 1 2 4 2 XII (does not exist) XIII / (2)(f)(i) 10 10 1 1 4 4 XIV / (1)(a)(iii) Q8 8 1 1 6 XV (does not exist) XVI / (5)(i) 21+6 + 128 1 70 1 56 XVII / (4)(i) 6 6 1 1 2 2 XVIII / (1)(a)(ii), (1)(e)(i), (3)(iv) 6 6 1 1 2 2 XIX / (1)(a)(i), (1)(d)(i) 4 4 1 1 2 XX / (2)(a)(i), (2)(e)(i), (3)(i) 22 4 1 2 1 XX’ / (2)(d)(i) 24 16 1 14 1 XXI 2 2 1 1 TABLE 4. Automorphism groups of del Pezzo surfaces of degree 1, cf. Section 6.3 AUTOMORPHISMS OF DEL PEZZO SURFACES IN CHARACTERISTIC 2 45 REFERENCES M. Aschbacher, Finite group theory, Second edition. Cambridge studies in Advanced Mathematics, 10. Cambridge University Press, Cambridge, 2000. xii+304 pp.
J. Conway, R. Curtis, S. Norton, R. Parker, R. Wilson, Atlas of finite groups, Oxford Univ. Press. 1985.
W. Barth, Moduli of vector bundles on the projective plane. Inventiones Mathematicae 42 (1977), 63–91.
A. Beauville, Prym Varieties and the Schottky Problem. Inventiones Mathematicae 41 (1977), 149–196.
S. Bosch, W. Lütkebohmert, M. Raynaud, Néron models. Ergebnisse der Mathematik und ihrer Grenzgebiete (3), 21. Springer-Verlag, Berlin, 1990.
R. Carter, Conjugacy classes in the Weyl group. Compositio Mathematica, Volume 25 (1972) no. 1, pp. 1-59 M. Demazure, Surfaces de del Pezzo, I-V, in “Séminaire sur les Singularités des Surfaces”, ed. by M. Demazure, H.
Pinkham and B. Teissier. Lecture Notes in Mathematics, 777. Springer, Berlin, 1980, pp. 21–69.
I. Dolgachev, V. Iskovskikh, Finite subgroups of the plane Cremona group, Algebra, arithmetic, and geometry: in honor of Yu. I. Manin. Vol. I, 443–548, Progr. Math., 269, Birkhäuser Boston, Inc., Boston, MA, 2009.
I. Dolgachev, Classical algebraic geometry:a modern view, Cambridge Univ. Press, 2012.
I. Dolgachev, Finite subgroups of the plane Cremona group, Algebraic geometry in East Asia—Seoul 2008, 1-–49, Adv. Stud. Pure Math., 60, Math. Soc. Japan, Tokyo, 2010.
I. Dolgachev, A. Duncan, Automorphisms of cubic surfaces in positive characteristic. Izv. Ross. Akad. Nauk Ser.
Mat. 83 (2019), no. 3, 15–92.
I. Dolgachev, G. Martin, Automorphisms of del Pezzo surfaces in odd characteristic, arXiv:2303.16170.
T. Ekedahl, Canonical models of surfaces of general type in positive characteristic. Publications Mathématiques de l’IHÉS, Tome 67 (1988), 97–144.
N. Katz, P. Sarnak Random matrices, Frobenius eigenvalues, and monodromy, American Mathematical Society Colloquium Publications, Volume 45, 1999, xii+419 pp.
W. Lang, Configurations of singular fibres on rational elliptic surfaces in characteristic two. Special issue in honor of Robin Hartshorne. Comm. Algebra 28 (2000), no. 12, 5813-–5836.
M. Neusel, L. Smith, Invariant theory of finite groups. Mathematical Surveys and Monographs, 94. American Mathematical Society, Providence, RI, 2002.
K. Oguiso, T. Shioda, The Mordell-Weil lattice of a rational elliptic surface. Comment. Math. Univ. St. Paul. 40 (1991), no. 1, 83–99.
R. Richardson, Conjugacy classes of involutions in Coxeter groups. Bull. Austral. Math. Soc. 26 (1982), 1–15.
DEPARTMENT OF MATHEMATICS UNIVERSITY OF MICHIGAN EAST HALL, 530 CHURCH STREET, ANN ARBOR, MI 48109-1043 USA Email address: idolga@umich.edu MATHEMATISCHES INSTITUT UNIVERSITÄT BONN ENDENICHER ALLEE 60 53115 BONN GERMANY Email address: gmartin@math.uni-bonn.de |
4166 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10864763/ | Imaging of pancreatic ductal adenocarcinoma – An update for all stages of patient management - PMC
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. 2024 Feb 8;12:100553. doi: 10.1016/j.ejro.2024.100553
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Imaging of pancreatic ductal adenocarcinoma – An update for all stages of patient management
Carlos Bilreiro
Carlos Bilreiro
a Radiology Department, Champalimaud Foundation, Lisbon, Portugal
b Champalimaud Research, Champalimaud Foundation, Lisbon, Portugal
c Nova Medical School, Lisbon, Portugal
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a,b,c,⁎, Luísa Andrade
Luísa Andrade
a Radiology Department, Champalimaud Foundation, Lisbon, Portugal
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a, Inês Santiago
Inês Santiago
a Radiology Department, Champalimaud Foundation, Lisbon, Portugal
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a, Rui Mateus Marques
Rui Mateus Marques
c Nova Medical School, Lisbon, Portugal
d Radiology Department, Hospital de S. José, Lisbon, Portugal
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c,d, Celso Matos
Celso Matos
a Radiology Department, Champalimaud Foundation, Lisbon, Portugal
b Champalimaud Research, Champalimaud Foundation, Lisbon, Portugal
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a,b
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a Radiology Department, Champalimaud Foundation, Lisbon, Portugal
b Champalimaud Research, Champalimaud Foundation, Lisbon, Portugal
c Nova Medical School, Lisbon, Portugal
d Radiology Department, Hospital de S. José, Lisbon, Portugal
⁎
Corresponding author at: Radiology Department, Champalimaud Foundation, Lisbon, Portugal. carlos.bilreiro@fundacaochampalimaud.pt
Received 2023 Oct 15; Revised 2024 Feb 2; Accepted 2024 Feb 3; Collection date 2024 Jun.
© 2024 The Authors. Published by Elsevier Ltd.
This is an open access article under the CC BY-NC-ND license (
PMC Copyright notice
PMCID: PMC10864763 PMID: 38357385
Abstract
Background
Pancreatic ductal adenocarcinoma (PDAC) is a common and lethal cancer. From diagnosis to disease staging, response to neoadjuvant therapy assessment and patient surveillance after resection, imaging plays a central role, guiding the multidisciplinary team in decision-planning.
Review aims and findings
This review discusses the most up-to-date imaging recommendations, typical and atypical findings, and issues related to each step of patient management. Example cases for each relevant condition are presented, and a structured report for disease staging is suggested.
Conclusion
Despite current issues in PDAC imaging at different stages of patient management, the radiologist is essential in the multidisciplinary team, as the conveyor of relevant imaging findings crucial for patient care.
Keywords: Pancreatic carcinoma; Neoplasm staging; Neoadjuvant therapy; Patient care team; Neoplasm recurrence, Local
Highlights
•
Imaging is crucial for patient management in pancreatic ductal adenocarcinoma.
•
Contrast-enhanced CT is preferred for diagnosis and staging; MRI for hepatic staging.
•
Imaging performs poorly for lymph node staging and response to neoadjuvant therapy.
•
Anatomic findings and disease biology should both be considered for patient staging.
•
A multidisciplinary team is essential for obtaining the best patient outcomes.
1. Introduction
Pancreatic ductal adenocarcinoma (PDAC) is one of the leading causes of cancer-related death in the modern world, with dismal survival rates . Resection remains the only potentially curative option and, combined with systemic chemotherapy, provides the best therapeutic intervention for improving survival , , . However, tumor resectability relies on an early diagnosis, which is often not possible, due to patients presenting with advanced disease and also diagnostic imaging challenges to detect subclinical disease . Furthermore, defining which patients are best suited for surgery, and which patients should be offered neoadjuvant or palliative therapy is not a simple task. Several staging systems have been developed, with different resectability criteria, and are used in different institutions; the National Cancer Comprehensive Network (NCCN) is the most widely used classification , , . Lastly, there are unresolved issues and difficulties when evaluating patients in the post-neoadjuvant therapy, and post-surgical follow-up periods , , , . Imaging plays a central role in all stages of patient management, and the radiologist’s assessment is the cornerstone of surgical and medical decision-planning. This review aims to present and discuss the most up-to-date imaging issues on pancreatic cancer diagnosis, staging, response to neoadjuvant therapy and patient follow-up.
2. Diagnosis of pancreatic ductal adenocarcinoma
2.1. Imaging technique and typical findings
CT is the preferred technique for PDAC diagnosis, due to its wide availability and high diagnostic accuracy, with a reported 89–97% sensitivity , . CT equipments with dual-energy / spectral technology further improve lesion conspicuity and detection, and the newer photon-counting CT equipments have also been reported to do so , , . Lower energy-level (usually 40–50 keV) monoenergetic images provide greater attenuation by iodine contrast material, leading to improved image contrast between the hypovascular PDAC and highly vascularized pancreatic parenchyma. Transabdominal ultrasound maintains some relevancy, as it is the most accessible and least expensive imaging technique, and many pancreatic tumors are first identified on abdominal ultrasound scans . However, due to its low sensitivity and operator dependence, CT should still be performed as a first-line imaging test when PDAC is suspected, or for diagnostic confirmation and tumor staging when a tumor has been identified in ultrasound. MRI is currently a second-line diagnostic technique due to its higher cost and lower availability, with a similar sensitivity to CT for diagnosing PDAC: 83–93.5% . Its use is mainly reserved for diagnostic problem-solving. PET-CT has an estimated 90% sensitivity in detecting pancreatic cancer, but it does not offer clear benefits over CT or MRI for this purpose . Finally, endoscopic ultrasound (EUS) has a reported 91–100% sensitivity for PDAC detection, while also allowing the collection of pancreatic tissue with fine-needle biopsy for histological diagnosis , . Due to its lower accessibility when compared with other imaging alternatives, EUS is usually reserved for tissue collection for histological analysis.
For PDAC diagnosis, an adequate CT protocol is of paramount importance, including pancreatic, portal venous and delayed phases (Table 1, Fig. 1). Sufficient iodine contrast dose and rate of injection, as well as a saline flush, are necessary to provide adequate lesion conspicuity in dedicated pancreatic multi-phase acquisitions , . The use of bolus-tracking technique allows acquisition timing to be individualised to each patient circulatory dynamics, ensuring optimal post-contrast phases, and is therefore recommended . When using MRI, an adequate protocol is equally important (Table 2). Typical PDAC diagnostic findings include: (1) pancreatic hypovascular mass with dilation of the upstream main pancreatic duct; (2) simultaneous dilation of the common bile duct and main pancreatic duct if the tumor is located in the head – also known as “double duct sign”; (3) main pancreatic duct stenosis with signs of chronic pancreatitis in the upstream pancreas (Fig. 2). Imaging staging should be performed before a biliary stent is placed, so ductal anatomy is clearly defined and inflammatory changes do not overlap with tumoral tissue.
Table 1.
CT protocol used in our Institution for pancreatic ductal adenocarcinoma diagnosis and staging.
| Oral contrast | Neutral (water, 500 mL ingested over 10 min) |
--- |
| Intravenous contrast | 300 mg/mL; 1.5 mL/kg @3-5 mL/s; followed by 20 mL saline flush |
| Acquisition phases | Without contrast | Useful for detection of calcifications |
| Pancreatic | 35-50 s Pancreatic tumor detection; Arterial staging |
| Portal venous | 70-80 s Metastases detection; Venous staging |
| Delayed | 3-5 min Isoattenuating tumor detection |
| Acquisition | Volumetric (isotropic) | Thin slices (≤ 1 mm) |
| Monoenergetic (40 keV) |
| MPR (tumor-organ relations) |
| MIP (tumor-vascular relations) |
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CT equipment with new-generation spectral / dual-energy / photon-counting technology for improved image contrast is recommended. MPR: multiplanar reformation; MIP: maximum intensity projection.
Fig. 1.
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Importance of the pancreatic and delayed phases for tumor detection. In the pancreatic phase (A), there is a clear distinction (dashed line) between tumoral tissue (yellow arrow) and pancreatic parenchyma (white arrow). In the portal venous phase (B), the distinction between tumor and normal pancreas is not clear. (C - E) depicts another case, where a pancreatic tumor is ill-defined on the pancreatic (C) and portal venous (D) phases, but well-defined on the delayed phase (E, arrows).
Table 2.
MRI protocol used in our Institution for pancreatic ductal adenocarcinoma diagnosis and staging.
| Acquisitions | T2 TSE | Axial and coronal |
---
| T1 GRE | In- and opposed-phase |
| DWI | b =10 s/mm 2 |
| DWI | b ≥800 s/mm 2; ADC map |
| MRCP | 3D or 2D thick slab |
| Intravenous contrast | THRIVE (fat-saturated T1 GRE) | Without contrast
| Pancreatic | 35-50 s |
| Portal venous | 70-80 s |
| Delayed | 3-5 min |
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Fat-suppressed T2 TSE can be acquired as an alternative depending on local preference and MR equipment perfomance. ADC: apparent diffusion coefficient; DWI: diffusion-weighted imaging; GRE: gradient recalled echo; MRCP: magnetic resonance cholangiopancreatography; THRIVE: T1-weighted High Resolution Isotropic Volume Examination; TSE: turbo spin-echo.
Fig. 2.
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Typical findings of PDAC. (A) depicts a double duct sign, where both the main pancreatic duct and the biliary tree are dilated (arrows) due to an obstructing tumor in the pancreatic head (dashed line). In (B) there is parenchymal atrophy of the pancreatic tail (arrow), due to an obstructing tumor in the pancreatic body (dashed line). (C) shows a double duct sign in MRCP (white arrows), caused by a presumed obstructing lesion (yellow arrow). (D) reveals an obstructing (white arrow) pancreatic head tumor in a T2-weighted image (yellow arrow).
Notably, new developments in artificial intelligence applications may provide new opportunities for PDAC diagnosis, such as risk assessment using clinical records and accurate detection using only non-contrast-enhanced CT, as demonstrated in a large multicentric study , .
2.2. Small and isoattenuating tumors
In 5.4–11% of cases, and most commonly with < 20 mm tumors, PDAC is isoattenuating: it does not stand out from the pancreatic parenchyma in the pancreatic or portal venous phases , , , . These tumors have been reported as a frequent cause of missed diagnosis for early PDAC, especially when found incidentally in imaging studies . Indirect signs such as main pancreatic duct dilation upstream of an abrupt stenosis, a double duct sign, segmental pancreatic atrophy or faint enhancement are key to the diagnosis and should prompt further investigation , . This is also where a delayed acquisition can be very useful, as it has been shown to improve sensitivity for these lesions, which often enhance in delayed scans . MRI is also helpful for detecting small and isoattenuating tumors (Fig. 3), with a reported sensitivity of 79.2% . A reduced field-of-view DWI acquisition and MR perfusion techniques have been reported to improve image quality and pancreatic focal lesion characterization, and could therefore be useful in this context , . Another useful imaging technique for characterizing small and indeterminate tumors on CT is EUS, with a reported sensitivity of 87.3%, which can be improved with the use of contrast enhancement , , .
Fig. 3.
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Isodense PDAC. This patient presented with dilated main pancreatic duct (A and B, white arrows) and slightly dilated biliary tree, but a pancreatic tumor was not clearly seen on CT on both pancreatic (A) and portal venous (B) phases. MRI clearly identifies the obstructing tumor in the pancreatic head, as observed on DWI (C, yellow arrow) and a double duct sign on MRCP (D, white arrows).
2.3. Diffuse tumoral infiltration and mass-forming pancreatitis
An uncommon diagnostic challenge is the diffuse neoplastic infiltration, mainly in the tail of the pancreas, without dilation of the main pancreatic duct . The main differential diagnosis is autoimmune pancreatitis, which presents similar imaging findings. Likewise, mass-forming pancreatitis can mimic a neoplastic lesion and be impossible to differentiate from PDAC on CT. In both cases, magnetic resonance cholangiopancreatography (MRCP) may be useful, showing the “duct penetrating sign” - characteristic of benign lesions and enhanced with secretin administration , . Also, MRI can show multi-organ involvement in cases of autoimmune pancreatitis, often in the kidneys in the context of IgG4-related disease, facilitating this diagnosis (Fig. 4) , . Additional useful findings favoring auto-immune pancreatitis instead of PDAC include the typical diffuse enlargement with loss of lobulation of pancreatic parenchyma (sausage-shape), homogeneous enhancement, absence of upstream main pancreatic duct dilatation, capsulelike rim and absence of vascular invasion, .
Fig. 4.
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Diagnostic problem-solving with MRI. This patient presented with dilated main pancreatic duct and biliary tree (A, white arrows), highly suspicious for a pancreatic head tumor (yellow arrow). However, DWI revealed not only a restricting pancreatic head (B, yellow arrow), but also areas of restriction in both kidneys (B, white arrows). These findings prompted the diagnosis of auto-immune pancreatitis, which was confirmed after a 2-week steroid trial.
3. Staging of PDAC
When deciding what treatment to offer a patient with PDAC, either resection, neoadjuvant therapy or palliative care, imaging plays the most central role. Patients’ disease is initially staged as resectable, borderline resectable, locally advanced, or metastatic. This review will focus on the latest NCCN PDAC staging criteria, as it is the most widely used classification . The goal is to obtain an R0 resection, which provides the best chance for patient survival, and the surgical procedure is dependent on tumor location, vascular and adjacent organ involvement.
3.1. Local and vascular staging
Vascular involvement is the major resectability determinant in PDAC and is best assessed with contrast-enhanced CT (Fig. 5). Multiplanar reformations (MPR) and maximum intensity projections (MIP) are extremely useful, both for diagnosis and for showcasing findings to the surgeon. The terms “abutment” and “encasement” refer to ≤ 180º and > 180º tumoral contact with a blood vessel, respectively, the latter predicting the presence of vascular invasion , , . Blood vessel deformity and the “teardrop sign”, as well as occlusion, are also highly predictive of vascular invasion (Fig. 6, Fig. 7) .
Fig. 5.
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(A) is a schematic representation of relevant anatomical structures for PDAC staging. If a tumor is located on the pancreatic head (1), a cephalic duodenopancreatectomy is performed, and if the tumor is located on the pancreatic body and/or tail (2), then a distal pancreatectomy is performed. More extensive tumors might require a total pancreatectomy. (B) is a coronal MIP displaying the relevant blood vessels for PDAC staging, an extremely useful tool for showcasing findings in multidisciplinary team meetings. CT: celiac trunk; HA: hepatic artery (comprises both common hepatic and hepatic proper); LGA: left gastric artery; PV: portal vein; SA: splenic artery; SMA: superior mesenteric artery; SMV: superior mesenteric vein; SV: splenic vein.
Fig. 6.
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Schematic representation of blood vessel involvement in PDAC staging. (A) tumor-vein contact ≤ 180º (abutment); (B) tumor-vein contact > 180º (encasement); (C) teardrop sign; (D) vessel contour deformity; (E) and (F) arterial abutment and encasement (≤ and > 180º), respectively.
Fig. 7.
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Venous involvement in PDAC. (A) Tumor-superior mesenteric vein contact < 180º (abutment). (B) Teardrop sign of the superior mesenteric vein (arrow). (D) Portal vein stenosis and deformity (arrows). (D) Superior mesenteric vein occlusion (arrows).
As summarized in Table 3, resectable tumors have no arterial contact, and no / limited venous contact (abutment). Patients in this category can be proposed for upfront surgery. Borderline resectable tumors include tumors with limited vascular involvement, or vascular involvement that can be surgically repaired. Patients with borderline resectable disease are offered neoadjuvant therapy; surgical resection can be later performed, depending on response to neoadjuvant therapy and local surgical expertise. Patients with locally advanced disease have extensive vascular involvement. These patients are offered systemic chemotherapy, chemoradiation, or inclusion in clinical trials. Depending on response assessment and performance status, resection surgery can still be performed after neoadjuvant therapy , . If disease progression is evident with vascular involvement precluding surgery or metastatic disease, then the best palliative care or inclusion in clinical trials are the remaining options.
Table 3.
NCCN V2.2023 resectability criteria adapted from .
| Resectable | Arteries | No tumoral contact |
---
| Veins | No tumoral contact / ≤ 180º PV or SMV contact without deformity |
| Borderline | Arteries | ≤ 180º SMA or CT contact |
| Limited CHA contact (not extending to its bifurcation nor the CT) |
| Anatomical variant with arterial involvement allowing for surgical management |
| Veins | ≤ 180º PV or SMV contact with deformity |
| > 180º PV or SMV contact but surgically repairable |
| Tumoral contact with IVC |
| Unresectable | Arteries | > 180º SMA or CT contact; extensive HA involvement |
| Veins | > 180º PV or SMV contact non-surgically repairable |
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CHA: common hepatic artery; CT: celiac trunk; IVC: inferior vena cava; HA: hepatic artery; NCCN: National Comprehensive Cancer Network; PV: portal vein; SMA: superior mesenteric artery; SMV: superior mesenteric vein. Abutment and encasement are considered synonyms with < 180º and > 180º tumoral contact, respectively.
Since being first introduced in 1997, the criteria for borderline resectable staging category have been updated over the years, reflecting accumulating evidence and international consensus , . These patients have limited vascular involvement and surgical resection is possible, but a margin-free resection is unlikely, therefore benefiting clearly from neoadjuvant therapy. The definition of vascular involvement with objective criteria (vessel circumference involvement and contour irregularity) has contributed to a more standardized classification. More recently, the involvement of proximal jejunal branches was categorized as unresectable disease in 2017, but this involvement is not included in the latest versions , . This led to a reported improved accuracy in disease staging and better patient stratification . It is important to be aware of the evolving nature of these staging criteria, with consequences both for clinical practice over time and for the validity of reported data, when comparing with current guidelines.
Vascular anatomical variants are common and can have surgical implications for PDAC , , , . Not only should replaced or accessory hepatic arteries be identified to avoid surgical complications, these variants can also provide surgical opportunities, when providing an alternative hepatic arterial supply without tumoral involvement (Fig. 8). Anatomical variants can also preclude surgery or determine a hepatic resection, especially if a replaced hepatic artery is involved by tumor and there is no collateral arterial inflow.
Fig. 8.
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Arterial variants with surgical implications. (A) depicts the anatomy of a patient with a replaced right hepatic artery (RRHA) originating from the superior mesenteric artery (SMA) and a replaced left hepatic artery (RLHA) originating from the left gastric artery (LGA). The celiac trunk (CT) was encased by the tumor, as represented in (B, dashed line). As arterial collaterals were present between both replaced hepatic arteries, a celiac trunk resection along with tumor resection were performed, and the patient maintained a preserved liver vascularization. (C) represents another patient with the following anatomy: RLHA from LGA, middle hepatic artery (MHA) from CT, and RRHA from SMA. The RRHA was encased by tumor, as shown in (D, dashed line). This allowed tumor resection along with the RRHA, after an embolization procedure for facilitating the development of collaterals.
Celiac trunk stenosis, either due to atherosclerosis or arcuate ligament syndrome, can predispose patients to hepatic ischemia, as the collateral arterial flow to the liver is disrupted during surgery (Fig. 9) , , . CT is the preferred method for diagnosing this condition, allowing preventive surgical procedures, such as arcuate ligament release and celiac artery stenting , .
Fig. 9.
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Arcuate ligament syndrome. Proximal celiac trunk (CT) stenosis with a hooked appearance (arrow), with pos-stenotic dilatation.
Besides the vascular staging, involvement of adjacent organs should be identified, most often the duodenum, stomach, left adrenal, left kidney, spleen and colon, as these will have to be resected during surgery if involved by the tumor . Also, tumoral extension to the mesocolon, most often the transverse mesocolon, should be identified at this stage for surgical planning, possibly determining a colon resection.
Of mention, EUS has been reported to improve staging accuracy, especially when compared with older CT technology, identifying 14% unresectable patients previously staged as resectable on CT , , . However, newer CT technology has improved diagnostic and staging accuracy, and comparative studies with EUS are not yet available.
3.2. Hepatic, peritoneal, and pulmonary staging
Liver metastases can be detected by CT, however with a relatively low sensitivity: 69% , . MRI is superior for this purpose, especially when performed with gadoxetic acid, with a reported sensitivity of 85% . Furthermore, DWI has been reported to detect undiagnosed liver metastases on CT in 10–12.9% of patients and prevent futile surgeries in metastatic patients , , . As such, it is recommended that liver staging is performed using MRI (Fig. 10). PET-CT has a reported 93% sensitivity in the detection of liver metastases for lesions > 1 cm, which is reduced to 70% when considering all lesion sizes .
Fig. 10.
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Importance of MRI for hepatic staging. In a contrast-enhanced CT for PDAC staging (A), no liver lesions were found. The staging MRI clearly revealed hepatic metastases (arrows) with restricted diffusion on DWI (B).
There is no evidence determining the best timing for liver staging MRI (with or without gadoxetic acid) in the case of biliary obstruction, if before or after stent placement. Performing MRI before stent placement, will avoid artifacts in the case of metallic stents and allow better biliary anatomy depiction. Also, biliary stent placement may be associated with cholangitis, which may cause liver perfusion changes and course with small hepatic abscesses , . For these reasons, we perform hepatic staging MRI as soon as possible.
Finally, differentiating small hepatic abscesses from metastases can be a difficult task. Contrast-enhanced CT findings have been reported to help in this regard, favoring the diagnosis of abscesses: patchy parenchymal enhancement, perilesional hyperemia and, especially, arterial rim enhancement persistent through portal venous phase . Dual-energy CT may improve on the technique’s ability to distinguish both lesions . DWI may also be helpful, as hepatic metastases tend to present lower ADC values at their periphery, when compared with abscesses .
Peritoneal metastases usually manifest as nodular lesions, detectable by both CT and MRI (Fig. 11). However, early peritoneal disease can be hard to identify, and evidence is lacking into imaging effectiveness for its diagnosis , . Staging laparoscopy can be performed for this purpose, and has been shown to be more sensitive than CT, but its usefulness has been debated due to poor cost-effectiveness when staging MRI is performed , . Nevertheless, recent data reporting high rates of positive staging laparoscopies has prompted some authors to advocate its use in the majority of patients prior to resection .
Fig. 11.
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Metastatic PDAC with peritoneal metastases. (A) reveals a bulky PDAC of the body and tail (dashed line), with hepatic metastases (arrows). There are also peritoneal metastases, as observed with an “omental cake” appearance in the pelvis (B, arrows).
PDAC can also metastasize to the lung, but the usefulness of including chest CT in tumoral staging and follow-up has been debated , . Nevertheless, we include a chest CT in our staging protocol, which may serve as a baseline for comparison in future studies, should indeterminate or suspicious lung nodules be identified.
3.3. Lymph node staging
Lymph node staging is still poorly performed by imaging. CT has a reported sensitivity of only 14–44% for diagnosing nodal tumor infiltration when 10 mm short-axis is used as diagnostic criterion , . Combining size criteria with contour irregularity, heterogeneous signal intensity and/or density and increased number of visible nodes has been reported to improve sensitivity . EUS has been reported to be superior in identifying suspicious lymph nodes in loco-regional and para-aortic locations, and is currently regarded as the most sensitive imaging technique for this purpose , . Nevertheless, it is not routinely included in lymph node staging recommendations . Gallium-68-labeled fibroblast activation protein inhibitor PET-CT was reported to improve detection of pathologic lymph nodes in a preliminary study, but more data is needed before recommending its use . A promising artificial intelligence model outperformed radiologists in lymph node staging using CT, while also providing prognostic information for patients . Enlarged lymph nodes should be mentioned in the radiological CT or MRI report, especially when outside of loco-regional location, but this low diagnostic performance with current imaging methods must be kept in mind .
3.4. Radiological structured report
The use of a radiological structured report is recommended, to provide a clear and standardized communication of imaging findings to surgeons and oncologists , . This also ensures that no relevant structures for tumor staging are forgotten or overlooked in the report. An example structured reporting template is suggested in Table 4, adapted from . The reporting template should be tailored in each center to provide the multidisciplinary team with the information needed in an easily accessible way. Note that extensive descriptions of vascular involvement in patients with metastatic disease are usually not necessary, unless there is a specific motive for it, such as inclusion in a clinical trial.
Table 4.
Structured reporting template.
| Pancreatic tumor | Location | Head / Neck / Body / Tail |
---
| Size | Largest axis; approximate measurement if ill-defined |
| MPD and biliary tree | Dilated / Not dilated |
| Adjacent organ involvement | NA / Describe involvement (stomach, adrenal, colon, etc.) |
| Arteries | CT | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) Mention stenosis or atherosclerosis |
| CHA | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) Extension to CT or bifurcation |
| SA | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) |
| SMA | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) |
| SMA branches | Free Mention involved branches (extension in mm) |
| Variant anatomy | NA / Replaced CHA / Accessory RHA / Replaced RHA / Accessory LHA / Replaced LHA / Other_ |
| Variant involvement | _NA / Free / ≤ 180º / > 180º / Deformity / Occlusion |
| Veins | PV | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) |
| SMV | Free ≤ 180º / > 180º / Deformity / Occlusion (extension in mm) |
| Collaterals | NA / pancreatic head / hepatic hilum / mesenteric / LUQ |
| Metastases | Liver | Absent / Present / Indeterminate |
| Peritoneum | Absent / Present / Indeterminate Ascites Absent / Present |
| Lung | Absent / Present / Indeterminate |
| Lymph nodes | If suspicious, mention location(s) | NA / Peri-pancreatic / Hepatic hilum / Celiac / Aortocaval / Para-aortic / Splenic |
| Other findings | Free writing |
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One or more options presented in italics should be chosen in each field according to imaging findings, and longitudinal extension of vessel involvement provided in millimeters. Further description can be added at the end of the report, when variant anatomy or involvement cannot be adequately described in the template. CHA: common hepatic artery; CT: celiac trunk; LHA: left hepatic artery; LUQ: left upper quadrant; MPD: main pancreatic duct; NA: non-applicable; PV: portal vein; RHA: right hepatic artery; SA: splenic artery; SMA: superior mesenteric artery; SMV: superior mesenteric vein.
3.5. Current issues with PDAC staging
Some aspects of the current widely used NCCN staging classification for PDAC have been debated . Mainly, the dependence on strict anatomic criteria for defining treatment strategies for patients with PDAC, directed at surgical feasibility, often fails to consider disease biology.
Starting with anatomic staging issues, the 180º threshold can be insufficient for vascular staging, as the longitudinal extension of vessel involvement also determines surgical technique and has been shown to predict survival: patients with more than 20 mm of vascular involvement have a worse prognosis and may be candidates for neoadjuvant therapy . Another issue is the involvement of the splenic artery which may fit in the resectable category, while the involvement of the celiac trunk, hepatic artery or superior mesenteric artery determine borderline resectable or locally advanced disease . Patients may therefore be proposed for upfront surgery with a resectable although aggressive disease, missing potential survival benefits from neoadjuvant therapy. Finally, large tumor size, which is not category-defining for resectability, has also been associated with poor prognosis, and should be considered when deciding which treatment to offer a patient , .
The presence of para-aortic lymph node metastases, which may easily be missed on imaging as previously discussed, is another predictor of poor prognosis , , , . Despite these diagnostic difficulties and the fact that they are not routinely removed on surgery, these nodes are considered metastatic disease, and adjuvant chemotherapy has been recommended for improving outcomes in these patients , .
Surgical resection is not recommended for metastatic patients according to current guidelines; however, it is performed after neoadjuvant therapy in some institutions, for selected patients with oligometastatic disease. This has been reported to improve survival, but more evidence is needed before widespread acceptance , , .
Recurrence risk is also not sufficiently addressed in the anatomic classification. In a large series evaluating early recurrence, pre-operative risk factors were identified, including: CA 19–9 > 210 U/mL, Charlson age-comorbidity index ≥ 4, tumor size > 3 cm; post-operative risk factors were also identified, including: poor tumor differentiation, microscopic lymphovascular invasion, lymph node ratio > 0.2, and CA 19–9 > 37 U/mL . None of these are considered in the anatomic resection criteria. Furthermore, the inclusion of neoadjuvant therapy based on risk factors for early recurrence was shown to improve survival . Local recurrence has additionally been associated with high rates of post-venous reconstruction thrombosis, rather than anatomic or technical factors, further demonstrating the relevance of disease biology in determining patient outcomes .
Management of patients with PDAC with resectable disease is also currently being discussed. Neoadjuvant therapy has been proposed in purely resectable patients, showing some improvement in disease-free survival in a small number of trials . However, conflicting evidence showed no clear improvement in survival, and more data on this issue is needed before issuing favorable recommendations for clinical practice , . More recently, a randomised clinical trial found no survival benefit from the use of neoadjuvant therapy in resectable patients with PDAC, when compared to upfront resection, further disfavoring its use .
In summary, accumulating data tend to consider PDAC as a systemic disease and favor the use of neoadjuvant therapy even in selected high-risk resectable patients, while also selecting patients previously categorized as unresectable or even oligometastatic, as candidates for surgical resection with favorable outcomes. Patient management decisions should therefore be made in a multidisciplinary team setting, considering both the anatomic staging and additional prognostic factors.
3.6. Alternative staging classifications
Other staging classifications have been developed, including additional prognostic factors in the decision-making process. The MD Anderson Cancer Center developed a well-known classification incorporating anatomic criteria, tumor biology and patient performance status for patient categorization , . This type of classification has been used later in a consensus meeting to categorize borderline resectable and locally advanced patients in different types, according to anatomic, biological, and conditional factors . Here, in addition to the anatomic classification (resectable, borderline resectable and locally advanced), biological factors (serum CA 19–9 >500 IU/mL and/or positive regional lymph node metastases) and conditional factors (performance status) were added for an integrative patient stratification.
Additionally, other classifications were developed based on anatomic criteria, but without such widespread use as the NCCN classification. These include the American Society of Clinical Oncology, Intergroup Alliance, and Americas Hepato-Pancreato-Biliary Association / Society of Surgical Oncology / Society for Surgery of the Alimentary Tract , , . These classifications also use the resectable, borderline resectable and unresectable / locally advanced categories, but with different criteria, which were highlighted in a consensus meeting .
4. Response to neoadjuvant therapy
Assessing response to neoadjuvant therapy for PDAC is a difficult task in clinical practice. Tumor size criteria are not useful for this purpose, as the fibrotic tumoral stroma remains, even when a response has occurred in cancer cells , , . The current methods for evaluating response to neoadjuvant therapy reside on tumoral markers evolution (serum CA19–9 concentration) and the absence of disease progression as evidenced by imaging studies , , . If progression is not apparent – the pancreatic tumor is stable or has reduced in size, and tumor markers have not increased – then a response is presumed and the patient can be proposed for surgery, in the case of initial borderline or even locally advanced disease (Fig. 12). Also, standard criteria for vascular resectability based on the amount of tumor-vessel contact are not useful post-neoadjuvant therapy, and an improvement of tumor-vascular involvement is highly predictive of an R0 resection, even if partial . Furthermore, the presence of tumor regression after neoadjuvant therapy was not correlated with survival, in a cohort of patients with R0 resections . Disease extension as observed on CT should not just be disregarded, however. Smaller tumor size and reduction of arterial contact, as well as imaging scores based on arterial involvement and resectability status, performed both before and after neoadjuvant therapy, were correlated with probability of R0 resection , .
Fig. 12.
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Response to neoadjuvant therapy. (A) reveals a locally advanced PDAC (dashed line) with common hepatic artery (CHA) encasement up to the celiac trunk bifurcation (CT). After the completion of neoadjuvant chemotherapy, the tumor had reduced in size (B, dashed line), but maintained a long encasement of the CHA. The patient underwent surgery, with an R0 resection.
Some studies have evaluated the role of DWI for response assessment, reporting a correlation of apparent diffusion coefficient (ADC) with histopathological response, but others have not found it useful , , , , . The value of DWI in this context remains undetermined, and further studies are needed to provide recommendations for clinical practice .
The role of FDG-PET for neoadjuvant therapy response assessment has also been evaluated. Both CT and MRI FDG-PET’s metabolic response has been reported to be associated with pathologic response, being superior than CA19–9 for this purpose , . FDG-PET MRI added to contrast-enhanced CT showed improved accuracy for determining resectability status when compared with contras-enhanced CT alone, in a retrospective study . A recent systematic review reported FDG-PET’s potential for predicting and assessing response to neoadjuvant therapy; however, the small numbers of patients enrolled in each study and methodological heterogeneity prompt the need for well-designed prospective trials .
Recent developments in molecular imaging provided potential biomarkers for response to neoadjuvant therapy evaluation, as well as improving diagnostic and staging accuracy, with encouraging results . However, these require validation and thus cannot yet be recommended for clinical practice.
Considering current limited tools for response assessment, the therapeutic decision after neoadjuvant therapy should always be made by a multidisciplinary team, taking into account imaging, biochemical and patient fitness data. The surgical decision is heavily dependent on the surgical team expertise, especially in cases of initially advanced disease or extensive vascular involvement , .
5. Patient surveillance after resection
First, some important issues regarding pathologic assessment of PDAC resection margin should be discussed. The R0 definition for PDAC resection has been debated, and controversy regarding microscopic margin involvement has contributed to variability in reported R1 rates , , . Although UICC’s classification considers R0 as absence of tumor cells at the resection margin, a free resection margin of ≥ 1 mm defining R0 is also frequently used, and was associated with better survival when compared with < 1 mm , . Despite this, in the presence of lymph node positive disease, recurrence patterns were reportedly similar for both R0 and R1 patients . Also, free margins larger than 1.5 mm have been reported to correlate with survival, as explained by the dispersed growth pattern of PDAC , . Other issues including challenges in specimen grossing and sampling, and the variable R0 definitions used in published literature, have been considered as important factors undermining the data quality of reported clinical outcomes of PDAC resection .
Evidence for recommending specific surveillance methods after PDAC resection is lacking, and the value of surveillance has been debated , . Nevertheless, patients undergoing post-resection surveillance have been reported to have improved detection of disease recurrence at asymptomatic stages, were more likely to receive treatment for the recurrence and presented longer overall survival , . A tumoral recurrence is usually found in the surgical bed, in the mesenteric root or the pancreatic remnant (Fig. 13). Local perivascular densification is common post-resection; however, a perivascular densification increasing in size in serial imaging or leading to a de novo change in the caliber or regularity of the vessel should be considered suspicious for local recurrence . Distant recurrence is most commonly found in the liver or lungs . A surveillance scheme including serum CA19–9 and CT (chest, abdomen and pelvis) every 3–6 months for 2 years after surgical resection has been recommended, and is followed in our institution . MRI is usually not included in a surveillance setting but can be used for disease staging if a recurrence is suspected.
Fig. 13.
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Recurrence of PDAC. Two cases of local recurrence of PDAC (dashed lines), in (A) encasing the celiac trunk (CT) and common hepatic artery (CHA), and in (B) encasing the superior mesenteric artery (SMA) and invading the superior mesenteric vein (SMV) with a teardrop sign.
6. Conclusion
Imaging plays a central role in all stages of patient management for PDAC: diagnosis, staging, response to neoadjuvant therapy and post-surgical surveillance. Diagnostic and staging performance of current imaging modalities is high; however, there are unresolved issues when staging patients and assessing response to neoadjuvant therapy, and evidence is lacking for dedicated patient surveillance protocols. Nevertheless, an experienced radiologist remains essential in the multidisciplinary team caring for patients with PDAC.
CRediT authorship contribution statement
Matos Celso: Writing – review & editing, Methodology, Conceptualization. Marques Rui Mateus: Writing – review & editing, Methodology. Santiago Inês: Writing – review & editing. Andrade Luísa: Writing – review & editing, Methodology, Investigation, Data curation, Conceptualization. Bilreiro Carlos: Writing – review & editing, Writing – original draft, Visualization, Resources, Methodology, Formal analysis, Data curation, Conceptualization.
Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
This work did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
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1. Introduction
2. Diagnosis of pancreatic ductal adenocarcinoma
3. Staging of PDAC
4. Response to neoadjuvant therapy
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4167 | https://ceur-ws.org/Vol-1949/ICTCSpaper05.pdf | Distributed approximation algorithms for k-dominating set in graphs of bounded genus and linklessly embeddable graphs Regular Submission Andrzej Czygrinow1⋆, Micha l Han´ ckowiak2, Wojciech Wawrzyniak2, and Marcin Witkowski2 1 Faculty of Mathematics and Computer Science Adam Mickiewicz University, Pozna´ n, Poland mhanckow,wwawrzy, mw@amu.edu.pl 2 School of Mathematical and Statistical Sciences Arizona State University, Tempe, AZ,85287-1804, USA.
aczygri@asu.edu Abstract. A k-dominating set in a graph G = (V, E) is a set U ⊆V such that ever vertex of G is either in U or has at least k neighbors in U. In this paper we give simple distributed approximation algorithms in the local model for the minimum k-dominating set problem for k ≥2 in graphs with no K3,h-minor and graphs with no K4,4-minor. In par-ticular, this gives fast distributed approximations for graphs of bounded genus and linklessly embeddable graphs. The algorithms give a constant approximation ratio and run in a constant number of rounds. In addi-tion, we will give a (1 + ϵ)-approximation for an arbitrary fixed ϵ > 0 which runs in O(log∗n) rounds where n is the order of a graph.
1 Introduction The minimum dominating set (MDS) problem is one of the most extensively studied graph-theoretic questions. On one hand, it plays an important role in graph theory, on the other, it admits many important applications. Its complex-ity has been studied in many different computational models and for various classes of graphs. In general, the problem is NP-hard , and it is even NP-hard to find a (C log ∆(G))-approximation for some constant C . In addition to the above hardness results for the sequential model, similar restrictions apply to the distributed complexity of the problem in general graphs. At the same time, there has been much more success in devising efficient approximations for spe-cific classes of graphs. In particular, the problem is much more tractable when restricted to the family of planar graphs and an efficient deterministic approx-imation algorithm for the minimum dominating set problem in planar graphs was given in . The algorithm from gives a constant approximation of a ⋆Research supported in part by Simons Foundation Grant # 521777.
minimum dominating set, runs in a constant number of rounds but uses long messages (more than O(log n) bits).In , Wawrzyniak improved the analysis from and showed that the algorithm from gives a 52-approximation. In addition, in , Wawrzyniak proposed a local (constant-time) algorithm which gives a 694-approximation of the minimum dominating set problem in planar graphs which uses messages of length at most O(log n) and works in the port numbering model. (In this model, which is similar to the Local model, no unique identifiers are required but edges which are incident to a vertex are numbered, see .) In , the authors showed that essentially the same algorithm gives a constant approximation of a minimum dominating set in graphs of bounded genus. In fact, as proved in , no geometrical considerations are needed when analyzing the procedure from as the only requirement is that the underlying graph has no K3,h as a minor.
Often a stronger notion of a domination is needed. For example, in some ap-plications it can be desirable to require that every vertex in a graph is dominated not by one vertex from a set but by k vertices for some k ∈Z+. This can be interpreted as a fault tolerance requirement; Servers are placed in a network so that each vertex/client has direct access to k servers.
To define the problem formally we need some notation and terminology. Let G = (V, E) be a graph with |V | = n. By NG(v) = {u ∈V : uv ∈E} we denote the neighborhood of a vertex v ∈V and by NG[v] = NG(v) ∪{v} the closed neighborhood of v. We use dG(v) (or d(v) if G is clear from the context) to denote |NG(v)|. Set D ⊆V is called a dominating set in G if every vertex in V \D is adjacent to a vertex in D. A minimum dominating set in a graph G is a dominating set in G of the smallest size. For a positive integer k, a k-dominating set in G is a subset D ⊆V such that for every v ∈V \ D, |N(v) ∩D| ≥k, that is, for every vertex v ∈V , v is in D or v has at least k neighbors in D.
In particular, a 1-dominating set is simply a dominating set. In addition, we will say that a set W k-dominates set U in G, if every vertex from U \ W has at least k neighbors in W. Note that other notions of k-domination have been considered and many variants of the problem have been studied (see ).
The k-domination problem was proposed in as a natural generalization of the minimum dominating set problem. It has been studied extensively in graph theory, where the main line of research is to establish upper bounds for γk(G), the minimum size of a k-dominating set. It is known that the problem is NP-complete in general graphs but can be solved in linear time in trees and series-parallel graphs (). Further generalizations of the k-dominating set were considered and many slightly different variants have been proposed. The distributed complexity of finding γk(G) (the k-MDS problem) were studied in and . In addition, in distributed approximation algorithms for the minimum k-dominating set problem in planar graphs are proposed. In this paper, we will study distributed complexity of the problem in two classes of graphs, graphs of bounded genus and linklessly embeddable graphs. Both of these classes are proper minor-closed families of graphs and in our algorithms and analysis we will not need any geometric properties of the graphs as we will work in a slightly more general setting by forbidding some graphs as minors. Let G be a graph, we say that H is a minor of G (or that G has an H-minor) if H can be obtained from a subgraph of G by sequence of edge deletions and contractions. We say that G contains a subdivision of H (or that TH is a subgraph of G) if G contains a graph obtained from H by replacing edges of H by independent paths of length at least one.
Let Sh denote the surface obtained from the sphere by adding h handles (see for a formal definition of this process). The genus of a graph G, g(G), is the least integer h such that there is an embedding of a graph G into Sh. In what follows we will not make any geometric arguments as our argument applies to a bigger class of graphs than the class of graphs of genus at most g. Note that if G is embeddable in a surface S, then the graphs obtained from G by deleting an edge or contracting an edge are embeddable as well. Consequently the class of graphs of genus at most g is a proper minor-closed family of graphs and can be characterized by a finite set of forbidden minors. In addition, since g(Km,n) = ⌈(m −2)(n −2)/4⌉(see ), we have that if g(G) ≤g, then G has no K3,4g+3-minor. As in , we will work with the family of graphs Bh which have no K3,h-minor.
In addition to the graphs with no K3,h-minor we will give a distributed approximation algorithm for graphs with no K4,4-minor. This, in particular, yields an algorithms for linklessly embeddable graphs which are sometimes viewed as 3-dimensional analogs of planar graphs. An embedding of a graph in R3 is called linkless if every pair of (vertex) disjoint cycles are unlinked closed curves.
It is not difficult to see that K4,4 is not linklessly embeddable (the only fact that will be needed).
We will consider the Local model (), which is a synchronous message-passing model in which vertices have unique identifiers from {1, . . . , n} where n is the order of the graph. Computations proceed in rounds and in each round a vertex can send messages to all its neighbors, can receive messages from its neighbors and can perform some local computations. Neither the size of the messages nor the amount of local computations is restricted in any way. An algorithm is executed by the vertices of a network and its objective is to find a solution to a problem (in our case the k-MDS problem) in this underlying network.
Our contribution is twofold. First, we show that for graphs of bounded genus there is an easy distributed algorithm which gives a constant approximation of the minimum k-dominating set for k ≥2. This mirrors the corresponding result for planar graphs. In addition, the same approach gives a constant approximation for k ≥3 in linklessly embeddable graphs. The case k = 2 is different, and to address it we use a modification of the algorithm from and some ideas from and . The above approximation algorithms run in a constant number of rounds and give a constant approximation ratio. Further improvement of the approximation ratio are left for future work. Second, we propose a distributed approximation algorithm with approximation ratio of (1+ϵ) for every fixed ϵ > 0.
This algorithm runs in O(log∗n) rounds where n is the order of the graph. This algorithm combines the main procedure from with some additional ideas.
The paper is structured as follows. In the next section, we state some pre-liminary facts. In the following section we give the first (simple) approximation algorithm for the minimum k-dominating set in graphs with no TKh,k+1. This yields an algorithm for graphs of bounded genus for k ≥2 and for linklessly embeddable graphs when k ≥3. In Section 4 we address the 2MDS problem for graphs with no TK4,4 which gives, in particular, an algorithm for linklessly embeddable graphs when k = 2. Finally, in Section 5, we discuss the (1 + ϵ)-approximation.
2 Preliminaries We will start our technical discussion with a few facts which will be used later.
We have the following bound for the number of edges in a graph with TKh (see ).
Lemma 1. If G is a graph such that |E(G)| ≥5h2n, then G has TKh.
We will also need a more general bound for E(G) in the case G has no Kp-minor.
The following was proved by Kostochka in .
Lemma 2 (Kostochka). There exists c ∈R such that for every r ∈Z+ if G is a graph on n vertices with no Kr-minor, then |E(G)| ≤cr√log rn.
One of the main tools used in the proofs will be lemmas from . Let v be a vertex in G and let S ⊆V (G) \ {v}. Then a set of paths is called a v, S-fan if every path starts at v, ends in a vertex from S which is the only vertex from S on the path, and any two distinct paths have only v in common. For a dominating set D in a graph G let Dk,l be the set of vertices v ∈V (G) \ D such that there is a v, D-fan subgraph in G consisting of k paths, each of length at most l. The following two facts are proved in .
Lemma 3. For h, l ∈Z+ there is c such that the following holds. Let G be a graph with no TKh and let D be a dominating set in G. Then |Dh−1,l| ≤c|D|.
In addition, we have the following fact.
Lemma 4. Let h ∈Z+. For l, m ∈Z+ there is c = c(m, l) such that if G is graph with no TKm,h and D is a dominating set in G then |Dm,l| ≤c|D|.
3 First algorithm for graphs with bounded genus and linklessly embeddable graphs In our first algorithm (Procedure 1) we simply add to the solution all vertices of sufficiently large degree and claim that this gives a constant approximation of a k-dominating set. Specifically, we will show that in the case when k ≥2 and G has bounded genus or k ≥3 and G is linklessly embeddable, this gives a constant approximation of a minimum k-dominating set. We will need the following lemma in the analysis.
Lemma 5. Let k, h, q ∈Z+ be such that k ≤q. If H = H[U, X] is a bipartite graph such that |U| ≥ q k (h −1) + 1, |X| ≤q and for every vertex u ∈U, dH(u) ≥k, then H contains Kh,k.
Proof. The proof is similar to a proof of an upper bound for the extremal number of a complete bipartite graph. Suppose H has no Kh,k. We will count the number of k-stars (stars of degree k) with centers in U. On one hand, the number of such k-stars is at least P u∈U d(u) k ≥|U|. On the other hand, since H has no Kh,k, the number of k-stars is at most |X| k (h −1). Consequently, if |U| ≥ |X| k (h −1) + 1, then H contains Kh,k. □ Let h, k ∈Z+, set L := k+h k (h−1)+2(k +h)+1 and consider the following procedure.
Procedure 1 1. For every v ∈V (G), if d(v) ≥L, then add v to D.
2. For every v ∈V (G) \ D, if |N(v) ∩D| < k, then add v to D.
3. Return D.
In fact we will proves something slightly more general.
Fact 1 For h, k ∈Z+ let L := k+h k (h −1) + 2(k + h) + 1. If G is a graph with no TKh+k+1 and no Kh,k+1, then Procedure 1 applied to G with L returns a k-dominating set of size O(γk(G)).
Proof. Let W := {v ∈V (G) : d(v) ≥L} and let D∗be an optimal k-dominating set. Then the number of vertices in G which have a neighbor in D∗\ W is at most (L −1)|D∗|. Suppose that v ∈V (G) \ W has |N(v) ∩W| < k and so it is added to D in the second step of the algorithm. Then v is either in D∗ or has at least one neighbor in D∗\ W and so the number of vertices added in step 2 is at most L|D∗|. Thus to show that |D| = O(γk(G)) it is enough to show that |W| = O(|D∗|). To that end, note that for every v ∈W \ D∗, N(v) \ D∗is k-dominated by D∗. If |N(v) ∩D∗| ≥h + k, then v ∈D∗ h+k,1.
Since G has no TKh+k+1, D∗is a k-dominating set and so a 1-dominating set, by Lemma 3, the number of such vertices is O(|D∗|). Thus we may assume that |N(v) \ D∗| > L −(h + k) and recall that N(v) \ D∗is k-dominated by D∗. Consider a maximum matching M in G[N(v) \ D∗, D∗] and note that if |M| ≥h + k, then v ∈D∗ h+k,2. Thus we may assume that |M| < h + k. Since M is a maximum matching in G[N(v)\D∗, D∗], if w ∈N(v)(D∗∪V (M), then all neighbors of w which are in D∗are in V (M) as otherwise we could increase M.
Consequently, |N(N(v)(D∗∪V (M)))∩D∗| ≤|V (M)∩D∗| < h+k. In addition, |E(N(v)(D∗∪V (M)), V (M)∩D∗)| ≥k|N(v)(D∗∪V (M))| because D∗is a k-dominating set,and , by definition of L, |N(v)(D∗∪V (M))| ≥ k+h k (h−1)+1.
Thus, by Lemma 5, G[N(v) \ (D∗∪V (M)), V (M) ∩D∗] contains Kh,k, which in connection with v gives Kh,k+1 contradicting the fact that G has no TKh,k+1.
The contradiction shows that |M| ≥h + k and so v ∈D∗ h+k,2. By Lemma 3, |D∗ h+k,2| = O(γk(G)). □ We immediately get constant approximations for γk(G) in linklessly embed-dable graphs for k ≥3.
Corollary 2 Let G be a linklessy embeddable graph and let k ≥3. Then Proce-dure 1 applied to G returns a k-dominating set of size O(γk(G)) when applied with h = 4.
Proof. Since G is linklessly embeddable, G has no K4,4-minor and therefore no K4,4 and no TK8. We can apply Procedure 1 with h = 4. □ If G is a graph with g(G) ≤g, then G has no K4g+3,3-minor. Consequently Procedure 1 can be applied to G for every k ≥2 gives a constant approximation.
Corollary 3 Let g ∈Z+ and k ≥2. For every graph G with g(G) ≤g, Proce-dure 1 returns a k-dominating set of size O(γk(G)) when applied with h = 4g+3.
Proof. Since G has genus bounded by g, G ∈B4g+3, and so G has no TK4g+3,3.
□ 4 Graphs with no T K4,4 for k = 2 In the case G has no K4,4-minor and k = 2 it is no longer possible to claim that Procedure 1 gives a constant approximation of γk(G) and a slightly more involved algorithm is needed. We will show that a modification of the algorithm for the minimum dominating set in planar graphs from gives a constant approximation in this case. Recall that we say that a vertex v is 2-dominated by a set S if either v ∈S or v has at least two neighbors in S.
We will start with the following observation.
Lemma 6. Let K ∈Z+. If U is a subset of V (G) such that |U| ≥4K2 and there is a set S ⊆V (G) \ U of size |S| ≤K which 2-dominates U, then there exist s1, s2 ∈S such that s1 ̸= s2 and |N(s1) ∩N(s2) ∩U| ≥4.
Proof. Clearly there exists s1 ∈S such that |N(s1) ∩U| ≥4K2/K = 4K and there exists s2 ∈S \ {s1} such that |N(s2) ∩(N(s1) ∩U)| ≥4K/K ≥4. □ In what follows, let K := 642, C := 4K2 + K and L := 4C + 3. We will assume that G has no TK4,4.
Procedure 2 1. For every v ∈V (G): If there is no set S ⊆V \ {v} such that |S| ≤K and S 2-dominates N(v), then add v to D1. In addition if d(v) < 2, then add v to D1.
2. Let Z be the set of vertices which are 2-dominated by D1 and let d′(v) := |N(v) \ Z| for v ∈V .
3. For every u ∈V \ Z: – If |N(u)∩D1| = 1, then choose v ∈N(u)\D1 such that d′(v) is maximum and add it to D2.
– If |N(u)∩D1| = 0, then choose two distinct v, v′ ∈N(u) with the largest degrees d′ and add them to D2.
4. Return D1 ∪D2.
Note that Procedure 2 works in the Local model as in the second step a vertex v can obtain locally information about the ball of radius two centered at v to determine if it should be added to D1.
We first note that D1 ∪D2 is a 2-dominating set.
Fact 4 The set D1 ∪D2 is a 2-dominating set in G.
Proof. Suppose u / ∈D1 and |N(u) ∩D1| = i for i = 0, 1. Then d(u) ≥2, and so u has at least 2 −i neighbors in V \ D1. In the second step of the procedure 2 −i of them will be added to D2. □ Let D∗be an optimal 2-dominating set in G. We will now show that |D1 ∪ D2| ≤C|D∗| for some constant C.
Fact 5 |D1| = O(|D∗|).
Proof. First note that if d(v) < 2, then v ∈D∗. In addition, for v ∈D1 \D∗, we certainly have |N(v)| > K because N(v) is an option for S. Let V1 := {v ∈D1 \ D∗: |N(v) ∩D∗| ≥K/2}. We have K|V1|/2 ≤|EG(V1, D∗)| < 5 · 64(|V1| + |D∗|) by Lemma 1, and so |V1| < 320|D∗|. Let V2 := {v ∈D1 \ D∗: |N(v) ∩D∗| < K/2}. Take v ∈V2 and let W ⊆D∗be a minimum set which 2-dominates N(v)\D∗. We have |W| ≥K/2 because W ∪(N(v)∩D∗) 2-dominates N(v). In addition, |N(v) \ D∗| = |N(v)| −|N(v) ∩D∗| > K/2. Therefore, by minimality of W, there is a matching of size at least |W|/2 ≥K/4 ≥4 between W and N(v) \ D∗, and consequently there is a v, W2-fan of size four. Therefore, v ∈D∗ 4,2 and, by Lemma 4, |V2| = O(|D∗|). □ To show that |D1 ∪D2| = O(|D∗|) we will now prove that |D2| = O(|D∗|).
Fact 6 |D2| = O(|D∗|).
Proof. If D∗\D1 = ∅, then every vertex in V \D1 has two neighbors in D1 and D2 = ∅. For a vertex u ∈V \ (Z ∪D∗) which has one neighbor in D1, there is at least one v ∈D∗\ D1 such that u ∈N(v). We fix one such v and say that u belongs to v. If u has no neighbors in D1, then there exist at least two vertices v, v′ ∈D∗\ D1 such that u ∈N(v) ∩N(v′). We fix such a pair (v, v′) and say that u belongs to v if d′(v) < d′(v′) (if the degrees are equal then we select one of v, v′ arbitrarily.) Note that every vertex u ∈V \ Z is either in D∗\ D1 or belongs to exactly one vertex from D∗\ D1.
In order to show that |D2| = O(|D∗|) we will argue that for every vertex v ∈D∗\ D1 there can be at most a constant number of vertices added to D2 by v and by vertices which belong to v. Let v ∈D∗\ D1. Note that v can select at most two vertices to be added to D2. In addition, there exists a set S ⊆V \ {v} such that |S| ≤K and S 2-dominates N(v). Fix such a set S and call it Sv. Let ˜ N(v) be the set of u ∈N(v) such that u belongs v. First suppose | ˜ N(v)| ≤L. Since each u ∈˜ N(v) selects at most two vertices in step 3, the number of vertices added to D2 by vertices from ˜ N(v) is at most 2L. Now assume | ˜ N(v)| > L. Let w ∈V (D1 ∪D∗∪Sv) and suppose w has a neighbor in ˜ N(v). If |N(w) ∩˜ N(v)| > C, then Sv 2-dominates (N(w) ∩˜ N(v)) \ Sv which has size at least 4K2, and so by Lemma 6, G contains K4,4 which is not possible. Thus 0 < |N(w) ∩˜ N(v)| ≤C for every vertex v. For a vertex x, if u ∈N(w) ∩˜ N(x), u belongs to x and d′(x) > L. Thus if u selects w in step 3, then d′(w) > L.
All neighbors of w in G[V \ Z] are either in D∗\ D1 or in one of the sets ˜ N(x).
If w has four neighbors in D∗, then w ∈D∗ 4,1. Otherwise, w has at least one neighbor in ˜ N(x) for at least (L −3)/C ≥4 vertices x ∈D∗\ D1. Consequently, by Lemma 4, the number of such vertices is O(|D∗|). Finally, since |Sv| ≤K, there are at most K vertices in Sv which can be added to D2 by vertices in ˜ N(v) in step 3. □ Thus we have the following theorem.
Theorem 7. There exists C ∈Z+ such that the following holds. Let G be a graph with no TK4,4. Then Procedure 2 finds a 2-dominating set D such that |D| ≤Cγ2(G).
From Theorem 7, we get a constant approximation for the 2-minimum dominat-ing set problem in linklessly embeddable graphs.
Corollary 8 Let G be a linklessy embeddable graph. Then Procedure 2 applied to G returns a 2-dominating set of size O(γ2(G)).
5 (1 + ϵ)-approximation In this section we will show that a modification of the approach from gives a (1 + ϵ)-approximation of the k-dominating set for k ≥2 in the class of graphs of bounded genus as well as in the class of linklessly embeddable graphs. In fact, our main algorithm is much more general and applies to the class of graphs with no Kp-minor for some fixed positive integer p. Let G be a graph on n vertices with no Kp as a minor and suppose ω : E(G) →Z+ is a weight function. For a partition (V1, . . . , Vl) of V (G), we will denote by ˜ G = ˜ G(V1, . . . , Vl) the graph obtained from G by contracting each of the sets Vi to a vertex vi and setting the weight ω(vivj) = P e∈E(Vi,Vj) ω(e) when E(Vi, Vj), the set of edges with one endpoint in Vi and another in Vj, is non-empty. A straightforward generalization of the main clustering procedure from gives the following theorem.
Theorem 9. Let p ∈Z+ and let ϵ > 0. There exists C such that the following holds. Suppose G is a graph on n vertices with no Kp-minor and let ω : E(G) → Z+. There is a distributed algorithm which finds a partition (V1, . . . , Vl) such that G[Vi] has diameter O(C) and X e∈˜ G ω(e) ≤ϵ X e∈G ω(e).
The algorithm runs in C log∗n rounds.
By combining Theorem 9 with some additional analysis we obtain the following fact.
Theorem 10. Let p, k ∈Z+ and let δ > 0. Let G be a graph on n vertices with no Kp-minor and suppose Q is a k-dominating set for G. There is a distributed algorithm which finds a k-dominating set D in G such that |D| ≤γk(G) + δ|Q|.
The algorithm runs in C log∗n rounds where C depends on p, k and δ only.
Note that the case k = 1 is proved in but the algorithm from doesn’t address the situation k > 1 and a slightly different approach is needed. Before proving Theorem 10 let us note that in view of the results from the previous section we have the following corollaries.
Corollary 11 Let g, k ∈Z+ such that k ≥2 and let ϵ > 0. Let G be a graph on n vertices with genus g. There is a distributed algorithm which finds a k-dominating set D in G such that |D| ≤(1 + ϵ)γk(G). The algorithm runs in C log∗n rounds where C depends only on g, k and ϵ.
Proof. In view of Corollary 3, there is a distributed algorithm which in a con-stant number of rounds finds a k-dominating set Q in G such that |Q| ≤LγK(G) for some constant L. Let δ := ϵ/L. The by Theorem 10 finds a dominating set D such that |D| ≤(1 + ϵ)γk(G). □ In addition, using the algorithms for linklessly embeddable graphs from pre-vious sections we obtain the following fact.
Corollary 12 Let k ≥2 and let ϵ > 0. Let G be a linklessly embeddable graph on n vertices. There is a distributed algorithm which finds a k-dominating set D in G such that |D| ≤(1 + ϵ)γk(G). The algorithm runs in C log∗n rounds where C depends only on k and ϵ.
Proof of Theorem 10. Let Q be a k-dominating set in G. Let G′ be the following oriented graph obtained form the bipartite graph G[Q, V (G) \ Q]. We put the arc from v to w for every edge vw with v ∈V (G) \ Q and w ∈Q. Since Q is k-dominating, every vertex v ∈V (G)\Q has at least k out-neighbors in G′.
By choosing one such out-neighbor arbitrarily, we obtain a set of stars S1, . . . , Sl with centers in vertices from Q. Let Uj := V (Sj) and let cUj denote the center of Sj, that is {cUj} = Uj ∩Q. We will refer to sets Uj as small clusters. Let H0 be the digraph obtained from G′ by contracting each Sj to a vertex. (Note that it is possible to have both arcs UjUi and UiUj in H0.) Let D := ∅and let K := 2c · p√log pk/δ where c is the constant in Lemma 2.
We will modify the original small clusters and construct a sequence of di-graphs H1, . . . , Hk−1 by increasing D in each step so that |D| is small with respect to |Q| and clusters in Hk−1 obtained by deleting vertices which are k-dominated by D have out-neighbors in at most K other vertices from Q. Some of the centers of small clusters will be added to D, some vertices will be reas-signed from one cluster to another, and, in addition, some vertices will become k-dominated by D and will become inactive. Although we no longer need to k-dominate inactive vertices (as they are k-dominated by D), they can still play an important role because they themselves can be inside an optimal k-dominating set. Although the sets Uj will change as we modify them, we will use Uj to re-fer to the cluster determined by cUj during the execution of the algorithm. For cluster U we use U ′′ to denote the set of inactive vertices in U \ {cU} and we set U ′ := U \ U ′′. Initially, U ′′ := ∅for every cluster U and we set B0 := V (G) \ Q.
We will now describe our procedure which modifies the small clusters. We will use d+ F (X) to denote the number of out-neighbors of a cluster X in F. Recall that clusters are associated with their centers and so cluster X at step i can be different than X at step j for i ̸= j.
For the general step, assume i ≥1 and let Xi−1 := {U|d+ Hi−1(U ′) > K}. If U ∈Xi−1, then add cU to D. Consider v ∈Bi−1 ∩U ′. If v becomes k-dominated by D, then move v to U ′′ and otherwise, add v to Bi. Every vertex v ∈Bi has at most k −1 neighbors in D and so there exists a small cluster W such that cW / ∈D and vcW ∈G′. Let v join one such cluster W.
Let Hi be obtained by contracting each small cluster U to a vertex and by adding the arc UW if the set of arcs from U ′ to W ′ in G′ is non-empty.
We will prove the following lemma.
Lemma 7. The following holds: (a) Pk i=1 |Xi−1| < δ|Q|/2.
(b) For every v ∈Bk, |NG(v) ∩D| ≥k.
(c) For every small cluster U there exists a set of small clusters Q′ ⊆Q such that |Q′| ≤K and every vertex v ∈U (Bk ∪{cU}) has NG(v)∩Q ⊆Q′ ∪D.
To prove the lemma we will show a sequence of claims. Part (a) follows from the following observation.
Claim. |Xi−1| < δ|Q|/(2k).
Proof. From Lemma 2, P U∈Hi−1 d+ Hi−1(U) = |E(Hi−1)| ≤cp√log p|Q|. Thus |Xi−1| < cp√log p|Q|/K = δ|Q|/(2k). □ We will now verify part (b).
Claim. If v ∈Bi, then |NG(v) ∩D| ≥i.
Proof. This is certainly true when i = 0. Suppose i ≥1. If v ∈Bi, then v ∈Bi−1 and in the ith step of the procedure v ∈U ′ for some U ∈Xi−1. Since cU is added to D, |NG(v) ∩D| increases by at least one. □ Part (c) is slightly more involved and we split the argument into two claims.
First note the following.
Claim. If U ∈Xi, then U / ∈Xi+1.
Proof. We proceed by induction on i. If U ∈X0, then after adding cu to D and partitioning U into U ′ and U ′′, every vertex from U ′ is reassigned and so in the next step U ′ is empty. Suppose U ∈Xi. Then, by induction, U / ∈Xi−1. Thus some vertices from Bi−1 joined U, as this is the only way for the out-degree of a cluster to increases. Each such vertex is either inactive or is in Bi after cU is added to D in the ith step, and every vertex from Bi ∩U joins a cluster different than U with center which is not in D. Consequently, none of these added vertices is in U ′ in the iteration i + 1, and we have d+ Hi+1(U ′) ≤d+ Hi−1(U ′). □ Claim. Let i ≥1. For every U there exists a set Q′ ⊆Q such that |Q′| ≤K and for every vertex v ∈U \ (Bi ∪{cU}), NG(v) ∩Q ⊆Q′ ∪D.
Proof. If U / ∈Xi then d+ Hi(U ′) ≤K and so NG(U ′ \ (Bi ∪{cU})) ∩Q ⊆ Q′ for some set Q′ of size at most K. Suppose U ∈Xi. Then U / ∈Xi−1 by Claim 5. Therefore some vertices from Bi−1 joined U and each of them is either inactive (and dominated by D) or is in Bi. Consider U ′ in the ith iteration of the procedure. The set U ′ is a subset of the union of X and Y , where X is the set U ′ \ Bi−1 from the iteration i −1 and Y is the set of vertices added to the small cluster U in the ith iteration. The number of distinct out-neighbors of X in Q is therefore at most K, and the vertices from Y are either dominated by D or are in Bi. □ This proves part (c) of the lemma.
We will now continue with the proof of Theorem 10.
Let U1, . . . , Ul denote the clusters in Hk and recall that l = |Q|. Let G∗ be obtained from G by contracting each Ui to a vertex ui. In addition, set ω(uiuj) = 1 for every edge uiuj ∈G∗. Let ϵ := δ/(6c · p√log pK) and use the clustering procedure from Theorem 9 to find a partition P of {u1, . . . , ul} into V1, . . . , Vs. Then ω( ˜ G) ≤ϵ|E(G)| ≤ϵ · cp p log pl < δl/(6K).
Let ∂P be the set of u ∈{u1, . . . , ul} such that for some i ̸= j, u ∈Vi and NG∗(u) ∩Vj ̸= ∅. We have |∂P| ≤2ω( ˜ G) ≤δl/(3K).
Let u ∈∂P and consider the small cluster U which was contracted to u. Add cU to D and note that by Lemma 7 (b) every vertex v ∈Bk ∩U has k-neighbors in D. In addition, by Lemma 7 (c), there is a set Q′ of at most K vertices in Q such that every vertex v ∈U \ (Bk ∪{cU}) has NG(v) ∩Q ⊆Q′ ∪D. Since every vertex v ∈V (G) \ Q has at least k neighbors in Q, by adding Q′ ∪{cU} to D we k-dominate all vertices from U. Add Q′ ∪{cU} to D for every U ∈∂P. By Lemma 7 (a), we have |D| ≤δ|Q|/2 + δl(K + 1)/(3K) < δ|Q|.
Let Wi := S U∈Vi\∂P U. Since the diameter of G[Vi] is O(1), it is possible to find in O(1) rounds an optimal set Di ⊆Vi such that Di ∪D k-dominates Wi. Let D′ := Sl i=1 Di ∪D and let D∗denote an optimal k-dominating set in G. Then |D′| ≤ X |Di| + |D| ≤ X |D∗∩Vi| + |D| ≤|D∗| + |D| < γk(G) + δ|Q| where the second inequality follows from the fact that every vertex in Wi can be k-dominated only by vertices in Vi ∪D and every vertex in S U∈∂P U is k-dominated by D. □ References 1. N. Alon, S. Gutner: Kernels for the Dominating Set Problem on Graphs with an Excluded Minor, Electronic Cooloquium on Computational Complexity, Report No.
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2. S. A. Amiri, S. Schmid, S. Siebertz: A Local Constant Factor MDS Approximation for Bounded Genus Graphs, PODC 2016, Proceedings of the 2016 ACM Symp. on Principles of Distributed Computing, (2016), 227–233.
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4. A. Czygrinow, M. Han´ ckowiak, and W. Wawrzyniak: Fast distributed approxima-tions in planar graphs, International Symposium on Distributed Computing, DISC, Arcachon, France, September 2008, LNCS 5218, pages 78–92.
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Lower bounds for local approximation, J. ACM 60(5): 39, 2013, pages 1–23.
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4168 | https://archive.org/details/vogelsqualitativ0000voge | Vogel's qualitative inorganic analysis : Vogel, Arthur Israel : Free Download, Borrow, and Streaming : Internet Archive
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Vogel's qualitative inorganic analysis
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Vogel's qualitative inorganic analysis
byVogel, Arthur Israel
Publication date1987TopicsChemistry, Analytic -- Qualitative, Chemistry, Inorganic, 11030 inorganic 21030 qualitative analysis, Chemistry, QualitativePublisher Harlow, Essex, England : Longman Scientific & Technical ; New York : Wiley Collectioninternetarchivebooks; inlibrary; printdisabledContributorInternet ArchiveLanguageEnglishItem Size 740.5M
ix, 310 pages : 24 cm
Previous ed. as: Vogel's textbook of macro and semimicro qualitative inorganic analysis. 1979
Includes bibliographical references and index
Notes
cut text due to tight binding
page 56 cut text
Access-restricted-item true Addeddate 2020-09-16 14:03:33 Associated-names Svehla, G; Vogel, Arthur Israel. Textbook of macro and semimicro qualitative inorganic analysis Boxid IA1933109 Camera USB PTP Class Camera Collection_set printdisabled External-identifierurn:oclc:record:1200622187
urn:lcp:vogelsqualitativ0000voge:lcpdf:b48c412b-4011-4caf-a326-74153fe68373
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4169 | https://www.geeksforgeeks.org/dsa/find-last-2-survivors-in-n-persons-standing-in-a-circle-after-killing-next-to-immediate-neighbour/ | Find last 2 survivors in N persons standing in a circle after killing next to immediate neighbour - GeeksforGeeks
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Find last 2 survivors in N persons standing in a circle after killing next to immediate neighbour
Last Updated : 12 Jul, 2025
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Given an integer N representing N persons standing in a circle, the task is to find the last 2 persons remaining when a person kills their next to the immediate neighbor in a clockwise direction.
Examples:
Input: N = 5
Output: 1 4
Explanation:
Initially: 1 2 3 4 5
=> 1 kills 3
Standing: 1 2 4 5
=> 2 kills 5
Standing: 1 2 4
=> 4 kills 2
Final Standing: 1 4
Input: N = 2
Output: 1 2
Naive Approach: A simple approach is to keep a bool array of size N to keep track of whether a person is alive or not.
Initially, the boolean array will be true for all persons.
Keep two pointers, one at the current alive person and second to store previous current person.
Once found a second alive neighbour from the current person, change its boolean value to false.
Then again current is updated to next alive from previous.
This process will continue till last two persons survive.
Time Complexity: O(N 2)
Auxiliary Space: O(N)
Efficient Approach: An efficient approach is to remove the person, if dead, from the data structure so that it is not traversed again.
After one complete round only, there will be only N/2 person, at max.
Then in the next round it will be left with N/4 person and so on until a number of alive people become 2.
Below is the implementation of the above approach:
C++
```CPP
// C++ implementation of the approach
include
using namespace std;
// Node for a Linked List
struct Node {
int val;
struct Node next;
Node(int _val)
{
val = _val;
next = NULL;
}
};
// Function to find the last 2 survivors
void getLastTwoPerson(int n)
{
// Total is the count
// of alive people
int total = n;
struct Node head = new Node(1);
struct Node temp = head;
// Initiating the list of n people
for (int i = 2; i <= n; i++) {
temp->next = new Node(i);
temp = temp->next;
}
temp->next = head;
temp = head;
struct Node del;
// Total != 2 is terminating
// condition because
// at last only two-person
// will remain alive
while (total != 2) {
// Del represent next person
// to be deleted or killed
del = temp->next->next;
temp->next->next
= temp->next->next->next;
temp = temp->next;
free(del);
total -= 1;
}
// Last two person to
// survive (in any order)
cout << temp->val << " "
<< temp->next->val;
}
// Driver code
int main()
{
int n = 2;
getLastTwoPerson(n);
return 0;
}
```
// C++ implementation of the approach
#include using namespace std;// Node for a Linked List struct Node { int val; struct Node next; Node(int _val) { val = _val; next = NULL; }};// Function to find the last 2 survivors void getLastTwoPerson(int n){ // Total is the count // of alive people int total = n; struct Node head = new Node(1); struct Node temp = head; // Initiating the list of n people for (int i = 2; i <= n; i++) { temp->next = new Node(i); temp = temp->next; } temp->next = head; temp = head; struct Node del; // Total != 2 is terminating // condition because // at last only two-person // will remain alive while (total != 2) { // Del represent next person // to be deleted or killed del = temp->next->next; temp->next->next = temp->next->next->next; temp = temp->next; free(del); total -= 1; } // Last two person to // survive (in any order) cout << temp->val << " " << temp->next->val;}// Driver code int main(){ int n = 2; getLastTwoPerson(n); return 0;}
Java
```java
// Java implementation of the approach
class GFG{
// Node for a Linked List
static class Node
{
int val;
Node next;
Node(int _val)
{
val = _val;
next = null;
}
};
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
// Total is the count
// of alive people
int total = n;
Node head = new Node(1);
Node temp = head;
// Initiating the list of n people
for(int i = 2; i <= n; i++)
{
temp.next = new Node(i);
temp = temp.next;
}
temp.next = head;
temp = head;
Node del;
// Total != 2 is terminating
// condition because
// at last only two-person
// will remain alive
while (total != 2)
{
// Del represent next person
// to be deleted or killed
del = temp.next.next;
temp.next.next = temp.next.next.next;
temp = temp.next;
del = null;
System.gc();
total -= 1;
}
// Last two person to
// survive (in any order)
System.out.print(temp.val + " " +
temp.next.val);
}
// Driver code
public static void main(String[] args)
{
int n = 2;
getLastTwoPerson(n);
}
}
// This code is contributed by Rajput-Ji
Python3python3
Python3 implementation of the approach
Node for a Linked List
class newNode:
def __init__(self, val):
self.val = val
self.next = None
Function to find the last 2 survivors
def getLastTwoPerson(n):
# Total is the count
# of alive people
total = n
head = newNode(1)
temp = head
# Initiating the list of n people
for i in range(2, n + 1, 1):
temp.next = newNode(i)
temp = temp.next
temp.next = head
temp = head
de = None
# Total != 2 is terminating
# condition because
# at last only two-person
# will remain alive
while (total != 2):
# de represent next person
# to be deleted or killed
de = temp.next.next
temp.next.next = temp.next.next.next
temp = temp.next
del de
total -= 1
# Last two person to
# survive (in any order)
print(temp.val, temp.next.val)
Driver code
if name == 'main':
n = 2
getLastTwoPerson(n)
This code is contributed by SURENDRA_GANGWAR
C#csharp
// C# implementation of the approach
using System;
class GFG{
// Node for a Linked List
class Node
{
public int val;
public Node next;
public Node(int _val)
{
val = _val;
next = null;
}
};
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
// Total is the count
// of alive people
int total = n;
Node head = new Node(1);
Node temp = head;
// Initiating the list of n people
for(int i = 2; i <= n; i++)
{
temp.next = new Node(i);
temp = temp.next;
}
temp.next = head;
temp = head;
Node del;
// Total != 2 is terminating
// condition because
// at last only two-person
// will remain alive
while (total != 2)
{
// Del represent next person
// to be deleted or killed
del = temp.next.next;
temp.next.next = temp.next.next.next;
temp = temp.next;
del = null;
total -= 1;
}
// Last two person to
// survive (in any order)
Console.Write(temp.val + " " +
temp.next.val);
}
// Driver code
public static void Main(String[] args)
{
int n = 2;
getLastTwoPerson(n);
}
}
// This code is contributed by Amit Katiyar
JavaScriptjavascript
// Javascript implementation of the approach
// Node for a Linked List
class Node {
constructor(val)
{
this.val = val;
this.next = null;
}
};
// Function to find the last 2 survivors
function getLastTwoPerson(n)
{
// Total is the count
// of alive people
var total = n;
var head = new Node(1);
var temp = head;
// Initiating the list of n people
for (var i = 2; i <= n; i++) {
temp.next = new Node(i);
temp = temp.next;
}
temp.next = head;
temp = head;
var del;
// Total != 2 is terminating
// condition because
// at last only two-person
// will remain alive
while (total != 2) {
// Del represent next person
// to be deleted or killed
del = temp.next.next;
temp.next.next
= temp.next.next.next;
temp = temp.next;
total -= 1;
}
// Last two person to
// survive (in any order)
document.write( temp.val + " "
+ temp.next.val);
}
// Driver code
var n = 2;
getLastTwoPerson(n);
// This code is contributed by noob2000.
```
Output:
1 2
Time Complexity: O(Nlog N)
Auxiliary Space: O(N)
Comment
More info
S
souravrawat191
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4170 | https://akuli.github.io/math-derivations/plane-geometry/inscribed-circle.html | Inscribed Circle - Math Derivations
Front page- [x] Equations and functions
Inverse functionsHow solving equations worksStrictly increasing/decreasing functionsHow solving inequalities works
[x] Vectors
[x] Dot product
ProjectionAngle between vectorsi→'s with i→'s, j→'s with j→'s
[x] Cross product
Definition
[x] Discrete math
Sum formulasBinomial coefficients
[x] Plane geometry
[x] Circle
Inscribed angle theorem
[x] Triangle
Law of sinesLaw of cosinesInscribed circle
[x] Analytic plane geometry
[x] Transforming curves
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[x] Parabola
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Why is y=1 x a hyperbola?
[x] Calculus
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"Definition" by plugging in numbersDefinition of limitUniquenessLimit of sumLimit of productLimit of differenceLimit of divisionOne-sided limitsInequality and limitLimit of vector
[x] Continuity
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[x] Derivative
Definition of derivativeNotation for derivativesPower rule, part 1Basic derivative rulesChain rulePower rule, part 2Product rule and quotient ruleDerivative of vectorDerivatives of sin, cos and tan
[x] Integral
Introduction to integralsRefinement of partitionDefining the integralAverageFundamental theorem of calculus, part 1Fundamental theorem of calculus, part 2More notation for integralsIntegral rulesU substitutionVolume of solid of revolutionArea of surface of revolution
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[x] Subspace and basis
Defining subspaces and comparing to spansBasis
Inscribed Circle¶
As a part of this derivation, we got a formula for the radius of a circle drawn around a triangle. In this derivation, we will find a formula for a circle drawn inside a triangle.
Consider any triangle with any side lengths a, b and c, and draw a circle inside it so that it touches each side as in the picture below. A circle like this is called the inscribed circle of the triangle. Our goal is to find its radius r.
Let's split the triangle into 3 smaller triangles by connecting the corners to the center of the circle:
Because the radius r is perpendicular to the side with length a, the area of the green part is a r/2. Areas of the other parts are similarly b r/2 and c r/2. Letting A denote the area of the whole triangle, we get A=a r 2+b r 2+c r 2=(a+b+c)r 2. Multiplying both sides by 2 and dividing by a+b+c gives the following result.
Consider any triangle. Let a, b and c denote its side lengths, let A denote its area, and let r denote the radius of the inscribed circle. Then r=2 A a+b+c. |
4171 | http://www.stat.ucla.edu/~vlew/stat11/FA01/lectures/fa011118.pdf | Statistics M11/Economics 40 Lew Chapter 9.1 Chapter 9.1 – Inference for Two-Way Tables Introduction • The methods introduced in this chapter allow us to compare scenarios when the categorical data implies more than two proportions – more than two “p”s. • A single statistical test will handle two different scenarios: • Comparing two or more populations when the response variable has two or more possible values. • Testing to see if two categorical variables are independent. (9.2 optional) • Will use the methods for summarizing categorical data (two-way tables, marginal distributions, conditional distributions) Inference for Two-Way Tables • Categorical Variables (optional 2.6) These are inherently qualitative variables OR quantitative variables that have been grouped into categories. The analysis of categorical variables uses counts (frequencies) or percentages. • The two-way table To summarize the relationship between two categorical variables, we construct a two-way table. The size of the table is given by the notation r x c where r is the number of rows and c is the number of columns. Each row defines a possible value for the first categorical variable, and each column defines a possible value for the second categorical value. A two-way table with r rows and c columns has r x c cells. See Example 9.3, a 4 x 2 table for gender and athletic goals. • Describing relations in two-way tables Use the table to calculate marginal distributions, joint distributions and conditional distributions. See Example 9.4. A marginal distribution is the distribution of the row variable alone or of the column variable alone. Sometimes referred to as the "marginals" because they are the information at the margins of a table. The conditional distribution is the distribution of one row against all the columns or one column against all of the rows. It is conditional because one is looking at a single "condition". The joint distribution involves examining a specific row value with a specific column value simultaneously, it just the pairing of values. • The (null) hypothesis: no association The null hypothesis – a statement conveying that there is “nothing going on” – for two-way table inference is that there is no association between the row and column variables. The alternative hypothesis – a statement conveying what must be true if Statistics M11/Economics 40 Lew Chapter 9.1 the null is not true – for two-way table inference is that there is an association between the two variables. Note that this form of the alternative hypothesis does not imply any particular direction for the association, just that the association exists. As it states in the text, “[w]e cannot describe Ha [the alternative hypothesis] as either one-sided or two-sided, because it includes all of the many kinds of association that are possible.” (IPS, p. 628) • Expected cell counts In order to see if there is an association between the two variables, we will first re-calculate the table reflecting what we would expect to be true assuming no association exists. This means calculating the expected cell counts. We will then use these expected cells counts to calculate a test statistic, which will compare these expected cell counts against the actual cell counts. We will use this test statistic answer to judge how different these two sets of counts are. If there is a “big” difference [as with the other significance tests, we’ll be using what we know about the probability distribution for the test statistic to define how big “big” is], then we can say that there is evidence that an association exists. If there is not a “big” difference, then we will conclude that there is not sufficient evidence to say that an association exists. How do we calculate these expected cell counts? We will use the marginal distributions for one of the variables and apply them to the row (or column) totals of the other variable, using this formula: expected cell count = n total column total row × where n is the total number of observations for the table. • The chi-square test The test statistic we will use for this inference exercise is the chi-square statistic, X 2, where ∑ − = count expected ) count expected count observed ( 2 2 X Statistics M11/Economics 40 Lew Chapter 9.1 As we stated above, in order to use this test statistic, we need to know its probability distribution. This test statistic is related to the binomial approximation we used in Chapter 8, but if the null hypothesis is true, it has its own approximate distribution, called the chi-square distribution, which we denote using the Greek symbol for chi: χ2. As was the case the with Student’s t distribution, the chi-square distribution is actually a family of distributions, each completely specified by its degrees of freedom, which is equal to (r – 1)(c – 1). Note that chi-square distributions are different from the distributions we have used thus far in that all chi-square values are always positive (the square of any real number is always positive) and the distributions are always skewed to the right. See Figure 9.3 (below): Chi-Square Test for Two-Way Tables Chi-Square tests for two-way tables always have the following null and alternative hypotheses (or some version thereof): H0: There is no relationship between the two variables being examined. (This is often shorten to the statement, “The variables are not related.”) Ha: There is a relationship between the two variables being examined. (This is often shorten to the statement, “The variables are related.”) If H0 is true, the chi-square test statistic X 2 has approximately a χ 2 distribution with (r – 1)(c – 1) degrees of freedom. The P-value for the chi-square test is P(χ 2 ≥ X 2) where X 2 is a random variable having the χ2(df) distribution with df = (r – 1)(c – 1). You can safely use the chi-square test with critical values from the chi-square distribution when all expected cell counts are ≥ 1, and no more than 20% of the expected cell counts are < 5. For a 2 x 2 table, all four expected cell counts should be ≥ 5. Statistics M11/Economics 40 Lew Chapter 9.1 • The chi-square test and the z test for two proportions Note that the data used for a z test for two proportions can be summarized as a 2 x 2 table. Thus we can compare these two population proportions either using the z test (from Section 8.2) or using the chi-square test above. As the test states, “…these tests always give the same result, because the chi-square statistic is equal to the square of the z statistic, and the χ 2(1) critical values are equal to the squares of the corresponding N(0,1) critical values. … The advantage of the z test is that we can test either one-sided or two-sided alternatives. The chi-square test always tests the two-sided alternative.” (IPS, p. 632; italics used in the original) Æ You can only use the z test when you have a 2 x 2 table – if the table is larger, e.g. 2 x 3, 3 x 4, you must use a chi-square test. |
4172 | https://www.reddit.com/r/puzzles/comments/1gypkjg/you_have_a_cube_a_beautiful_logic_puzzle/ | You have a cube. A beautiful logic puzzle. : r/puzzles
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You have a cube. A beautiful logic puzzle.
[SOLVED]
You have a universally white cube. You paint the outside of the cube black. You cut the cube into 3x3x3 so that there are 27 cubes. You disassemble the cube and put all 27 cubes into a bag. At random, a cube is selected from the bag and randomly placed on the table in front of you. You can only see five sides of this small cube and cannot see the underside. The five sides that you see are all white. What is the chance that the underside is black?
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How do you create a probability distribution of rolling two dice, where the random variable x is the sum of the dice?
Sushant Kala
M.Sc in Mathematics, Indian Institute of Technology Ropar (IITRPR) (Graduated 2020)
·
4y
The sample space for your experiment is S={(m,n):m,n∈Zand1≤m,n≤6}.
A random variable associated with an experiment is a real-valued function X on the sample space S. Here, X(m,n)=m+n taking values from 2 to 12.
We then say X is a random variable on S, taking values 2,3,4,….,12 and by P(X=k) we mean P(X^{-1}(k)), where X^{-1}(k) contain those (m,n) whose sum is k. (It is generally probability of getting sum k.)
Now we know for a finite sample space P(X=k)=#X^{-1}(k)/#S, from this you will get all the probabilities by counting cardinality of these sets.
(See the picture: from Wikipedia)
The probabili
The sample space for your experiment is S={(m,n):m,n∈Zand1≤m,n≤6}.
A random variable associated with an experiment is a real-valued function X on the sample space S. Here, X(m,n)=m+n taking values from 2 to 12.
We then say X is a random variable on S, taking values 2,3,4,….,12 and by P(X=k) we mean P(X^{-1}(k)), where X^{-1}(k) contain those (m,n) whose sum is k. (It is generally probability of getting sum k.)
Now we know for a finite sample space P(X=k)=#X^{-1}(k)/#S, from this you will get all the probabilities by counting cardinality of these sets.
(See the picture: from Wikipedia)
The probability distribution of a discrete random variable is generally a function, mapping values of a random variable to their respective probabilities. Defining, f: X → [0,1] such that f(k) = P(X=k). Note: here domain of f is values taken by random variable X.
The formula for expectation is given by,
E(X)=∑k∈X k.P(X=k)
Related questions
Two dice are rolled. Let X be a random variable denoting the sum of the numbers on the two dice. What is the probability distribution of X? What is its expected value?
What is the probability distribution of X if you roll two fair dice and let X denote the sum if the score obtained on the two dice?
What is the probability of rolling two dice?
In a single throw of two dice, what is the probability of getting a sum of 9?
What is the probability of rolling a sum of 8 with two fair dice?
Steven Smith
Earned 98% or higher in all my math classes at UCMO.
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Author has 3.4K answers and 9.1M answer views
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8y
Originally Answered: What is the formula for the probability of rolling two dice with a sum of x?
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If x≤6, this can be done simply using stars and bars. Stars and bars (combinatorics) - Wikipedia
You have x pips to place onto 2 dice, where each die must have from 1 to 6 pips.
First, you can place the minimum required pips, 1 on each die, which would be 2. That leaves you with x−2 pips that can be placed randomly. There are 2 dice the pips can be placed on, to separate those we need 1 bar. The pips will be the stars.
Using the stars and bars algorithm, that means there are ((x−2)+11)=x−1 ways to place the remaining pips.
There are 6⋅6=36 possible results of rolling 2 dice, so
If x≤6, this can be done simply using stars and bars. Stars and bars (combinatorics) - Wikipedia
You have x pips to place onto 2 dice, where each die must have from 1 to 6 pips.
First, you can place the minimum required pips, 1 on each die, which would be 2. That leaves you with x−2 pips that can be placed randomly. There are 2 dice the pips can be placed on, to separate those we need 1 bar. The pips will be the stars.
Using the stars and bars algorithm, that means there are ((x−2)+11)=x−1 ways to place the remaining pips.
There are 6⋅6=36 possible results of rolling 2 dice, so that gives us a probability of:
x−136
Next we need to determine the probability for 7–12. There are 2 ways of doing this, subtracting the counts that are 7 or above, or taking advantage of symmetry.
I like symmetry, so I will use that method.
There is only 1 way to roll 2, double 1s. There is only 1 way to roll 12, double 6s.
There are 2 ways to roll 3, 1 and 2, 2 and 1. There are 2 ways to roll 11, 5 and 6, 6 and 5.
This symmetry continues.
Therefore, P(X=x)=P(X=(14−x))
Plugging this into the earlier solution:
(14−x)−136=13−x36
That gives this solution:
pX(x)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩x−136for x=1,2,3,4,5,613−x36for x=7,8,9,10,11,120otherwise
The other way is this:
That method can be expanded for larger numbers of dice as well.
Assistant
Bot
·
1y
To create a probability distribution for the random variable , which represents the sum of the values when rolling two six-sided dice, follow these steps:
Step 1: Identify the Possible Outcomes
When rolling two dice, each die has 6 faces, leading to a total of possible outcomes. Each outcome can be represented as an ordered pair , where is the result of the first die and is the result of the second die.
Step 2: Determine the Sums
The possible sums can range from 2 (when both dice show 1) to 12 (when both dice show 6). Here are the possible sums and how they can
To create a probability distribution for the random variable , which represents the sum of the values when rolling two six-sided dice, follow these steps:
Step 1: Identify the Possible Outcomes
When rolling two dice, each die has 6 faces, leading to a total of possible outcomes. Each outcome can be represented as an ordered pair , where is the result of the first die and is the result of the second die.
Step 2: Determine the Sums
The possible sums can range from 2 (when both dice show 1) to 12 (when both dice show 6). Here are the possible sums and how they can be obtained:
Sum = 2: (1, 1) → 1 way
Sum = 3: (1, 2), (2, 1) → 2 ways
Sum = 4: (1, 3), (2, 2), (3, 1) → 3 ways
Sum = 5: (1, 4), (2, 3), (3, 2), (4, 1) → 4 ways
Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → 5 ways
Sum = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 ways
Sum = 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → 5 ways
Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 ways
Sum = 10: (4, 6), (5, 5), (6, 4) → 3 ways
Sum = 11: (5, 6), (6, 5) → 2 ways
Sum = 12: (6, 6) → 1 way
Step 3: Count the Ways for Each Sum
Summarizing the counts of ways to achieve each sum:
| Sum (x) | Number of Ways |
----------------|
| 2 | 1 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
| 6 | 5 |
| 7 | 6 |
| 8 | 5 |
| 9 | 4 |
| 10 | 3 |
| 11 | 2 |
| 12 | 1 |
Step 4: Calculate the Probability for Each Sum
To find the probability of each sum, divide the number of ways to achieve that sum by the total number of outcomes (36):
Step 5: Create the Probability Distribution Table
Now, we can construct the probability distribution:
| Sum (x) | Number of Ways | Probability |
----------------
| 2 | 1 | |
| 3 | 2 | |
| 4 | 3 | |
| 5 | 4 | |
| 6 | 5 | |
| 7 | 6 | |
| 8 | 5 | |
| 9 | 4 | |
| 10 | 3 | |
| 11 | 2 | |
| 12 | 1 | |
Conclusion
This table represents the probability distribution of the sum of two dice rolls, showing the likelihood of each possible sum occurring.
Traruh Synred
Ph.D. in Particle Physics, University of Illinois at Urbana-Champaign (Graduated 1974)
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Author has 16.2K answers and 3M answer views
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3y
It simple. Just count the number combinations that give a given su...
Related questions
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What is the probability of rolling a double (both dice have the same number) given the sum of the 2 dice is 4?
Fred Oberlander
Graduate Degrees Compsci, Math, Stat, Finance, Law
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Author has 1.9K answers and 483.4K answer views
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Updated 3y
Originally Answered: Using the sample space for rolling two dice, what is a probability distribution for the random variable X representing the sum of the numbers that appear?
·
it is the discrete distribution you know perfectly well which has values from 1 to 12 with the most likely result a 7 which can occur any of 6 ways so is 6/12 and the others symmetric about that mean and median.
EDIT: my bad. the SPACE is 1 to 12 however it contains “duplicates” meaning there are 36 actual outcomes tho there are only 12 distinct values. You asked for the SPACE in terms of the totals not the results, big difference. Think of it this way. The domain of possible results is 1,…,12. The number of EVENTS that can occur that are not identical is 36.
Thus the chance of getting a 7 is 6/
it is the discrete distribution you know perfectly well which has values from 1 to 12 with the most likely result a 7 which can occur any of 6 ways so is 6/12 and the others symmetric about that mean and median.
EDIT: my bad. the SPACE is 1 to 12 however it contains “duplicates” meaning there are 36 actual outcomes tho there are only 12 distinct values. You asked for the SPACE in terms of the totals not the results, big difference. Think of it this way. The domain of possible results is 1,…,12. The number of EVENTS that can occur that are not identical is 36.
Thus the chance of getting a 7 is 6/36 not 6/12. Obviously I knew that I mean I have played dice and have known it since I was 8. My apologies.
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Bill Bell
Former Software Developer (1972–2010)
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Author has 2.1K answers and 1.2M answer views
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4y
Originally Answered: Using the sample space for rolling two dice, what is a probability distribution for the random variable X representing the sum of the numbers that appear?
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Draw a six by six table for yourself where the columns represent the possible outcomes for one die and the rows represent the outcomes for the other die.
There will be thirty-six cells in the table, all equally likely. You can calculate the probability of each.
In each cell put the sum of the value for row die and the column die.
You can now look through the table and see that some of the sums occur once, some twice, some three times, and so on.
You can now summarise this information in the form of a distribution: a sum of one has a probability equal to the number of cells in which this sum appear
Draw a six by six table for yourself where the columns represent the possible outcomes for one die and the rows represent the outcomes for the other die.
There will be thirty-six cells in the table, all equally likely. You can calculate the probability of each.
In each cell put the sum of the value for row die and the column die.
You can now look through the table and see that some of the sums occur once, some twice, some three times, and so on.
You can now summarise this information in the form of a distribution: a sum of one has a probability equal to the number of cells in which this sum appears multiplied by the probability represented by one cell.
Just keep going.
Mario Luis Iovaldi
MD, Surgeon, not Statistician. Argentina
·
Author has 2.9K answers and 965.3K answer views
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3y
In R: the tab and the plot
tab
x frequency prob.x
[1,] 2 1 0.02777778
[2,] 3 2 0.05555556
[3,] 4 3 0.08333333
[4,] 5 4 0.11111111
[5,] 6 5 0.13888889
[6,] 7 6 0.16666667
[7,] 8 5 0.13888889
[8,] 9 4 0.11111111
[9,] 10 3 0.08333333
[10,] 11 2 0.05555556
[11,] 12 1 0.02777778
plot(tab, type ="h", col= "red", lwd = 4)
In R: the tab and the plot
tab
x frequency prob.x
[1,] 2 1 0.02777778
[2,] 3 2 0.05555556
[3,] 4 3 0.08333333
[4,] 5 4 0.11111111
[5,] 6 5 0.13888889
[6,] 7 6 0.16666667
[7,] 8 5 0.13888889
[8,] 9 4 0.11111111
[9,] 10 3 0.08333333
[10,] 11 2 0.05555556
[11,] 12 1 0.02777778
plot(tab, type ="h", col= "red", lwd = 4)
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Rachel Townsend
ssa in Business Management, Northeast State Technical Community College (Graduated 2023)
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3y
Originally Answered: What is the formula for the probability of rolling two dice with a sum of x?
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This means the smallest # is 1, the largest # is 6, and there are 2 dice. When you press Enter, you will
get the two numbers rolled on the dice. Keep pressing enter to get additional rolls)
1) Roll 2 dice 50 times each, and record the #’s on each die in a chart. It will look like this:
3, 5 4,1 2,3
You will have 50 blocks where you record your numbers that are rolled (this is just an example)
2) Make a Probability distribution for the SUM of the numbers in each block that you recorded
from your rolls of the dice
(It will look like:
X = sum P(x) = # of times that # occurred/50
(Note: The sum co
This means the smallest # is 1, the largest # is 6, and there are 2 dice. When you press Enter, you will
get the two numbers rolled on the dice. Keep pressing enter to get additional rolls)
1) Roll 2 dice 50 times each, and record the #’s on each die in a chart. It will look like this:
3, 5 4,1 2,3
You will have 50 blocks where you record your numbers that are rolled (this is just an example)
2) Make a Probability distribution for the SUM of the numbers in each block that you recorded
from your rolls of the dice
(It will look like:
X = sum P(x) = # of times that # occurred/50
(Note: The sum comes from each block or rolls. For instance, my sums above are:
3+5=8; 4+1=5; 2+3=5 Yours will be different. I will only list the 5 one time, even though it
occurred more than once)
3) Make another Probability Distribution for SUM of the numbers in each block that you would
EXPECT to occur using Classical Probability. (this is not based on your rolls, but what should
happen)
(It will look like:
X=sum P(x) = the probability of that sum occurring
4) From #2, find:
a. Mean = _________________
b. Standard Deviation = ________________
c. Variance = ___________________
5) From #3, find:
a. Mean = _________________
b. Standard Deviation = ________________
c. Variance = ___________________
6) Describe how close your probability calculations from #2 are to the expected probability
calculations in #3:
7) Why would we expect the probabilities to be different in #2 and #3, and what could we do to
make the probabilities equal in #2 and #3?
Bernard Montaron
Studied Mathematics & Counting (Combinatorics) at Pierre and Marie Curie University (Graduated 1980)
·
Author has 3.2K answers and 2.1M answer views
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4y
Originally Answered: Two dice are rolled. Let X be a random variable denoting the sum of the numbers totaling on the two dice. What is the probability distribution of X? What are the expected value of X and its variance?
·
The range for [math]X[/math] is 2 to 12. [math]X=2[/math] only if the two dices are (1,1) and the probability for this is math^2=1/36[/math]. In a similar way [math]X=12[/math] only if the two dices are (6,6) and the probability for this is also [math]1/36[/math].
If [math]X=(2+12)/2=7[/math] there are 6 possible dice combinations: (1,6),(6,1),(2,5)(5,2),(3,4),(4,3) giving a probability [math]6/36=1/6[/math] for [math]X=7[/math]. You can do this for all possible values of [math]X[/math], this is summarized in the table below.
As you can observe, the distribution has the shape of a triangle, and it is symmetrical.
The range for [math]X[/math] is 2 to 12. [math]X=2[/math] only if the two dices are (1,1) and the probability for this is math^2=1/36[/math]. In a similar way [math]X=12[/math] only if the two dices are (6,6) and the probability for this is also [math]1/36[/math].
If [math]X=(2+12)/2=7[/math] there are 6 possible dice combinations: (1,6),(6,1),(2,5)(5,2),(3,4),(4,3) giving a probability [math]6/36=1/6[/math] for [math]X=7[/math]. You can do this for all possible values of [math]X[/math], this is summarized in the table below.
As you can observe, the distribution has the shape of a triangle, and it is symmetrical.
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Harry Kanigel
Former Software Engineer & IT Officer (Ret) at UBS, Schwab, Metallgesellschaft
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Author has 126 answers and 105.3K answer views
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3y
First construct a 6 x 6 grid showing all 36 ways that two dice can express a sum. The diagonals of this grid show the incidences of each total. So, e.g., the longest diagonal shows six occurrences of the value 7. The second-to-longest diagonal shows five occurrences of the value 6. And so on (the values 12 and 2 each have only one occurrence). The values could be rearranged into a normal curve where 7 has the tallest bar (centered) and 2 has a short bar on the left hand side and 12 has a short bar on the right hand side.
Manie, Ph.D.
Applied Scientist at Amazon
·
Author has 550 answers and 651.6K answer views
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4y
First, let [math]X_1[/math] be the outcome for the first dice, [math]X_2[/math] be the outcome for the second dice. Then we are looking for the distribution of [math]X_1 + X_2[/math].
For example:
[math]P(X_1 + X_2 = 1) = 0[/math]
[math]P(X_1 + X_2 = 2) = \dfrac{1}{36}[/math]
The table simulated in R will be like this:
[code]Prob<- matrix(c(0,1/36,2/36,3/36,4/36,5/36,6/36,5/36,4/36,3/36,2/36,1/36),nrow=12,byrow = FALSE)
colnames(Prob)<- c("P...
Ian Rayner
I use statistics daily to build hedge fund portfolios.
·
Author has 390 answers and 442.1K answer views
·
4y
Originally Answered: Construct the discrete probability distribution table for random variable X, which would be the sum of 2 rolled dice?
·
This is straightforward if you understand the connection between a frequency table and a probability table.
Your question asks for a probability table. This means a list of the probabilities of getting each of the (discrete) possible outcomes from rolling 2 dice and summing their result.
A frequency table is a list of the counts of the ways of getting each of the outcomes. To translate from frequency to probability, we simply divide each frequency by the total count of all the frequencies.
At first glance, it might seem silly to think about the frequencies. After all, the only possibilities from
This is straightforward if you understand the connection between a frequency table and a probability table.
Your question asks for a probability table. This means a list of the probabilities of getting each of the (discrete) possible outcomes from rolling 2 dice and summing their result.
A frequency table is a list of the counts of the ways of getting each of the outcomes. To translate from frequency to probability, we simply divide each frequency by the total count of all the frequencies.
At first glance, it might seem silly to think about the frequencies. After all, the only possibilities from 2 dice are 2, 3, 4, …, 12. But then you notice you can only get 2 one way: 1 + 1, but you can get 5 several different ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1.
So, make a list of the numbers from 2 to 12, and for each count the different ways you can get each number from 2 dice. Remember to count both 2+3 and 3+2.
I think I have given you enough to answer the question!
Remember to do a reality check once you have constructed your frequency table: how many different possible outcomes are there from rolling 2 six-sided dice? The total of all the frequencies in your table should equal this number.
Partha Chattopadhyay
Taught while teaching and still practicing
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Maria Soosai David
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Let [math]x=y+z, y=1,2,3,4,5,6, z=1,2,...6,. x=2,3,...12[/math]
& [math]\sum_{i=2}^{12}P(x=i)=1[/math]
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Reviewing the options for local estrogen treatment of vaginal atrophy
Sarah H Lindahl
Correspondence: Sarah H Lindahl, Sutter East Bay Medical Foundation, 20101 Lake Chabot Blvd, Castro Valley, CA 94546, USA, Tel +1 510 886 3400, Fax +1 510 581 6517, Email shlindobgyn@yahoo.com
Collection date 2014.
The full terms of the License are available at Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed.
Abstract
Background
Vaginal atrophy is a chronic condition with symptoms that include vaginal dryness, pain during sex, itching, irritation, burning, and discharge, as well as various urinary problems. Up to 45% of postmenopausal women may be affected, but it often remains underreported and undertreated. This article aims to review the current recommendations for treatment of vaginal atrophy, and current data on the effectiveness and safety of local vaginal estrogen therapies.
Methods
Literature regarding vaginal atrophy (2007–2012) was retrieved from PubMed and summarized, with emphasis on data related to the treatment of vaginal atrophy with local vaginal estrogen therapy.
Results
Published data support the effectiveness and endometrial safety of low-dose local estrogen therapies. These results further support the general recommendation by the North American Menopause Society that a progestogen is not needed for endometrial protection in patients using low-dose local vaginal estrogen. Benefits of long-term therapy for vaginal atrophy include sustained relief of symptoms as well as physiological improvements (eg, decreased vaginal pH and increased blood flow, epithelial thickness, secretions).
Conclusion
Currently available local vaginal estrogen therapies are well tolerated and effective in relieving symptoms of vaginal atrophy. Recent data support the endometrial safety of low-dose regimens for up to 1 year.
Keywords: menopause, estrogen, local estrogen therapy, vaginal atrophy
Introduction
Before menopause, estrogen stimulates the exfoliation of vaginal cells. As these cells exfoliate and die, they release large amounts of glycogen. This glycogen is hydrolyzed to glucose and lactobacillus converts the glucose to lactic acid, which helps maintain vaginal pH between 3.5 and 4.5.1 After menopause, with declining levels of estrogen, this pathway is less active, which may allow postmenopausal pH to increase to the range of 5.0–7.5.1 At these higher pH levels, lactobacillus survival is hindered, and vaginal colonization by other bacterial species predisposes women to symptomatic vaginal infection or urinary tract infection.2
Vaginal atrophy is a chronic condition resulting from a decline in estrogen in the urogenital tissues that affects up to 45% of postmenopausal women.3,4 Vaginal atrophy presents as thin, pale, and dry vaginal walls; reduced elasticity of vaginal tissue; loss of rugae; and a shortened and narrowed vagina. Symptoms include vaginal dryness, pain during sex, bleeding during intercourse/touching, itching, irritation, burning, and discharge, as well as various urinary problems, including increased frequency, urgency, incontinence, dysuria, and infection.2,5 These signs and symptoms can have an enormous impact on a patient’s well-being, especially in the context of sexual function.6 Lack of lubrication with sexual intercourse causes dyspareunia, which can lead to avoidance of healthy sexual activity and to further atrophy. A woman with vaginal atrophy may also experience bleeding with minimal trauma, such as during a medical examination or exercise.
The lower urinary tract is also directly affected by decreased estrogen, as estrogen receptors are found throughout the epithelial tissues of the urethra, trigone, and bladder, and in some of the surrounding muscles and ligaments.7,8 Thinning of the urinary epithelium and weakening of the surrounding tissue may promote reduced urethral closure pressure, reduced sensory threshold in the bladder, and, in some cases, recurrent urinary tract infection.8
Typically, a diagnosis of vaginal atrophy can be made from clinical examination, with many women presenting with dry, glazed-looking vaginal epithelium; a thinning cervix; a loss of labial fat pad; or a vagina that has lost elasticity, has shortened, has narrowed, has become less distensible, and can be easily traumatized and irritated.5,9 Diagnosis may be further supported by pH or the vaginal maturation index (VMI), which indicates the relative proportion of parabasal cells, intermediate cells, and mature superficial cells of the vaginal squamous epithelium, and is calculated based the following formula: VMI =0.5(X2) + 1(X3), where X2=% intermediate cells and X3=% superficial cells.10 The VMI is 0%–49% in patients with an absent or low estrogenic effect, 50%–64% in patients with a moderate estrogenic effect, and 65%–100% in patients with a high estrogenic effect.10
While the vasomotor symptoms of menopause generally improve over time, vaginal symptoms usually worsen and do not improve without treatment.5 Women may attribute these changes to be a part of “normal aging” and may not be aware of safe and simple treatments. Only about 25% of women who experience symptoms of vaginal atrophy seek medical help.9 As vaginal infections should be part of the differential diagnosis,1 vaginal atrophy may be misdiagnosed by patients and providers as a vaginal infection caused by yeast or bacteria. Obtaining a thorough history and performing an examination should help to avoid misdiagnosis and delaying treatment in postmenopausal patients experiencing some or all of these symptoms.
The goal of this article is to provide an overview of the current recommendations for treatment of vaginal atrophy, followed by a succinct review of the effectiveness and safety of local vaginal estrogen therapies.
Methods
PubMed was initially searched using the terms “vaginal” and “atrophy.” To find recent publications focused on the influence of estrogen in the lower urinary tract that may not have been located in the main search, the terms “urogenital,” “atrophy,” and “estrogen” were used. Relevant literature (restricted to works in English and published after January 1, 2007) was reviewed and summarized, with emphasis on data related to treatment with local vaginal estrogen therapy. In addition, references cited within the relevant literature sources identified in the literature search were reviewed for content related to estrogen therapy and considered for use as original data sources. Recommendations from the American Congress of Obstetricians and Gynecologists (ACOG) and the North American Menopause Society (NAMS) were also consulted.
Recommendations for the treatment of vaginal atrophy
According to NAMS, first-line treatment for vaginal atrophy should include nonhormonal vaginal lubricants and moisturizers, as well as continued sexual activity.5 Regular sexual activity helps maintain active blood flow to the vagina and helps produce vaginal lubrication. Moisturizers, such as Replens (Lil’ Drug Store Products, Inc., Cedar Rapids, IA, USA), are intended for regular use, while lubricants, such as K-Y (McNeil-PPC, Inc., Fort Washington, PA, USA) or Astroglide (BioFilm, Inc., Vista, CA, USA), are intended for use during sexual activity. Replens has been shown to increase vaginal moisture, fluid volume, and elasticity, while decreasing pH.11
In addition, Osphena (Shionogi Inc., Florham Park, NJ, USA), an estrogen agonist/antagonist, is an oral treatment for moderate to severe dyspareunia that was approved by the US Food and Drug Administration in February 2013.12
The Vaginal Health: Insights, Views and Attitudes (VIVA) survey was designed to assess the knowledge of vaginal atrophy in postmenopausal women with the goal of increasing awareness of vaginal atrophy, aiding in increasing physician-patient dialogue, and therefore allowing women to make better-informed treatment decisions.13 Among the 1,578 respondents who reported experiencing vaginal discomfort, 49% reported trying lubricating gels and creams; 22% reported not using any treatment.13
Although over-the-counter treatments may work for women with mild symptoms, they are often inadequate for women with moderate to severe symptoms. If nonhormonal options are not sufficient, ACOG has also noted that estrogen therapy, whether topical (local) or systemic, is an effective treatment for vaginal dryness or atrophic changes that impede sexual function.14
Although many systemic estrogen hormone products are approved for the prevention of osteoporosis and the treatment of moderate to severe vasomotor symptoms associated with menopause, there is a risk for deep vein thrombosis, pulmonary embolism, coronary heart disease, and endometrial and breast cancer.15–18 Long-term systemic hormone therapy is no longer recommended if considered solely for the treatment of vaginal atrophy, and therefore local vaginal estrogen therapy is advised.5 Low-dose, local vaginal estrogen therapy reduces the risk associated with long-term systemic estrogen exposure, and is effective and well tolerated in most women.5
Efficacy and safety of local vaginal estrogen treatment
Four local vaginal estrogen treatments are available in the United States for the treatment of vaginal atrophy (two vaginal creams, one vaginal ring, and one vaginal tablet). They are well tolerated and considered equally effective by NAMS.5 In the VIVA survey, 13% of respondents who reported experiencing vaginal discomfort reported they had used vaginal hormone creams, 12% reported they had used vaginal hormone tablets, and 1% reported having used a vaginal hormone ring.13 Because vaginal estrogens fall into the general category of estrogen therapies, package inserts on local vaginal estrogen treatments carry the same warnings as systemic estrogens. The following sections aim to provide a concise review of the most recent and relevant data concerning the efficacy and safety of local vaginal estrogen treatments.
Vaginal creams
Two vaginal creams are available: Premarin19 (Wyeth Pharmaceuticals Inc., part of Pfizer, Inc., New York, NY, USA) and Estrace20 (Warner Chilcott [US], LLC, Rockaway, NJ, USA). Premarin vaginal cream contains 0.625 mg conjugated equine estrogens (CEE) per gram of cream. Its composition is unique in that it does not solely contain estradiol as the active ingredient. CEE cream contains a mixture of estrogenic compounds, predominantly estrone, equilin, 17 α-dihydroequilin, 17 α-estradiol, and 17 β-dihydroequilin.19 Recommended dosing is 0.5–2.0 g once daily for 21 days, followed by 7 days without treatment (repeated in a 28-day cycle) or 0.5 g twice weekly. In a randomized, placebo-controlled study, postmenopausal women with moderate to severe vaginal atrophy who received low-dose CEE cream (0.3 mg CEE, equivalent to 0.5 g Premarin vaginal cream) according to either the cyclic or the twice-weekly regimen exhibited improved VMI, improved vaginal pH, and improved most bothersome symptom scores, including those for dyspareunia, at week 12 of treatment.21 After the initial 12 weeks, open-label treatment was continued for 40 weeks, consistent with the patient’s prior regimen. Endometrial hyperplasia or carcinoma did not develop in any patients, based on 155 biopsy samples, and the CEE and placebo groups reported adverse events (AEs) of similar type and frequency. None of the serious AEs were considered related to the study drug.
Estrace20 contains 0.1 mg estradiol per gram of cream and recommended dosing is 2–4 g daily for 1–2 weeks, a gradual reduction to half the initial dose over 1–2 weeks, and then 1 g 1–3 times weekly for maintenance. For the initial period of treatment, it is important to adhere to the medication schedule while waiting for a response. No randomized controlled trials have been performed, and no results of observational studies have been published in the past 10 years.
Vaginal ring
The Estring22 vaginal ring (Pharmacia and Upjohn Company, Division of Pfizer, Inc., New York, NY, USA) contains a 2 mg estradiol reservoir. When inserted into the upper third of the vagina, the ring releases approximately 7.5 μg per day for 90 days. In an open-label study, women using Estring for 12 months experienced an improvement in qualitative cytological evaluations (generally atrophic at baseline to proliferative [moderate estrogen effect] or highly proliferative [high estrogen effect] at week 48).23 Patient-reported symptoms and clinical signs on inspection improved greatly. The primary endpoint—endometrial safety—was assessed at baseline and at week 48 by ultrasound measurement of endometrial thickness and a progestogen challenge test.23 In women using Estring, the mean endometrial thickness remained constant, and no women experienced withdrawal bleeding or spotting during the progestogen challenge test. Only two of 126 women had significant endometrial thickening (13.0 and 19.0 mm).23 One of these women had a uterine myoma, while the other experienced a rapid reduction in endometrial thickness to 4 mm without bleeding; neither had evidence of proliferation on endometrial biopsy. The low-dose Estring should not be confused with Femring24 (Warner Chilcott [US]), a vaginal ring that delivers systemic levels of estrogen replacement. Less than 2% of women experience ulcers with vaginal rings (2 of 136 women in a 1-year study).25
Vaginal tablet
The Vagifem26 vaginal tablet (Novo Nordisk A/S, Bagsvaerd, Denmark) contains 10 μg estradiol. This ultra-low-dose vaginal tablet, introduced in 2010, replaced a 25 μg vaginal tablet that was previously available. As both regulatory authorities and therapeutic experts from menopause societies agree that the minimum effective dose of estrogen should be used for the treatment of vaginal atrophy,5,27,28 the ultra-low-dose (10 μg) vaginal tablets were developed to provide symptom relief while reducing systemic exposure to estrogen. The recommended dosing schedule of Vagifem 10 μg is one tablet daily for 2 weeks, followed by one tablet twice weekly for maintenance, with an annual estrogen exposure of 1.14 mg.26 The efficacy and safety of this ultra-low-dose regimen for the treatment of vaginal atrophy in postmenopausal women has been established in recent studies.29–31 At 12 weeks of treatment, significant improvement in VMI, maturation value, grading of vaginal health, vaginal pH, and most bothersome symptom score was seen.29 Improvement over baseline was maintained through 52 weeks. One participant (from 164 treated participants completing the study) lacked an evaluable biopsy at baseline and was found to have endometrial adenocarcinoma stage II, grade 2 after approximately 10 months of treatment; de novo development was considered unlikely because of the clinical time frame.29 To further assess safety, endometrial biopsy was performed at the beginning and end of a 52-week study.30 No endometrial hyperplasia or endometrial cancer was found. Over the course of the study, endometrial thickness measured by transvaginal ultrasound decreased from 2.04 mm (n=334) to 1.94 mm (n=293), which may reflect the expected decrease in thickness with age, with no interfering endometrial stimulation.30 None of the serious AEs were considered related to the study drug. A larger pooled analysis (based on 386 evaluable biopsy samples) determined that ultra-low-dose estradiol vaginal tablets did not increase the risk for endometrial hyperplasia or carcinoma in postmenopausal women for 1 year under study conditions.31
Lack of need for progesterone
In its 2013 position statement, NAMS stated that the primary indication for progesterone is to reduce the risk of endometrial cancer and should be used in patients who are on systemic estrogen therapy and have an intact uterus.5 However, progesterone is “generally not indicated when low-dose vaginal estrogen is administered.”5 Because low doses of vaginally administered estrogen, such as those delivered by a low-dose CEE regimen,21 the vaginal ring,23 and the vaginal tablet,31 have not been associated with increased risk for endometrial hyperplasia, progestogen is not generally indicated when local estrogen products are administered.5 However, no recommendations have been made regarding the need for progesterone with creams that deliver higher doses of estrogen. Key safety information from recent clinical trials is summarized in Table 1.
Table 1.
Recent clinical trials assessing the safety of local vaginal estrogen therapies
Product | N | Duration and design | Treatment groups | Safety assessment | Reference
Premarin vaginal cream | 423 | 12-week double-blind phase, followed by40-week open-label phase consistent with previous regimen | Premarin cream (0.3 mg CE) or placebo cream once daily (21 days on/7 days off) or twice weekly (first 12 weeks), then all women taking Premarin cream according to prior regimen | No hyperplasia or carcinoma, as assessed by transvaginal ultrasound and endometrial biopsyAdverse event frequency similar to that of placebo during double-blind phase | 21
Estrace vaginal cream | No recent trials
Estring vaginal ring | 126 | 12-month open-label | Estring vaginal ring (7.5 μg estradiol released daily) compared with Vagifem vaginal tablet (25 μg estradiol, which has since been discontinued and replaced by a 10-μg estradiol version) daily for 2 weeks, then twice weekly | No increase in average endometrial thickness No bleeding/spotting in Estring group with PCT | 23
Vagifem vaginal tablet | 309 | 52-week double-blind | Vagifem vaginal tablet (10 μg estradiol) or placebo tablet assigned in 2:1 ratio daily for 2 weeks, then twice weekly | No major safety findings regarding physical, gynecologic, or laboratory assessments | 29
336 | 52-week open-label | Vagifem vaginal tablet (10 μg estradiol) daily for 2 weeks, then twice weekly | No cases of endometrial hyperplasia or endometrial cancerDecrease in mean endometrial thickness | 30
Benefits of long-term therapy for vaginal atrophy
All of the aforementioned treatments, when used as directed, are well tolerated and effective for relieving symptoms of vaginal atrophy.5 They provide vaginal estrogen, while minimizing systemic exposure,32–34 and result in increased blood flow, increased epithelial thickness, and increased secretions, as well as reduced pH.5 These physiological improvements represent a reversal of atrophy and lead to a positive clinical outcome for most women.
The urogenital symptoms of estrogen deficiency that present in the postmenopausal years may require long-term therapy, since women have reported the return of symptoms during washout periods between different treatments, even after just a few weeks.32 Given the comparable efficiency of the low-dose, locally administered estrogen products, NAMS recommends that severity of symptoms, effectiveness and safety of therapy for the individual patient, and patient preference guide the choice of treatment.5 It is important that patients accept and adhere to their treatment in order to fully realize the benefits. Clinicians can be instrumental in educating patients about vaginal atrophy as a chronic condition; the advantages of low-dose, local vaginal estrogen therapy; and the importance of long-term management.
Conclusion
Vaginal atrophy is a common, underreported, and under-treated chronic condition that results from a decrease in estrogen levels during menopause.1 When over-the-counter treatments fail to relieve symptoms, NAMS recommends low-dose, local vaginal estrogen therapy. A variety of formulations (ie, creams, ring, and tablet) are available, and based on the collective results of many studies conducted over the past two decades, NAMS considers all of these effective and well tolerated.5
Because vaginal atrophy is a chronic condition, treatment should be continued for as long as symptoms persist and may need to be continued indefinitely. Long-term studies (up to 1 year) have shown that low-dose, local vaginal estrogen therapy has an excellent safety profile marked by no increased risk for endometrial hyperplasia, which is a common concern with estrogen-alone therapies. Such data should provide reassurance to women with vaginal atrophy who may otherwise hesitate to accept a hormonal form of treatment.
Finally, many women remain untreated because they avoid reporting symptoms of vaginal atrophy to their health care provider. Clinicians may take an active role in encouraging such discussions by asking their postmenopausal patients whether they have experienced any vaginal dryness or discomfort, with or without sexual activity. By initiating the conversation and then following up on any complaints with an examination or a referral, it may become easier to diagnose and appropriately treat postmenopausal women with vaginal atrophy.
Acknowledgments
Medical writing assistance was provided by Pamela Barendt, PhD, of ETHOS Health Communications, Newtown, Pennsylvania, with financial assistance from Novo Nordisk, Inc., in compliance with international guidelines for Good Publication Practice. The author received no remuneration of any kind for the development of this manuscript.
Footnotes
Disclosure
The author has no conflict of interest to report.
References
Articles from International Journal of Women's Health are provided here courtesy of Dove Press
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Using Number Lines in Math Class with Middle Schoolers - Cognitive Cardio Math
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Using Number Lines in Math Class with Middle Schoolers
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I’m a huge fan of giving our students the right tools they need to succeed. I believe they’re never too old to use them! When I discover tools that help my students with math, I feel like I’m setting them up for success, both mentally and academically. It’s amazing to see how much more confident they become when they have resources to guide them. One tool that’s always available in my classroom is number lines. Yes, even in middle school! Today, I’m excited to show you just how beneficial using number lines can be for our middle schoolers!
How Number Lines Help Middle Schoolers
Using number lines in middle school math is a game-changer! When we first think about number lines, we automatically think about our younger learners. While we traditionally think of number lines as a tool for elementary students, I want to challenge you to get the idea of a 0 to 10 number line out of your head and instead think of number lines in relation to the skills and concepts you teach. They’re incredibly valuable for our older students, too. When we introduce them to middle schoolers, we give them a visual tool that helps make abstract concepts more concrete. They are fantastic for introducing new concepts, reviewing previously taught ones, or as part of intervention.
Number lines boost confidence and comprehension in middle school math. When our students see math problems visually laid out, it reduces anxiety and makes learning more interactive and engaging. I’ve seen firsthand how my students light up when they “get it” after using a number line. It’s a tool that truly supports their mathematical journey and helps them thrive!
Middle School Concepts & Number Lines
Number lines are fantastic for helping our students understand positive and negative numbers. They can see the relationship between these numbers and grasp the idea of moving left or right on the line to add or subtract. This visual representation makes a world of difference, especially for our students who struggle with traditional methods. It’s a great way to differentiate learning for your students as well.
These tools also simplify the process of finding common denominators and understanding fractions. By seeing fractions on a number line, our students can easily compare sizes and understand equivalences. This visual approach takes the mystery out of fractions and makes them much more approachable.
Number lines are also great for reinforcing the concept of absolute value. Students can quickly see the distance between numbers and zero, which solidifies their understanding. It’s one thing to memorize a rule, but it’s entirely different (and more effective) to visualize it on a number line.
Three Ways to Use Number Lines in Your Classroom
I want to share three specific ways you can use number lines in the classroom to help students excel in math. These methods have been impactful for my students, making math more interactive and easier to understand and I know they can help your students too. Let’s check out these three strategies and see how we can bring number lines to life in our lessons!
1. Classroom Decor
One of my favorite ways to use number lines in the classroom is by making them an integral part of classroom decor that stays up all year round. This is a perfect way to make decorating functional and useful! I like to refer to them as tools that my students can reference all year long.
In my classroom, I have a fraction, decimal, and percent number line displayed above the whiteboard at the front. It shows the most common fractions and their decimal and percent equivalents. This visual aid is incredibly helpful for my students, as they frequently refer to it to learn and remember these common equivalences. It’s like having a cheat sheet right there in front of them every day!
I also have a positive and negative integers number line displayed at the top of another wall. When we start exploring negative numbers, this huge visual helps us discuss moving in the positive or negative direction when combining numbers. It’s also super handy when we dive into absolute value. Seeing that both -5 and 5 are the same distance from zero makes it clear why they have the same absolute value.
2. Double Number Lines
I love finding different ways to use this tool in my classroom. One of my favorite methods is using double number lines (or two number lines side by side) to explore the relationship between different amounts. It’s an easy way for students to see how numbers in different formats relate to each other. This is especially useful when we’re working with percentages or ratios.
For instance, when finding the percent of a number, I place the number we’re working with on one number line and the percentages from 1% to 100% on another. This makes it easy for my students to visually grasp how percentages relate to the original number. It’s a great way to make abstract concepts more tangible!
Teaching ratios are made easy with these tools. My students can see the relationship between two quantities side by side. This visual representation helps them understand how ratios work and how to find equivalent ratios. It’s a powerful way to make the concept of ratios more concrete and less intimidating.
3. Vertical Number Lines
I also like to use vertical number lines, similar to thermometers, to visualize patterns in numbers. These are particularly helpful when we’re looking at the increasing tens and ones in two-digit numbers. My students can clearly see the sequence and repetition of digits. This makes it easier for them to understand the value of numbers. They quickly get the hang of counting up and counting down when adding or subtracting, seeing how sums increase and differences decrease in value.
A key benefit of number lines is how they help students connect the concepts of rounding up and rounding down. When they see numbers on the line, it becomes much clearer why we round up or down to the nearest five or ten. The visual representation really solidifies their understanding.
Helpful Resources for Middle School Math
I love using the Fraction, Decimal, and Percent Number Line to help my students keep their conversions straight! This resource is a one-page number line that’s packed with useful information. It includes common decimals placed right on the number line, along with their equivalent fractions in their lowest terms and the corresponding percentages.
I have my students place this number line in their math notebooks. It becomes their go-to reference for quick conversions. Having this tool in their toolbox helps make math problems a lot less daunting.
I also post a larger version of this number line as part of my classroom decor. This way, my students can easily glance up and find the information they need. It’s amazing how much this visual tool can boost their confidence and understanding of fractions, decimals, and percentages!
If you want to ensure that your students have this resource at their fingertips, I also have a Fraction, Decimal, Percent Number Line Bookmark. Have your students place them in their book they are reading or even their math textbook!
To read more about number lines, check out my post on Fraction, Decimal, and Percent Conversions. You can also check out the Using Number Lines in the Math Classroom episode on The Teaching Toolbox podcast. It is filled with examples and ideas on how you can use number lines in your math classroom.
Embrace Number Lines in the Middle School Math Classroom
Bringing number lines into our middle school math classroom will change how your students approach math. From helping them understand complex concepts to serving as a constant visual aid, number lines are invaluable tools. Whether we’re using them for fractions, decimals, percents, or even decorating the classroom with them, they make math more accessible and less intimidating for our students. By providing these resources, we’re not only boosting their confidence but also paving the way for their academic success.
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Remember to save this post to your favorite math Pinterest board for ideas on how to use number lines!
Ellie
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Teaching the Properties of Addition and Multiplication in 3rd Grade
Tips and Activities for Introducing Division Concepts in 3rd Grade
The Keys to Teaching Integer Operations
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4176 | https://www.idc-online.com/technical_references/pdfs/electrical_engineering/Wye_Delta_OR_Delta_Wye_Transformations.pdf | Module 2 DC Circuit Version 2 EE IIT, Kharagpur Lesson 6 Wye (Y) - Delta (∆) OR Delta (∆)-Wye (Y) Transformations Version 2 EE IIT, Kharagpur Objectives • A part of a larger circuit that is configured with three terminal network Y (orΔ ) to convert into an equivalent Δ (or Y ) through transformations. • Application of these transformations will be studied by solving resistive circuits. L.6.1 Introduction There are certain circuit configurations that cannot be simplified by series-parallel combination alone. A simple transformation based on mathematical technique is readily simplifies the electrical circuit configuration. A circuit configuration shown below is a general one-port circuit. When any voltage source is connected across the terminals, the current entering through any one of the two terminals, equals the current leaving the other terminal. For example, resistance, inductance and capacitance acts as a one-port. On the other hand, a two-port is a circuit having two pairs of terminals. Each pair behaves as a one-port; current entering in one terminal must be equal to the current living the other terminal. Fig.6.1.(b) can be described as a four terminal network, for convenience subscript 1 to refer to the variables at the input port (at the left) and the subscript 2 to refer to the variables at the output port (at the right). The most important subclass of two-port networks is the one in which the minus reference terminals of the input and output ports are at the same. This circuit configuration is readially possible to consider the ‘ or π Δ ’ – network also as a three-terminal network in fig.6.1(c). Another frequently encountered circuit configuration that shown in fig.6.1(d) is approximately refered to as a three-terminal Y connected circuit as well as two-port circuit. Version 2 EE IIT, Kharagpur The name derives from the shape or configuration of the circuit diagrams, which look respectively like the letter Y and the Greek capital letter Δ. L.6.1.1 Delta (Δ) – Wye (Y) conversion These configurations may often be handled by the use of a Δ −Y or −Δ Y transformation. One of the most basic three-terminal network equivalent is that of three resistors connected in “Delta ” and in “Wye( ”. These two circuits identified in fig.L6.1(e) and Fig.L.6.1(f) are sometimes part of a larger circuit and obtained their names from their configurations. These three terminal networks can be redrawn as four-terminal networks as shown in fig.L.6.1(c) and fig.L.6.1(d). We can obtain useful expression for direct ( ) Δ ) Y Version 2 EE IIT, Kharagpur transformation or conversion from Δ to Y or Y to Δ by considering that for equivalence the two networks have the same resistance when looked at the similar pairs of terminals. L.6.2 Conversion from Delta (Δ) to Star or Wye (Y) Let us consider the network shown in fig.6.1(e) (or fig.
) and assumed the resistances ( ) in 6.1( ) c → , , AB BC CA R R and R Δ network are known. Our problem is to find the values of in Wye (Y ) network (see fig.6.1(e)) that will produce the same resistance when measured between similar pairs of terminals. We can write the equivalence resistance between any two terminals in the following form. , , A B C R R and R Between & A C terminals: ( ) CA AB BC A C AB BC C R R R R R A R R R + + = + + (6.1) Between terminals: & C B ( ) BA AB CA C B AB BC C R R R R R A R R R + + = + + (6.2) Between & B A terminals: ( ) AB CA BC B A AB BC C R R R R R A R R R + + = + + (6.3) By combining above three equations, one can write an expression as given below. AB BC BC CA CA AB A B C AB BC CA R R R R R R R R R R R R + + + + = + + (6.4) Subtracting equations (6.2), (6.1), and (6.3) from (6.4) equations, we can write the express for unknown resistances of Wye (Y ) network as AB CA A AB BC CA R R R R R R = + + (6.5) AB BC B AB BC CA R R R R R R = + + (6.6) BC CA C AB BC CA R R R R R R = + + (6.7) Version 2 EE IIT, Kharagpur L.6.2.1 Conversion from Star or Wye (Y) to Delta (Δ) To convert a Wye (Y ) to a Delta (Δ ), the relationships must be obtained in terms of the Wye (Y 3 , , AB BC R R and R ) resistances (referring to fig.6.1 (f)). Considering the Y connected network, we can write the current expression through , , A B C R R and R A R resistor as ( ) A N A A V V I R − = (for Y network) (6.8) Appling KCL at ‘ ’ for Y connected network (assume N , , A B C terminals having higher potential than the terminal ) we have, N ( ) ( ) ( ) 0 A N B N C N A B C V V V V V V R R R − − − + + = 1 1 1 C A B N A B C A B C V V V V R R R R R R ⎛ ⎞⎛ ⇒ + + = + + ⎜ ⎟⎜ ⎝ ⎠⎝ ⎞ ⎟ ⎠ or, 1 1 1 C A B A B C N A B C V V V R R R V R R R ⎛ ⎞ + + ⎜ ⎟ ⎝ ⇒ = ⎛ ⎞ + + ⎜ ⎟ ⎝ ⎠ ⎠ (6.9) For -network (see fig.6.1.(f)), Δ Current entering at terminal A = Current leaving the terminal ‘ A’ AC AB A AB A V V I C R R = + (for network) (6.10) Δ From equations (6.8) and (6.10), ( ) A N AC AB A AB AC V V V V R R R − = + Using the expression in the above equation, we get N V 1 1 1 C A B A B C A A B C AC AB A AB AC V V V R R R V R R R V V R R R ⎛ ⎞ ⎛ ⎞ + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ − ⎜ ⎟ ⎛ ⎞ + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= + ⇒ 1 1 1 A B A C B C A B C AC AB A AB AC V V V V R R R R R V V R R R ⎛ ⎞ − ⎛ ⎞ − + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= + or 1 1 1 AB AC B C A B C AC AB A AB AC V V R R R R R V V R R R ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= + (6.11) Equating the coefficients of and in both sides of eq.(6.11), we obtained the following relationship. AB V AC V Version 2 EE IIT, Kharagpur 1 1 1 1 1 A B AB A B AB C A B A B C R R R R R R R R R R R R = ⇒ = + ⎛ ⎞ + + ⎜ ⎟ ⎝ ⎠ + (6.12) 1 1 1 1 1 A C AC A C AC B A C A B C R R R R R R R R R R R R = ⇒ = + ⎛ ⎞ + + ⎜ ⎟ ⎝ ⎠ + (6.13) Similarly, B I for both the networks (see fig.61(f)) are given by ( ) B N B B V V I R − = (for Y network) BC BA B BC BA V V I R R = + (for network) Δ Equating the above two equations and using the value of (see eq.(6.9), we get the final expression as N V 1 1 1 BC BA C A A B C BC BA B BC V V R R R R R V V BA R R R ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= + Equating the coefficient of in both sides of the above equations we obtain the following relation BC V 1 1 1 1 1 B C BC B C BC A B C A B C R R R R R R R R R R R R = ⇒ = + ⎛ ⎞ + + ⎜ ⎟ ⎝ ⎠ + (6.14) When we need to transform a Delta (Δ ) network to an equivalent Wye (Y ) network, the equations (6.5) to (6.7) are the useful expressions. On the other hand, the equations (6.12) – (6.14) are used for Wye (Y ) to Delta (Δ ) conversion. Observations In order to note the symmetry of the transformation equations, the Wye (Y ) and Delta (Δ ) networks have been superimposed on each other as shown in fig. 6.2. Version 2 EE IIT, Kharagpur • The equivalent star (Wye) resistance connected to a given terminal is equal to the product of the two Delta (Δ) resistances connected to the same terminal divided by the sum of the Delta (Δ) resistances (see fig. 6.2). • The equivalent Delta ( ) resistance between two-terminals is the sum of the two star (Wye) resistances connected to those terminals plus the product of the same two star (Wye) resistances divided by the third star (Wye (Y )) resistance (see fig.6.2). Δ L.6.3 Application of Star ( ) Y to Delta Δ ( ) or Delta Δ ( ) to Star ( ) Y Transformation Example: L.6.1 Find the value of the voltage source ( ) that delivers 2 Amps current through the circuit as shown in fig.6.3. S V Version 2 EE IIT, Kharagpur Solution: Convert the three terminals Δ-network (a-c-d & e-f-g) into an equivalent Y -connected network. Consider the Δ-connected network ‘a-c-d’ and the corresponding equivalent Y -connected resistor values are given as Version 2 EE IIT, Kharagpur 3 1 1 1 3 1 0.6 ; 0.2 ; 0.6 5 5 5 ao co do R R R × × × = = Ω = = Ω = = Ω Similarly, for the -connected network ‘e-f-g’ the equivalent the resistances of Y -connected network are calculated as Δ 3 1 3 1 1 1 0.6 ; 0.6 ; 0.2 5 5 5 eo go fo R R R ′ ′ ′ × × × = = Ω = = Ω = = Ω Now the original circuit is redrawn after transformation and it is further simplified by applying series-parallel combination formula. Version 2 EE IIT, Kharagpur The source s V that delivers 2A current through the circuit can be obtained as . 3.2 2 3.1 6.2 s V I Volts = × = × = Example: L.6.2 Determine the equivalent resistance between the terminals A and B of network shown in fig.6.4 (a). Version 2 EE IIT, Kharagpur Solution: A ‘ ’ is substituted for the ‘Y ’ between points c, d, and e as shown in fig.6.4(b); then unknown resistances value for Y t Δ o Δ transformation are computed below. 2 4 2 4 8.66 3 cB R × = + + = Ω ; 4 3 3 4 13 2 eB R × = + + = Ω ; 2 3 2 3 6.5 4 ce R × = + + = Ω Next we transform ‘Δ’connected 3-terminal resistor to an equivalent ‘Y ’ connected network between points ‘A’; ‘c’ and ‘e’ (see fig.6.4(b)) and the corresponding Y connected resistances value are obtained using the following expression. Simplified circuit after conversion is shown in fig. 6.4(c). Version 2 EE IIT, Kharagpur 4 2 0.64 ; 4 2 6.5 Ao R × = = Ω + + 4 6.5 2.08 ; 4 2 6.5 co R × = = Ω + + 6.5 2 1.04 ; 4 2 6.5 eo R × = = Ω + + & The circuit shown in fig.6.5(c) can further be reduced by considering two pairs of parallel branches and the corresponding simplified circuit is shown in fig.6.4(d). 3 8.66 13 1 and & Now one can find the equivalent resistance between the terminals ‘ A ’ and ‘ B ’ as ( ) ( ) 2.23 2.08 1.04 0.93 0.64 2.21 AB R = + + + = & Ω . Example: L.6.3 Find the value of the input resistance of the circuit. in R Version 2 EE IIT, Kharagpur Solution: Y connected network formed with the terminals a-b-o is transformed into Δ connected one and its resistance values are given below. Ω = × + + = 61 .
64 26 12 36 12 36 ab R ; Ω = × + + = 66 .
46 36 12 26 26 12 bo R Ω = × + + = 140 12 36 26 36 26 ao R Similarly, connected networks formed with the terminals ‘b-c-o’ and ‘c-a-o’ are transformed to connected networks. Y Δ Ω = × + + = 738 .
6 26 6 .
0 6 6 .
0 6 bc R ; Ω = × + + = 2 .
29 6 26 6 .
0 26 6 .
0 co R Ω = × + + = 60 .
34 6 .
0 26 6 26 6 0 b R and, Ω = × + + = 00 .
54 30 26 15 26 15 co R ; Ω = × + + = 108 15 26 30 26 30 ao R Version 2 EE IIT, Kharagpur Ω = × + + = 31 .
62 26 15 30 15 30 ac R Note that the two resistances are connected in parallel (140 108 ) between the points ‘a’ and ‘o’. Similarly, between the points ‘b’ and ‘o’ two resistances are connected in parallel ( 46.66 34.6) and resistances 54.0Ω and 29.2Ω are connected in parallel between the points ‘c’ and ‘o’. Now Y connected network formed with the terminal ‘a-b-c’ is converted to equivalent connected network. Δ Version 2 EE IIT, Kharagpur Now, Ω = + + + = 37 .
19 ) ( ca bc ab ab bc ac in R R R R R R R Remarks: • If the connected network consists of inductances (assumed no mutual coupling forms between the inductors) then the same formula can be used for conversion (see in detail 3-phase ac circuit analysis in Lesson-19). or Y Δ Y to or to Y Δ Δ • On the other hand, the connected network consists of capacitances can be converted to an equivalent or Y Δ Δ Y or network provided the capacitance value is replaced by its reciprocal in the conversion formula (see in detail 3-phase ac circuit analysis in Lesson-19). Example: L.6.4 Find the equivalent inductance of the network (see fig.6.5(a)) at the terminals ‘a’ & ‘b’ using transformations. eq R & Y −Δ Δ −Y Version 2 EE IIT, Kharagpur Solution: Convert the three terminals (c-d-e) Δ network (see fig.6.5(a)) comprising with the resistors to an equivalent Y -connected network using the following Y Δ − conversion formula. 6 4 6 2 2 4 2 ; 1 ; 0.666 12 12 12 co do eo R R and R × × × = = Ω = = Ω = = Ω Version 2 EE IIT, Kharagpur Similarly, the -connected network ( f-e-b) is converted to an equivalent Y -connected Network. Δ 1 9 5 9 1 5 0.6 ; 3 ; 0.333 15 15 15 fo eo bo R R and R ′ ′ ′ × × × = = Ω = = Ω = = Ω After the conversions, the circuit is redrawn and shown in fig.6.5(b). Next the series-parallel combinations of resistances reduces the network configuration in more simplified form and it is shown in fig.6.5(c). This circuit (see fig.6.5(c)) can further be simplified by transforming Y connected network comprising with the three resistors (2 Y Δ − Ω , 4Ω , and 3.666 ) to a -connected network and the corresponding network parameters are given below: Ω Δ 2 3.666 2 4 2 3.666 7.5 ; 2 4 8.18 ; 4 3.666 4 3.666 4 3.666 15 2 co cb bo R R and R ′ ′ × × = + + = Ω = + + = Ω × = + + = Ω Version 2 EE IIT, Kharagpur Simplified form of the circuit is drawn and shown in fig.6.5(d) and one can easily find out the equivalent resistance between the terminals ‘a’ and ‘b’ using the series- parallel formula. From fig.6.5(d), one can write the expression for the total equivalent resistance at the terminals ‘a’ and ‘b’ as eq R eq R ( ) ( ) [ ] ( ) 5 4.6 || 7.5 0.333||15 ||8.18 5 2.85 0.272 ||8.18 5 3.122 ||8.18 7.26 eq R ⎡ ⎤ = + + ⎣ ⎦ = + + = + = Ω L.6.3 Test Your Understanding [Marks: 40] T.1 Apply transformations only to find the value of the Current that drives the circuit as shown in fig.6.6. Y or Y −Δ Δ − I Version 2 EE IIT, Kharagpur (ans: 10.13 ) Ω T.2 Find the current through I 4Ω resistor using Y −Δ or Y Δ − transformation technique only for the circuit shown in fig.6.7. (ans: 7.06 A ) T.3 For the circuit shown in fig.6.8, find without performing any conversion. eq R (Ans.6 ) Ω Version 2 EE IIT, Kharagpur T.4 For the circuit shown in fig.6.9, calculate the equivalent inductance for each circuit and justify your answer conceptually. eq R (ans. ) 1 2 eq eq R R = T.5 Find the value of for the circuit of fig.6.10 when the switch is open and when the switch is closed. eq R (Ans. ; ) 8.75 eq R = Ω 7.5 eq R = Ω T.6 For the circuit shown in fig.6.11, find the value of the resistance ‘ ’ so that the equivalent capacitance between the terminals ‘a’ and b’ is 20.
R 57Ω . (Ans.30 ) Ω Version 2 EE IIT, Kharagpur T.7 conversion is often useful in reducing the ------------ of a resistor network ---------- to the beginning nodal or mesh analysis. Y or Y −Δ Δ − T.8 Is it possible to find the current through a branch or to find a voltage across the branch using conversions only? If so, justify your answer. / Y −Δ Δ −Y ___________ Version 2 EE IIT, Kharagpur Source: Basic%20Electrical%20Technology/pdf/L-06(GDR)(ET)%20((EE)NPTEL).pdf |
4177 | https://www.reddit.com/r/Mcat/comments/tedq3r/electron_pair_geometry_vs_molecular_geometry/ | Electron pair geometry vs Molecular geometry? : r/Mcat
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Electron pair geometry vs Molecular geometry?
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Can someone please confirm, electron pair geometry does not account for lone pairs while molecular geometry does take lone pairs into account? I always thought the opposite since it was more intuitive to me that way, but I guess not? Thanks.
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4178 | http://kirkmcd.princeton.edu/examples/ellipsoid.pdf | Conducting Ellipsoid and Circular Disk Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (September 21, 2002; updated December 27, 2019) 1 Problem Show that the surface charge density σ on a conducting ellipsoid, x2 a2 + y2 b2 + z2 c2 = 1, (1) can be written, σellipsoid = Q 4πabc x2 a4 + y2 b4 + z2 c4 , (2) where Q is the total charge.
Show that if the charge distribution of the ellipsoid is projected onto any of its symmetry planes, the result is independent of the extent of the ellipse perpendicular to the plane of projection (i.e., σxy, the projection of σellipsoid on the x-y plane, is independent of parameter c). Thus, the projection of the charge distribution of a conducting oblate or prolate spheroid onto its equatorial plane is the same as the projected charge distribution of a conducting sphere.
By considering a thin conducting circular disk of radius a as a special case of an ellipsoid, show that its surface charge density (summed over both sides) can be written as, σcircular disk = V0 2π2√ a2 −r2, (3) where the electric potential V0 of the disk is related by V0 = πQ/2a.
Show also that if the charge distribution on the conducting ellipsoid is projected onto any of the coordinate axes, the result is uniform (i.e., the charge distribution projected onto the x-axis is σx = Q/2a). In particular, we expect that the charge distribution along a conducting needle will be uniform, since the needle can be considered as the limit of a conducting ellipsoid, two of whose three axes have shrunk to zero.
2 Solution The charge distribution (3) on a conducting ellipsoid can be deduced in a variety of ways -. We record here a highly geometric derivation following Thomson (1869) .1,2 1The charge distribution on the surface of a conducting, prolate spheroid was deduced by Green (1828), pp. 68-69 of , by noting that the equipotentials of a uniformly charged needle are spheroids. He stated that his results “agree with what has been long known”.
2Thomson’s method was stated in a textbook by Murphy in 1833 (with no equations), which suggests that it was already well known.
1 The starting point is the “elementary” result that the electric field is zero in the interior of a spherical shell of any thickness that has a uniform volume charge density between the inner and outer surfaces of the shell. A well-known geometric argument (due to Newton, book 1, prop. 70, p. 218 of ) for this is illustrated in Fig. 1.
Figure 1: For any point r0 in the interior of a uniformly charged shell of charge, the axis of a narrow bicone intercepts the inner surface of the shell at points r1 and r2. The corresponding areas on the inner surface of the shell intercepted by the bicone are A1 and A2. In the limit of small areas, A1/R2 01 = A2/R2 02.
The electric field at point r0 in the interior of the shell due to a lamina of thickness δ and area A1 centered on point r1 that lies within a narrow cone whose vertex is point 0 is given by, E1 = ρ dVol1 R2 01 ˆ R01, (4) where ρ is the volume charge density, dVol1 = A1δ, R01 = r0−r1, and the center of the sphere is taken to be at the origin. Likewise, the electric field from a lamina of area A2 centered on point r2 defined by the intercept with the shell of the same narrow cone extended in the opposite direction (forming a bicone) is given by, E2 = ρ dVol2 R2 02 ˆ R02, (5) In the limit of bicones of small half angle, the two parts of the bicone as truncated by the shell are similar, so that, A1 R2 01 = A2 R2 02 , dVol1 R2 01 = dVol2 R2 02 , (6) and, of course, ˆ R02 = −ˆ R01. Hence E1 + E2 = 0. Since this construction can be applied to all points in the material of the spherical shell, and for all pairs of surface elements subtended by (narrow) bicones, the total electric field in the interior of the shell is zero.
2 We now reconsider the above argument after arbitrary scale transformations have been applied to the rectangular coordinate axes, x →k1x, y →k2y, z →k3z.
(7) A spherical shell of radius s is thereby transformed into an ellipsoid, x2 s2 + y2 s2 + z2 s2 = 1 → x2 s2/k2 1 + y2 s2/k2 2 + z2 s2/k2 3 = 1.
(8) As parameter s is varied, one obtains a set of similar ellipsoids, centered on the origin.
A small volume element obeys the transformation, dVol = dxdydz →k1k2k3 dxdydz = k1k2k3 dVol.
(9) The three points 0, 1, and 2 in Fig. 1 lie along a line, so that, R01 = r0 −r1 = CR02 = C(r0 −r2), (10) where C is a (negative) constant. This relation is invariant under the scale transformation (7), so that together with eq. (9) the relation, dVol1 R2 01 = dVol2 R2 02 , (11) is also invariant. Hence, if the ellipsoidal shell, which is the transform of the spherical shell of Fig. 1, contains a uniform volume charge density, the relation E1 + E2 = 0 remains true at the vertex of any bicone in the interior of the shell, which implies that the total electric field is zero there.3 This proof is based on the premise that the ellipsoidal shell is bounded by two similar ellipsoids, and that the volume charge density in the shell is uniform.
If we let the outer ellipsoid of the shell approach the inner one, always remaining similar to the latter, we reach a configuration that is equivalent to a thin, conducting ellipsoid, since in both cases the electric field is zero in the interior. Hence, the surface charge distribution on a thin, conducting ellipsoid must the same as the projection onto its surface of a uniform charge distribution between that surface and a similar, but slightly larger ellipsoidal surface.
The charge σ per unit area on the surface of a thin, conducting ellipsoid is therefore proportional to the thickness, which we write as δd, of the ellipsoidal shell formed by that surface and a similar, but slightly larger ellipsoid, σ = ρ δd, (12) where constant ρ is to be determined from a knowledge of the total charge Q on the con-ducting ellipsoid.
3This result was obtained by Newton (for the gravtitational attraction) at points within spheroids shells, book 1, prop. 91, cor. 3, p. 239 of . The results for ellipsoidal shells may have first been given by Ivory, p. 364 of (1809).
3 The thickness δd of a thin ellipsoidal shell at some point on its inner surface is the distance between the plane that is tangent to the inner surface at the specified point, and the plane that is tangent to the outer surface at the point similar to the specified point. These planes are parallel since the ellipsoids are parallel. In particular, if the semimajor axes of the inner ellipsoid are called a, b, and c, then those of the outer ellipsoid can be written a + δa, b + δb and c+δc. Let the (perpendicular) distance from the plane tangent to the specified point on the inner ellipsoid to its center be called d, and the corresponding distance from the outer tangent plane be d+δd, so that δd is the desired thickness of the shell at the specified point.
Then, the condition of similarity is that, δa a = δb b = δc c = δd d .
(13) Since the volume of an ellipsoid with semimajor axes a, b, and c is 4πabc/3, the volume of the ellipsoidal shell is 4π(a+δa)(b+δb)(c+δc)/3−4πabc/3 = 4πabc(δd/d), using eq. (13).
As the constant ρ has an interpretation as the uniform charge density within the material of the ellipsoidal shell, we find that the total charge Q on the conducting ellipsoid is related by, Q = ρ Volshell = 4πabc d ρ δd, (14) and hence, σ = ρ δd = Qd 4πabc .
(15) It remains to find an expression for the distance d to the tangent plane. If we write the equation for the ellipsoid in the form, f(x, y, z) = x2 a2 + y2 b2 + z2 c2 −1 = 0, (16) then the gradient of f is perpendicular to the tangent plane. Thus, the vector d from the center of the ellipsoid to the tangent plane is proportional to ∇f. That is, d ∝∇f = x a2 , y b2 , z c2 , (17) noting that x2/a2 + y2/b2 + z2/c2 = 1 on the ellipsoid. The unit vector ˆ d is therefore, ˆ d = x a2 , y b2 , z c2 x2 a4 + y2 b4 + z2 c4 .
(18) The magnitude d of the vector d is related to the vector r = (x, y, z) of the specified point on the ellipse by,4 d = r · ˆ d = x2 a2 + y2 b2 + z2 c2 x2 a4 + y2 b4 + z2 c4 = 1 x2 a4 + y2 b4 + z2 c4 .
(19) 4Nov. 15, 2019. It was pointed out in that d = 1/(ˆ d·∇f) = 1/d f/du = d f δ(f)/d f/du = du δ(f), where u is a coordinate perpendicular to surface of the ellipsoid, with u = 0 on the surface, and δ(f) is the Dirac delta function.
4 At length, we have found the charge density on the surface of a conducting ellipsoid to be, σellipsoid = Q 4πabc x2 a4 + y2 b4 + z2 c4 , (20) where Q is the total charge.
To deduce the projection of the charge distribution of the conducting ellipse onto one of its symmetry planes, say the x-y plane, note that a projected area element dx dy corresponds to area dA on the surface of the ellipsoid that is related by, dx dy = dA · ˆ z = dA dz, (21) where the unit vector ˆ d that is normal to the surface of the ellipsoid at the point (x, y) is given by eq. (19). Thus, dA = dx dy ˆ dz = c2 dx dy z x2 a4 + y2 b4 + z2 c4 .
(22) The charge projected onto the x-y plane from both z > 0 and z < 0 is, dQxy = 2σellipsoiddA = Qc dx dy 2πabz = Q dx dy 2πab 1 −x2 a2 −y2 b2 , (23) combining eqs. (1), (20) and (22). The projected charge density (due to both halves of the ellipsoid), σxy = dQxy/dx dy, is independent of the parameter c that specifies the size of the ellipsoid in z. For example, the charge distributions of a sphere, a disk, and both an oblate and prolate spheroid, all of the same equatorial diameter, are the same when projected onto the equatorial plane.
If the project the charge dQxy of eq. (23) onto the x axis, the result is, dQx = Q dx 2πa b2−b2x2 a2 − b2−b2x2 a2 dy b2 −b2x2 a2 −y2 = Q 2adx, (24) which is uniform in x! In particular, the uniform charge distribution on a conducting sphere projects to a uniform charge distribution on any diameter; and the charge distribution is uniform along a conducting needle that is the limit of conducting ellipsoid.
The case of a thin, conducting elliptical disk in the x-y plane can be obtained from eq. (20) by letting c go to zero.5 For this we note that eq. (16) for a general ellipsoid permits us to write, c x2 a4 + y2 b4 + z2 c4 = c2 x2 a4 + y2 b4 + 1 −x2 a2 −y2 b2 → 1 −x2 a2 −y2 b2 .
(25) 5For a rather different method of solution, see .
5 The charge density on each side of a conducting elliptical disk is therefore, σelliptical disk = Q 4πab 1 −x2 a2 + y2 b2 .
(26) The charge density on each side of a conducting circular disk of radius a follows imme-diately as, σcircular disk = Q 4πa √ a2 −r2 , (27) where r2 = x2 + y2. Such a disk has potential V0, which can be found by calculating the potential at the center of the disk according to, V0 = V (r = 0, z = 0) = a 0 2σ(r) r 2πr dr = Q a a 0 dr √ a2 −r2 = πQ 2a .
(28) Hence, a conducting disk of radius a at potential V0 has charge density, σcircular disk = V0 2π2√ a2 −r2.
(29) References R. Murphy, Elementary Principles of the Theories of Electiricity, Heat and Molecular Actions, Part 1 (Pitt Press, 1833), pp. 68-70, W. Thomson (Lord Kelvin), Papers on Electricity and Magnetism, 2nd ed. (Macmillan, 1884), pp. 7 and 178-179, Lord Kelvin and P.G. Tait, Treatise on Natural Philosophy, Vol. 2 (Cambridge U. Press, 1903), secs. 519-520, O.D. Kellogg, Foundations of Potential Theory (Springer, 1929), pp. 188-191, E.J. Routh, A Treatise on Analytical Statics with Illustrations Taken from the Theories of Electricity and Magnetism, Vol. 2 (Cambridge U. Press, 1922), Arts. 71 and 376, E. Durand, ´ Electrostatique, Vol. 2 (Masson, 1966), pp. 415-417, W.R. Smythe, Static and Dynamic Electricity, 3rd ed. (McGraw-Hill, 1968), sec. 5.02, G. Green, Mathematical Papers (Macmillan, 1871), 6 I. Newton, Philosophiæ Naturalis Principia Mathematica (London, 1726), J. Ivory, On the Attractions of homogeneous Ellipsoids, Phil. Trans. Roy. Soc. London 99, 345 (1809), T.L. Curtright et al., Charge Densities for Conducting Ellipsoids, Eur. J. Phys.
41 035204 (2020), 20.pdf E.T. Copson, On the problem of the electrified disc, Proc. Edin. Math. Soc. 8, 14 (1947), 7 |
4179 | http://www.few.vu.nl/~sdn249/BeChDaYa-misc.pdf | GENERALIZED FERMAT EQUATIONS: A MISCELLANY MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI Abstract. This paper is devoted to the generalized Fermat equation xp + yq = zr, where p, q and r are positive integers, and x, y and z are nonzero coprime integers. We begin by surveying the exponent triples (p, q, r), including a number of infinite families, for which the equation has been solved to date, detailing the techniques involved. In the remainder of the paper, we attempt to solve the remaining infinite families of generalized Fermat equations that appear amenable to current techniques. The main tools we employ are based upon the modularity of Galois representations (as is indeed true with all previously solved infinite families).
Contents 1.
Introduction 2 2.
The Euclidean case 5 3.
Multi-Frey techniques 6 3.1.
The equation a3 + b6 = cn 6 3.2.
The equations a2 ± cn = b6 8 4.
Covers of spherical equations 11 4.1.
The equation x2 + y2 = z3 11 4.2.
Quadratic reciprocity 12 4.3.
The equation x2 + y2 = z5 16 4.4.
The equation x2 + y4 = z3 17 4.5.
The equation x3 + y3 = z2 22 4.6.
Other spherical equations 23 5.
Historical notes on the equations a4 ± b4 = c3 23 5.1.
Reduction to elliptic generalized Fermat equations 24 5.2.
Relation to work of Lucas 25 Date: July 2013.
2000 Mathematics Subject Classification. Primary: 11D41, Secondary: 11D61, 11G05, 14G05.
Key words and phrases. Generalized Fermat equations, Galois representations, multi-Frey techniques.
Research supported by NSERC, the third-named author is supported by an NWO-Veni grant.
1 2 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI 6.
Future work 25 References 26 1. Introduction Since Wiles’ remarkable proof of Fermat’s Last Theorem, a number of techniques have been developed for solving various generalized Fermat equations of the shape (1) ap + bq = cr, with 1 p + 1 q + 1 r ≤1, where p, q and r are positive integers, and a, b and c are coprime integers. The Euclidean case, when 1/p + 1/q + 1/r = 1, is well understood (see e.g. Proposition 4) and hence the main topic of interest is when 1/p + 1/q + 1/r < 1, the hyperbolic case. The number of solutions (a, b, c) to such an equation is known to be finite, via work of Darmon and Granville , provided we fix the triple (p, q, r). It has, in fact, been conjectured that there are only finitely many nonzero coprime solutions to equation (1), even allowing the triples (p, q, r) to be variable (counting solutions corresponding to 1p + 23 = 32 just once). Perhaps the only solutions are those currently known; i.e. (a, b, c, p, q, r) coming from the solution to Catalan’s equation 1p + 23 = 32, for p ≥6, and from the following nine identities: 25 + 72 = 34, 73 + 132 = 29, 27 + 173 = 712, 35 + 114 = 1222, 177 + 762713 = 210639282, 14143 + 22134592 = 657, 92623 + 153122832 = 1137, 438 + 962223 = 300429072, and 338 + 15490342 = 156133.
Since all known solutions have min{p, q, r} ≤2, a similar formulation of the aforementioned conjecture is that there are no nontrivial solutions in coprime integers to (1), once min{p, q, r} ≥3. For references on the history of this problem, the reader is directed to the papers of Beukers , , Darmon and Granville , Mauldin and Tijdeman , and, for more classical results along these lines, to the book of Dickson .
Our goals in this paper are two-fold. Firstly, we wish to treat the remaining cases of equation (1) which appear within reach of current technology (though, as a caveat, we will avoid discussion of exciting recent developments involving Hilbert modular forms , , in the interest of keeping our paper reasonably self-contained). Secondly, we wish to take this opportunity to document what, to the best of our knowledge, is the state-of-the-art for these problems. Regarding the former objective, we will prove the following two theorems.
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 3 Theorem 1. Suppose that (p, q, r) are positive integers with 1 p + 1 q + 1 r < 1 and (p, q, r) ∈{(2, n, 6), (2, 2n, 9), (2, 2n, 10), (2, 2n, 15), (3, 3, 2n), (3, 6, n), (4, 2n, 3)} for some integer n. Then equation (1) has no solutions in coprime nonzero integers a, b and c.
Proof. These seven cases will be dealt with in Propositions 11, 17, 19, 20, 25, 7 and 21 respectively.
□ Theorem 2. Suppose that (p, q, r) are positive integers with 1 p + 1 q + 1 r < 1 and (p, q, r) = (2m, 2n, 3), n ≡3 (mod 4), m ≥2, or (2, 4n, 3), n ≡±2 (mod 5) or n ≡±2, ±4 (mod 13) for some integers n and m. Then the only solution to equation (1) in coprime nonzero integers a, b and c is with (p, q, r, |a|, |b|, c) = (2, 8, 3, 1549034, 33, 15613).
Proof. The first case will be dealt with in Proposition 16, the second case in Proposition 22.
□ Taking these results together with work of many other authors over the past twenty years or so, we currently know that equation (1) has only the known solutions for the following triples (p, q, r); in the first table, we list infinite families for which the desired results are known without additional conditions.
(p, q, r) reference(s) (n, n, n), n ≥3 Wiles , Taylor-Wiles (n, n, 2), n ≥4 Darmon-Merel , Poonen (n, n, 3), n ≥3 Darmon-Merel , Poonen (2n, 2n, 5), n ≥2 Bennett (2, 4, n), n ≥4 Ellenberg , Bennett-Ellenberg-Ng , Bruin (2, 6, n), n ≥3 Bennett-Chen , Bruin (2, n, 4), n ≥4 immediate from Bennett-Skinner , Bruin (2, n, 6), n ≥3 Theorem 1, Bruin (3j, 3k, n), j, k ≥2, n ≥3 immediate from Kraus (see Remark 6) (3, 3, 2n), n ≥2 Theorem 1 (3, 6, n), n ≥2 Theorem 1 (2, 2n, k), n ≥2, k ∈{9, 10, 15} Theorem 1 (4, 2n, 3), n ≥2 Theorem 1 4 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI (p, q, r) reference(s) (3, 3, n)∗ Chen-Siksek , Kraus , Bruin , Dahmen (2, 2n, 3)∗ Chen , Dahmen , , Siksek , (2, 2n, 5)∗ Chen (2m, 2n, 3)∗ Theorem 2 (2, 4n, 3)∗ Theorem 2 (3, 3n, 2)∗ Bennett-Chen-Dahmen-Yazdani (2, 3, n), 6 ≤n ≤10 or n = 15 Poonen-Schaefer-Stoll , Bruin , , , Brown , Siksek , Siksek-Stoll (3, 4, 5) Siksek-Stoll (5, 5, 7), (5, 5, 19), (7, 7, 5) Dahmen-Siksek The (∗) in the second table indicates that the result has been proven for a family of exponents of natural density one (but that there remain infinitely many prime exponents of positive Dirichlet density untreated). The following table provides the exact conditions that the exponents must satisfy.
(p, q, r) n (3, 3, n) 3 ≤n ≤104, or n ≡2, 3 (mod 5), n ≡17, 61 (mod 78), n ≡51, 103, 105 (mod 106), or n ≡43, 49, 61, 79, 97, 151, 157, 169, 187, 205, 259, 265, 277, 295, 313 367, 373, 385, 403, 421, 475, 481, 493, 511, 529, 583, 601, 619, 637, 691, 697, 709, 727, 745, 799, 805, 817, 835, 853, 907, 913, 925, 943, 961, 1015, 1021, 1033, 1051, 1069, 1123, 1129, 1141, 1159, 1177, 1231, 1237, 1249, 1267, 1285 (mod 1296) (2, 2n, 3) 3 ≤n ≤107 or n ≡−1 (mod 6) (2m, 2n, 3) m ≥2 and n ≡−1 (mod 4) (2, 4n, 3) n ≥2 and either n ≡±2 (mod 5) or n ≡±2, ±4 (mod 13) (2, 2n, 5) n ≥17 and n ≡1 (mod 4) prime (3, 3n, 2) n ≡1 (mod 8) prime GENERALIZED FERMAT EQUATIONS: A MISCELLANY 5 Remark 3. We do not list in the above tables examples of equation (1) which can be solved under additional local conditions (such as, for example, the case (p, q, r) = (5, 5, n) with c even, treated in an unpublished note of Darmon and Kraus). We will also not provide information on generalized versions of (1) such as equations of the shape Aap +Bbq = Ccr, where A, B and C are integers whose prime factors lie in a fixed finite set. Regarding the latter, the reader is directed to , , , , , , (for general signatures), , (for signature (p, p, p)), , , , , , (for signature (p, p, 2)), , (for signature (p, p, 3)), and to , , , , and (for various signatures of the shape (n, n, p) with n fixed).
In each of these cases, where equation (1) has been treated for an infinite family of exponents, the underlying techniques have been based upon the modularity of Galois representations. The limitations of this approach are unclear at this time, though work of Darmon and Granville (e.g. Proposition 4.2 of ; see also the discussion in ) suggests that restricting attention to Frey-Hellegouarch curves over Q (or, for that matter, to Q-curves) might enable us to treat only signatures which can be related via descent to one of (2) (p, q, r) ∈{(n, n, n), (n, n, 2), (n, n, 3), (2, 3, n), (3, 3, n)} .
Of course, as demonstrated by the striking work of Ellenberg (and, to a lesser degree, by Theorems 1 and 2), there are some quite nontrivial examples of equations of the shape (1) which may be reduced to the study of the form Aap + Bbq = Ccr for signatures (p, q, r) in (2).
For more general signatures, an ambitious program of Darmon , based upon the arithmetic of Frey-Hellegouarch abelian varieties, holds great promise for the future, though, in its full generality, perhaps not the near future.
As for notation, by a newform f, we will always mean a cuspidal newform of weight 2 with respect to Γ0(N) for some positive integer N. This integer N will be called the level of f.
2. The Euclidean case For convenience in the sequel, we will collect together a number of old results on the equation ap + bq = cr in the Euclidean case when 1 p + 1 q + 1 r = 1.
Proposition 4. The equations a2 + b6 = c3, a2 + b4 = c4, a4 + b4 = c2 and a3 + b3 = c3 have no solutions in coprime nonzero integers a, b and c. The only solutions to the equation a2+b3 = c6 in coprime nonzero integers a, b and c are with (|a|, b, |c|) = (3, −2, 1).
6 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI Proof. This is standard (and very classical). The equations correspond to the elliptic curves E/Q denoted by 144A1, 32A1, 64A1, 27A1 and 36A1 in Cremona’s notation, respectively. Each of these curves is of rank 0 over Q; checking the rational torsion points yields the desired result.
□ 3. Multi-Frey techniques In , the first two authors applied multi-Frey techniques pioneered by Bugeaud, Mignotte and Siksek to the generalized Fermat equation a2 + b6 = cn. In this approach, information derived from one Frey-Hellegouarch curve (in this case, a Q-curve specific to this equation) is combined with that coming from a second such curve (corresponding, in this situation, to the generalized Fermat equation x2 + y3 = zn, with the additional constraint that y is square).
In this section, we will employ a similar strategy to treat two new families of generalized Fer-mat equations, the second of which is, in some sense, a “twisted” version of that considered in (though with its own subtleties). A rather more substantial application of such techniques is published separately in , where we discuss the equation a3 + b3n = c2.
3.1. The equation a3 + b6 = cn. Here, we will combine information from Frey-Hellegouarch curves over Q, corresponding to equation (1) for signatures (2, 3, n) and (3, 3, n). We begin by noting a result of Kraus on (1) with (p, q, r) = (3, 3, n).
Proposition 5 (Kraus). If a, b and c are nonzero, coprime integers for which a3 + b3 = cn, where n ≥3 is an integer, then c ≡3 (mod 6) and v2(ab) = 1.
Actually, Kraus proves this only for n ≥17 a prime. The remaining cases of the above proposition follow from Proposition 4 and the results of and , which yield that there are no nontrivial solutions to the equation above when n ∈{3, 4, 5, 7, 11, 13}.
Remark 6. Proposition 5 trivially implies that the equation a3j + b3k = cn has no solutions in coprime nonzero integers a, b and c, provided n ≥3 and the integers j and k each exceed unity. The case with n = 2 remains, apparently, open.
Returning to the equation a3 + b6 = cn, we may assume that n > 163 is prime, by appealing to work of Dahmen (for n ∈{5, 7, 11, 13}) and Kraus (for primes n with 17 ≤n ≤163). The GENERALIZED FERMAT EQUATIONS: A MISCELLANY 7 cases n ∈{3, 4} follow from Proposition 4 (alternatively, if n ∈{4, 5}, one can also appeal to work of Bruin ). Applying Proposition 5, we may suppose further that c ≡3 (mod 6) and v2(a) = 1. We begin by considering the Frey-Hellegouarch curve E1 : Y 2 = X3 + bX2 + b2 + a 3 X + b(b2 + a) 9 , essentially a twist of the standard curve for signature (2, 3, n) (see page 530 of Darmon and Granville ). Since 3 | c, noting that v3(a2 −ab2 + b4) ≤1, we thus have v3(a + b2) ≥n −1. It follows, from a routine application of Tate’s algorithm, that E1 has conductor 26 · 3 · Q p, where the product runs through primes p > 3 dividing c (the fact that 3 divides c ensures multiplicative reduction at 3).
Here and henceforth, for an elliptic curve E/Q and prime l, we denote by ρE l : Gal(Q/Q) →GL2(Fl) the Galois representation induced from the natural action of Gal(Q/Q) on the l-torsion points of E.
Since n > 163, by work of Mazur [54, Theorem 7.1], the representation ρE1 n is irreducible. Appealing to modularity and Ribet’s level lowering , , it follows that the newform attached to E1 is congruent to a newform g of level 26 · 3 = 192. All such newforms are integral and, in particular, have Fourier coefficients satisfying a7(g) ∈{0, ±4}.
Considering the curve E1 modulo 7, we find that either 7 | c or that a7(E1) ≡−b3 (mod 7). In the first case, a7(g) ≡±8 (mod n), contradicting the fact that n > 163. If 7 ∤bc, then a7(E1) ≡ ±1 (mod 7) and hence, by the Weil bounds, a7(E1) = ±1, which is incongruent modulo n to any of the choices for a7(g). We therefore conclude that 7 | b.
We turn now to our second Frey-Hellegouarch curve, that corresponding to signature (3, 3, n).
Following Kraus , we consider E2 : Y 2 = X3 + 3ab1X + a3 −b3 1, b1 = b2.
Arguing as in , the newform attached to E2 is congruent modulo n to the unique newform g′ of level 72. Since 7 | b, we find that a7(E2) = ±4, while a7(g′) = 0, an immediate contradiction. We thus may conclude as follows.
Proposition 7. If n ≥2 is an integer, then the only solutions to the equation a3 +b6 = cn in nonzero coprime integers a, b and c are given by (n, a, |b|, |c|) = (2, 2, 1, 3).
8 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI 3.2. The equations a2 ± cn = b6. We begin by noting that the cases with n = 3 follow from Proposition 4, while those with n = 4 were treated by Bruin (Theorems 2 and 3). The desired result with n = 7 is immediate from . We will thus suppose, without loss of generality, that there exist coprime nonzero integers a, b and c, with (3) a2 + cn = b6, for n = 5 or n ≥11 prime.
We distinguish two cases depending upon the parity of c.
Assume first that c is odd. In the factorization b6 −a2 = (b3 −a)(b3 + a), the factors on the right hand side must be odd and hence coprime. We deduce, therefore, the existence of nonzero integers A and B for which b3 −a = An and b3 + a = Bn, where gcd(b, A, B) = 1. This leads immediately to the Diophantine equation An + Bn = 2b3, which, by Theorem 1.5 of , has no coprime solutions for primes n ≥5 and |AB| > 1. It follows that there are no nonzero coprime solutions to equation (3) with c odd.
Remark 8. If we write the Frey-Hellegouarch curve used to prove Theorem 1.5 of in terms of a and b, i.e. substitute An = b3 + a, we are led to consider E : Y 2 + 6 b XY + 4(b3 + a)Y = X3.
This model has the same c-invariants as, and hence is isomorphic to, the curve given by (4) Y 2 = X3 −3(5b3 −4a)bX + 2(11b6 −14b3a + 2a2).
On replacing a by −ia in (4), one obtains the Frey-Hellegouarch Q-curve used for the equation a2+b6 = cn in .
Next, assume that c is even. In this case, we can of course proceed as previously, i.e. by factoring b6 −a2, reducing to a generalized Fermat equation (5) An + 2n−2Bn = b3, and considering the Frey-Hellegouarch elliptic curve E1 : Y 2 + 3bXY + AnY = X3.
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 9 This approach, as it transpires, again yields a curve isomorphic to (4). By Lemma 3.1 of , the Galois representation on the n-torsion points of E1 is absolutely irreducible for n ≥5, whereby we can apply the standard machinery based on modularity of Galois representations. If one proceeds in this direction, however, it turns out that one ends up dealing with (after level lowering, etc.) newforms of level 54; at this level, we are apparently unable to obtain the desired contradiction, at least for certain n. One fundamental reason why this level causes such problems is the fact that the curve (4), evaluated at (a, b) = (3, 1) or (a, b) = (17, 1), is itself, in each case, a curve of conductor 54.
It is, however, still possible to use this approach to rule out particular values of n, appealing to the method of Kraus – we will do so for n = 5 and n = 13. In case n = 5, considering solutions modulo 31 to (5), we find that if 31 ∤AB, then necessarily a31(E1) ∈{−7, −4, 2, 8}, whereby we have, for F1 a newform of level 54, that a31(F1) ≡−7, −4, 2, 8 (mod 5) or a31(F1) ≡±32 (mod 5). Since each such newform is one-dimensional with a31(F1) = 5, we arrive at a contradiction, from which we conclude that equation (3) has no nonzero coprime solutions with n = 5.
Similarly, if n = 13 and we consider solutions modulo 53 to (5), we find that a53(E1) ∈{−6, 3, 12} or E1 has multiplicative reduction at 53. This implies that for F1 a newform of level 54, we have a53(F1) ≡−6, 3, 12 (mod 13) or a53(F1) ≡±54 (mod 13). On the other hand, for every such newform F1, a53(F1) = ±9, a contradiction. Equation (3) thus has no nonzero coprime solutions with n = 13.
To treat the remaining values of n, we will employ a second Frey-Hellegouarch curve (that for the signature (2, 3, n)). Specifically, to a potential solution (a, b, c) to (3) with n ≥11 and n ̸= 13 prime, we associate the curve given by the Weierstrass equation (6) E2 : Y 2 = X3 −3b2X −2a.
This model has discriminant ∆= 2633cn. Note that since c is even, both a and b are odd, whereby it is easy to show that v2(c4) = 4, v2(c6) = 6 and v2(∆) > 12 (since n > 6). These conditions alone are not sufficient to ensure non-minimality of the model at 2 (in contrast to like conditions at an odd prime p). A standard application of Tate’s algorithm, however, shows that for a short Weierstrass model satisfying these conditions either the given model or that obtained by replacing a6 by −a6 (i.e. twisting over Q(√−1)) is necessarily non-minimal. Without loss of generality, replacing a by −a if necessary, we may thus assume that E2 is not minimal at 2. It follows that a minimal model for this curve has v2(c4) = v2(c6) = 0 and v2(∆) > 0, whereby the conductor N(E2) of E2 satisfies v2(N(E2)) = 1. If 3 ∤c, then v3(∆) ≤3 and so v3(N(E2)) ≤3. If 3 | c, then v3(c4) = 2, v3(c6) = 3 and v3(∆) > 6 (since n > 3), which implies that the twist of E2 over either of Q(√±3) has multiplicative reduction at 3, whereby v3(N(E2)) = 2. For any prime p > 3, we see that the model for E2 is minimal 10 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI at p. In particular n | vp(∆min(E2)) for primes p > 3. In conclusion, N(E2) = 2 · 3α Y p|c, p>3 p, α ≤3.
In order to apply level lowering, it remains to establish the irreducibility of the representation ρE2 n .
Lemma 9. If n ≥11, n ̸= 13 is prime, then ρE2 n is irreducible.
Proof. As is well-known (see e.g. [27, Theorem 22]) by the work of Mazur et al, ρE2 n is irreducible if n = 11 or n ≥17, and j(E2) is not one of −215, −112, −11 · 1313, −17 · 3733 217 , −172 · 1013 2 , −215 · 33, −7 · 1373 · 20833, −7 · 113, −218 · 33 · 53, −215 · 33 · 53 · 113, −218 · 33 · 53 · 233 · 293.
Since j(E2) = 26 ·33b6/cn, one quickly checks that none of these j-values leads to a solution of (3).
□ Remark 10. We note that proving irreducibility of ρE2 n for n = 5, 7, 13 is reduced to studying the Diophantine equation j(E2) = jn(x), where jn(x) is the j-map from X0(n) to X(1). For example, when n = 13, this amounts (after introducing y = a/b3) to finding rational points on a hyperelliptic curve of genus 3 that we can solve (with some work) using standard Chabauty-type techniques. The previous argument then shows that the Frey-Hellegouarch curve E2 can be used as well to solve (5) for n = 13. We leave the details to the interested reader.
Using Lemma 9, modularity and level lowering , , we thus arrive at the fact that ρE2 n is modular of level 2 · 3α with α ≤3 (and, as usual, with weight 2 and trivial character). At levels 2, 6 and 18, there are no newforms whatsoever, while at level 54 there are only rational newforms.
It follows that there exists a newform f of level 54, with ρE2 n ≃ρf n (equivalently, an elliptic curve F2 of conductor 54 with ρE2 n ≃ρF2 n ). If 5 | c, then E2 has multiplicative reduction at 5 and hence a5(f) ≡±6 (mod n). Since we are assuming that n ≥11, and since a5(f) = ±3, this leads to a contradiction. If 5 ∤c, then E2 has good reduction at 5 and, considering all possible solutions of equation (3) modulo 5, we find that a5(E2) ∈{±4, ±1, 0}. Since a5(f) ≡a5(E2) (mod n) and n ≥11, the resulting contradiction finishes our proof. We have shown Proposition 11. The only solutions to the generalized Fermat equation a2 + δ cn = b6, in coprime nonzero integers a, b and c, with n ≥3 an integer and δ ∈{−1, 1}, are given by (n, |a|, |b|, δc) = (3, 3, 1, −2) (i.e. the Catalan solutions).
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 11 Remark 12. In the preceding proof, we saw that the possibilities for ap(f) and ap(E2) are disjoint for p = 5. This does not appear to be the case for any prime p > 5 (and we cannot use p = 2, 3 in this fashion), so, insofar as there is ever luck involved in such a business, it appears that we have been rather lucky here.
4. Covers of spherical equations The spherical cases of the generalized Fermat equation xp+yq = zr are those with signature (p, q, r) satisfying 1 p + 1 q + 1 r > 1 (for integers p, q and r, each exceeding unity). To be precise, they are, up to reordering (i.e. permuting x, y and z and changing their signs), (p, q, r) ∈{(2, 3, 3), (2, 3, 4), (2, 4, 3), (2, 3, 5)} and (p, q, r) = (2, 2, n) or (2, n, 2), for some n ≥2. In each case, the corresponding equations possess infinitely many coprime nonzero integer solutions, given by a finite set of 2-parameter families (see e.g. and ). The explicit parameterizations (with proofs) can be found in Chapter 14 of Cohen . We will have need of those for (p, q, r) = (2, 2, 3), (2, 2, 5), (2, 4, 3) and (3, 3, 2).
4.1. The equation x2 + y2 = z3. If x, y and z are coprime integer solutions to this equation, then we have (see page 466 of ) (7) (x, y, z) = (s(s2 −3t2), t(3s2 −t2), s2 + t2), for coprime integers s and t, of opposite parity. We begin this subsection with some remarks on the Diophantine equation a2 + b2n = c3. This particular family is treated in and in , where, using techniques of Kraus and Chen-Siksek , the following is proved.
Theorem 13 (Dahmen ). If n is a positive integer satisfying 3 ≤n ≤107 or n ≡−1 (mod 6), then the Diophantine equation a2 + b2n = c3 has no solutions in nonzero coprime integers a, b and c.
Here we recall part of the proof of this theorem for completeness (and future use).
Proposition 14. If a, b and c are nonzero coprime integers for which a2 + b2n = c3, where n ≥3 is an integer, then b ≡3 (mod 6).
Proof. We may suppose that n ≥7 is prime, since for n = 3, 4 and 5 there are no solutions (see Proposition 4, and ). From (7), if we have coprime integers a, b and c with a2 + b2n = c3, 12 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI there exist coprime integers s and t, of opposite parity, for which bn = t (3s2 −t2), and hence coprime integers B and C, and δ ∈{0, 1}, with t = 3−δBn and 3s2 −t2 = 3δ Cn.
If δ = 0 (this is the case when 3 ∤b), it follows that Cn + B2n = 3s2 which, via Theorem 1.1 of , implies a = 0 (and so s = 0). If, on the other hand, we have δ = 1 (so that 3 | b) and b (and hence t) even, then, writing B = 3B1, we have that Cn + 32n−3B2n 1 = s2, with B1 even. Arguing as in , there thus exists a newform of level 6, an immediate contradiction.
□ Note that when b ≡3 (mod 6) we are led to the Diophantine equation (8) Cn + 1 27B2n = A2, with A even and BC odd, and hence, via (say) the Frey-Hellegouarch curve (see e.g. ) (9) E : Y 2 = X3 + 2AX2 + B2n 27 X, to a newform of level 96, which we are (presently) unable to rule out for certain n. However, arguing as in and , we can resolve this case for a family of exponents n of natural density 1. We recall these techniques here.
4.2. Quadratic reciprocity. In what follows, we will employ the Hilbert symbol instead of the Legendre symbol, to enable us to treat the prime 2 without modification of our arguments. Recall the (symmetric, multiplicative) Hilbert symbol (, )K : K∗× K∗→{±1} defined by (A, B)K = 1 if z2 = Ax2 + By2 has a nonzero solution in K, −1 otherwise.
For concision, we let (, )p, (, ) and (, )∞denote (, )Qp, (, )Q and (, )R, respectively. Note that we have the reciprocity law (10) Y p≤∞ (a, b)p = 1, valid for all nonzero rationals a and b. For an odd prime p, if A = pαu and B = pβv with u and v p-adic units, we further have (11) (A, B)p = (−1)αβ (p−1) 2 u p β v p α .
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 13 In particular, for an odd prime p where vp(A) and vp(B) are even, it follows that (A, B)p = 1. When p = 2, we have the following analogous formula: if we write A = 2αu and B = 2βv with u and v 2-adic units, then (A, B)2 = (−1) u−1 2 v−1 2 +α v2−1 8 +β u2−1 8 .
Proposition 15. Let r and s be nonzero rational numbers. Assume that vl(r) = 0 for all l | n and that the Diophantine equation A2 −rB2n = s(Cn −B2n) has a solution in coprime nonzero integers A, B and C, with BC odd. Then (r, s(C −B2))2 Y vp(r) odd vp(s) odd 2<p<∞ (r, s(C −B2))p = 1.
Proof. Let us begin by noting that, by the reciprocity law (10), we have (12) (r, s(C −B2))2 Y vp(r) odd, vp(s) odd, 2<p<∞ (r, s(C −B2))p = (r, s(C −B2))∞ Y vp(r) even, vp(s) odd, 2<p<∞ (r, s(C −B2))p.
Since we suppose A2 = rB2n + s(Cn −B2n) · 12, it follows that (r, s(Cn −B2n))p = 1, for all primes p ≤∞. Therefore (r, s(C −B2))p = (r, Cn−1 + · · · + B2n−2)p.
Since Cn−1 + · · · + B2n−2 > 0, we also have (r, s(C −B2))∞= 1. Now, assume that vp(r) is even for an odd prime p. If vp(s(C −B2)) is also even then, by equation (11), we have that (r, s(C −B2))p = 1.
If vp(s(C −B2)) is odd, but p ∤n then vp(Cn−1 + · · · + B2n−2) = 0, which implies that (r, s(C −B2))p = (r, Cn−1 + · · · + B2n−2)p = 1.
When p | n and vp(s(C −B2)) is odd, since we are assuming that r is a p-unit and since A, B and C are coprime, it follows that A2 ≡rB2n (mod p), and hence r p = 1. Appealing again to equation (11), we conclude that (r, s(C −B2))p = 1, as desired.
□ 14 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI This proposition provides us with an extra constraint upon C/B2 (mod r) to which we can appeal, at least on occasion, to rule out exponents n in certain residue classes. If we suppose that we have a solution to equation (8) in integers A, B, C and n with A even and BC odd, we can either add or subtract B2n from both sides of the equation in order to apply the above proposition. Subtracting B2n (this is the case treated in ), we obtain A2 −28 27B2n = Cn −B2n.
Here we have r = 28 27 and s = 1, and, via Proposition 15 (supposing that n ≡−1 (mod 6) and appealing to to treat the cases with 7 | n), may conclude that (28/27, C −B2)2(28/27, C −B2)3(28/27, C −B2)7 = 1.
Since 3 | B, the quantity Cn is a perfect square modulo 3 and so (28/27, C −B2)3 = 1. Also, since Cn−1 + · · · + B2n−2 is odd, we may compute that (28/27, C −B2)2 = (28/27, Cn −B2n)2 = 1.
If 7 | C −B2 then necessarily 7 | A, whereby 0 = a7(E) ̸≡−4 (mod n), an immediate contradiction.
Therefore 7 ∤C −B2, and so (13) 1 = (28/27, C −B2)7 = C −B2 7 .
On the other hand, since each elliptic curve E/Q of conductor 96 has a7(E) = ±4, computing the corresponding Fourier coefficient for our Frey-Hellegouarch curve (9), we find that A2 ≡B2n (mod 7) (where A ̸≡0 (mod 7)).
It follows from (8) that (C/B2)n ≡2 (mod 7). Since n ≡−1 (mod 6), we therefore have C/B2 −1 ≡ 3 (mod 7), contradicting (13). This proves the second part of Theorem 13.
Similarly, adding B2n to both sides of equation (8), we have A2 + 26 27B2n = Cn + B2n = −((−C)n −B2n), where we suppose that n ≡3 (mod 4) is prime (so that, via Theorem 13, n ≡7 (mod 12)). We may thus apply Proposition 15 with r = −26/27 and s = −1 to conclude that Y p|78 (−26/27, s((−C) −B2))p = Y p|78 (−78, C + B2)p = 1.
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 15 As before, we find that (−78, C + B2)3 = 1.
Since each elliptic curve E/Q of conductor 96 has a13(E) = ±2, we thus have, via (9), A ≡±1 (mod 13), B2n ≡4, 9, 10 or 12 (mod 13), or A ≡±2 (mod 13), B2n ≡1, 3, 9 or 10 (mod 13), or A ≡±3 (mod 13), B2n ≡3, 4, 10 or 12 (mod 13), or A ≡±4 (mod 13), B2n ≡1, 4, 10 or 12 (mod 13), or A ≡±5 (mod 13), B2n ≡1, 3, 4 or 9 (mod 13), or A ≡±6 (mod 13), B2n ≡1, 3, 9 or 12 (mod 13), whereby (C/B2)n ≡2, 3, 9 or 11 (mod 13) and hence from n ≡7 (mod 12), C/B2 ≡2, 3, 9 or 11 (mod 13).
It follows that (−78, C + B2)13 = 1, whereby (−78, C + B2)2 = 1.
We also know that A is even, while BC is odd, whence C + B2 ≡2 (mod 4). It follows from (−78, C + B2)2 = 1 that C/B2 + 1 ≡±2 (mod 16), and so C/B2 ≡1 or 13 (mod 16). If we now assume that v2(A) > 1, then equation (8) implies that Cn ≡−3B2n (mod 16) (and so necessarily C/B2 ≡13 (mod 16)). Our assumption that n ≡3 (mod 4) thus implies (C/B2)n ≡5 (mod 16), a contradiction. In conclusion, appealing to Proposition 4 in case 3 | n, we have Proposition 16. If n ≡3 (mod 4) and there exist nonzero coprime integers a, b and c for which a2 + b2n = c3, then v2(a) = 1. In particular, if m ≥2 is an integer and n ≡3 (mod 4), then the equation a2m + b2n = c3 has no solution in nonzero coprime integers a, b and c.
4.2.1. The equation a2 + b2n = c9. The case n = 2 was handled previously by Bennett, Ellenberg and Ng , while the case n = 3 is well known (see Proposition 4). We may thereby suppose that n ≥5 is prime. Applying Proposition 14 and (7), there thus exist coprime integers s and t, with s even and t ≡3 (mod 6), for which bn = t (3s2 −t2) and c3 = s2 + t2.
We can therefore find coprime A, B ∈Z with t = 3n−1 An and 3s2 −t2 = 3 Bn, whence Bn + 4 · 32n−3A2n = c3.
Via Lemma 3.4 of , for prime n ≥5 this leads to a newform of level 6, a contradiction. We thus have Proposition 17. If n is an integer with n ≥2, then the equation a2 + b2n = c9 has no solutions in nonzero coprime integers a, b and c.
16 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI 4.3. The equation x2 + y2 = z5. If x, y and z are coprime integers satisfying x2 + y2 = z5, then (see page 466 of ) there exist coprime integers s and t, of opposite parity, with (14) (x, y, z) = (s(s4 −10s2t2 + 5t4), t(5s4 −10s2t2 + t4), s2 + t2).
The following result is implicit in ; we include a short proof for completeness.
Proposition 18. If a, b and c are nonzero coprime integers for which a2 + b2n = c5, where n ≥2 is an integer, then b ≡1 (mod 2).
Proof. The cases n = 2, 3 and 5 are treated in , and , respectively. We may thus suppose that n ≥7 is prime. From (14), there are coprime integers s and t, of opposite parity, for which bn = t (5s4 −10s2t2 + t2). There thus exist integers A and B, and δ ∈{0, 1}, with t = 5−δAn and 5s4 −10s2t2 + t4 = 5δ Bn.
It follows that (15) 5δ Bn + 4 · 5−4δA4n = 5(s2 −t2)2.
If b is even (whereby the same is true of t and A) and δ = 1, then again arguing as in , we deduce the existence of a newform of level 10, a contradiction. If, however, b is even and δ = 0, the desired result is an immediate consequence of Theorem 1.2 of .
□ 4.3.1. The equation a2 + b2n = c10. As noted earlier, we may suppose that n ≥7 is prime and, from Proposition 18, that b is odd. Associated to such a solution, via the theory of Pythagorean triples, there thus exist coprime integers u and v, of opposite parity, with (16) bn = u2 −v2 and c5 = u2 + v2.
Hence, we may find integers A and B with u −v = An and u + v = Bn.
From the second equation in (16) and from (14), there exist coprime integers s and t, of opposite parity, with u −v = (s −t) s4 −4s3t −14s2t2 −4st3 + t4 .
Since (s −t)4 − s4 −4s3t −14s2t2 −4st3 + t4 = 20s2t2, GENERALIZED FERMAT EQUATIONS: A MISCELLANY 17 it follows that gcd(s −t, s4 −4s3t −14s2t2 −4st3 + t4) | 5. Similarly, we have u + v = (s + t) s4 + 4s3t −14s2t2 + 4st3 + t4 where gcd(s+t, s4 +4s3t−14s2t2 +4st3 +t4) also divides 5. Since s and t are coprime, we cannot have s−t ≡s+t ≡0 (mod 5) and so may conclude that at least one of gcd(s−t, s4−4s3t−14s2t2−4st3+t4) or gcd(s + t, s4 + 4s3t −14s2t2 + 4st3 + t4) is equal to 1. There thus exist integers X and Y such that either (Xn, Y n) = (s −t, s4 −4s3t −14s2t2 −4st3 + t4) or (s + t, s4 + 4s3t −14s2t2 + 4st3 + t4). In either case, X4n −Y n = 5(2st)2 which, with Theorem 1.1 of , contradicts st ̸= 0. In conclusion, Proposition 19. If n is an integer with n ≥2, then the equation a2 + b2n = c10 has no solutions in nonzero coprime integers a, b and c.
4.3.2. The equation a2+b2n = c15. As before, we may suppose that n ≥7 is prime. Using Proposition 14, we may also assume that b ≡3 (mod 6). Appealing to our parametrizations for x2 + y2 = z5 (i.e.
equation (14)), we deduce the existence of a coprime pair of integers (s, t) for which bn = t (5s4 −10s2t2 + t4) and c3 = s2 + t2.
Since s and t are coprime, it follows that 5s4−10s2t2+t4 ≡±1 (mod 3), whereby 3 | t. There thus exist integers A and B, and δ ∈{0, 1} satisfying equation (15), with the additional constraint that 3 | A. It follows that the corresponding Frey-Hellegouarch curve has multiplicative reduction at the prime 3, but level lowers to a newform of level N = 40 or 200 (depending on whether δ = 1 or 0, respectively).
This implies the existence of a form f at one of these levels with a3(f) ≡±4 (mod n). Since all such forms are one dimensional and have a3(f) ∈{0, ±2, ±3}, it follows that n = 7, contradicting the main result of . We thus have Proposition 20. If n is an integer with n ≥2, then the equation a2 + b2n = c15 has no solutions in nonzero coprime integers a, b and c.
4.4. The equation x2 +y4 = z3. Coprime integer solutions to this equation satisfy one of (see pages 475–477 of ) (17) x = 4ts s2 −3t2 s4 + 6s2t2 + 81t4 3s4 + 2s2t2 + 3t4 y = ±(s2 + 3t2) s4 −18s2t2 + 9t4 z = (s4 −2t2s2 + 9t4)(s4 + 30t2s2 + 9t4), 18 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI (18) x = ±(4s4 + 3t4)(16s8 −408t4s4 + 9t8) y = 6ts(4s4 −3t4) z = 16s8 + 168t4s4 + 9t8, (19) x = ±(s4 + 12t4)(s8 −408t4s4 + 144t8) y = 6ts(s4 −12t4) z = s8 + 168t4s4 + 144t8, or (20) x = ±2(s4 + 2ts3 + 6t2s2 + 2t3s + t4) 23s8 −16ts7 −172t2s6 −112t3s5 −22t4s4 −112t5s3 −172t6s2 −16t7s + 23t8 y = 3(s −t)(s + t)(s4 + 8ts3 + 6t2s2 + 8t3s + t4) z = 13s8 + 16ts7 + 28t2s6 + 112t3s5 + 238t4s4 +112t5s3 + 28t6s2 + 16t7s + 13t8.
Here, s and t are coprime integers satisfying s ̸≡t (mod 2) and s ≡±1 (mod 3), in case (17), t ≡1 (mod 2) and s ≡±1 (mod 3), in case (18), s ≡1 (mod 2) and s ≡±1 (mod 3), in case (19), s ̸≡t (mod 2) and s ̸≡t (mod 3), in case (20).
Since work of Ellenberg (see also ) treats the case where z is an nth power (and more), we are interested in considering equations corresponding to x = an or y = bn. We begin with the former.
4.4.1. The equation a2n+b4 = c3. The case n = 2 follows (essentially) from work of Lucas; see Section 5. We may thus suppose that n ≥3. We appeal to the parametrizations (17) – (20), with x = an.
In (17) and (20), we have a even, while, in (18) and (19), a is coprime to 3. Applying Proposition 14 leads to the following desired conclusion.
Proposition 21. If n is an integer with n ≥2, then the equation a2n + b4 = c3 has no solutions in nonzero coprime integers a, b and c.
4.4.2. The equation a2 + b4n = c3. For this equation, with n ≥2 an integer, Proposition 14 implies that we are in case (20), i.e. that there exist integers s and t for which (21) bn = 3(s −t)(s + t)(s4 + 8s3t + 6s2t2 + 8st3 + t4), s ̸≡t (mod 2), s ̸≡t (mod 3).
GENERALIZED FERMAT EQUATIONS: A MISCELLANY 19 Assuming n is odd, we thus can find integers A, B and C with s −t = An, s + t = 1 3Bn and s4 + 8s3t + 6s2t2 + 8st3 + t4 = Cn.
It follows that (22) A4n −1 27B4n = −2Cn, with ABC odd and 3 | B. There are (at least) three Frey-Hellegouarch curves we can attach to this Diophantine equation : E1 : Y 2 = X(X −A4n) X −B4n 27 , E2 : Y 2 = X3 + 2A2nX2 −2CnX, E3 : Y 2 = X3 −2B2n 27 X2 + 2Cn 27 X.
Although the solution (A4n, B4n, Cn) = (1, 81, 1) does not persist for large n, it still appears to cause an obstruction to resolving this equation fully using current techniques: none of the Ei have complex multiplication, nor can we separate out this solution using images of inertia at 3 or other primes dividing the conductor. In terms of the original equation, the obstructive solution is (±46)2 +(±3)4 = 133. Incidentally, this is the same obstructive solution which prevents a full resolution of a2+b2n = c3.
By Theorem 13, we may assume that every prime divisor l of n exceeds 106, which implies that ρEi l is, in each case, irreducible. Applying level lowering results, we find that the newform attached to Ei is congruent to a newform fi of level Ni, where Ni = 96 i = 1, 384 i = 2, 1152 i = 3.
The latter two conductor calculations can be found in and the former in . Since l > 106, all the fi’s with noninteger coefficients can be ruled out, after a short computation. This implies that there is an elliptic curve Fi with conductor Ni such that ρFi l ≃ρEi l . Furthermore, Ei must have good reduction at primes 5 ≤p ≤53 (again after a short calculation using the fact that l > 106).
Adding 2B4n to both sides of equation (22), we have A4n + 53 27B4n = 2(−Cn + B4n) and hence, via Proposition 15, (−53/27, 2(−C + B4))2(−53/27, 2(−C + B4))3(−53/27, 2(−C + B4))53 = 1.
20 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI Since −2Cn ≡A4n (mod 3), we have (−53/27, 2(−C + B4))3 = 1. Also, since −53/27 ≡1 (mod 8), it is a perfect square in Q2, which implies that (−53/27, 2(−C + B4))2 = 1. Therefore (−53/27, 2(−C + B4))53 = 1, i.e. −C/B4 + 1 is a quadratic nonresidue modulo 53. Since all the elliptic curves F1 of conductor 96 have a53(F1) = 10 (whereby, from l > 106, a53(E1) = 10), if follows that (A4/B4)n ≡36 (mod 53).
Therefore (−C/B4)n ≡17 (mod 53).
If n ≡±9, ±11, ±15, ±17 (mod 52) then (−C/B4)n = (α −1)n ̸≡17 (mod 53) for any choice of quadratic nonresidue α. It follows that if n ≡±2, ±4 (mod 13), equation (22) has no solution with ABC odd and 3 | B.
Similarly, if we subtract 2B2n from both sides of equation (22), we obtain A4n −55 27B4n = 2(−Cn −B4n).
Proposition 15 thus implies (55/27, 2(−C −B4))2 (55/27, 2(−C −B4))3 (55/27, 2(−C −B4))5 (55/27, 2(−C −B4))11 = 1.
As before we have (55/27, 2(−C −B4))3 = 1 and since 55/27 ≡1 (mod 4) and Cn−1 + · · · + B4n−4 is odd, also (55/27, Cn−1 + · · · + B4n−4)2 = (55/27, 2(−C −B4))2 = 1. Similarly, since (A4/B4)n ≡ 1 (mod 5), it follows that (−C/B4)n ≡−1 (mod 5), whereby, since n is odd, −C/B4 ≡−1 (mod 5).
Therefore (55/27, 2(−C−B4))5 = 1, which implies (55/27, 2(−C−B4))11 = 1. In particular −C/B4 ≡ λ (mod 11) where (23) λ ∈{0, 1, 3, 7, 8, 9}.
We can rule out λ = 0 and 1 since we are assuming that 11 ∤ABC. To treat the other cases, we will apply the Chen-Siksek method to the curves E2 and E3. We first show that a11(E2) = −4.
Notice that considering all possible solutions to equation (22) modulo 13, necessarily a13(E2) = −6 (since we assume that 13 ∤ABC). Observe also that E2 has nonsplit multiplicative reduction at 3.
Therefore ρE2 n |G3 ≃ρF2 n |G3 ≃ χϵ ∗ 0 ϵ where χ is the cyclotomic character and ϵ : G3 →F∗ n is the unique unramified quadratic character (see, for example, ). It follows that F2 must have nonsplit multiplicative reduction at 3 and hence GENERALIZED FERMAT EQUATIONS: A MISCELLANY 21 F2 must be isogenous to the elliptic curve 384D in the Cremona’s database. In particular,we have a11(E2) = −4 and so A2n B2n ≡1, 5 (mod 11).
When A2n/B2n ≡1 (mod 11), we find, by direct computation, that a11(E3) = 0, contradicting the fact that a11(E) ∈{±2, ±4} for every elliptic curve E/Q of conductor 1152. Therefore, we necessarily have A2n B2n ≡5 (mod 11) = ⇒ −C B4 n ≡8 (mod 11), whereby λn ≡8 (mod 11).
A quick calculation shows, however, that for n ≡±3 (mod 10), this contradicts (23). Collecting all this together, we conclude as follows (noting the solution coming from (p, q, r) = (2, 8, 3); see Bruin ).
Proposition 22. If n is a positive integer with either n ≡±2 (mod 5) or n ≡±2, ±4 (mod 13), then the equation a2 +b4n = c3 has only the solution (a, b, c, n) = (1549034, 33, 15613, 2) in positive coprime integers a, b and c.
Remark 23. The table of results regarding equation (1) given on page 490 of Cohen lists the case of signature (2, 4n, 3) as solved, citing work of the first two authors. This is due to an over-optimistic communication of the first author to Professor Cohen. We regret any inconvenience or confusion caused by this mistake.
Remark 24. Regarding the Diophantine equation (22) as an equation of signature (2, 4, n), we can attach the Frey-Hellegouarch Q-curves E4 : Y 2 = X3 + 4BnX2 + 2(B2n + 3 √ 3A2n)X and E5 : Y 2 = X3 + 4AnX2 + 2 A2n + 1 3 √ 3B2n X.
One further Frey-Hellegouarch Q-curve can be derived as follows. Defining U = (A4n + B4n/27)/2 = 2(s4 + 2s3t + 6s2t2 + 2st3 + t4) we have U 2 −1 27A4nB4n = C2n.
22 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI Considering this as an equation of signature (n, n, 2) turns out to give us the Frey-Hellegouarch curve E1 again.
Writing V = AnBn/3 and W = C2, we arrive at the generalized Fermat equation of signature (2, 4, n) U 2 −3V 4 = W n in nonzero coprime integers U, V and W, with 3 | V and v2(U) = 1. As before, we can associate a Q-curve to this equation.
Note that the solution (A4n, B4n, Cn) = (1, 81, 1) does not satisfy our desired 3-adic properties.
However, it still apparently forms an obstruction using current techniques to solving this equation in full generality for all Frey-Hellegouarch (Q-)curves we have considered.
4.5. The equation x3 + y3 = z2. From (pages 467 – 470), the coprime integer solutions to this equation satisfy one of (24) x = s(s + 2t)(s2 −2ts + 4t2) y = −4t(s −t)(s2 + ts + t2) z = ±(s2 −2ts −2t2)(s4 + 2ts3 + 6t2s2 −4t3s + 4t4), (25) x = s4 −4ts3 −6t2s2 −4t3s + t4 y = 2(s4 + 2ts3 + 2t3s + t4) z = 3(s −t)(s + t)(s4 + 2s3t + 6s2t2 + 2st3 + t4), or (26) x = −3s4 + 6t2s2 + t4 y = 3s4 + 6t2s2 −t4 z = 6st(3s4 + t4).
Here, the parametrizations are up to exchange of x and y, and s and t are coprime integers with s ≡1 (mod 2) and s ̸≡t (mod 3), in case (24), s ̸≡t (mod 2) and s ̸≡t (mod 3), in case (25), s ̸≡t (mod 2) and t ̸≡0 (mod 3), in case (26).
4.5.1. The equation a3 +b3 = c2n. The cases n ∈{2, 3, 5} follow from and Proposition 4. We may thus suppose that n ≥7 is prime. Since Proposition 5 implies c ≡3 (mod 6), it follows that cn = 3(s −t)(s + t)(s4 + 2s3t + 6s2t2 + 2st3 + t4), GENERALIZED FERMAT EQUATIONS: A MISCELLANY 23 for s and t coprime integers with s ̸≡t (mod 2) and s ̸≡t (mod 3). There thus exist integers A, B and C with s −t = An, s + t = 3n−1 Bn and s4 + 2s3t + 6s2t2 + 2st3 + t4 = Cn, whereby A4n + 34n−3B4n = 4 Cn, which we rewrite as 4 Cn −A4n = 3 32n−2 B2n2 .
Applying Theorem 1.2 of to this last equation, we may conclude, for n ≥7 prime, that either ABC = 0 or 32n−2 B2n = ±1, in either case a contradiction. We thus have Proposition 25. If n is an integer with n ≥2, then the equation a3 + b3 = c2n has no solutions in nonzero coprime integers a, b and c.
4.5.2. The equation a3 + b3n = c2. The techniques involved in this case require some of the most elaborate combination of ingredients to date, including Q-curves and delicate multi-Frey and image of inertia arguments. For this reason, we have chosen to publish this separately in . Our main result there is as follows.
Theorem 26 (). If n is prime with n ≡1 (mod 8), then the equation a3+b3n = c2 has no solutions in coprime nonzero integers a, b and c, apart from those given by (a, b, c) = (2, 1, ±3).
4.6. Other spherical equations. Solutions to the generalized Fermat equation with icosahedral signature (2, 3, 5) correspond to 27 parametrized families, in each case with parametrizing forms of degrees 30, 20 and 12 (see e.g. ). We are unable to apply the techniques of this paper to derive much information of value in this situation (but see ).
5. Historical notes on the equations a4 ± b4 = c3 In , it is proved that the generalized Fermat equation (1) with (p, q, r) = (n, n, 3), has no coprime, nonzero integer solutions a, b and c, provided n ≥7 is prime (and assuming the modularity of elliptic curves over Q with conductor divisible by 27, now a well known theorem ). To show the nonexistence of solutions for all integers n ≥3, it suffices, in addition, to treat the cases n = 3, 4 and 5; the first of these is classical and was (essentially) solved by Euler (see Proposition 4), while the last was handled by Poonen in . The case n = 4 is attributed in and , citing [34, p. 630], to the French mathematician ´ Edouard Lucas (1842–1891), in particular to and [53, Chapitre III].
24 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI In these two papers, as well as in other work of Lucas [42, pp. 282–288], there does not appear to be, however, any explicit mention of the equation (27) a4 + b4 = c3, a, b, c ∈Z, abc ̸= 0, gcd(a, b, c) = 1.
In this section, we will attempt to indicate why, despite this, the aforementioned attributions are in fact correct. It is worth mentioning that the equations a4 ±b4 = c3 are also explicitly solved in Cohen (as Proposition 14.6.6).
5.1. Reduction to elliptic generalized Fermat equations. First of all, it is quite elementary to reduce the nonexistence of solutions to (27) (or, analogously, the equation a4 −b4 = c3; in the sequel, we will not discuss this latter equation further) to the nonexistence of solutions to certain elliptic generalized Fermat equations of signature (4, 4, 2). To carry this out, we note that a solution in integers a, b and c to (27) implies, via (7) (and changing the sign of t), the existence of nonzero coprime integers s and t for which a2 = s(s2 −3t2) (28) b2 = t(t2 −3s2) (29) (and c = s2 + t2). Without loss of generality, we assume that 3 ∤s. Then gcd(s, s2 −3t2) = 1 and, from (28), we have s = ϵ1α2 (30) s2 −3t2 = ϵ1β2 (31) for some nonzero integers α, β and ϵ1 ∈{±1}. Using gcd(t, t2 −3s2) ∈{1, 3} and (29), we have t = ϵ2γ2 (32) t2 −3s2 = ϵ2δ2 (33) for nonzero integers γ, δ and ϵ2 ∈{±1, ±3}. Examining (31) or (33) modulo 4, shows that ϵ1, ϵ2 ̸≡ −1 (mod 4). Considering these equations simultaneously modulo 8, now shows that ϵ2 ̸= 1. It follows that we have ϵ1 = 1 and ϵ2 = −3.
Substituting (30) and (32) in (31) now yields (34) α4 −27γ4 = β2, GENERALIZED FERMAT EQUATIONS: A MISCELLANY 25 while substituting (30) and (32) in (33) yields (35) α4 −3γ4 = δ2.
5.2. Relation to work of Lucas. In the the preceding subsection, we showed that in order to prove that there are no solutions to (27), it suffices to demonstrate that one of the Diophantine equations (34) or (35) does not have solutions in nonzero integers. In [52, Chapitre I] and [53, Chapitre III], Lucas studied the Diophantine equation Ax4 +By4 = Cz2 in unknown integers x, y, z, using Fermat’s method of descent. Here, A, B and C are integers whose prime divisors are contained in {2, 3}. His chief concern, however, was not with explicitly showing that a given equation of this shape has no nontrivial solutions, but rather in describing nontrivial solutions in the cases where they exist. In [52, Chapitre I, §X] a description of all the equations Ax4 + By4 = Cz2 as above that do have nontrivial solutions is recorded, together with a reference to the explicit solutions.
For the other equations Ax4 + By4 = Cz2, including (34) and (35), it is simply stated that there are no nontrivial solutions, without explicit proof of this fact. In these references, however, Lucas clearly demonstrates his mastery of Fermat’s method of descent and one can check that this method indeed applies immediately to prove the nonexistence of nontrivial solutions in these cases. This provides convincing evidence that Lucas had proofs for his claims that there are no nontrivial solutions to (34) and (35), amongst others (which he failed to record, apparently as he considered these cases to be lacking in interest!).
6. Future work A problem of serious difficulty that likely awaits fundamentally new techniques is that of solving equation (1) for, say, fixed r and infinite, unbounded families p and q, with gcd(p, q) = 1. A truly spectacular result at this stage would be to solve an infinite family where p, q and r are pairwise coprime. Indeed, solving a single new equation of this form will likely cost considerable effort using current techniques.
A limitation of the modular method at present is that the possible exponents (p, q, r) must relate to a moduli space of elliptic curves. When this precondition holds, the modular method can be viewed as a method which reduces the problem of resolving (1) to that of certain rational points on these moduli spaces through Galois representations. The inability to carry out the modular method in such a situation relates to a lack of sufficiently strong methods for effectively bounding these rational points (i.e. Mazur’s method fails or has not been developed). We note however that irreducibility is easier because the Frey-Hellegouarch curves will have semi-stable reduction away from small primes - this allows and for instance to prove irreducibility without resort to a Mazur type result 26 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI (essentially, a method of Serre [61, p. 314, Corollaire 2] which predates and is used in suffices with some extra effort).
For general (p, q, r), constructs Frey-Hellegouarch abelian varieties of GL2-type over a totally real field and establishes modularity in some cases; the analogous modular curves are in general quotients of the complex upper half plane by non-arithmetic Fuchsian groups.
The ABC conjecture implies that there are only finitely many solutions to (1) in coprime integers once min {p, q, r} ≥3. In addition, an effective version of the ABC conjecture would imply an effective bound on the size of the solutions, though this effectivity needs to be within computational range to allow a complete quantitative resolution of (1).
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S. Dahmen. Classical and modular methods applied to Diophantine equations, PhD thesis, University of Utrecht, 2008. Permanently available at S. Dahmen. A refined modular approach to the Diophantine equation x2 + y2n = z3, Int. J. Number Theory 7 (2011), no. 5, 1303–1316.
S. Dahmen and S. Siksek. Perfect powers expressible as sums of two fifth or seventh powers, preprint.
H. Darmon. Rigid local systems, Hilbert modular forms, and Fermat’s Last Theorem, Duke. Math. J. 102 (2000), 413–449.
H. Darmon, F. Diamond and R. Taylor. Fermat’s Last Theorem, In Elliptic curves, modular forms & Fermat’s Last Theorem (Hong Kong, 1993), pages 2–140. International Press, 1997.
H. Darmon and A. Granville. On the equations zm = F(x, y) and Axp + Byq = Czr, Bull. London Math. Soc. 27 (1995), 513–543.
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L.V. Dieulefait and N. Freitas. Fermat-type equations of signature (13, 13, p) via Hilbert cuspforms, Math. Ann., to appear.
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28 MICHAEL A. BENNETT, IMIN CHEN, SANDER R. DAHMEN AND SOROOSH YAZDANI D. Harkin. On the mathematical work of Francois-´ Edouard-Anatole Lucas, Enseignement Math. (2) 3 1957, 276– 288.
W. Ivorra. Sur les ´ equations xp + 2βyp = z2 et xp + 2βyp = 2z2, Acta Arith. 108 (2003), 327–338.
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A. Kraus. Sur les ´ equations ap + bp + 15cp = 0 et ap + 3bp + 5cp = 0, C. R. Acad. Sci. Par. S´ er. I Math. 322 (1996), 809–812.
A. Kraus. Majorations effectives pour l’´ equation de Fermat g´ en´ eralis´ ee, Canad. J. Math. 49 (1997), no. 6, 1139– 1161.
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´ E. Lucas. Recherches sur l’analyse ind´ etermin´ ee et l’arithm etique de Diophante, extrait du Bulletin de la Soci´ et´ e d’´ emulation du D´ epartement de l’Allier, 12, 1873, 441–532; Moulins, 1873, imprimerie de C. Desrosiers. Online at ´ E. Lucas. Recherches sur plusieurs ouvrages de L´ eonard de Pise et sur diverses questions d’arithmtique suprieure, extrait du Bullettino di Bibliografia e di Sioria delle Scienze matematiche e fisiche, 10, 1877, 129–193 and 239–293; Rome, 1877, imprimerie des sciences math´ ematiques et physiques. Online at B. Mazur. Rational isogenies of prime degree, Invent. Math. 44 (1978), no. 2, 129–162.
R. Mauldin. A generalization of Fermat’s Last Theorem: the Beal conjecture and prize problem, Notices Amer.
Math. Soc. 44 (1997), 1436–1437.
L. Merel. Arithmetic of elliptic curves and diophantine equations, J. Th´ eor. Nombres Bordeaux 11 (1999), 173–200.
B. Poonen. Some Diophantine equations of the form xn + yn = zm, Acta Arith. 86 (1998), 193–205.
B. Poonen, E. Schaefer and M. Stoll. Twists of X(7) and primitive solutions to x2 + y3 = z7, Duke Math. J. 137 (2007), 103–158.
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K. Ribet. Report on mod l representations of Gal(Q/Q), in Motives, Proc. Symp. Pure Math. 55:2 (1994), 639–676.
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S. Siksek and M. Stoll. The generalized Fermat equation x2 + y3 = z15, preprint.
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GENERALIZED FERMAT EQUATIONS: A MISCELLANY 29 R. Tijdeman. Diophantine equations and diophantine approximations, Number theory and applications (Banff, AB, 1988), Kluwer Acad. Publ., Dordrecht, 1989, 215–243.
A. Wiles. Modular elliptic curves and Fermat’s Last Theorem, Ann. of Math. 141 (1995), no. 3, 443–551.
Michael A. Bennett, Department of Mathematics, University of British Columbia, Vancouver, British Columbia, V6T 1Z2, CANADA E-mail address: bennett@math.ubc.ca Imin Chen, Department of Mathematics, Simon Fraser University, Burnaby, British Columbia, CANADA E-mail address: ichen@math.sfu.ca Sander R. Dahmen, Mathematisch Instituut, Universiteit Utrecht, P.O. Box 80 010, 3508 TA Utrecht, The Netherlands, E-mail address: s.r.dahmen@uu.nl Soroosh Yazdani, Department of Mathematics and Computer Science, University of Lethbridge, Leth-bridge, Alberta, T1K 3M4, CANADA E-mail address: soroosh.yazdani@uleth.ca |
4180 | https://www.neetprep.com/ncert/1483-Wave-Optics-Wave-Optics--NCERT-Chapter-PDF | NCERT Ebook for Wave Optics - Wave Optics - Chapter 10 - NCERT Physics - XII
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NCERT Ebook - Wave Optics
Introduction
Huygens Principle
Refraction and Reflection of Plane Waves using Huygens Principle
Coherent and Incoherent Addition of Waves
Interference of Light Waves and Young’s Experiment
Diffraction
Polarisation
Mindmap
Basic understanding of waves is important before starting this chapter
10.1 Introduction
In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. It explained the laws of reflection and refraction of light at an interface. The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium. This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book, the corpuscular model is very often attributed to Newton.
In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter. As we will see, the wave model could satisfactorily explain the phenomena of reflection and refraction; however, it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium. This is in contradiction to the prediction made by using the corpuscular model of light. It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model; Foucault carried out this experiment in 1850.
The wave theory was not readily accepted primarily because of Newton’s authority and also because light could travel through vacuum and it was felt that a wave would always require a medium to propagate from one point to the other. However, when Thomas Young performed his famous interference experiment in 1801, it was firmly established that light is indeed a wave phenomenon. The wavelength of visible light was measured and found to be extremely small; for example, the wavelength of yellow light is about 0.6 µ m. Because of the smallness of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses), light can be assumed to approximately travel in straight lines. This is the field of geometrical optics, which we had discussed in the previous chapter. Indeed, the branch of optics in which one completely neglects the finiteness of the wavelength is called geometrical optics and a ray is defined as the path of energy propagation in the limit of wavelength tending to zero.
After the interference experiment of Young in 1801, for the next 40 years or so, many experiments were carried out involving the interference and diffraction of lightwaves; these experiments could only be satisfactorily explained by assuming a wave model of light. Thus, around the middle of the nineteenth century, the wave theory seemed to be very well established. The only major difficulty was that since it was thought that a wave required a medium for its propagation, how could light waves propagate through vacuum. This was explained when Maxwell put forward his famous electromagnetic theory of light. Maxwell had developed a set of equations describing the laws of electricity and magnetism and using these equations he derived what is known as the wave equation from which he predicted the existence of electromagnetic waves. From the wave equation, Maxwell could calculate the speed of electromagnetic waves in free space and he found that the theoretical value was very close to the measured value of speed of light. From this, he propounded that light must be an electromagnetic wave. Thus, according to Maxwell, light waves are associated with changing electric and magnetic fields; changing electric field produces a time and space varying magnetic field and a changing magnetic field produces a time and space varying electric field. The changing electric and magnetic fields result in the propagation of electromagnetic waves (or light waves) even in vacuum.
In this chapter we will first discuss the original formulation of the Huygens principle and derive the laws of reflection and refraction. In Sections 10.4 and 10.5, we will discuss the phenomenon of interference which is based on the principle of superposition. In Section 10.6 we will discuss the phenomenon of diffraction which is based on Huygens-Fresnel principle. Finally in Section 10.7 we will discuss the phenomenon of polarisation which is based on the fact that the light waves are transverse electromagnetic waves.
Maxwell had predicted the existence of electromagnetic waves around 1855; it was much later (around 1890) that Heinrich Hertz produced radiowaves in the laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian waves
Does light travel in a straight line?
Light travels in a straight line in Class VI; it does not do so in Class XII and beyond! Surprised, aren’t you?
In school, you are shown an experiment in which you take three cardboards with pinholes in them, place a candle on one side and look from the other side. If the flame of the candle and the three pinholes are in a straight line, you can see the candle. Even if one of them is displaced a little, you cannot see the candle.This proves, so your teacher says,that light travels in a straight line.
In the present book, there are two consecutive chapters, one on ray optics and the other on wave optics. Ray optics is based on rectilinear propagation of light, and deals with mirrors, lenses, reflection, refraction, etc. Then you come to the chapter on wave optics, and you are told that light travels as a wave, that it can bend around objects, it can diffract and interfere, etc.
In optical region, light has a wavelength of about half a micrometre. If it encounters an obstacle of about this size, it can bend around it and can be seen on the other side. Thus a micrometre size obstacle will not be able to stop a light ray. If the obstacle is much larger, however, light will not be able to bend to that extent, and will not be seen on the other side.
This is a property of a wave in general, and can be seen in sound waves too. The sound wave of our speech has a wavelength of about 5 0 cm to 1 m. If it meets an obstacle of the size of a few metres, it bends around it and reaches points behind the obstacle. But when it comes across a larger obstacle of a few hundred metres, such as a hillock, most of it is reflected and is heard as an echo.
Then what about the primary school experiment? What happens there is that when we move any cardboard, the displacement is of the order of a few millimetres, which is much larger than the wavelength of light. Hence the candle cannot be seen. If we are able to move one of the cardboards by a micrometer or less, light will be able to diffract, and the candle will still be seen.
One could add to the first sentence in this bo x: It learns how to bend as it grows u p!
10.2 Huygens Principle
We would first define a wavefront: when we drop a small stone on a calm pool of water, waves spread out from the point of impact. Every point on the surface starts oscillating with time. At any instant, a photograph of the surface would show circular rings on which the disturbance is maximum. Clearly, all points on such a circle are oscillating in phase because they are at the same distance from the source. Such a locus of points, which oscillate in phase is called a wavefron t;thus a wavefront is defined as a surface of constant phase. The speed with which the wavefront moves outwards from the source is called the speed of the wave. The energy of the wave travels in a direction perpendicular to the wavefront.
Figure 10.1(a) A diverging spherical wave emanating from a point source. The wavefronts are spherical.
If we have a point source emitting waves uniformly in all directions, then the locus of points which have the same amplitude and vibrate in the same phase are spheres and we have what is known as a spherical wave as shown in Fig. 10.1(a). At a large distance from the source, a small portion of the sphere can be considered as a plane and we have what is known as a plane wave [Fig. 10.1(b)].
Figure 10.1 (b) At a large distance from the source, a small portion of the spherical wave can be approximated by a plane wave.
Now, if we know the shape of the wavefront at t = 0, then Huygens principle allows us to determine the shape of the wavefront at a later time τ. Thus, Huygens principle is essentially a geometrical construction, which given the shape of the wafefront at any time allows us to determine the shape of the wavefront at a later time. Let us consider a diverging wave and let F 1 F 2 represent a portion of the spherical wavefront at t = 0 (Fig. 10.2). Now, according to Huygens principle,each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time.
Thus, if we wish to determine the shape of the wavefront at t = τ, we draw spheres of radius v τ from each point on the spherical wavefront where v represents the speed of the waves in the medium. If we now draw a common tangent to all these spheres, we obtain the new position of the wavefront at t = τ. The new wavefront shown as G 1 G 2 in Fig. 10.2 is again spherical with point O as the centre.
Figure 10.2 F 1 F 2 represents the spherical wavefront (with O as centre) at t= 0. The envelope of the secondary wavelets emanating from F 1 F 2 produces the forward moving wavefront G 1 G 2. The backwave D 1 D 2 does not exist.
Figure 10.3 Huygens geometrical construction for a plane wave propagating to the right. F 1 F 2 is the plane wavefront at t = 0 and G 1 G 2 is the wavefront at a later time τ. The lines A 1 A 2, B 1 B 2 … etc., are normal to both F 1 F 2 and G 1 G 2 and represent rays.
The above model has one shortcoming: we also have a backwave which is shown as D 1 D 2 in Fig. 10.2. Huygens argued that the amplitude of the secondary wavelets is maximum in the forward direction and zero in the backward direction; by making this adhoc assumption, Huygens could explain the absence of the backwave. However, this adhoc assumption is not satisfactory and the absence of the backwave is really justified from more rigorous wave theory.
In a similar manner, we can use Huygens principle to determine the shape of the wavefront for a plane wave propagating through a medium (Fig. 10.3).
10.3 Refraction and Reflection of Plane Waves using Huygens Principle
10.3.1 Refraction of a plane wave
We will now use Huygens principle to derive the laws of refraction. Let PP′ represent the surface separating medium 1 and medium 2, as shown in Fig. 10.4. Let v 1 and v 2 represent the speed of light in medium 1 and medium 2, respectively. We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle i as shown in the figure. Let τ be the time taken by the wavefront to travel the distance BC. Thus,
BC =v 1 τ
FIGURE 10.4 A plane wave AB is incident at an angle i on the surface PP¢ separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v2< v1 so that the refracted waves bends towards the normal.
Christiaan Huygens (1629 – 1695)
Dutch physicist, astronomer, mathematician and the founder of the wave theory of light. His book, Treatise on light, makes fascinating reading even today. He brilliantly explained the double refraction shown by the mineral calcite in this work in addition to reflection and refraction. He was the first to analyse circular and simple harmonic motion and designed and built improved clocks and telescopes. He discovered the true geometry of Saturn’s rings.
In order to determine the shape of the refracted wavefront, we draw a sphere of radius v 2 τ from the point A in the second medium (the speed of the wave in the second medium is v 2). Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE =v 2 τ and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain Figure 10.4.A plane wave AB is incident at an angle i on the surface PP′ s eparating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted wavefront. The figure corresponds to v 2< v 1 so that the refracted waves bends towards the normal.
sin i = (10.1)
and
sin r = (10.2)
where i and r are the angles of incidence and refraction, respectively. Thus we obtain
(10.3)
From the above equation, we get the important result that if r<i (i.e., if the ray bends toward the normal), the speed of the light wave in the second medium (v 2) will be less then the speed of the light wave in the first medium (v 1). This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed, the prediction of the wave theory is correct. Now, if c represents the speed of light in vacuum, then,
(10.4)
and
n 2 = (10.5)
are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as
n 1 sin i =n 2 sin r (10.6)
This is the Snell’s law of refraction. Further, if λ 1 and λ 2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to λ 1 then the distance AE will be equal to λ 2(because if the crest from B has reached C in time τ, then the crest from A should have also reached E in time τ); thus,
or
(10.7)
The above equation implies that when a wave gets refracted into a denser medium (v 1>v 2) the wavelength and the speed of propagation decrease but the frequency ν (= v/λ) remains the same.
10.3.2 Refraction at a rarer medium
We now consider refraction of a plane wave at a rarer medium, i.e.,v 2>v 1. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have n 1 sin i =n 2 sin r . We define an angle i c by the following equation
(10.8)
Thus, if i =i c then sin r = 1 and r = 90°. Obviously, for i>i c, there can not be any refracted wave. The angle i c is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection. The phenomenon of total internal reflection and its applications was discussed in Section 9.4.
FIGURE 10.5 Refraction of a plane wave incident on a rarer medium for which v2> v1. The plane wave bends away from the normal.
10.3.3 Reflection of a plane wave by a plane surface
We next consider a plane wave AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium and if τ represents the time taken by the wavefront to advance from the point B to C then the distance Figure 10.5 Refraction of a plane wave incident on a rarer medium for which v 2> v 1. The plane wave bends away from the normal.
BC = v τ
In order to construct the reflected wavefront we draw a sphere of radius v τ from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously
AE = BC = v τ
FIGURE 10.6 Reflection of a plane wave AB by the reflecting surface MN.
AB and CE represent incident and reflected wavefronts.
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the law of reflection.
Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood. These phenomena were discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F. In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors.
Figure 10.7 Refraction of a plane wave by (a) a thin prism, (b) a convex lens. (c) Reflection of a plane wave by a concave mirror.
From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.
10.3.4 The doppler effect
We should mention here that one should be careful in constructing the wavefronts if the source (or the observer) is moving. For example, if there is no medium and the source moves away from the observer, then later wavefronts have to travel a greater distance to reach the observer and hence take a longer time. The time taken between the arrival of two successive wavefronts is hence longer at the observer than it is at the source. Thus, when the source moves away from the observer the frequency as measured by the source will be smaller. This is known as the Doppler effect. Astronomers call the increase in wavelength due to doppler effect as red shift since a wavelength in the middle of the visible region of the spectrum moves towards the red end of the spectrum. When waves are received from a source moving towards the observer, there is an apparent decrease in wavelength, this is referred to as blue shift.
You have already encountered Doppler effect for sound waves in Chapter 15 of Class XI textbook. For velocities small compared to the speed of light, we can use the same formulae which we use for sound waves. The fractional change in frequency∆ν/ν is given by –v radial/c, where v radial is the component of the source velocity along the line joining the observer to the source relative to the observer;v radial is considered positive when the source moves away from the observer. Thus, the Doppler shift can be expressed as:
(10.9)
The formula given above is valid only when the speed of the source is small compared to that of light. A more accurate formula for the Doppler effect which is valid even when the speeds are close to that of light, requires the use of Einstein’s special theory of relativity. The Doppler effect for light is very important in astronomy. It is the basis for the measurements of the radial velocities of distant galaxies.
Example 10.1 What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm?
Solution Since νλ =c, (for small changes in ν and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [using Eq. (10.9)]
or,v radial
= 306 km/s
Therefore, the galaxy is moving away from us.
Example 10.2
(a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Solution
(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as
oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
(b) No. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time.
10.4 Coherent and Incoherent Addition of Waves
In this section we will discuss the interference pattern produced by the superposition of two waves. You may recall that we had discussed the superposition principle in Chapter 15 of your Class XI textbook. Indeed the entire field of interference is based on the superposition principle according to which at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.
(a)
(b)
Figure 10.8(a) Two needles oscillating in phase in water represent two coherent sources. (b) The pattern of displacement of water molecules at an instant on the surface of water showing nodal N (no displacement) and antinodal A (maximum displacement) lines.
Consider two needles S 1 and S 2 moving periodically up and down in an identical fashion in a trough of water [Fig. 10.8(a)]. They produce two water waves, and at a particular point, the phase difference between the displacements produced by each of the waves does not change with time; when this happens the two sources are said to be coherent. Figure 10.8(b) shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time. Consider a point P for which
S 1 P = S 2 P
Since the distances S 1 P and S 2 P are equal, waves from S 1 and S 2 will take the same time to travel to the point P and waves that emanate from S 1 and S 2 in phase will also arrive, at the point P, in phase.
Thus, if the displacement produced by the source S 1 at the point P is given by
y 1 =a cos ω t
then, the displacement produced by the source S 2 (at the point P) will also be given by
y 2 =a cos ω t
Thus, the resultant of displacement at P would be given by
y = y 1 +y 2 = 2 a cos ω t
Since the intensity is proportional to the square of the amplitude, the resultant intensity will be given by
I = 4 I 0
where I 0 represents the intensity produced by each one of the individual sources;I 0 is proportional to a 2. In fact at any point on the perpendicular bisector of S 1 S 2, the intensity will be 4 I 0. The two sources are said to interfere constructively and we have what is referred to as constructive interference. We next consider a point Q [Fig. 10.9(a)]
for which
S 2 Q –S 1 Q = 2 λ
The waves emanating from S 1 will arrive exactly two cycles earlier than the waves from S 2 and will again be in phase [Fig. 10.9(a)]. Thus, if the displacement produced by S 1 is given by
y 1 =a cos ω t
then the displacement produced by S 2 will be given by
y 2 =a cos (ω t – 4 π) =a cos ω t
where we have used the fact that a path difference of 2 λ corresponds to a phase difference of 4 π. The two displacements are once again in phase and the intensity will again be 4 I 0 giving rise to constructive interference. In the above analysis we have assumed that the distances S 1 Q and S 2 Q are much greater than d (which represents the distance between S 1 and S 2) so that although S 1 Q and S 2 Q are not equal, the amplitudes of the displacement produced by each wave are very nearly the same.
Figure 10.9
(a) Constructive interference at a point Q for which the path difference is 2 λ. (b) Destructive interference at a point R for which the path difference is
We next consider a point R [Fig. 10.9(b)] for which
S 2 R – S 1 R = –2.5 λ
The waves emanating from S 1 will arrive exactly two and a half cycles later than the waves from S 2 [Fig. 10.10(b)]. Thus if the displacement produced by S 1 is given by
y 1 =a cos ω t
then the displacement produced by S 2 will be given by
y 2 =a cos (ω t + 5 π) = –a cos ω t
where we have used the fact that a path difference of 2.5 λ corresponds to a phase difference of 5 π. The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. This is referred to as destructive interference.
To summarise: If we have two coherent sources S 1 and S 2 vibrating in phase, then for an arbitrary point P whenever the path difference,
S 1 P ~S 2 P =n λ (n = 0, 1, 2, 3,...) (10.10)
we will have constructive interference and the resultant intensity will be 4 I 0; the sign~between S 1 P and S 2 P represents the difference between S 1 P and S 2 P. On the other hand, if the point P is such that the path difference,
S 1 P ~S 2 P = (n+) λ (n = 0, 1, 2, 3, ...) (10.11)
we will have destructive interference and the resultant intensity will be zero.Now, for any other arbitrary point G (Fig. 10.10) let the phase difference between the two displacements be φ. Thus, if the displacement produced by S 1 is given by
Figure 10.10 Locus of points for which S 1 P – S 2 P is equal to zero,±λ, ± 2 λ, ± 3 λ.
y 1 =a cos ω t
then, the displacement produced by S 2 would be
y 2 =a cos (ω t +φ)
and the resultant displacement will be given by
y = y 1 + y 2
= a [cos ω t + cos (ω t +φ)]
= 2 a cos (φ/2) cos (ω t +φ/2)
The amplitude of the resultant displacement is 2 a cos (φ/2) and therefore the intensity at that point will be
I = 4 I 0 cos 2 (φ/2) (10.12)
If φ = 0,± 2 π, ± 4 π,… which corresponds to the condition given by Eq. (10.10) we will have constructive interference leading to maximum intensity. On the other hand, if φ =±π, ± 3 π, ± 5 π … [which corresponds to the condition given by Eq. (10.11)] we will have destructive interference leading to zero intensity.
Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference φ at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time. However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by
(10.13)
where angular brackets represent time averaging. Indeed it is shown in Section 7.2 that if φ(t) varies randomly with time, the time-averaged quantity< cos 2 (φ/2)> will be 1/2. This is also intuitively obvious because the function cos 2 (φ/2) will randomly vary between 0 and 1 and the average value will be 1/2. The resultant intensity will be given by
I = 2 I 0 (10.14)
at all points.
When the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This is indeed what happens when two separate light sources illuminate a wall.
10.5 Interference of Light Waves and Young’s Experiment
We will now discuss interference using light waves. If we use two sodium lamps illuminating two pinholes (Fig. 10.11) we will not observe any interference fringes. This is because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10–10 seconds. Thus the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this happens, as discussed in the previous section, the intensities on the screen will add up.
The British physicist Thomas Young used an ingenious technique to “lock” the phases of the waves emanating from S 1 and S 2. He made two pinholes S 1 and S 2 (very close to each other) on an opaque screen [Fig. 10.12(a)]. These were illuminated by another pinholes that was in turn, lit by a bright source. Light waves spread out from S and fall on both S 1 and S 2. S 1 and S 2 then behave like two coherent sources because light waves coming out from S 1 and S 2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S 1 and S 2. Thus, the two sources S 1 and S 2 will be locked in phase; i.e., they will be coherent like the two vibrating needle in our water wave example [Fig. 10.8(a)].
FIGURE 10.11 If two sodium lamps illuminate two pinholes S1 and S2, the intensities will add up and no interference fringes will be observed on the screen.
FIGURE 10.12 Young’s arrangement to produce interference pattern.
Thus spherical waves emanating from S 1 and S 2 will produce interference fringes on the screen GG′, as shown in Fig. 10.12(b). The positions of maximum and minimum intensities can be calculated by using the analysis given in Section 10.4 where we had shown that for an arbitrary point P on the line GG′ [Fig. 10.12(b)] to correspond to a maximum, we must have
S 2 P – S 1 P =n λ; n = 0, 1, 2 ... (10.15)
Now,
(a) (b)
Figure 10.12 Young’s arrangement to produce interference pattern.
(S 2 P)2 – (S 1 P)2 == 2 x d
where S 1 S 2 =d and OP =x. Thus
S 2 P – S 1 P = (10.16)
If x, d<<D then negligible error will be introduced if
S 2 P +S 1 P (in the denominator) is replaced by 2 D. For example, for d = 0.1 cm,D = 100 cm, OP = 1 cm (which correspond to typical values for an interference experiment using light waves), we have
S 2 P + S 1 P = [(100)2 + (1.05)2]½ + [(100)2 + (0.95)2]½
200.01 cm
Thus if we replace S 2 P + S 1 P by 2 D, the error involved is about 0.005%. In this approximation, Eq. (10.16) becomes
Thomas Young (1773 – 1829)
Thomas Young (1773 – 1829)English physicist, physician and Egyptologist. Young worked on a wide variety of scientific problems, ranging from the structure of the eye and the mechanism of vision to the decipherment of the Rosetta stone. He revived the wave theory of light and recognised that interference phenomena provide proof of the wave properties of light.
S 2 P – S 1 P ≈ (10.17)
Hence we will have constructive interference resulting in a bright region when = n λ [Eq. (10.15)]. That is,
x =x n =; n = 0,± 1,± 2, ... (10.18)
On the other hand, we will have destructive interference resulting in a dark region when= (n+)λ that is
x =x n = (n+) (10.19)
Thus dark and bright bands appear on the screen, as shown in Fig. 10.13. Such bands are called fringes. Equations (10.18) and (10.19) show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by
β =x n+1 –x n or β =(10.20)
which is the expression for the fringe width. Obviously, the central point O (in Fig. 10.12) will be bright because S 1 O = S 2 O and it will correspond to n = 0 [Eq. (10.18)]. If we consider the line perpendicular to the plane of the paper and passing through O [i.e., along the y-axis] then all points on this line will be equidistant from S 1 and S 2 and we will have a bright central fringe which is a straight line as shown in Fig. 10.13. In order to determine the shape of the interference pattern on the screen we note that a particular fringe would correspond to the locus of points with a constant value of S 2 P – S 1 P. Whenever this constant is an integral multiple of λ, the fringe will be bright and whenever it is an odd integral multiple of λ/2 it will be a dark fringe. Now, the locus of the point P lying in the x-y plane such that S 2 P – S 1 P (=∆) is a constant, is a hyperbola. Thus the fringe pattern will strictly be a hyperbola; however, if the distance D is very large compared to the fringe width, the fringes will be very nearly straight lines as shown in Fig. 10.13.
Figure 10.13 Computer generated fringe pattern produced by two point source S 1 and S 2 on the screen GG′(Fig. 10.12); (a) and (b) correspond to d= 0.005 mm and 0.025 mm, respectively (both figures correspond to D= 5 cm and λ= 5 × 10–5 cm.) (Adopted from OPTICS by A. Ghatak, Tata McGraw Hill Publishing Co. Ltd., New Delhi, 2000.)
In the double-slit experiment shown in Fig. 10.12(b), we have taken the source hole S on the perpendicular bisector of the two slits, which is shown as the line SO. What happens if the source S is slightly away from the perpendicular bisector. Consider that the source is moved to some new point S′ and suppose that Q is the mid-point of S 1 and S 2. If the angle S′QS is φ, then the central bright fringe occurs at an angle –φ, on the other side. Thus, if the source S is on the perpendicular bisector, then the central fringe occurs at O, also on the perpendicular bisector. If S is shifted by an angle φ to point S′, then the central fringe appears at a point O′ at an angle –φ, which means that it is shifted by the same angle on the other side of the bisector. This also means that the source S′, the mid-point Q and the point O′ of the central fringe are in a straight line.
We end this section by quoting from the Nobel lecture of Dennis Gabor
The wave nature of light was demonstrated convincingly for the first time in 1801 by Thomas Young by a wonderfully simple experiment. He let a ray of sunlight into a dark room, placed a dark screen in front of it, pierced with two small pinholes, and beyond this, at some distance, a white screen. He then saw two darkish lines at both sides of a bright line, which gave him sufficient encouragement to repeat the experiment, this time with spirit flame as light source, with a little salt in it to produce the bright yellow sodium light. This time he saw a number of dark lines, regularly spaced; the first clear proof that light added to light can produce darkness. This phenomenon is called interference. Thomas Young had expected it because he believed in the wave theory of light.
Dennis Gabor received the 1971 Nobel Prize in Physics for discovering theprinciples of holography
We should mention here that the fringes are straight lines although S 1 and S 2 are point sources. If we had slits instead of the point sources (Fig. 10.14), each pair of points would have produced straight line fringes resulting in straight line fringes with increased intensities.
Figure 10.14 Photograph and the graph of the intensity distribution in Young’s double-slit experiment.
Example 10.3 Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when blue-green light of wavelength 500 nm is used?
Solution Fringe spacing
= 5 × 10–4 m = 0.5 mm
Example 10.4 What is the eff ect on the interference fringes in a Young’s double-slit experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved closer to the double-slit plane;
(e) the width of the source slit is increased;
(f) the monochromatic source is replaced by a source of white
light?
(In each operation, take all parameters, other than the one specified, to remain unchanged.)
Solution
(a) Angular separation of the fringes remains constant (=λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the
two slits.
(b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.
(c) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.
(d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition
s/S<λ/d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.
(e) Same as in (d). As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s/S≤ λ/d is not satisfied, the interference pattern disappears.
(f) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S 2 P –S 1 P =λ b/2, where λ b
(≈4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour.Slightly farther away where S 2 Q–S 1 Q =λ b=λ r/2 where λ r(≈8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue.
Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.
10.6 Diffraction
If we look clearly at the shadow cast by an opaque object, close to the region of geometrical shadow, there are alternate dark and bright regions just like in interference. This happens due to the phenomenon of diffraction. Diffraction is a general characteristic exhibited by all types of waves, be it sound waves, light waves, water waves or matter waves. Since the wavelength of light is much smaller than the dimensions of most obstacles; we do not encounter diffraction effects of light in everyday observations. However, the finite resolution of our eye or of optical instruments such as telescopes or microscopes is limited due to the phenomenon of diffraction. Indeed the colours that you see when a CD is viewed is due to diffraction effects. We will now discuss the phenomenon of diffraction.
10.6.1 The single slit
In the discussion of Young’s experiment, we stated that a single narrow slit acts as a new source from which light spreads out. Even before Young, early experimenters – including Newton – had noticed that light spreads out from narrow holes and slits. It seems to turn around corners and enter regions where we would expect a shadow. These effects, known as diffraction, can only be properly understood using wave ideas. After all, you are hardly surprised to hear sound waves from someone talking around a corne r!
When the double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source), a broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre (Fig. 10.16). To understand this, go to Fig. 10.15, which shows a parallel beam of light falling normally on a single slit LN of width a. The diffracted light goes on to meet a screen. The midpoint of the slit is M.
A straight line through M perpendicular to the slit plane meets the screen at C. We want the intensity at any point P on the screen. As before, straight lines joining P to the different points L,M,N, etc., can be treated as parallel, making an angle θ with the normal MC.
The basic idea is to divide the slit into much smaller parts, and add their contributions at P with the proper phase differences. We are treating different parts of the wavefront at the slit as secondary sources. Because the incoming wavefront is parallel to the plane of the slit, these sources are in phase.
The path difference NP – LP between the two edges of the slit can be calculated exactly as for Young’s experiment. From Fig. 10.15,
NP – LP = NQ
=a sin θ
≈ a θ (for smaller angles)(10.21)
Similarly, if two points M 1 and M 2 in the slit plane are separated by y, the path difference M 2 P – M 1 P ≈y θ. We now have to sum up equal, coherent contributions from a large number of sources, each with a different phase. This calculation was made by Fresnel using integral calculus, so we omit it here. The main features of the diffraction pattern can be understood by simple arguments.
At the central point C on the screen, the angle θ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C. Experimental observation shown in Fig. 10.15 indicates that the intensity has a central maximum at θ = 0 and other secondary maxima at θ ≈(n+1/2) λ/a, and has minima (zero intensity) at θ≈n λ/a,
n = ±1, ±2, ±3, .... It is easy to see why it has minima at these values of angle. Consider first the angle θ where the path difference a θ is λ. Then,
. (10.22)
Now, divide the slit into two equal halves LM and MN each of size a/2. For every point M 1 in LM, there is a point M 2 in MN such that M 1 M 2 = a/2. The path difference between M 1 and M 2 at P = M 2 P – M 1 P=θ a/2 = λ/2 for the angle chosen. This means that the contributions from M 1 and M 2 are 180° out of phase and cancel in the direction θ =λ/a. Contributions from the two halves of the slit LM and MN, therefore, cancel each other. Equation (10.22) gives the angle at which the intensity falls to zero. One can similarly show that the intensity is zero for θ =n λ/a, with n being any integer (except zero!). Notice that the angular size of the central maximum increases when the slit width a decreases.
Figure 10.15 The geometry of path differences for diffraction by a single slit.
It is also easy to see why there are maxima at θ = (n + 1/2)λ/a and why they go on becoming weaker and weaker with increasing n. Consider an angle θ = 3 λ/2 a which is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be
(10.23)
The first two-thirds of the slit can therefore be divided into two halves which have a λ/2 path difference. The contributions of these two halves cancel in the same manner as described earlier. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. Clearly, this will be much weaker than the central maximum (where the entire slit contributes in phase). One can similarly show that there are maxima at (n + 1/2) λ/a with n = 2, 3, etc. These become weaker with increasing n, since only one-fifth, one-seventh, etc., of the slit contributes in these cases. The photograph and intensity pattern corresponding to it is shown in Fig. 10.16.
Figure 10.16 Intensity distribution and photograph of fringes due to diffraction
at single slit.
There has been prolonged discussion about difference between intereference and diffraction among scientists since the discovery of these phenomena. In this context, it is interesting to note what Richard Feynman has said in his famous Feynman Lectures on Physics:
No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between the m. The best we can do is, roughly speaking, is to say that when there are only a few sources, say two interfering sources, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used.
In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern. This is shown in
Fig. 10.17. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a, that is the ratio of the distance between the two slits to the width of a slit. In the limit of a becoming very small, the diffraction pattern will become very flat and we will obsrve the two-slit interference pattern [see Fig. 10.13(b)].
FIGURE 10.17 The actual double-slit interference pattern The envelope shows the single slit diffraction..
Example 10.5 In Example 10.3, what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern?
Solution We want
Notice that the wavelength of light and distance of the screen do not enter in the calculation of a.
In the double-slit interference experiment of Fig. 10.12, what happens if we close one slit? You will see that it now amounts to a single slit. But you will have to take care of some shift in the pattern. We now have a source at S, and only one hole (or slit) S 1 or S 2. This will produce a single-slit diffraction pattern on the screen. The centre of the central bright fringe will appear at a point which lies on the straight line SS 1 or SS 2, as the case may be.
We now compare and contrast the interference pattern with that seen for a coherently illuminated single slit (usually called the single slit diffraction pattern).
(i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.
(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
Richand Feynman was one of the recipients of the 1965 Nobel Prize in Physics for his fundamental work in quantum electrodynamics.
(iii) For a single slit of width a, the first null of the interference pattern occurs at an angle of λ/a. At the same angle of λ/a, we get a maximum (not a null) for two narrow slits separated by a distance a.
One must understand that both d and a have to be quite small, to be able to observe good interference and diffraction patterns. For example, the separation d between the two slits must be of the order of a milimetre or so. The width a of each slit must be even smaller, of the order of 0.1 or 0.2 mm.
In our discussion of Young’s experiment and the single-slit diffraction, we have assumed that the screen on which the fringes are formed is at a large distance. The two or more paths from the slits to the screen were treated as parallel. This situation also occurs when we place a converging lens after the slits and place the screen at the focus. Parallel paths from the slit are combined at a single point on the screen.Note that the lens does not introduce any extra path differences in a parallel beam. This arrangement is often used since it gives more intensity than placing the screen far away. If f is the focal length of the lens, then we can easily work out the size of the central bright maximum. In terms of angles, the separation of the central maximum from the first null of the diffraction pattern is λ/a. Hence, the size on the screen will be f λ/a.
10.6.2 Seeing the single slit diffraction pattern
It is surprisingly easy to see the single-slit diffraction pattern for oneself. The equipment needed can be found in most homes –– two razor blades and one clear glass electric bulb preferably with a straight filament. One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This is easily done with the thumb and forefingers (Fig. 10.18).
Keep the slit parallel to the filament, right in front of the eye. Use spectacles if you normally do. With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands. Since the position of all the bands (except the central one) depends on wavelength, they will show some colours. Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue can be seen.
In this experiment, the filament plays the role of the first slit S in Fig. 10.16. The lens of the eye focuses the pattern on the screen (the retina of the eye).
Figure 10.18 Holding two blades to form a single slit. A bulb filament viewed through this shows clear diffraction bands.
With some effort, one can cut a double slit in an aluminium foil with a blade. The bulb filament can be viewed as before to repeat Young’s experiment. In daytime, there is another suitable bright source subtending a small angle at the eye. This is the reflection of the Sun in any shiny convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle
of (1/2)°.
In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.
10.6.3 Resolving power of optical instruments
In Chapter 9 we had discussed about telescopes. The angular resolution of the telescope is determined by the objective of the telescope. The stars which are not resolved in the image produced by the objective cannot be resolved by any further magnification produced by the eyepiece. The primary purpose of the eyepiece is to provide magnification of the image produced by the objective.
Consider a parallel beam of light falling on a convex lens. If the lens is well corrected for aberrations, then geometrical optics tells us that the beam will get focused to a point. However, because of diffraction, the beam instead of getting focused to a point gets focused to a spot of finite area. In this case the effects due to diffraction can be taken into account by considering a plane wave incident on a circular aperture followed by a convex lens (Fig. 10.19). The analysis of the corresponding diffraction pattern is quite involved; however, in principle, it is similar to the analysis carried out to obtain the single-slit diffraction pattern. Taking into account the effects due to diffraction, the pattern on the focal plane would consist of a central bright region surrounded by concentric dark and bright rings (Fig. 10.19). A detailed analysis shows that the radius of the central bright region is approximately given by
(10.24)
where f is the focal length of the lens and 2 a is the diameter of the circular aperture or the diameter of the lens, whichever is smaller. Typically if
λ≈ 0.5 µ m, f≈ 20 cm and a≈ 5 cm
we have
r 0≈ 1.2 µ m
Although the size of the spot is very small, it plays an important role in determining the limit of resolution of optical instruments like a telescope or a microscope. For the two stars to be just resolved
Figure 10.19 A parallel beam of light is incident on a convex lens.Because of diffraction effects, the beam gets focused to a spot of radius≈0.61 λ f/a.
implying
(10.25)
Thus ∆θ will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if a is large. It is for this reason that for better resolution, a telescope must have a large diameter objective.
Example 10.6 Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Solution A 100 inch telescope implies that 2 a = 100 inch= 254 cm. Thus if,
λ ≈6000Å = 6×10–5 cm
then
radians
We can apply a similar argument to the objective lens of a microscope. In this case, the object is placed slightly beyond f, so that a real image is formed at a distance v [Fig. 10.20]. The magnification (ratio of image size to object size) is given by m ≈ v/f. It can be seen from Fig. 10.20 that
D/f ≈ 2 tan β (10.26)
where 2 β is the angle subtended by the diameter of the objective lens at the focus of the microscope.
FIGURE 10.20 Real image formed by the objective lens of the microscope.
Determine the resolving power of your eye
You can estimate the resolving power of your eye with a simple experiment. Make black stripes of equal width separated by white stripes; see figure here. All the black stripes should be of equal width, while the width of the intermediate white stripes should increase as you go from the left to the right. For example, let all black stripes have a width of 5 mm. Let the width of the first two white stripes be 0.5 mm each, the next two white stripes be 1 mm each, the next two 1.5 mm each, etc. Paste this pattern on a wall in a room or laboratory, at the height of your eye.
Now watch the pattern, preferably with one eye. By moving away or closer to the wall, find the position where you can just see some two black stripes as separate stripes. All the black stripes to the left of this stripe would merge into one another and would not be distinguishable. On the other hand, the black stripes to the right of this would be more and more clearly visible. Note the width d of the white stripe which separates the two regions, and measure the distance D of the wall from your eye. Then d/D is the resolution of your eye.
You have watched specks of dust floating in air in a sunbeam entering through your window. Find the distance (of a speck) which you can clearly see and distinguish from a neighbouring speck. Knowing the resolution of your eye and the distance of the speck, estimate the size of the speck of dust.
When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light, the diffraction effects become important. The image of a point object will again be a diffraction pattern whose size in the image plane will be
(10.27)
Two objects whose images are closer than this distance will not be resolved, they will be seen as one. The corresponding minimum separation,d min, in the object plane is given by
d min=
=
or, since
=(10.28)
Now, combining Eqs. (10.26) and (10.28), we get
≈(10.29)
If the medium between the object and the objective lens is not air but a medium of refractive index n, Eq. (10.29) gets modified to
(10.30)
The product n sin β is called the numerical aperture and is sometimes marked on the objective.
The resolving power of the microscope is given by the reciprocal of the minimum separation of two points seen as distinct. It can be seen from Eq. (10.30) that the resolving power can be increased by choosing a medium of higher refractive index. Usually an oil having a refractive index close to that of the objective glass is used. Such an arrangement is called an‘oil immersion objective’. Notice that it is not possible to make sin β larger than unity. Thus, we see that the resolving power of a microscope is basically determined by the wavelength of the light used.
There is a likelihood of confusion between resolution and magnification, and similarly between the role of a telescope and a microscope to deal with these parameters. A telescope produces images of far objects nearer to our eye. Therefore objects which are not resolved at far distance, can be resolved by looking at them through a telescope. A microscope, on the other hand, magnifies objects (which are near to us) and produces their larger image. We may be looking at two stars or two satellites of a far-away planet, or we may be looking at different regions of a living cell. In this context, it is good to remember that a telescope resolves whereas a microscope magnifies.
10.6.4 The validity of ray optics
An aperture (i.e., slit or hole) of size a illuminated by a parallel beam sends diffracted light into an angle of approximately≈ λ/a. This is the angular size of the bright central maximum. In travelling a distance z, the diffracted beam therefore acquires a width z λ/a due to diffraction. It is interesting to ask at what value of z the spreading due to diffraction becomes comparable to the size a of the aperture. We thus approximately equate z λ/a with a. This gives the distance beyond which divergence of the beam of width a becomes significant. Therefore,
(10.31)
We define a quantity z F called the Fresnel distance by the following equation
Equation (10.31) shows that for distances much smaller than z F, the spreading due to diffraction is smaller compared to the size of the beam. It becomes comparable when the distance is approximately z F. For distances much greater than z F, the spreading due to diffraction dominates over that due to ray optics (i.e., the size a of the aperture). Equation (10.31) also shows that ray optics is valid in the limit of wavelength tending to zero.
Example 10.7 For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?
Solution
This example shows that even with a small aperture, diffraction spreading can be neglected for rays many metres in length. Thus, ray optics is valid in many common situations.
10.7 Polarisation
Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.21). Such a wave could be described by the following equation
Figure 10.21(a) The curves represent the displacement of a string at t= 0 and at t=∆t, respectively when a sinusoidal wave is propagating in the +x-direction. (b) The curve represents the time variation of the displacement at x= 0 when a sinusoidal wave is propagating in the +x-direction. At x=∆x, the time variation of the displacement will be slightly displaced to the right.
y (x,t) =a sin (kx –ω t)(10.32)
where a and ω (= 2 π ν) represent the amplitude and the angular frequency of the wave, respectively; further,
(10.33)
represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 15 of Class XI textbook. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised wave. Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave.
In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by
z (x,t) =a sin (kx –ω t) (10.34)
It should be mentioned that the linearly polarised waves [described by Eqs. (10.33) and (10.34)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.
Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid.
Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P 1, it is observed that its intensity is reduced by half. Rotating P 1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P 2 be placed before P 1. As expected, the light from the lamp is reduced in intensity on passing through P 2 alone. But now rotating P 1 has a dramatic effect on the light coming from P 2. In one position, the intensity transmitted by P 2 followed by P 1 is nearly zero. When turned by 90° from this position,P 1 transmits nearly the full intensity emerging from P 2 (Fig. 10.22).
The above experiment can be easily understood by assuming that light passing through the polaroid P 2 gets polarised along the pass-axis of P 2. If the pass-axis of P 2 makes an angle θ with the pass-axis of P 1, then when the polarised beam passes through the polaroid P 2, the component E cos θ (along the pass-axis of P 2) will pass through P 2. Thus, as we rotate the polaroid P 1 (or P 2), the intensity will vary as:
I =I 0 cos 2 θ (10.35)
where I 0 is the intensity of the polarized light after passing through P 1.This is known as Malus’ law. The above discussion shows that the intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids.
Figure 10.22(a) Passage of light through two polaroids P 2 and P 1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0° to 90°. Notice that the light seen through a single polaroid P 1 does not vary with angle. (b) Behaviour of the electric vector when light passes through two polaroids. The transmitted polarisation is the component parallel to the polaroid axis.The double arrows show the oscillations of the electric vector.
Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras.
Example 10.8 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?
Solution Let I 0 be the intensity of polarised light after passing through the first polariser P 1. Then the intensity of light after passing through second polariser P 2 will be
where θ is the angle between pass axes of P 1 and P 2. Since P 1 and P 3 are crossed the angle between the pass axes of P 2 and P 3 will be
(π/2–θ). Hence the intensity of light emerging from P 3 will be
=I 0 cos 2 θ sin 2 θ=(I 0/4) sin 2 2 θ
Therefore, the transmitted intensity will be maximum when θ=π/4.
10.7.1 Polarisation by scattering
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is nothing but sunlight, which has changed its direction (having been scattered) on encountering the molecules of the earth’s atmosphere. As Fig. 10.23(a) shows, the incident sunlight is unpolarised. The dots stand for polarisation perpendicular to the plane of the figure. The double arrows show polarisation in the plane of the figure. (There is no phase relation between these two in unpolarised light). Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have drawn an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel to the double arrows do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore represented by dots. It is polarised perpendicular to the plane of the figure. This explains the polarisation of scattered light from the sky.
FIGURE 10.23 (a) Polarisation of the blue scattered light from the sky. The incident sunlight is unpolarised (dots and arrows). A typical molecule is shown. It scatters light by 90° polarised normal to the plane of the paper (dots only). (b) Polarisation of light reflected from a transparent medium at the Brewster angle (reflected ray perpendicular to refracted ray).
The scattering of light by molecules was intensively investigated by C.V. Raman and his collaborators in Kolkata in the 1920s. Raman was awarded the Nobel Prize for Physics in 1930 for this work.
A SPECIAL CASE OF TOTAL TRANSMISSION
When light is incident on an interface of two media, it is observed that some part of it gets reflected and some part gets transmitted. Consider a related question: Is it possible that under some conditions a monochromatic beam of light incident on a surface (which is normally reflective) gets completely transmitted with no reflection? To your surprise, the answer is yes.
Let us try a simple experiment and check what happens. Arrange a laser, a good polariser, a prism and screen as shown in the figure here.
Let the light emitted by the laser source pass through the polariser and be incident on the surface of the prism at the Brewster’s angle of incidence i B. Now rotate the polariser carefully and you will observe that for a specific alignment of the polariser, the light incident on the prism is completely transmitted and no light is reflected from the surface of the prism. The reflected spot will completely vanish.
10.7.2 Polarisation by reflection
Figure 10.23(b) shows light reflected from a transparent medium, say, water. As before, the dots and arrows indicate that both polarisations are present in the incident and refracted waves. We have drawn a situation in which the reflected wave travels at right angles to the refracted wave. The oscillating electrons in the water produce the reflected wave. These move in the two directions transverse to the radiation from wave in the medium, i.e., the refracted wave. The arrows are parallel to the direction of the reflected wave. Motion in this direction does not contribute to the reflected wave. As the figure shows, the reflected light is therefore linearly polarised perpendicular to the plane of the figure (represented by dots). This can be checked by looking at the reflected light through an analyser. The transmitted intensity will be zero when the axis of the analyser is in the plane of the figure, i.e., the plane of incidence.
When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. Thus we have seen that when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarised wave. The angle of incidence in this case is called Brewster’s angle and is denoted by i B. We can see that i B is related to the refractive index of the denser medium. Since we have i B+r =π/2, we get from Snell’s law
(10.36)
This is known as Brewster’s law.
Example 10.9 Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other?
Solution For i +r to be equal to π/2, we should have tan i B =µ = 1.5. This gives i B = 57°. This is the Brewster’s angle for air to glass
interface.
For simplicity, we have discussed scattering of light by 90°, and reflection at the Brewster angle. In this special situation, one of the two perpendicular components of the electric field is zero. At other angles, both components are present but one is stronger than the other. There is no stable phase relationship between the two perpendicular components since these are derived from two perpendicular components of an unpolarised beam. When such light is viewed through a rotating analyser, one sees a maximum and a minimum of intensity but not complete darkness. This kind of light is called partially polarised.
Let us try to understand the situation. When an unpolarised beam of light is incident at the Brewster’s angle on an interface of two media, only part of light with electric field vector perpendicular to the plane of incidence will be reflected. Now by using a good polariser, if we completely remove all the light with its electric vector perpendicular to the plane of incidence and let this light be incident on the surface of the prism at Brewster’s angle, you will then observe no reflection and there will be total transmission of light.
We began this chapter by pointing out that there are some phenomena which can be explained only by the wave theory. In order to develop a proper understanding, we first described how some phenomena like reflection and refraction, which were studied on this basis of Ray Optics in Chapter 9, can also be understood on the basis of Wave Optics. Then we described Young’s double slit experiment which was a turning point in the study of optics. Finally, we described some associated points such as diffraction, resolution, polarisation, and validity of ray optics. In the next chapter, you will see how new experiments led to new theories at the turn of the century around 1900 A.D.
Summary
Huygens’ principle tells us that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time.
Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. When the speed of light is independent of direction, the secondary waves are spherical. The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. This principle leads to the well known laws of reflection and refraction.
The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. When we consider the intensity of light due to these sources at the given point, there is an interference term in addition to the sum of the individual intensities. But this term is important only if it has a non-zero average, which occurs only if the sources have the same frequency and a stable phase difference.
Young’s double slit of separation d gives equally spaced fringes of angular separation λ/d. The source, mid-point of the slits, and central bright fringe lie in a straight line. An extended source will destroy the fringes if it subtends angle more than λ/d at the slits.
A single slit of width a gives a diffraction pattern with a central maximum. The intensity falls to zero at angles ofetc., with successively weaker secondary maxima in between. Diffraction limits the angular resolution of a telescope to λ/D where D is the diameter. Two stars closer than this give strongly overlapping images. Similarly, a microscope objective subtending angle 2 β at the focus, in a medium of refractive index n, will just separate two objects spaced at a distance λ/(2 n sin β), which is the resolution limit of a microscope. Diffraction determines the limitations of the concept of light rays. A beam of width a travels a distance a 2/λ, called the Fresnel distance, before it starts to spread out due to diffraction.
Natural light, e.g., from the sun is unpolarised. This means the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polaroid transmits only one component (parallel to a special axis). The resulting light is called linearly polarised or plane polarised. When this kind of light is viewed through a second polaroid whose axis turns through 2 π, two maxima and minima of intensity are seen. Polarised light can also be produced by reflection at a special angle (called the Brewster angle) and by scattering through π/2 in the earth’s atmosphere.
Points to Ponder
Waves from a point source spread out in all directions, while light was seen to travel along narrow rays. It required the insight and experiment of Huygens, Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light.
The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive, as shown in Young’s experiment.
Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light.
Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarisation phenomena are special to transverse waves like light waves.
Exercises
10.1 Monochromatic light of wavelength 589 nm is incident from the air on a water surface. What are the wavelength, frequency, and speed of(a) reflected, and (b) refracted light? The Refractive index of water is 1.33.
NEETprep Answer
10.2 What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of the light from a distant star intercepted by the Earth.
NEETprep Answer
10.3(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in a vacuum is 3.0 × 10 8 m s–1)
(b) Is the speed of light in glass independent of the color of light? If not, which of the two colors red and violet travels slower in a glass prism?
NEETprep Answer
10.4 In Young’s double-slit experiment, the slits are separated by
0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
NEETprep Answer
10.5 In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?
NEETprep Answer
10.6 A beam of light consisting of two wavelengths, 650 nm, and 520 nm, is used to obtain interference fringes in Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
NEETprep Answer
10.7 In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be 4/3.
NEETprep Answer
10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
NEETprep Answer
10.9 Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ra y?
NEETprep Answer
10.10 Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.
NEETprep Answer
Additional Exercises
10.11 The 6563 Å H α line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
NEETprep Answer
10.12 Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in a vacuum. Is the prediction confirmed by experimental determination of the speed of light in wate r? If not, which alternative picture of light is consistent with the experiment?
NEETprep Answer
10.13 You have learned in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
NEETprep Answer
10.14 Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) the direction of propagation.
(iii) the motion of the source and/or observer.
(iv) wavelength.
(v) the intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depends?
NEETprep Answer
10.15 For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical to the two situations in case of light traveling in a mediu m?
NEETprep Answer
10.16 In a double-slit experiment using the light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slit s?
NEETprep Answer
10.17 Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction ban d?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room
10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in an understanding location and several other properties of images in optical instruments. What is the justification?
NEETprep Answer
10.18 Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
NEETprep Answer
10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the center of the screen. Find the width of the slit.
NEETprep Answer
10.20 Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learned in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
NEETprep Answer
10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n λ/a. Justify this by suitably dividing the slit to bring out the cancellation.
NEETprep Answer
Exemplar
Q. 1 Consider a tight beam incident from air to a glass slab at Brewster's angle as shown in figure. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid,
(a) For a particular orientation, there shall be darkness as observed
through the polaroid
(b) The intensity of tight as seen through the polaroid shall be
independent of the rotation
(c) The intensity of light as seen through the polaroid shall go through a
minimum but not zero for two orientations of the polaroid
(d) The intensity of light as seen through the polaroid shall go through a
minimum for four orientations of the polaroid
NEETprep Answer
Q. 2 Consider sunlight incident on a slit of width 10 4 4∘A A∘, The image seen
through the slit shall
(a) be a fine sharp slit white in colour at the centre
(b) a bright slit white at the centre diffusing to zero intensities at the edges
(c) a bright slit white at the centre diffusing to regions of different colours
(d) only be a diffused slit white in colour
NEETprep Answer
Q.3 Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is.
(1)4 π d λ(1−1 n 2 sin 2 θ)1/2+π(2)4 π d λ(1−1 n 2 sin 2 θ)1/2(3)4 π d λ(1−1 n 2 sin 2 θ)1/2+π 2(4)4 π d λ(1−1 n 2 sin 2 θ)1/2+2 π(1)4 π d λ 1-1 n 2 sin 2 θ 1/2+π(2)4 π d λ 1-1 n 2 sin 2 θ 1/2(3)4 π d λ 1-1 n 2 sin 2 θ 1/2+π 2(4)4 π d λ 1-1 n 2 sin 2 θ 1/2+2 π
NEETprep Answer
Q. 4 Ina Young's double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case,
(1) there shall be alternate interference patterns of red and blue
(2) there shall be an interference pattern for red distinct from that for blue
(3) there shall be no interference fringes
(4) there shall be an interference pattern for red mixing with one for blue
NEETprep Answer
Q. 5 Figure shows a standard two-slit arrangement with slits S 1,S 2,P 1,P 2 S 1,S 2,P 1,P 2 5 are the two minima points on either side of P (figure).
At P 2 P 2 on the screen, there is hole and behind is a second 2-slit arrangement with slits S 3,S 4 S 3,S 4 and a second screen behind them.
(1) There would be no interference pattern on the second screen but it would be lighted
(2) The second screen would be totally dark
(3) There would be a single bright point on the second screen
(4) There would be a regular two slit pattern on the second screen
NEETprep Answer
Q.6 Two Sources S 1 a n d S 2 S 1 a n d S 2 of intensity I 1 a n d l 2 I 1 a n d l 2 are front of a screen [Fig.(a)]. The pattern of intensity distribution seen in the central potion i given by Fig.(b).
In this case, which of the following statements are true?
(1)S 1 a n d S 2 S 1 a n d S 2 have the same intensities
(2)S 1 a n d S 2 S 1 a n d S 2 have a constant phase difference
(3)S 1 a n d S 2 S 1 a n d S 2 have the same phase
(4)S 1 a n d S 2 S 1 a n d S 2 have the same wavelength
NEETprep Answer
Q. 7 Consider sunlight incident on a pinhole of width 10 3 3∘A A∘. The image of the pinhole seen on a screen shall be
(1) a sharp white ring
(2) different from a geometrical image
(3) a diffused central spot, white in colour
(4) diffused coloured region around a sharp central white spot
NEETprep Answer
Q. 8 Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(1) the size decreases
(2) the intensity increases
(3) the size increases
(4) the intensity decreases
NEETprep Answer
Q. 9 for light diverging from a point source,
(a) the wavefront is spherical
(b) the intensity decreases in proportion to the distance squared
(c) the wavefront is parabolic
(d) the intensity at the wavefront does not depend on the distance
NEETprep Answer
Q.18 Consider a two-slit interference arrangement (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtanin the value of D in terms of λ λ such that the first minima on the screen falls at a distance D from the centre0.
NEETprep Answer
Q 19. Figure shown a two-slit arrangement with a source which emits unpolarised Ligt. P is a polariser with axis whose direction is not given. lf l 0 0 is the intensity of the principal maxima when no polariser is present, calculate in the case, the intensity of the principal maxima as well as of the first minima.
NEETprep Answer
Q.20
A C=C 0=D,S 1 C=S 2 C=d<<D A C=C 0=D,S 1 C=S 2 C=d<<D
A small transparent slab containing material of u=1.5 is placed along AS (figure) . What will be the distance from 0 of the principal maxma and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?
NEETprep Answer
Q. 21 Four identical monochromatic sources A, B, C, D as shown in the (figure) produce waves of the same wavelength λ λ, and are coherent. Two receiver, and R, are at great but equal distances from B. (i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when 8 is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?
R 1 B=d−R 2 B A B=B C=B D=λ/2 R 1 B=d-R 2 B A B=B C=B D=λ/2
NEETprep Answer
Q. 22 The optical properties of a medium are governed by the relative permittivity (ε r ε r) and relative permeability (u r u r) . The refractive index is defined as√u r ε r=n u r ε r=n. For ordinary material,ε r>0 a n d u r>0 ε r>0 a n d u r>0 and 4,> 0 and the positive sign is taken for the square root.
In 1964, a Russian scientist V. Veselago postulated the existence of
material with ε r ε r<0 and u r u r<0. Since, then such metamaterials have
been produced in the laboratories and their optical properties studied,
For such materials√u r ε r.u r ε r.As light enters a medium of such
refractive index the phases travel away from the direction of
propagation.
(i) According to the description above show that if rays of light enter
such a medium from air (refractive index = 1) at an angle 8 in 2nd
quadrant, then the refracted beam is in the 3rd quadrant.
(ii) Prove that Snell's law holds for such a medium.
NEETprep Answer
Q. 23 To ensure almost 100% transmissivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of
this material is intermediated between that of air and glass (which
makes the optical element of the lens). A typically used dielectric film is
MgF 2 2 (n = 1.38). What should the thickness of the film be so that at the
centre of the visible spectrum (5500∘A A∘) there is maximum transmission.
NEETprep Answer
Q. 10 Is Huygen’s principle valid for longitudinal sound waves?
NEETprep Answer
Q. 11 Consider a point at the focal point of a convergent lens. Another
convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
NEETprep Answer
Q.12 What is the shape of the wavefront on earth for sunlight?
NEETprep Answer
Q. 13 Why is the diffraction of sound waves more evident in daily experience than that of light wave?
NEETprep Answer
Q. 14 The human eye has an approximate angular resolution of ϕ=5.8×10−4 ϕ=5.8×10-4 rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.
NEETprep Answer
Q.15 A polaroid (1) is placed infront of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (II) and (II). In this case, wil light emerge from (II). Explain.
NEETprep Answer
Q. 16 Can reflection result in plane polarised tight if the light is incident on the interface from the side with higher refractive index?
NEETprep Answer
Q. 17 For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 A and
electrons accelerated through 100V used as the illuminating substance.
NEETprep Answer
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4181 | https://www.freemathhelp.com/forum/threads/similar-pyramids-find-ratios-of-volumes-bases-areas.50986/ | similar pyramids: find ratios of volumes, bases, areas | Free Math Help Forum
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similar pyramids: find ratios of volumes, bases, areas
Thread starterducksmeowtoo
Start dateApr 18, 2007
D
ducksmeowtoo
New member
Joined Apr 18, 2007 Messages 3
Apr 18, 2007
#1
two similar pyramids have heights 12 and 18. Find the ratios of the following:
a. base area
b. lateral area
c. total area
d. volumes
J
jonboy
Full Member
Joined Jun 8, 2006 Messages 547
Apr 18, 2007
#2
Is this a regular pyramid?
skeeter
Elite Member
Joined Dec 15, 2005 Messages 3,216
Apr 18, 2007
#3
linear dimensions will be in the ratio 12/18 = 2/3
areas (all of them) will have the ratio 4/9
volumes will have the ratio 8/27
D
ducksmeowtoo
New member
Joined Apr 18, 2007 Messages 3
Apr 18, 2007
#4
volumes will have the ratio 8/27
^ how does it get to that? :?:
M
Mrspi
Senior Member
Joined Dec 17, 2005 Messages 2,128
Apr 18, 2007
#5
ducksmeowtoo said:
volumes will have the ratio 8/27
^ how does it get to that? :?:
Click to expand...
If two figures are similar, then the ratio of related AREAS is the square of the ratio of any pair of corresponding lengths. If the ratio of the lengths is 2 / 3, then the ratio of areas would be (2 / 3)2.
If two figures are similar, then the ratio of their VOLUMES is the cube of the ratio of any pair of corresponding lengths. If the ratio the lengths is 2 / 3, then the ratio of the volumes would be (2 / 3)3.
(2 / 3)3 = (2/3)(2/3)(2/3), or 8/27
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4182 | https://www.bbc.co.uk/bitesize/articles/zsmnsk7 | KS3
How to convert decimals to percentages
Part of MathsFractions
Save to My Bitesize
Key points
Calculations often give a decimal answer.
It can be useful to convert a decimal to a percentageA fraction expressed as the number of parts per hundred and recorded using the notation % as a percentage is often easier to understand.
To do these conversions, you need to be able to multiply by 100
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How to convert a decimal to a percentage
Multiply the decimal by 100 to get the percentage value.
Write the percentage value and add the percentage symbol %
When the decimal has a value greater than one whole, the percentage will be greater than 100%. Eg the decimal 1∙2 converts to 120%
Examples
Image gallery
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1 of 6
Slide 1 of 6, Example 1: Zero point four five., Convert 0∙45 to a percentage.
End of image gallery
Questions
What is 0∙8 as a percentage?
Multiply 0∙8 by 100 to find the percentage value.
0∙8 x 100 = 80
0∙8 is 80%
What is 1∙6 as a percentage?
Multiply 1∙6 by 100 to find the percentage value.
1∙6 x 100 = 160
1∙6 is 160%
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Practise converting decimals to percentages
Try this quiz to practise converting decimals to percentages.
Quiz
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Real-world maths
Converting decimals to percentages can be a valuable skill for anyone who publishes or presents data. This could include scientists, journalists, retailers and engineers.
A journalist may investigate employment statistics and find that a company has given a pay rise to 0∙8 of its staff. The figure is expressed as a decimal (0∙8) and so the journalist needs to converts this to a percentage (80%) to make the information easier for their readers to understand. The published article states that: “80% of staff received a pay rise.”
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Play Sudoku with BBC Bitesize!
Every weekday we release brand new easy, medium and hard Sudoku puzzles. Perfect for testing your skill with numbers and logic.
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4183 | https://bbs.pinggu.org/thread-417270-1-1.html | 多元线性回归在什么情况下可以不用截距项 - 经管之家
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人大经济论坛›论坛›计量经济学与统计论坛 五区›计量经济学与统计软件›多元线性回归在什么情况下可以不用截距项
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多元线性回归在什么情况下可以不用截距项[推广有奖]
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[LV.1]初来乍到楼主 zhudic发表于 2009-2-14 22:30:00|AI写论文 是 否+2 论坛币 k人 参与回答 ### 经管之家送您一份 应届毕业生专属福利! 求职就业群 赵安豆老师微信：zhaoandou666 ### 经管之家联合CDA 送您一个全额奖学金名额~ ! 立即领取 ### 感谢您参与论坛问题回答 经管之家送您两个论坛币！ +2 论坛币 rt 用截距项和不用截距项具体有什么样的区别？ 万分感谢！ ### 扫码加我 拉你入群 请注明：姓名-公司-职位 以便审核进群资格，未注明则拒绝 分享 0收藏 2回帖 点赞 00 关键词：多元线性回归线性回归截距项万分感谢线性回归 ### 相关帖子 • 不含截距项多元线性回归 • 多元线性回归中常数的经济意义 • 多元线性回归，时间序列需要标准化处理么？ • 求救 • 【连载4】 如何用spss做一般（含虚拟变量）多元线性回归 • 求助：多元线性回归 • [求助]多元线性回归拟合优度是不是很重要？ • 用proc reg做出多元线性回归，想要得到回归系数的标准差要用什么命令啊 • 贴内的数据能做多元线性回归吗？自变量是二分变量，因变量是连续变量。 • 多元线性回归结果已做出来了，请大家帮忙评价一下！ ### 回帖推荐 swjswj11 发表于3楼 查看完整内容 一般情况下，截距项是必须包括的，不管截距项是不是预期为零。因为如果不包括截距项则意味着你默认估计出来的等式是通过原点的，这会导致用最小二乘的估计值不是无偏的。 ### 本帖被以下文库推荐 · 计量.统计精彩问答|主题: 12506, 订阅: 52
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沙发 lirry123发表于 2009-2-14 22:46:00 if the expected value of dependent variable Y has a zero mean, then an intercept is not necessary included. otherwise, you must include an intercept in this linear regression.
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藤椅 swjswj11发表于 2009-2-15 14:11:00 一般情况下，截距项是必须包括的，不管截距项是不是预期为零。因为如果不包括截距项则意味着你默认估计出来的等式是通过原点的，这会导致用最小二乘的估计值不是无偏的。 | 已有 2 人评分 | 经验 | 论坛币 | 学术水平 | 收起理由 | --- --- | qdqhb | | | + 1 | 精彩帖子 | | 胖胖小龟宝 | + 10 | + 10 | | 热心帮助其他会员 | 总评分:经验 + 10 论坛币 + 10 学术水平 + 1查看全部评分
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板凳 zhudic发表于 2009-2-15 14:17:00 谢谢！
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报纸 QueenCi.Shine发表于 2014-2-9 18:13:39 > swjswj11 发表于 2009-2-15 14:11 > > 一般情况下，截距项是必须包括的，不管截距项是不是预期为零。因为如果不包括截距项则意味着你默认估计出来 ... 什么样的实证研究适合用多远回归，也就是利用多元回归的依据和理由一般是什么？谢谢 | 已有 1 人评分 | 学术水平 | 收起理由 | --- | Alfred_G | + 1 | 问题很有启发意义 | 总评分:学术水平 + 1查看全部评分
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地板 dps2000发表于 2014-2-10 11:07:03 配方设计时。
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7 楼 Alfred_G发表于 2014-2-10 11:17:29 二三楼的解释已经很清晰了，稍微做个补充吧：多元线性回归在标准化之后也是没有截距的。道理比较简单，就是把变量都进行“趋中化”处理，得到一个个标准正态分布，都是经过零点的，这样会消除截距项。 不过，我们一般都是看非标准化的统计结果，解释不同变量的偏作用
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8 楼 Alfred_G发表于 2014-2-10 11:37:24 > QueenCi.Shine 发表于 2014-2-9 18:13 > > 什么样的实证研究适合用多远回归，也就是利用多元回归的依据和理由一般是什么？谢谢 朋友，这是一个好问题。 回归并不单单是一种方法，更是一种思想，关于因果关系的一种解释。无论是计量经济学，还是社会学等其他学科里面的因果解释，本质上的方法都是回归。因为，影响某一个事件的原因可能有许许多多，而并非单一原因，所以，在具体分析中，肯定不会使用一元。 举个例子吧，面板数据，时间序列，逻辑模型，概率模型，无一例外都是用的GLS或者OLS，或者就是OLS衍生出来的其他模型，都是建立在最小二乘法基础上的。 建议可以去读一些专门讲回归分析或者模型的书籍，可以加深理解~
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9 楼 QueenCi.Shine发表于 2014-2-21 13:36:06 > Alfred_G 发表于 2014-2-10 11:37 > > 朋友，这是一个好问题。 > > 回归并不单单是一种方法，更是一种思想，关于因果关系的一种解释。无论是计量经 ... 谢谢你，这个问题是好，，主要是是我导师问我的，瞬间我就不会了。恩恩，我懂了，大概就是当英雄被解释变量的因素很多，我们只是想具体研究其中一个解释变量与被解释变量之间关系的时候，会采用多元回归，因为这样可以控制其他变量的影响。。谢谢哦
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10 楼 runman发表于 2015-12-16 15:41:47 > Alfred_G 发表于 2014-2-10 11:17 > > 二三楼的解释已经很清晰了，稍微做个补充吧：多元线性回归在标准化之后也是没有截距的。道理比较简单，就是 ... “多元线性回归在标准化之后也是没有截距的。”请教一下，这句话如果落实到具体的模型中，如何理解呢？可以理解为多元线性回归不需要截距项吗？
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4184 | https://chemistry.stackexchange.com/questions/72284/why-is-the-ortho-isomer-a-major-product-in-the-nitration-of-toluene | Skip to main content
Why is the ortho isomer a major product in the nitration of toluene?
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Usually when an ortho-para directing substituent is present on the benzene ring for an electrophilic aromatic substitution reaction, the para product is the major product (exceptions can be there when hydrogen bonding or ortho effect of COOH group makes the ortho product a major one.) But I don't understand why the ortho product in nitration of toluene is the major product (and the para product isn't). The ortho position is supposed to have more steric crowding, right?
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edited Mar 3, 2018 at 14:38
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asked Apr 9, 2017 at 18:14
UtkarshUtkarsh
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Now I know nothing about this particular nitration but one thing that tends to make ortho products favorable is the inductive effect, due to which ortho position of toulene is slightly more activated than para.
– Sawarnik
Commented
Apr 9, 2017 at 18:49
i read the answers and now I'm confused . If this is the reason then what if there is an ethyl group instead of methyl . Ortho position should be major because of +I effect but this is not the case as para is major .why is that
– utkarsh
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Apr 10, 2017 at 12:19
@utkarsh Then steric hindrance increases and the nitro group will go to the para position.
– dr.drizzy
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Mar 2, 2018 at 20:29
Related: chemistry.stackexchange.com/questions/88752/…
– Avyansh Katiyar
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Mar 3, 2018 at 3:43
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A methyl-group is not t-butyl, so it isn't all that crowded.
There are two ortho-positions.
There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.
Given all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.
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edited Apr 10, 2017 at 12:26
Martin - マーチン♦
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answered Apr 9, 2017 at 19:29
permeakrapermeakra
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There are three effects: sterics, electronics and statistics.
Yes, there is more steric crowding in the ortho position, but a methyl group isn't that big. In addition, electronic effects are stronger in ortho position than in para, since it is much closer to the +I substituent. And last but not least: there are two ortho positions which can react, but only one para-position, which can have a rather big influence if reactivity towards ortho and para position is similar.
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answered Apr 9, 2017 at 19:29
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One other factor favoring the ortho isomer: hydrogen bonding. Yes, hydrogen bonding to the methyl group.
Following the molecular orbital description of hydrogen bonding, an electron pair from the nitro group oxygen atoms can overlap the antibonding orbitals in the methyl groups, delocalizing what are formally the carbon-hydrogen bonds in the methyl group. We can represent this interaction with an additional contributing structure that has no octet-deficient atoms:
A para, or for that matter meta attack would not allow this delocalization to take place due to the remoteness of the nitro and methyl functions.
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edited Jan 5 at 18:20
answered Oct 10, 2023 at 12:38
Oscar LanziOscar Lanzi
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Could you mention the source? Wasnt H bonding only when atoms had sufficient EN diff
– Piyush
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@Plyush This is far from true. Several examples in this site point to hydrogen bonding even with relatively low electronegativity nonmetals, including another hydrogen atom as in this answer.
– Oscar Lanzi
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This is so different from what we're being taught thanks tho
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Ortho nitrotoluene is a major product while para nitrotoluene is a minor product .
Reasons:
In case of reactivity:
Ortho product is more reactive comparing with para product because electron density is more enriched at ortho position compared to the para position.
In case of stability:
Para product is more stable because is away from substituent group so it does not affected by steric hinderence as the ortho position I'd more steric hindered.
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answered Jun 21, 2020 at 14:29
Saleh NassorSaleh Nassor
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4185 | https://www.youtube.com/watch?v=17CBLBJNgGo | Combining radical expressions
Brian McLogan
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👉 Learn how to add or subtract radicals. A radical is a number or an expression under the root symbol. Radicals can only be added or subtracted if the numbers or expressions under the roots are the same for all terms. To add or subtract radicals, we reduce/simplify the radicals and then add/subtract the coefficients and retain the root part.
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Transcript:
um so on this problem here what I have is a sare < TK of two plus a of8 plus of 50 now remember ladies and gentlemen we could only add or subtract our square roots or either any of our radicals any of our Roots as long as they were the same term so right now I have square of 2qu of 8 50 I cannot add these this is not equal to square otk of 60 okay they have to be like terms just like we add x's and y's you had you could only add X's with x's right and y's with y's x squ with X squ so we can't combine these the only thing we can do is say if we can simplify this so I look at the of two can I do I know the value of of two no can I simplify the square root of two no then I look at the square root of eight sorry then I look at the square Ro of eight and I say all right can I Square take the square Ro of eight no because I know the square Ro of n is or Square of three is nine so I don't don't have an immediate value for the square of 8 even though approximate value would be 2.8 28 4271 but we we don't want an approximate value we want the exact value but I want to see what square number can I divide what square number can I rewrite with eight and then so code you'd say the largest square number that divides into eight is four four so I rewrite this as the square root of four 2 it is the largest square number and four is my largest square number now I look at 50 and say what's the largest square number that divides into 50 25 now square of two is still square of two then I say all right do I know the square of four now that's two s of two do I know the square of 25 yes five sare of two and now you guys notice I'm taking the square root of the same number correct so now I can combine them up so I have 1 of two plus 2 2s plus 5 2 equals 11 how you square of 25 is five 5 + 7 5 + 2 is 7 + 1 is 8 remember there's a one this right so we got to remember that |
4186 | https://www.youtube.com/watch?v=bYTs5kWQVbY | Find the fourth proportional to 72, 168, and 150. | CLASS 8 | RATIO, PROPORTION AND VARIATION | ...
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Find the fourth proportional to 72, 168, and 150.
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Transcript:
आज राउत नात आप डाउनलोड करें डाउनलोड पर होगा अब आपके सभी मैथ्स केमिस्ट्री फिजिक्स और बायोलॉजी डाउट का सफाया बस अपने क्वेश्चन की फोटो खींचे उसे क्रॉप करो और तुरंत वीडियो सॉल्यूशन पाओ| डाउनलोड नौ प्रोपोर्शनल तू 72168 एंड 150 ओके तो प्रोपोर्शनल का मतलब क्या होता है देखो हम इसे लिख सकते हैं 72 रेश्यो 168 फिर प्रोपोर्शनल का साइन ए गया ये वाला डबल रेश्यो होता है जो फिर 150 एंड फोर्थ हमें फाइंड आउट करना है तो हमने उसे ले लिया एक्स ओके ये हमें फाइंड आउट करना है ओके एक्स ये है क्या है सी हैव तू फाइंड दिस ओके शिवलाल जी जो प्रोपोर्शन का साइन है यह इक्वल तू में चेंज होता है अगेन यह 150 पीछे वाली वैल्यू डिनॉमिनेटर में अब तो हो जाएगा ये क्वेश्चन एक्स की वैल्यू फाइंड आउट करनी है क्रॉस मल्टीप्लाई कर लो 72 एक्स की वैल्यू क्या हो जाएगी 150 168 एक्स की वैल्यू 6 है तो एक्स को उधर छोड़ दो बस 150 168 और डिनॉमिनेटर में ए जाएगा 72 जो उधर से आया ओके नौ कैंसिल दिस देखो 4 से कैंसिल हो सकता है फॉर मल्टीप्लाई वैन फोर एंड 32 4 8 इस 32 ये हो जाएगा 4 एंड 2 ओके सो 18 देखो ये सिक्स हो जाएगा 63 एंड ये हो जाएगा 6 7 42 अब देखो 3 5.0 तो बचा क्या एक्स = 15 7 5.0 एक और है तो लगा दो व्हाट्सएप कीजिए |
4187 | https://askfilo.com/user-question-answers-smart-solutions/in-how-many-ways-can-three-persons-stay-in-5-hotel-in-how-3133373133393736 | Question asked by Filo student
In how many ways can three persons stay in 5 hotel ? In how many of these cases each person stays in a different hotel?
Views: 5,372 students
Updated on: Nov 14, 2024
Text SolutionText solutionverified iconVerified
Concepts:
Combinations, Permutations
Explanation:
To solve the problem, we will break it down into two parts: 1. The total number of ways three persons can stay in 5 hotels. 2. The number of ways in which each person stays in a different hotel.
Part 1: Total Ways
Each of the 3 persons can choose any of the 5 hotels. Therefore, the total number of ways is:
53=125
Part 2: Each Person in Different Hotels
To find the number of ways in which each person stays in a different hotel, we first choose 3 hotels out of the 5, and then assign each person to one of these hotels. The number of ways to choose 3 hotels from 5 is given by the combination formula:
C(5,3)=3!(5−3)!5!=10
After choosing the hotels, we can assign the 3 persons to these hotels in 3! ways. Thus, the total number of ways is:
C(5,3)×3!=10×6=60
So, the answers are:
Step by Step Solution:
Step 1
Calculate the total number of ways three persons can stay in 5 hotels: 53=125.
Step 2
Calculate the number of ways each person can stay in a different hotel: C(5,3)×3!=10×6=60.
Final Answer:
Total ways: 125; Ways with each in different hotels: 60
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| Question Text | In how many ways can three persons stay in 5 hotel ? In how many of these cases each person stays in a different hotel? |
| Updated On | Nov 14, 2024 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 11 |
| Answer Type | Text solution:1 |
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4188 | https://www.expii.com/t/isosceles-triangle-properties-983 | Expii
Isosceles Triangle Properties - Expii
In an isosceles triangle, the base angles are congruent. In other words, iso-lateral (isosceles) triangles are iso-angular. This is sometimes called the isosceles triangle theorem, or pons asinorum.
Explanations (4)
Aryaman Khandelwal
Text
5
The isosceles triangle theorem simply states that "If two sides of a triangle are congruent, the angles opposite them are congruent." This is quite obvious to most, as many people take it for granted that the two base angles of an isosceles triangle are congruent.
The proof for this theorem is also quite straightforward.
The first step requires you to draw the angle bisector AD.
Now, because of the triangle congruency postulate SAS, we can say that triangle CAD is congruent to triangle BAD. Because Congruent Parts of Congruent Triangles are Congruent(CPCTC), we can say that angle ACD is congruent to angle ABD.
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Related Lessons
Proving a Triangle Is Isosceles
Acute, Right, and Obtuse Triangles
Isosceles and Equilateral Symmetry and Special Lines
Side-Angle Inequalities in a Triangle
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Emma Wong
Text
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Isosceles triangle theorem
The isosceles triangle theorem is a very important theorem. It states that the angles opposite the two equal sides of an isosceles triangle are equal. Aryaman Khandelwal proved it by defining an angle bisector. I will show you a different proof which doesn't require you to draw any new points.
Let △ABC be a triangle such that |AB|=|AC|. △BAC≅△CAB, because |BA|=|CA|, |AC|=|AB| and ∠BAC=∠CAB. So, ∠ACB=∠ABC.
We can prove the converse of this theorem similarly.
Let △ABC be a triangle such that ∠ACB=∠ABC. △BAC≅△CAB, because |BC|=|CB|, ∠ACB=∠ABC and ∠BAC=∠CAB. So, |BA|=|CA|.
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Ethan Tan
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The Isosceles Triangle Theorem States that if a triangle contains 2 congruent sides, then the angles opposite those sides are congruent.
This theorem can be proved using the Law of Sines:
The Law of Sines states that: asinA=bsinB=csinC
If a=c, making △ABC an isosceles triangle, then asinA=asinC!
By simply dividing both sides of the equation by a and raising to the power of negative 1, we are left with sinA=sinC
And therefore m∠A=m∠C.
The converse of the Isosceles Triangle Theorem can be proved in a similar manner.
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Freya Edholm
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On each of the worksheets just linked to, please try problems 4-10. Have fun!
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4189 | https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/17%3A_Physics_of_Hearing/17.05%3A_Sound_Interference_and_Resonance-_Standing_Waves_in_Air_Columns | Skip to main content
17.5: Sound Interference and Resonance- Standing Waves in Air Columns
Last updated
: Feb 20, 2022
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17.4: Doppler Effect and Sonic Booms
17.6: Hearing
Page ID
: 1618
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
By the end of this section, you will be able to:
Define antinode, node, fundamental, overtones, and harmonics.
Identify instances of sound interference in everyday situations.
Describe how sound interference occurring inside open and closed tubes changes the characteristics of the sound, and how this applies to sounds produced by musical instruments.
Calculate the length of a tube using sound wave measurements.
Interference is the hallmark of waves, all of which exhibit constructive and destructive interference exactly analogous to that seen for water waves. In fact, one way to prove something “is a wave” is to observe interference effects. So, sound being a wave, we expect it to exhibit interference; we have already mentioned a few such effects, such as the beats from two similar notes played simultaneously.
Figure 17.5.2 shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference are contemplated for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and consistent with Pascal’s principle; pressures from two different sources add and subtract like simple numbers; that is, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique.
Where else can we observe sound interference? All sound resonances, such as in musical instruments, are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle, to the characteristic flavor of a violin’s sounding box, to the recognizability of a great singer’s voice, resonance and standing waves play a vital role.
Interference
Interference is such a fundamental aspect of waves that observing interference is proof that something is a wave. The wave nature of light was established by experiments showing interference. Similarly, when electrons scattered from crystals exhibited interference, their wave nature was confirmed to be exactly as predicted by symmetry with certain wave characteristics of light.
Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 17.5.3, Figure 17.5.4, Figure 17.5.5, and Figure 17.5.6. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This observation just means that the air column has only certain natural frequencies. The figures show how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown.
The standing wave formed in the tube has its maximum air displacement (an antinode) at the open end, where motion is unconstrained, and no displacement (a node) at the closed end, where air movement is halted. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; thus, λ=4L. This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 17.5.7. It is best to consider this a natural vibration of the air column independently of how it is induced.
Given that maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube, such as the one shown in Figure 17.5.8. Here the standing wave has three-fourths of its wavelength in the tube, or L=(3/4)λ′, so that λ′=4L/3. Continuing this process reveals a whole series of shorter-wavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. All resonant frequencies are integral multiples of the fundamental, and they are collectively called harmonics. The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 17.5.9shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end.
The fundamental and overtones can be present simultaneously in a variety of combinations. For example, middle C on a trumpet has a sound distinctively different from middle C on a clarinet, both instruments being modified versions of a tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different and subject to shading by the musician. This mix is what gives various musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, sounding boxes, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Simple resonant cavities can be made to resonate with the sound of the vowels, for example. (See Figure 17.5.10.) In boys, at puberty, the larynx grows and the shape of the resonant cavity changes giving rise to the difference in predominant frequencies in speech between men and women.
Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has λ=4L, and frequency is related to wavelength and the speed of sound as given by:
vw=fλ.(17.5.1)
Solving for f in this equation gives
f=vwλ=vw4L,(17.5.2)
where vw is the speed of sound in air. Similarly, the first overtone has λ′=4L/3 (see Figure 17.5.9), so that
f′=3vw4L=3f.(17.5.3)
Because f′=3f we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are
fn=nvw4L,n=1,3, 5,(17.5.4)
where f1 is the fundamental, f3 is the first overtone, and so on. It is interesting that the resonant frequencies depend on the speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them.
Example17.5.1: Find the Length of a Tube with a 128 Hz Fundamental
What length should a tube closed at one end have on a day when the air temperature, is 22oC, if its fundamental frequency is to be 128 Hz (C below middle C)?
What is the frequency of its fourth overtone?
Strategy
The length L can be found from the relationship in fn=nvw4L, but we will first need to find the speed of sound vw.
Solution for (a)
(1) Identify knowns:
the fundamental frequency is 128 Hz
the air temperature is 22oC
(2) Use fn=nvw4L to find the fundamental frequency (n = 1).
f1=vw4L(17.5.5)
(3) Solve this equation for length.
L=vw4f1(17.5.6)
(4) Find the speed of sound using vw=(331m/s)T273K−−−−√.
vw=(331m/s)295K273K−−−−−−√=344m/s(17.5.7)
(5) Enter the values of the speed of sound and frequency into the expression for L.
L=vw4f1=344m/s4(128Hz)=0.672m(17.5.8)
Discussion on (a)
Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and hence, the frequency of the note played. Horns producing very low frequencies, such as tubas, require tubes so long that they are coiled into loops.
Solution for (b)
(1) Identify knowns:
the first overtone has n=3
the second overtone has n=5
the third overtone has n=7
the fourth overtone has n=9
(2) Enter the value for the fourth overtone into fn=nvw4L.
f9=9vw4L=9f1=1.15kHz(17.5.9)
Discussion on (b)
Whether this overtone occurs in a simple tube or a musical instrument depends on how it is stimulated to vibrate and the details of its shape. The trombone, for example, does not produce its fundamental frequency and only makes overtones.
Another type of tube is one that is open at both ends. Examples are some organ pipes, flutes, and oboes. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends, as illustrated in Figure 17.5.11. Standing waves form as shown.
Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using Figure 17.5.11 as a guide, we can see that the resonant frequencies of a tube open at both ends are:
fn=nvw2L,n=1,2,3...,(17.5.10)
where f1 is fundamental, f2 is the overtone, f3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube, because it has even multiples of the fundamental as well as odd. A closed tube has only odd multiples.
REAL WORLD APPLICATIONS: RESONANCE IN EVERYDAY
Resonance occurs in many different systems, including strings, air columns, and atoms. Resonance is the driven or forced oscillation of a system at its natural frequency. At resonance, energy is transferred rapidly to the oscillating system, and the amplitude of its oscillations grows until the system can no longer be described by Hooke’s law. An example of this is the distorted sound intentionally produced in certain types of rock music.
Wind instruments use resonance in air columns to amplify tones made by lips or vibrating reeds. Other instruments also use air resonance in clever ways to amplify sound. Figure 17.5.12shows a violin and a guitar, both of which have sounding boxes but with different shapes, resulting in different overtone structures. The vibrating string creates a sound that resonates in the sounding box, greatly amplifying the sound and creating overtones that give the instrument its characteristic flavor. The more complex the shape of the sounding box, the greater its ability to resonate over a wide range of frequencies. The marimba, like the one shown in Figure 17.5.13uses pots or gourds below the wooden slats to amplify their tones. The resonance of the pot can be adjusted by adding water.
We have emphasized sound applications in our discussions of resonance and standing waves, but these ideas apply to any system that has wave characteristics. Vibrating strings, for example, are actually resonating and have fundamentals and overtones similar to those for air columns. More subtle are the resonances in atoms due to the wave character of their electrons. Their orbitals can be viewed as standing waves, which have a fundamental (ground state) and overtones (excited states). It is fascinating that wave characteristics apply to such a wide range of physical systems.
Exercise 17.5.1
Describe how noise-canceling headphones differ from standard headphones used to block outside sounds.
Answer
: Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds.
Exercise 17.5.2
How is it possible to use a standing wave's node and antinode to determine the length of a closed-end tube?
Answer
: When the tube resonates at its natural frequency, the wave's node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube.
PHET EXPLORATIONS: SOUND
This simulation lets you see sound waves. Adjust the frequency or volume and you can see and hear how the wave changes. Move the listener around and hear what she hears.
Summary
Sound interference and resonance have the same properties as defined for all waves.
In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics.
The resonant frequencies of a tube closed at one end are: fn=nvw4L,n=1,3,5..., f1 is fundamental and L is the length of the tube.
The resonant frequencies of a tube open at both ends are: fn=nvw2L,n=1,2,3...
Glossary
antinode
: point of maximum displacement
node
: point of zero displacement
fundamental
: the lowest-frequency resonance
overtones
: all resonant frequencies higher than the fundamental
harmonics
: the term used to refer collectively to the fundamental and its overtones
17.4: Doppler Effect and Sonic Booms
17.6: Hearing |
4190 | https://brilliant.org/wiki/spanning-trees/ | Spanning Trees
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Karleigh Moore,
Timmy Jose,
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Jimin Khim
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Spanning trees are special subgraphs of a graph that have several important properties. First, if T is a spanning tree of graph G, then T must span G, meaning T must contain every vertex in G. Second, T must be a subgraph of G. In other words, every edge that is in T must also appear in G. Third, if every edge in T also exists in G, then G is identical to T.
Spanning trees are important in path-finding algorithms such as Dijkstra's shortest path algorithm and A search algorithm. Spanning trees are calculated as sub-parts in those algorithms. It is also used in network routing protocols. The spanning tree protocol is one example.
Graph with a spanning tree highlighted in blue
Contents
Properties
Spanning Trees and Graph Types
Finding Spanning Trees
Minimum Spanning Trees
References
Properties
There are a few general properties of spanning trees.
A connected graph can have more than one spanning tree. They can have as many as ∣v∣∣v∣−2, where ∣v∣ is the number of vertices in the graph.
All possible spanning trees for a graph G have the same number of edges and vertices.
Spanning trees do not have any cycles.
Spanning trees are all minimally connected. That is, if any one edge is removed, the spanning tree will no longer be connected.
Adding any edge to the spanning tree will create a cycle. So, a spanning tree is maximally acyclic.
Spanning trees have ∣n∣−1 edges, where ∣n∣ is the number of vertices.
Spanning Trees and Graph Types
Different types of graphs have different numbers of spanning trees. Here are a few examples.
1) Complete Graphs
A complete graph is a graph where every vertex is connected to every other vertex. The number of spanning trees for a graph G with ∣v∣ vertices is defined by the following equation: T(Gcomplete)=∣v∣∣v∣−2.
Complete Graph
2) Connected Graphs
For connected graphs, spanning trees can be defined either as the minimal set of edges that connect all vertices or as the maximal set of edges that contains no cycle.
A connected graph is simply a graph that necessarily has a number of edges that is less than or equal to the number of edges in a complete graph with the same number of vertices. Therefore, the number of spanning trees for a connected graph is T(Gconnected)≤∣v∣∣v∣−2.
Connected Graph
3) Trees
If a graph G is itself a tree, the only spanning tree of G is itself. So a tree with ∣v∣ vertices is defined as T(Gtree)=∣v∣∣v∣−2.
Tree Graph
4) Complete Bipartite Graph
A bipartite graph is a graph where every node can either be associated with one of two sets, m or n. Vertices within these sets only connect to vertices in the other. There are no intra-set edges. A complete bipartite graph then is a bipartite graph where every vertex in set m is connected to every vertex in set n, and vice versa.
The number of spanning trees for a bipartite graph is defined by T(Gcomplete-bipartite)=mn−1⋅nm−1.
Complete Bipartite Graph
5) General Graph
To calculate the number of spanning trees for a general graph, a popular theorem is Kirchhoff's theorem.
To perform this theorem, a two-dimensional matrix must be constructed that can be indexed via both row and column by the graphs' vertices. The cell in the ith row and jth column has a value that is determined by three things. If i=j, then the value in the cell will be equal to the degree of i. If i and j are adjacent, then the value will be −1. Otherwise, the value will be 0.
From here, an arbitrary vertex is chosen and its corresponding row and column is removed from the matrix. The determinant of this new matrix is a spanning tree T(G).
Finding Spanning Trees
Spanning trees can be found in linear O(V+E) time by simply performing breadth-first search or depth-first search. These graph search algorithms are only dependent on the number of vertices in the graph, so they are quite fast.
Breadth-first search will use a queue to hold vertices to explore later, and depth-first search will use a stack. In either case, a spanning tree can be constructed by connecting each vertex v with the vertex that was used to discover it.
Unfortunately, these search algorithms are not well suited for parallel or distributed computing, an area in which spanning trees are popular. There are, however, algorithms that are designed to find spanning trees in a parallel setting.
For complete graphs, there is an exact number of edges that must be removed to create a spanning tree. For a complete graph G, a spanning tree can be calculated by removing ∣e∣−∣v∣+1 edges. In this equation, ∣e∣ is the number of edges, and ∣v∣ is the number of vertices.
Minimum Spanning Trees
Minimum spanning trees are a variant of the spanning tree.
A minimum spanning tree for an unweighted graph G is a spanning tree that minimizes the number of edges or edge weights.
A minimum spanning tree for a weighted graph G is a spanning tree that minimizes the weights of the edges in the tree.
These two images show the difference between a spanning tree and minimum spanning tree. The edges that are grayed out are left out of their respective trees, but they're left in the images to show their weights.
Weighted spanning tree
Weighted minimum spanning tree
Minimum spanning trees are very helpful in many applications and algorithms. They are often used in water networks, electrical grids, and computer networks. They are also used in graph problems like the traveling salesperson problem, and they are used in important algorithms such as the min-cut max-flow algorithm.
There are many ways to find the minimum spanning trees, but Kruskal's algorithm is probably the fastest and easiest to do by hand.
Find the minimum spanning tree for the graph below. What is its total weight?
Example Graph
The minimum spanning tree is shown below. Its total weight is 31.
Minimum Spanning Tree
Eppstein, D.
Spanning Trees.
Retrieved
April 10, 2016,
from
Benbennick, D.
Wikipedia Complete Graph.
Retrieved
May 21, 2016,
from
A, L.
Wikipedia Connected Graph.
Retrieved
May 21, 2016,
from
A, L.
Wikipedia Tree.
Retrieved
May 21, 2016,
from
A, K.
Wikipedia Complete Bipartite Graph.
Retrieved
May 21, 2016,
from
Cite as:
Spanning Trees.
Brilliant.org.
Retrieved
16:50, September 17, 2025,
from |
4191 | https://www.youtube.com/watch?v=ZrEJKHlcm4w | Right Triangle Definitions of Trigonometric Functions SOH CAH TOA
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Right Triangle Definitions of Trigonometric Functions SOH CAH TOA
Transcript:
hey everyone in this video we're going to define the trigonometric functions in terms of a right triangle so here we'll draw one leg of the triangle here's the hypotenuse and then here is another leg so this will be a right triangle meaning that this is 90 degrees and then here we have our angle theta so according to theta this is called the adjacent side so this is the side adjacent to theta so a a for adjacent this is the side opposite of theta so this is called opposite and so we'll use oh for opposite and this is of course the hypotenuse so hypotenuse and we'll use H for hypotenuse and so using this triangle we can define the trig functions using an ancient method so so so call Toa sohcahtoa so sine of theta is so sine is so so so it's opposite over hypotenuse okay opposite over hypotenuse that means that the reciprocal function which is cosecant of theta it's hoe so H over Oh cosine of theta so it's sohcahtoa so it's a over H a over H and then secant is the reciprocal so it's ha and then tangent so it's so called Toa Oh wha so it's over a and so cotangent is aoao so a over o so pretty easy to memorize this extremely easy I think so it's so caught oh ah so really really simple very very very useful let's go ahead and do a really simple example so you see how this works so we'll be given a right triangle and we'll be asked we'll find the the trig function values of theta so here's theta here's the number 28 35 and save 45 and based off of this we can write down all of the trig functions so let's see sine of theta so let me label everything this is going to be the adjacent side this is the opposite side and this one here is the hypotenuse so sohcahtoa so over H so would be 35 over 45 and if you simplify this you just get 7 over 9 okay then the reciprocal of sine is cosecant it's the one that starts with the other letter so I'm just going to flip it and just do 9 over 7 ok so two down let's do cosine no so cosine so it's so cool so it's a over H so a is 28 and then H is 45 okay and the reciprocal of this would be the secant function so secant of theta would be 45 over 28 and finally we can do tangent so sohcahtoa so tangent is o over a so o is 35 and a is 28 and the reciprocal of tangent is simply cotangent so that would be you just flip it so you get 28 over 35 so just a quick video to show you probably one of the most useful things that you can learn in trig sohcahtoa so kah Toa super super powerful stuff hope that helped |
4192 | https://www.wyzant.com/resources/answers/29153/use_the_angle_sum_or_difference_identity_to_find_the_exact_value_of_sin_15_degrees | use the angle sum or difference identity to find the exact value of sin 15 degrees | Wyzant Ask An Expert
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Algebra 2Trigonometry
Steven P.
asked • 03/10/14
use the angle sum or difference identity to find the exact value of sin 15 degrees
I don't understand what to do after i the two numbers like (45-30)
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Steve S.answered • 03/11/14
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Find the EXACT value of sin(15°) using a sum or difference formula.
In this solution the following will be strictly adhered to:
Trig Functions are FUNCTIONS and proper functional notation should always be used.
Parentheses will be used to create non-ambiguous expressions.
The value will be expressed as requested; i.e., in EXACT form (NO CALCULATORS WILL BE USED).
sin(15°) = sin(45° – 30°)
= sin(45°)cos(30°) – cos(45°)sin(30°)
Recall the two Special Triangles: 45°-45°-90° and 30°-60°-90°.
Their sides across from the angles are in the ratios: 1:1:√(2) & 1:√(3):2.
Draw and label diagrams of the two Special Triangles to guide you.
sin(15°) = sin(45°)cos(30°) – cos(45°)sin(30°)
= (1/√(2))(√(3)/2) – (1/√(2))(1/2)
= (1/√(2)) ( (√(3)/2) – (1/2) )
= (√(2)/2) ( (√(3) – 1)/2) )
= ( √(6) – √(2) )/4
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Amarjeet K.answered • 03/10/14
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sin(45-30) = sin45cos30 - cos45sin30 = (1/√2 x √3/2) - (1/2 x 1/2) = √3/2√2 - 1/4
:)
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Mozhgan Z.answered • 03/10/14
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sin(15)=sin(45-30)
knowing the following equation:
sin(α – β) = sin(α)cos(β) – cos(α)sin(β)
we can say:
sin(45-30)=sin45cos30 - cos45sin30
sin45=cos45=2^0.5/2
sin30=1/2, cos30=3^0.5/2
plugging in these values:
sin(45-30)=(2^0.5/23^0.5/2)-(2^0.5/21/2)
doing the calculations:
sin(15) = 0.259
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Steve S.
Problem asked for "exact value". A calculator only gives approximate values.
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4193 | https://scienceandculture.com/2024/05/astounding-the-role-of-motor-proteins-in-cell-division/ | Astounding: Motor Proteins in Cell Division | Science and Culture Today
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Astounding: The Role of Motor Proteins in Cell Division
Jonathan McLatchieMay 8, 2024Engineering, Intelligent Design, Life Sciences8 CategoriesEngineeringIntelligent DesignLife Sciences
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One of the most astounding examples of molecular machinery in the cell is the mitotic spindle that governs the segregation of chromosomes during cell division. You can watch the astonishing process of mitotic cell division unfold in this animation:
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A crucial aspect of mitosis is the movement of chromosomes — initially the alignment of as paired sister chromatids during prometaphase and metaphase, and then the separation of these sister chromatids during anaphase. But how does the spindle exert force on the chromosomes? And how do the chromosomes attach to the spindle such that they can be moved to the proper place at the proper time? Here, I will give an introduction to this astounding process.
Microtubules
Two structural components make up the mitotic spindle — two large structures, found at the poles, called centrosomes, and protein filaments known as microtubules. From each of the centrosomes radiate the spindle microtubules, which bind to a chromosome (kinetochore microtubules), or a microtubule emanating from the other centrosome (polar microtubules), or the cell’s plasma membrane (astral microtubules).
The subunit of microtubules are heterodimers of tubulin, comprised of α-tubulin and β-tubulin. These heterodimers fit together to create linear protofilaments, 13 of which are aligned side-by-side to form a microtubule. Due to the asymmetry of tubulin heterodimers, the microtubule has intrinsic polarity. The plus end (which extends towards the chromosomes) is comprised of β-tubulin subunits, whereas the minus end (which is anchored at the centrosome) is comprised of α-tubulin subunits.
In the early mitotic stages, microtubules assemble and disassemble rapidly (mediated by the addition and removal of tubulin heterodimers), a phenomenon known as dynamic instability (the processes of depolymerization and polymerization are referred to as catastrophe and rescue respectively). By this process, microtubules randomly probe the cell, until they by chance encounter their target. The regulation of microtubule catastrophe and rescue is linked to the hydrolysis of guanosine triphosphate (GTP), a nucleotide bound to tubulin within the microtubule lattice. Each tubulin subunit is able to bind to one molecule of GTP. During polymerization, tubulin subunits bound to GTP are added to the growing end of the microtubule, forming a GTP cap which stabilizes the microtubule and promotes growth.
As the microtubule continues to polymerize, the GTP-bound tubulin subunits undergo hydrolysis, converting GTP to guanosine diphosphate (GDP). Once hydrolyzed, the tubulin subunits become less stable, and the microtubule is more prone to depolymerization. Upon reaching a critical concentration of GDP-bound tubulin at the microtubule end, catastrophe occurs, leading to rapid depolymerization. Microtubule rescue involves the exchange of GDP-bound tubulin for GTP-bound tubulin, thereby promoting microtubule growth. Certain cellular factors, such as microtubule-associated proteins (MAPs) and motor proteins, can facilitate this process by promoting the incorporation of GTP-bound tubulin or by protecting the GTP cap from hydrolysis. For example, kinesin-8 proteins are plus-end directed motor proteins that destabilize microtubules.1 1 Shrestha S, Hazelbaker M, Yount AL, Walczak CE. Emerging Insights into the Function of Kinesin-8 Proteins in Microtubule Length Regulation. Biomolecules. 2018 Dec 20;9(1):1.Another motor protein, kinesin-13, is bidirectional (being able to move in the direction of either the plus or minus ends of microtubules), in contrast to most motors that are unidirectional. They are thereby able to promote the depolymerization of a microtubule from both ends.2 2 Ems-McClung SC, Walczak CE. Kinesin-13s in mitosis: Key players in the spatial and temporal organization of spindle microtubules. Semin Cell Dev Biol. 2010 May;21(3):276-82.The activity of kinesin-13 proteins is “regulated by distinct targeting to regions of the spindle, by regulatory phosphorylation events, and by interactions with different binding partners.”3 3 Ibid.These remarkable kinesin-13 motors bind the end of the microtubule and trigger a conformational change that results in microtubule depolymerization.4 4 Desai A, Verma S, Mitchison TJ, Walczak CE. Kin I kinesins are microtubule-destabilizing enzymes.Cell. 1999 Jan 8;96(1):69-78.
Motor Proteins
In 2014, Discovery Institute published an animation of the molecular motor kinesin, which carries cellular cargo along microtubule tracks and is driven by ATP hydrolysis:
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Kinesin typically moves in the direction of the plus end of the microtubule, towards the periphery of the cell. Another form of motor protein, called dynein, generally moves towards the minus end of the microtubule, i.e., towards the cell center. Whereas kinesin motor proteins walk step-by-step by placing one foot in front of the other, dynein moves by a swinging crossbridge mechanism.
One of the most incredible functions of motor proteins in the cell is their role in the assembly and function of the mitotic spindle during eukaryotic cell division.5 5 Goshima G, Vale RD. The roles of microtubule-based motor proteins in mitosis: comprehensive RNAi analysis in the Drosophila S2 cell line.J Cell Biol. 2003 Sep 15;162(6): 1003-16.Envision a robotic factory that assembles and organizes the cell in preparation for undergoing division, helping to facilitate the controlled segregation of the genetic material into the daughter cells. This is represented in the (highly simplified) figure below:
As shown in the figure, the motor proteins involved in the organization of the mitotic spindle can be divided into four major classes: kinesin-5, kinesin-14, kinesins-4 and 10, and dynein. Kinesin-5 is directed towards the plus end and slides apart those microtubules that have their polarities oriented in opposite directions (i.e., antiparallel microtubules). Thus, kinesin-5 contributes to spindle elongation and bipolarity by pushing apart the spindle poles.6 6 Ferenz NP, Gable A, Wadsworth P. Mitotic functions of kinesin-5. Semin Cell Dev Biol. 2010 May;21(3): 255-9.On the other hand, kinesin-14 motors are directed towards the minus-end of microtubules. They each possess one motor domain in addition to other domains that associate with a different microtubule.7 7 Mitra A, Meißner L, Gandhimathi R, Renger R, Ruhnow F, Diez S. Kinesin-14 motors drive a right-handed helical motion of antiparallel microtubules around each other.Nat Commun. 2020 May 22;11(1): 2565.They cross-link antiparallel interpolar microtubules at the spindle midzone, and thereby pull the poles towards one another.
Another group of kinesins, kinesin-4 and kinesin-10 (collectively known as chromokinesins), play a critical role in the positioning and segregation of chromosomes during mitosis.8 8 Mazumdar M and Misteli T. Chromokinesins: multitalented players in mitosis.Trends Cell Biol. 2005 Jul;15(7): 349-55.The job of chromokinesins is to move the chromosomes to their proper positions such that each daughter cell will receive the right number of chromosomes.
The last group of motor proteins involved in mitosis are dyneins that have their motor domains associated with the microtubules, which emanate from the centrosomes, and their cargo-binding domains bound to proteins embedded in the cell cortex. The movement of dynein exerts a pulling force on the microtubule minus ends and pulls the centrosomes in the direction of the cell cortex.9 9 Raaijmakers JA and Medema RH (2014). Function and regulation of dynein in mitotic chromosome segregation.Chromosoma. 123(5): 407-22.This movement ensures that each daughter cell receives a complete set of chromosomes. Note that, although I have, for simplicity, only shown one microtubule emanating from each centrosome towards the cell cortex, in real life there would be many more than this.
Incredible Engineering
In any other realm of experience, we would, upon observing a robotic factory like this that operates at a nanoscale, immediately ascribe it to design. The engineering of the eukaryotic cell division cycle is astounding at every level. Moreover, these molecular motors are absolutely indispensable to successful cell division in eukaryotes and thus constitute an irreducibly complex system. It therefore points forcefully and unequivocally to conscious intent.
Notes
Shrestha S, Hazelbaker M, Yount AL, Walczak CE. Emerging Insights into the Function of Kinesin-8 Proteins in Microtubule Length Regulation. Biomolecules. 2018 Dec 20;9(1):1.↩︎
Ems-McClung SC, Walczak CE. Kinesin-13s in mitosis: Key players in the spatial and temporal organization of spindle microtubules. Semin Cell Dev Biol. 2010 May;21(3):276-82.↩︎
Ibid.↩︎
Desai A, Verma S, Mitchison TJ, Walczak CE. Kin I kinesins are microtubule-destabilizing enzymes.Cell. 1999 Jan 8;96(1):69-78.↩︎
Goshima G, Vale RD. The roles of microtubule-based motor proteins in mitosis: comprehensive RNAi analysis in the Drosophila S2 cell line.J Cell Biol. 2003 Sep 15;162(6): 1003-16.↩︎
Ferenz NP, Gable A, Wadsworth P. Mitotic functions of kinesin-5. Semin Cell Dev Biol. 2010 May;21(3): 255-9.↩︎
Mitra A, Meißner L, Gandhimathi R, Renger R, Ruhnow F, Diez S. Kinesin-14 motors drive a right-handed helical motion of antiparallel microtubules around each other.Nat Commun. 2020 May 22;11(1): 2565.↩︎
Mazumdar M and Misteli T. Chromokinesins: multitalented players in mitosis.Trends Cell Biol. 2005 Jul;15(7): 349-55.↩︎
Raaijmakers JA and Medema RH (2014). Function and regulation of dynein in mitotic chromosome segregation.Chromosoma. 123(5): 407-22.↩︎
Jonathan McLatchie
Resident Biologist and Fellow, Center for Science and Culture Dr. Jonathan McLatchie holds a Bachelor's degree in Forensic Biology from the University of Strathclyde, a Masters (M.Res) degree in Evolutionary Biology from the University of Glasgow, a second Master's degree in Medical and Molecular Bioscience from Newcastle University, and a PhD in Evolutionary Biology from Newcastle University. Previously, Jonathan was an assistant professor of biology at Sattler College in Boston, Massachusetts. Jonathan has been interviewed on podcasts and radio shows including "Unbelievable?" on Premier Christian Radio, and many others. Jonathan has spoken internationally in Europe, North America, South Africa and Asia promoting the evidence of design in nature.Follow JonathanProfileWebsite
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irreducibly complex systems
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Article Outline
Microtubules
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Incredible Engineering
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4194 | https://www.busymac.com/docs/busycal/70594-tasks/ | Tasks | Busy Apps by Beehive
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Tasks
BusyCal can manage all your tasks (also known as to dos or reminders). Tasks can be displayed in a Task List in the right sidebar and/or in the calendar on the date they are due.
Tasks come in three flavors:
Undated Tasks — Tasks without a due date are displayed in the Task List only.
Dated Tasks — Tasks with a due date can be displayed in the calendar, in the Task List, or both.
Timed Tasks — Tasks with a due date and time can be displayed in the calendar, in the Task List, or both.
Task List
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To show the Task List, choose View > Show Task List, press ⌘-Option-T, or click the Hide/Show Task List button on the right side of the toolbar and choose either Show Task List or Show Both from the pop-up menu. The Task List appears in the right sidebar.
The Task List can be sorted by Due Date, Priority, Title, Calendar, Tag, or Manually. To change the current sort order and other Task List settings, use the pop-up menu at the top of the Task List, or go to BusyCal > Settings > Tasks.
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You can show dated tasks, timed tasks, and completed tasks in the calendar on the date they are due or completed. If a task is not completed on its due date, it carries forward each day until completed (this behaviour can be changed from Settings > Tasks uncheck Carry forward uncompleted Tasks). (If dated and completed tasks don’t appear in your calendar, see Task Settings, just ahead.)
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Timed tasks are sorted inline with calendar events (e.g. a timed task with a due time of 9:00 AM will appear sandwiched in between events at 8:00 AM and 10:00 AM). Dated tasks are grouped together below events. Tasks that occur at the same date or time are sorted by priority and title. Completed tasks are sorted by completion time.
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The Task List supports manual sorting within each sorted section, giving you the flexibility to arrange tasks in your preferred order.
If your tasks aren’t appearing in the expected sort order, it’s possible they were manually reordered at some point.
To reset the manual order:
Select all tasks (click one task, then press CMD+A).
Control-click (right-click) any selected task and choose Reset Manual Task Ordering… from the context menu.
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Dated Tasks — To create a new dated task, choose File > New Task, press ⌘-K, or Control-click on a day and choose New Task from the contextual menu. A dated task appears in your calendar on the due date, with a checkbox in front of it.
Timed Tasks — To create a new timed task, choose File > New Timed Task, press ⌘-Option-K, or Control-click on a day and choose New Timed Task from the contextual menu. A timed task appears in your calendar on the due date and time, with a checkbox in front of it.
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To create a timed task in Week view, hold down ⌘ while double-clicking in the calendar.
Undated Tasks — To create a new undated task, choose File > New Undated Task, press ⌘-U, or Control-click in the Task List sidebar and choose New Undated Task from the contextual menu.
You can also create tasks using Quick Entry in the app or from the BusyCal Menu by preceding the text with a dash -, or the words todo, remember to, or remind me to. If a date or time is included, BusyCal creates a dated or timed task. Otherwise, an undated task is created. For example:
-Call Bobtodo Call Bob tomorrow-Call Bob tomorrow at 10amremember to Call Bob next Tuesday-Call Bob Nov 7remind me to pick up dry cleaning at 5pm
If you mark a task as completed by checking the “done” checkbox in the calendar or Info Panel (or pressing ⌘-/), the completed task appears with a checkmark in front of it on the date that it was completed.
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To customize how tasks are created and displayed, choose BusyCal > Settings > Tasks.
The following options are available for customizing tasks:
Default list for new Tasks — The default calendar / task-list new tasks are created in.
Default priority — The default priority assigned to newly created tasks.
Sort Task List by — The sort order for the Task List. Sorting can also be changed directly from the Task Panel, by clicking on the sorting drop-down button.
Show dated Tasks — The date range for tasks displayed in the Task List. The default displays tasks due by next month. You can select a different range if you want to view fewer or more tasks.
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Duration in day & week view — Default duration to use for tasks when a custom duration is not assigned.
Show dated Tasks in calendar — Whether to display dated tasks in the calendar on their due dates.
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Carry forward uncompleted Tasks — Whether to carry forward past due tasks and display them on the current date. Such Tasks appear with a small > arrow displayed in the middle of the checkbox in Month / Week / Day view.
Repeating Tasks
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You can create tasks that repeat based on either the due date (e.g., Pay Rent on the 1st of every month) or on the completion date of the previous task (e.g., Oil Change 3 months after completion of the last Oil Change).
Please note - unlike a repeating event, repeating Tasks do not display future occurrences on the calendar. A new occurrence is created only after you mark a Task complete.
To create a repeating task that repeats based on the due date (e.g., Pay Rent on the 1st of every month), select a repeat interval from the “repeat” pop-up menu on the Info Panel.
You can change the default regeneration / repeating style from under Settings > Tasks > Default Regeneration Style. See Tasks Settings for more information.
Also see Repeating Events for additional information on creating custom repeat intervals.
Completed occurrences of a repeating task appear in the calendar, but only the next uncompleted occurrence of a repeating task appears. When you complete an occurrence of a repeating task, the next occurrence of the repeating task is created automatically.
For example, you could create a task on Thursday to Take out trash that repeats weekly. Only the next uncompleted occurrence of the task appears in the calendar. When you complete it (or delete it), the next occurrence of the task is created on the following Thursday.
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When you mark a recurring task as completed, BusyCal sets the completed date as today (even if the event appears on an earlier date on your calendar) and then creates the next occurrence. Depending on your settings in BusyCal > Settings > Tasks, this may cause the task to disappear from your calendar on its earlier date, even if your intention was to indicate that you completed it on a date in the past. BusyCal’s default logic is based on the Change Oil model: if you plan to change your oil every month, and you skip a month or two, BusyCal ignores the repeats that would have occurred in the past and schedules the next occurrence forward from the date on which you mark the task as completed. You can change this behavior from Settings > Tasks > Default regeneration style.
Converting Tasks to Events
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You can convert tasks to events by ⌘-dragging them from the Task List to the calendar:
Hold down ⌘ while dragging a task from the Task List to the calendar in Month view to convert it to an event on that date, at the current time.
Hold down ⌘ while dragging a task from the Task List to the all-day portion of a day in Day or Week view to convert it to an all-day event.
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You can also drag an undated task to the calendar to set the due date, or drag it to a different date to change the due date.
Converting Events to Tasks
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You can convert events to tasks by simply dragging them from the calendar and dropping them into the Task List while at least one Task list calendar has been checked in the left panel (note: you should not hold down the ⌘ key when doing this).
iCloud and Reminders
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Tasks in BusyCal sync with the Reminders app in macOS and iOS through iCloud.
iCloud calendars are type-specific. Each iCloud calendar can contain either events or tasks, but not both. BusyCal indicates a task calendar by displaying a checkmark next to its name. The task calendars in BusyCal sync with the Reminder Lists in the Reminders app on macOS and iOS through iCloud.
Tasks with Location Alarms
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The Reminders app on iOS can display location alarms when arriving at or leaving a specified location. For example, you could create a task to “Pick up dry cleaning” that triggers when you leave the office. BusyCal supports location alarms that sync with the Reminders app, and you can even create and edit location alarms in BusyCal.
To create a location alarm in BusyCal, create a task with a custom Alarm Type of Location, click Location, and select a contact or search for a location using the embedded Google Maps interface.
Location alarms trigger on your iOS device each time you arrive at or leave a location until you complete the task or delete it. In addition, you can set a specific date and time to be reminded on both your Mac and iOS device, regardless of whether you reach the location.
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BusyCal supports location alarms only on iCloud task calendars.
Last updated on May 22, 2025
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4195 | https://testbook.com/question-answer/what-is-the-maximum-number-of-different-functions--68468efc5999e6a0b11f568e | [Solved] What is the maximum number of different functions involving
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What is the maximum number of different functions involving n Boolean variables?
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n 2
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2 n 2
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Answer (Detailed Solution Below)
Option 4 : 2 2 n
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The correct answer is: Option 4) 22n
Explanation:
For n Boolean variables, each variable can be either 0 or 1.
So, the total number of unique combinations of input values is:
2 n
Each of these input combinations can be mapped to either output 0 or 1 in a Boolean function.
So for each of the 2 n input combinations, we can choose the output in 2 ways (0 or 1).
Therefore, the total number of possible functions is:
2(2 n)
This is because we are choosing a value (0 or 1) for each of the 2 n input combinations, which gives us:
2 × 2 × 2 × ... (2 n times) = 2 2 n
✅ Hence, the correct answer is: Option 4
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More Boolean Functions Questions
Q1.What is the maximum number of different functions involving n Boolean variables?
Q2.Which combination of symbols represents an XOR (Exclusive-OR) gate?
Q3.The XOR operator is NOT
Q4.A = a1a0 and B = b1b0 are two 2-bit unsigned binary numbers. If F(a1, a0, b1, b0)is a Boolean function such that 𝐹 = 1 only when 𝐴 > 𝐵, and 𝐹 = 0 otherwise, then 𝐹 can be minimized to the form ______
Q5.Simplify the following Boolean function in sum of minterm: F(A, B, C, D) = Σ(0, 6, 8, 13, 14) d (A, B, C, D) = Σ(2, 4, 10) where, d stands for don't care condition
Q6.Simplify the following Boolean function: f (A, B,C) =A―B C+A B―C+A B C―+ ABC
Q7.In Boolean algebra A + (B.C) = ?
Q8.How many Boolean function is one variable are NOT idempotent: i.e. they do not satisfy ∀a.f(f(a)) = f(a).
Q9.How many different Boolean circuits of n variables are there?
More Logic Gates and Boolean Algebra Questions
Q1.Which of the following gate is known as building block in digital circuitry? I. NAND II. NOR
Q2.According to Boolean Theorems which of the following is correct? I. X.(Y + Z) = (X.Y) + (X.Z) II. X + X = X
Q3.A Karnaugh map (K-map) is an abstract form of a ___ organized as a matrix of squares.
Q4.Which of the following represents the logic function F(x, y, z) in the figure? The min terms are derived using x as MSB and z as LSB.
Q5.The expression (a.b)c' + (a.b')c + (a'.b)c was to be realised with 2-input AND gates and OR gates. However during realization all 2-input AND gates were mistakenly substituted by 2-input NAND gates. What is the function finally obtained ?
Q6.How would you implement the Boolean expression A + B̅ using basic logic gates?
Q7.What is the maximum number of different functions involving n Boolean variables?
Q8.Which of the following is a universal gate?
Q9.The resulting logic circuit obtained when both inputs of NOR and NAND gates are connected together is:
Q10.The circuit behaves as a:
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More Digital Electronics Questions
Q1.A decoder is a combinational circuit that converts binary information from n input lines to a maximum of _ unique output lines.
Q2.Which of the following statement is true regarding register? I. A register is a group of memory elements which stores a binary word and it may modify the stored word in a particular fashion as is desired by the application in which it is used. II. It is capable of shifting the stored binary word a step or more towards left or right.
Q3.Which of the following gate is known as building block in digital circuitry? I. NAND II. NOR
Q4.What is the octal equivalent of (15)10?
Q5.A combinational circuit that performs the addition of _ bits is called as half-adder.
Q6.What is the decimal equivalent of (1100)2?
Q7.According to Boolean Theorems which of the following is correct? I. X.(Y + Z) = (X.Y) + (X.Z) II. X + X = X
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Integral Calculus
Course: Integral Calculus>Unit 1
Lesson 19: Improper integrals
Introduction to improper integrals
Divergent improper integral
Improper integrals
Improper integrals review
Improper integral with two infinite bounds
Math>
Integral Calculus>
Integrals>
Improper integrals
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Improper integrals
AP.CALC: LIM‑6 (EU), LIM‑6.A (LO), LIM‑6.A.1 (EK), LIM‑6.A.2 (EK)
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Problem
Evaluate the improper integral if it exists.
∫1∞1 x d x
Choose 1 answer:
Choose 1 answer:
(Choice A) 1 A 1
(Choice B) 2 B 2
(Choice C) e C e
(Choice D) The improper integral diverges. D The improper integral diverges.
Related content
Video 3 minutes 52 seconds 3:52 Introduction to improper integrals
Video 2 minutes 42 seconds 2:42 Divergent improper integral
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4198 | https://en.wikipedia.org/wiki/Finger_binary | Published Time: 2005-07-19T16:49:51Z
Finger binary - Wikipedia
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1 MechanicsToggle Mechanics subsection
1.1 Examples
1.1.1 Right hand
1.1.2 Left hand
2 Negative numbers and non-integersToggle Negative numbers and non-integers subsection
2.1 Negative numbers
2.2 Fractions
2.2.1 Dyadic fractions
2.2.2 Rational numbers
2.3 Decimal fractions and vulgar fractions
3 Finger ternary
4 See also
5 References
6 External links
Finger binary
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From Wikipedia, the free encyclopedia
Finger-counting system
This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations.(January 2009) (Learn how and when to remove this message)
19 in finger binary: the pinkie finger is 16, added to the 2 of the index finger and the 1 of the thumb
Finger binary is a system for counting and displaying binary numbers on the fingers of either or both hands. Each finger represents one binary digit or bit. This allows counting from zero to 31 using the fingers of one hand, or 1023 using both: that is, up to 2 5−1 or 2 10−1 respectively.
Modern computers typically store values as some whole number of 8-bitbytes, making the fingers of both hands together equivalent to 1¼ bytes of storage—in contrast to less than half a byte when using ten fingers to count up to 10.
Mechanics
[edit]
Further information: Binary numeral system
In the binary number system, each numerical digit has two possible states (0 or 1) and each successive digit represents an increasing power of two.
Note: What follows is but one of several possible schemes for assigning the values 1, 2, 4, 8, 16, etc. to fingers, not necessarily the best. (see below the illustrations.): The rightmost digit represents two to the zeroth power (i.e., it is the "ones digit"); the digit to its left represents two to the first power (the "twos digit"); the next digit to the left represents two to the second power (the "fours digit"); and so on. (The decimal number system is essentially the same, only that powers of ten are used: "ones digit", "tens digit" "hundreds digit", etc.)
It is possible to use anatomical digits to represent numerical digits by using a raised finger to represent a binary digit in the "1" state and a lowered finger to represent it in the "0" state. Each successive finger represents a higher power of two.
With palms oriented toward the counter's face, the values for when only the right hand is used are:
| | Pinky | Ring | Middle | Index | Thumb |
--- --- --- |
| Power of two | 2 4 | 2 3 | 2 2 | 2 1 | 2 0 |
| Value | 16 | 8 | 4 | 2 | 1 |
When only the left hand is used:
| | Thumb | Index | Middle | Ring | Pinky |
--- --- --- |
| Power of two | 2 4 | 2 3 | 2 2 | 2 1 | 2 0 |
| Value | 16 | 8 | 4 | 2 | 1 |
When both hands are used:
| | Left hand | Right hand |
---
| Thumb | Index | Middle | Ring | Pinky | Pinky | Ring | Middle | Index | Thumb |
| Power of two | 2 9 | 2 8 | 2 7 | 2 6 | 2 5 | 2 4 | 2 3 | 2 2 | 2 1 | 2 0 |
| Value | 512 | 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
And, alternately, with the palms oriented away from the counter:
| | Left hand | Right hand |
---
| Pinky | Ring | Middle | Index | Thumb | Thumb | Index | Middle | Ring | Pinky |
| Power of two | 2 9 | 2 8 | 2 7 | 2 6 | 2 5 | 2 4 | 2 3 | 2 2 | 2 1 | 2 0 |
| Value | 512 | 256 | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
The values of each raised finger are added together to arrive at a total number. In the one-handed version, all fingers raised is thus 31 (16 + 8 + 4 + 2 + 1), and all fingers lowered (a fist) is 0. In the two-handed system, all fingers raised is 1,023 (512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1) and two fists (no fingers raised) represents 0.
It is also possible to have each hand represent an independent number between 0 and 31; this can be used to represent various types of paired numbers, such as month and day, X-Y coordinates, or sports scores (such as for table tennis or baseball). Showing the time as hours and minutes is possible using 10 fingers, with the hour using 4 fingers (0-23) and the minutes using 6 fingers (0-59).
Examples
[edit]
Right hand
[edit]
0 = empty sum
1 = 1
2 = 2
4 = 4
6 = 4 + 2
7 = 4 + 2 + 1
14 = 8 + 4 + 2
16 = 16
19 = 16 + 2 + 1
26 = 16 + 8 + 2
28 = 16 + 8 + 4
30 = 16 + 8 + 4 + 2
31 = 16 + 8 + 4 + 2 + 1
Left hand
[edit]
When used in addition to the right.
512 = 512
256 = 256
768 = 512 + 256
448 = 256 + 128 + 64
544 = 512 + 32
480 = 256 + 128 + 64 + 32
992 = 512 + 256 + 128 + 64 + 32
Negative numbers and non-integers
[edit]
Further information: Binary numeral system §Representing real numbers
Just as fractional and negative numbers can be represented in binary, they can be represented in finger binary.
Negative numbers
[edit]
Representing negative numbers is extremely simple, by using the leftmost finger as a sign bit: raised means the number is negative, in a sign-magnitude system. Anywhere between −511 and +511 can be represented this way, using two hands. Note that, in this system, both a positive and a negative zero may be represented.
If a convention were reached on palm up/palm down or fingers pointing up/down representing positive/negative, you could maintain 2 10 −1 in both positive and negative numbers (−1,023 to +1023, with positive and negative zero still represented).
Fractions
[edit]
Dyadic fractions
[edit]
Fractions can be stored natively in a binary format by having each finger represent a fractional power of two: 1 2 x{\displaystyle {\tfrac {1}{2^{x}}}}. (These are known as dyadic fractions.)
Using the left hand only:
| | Pinky | Ring | Middle | Index | Thumb |
--- --- --- |
| Value | 1/2 | 1/4 | 1/8 | 1/16 | 1/32 |
Using two hands:
| Left hand | Right hand |
--- |
| Pinky | Ring | Middle | Index | Thumb | Thumb | Index | Middle | Ring | Pinky |
| 1/2 | 1/4 | 1/8 | 1/16 | 1/32 | 1/64 | 1/128 | 1/256 | 1/512 | 1/1024 |
3/4, in fractional finger binary
The total is calculated by adding all the values in the same way as regular (non-fractional) finger binary, then dividing by the largest fractional power being used (32 for one-handed fractional binary, 1024 for two-handed), and simplifying the fraction as necessary.
For example, with thumb and index finger raised on the left hand and no fingers raised on the right hand, this is (512 + 256)/1024 = 768/1024 = 3/4. If using only one hand (left or right), it would be (16 + 8)/32 = 24/32 = 3/4 also.
The simplification process can itself be greatly simplified by performing a bit shift operation: all digits to the right of the rightmost raised finger (i.e., all trailing zeros) are discarded and the rightmost raised finger is treated as the ones digit. The digits are added together using their now-shifted values to determine the numerator and the rightmost finger's original value is used to determine the denominator.
For instance, if the thumb and index finger on the left hand are the only raised digits, the rightmost raised finger (the index finger) becomes "1". The thumb, to its immediate left, is now the 2s digit; added together, they equal 3. The index finger's original value (1/4) determines the denominator: the result is 3/4.
Rational numbers
[edit]
Combined integer and fractional values (i.e., rational numbers) can be represented by setting a radix point somewhere between two fingers (for instance, between the left and right pinkies). All digits to the left of the radix point are integers; those to the right are fractional.
Decimal fractions and vulgar fractions
[edit]
Dyadic fractions, explained above, have limited use in a society based around decimal figures. A simple non-dyadic fraction such as 1/3 can be approximated as 341/1024 (0.3330078125), but the conversion between dyadic and decimal (0.333) or vulgar (1/3) forms is complicated.
Instead, either decimal or vulgar fractions can be represented natively in finger binary. Decimal fractions can be represented by using regular integer binary methods and dividing the result by 10, 100, 1000, or some other power of ten. Numbers between 0 and 102.3, 10.23, 1.023, etc. can be represented this way, in increments of 0.1, 0.01, 0.001, etc.
Vulgar fractions can be represented by using one hand to represent the numerator and one hand to represent the denominator; a spectrum of rational numbers can be represented this way, ranging from 1/31 to 31/1 (as well as 0).
Finger ternary
[edit]
In theory, it is possible to use other positions of the fingers to represent more than two states (0 and 1); for instance, a ternary numeral system (base 3) could be used by having a fully raised finger represent 2, fully lowered represent 0, and "curled" (half-lowered) represent 1. This would make it possible to count up to 242 (3 5−1) on one hand or 59,048 (3 10−1) on two hands. In practice, however, many people will find it difficult to hold all fingers independently (especially the middle and ring fingers) in more than two distinct positions.
See also
[edit]
Chisanbop
Senary §Finger counting
References
[edit]
^Since computers typically store data in a minimum size of one whole byte, fractions of a byte are used here only for comparison.
Pohl, Frederik (2003). Chasing Science (reprint, illustrated ed.). Macmillan. p.304. ISBN978-0-7653-0829-0.
Pohl, Frederik (1976). The Best of Frederik Pohl. Sidgwick & Jackson. p.363. ISBN978-0-283-98341-2.
Fahnestock, James D. (1959). Computers and how They Work. Ziff-Davis Pub. Co. p.228.
External links
[edit]
Binary Counting
| hide v t e Gestures |
| Friendly gestures | Air kiss Applause Biker wave Cheek kiss Dap Elbow bump Eskimo kiss Finger heart Fist bump Forehead kiss Hand heart Handshake Hand wave High five Hongi ILY sign Kiss Liberian snap handshake Lip kiss Namaste OK Pinky swear Pound hug Shaka Thumb signal |
| Gestures of respect | Adab Añjali Mudrā Bow Canonical digits Curtsy Fist-and-palm Gadaw Genuflection Hand-kiss Hat tip Kowtow Kuji-in Mano Mudra Namaste Ojigi Orans Pranāma Prostration Sampeah Sembah Schwurhand Wai Zolgokh |
| Salutes | Bellamy salute Nazi salute Raised fist Roman salute Scout sign and salute Three-finger salute (Serbian) Three-finger salute (pro-democracy) Two-finger salute Vulcan salute Wolf salute Zogist salute |
| Celebratory gestures | Applause Crossed hands Fist pump High five Low five Victory clasp V sign |
| Finger-counting | Chinese number gestures Chisanbop Finger binary |
| Obscene gestures | Anasyrma Bras d'honneur Fig sign Middle finger Mooning Mountza Nazi salute Reversed V sign Shocker Sign of the horns Thumb/index-finger ring Wanker |
| Taunts | Akanbe Loser Talk to the hand |
| Head motions | Head bobble Head shake Nod |
| Other gestures | Air quotes Allergic salute Aussie salute Awkward turtle Che vuoi? Crossed fingers Distress signal Duterte fist Eyelid pull Facepalm Finger gun Gang signal Hand-in-waistcoat Hand rubbing Jazz hands Laban sign Merkel-Raute Pointing Pollice verso Shrug Sign of the cross Lesser sign of the cross |
| Related | List of gestures Articulatory gestures Finger pinching conspiracy theory Hand signals Manual communication Mudra Nonverbal communication Sign language |
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Engage NY Grade 2 Module 5 Supplemental Printable and Digital Resource
Rated 4.91 out of 5, based on 177 reviews
4.9(177 ratings)
$11.00
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"Your Eureka Math products are so helpful! After a math lesson, my students need some extra practice and support with several concepts. These work perfectly! Thank you. "
Letitia E.
"One of my favorite resources! I use this along Eureka. Perfect for extra practice. I love how it states what lessons it aligns with. Great resource."
Diana D.
See 175 more reviews
Description
This resource is a must have if you are using the Grade 2 Engage NY Module 5 program. Module 5 focuses on Addition and Subtraction within 1,000 with Word Problems to 100. There are 20 lessons covered in 90pages. This resource is ready to Print-N-Go or use Digitally and is perfect to review key concepts, for homework, or small group instruction all in a kid friendly format.
This resource helps students learn the key concepts of addition and subtraction with regrouping using hands on elements. Children will learn how to use place value disks, and solve problems on a place value mat. Also contains Place Value Mat, Mid-Module Review and End of Module Assessment. Finally, a kid-friendly ready to print resource to use with Engage NY Math!
Check out the preview to see a sample of what is included.
This resource has 2 digital versions available.
You can access the TPT Digital Overlay which is located underneath the printable download. This version contains answer boxes and you can assign it directly to your Google Classroom.
This resource also contains a link in the printable version to download in Google Slides. You can add text boxes and directions to fit your specific needs.
If you try one of these Supplemental Resources I guarantee you will want them all!
TEACHERS LIKE YOU SAID...
⭐️⭐️⭐️⭐️⭐️Kimberly V. said, "Lori Flaglor provides the top notch product to correspond to Eureka Engage NY math program. Her pieces provide explanations that help out parents and kiddos to understand the strategies. The basic Eureka program is so dry and boring and provides NO SUPPORT but with these products allow us to support all our learners. AWESOME WORK LORI!!!"
⭐️⭐️⭐️⭐️⭐️Fairly Teach Mother said, "I love using these pages to work with my students on the Eureka lessons. They are also a great resource to use when giving homework! I used these pages in an interactive notebook. the students loved it and I loved that they could go back and review from previous lessons.
⭐️⭐️⭐️⭐️⭐️Marie I. said, "WOW. Thorough, aligned, and beautifully laid out. I love this math curriculum but the layout of the problem sets is sometimes busy and confusing. These worksheets are perfect to supplement the Eureka pages."
⭐️⭐️⭐️⭐️⭐️Chelsea D. said, "I LOVE this seller. She creates grate resources for Eureka math! I used her when I was teaching second grade last year and now have bought her resources for first grade as well! They are great! This was perfect for second grade to do as independent work during centers."
⭐️⭐️⭐️⭐️⭐️Elizabeth T. said, "I LOVE this seller. She creates grate resources for Eureka math! I used her when I was teaching second grade last year and now have bought her resources for first grade as well! They are great! This was perfect for second grade to do as independent work during centers."
You may also like:
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Module 1 Interactive Math Game
Module 1 Supplemental Printables
Module 1 Supplemental Resources
Module 1 Exit Tickets
Grade 2 Module 2 Resources
Module 2 Interactive Math Game
Module 2 Supplemental Printables
Module 2 Exit Tickets
Grade 2 Module 3 Resources
Module 3 Supplemental Printables
Engage NY Grade 2 Module 3 Interactive Math Game
Module 3 Exit Tickets
Grade 2 Module 4 Resources
Engage NY Grade 2 Module 4 Interactive Math Game
Engage NY Grade 2 Module 4 Printables
Module 4 Exit Tickets
Grade 2 Module 5 Resources
Engage NY Grade 2 Module 5 Interactive Math Game
Module 5 Exit Tickets
Grade 2 Module 6 Resources
Engage NY Grade 2 Module 6 Interactive Math Game
Engage NY Grade 2 Module 6 Printables
Module 6 Exit Tickets
Grade 2 Module 7 Resources
Engage NY Grade 2 Module 7 Interactive Math Game
Engage NY Grade 2 Module 7 Printables
Module 7 Exit Tickets
Grade 2 Module 8 Resources
Engage NY Grade 2 Module 8 Interactive Math Game
Engage NY Grade 2 Module 8 Printables
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The purchaser may use this resource for their personal classroom use only.This product is not to be shared with colleagueswithout purchasing the proper number of licenses.
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Engage NY Grade 2 Module 5 Supplemental Printable and Digital Resource
Rated 4.91 out of 5, based on 177 reviews
4.9(177 ratings)
Lori Flaglor
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7.3k Followers
$11.00
Add to cart
Wish List
Specs
What's Included
Grade
2 nd
Mostly used with 2nd
Subject
Basic Operations, Math
Standards
CCSS 2.NBT.B.7
CCSS 2.NBT.B.8
CCSS 2.NBT.B.9
Tags
Homeschool, Homework, Printables, Worksheets
Save even more with bundles
Engage NY Grade 2 BUNDLED Supplemental Printables & Digital Resource
If you are currently using the Engage NY Math Modules then you must have this resource! These printable pages are a game changer for this program! Pages are ready to Print N Go or use Digitally with Google Slides all in a kid friendly format. This bundled product is aligned to the Common Core Standa
$69.00 Price $69.00$84.50 Original Price $84.50 Save $15.50
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Rated 4.91 out of 5, based on 177 reviews
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Rated 5 out of 5
April 30, 2025
Your Eureka Math products are so helpful! After a math lesson, my students need some extra practice and support with several concepts. These work perfectly! Thank you.
Letitia E.
564 reviews
Grades taught:2nd
Rated 5 out of 5
March 7, 2025
One of my favorite resources! I use this along Eureka. Perfect for extra practice. I love how it states what lessons it aligns with. Great resource.
Diana D.
135 reviews
Grades taught:2nd
Student populations:Emerging bilinguals
Rated 5 out of 5
January 17, 2025
This is a great supplemental resource for small group instruction.
Tiffany Womack
(TPT Seller)
92 reviews
Grades taught:2nd
Student populations:Learning difficulties
Rated 5 out of 5
September 11, 2024
I've used this resource for many years as additional practice and assessment for the original Eureka Math curriculum. I still use it with Eureka Squared Module 4 as the lessons still align with the new curriculum.
2nd Grade on Parade
(TPT Seller)
252 reviews
Grades taught:2nd
Rated 5 out of 5
August 8, 2024
This resource has several activities for each lesson, and I love that they are able to be used if students need extra practice. Great resource for supplemental work and center rotations!
Educating with Em
(TPT Seller)
641 reviews
Grades taught:2nd
Student populations:Autism
Rated 5 out of 5
July 30, 2024
Great resource. Loved using this in my classroom! Thank you!
Shannon Pilato
(TPT Seller)
751 reviews
Rated 5 out of 5
July 23, 2024
Love these worksheets. So much funner than what comes with the curriculum.
Tabitha S.
312 reviews
Grades taught:2nd
Rated 5 out of 5
February 20, 2024
I love using your resources as a supplement to my math curriculum! I send the sheets home with the examples at the top for homework. Thank you for putting these together to close the gaps in our curriculum!
Robin R.
127 reviews
Grades taught:2nd
Show more reviews
Questions & Answers
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Standards
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to see state-specific standards (only available in the US).
CCSS 2.NBT.B.7
Add and subtract within 1000, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method. Understand that in adding or subtracting three-digit numbers, one adds or subtracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds.
CCSS 2.NBT.B.8
Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100–900.
CCSS 2.NBT.B.9
Explain why addition and subtraction strategies work, using place value and the properties of operations.
Meet the Teacher-Author
### Lori Flaglor
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I have 36 years of teaching experience and proud of it! The majority of my experience has been teaching 2nd grade. I taught my first four years in Inglewood, California and the rest in Akron, New York.
New York, United States
4.94
Store rating after 21.5k reviews
7.3k
Followers
Outstanding Student Teacher Award Nominated Teacher of the Year Inglewood California
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